Download 4. Solution Guide to Supplementary Exercises

Topic
5
Redox Reactions, Chemical Cells
and Electrolysis
Part A Unit-based exercise
Unit 18 Chemical cells in daily life
Fill in the blanks
1
chemical; electrical
2
electrolyte
3
voltmeter; multimeter
4
a) negative; positive
b) zinc; copper
5
a) anode
b) cathode
c) electrolyte
6
a) zinc
b) carbon
c) manganese(IV) oxide
d) ammonium chloride
e) 1.5
7
a) lithium; graphite
b) lithium cobalt oxide; lithium manganese oxide
c) lithium
d) 3.7
8
a) hydrogen absorbing
b) nickel(II) hydroxide
c) potassium hydroxide
d) 1.2
True or false
9
F
In a chemical cell, electrons flow from the negative electrode to the positive electrode in the external
circuit.
10 T
125
11 F
In a zinc-carbon cell, the carbon rod is the positive electrode (cathode). The cylinder casing is made of
zinc, which serves as the negative electrode (anode).
12 F
The electrolyte in an alkaline manganese cell is potassium hydroxide.
13 F
Compared with zinc-carbon cells, alkaline manganese cells give a more steady voltage during discharge.
14 T
15 F
Silver oxide cell is a primary cell. It is NOT rechargeable.
16 T
17 F
The cathode of a nickel metal hydride cell is made of nickel(II) hydroxide.
18 T
19 T
20 T
Multiple choice questions
21 C Option A — Electrons flow from the zinc strip to the copper strip in the external circuit.
Option B — Copper(II) ions in the solution near to the copper strip gain electrons and form copper
atoms. As a result, a deposit of copper forms on the copper strip. The mass of the
copper strip increases.
Cu2+(aq) + 2e–
Cu(s)
Option C — Zinc atoms lose electrons and form zinc ions. These ions then go into the copper(II)
sulphate solution. As a result, the mass of the zinc strip decreases.
Zn(s)
Zn2+(aq) + 2e–
Option D — Ions flow through the copper(II) sulphate solution.
22 C Option A — The copper strip is the positive electrode.
Option B — Electrons flow from the magnesium strip to the copper strip in the external circuit.
Option D — Chemical energy is converted into electrical energy.
23 C Options A & B — Pure water and ethanol are non-conductors.
Option C — A chemical cell consists of two different metals and an electrolyte. The chemical cell
makes the light bulb shine.
24 A
126
25 A Option C — The maximum voltage of a zinc-carbon cell is 1.5 V.
Option D — The service life of a zinc-carbon cell is relatively short.
26 D
27 D Option D — Lithium ion cells are commonly used in mobile phones.
28 A
29 B
30 B Option A — Silver oxide cells are NOT rechargeable.
31 B Option A — A lithium ion secondary cell has a high energy density.
Option C — The maximum voltage of a lithium ion secondary cell is 3.7 V.
Option D — The electrolyte of a lithium ion secondary cell is composed of a lithium salt dissolved in
an organic solvent.
The lithium salt is NOT dissolved in water because lithium reacts violently with water.
32 D Option A — Potassium hydroxide acts as the electrolyte of a nickel metal hydride cell.
Option B — A lead-acid accumulator is mainly used for automotive starting, lighting and ignition
applications.
Option C — The maximum voltage of a nickel metal hydride cells is 1.2 V.
33 D
34 B Option B — The voltage of a zinc-carbon cell drops rapidly during discharge.
35 D
36 A (1)
EJHJUBMNVMUJNFUFS
BTWPMUNFUFS
FMFDUSPOGMPX
FMFDUSPEF9QPTJUJWFFMFDUSPEF
FMFDUSPEF:OFHBUJWFFMFDUSPEF
FMFDUSPMZUF
(3) Ions flow through the electrolyte.
127
37 D (1) The magnesium strip is the negative electrode. Electrons flow from the magnesium strip to the
copper strip in the external circuit.
EJHJUBMNVMUJNFUFS
BTWPMUNFUFS
FMFDUSPOGMPX
DPQQFSTUSJQ
NBHOFTJVNTUSJQ
MFNPO
(3) The lemon juice acts as the electrolyte.
Potato juice can be used also.
38 B When the multimeter gives a positive voltage, the metal connected to the positive terminal of the
multimeter is the positive electrode while the metal connected to the negative terminal of the
multimeter is the negative electrode.
EJHJUBMNVMUJNFUFS
BTWPMUNFUFS
m
FMFDUSPEF9QPTJUJWFFMFDUSPEF
FMFDUSPEF:OFHBUJWFFMFDUSPEF
FMFDUSPMZUF
(2) Electrons flow from electrode Y to electrode X in the external circuit.
39 D
40 B (2) Potassium hydroxide acts as the electrolyte.
41 B (1) A silver oxide cell is NOT rechargeable.
(3) The voltage of a silver oxide cell remains steady during discharge.
42 A (2) In a lithium ion secondary cell, lithium atoms lying between graphite sheets act as the negative
electrode.
(3) A lithium ion secondary cell is not very robust and CANNOT take high charging currents. Using a
high charging current may overheat the pack and lead to explosion.
43 A (3) Potassium hydroxide acts as the electroyte of a nickel metal hydride cell.
128
44 A (3) A 12 V accumulator consists of six cells joined in series.
45 A (3) A nickel metal hydride cell uses hydrogen absorbing alloys to make the negative electrode.
46 C (1) Lithium ion cell uses lithium salt dissolved in an organic solvent as the electrolyte.
47 C (1) The maximum voltage of alkaline manganese cells and zinc-carbon cells are both 1.5 V.
48 B (1) Both nickel metal hydride cells and nickel-cadmium cells are rechargeable.
(3) Both nickel metal hydride cells and nickel-cadmium cells can maintain a constant voltage during
discharge.
49 B (1) Nickel metal hydride cells can deliver high discharge currents while lithium ion cells cannot.
3FMBUJWFFOFSHZEFOTJUZ
(2) Lithium ion cells have a higher energy density than nickel metal hydride cells.
-JUIJVNJPODFMM
/J.)DFMM
(3) A lithium ion cell is not very robust and CANNOT take high charging currents. Using a high
charging current may overheat the pack and lead to explosion.
50 C (1) Using new and used zinc-carbon cells at the same time in an electrical appliance is NOT dangerous,
but this gives poorer results.
(2) Charging a lithium ion secondary cell with a high current may overheat the pack and lead to
explosion.
51 C In a magnesium-copper chemical cell, electrons flow from the magnesium electrode to the copper
electrode in the external circuit.
52 D Zinc-carbon cells are NOT rechargeable.
Ammonium chloride acts as the electrolyte in zinc-carbon cells.
53 A
54 C The electrolyte of a zinc-carbon cell is a moist paste of ammonium chloride.
55 D A lithium ion cell is not very robust and cannot deliver high discharge currents. Loading the cell with
excess discharge current may overheat the pack and lead to explosion.
129
Unit 19 Simple chemical cells
Fill in the blanks
1
a) electrons; magnesium ions
b) electrons; copper
Mg2+(aq) + 2e–
c) i) Mg(s)
ii) Cu2+(aq) + 2e–
Cu(s)
d) magnesium; copper
e) decreases; increases
2
electrochemical
3
a) higher
b) lower
4
a) zinc
b) copper
c) i) Zn(s)
2+
2+
–
Zn (aq) + 2e
–
ii) Cu (aq) + 2e
Cu(s)
d) zinc strip; copper container
e) ions
True or false
5
T
6
F
7
T A copper-silver chemical cell is shown below:
In the electrochemical series, the position of calcium is higher than that of sodium. The order is
different from that in the reactivity series. This is because, when compared with sodium, a calcium
atom loses electrons more readily in cell reactions whereas in reactions with air, water and dilute acids
it loses electrons less readily.
EJHJUBMNVMUJNFUFS
BTWPMUNFUFS
FMFDUSPOGMPX
DPQQFSTUSJQ
OFHBUJWFFMFDUSPEF
Fm
Fm
$V
$V
TJMWFSTUSJQ
QPTJUJWFFMFDUSPEF
DPQQFS**
TVMQIBUF
TPMVUJPO
In the copper-silver chemical cell, copper atoms lose electrons and form copper(II) ions. These ions then
go into the copper(II) sulphate solution.
130
Cu(s)
Cu2+(aq) + 2e–
Electrons given up by copper atoms flow along the conducting wires to the silver strip. Copper(II) ions
in the copper(II) sulphate solution near to the silver strip gain these electrons and form copper atoms.
As a result, a deposit of copper forms on the silver strip.
Cu2+(aq) + 2e–
Cu(s)
Copper is transferred from the copper strip to the silver strip.
The concentration of copper(II) ions in the electrolyte remains the same. The blue colour of the solution
remains unchanged.
8
F
A salt bridge can be prepared by soaking a piece of filter paper in an ionic salt solution.
Sugar solution does NOT conduct electricity.
Hence sugar solution CANNOT be used to prepare salt bridges.
9
T
10 F
In a Daniell cell, electrons flow from the zinc strip to the copper container in the external circuit, i.e. a
current flows from the copper container to the zinc strip.
Multiple choice questions
11 A Zinc forms ions more readily than copper does.
Hence zinc atoms would lose electrons and form zinc ions.
Zn(s)
Zn2+(aq) + 2e–
The zinc strip is the negative electrode.
Copper(II) ions in the electrolyte near to the copper strip gain electrons and form copper atoms.
Cu2+(aq) + 2e–
Cu(s)
The copper strip is the positive electrode.
12 B Options A & C — Pure water and ethanol are non-conductors.
Option B — Zinc and copper are farther apart in the electrochemical series than lead and copper.
Hence the zinc-copper couple gives the highest voltage, making the light bulb the
brightest.
13 C Option C — The position of magnesium in the electrochemical series is the highest.
Hence magnesium loses electrons to form ions most readily.
14 D Option D — Electrons move from the zinc strip to the copper strip in the external circuit.
131
15 A
7
FMFDUSPOGMPX
TUSJQPGNFUBM9
DPQQFSTUSJQ
EJMVUFTVMQIVSJD
BDJE
In the above chemical cell, electrons flow from the strip of metal X to the copper strip in the external
circuit.
It can be deduced that metal X forms ions more readily than copper does.
Option A — The copper strip is the positive electrode and the strip made of metal X is the negative
electrode.
Option B — At the copper strip, hydrogen ions in the acid gain electrons and form hydrogen gas.
2H+(aq) + 2e–
H2(g)
Option C — Atoms of metal X lose electrons and form ions. As a result, the mass of the strip of
metal X decreases.
Option D — Metal X is at a higher position in the electrochemical series than copper.
16 D Magnesium forms ions more readily than zinc. Magnesium and copper are farther apart in the
electrochemical series than zinc and copper. Therefore the voltage of the cell would increase if the zinc
strip is replaced by a magnesium strip.
17 C
EJHJUBMNVMUJNFUFS
BTWPMUNFUFS
FMFDUSPOGMPX
NFUBM9SPE
JSPOSPE
FMFDUSPMZUF
Option A — In the above chemical cell, electrons flow from the metal X rod to the iron rod in the
external circuit.
It can be deduced that metal X forms ions more readily than iron does.
Option B — Atoms of metal X lose electrons and form ions. As a result, the metal X rod dissolves
gradually.
Option C — The iron rod is the positive electrode, i.e. the cathode.
18 D For the first two chemical cells, metal W is the positive electrode while metals X and Y are the
negative electrodes. Therefore metals X and Y form ions more readily than metal W.
The Y / W couple gives a higher voltage than the X / W couple. Therefore the difference in the
tendency to form ions between metal Y and metal W is greater than that between metal X and metal W.
Hence metal Y forms ions most readily.
132
For the third chemical cell, the voltmeter gives a negative voltage. Therefore metal W is the negative
electrode while metal Z is the positive electrode. Metal W forms ions more readily than metal Z.
This gives the descending order of reactivity of the four metals: Y, X, W, Z.
19 D For a simple chemical cell, the farther apart the two metals are in the electrochemical series (or the
reactivity series), the higher is the voltage of the cell.
20 C From the cell of the Mn / Fe couple, it can be deduced that Mn forms ions more readily than Fe.
From the cells of the Fe / Ag couple and Fe / Cu couple, it can be deduced that Fe forms ions more
readily than Ag and Cu.
The Fe / Ag couple gives a higher voltage than the Fe / Cu couple. Therefore the difference in the
tendency to form ions between Fe and Ag is greater than that between Fe and Cu. Hence Ag forms
ions least readily.
This gives the descending order of the tendency to form ions for the four metals: Mn, Fe, Cu, Ag.
21 D
Option
Metal X
Direction of electron flow in the external circuit
A
copper
from zinc to copper
B
iron
from zinc to iron
C
magnesium
from magnesium to zinc
D
silver
from zinc to silver
22 A X displaces Z from Z(NO3)2 solution. Hence X is more reactive than Z and forms ions more readily.
The following chemical cell uses X and Z as electrodes.
EJHJUBMNVMUJNFUFS
BTWPMUNFUFS
TUSJQNBEFPG9
TUSJQNBEFPG;
TPEJVNDIMPSJEFTPMVUJPO
Option A — Atoms of metal X lose electrons and form ions.
Electrons flow from X to Z in the external circuit.
Options C and D — Atoms of metal X lose electrons and form ions.
As a result, the strip made of X dissolves gradually and its mass decreases.
133
23 A
7
TJMWFSSPE
Fm
FMFDUSPOGMPX
Fm
/J
/J
OJDLFMSPE
OJDLFM**
TVMQIBUF
TPMVUJPO
Nickel is more reactive than silver. Hence nickel forms ions more readily than silver does.
In the above chemical cell, nickel atoms lose electrons and form nickel(II) ions.
Ni(s)
Ni2+(aq) + 2e–
Electrons given up by nickel atoms flow along the conducting wires to the silver rod. Nickel(II) ions in
the solution near to the silver rod gain these electrons and form nickel atoms. As a result, a deposit of
nickel forms on the silver rod.
Ni2+(aq) + 2e–
Ni(s)
Hence the mass of the silver rod (i.e. the cathode) increases.
Nickel was transferred from the nickel rod to the silver rod.
The concentration of nickel(II) ions in the electrolyte remains the same. The colour intensity of the
solution would not change.
24 B The order of reactivity of the metals is:
Q > T > R > S
Q and S are the furthest apart in the electrochemical series. Hence the Q / S couple would give the
highest voltage.
25 C Option C — The zinc plate reacts with the dilute sulphuric acid. Hence colourless gas bubbles appear
at the surface of the zinc plate.
26 B
FMFDUSPOGMPX
[JODQMBUF
"
Fm
Fm
)
DPQQFSQMBUF
EJMVUFTVMQIVSJDBDJE
;O
When the circuit is closed, electrons flow from the zinc plate to the copper plate in the external circuit.
Zinc atoms lose electrons and form ions. As a result, the mass of the zinc plate (i.e. the anode)
decreases.
Zinc ions are colourless. Hence these is NO change in the colour of the dilute sulphuric acid.
134
27 A
EJHJUBMNVMUJNFUFS
BTWPMUNFUFS
FMFDUSPOGMPX
TJMWFSSPE
TJMWFS
OJUSBUF
TPMVUJPO
Fm
Fm
OJDLFMSPE
TBMU
CSJEHF
"H
OJDLFM**
OJUSBUF
TPMVUJPO
/J
Nickel is more reactive than silver. Hence nickel forms ions more readily than silver does.
In the above chemical cell, electrons flow from the nickel rod to the silver rod in the external circuit.
Nickel is the negative electrode (i.e. the anode) of the chemical cell.
Nickel atoms lose electrons and form nickel(II) ions.
Ni(s)
Ni2+(aq) + 2e–
28 C Option C — Magnesium forms ions more readily than copper does.
Magnesium is the negative electrode (i.e. anode) of the chemical cell.
Magnesium atoms lose electrons and form magnesium ions.
Mg(s)
Mg2+(aq) + 2e–
29 D Option D — Magnesium atoms lose electrons and form magnesium ions. As a result, the mass of the
magnesium electrode decreases.
Electrons given up by magnesium atoms flow along the conducting wires to the copper
electrode. Copper(II) ions in the copper(II) sulphate solution near to the copper electrode
gain these electrons and form copper atoms. As a result, a deposit of copper forms on
the copper electrode.
Cu2+(aq) + 2e–
Cu(s)
Hence the mass of the copper electrode increases.
30 B
EJHJUBMNVMUJNFUFS
BTWPMUNFUFS
FMFDUSPOGMPX
DPQQFS
FMFDUSPEF
NBHOFTJVN
FMFDUSPEF
DPQQFS**
TVMQIBUF
TPMVUJPO
NBHOFTJVN
TVMQIBUF
TPMVUJPO
BQJFDFPGGJMUFSQBQFSTPBLFE
XJUITBUVSBUFEQPUBTTJVN
OJUSBUFTPMVUJPO
135
Electrons flow from the magnesium electrode towards the copper electrode in the external circuit.
LFZ
DPQQFS
FMFDUSPEF
NBHOFTJVN
FMFDUSPEF
TBMU
CSJEHF
NBHOFTJVN
TVMQIBUF
TPMVUJPO
DPQQFS**
TVMQIBUF
TPMVUJPO
.HJPOT
40mJPOT
,JPOTGSPN
UIFTBMUCSJEHF
/0mJPOTGSPN
UIFTBMUCSJEHF
The nitrate ions in the salt bridge move towards the magnesium electrode.
31 C Option D — Ammonium and nitrate ions are NOT involved in the reactions of the half-cells.
32 A
EJHJUBMNVMUJNFUFS
BTWPMUNFUFS
FMFDUSPOGMPX
DPQQFSFMFDUSPEF
Fm
Fm
JSPOFMFDUSPEF
TBMU
CSJEHF
$V/0
BR
$V
'F/0
BR
'F
Iron forms ions more readily than copper does.
Iron is the negative electrode of the chemical cell.
Iron atoms lose electrons and form iron ions.
Fe(s)
2+
–
Fe (aq) + 2e
33 B Nickel is more reactive than silver. Hence nickel forms ions more readily than silver does.
In the chemical cell, nickel atoms lose electrons and form nickel(II) ions.
Electrons given up by nickel atoms flow along the conducting wires to the silver electrode. Silver ions
in the silver nitrate solution near to the silver electrode gain these electrons and form silver atoms.
Ag+(aq) + e–
Ag(s)
The silver electrode is the positive electrode, i.e. the cathode.
34 D Option A —
FMFDUSPOGMPX
7
TJMWFSFMFDUSPEF
TBMU
CSJEHF
"H/0BR
"H
136
Fm
Fm
OJDLFMFMFDUSPEF
/J/0
BR
/J
Electrons flow from the nickel electrode to the silver electrode in the external circuit.
Options B and D — Nickel atoms lose electrons and form nickel(II) ions.
2+
–
Ni (aq) + 2e
Ni(s)
NO bubbles of gas are given off from the nickel electrode.
The concentration of nickel(II) ions in the nickel(II) nitrate solution increases.
Hence the green colour of the solution becomes more intense.
Option C — Silver ions in the silver nitrate solution near to the silver electrode gain electrons and form
silver atoms.
Hence the concentration of silver ions in the solution decreases.
35 B
Combination
X
Y
Z
Direction of electron flow in
the external circuit
A
magnesium
copper
magnesium sulphate solution
from X to Y
B
carbon
magnesium
magnesium sulphate solution
from Y to X, as shown in the
diagram
C
copper
silver
silver nitrate solution
from X to Y
D
magnesium
carbon
silver nitrate solution
from X to Y
Option
36 C
FMFDUSPOGMPX
7
TUSJQNBEFPG9
Fm
Fm
TBMU
CSJEHF
9/0
BR
9
TUSJQNBEFPG:
:/0
BR
:
Electrons flow from X to Y in the external circuit.
It can be deduced that X forms ions more readily than Y does.
Options A and C — In the above chemical cell, atoms of X lose electrons and form ions. These ions
then go into the X(NO3)2 solution.
Hence the strip made of X gradually dissolves.
The strip made of X is the negative electrode, i.e. the anode.
Option B — In the electrochemical series, the position of X is higher than that of Y.
Options A and D — Electrons given up by atoms of X flow along the conducting wires to the strip
made of Y. Y2+ ions in the Y(NO3)2 solution near to the strip made of Y gain
these electrons and form atoms. As a result, a deposit of Y forms on the strip.
Y2+(aq) + 2e–
Y(s)
137
37 C
FMFDUSPOGMPX
DPQQFSFMFDUSPEF9
7
Fm
Fm
TBMU
CSJEHF
m
NPMEN $V40BR
$V
DPQQFSFMFDUSPEF:
NPMENm
$V40BR
$V
The above chemical cell is made up of copper electrodes and copper(II) sulphate solution of different
concentrations.
Electrons flow in such a direction that the concentration of Cu2+(aq) ions in each half-cell becomes the
same evantually, i.e. electrons flow from X to Y in the external circuit (i.e. Option C is correct).
At electrode X
Cu(s)
2+
–
Cu (aq) + 2e
At electrode Y
Cu2+(aq) + 2e–
Cu(s)
Option A — Electrode Y is the positive electrode.
Option B — The mass of electrode X decreases.
Option D — Electrons flow in the external circuit, NOT via the salt bridge.
38 C
MPBE
FMFDUSPOGMPX
JOFSUFMFDUSPEF"
QPSPVTEFWJDF
JOFSUFMFDUSPEF#
TVMQIVS
TPEJVN
Option A — In the above chemical cell, electrons flow from electrode A to electrode B in the external
circuit.
At electrode A
Na(l)
Na+(aq) + e–
At electrode B
S(l) + 2e–
S2–(l)
Option D — A current flows from electrode B to electrode A in the external circuit.
138
39 D
EJHJUBMNVMUJNFUFS
BTWPMUNFUFS
FMFDUSPOGMPX
[JODTUSJQ
Fm
Fm
DPCBMUTUSJQ
m
NPMENm
$P/0
BR
NPMEN ;O/0
BR
;O
$P
BQJFDFPGGJMUFSQBQFSTPBLFE
XJUITBUVSBUFEQPUBTTJVN
OJUSBUFTPMVUJPO
The zinc strip gradually dissolves. It can be deduced that zinc atoms lose electrons and form zinc ions.
Zn(s)
Zn2+(aq) + 2e–
Option D — Electrons flow from the zinc strip to the cobalt strip in the external circuit.
Hence the cobalt strip is the positive electrode (i.e. the cathode) of the chemical cell.
Electrons given up by zinc atoms flow along the conducting wires to the cobalt electrode.
Cobalt(II) ions in the solution near to the cobalt electrode gain these electrons and form
cobalt atoms.
Co2+(aq) + 2e–
Co(s)
40 A (2) Electrons flow in the external circuit, NOT in the salt bridge.
(3) The salt bridge provides ions that can move into the half-cells to prevent the build-up of excess
positively or negatively charged ions in the solutions.
41 C Option C —
EJHJUBMNVMUJNFUFS
BTWPMUNFUFS
FMFDUSPOGMPX
OJDLFM
FMFDUSPEF
m
NPMEN /J BR
Fm
Fm
/J
1E
DBUJPONPWFNFOU
QBMMBEJVN
FMFDUSPEF
NPMENm
1EBR
BOJPONPWFNFOU
QPSPVTCBSSJFS
Cations (i.e. positive ions) move from the nickel half-cell to the palladium half-cell.
It can be deduced that nickel atoms lose electrons and form nickel(II) ions.
Ni(s)
Ni2+(aq) + 2e–
Hence the nickel electrode is the negative electrode, i.e. the anode.
139
42 A (1) The palladium electrode is the positive electrode, i.e. the cathode.
(3) Electrons flow in the external circuit, NOT through the porous barrier.
43 D Option A — Electrons flow from the zinc strip to the copper container in the external circuit, i.e. a
current flows from the copper container to the zinc strip in the external circuit.
Option B — The zinc strip is the negative electrode, i.e. the anode.
Option D — The copper container is the positive electrode. The copper(II) ions in the copper(II) sulphate
solution gain electrons from the external circuit to form copper metal.
Cu2+(aq) + 2e–
Cu(s)
The concentration of copper(II) ions in the solution decreases. Hence the blue colour of
the copper(II) sulphate solution becomes less intense.
44 B (2) The porous pot is different from a salt bridge. It does NOT provide ions to balance excess charges
in the solutions of the cell.
45 A (1) and (3) In the copper-silver chemical cell, copper atoms lose electrons and form copper(II) ions.
These ions then go into the silver nitrate solution.
Cu2+(aq) + 2e–
Cu(s)
Hence the mass of the copper electrode decreases.
The colour of the silver nitrate solution would change as the blue copper(II) ions go into
the solution.
(2) Electrons given up by copper atoms flow along the conducting wires to the silver electrode. Silver
ions in the solution near to the silver electrode gain these electrons and form silver atoms. As a
result, a deposit of silver forms on the silver electrode.
Hence the mass of the silver electrode increases.
46 B Electrons flow from the electrode made of metal X to the copper electrode in the external circuit.
It can be deduced that metal X forms ions more readily than copper does.
(1) The electrode made of metal X is the negative electrode, i.e. the anode.
(3) The dilute sulphuric acid would NOT turn blue as NO copper(II) ions are formed.
47 D
FMFDUSPOGMPX
7
TUSJQNBEFPG9
Fm
Fm
TBMU
CSJEHF
9/0
BR
9
TUSJQNBEFPG:
:/0
BR
:
Electrons flow from Y to X in the external circuit.
It can be deduced that Y forms ions more readily than X does.
140
(1) In the above chemical cell, atoms of Y lose electrons and form ions. These ions then go into
Y(NO3)2(aq).
Hence the strip made of Y gradually dissolves.
2+
Electrons given up by atoms of Y flow along the conducting wires to the strip made of X. X (aq)
ions in X(NO3)2(aq) near to the strip made of X gain these electrons and form atoms. As a result, a
deposit of X forms on the strip.
Hence the strip made of X does NOT dissolve.
(2)
LFZ
TUSJQNBEFPG9
TBMU
CSJEHF
9/0
BR
:JPOT
TUSJQNBEF
PG:
/0mJPOT
:/0
BR
QPTJUJWFJPOTGSPN
UIFTBMUCSJEHF
OFHBUJWFJPOTGSPN
UIFTBMUCSJEHF
Negative ions, i.e. anions, in the salt bridge migrate towards the strip made of Y.
(3) Y is more reactive than X.
Hence Y can displace X from X(NO3)2 solution.
48 D When nickel is placed in a zinc sulphate solution, no reaction occurs.
It can be deduced that zinc is more reactive than nickel, i.e. zinc forms ions more readily than nickel
does.
(1) and (3) Electrons flow from the zinc electrode to the nickel electrode in the external circuit.
The zinc electrode is the negative electrode, i.e. the anode.
(2)
LFZ
[JODFMFDUSPEF
OJDLFMFMFDUSPEF
TBMU
CSJEHF
OJDLFM**
TVMQIBUF
TPMVUJPO
[JODTVMQIBUF
TPMVUJPO
;OJPOT
40mJPOT
QPTJUJWFJPOTGSPN
UIFTBMUCSJEHF
OFHBUJWFJPOTGSPN
UIFTBMUCSJEHF
Positive ions, i.e. cations, in the salt bridge migrate towards the nickel half-cell.
141
49 B (1) X can displace Y from Y(NO3)2 solution.
It can be deduced that X is more reactive than Y.
(2) The position of X in the electrochemical series is higher than that of Y.
(3) The following chemical cell uses X and Y as electrodes.
EJHJUBMNVMUJNFUFS
BTWPMUNFUFS
FMFDUSPOGMPX
9
:
FMFDUSPMZUF
Atoms of X lose electrons and form ions.
Electrons flow from X to Y in the external circuit. Hence X is the negative electrode.
50 A (1) Copper forms ions more readily than silver does. Hence electrons flow from the copper electrode to
the silver electrode in the external circuit.
EJHJUBMNVMUJNFUFS
BTWPMUNFUFS
FMFDUSPOGMPX
DPQQFS
FMFDUSPEF
Fm
Fm
DPQQFS**
TVMQIBUF
TPMVUJPO
TJMWFS
FMFDUSPEF
QPSPVTQPU
$V "H
TJMWFSOJUSBUF
TPMVUJPO
The copper electrode is the negative electrode, i.e. the anode.
(2) A porous pot is different from a salt bridge. It does NOT provide ions to balance excess charges in
the solutions of the cell.
(3) The porous pot CANNOT be replaced by a glass cylinder.
A glass cylinder does NOT allow ions to move between the two solutions.
51 D Iron forms ions more readily than copper does.
In a chemical cell with a copper-iron couple, the copper is the positive electrode.
52 C The position of calcium is higher than that of sodium in the electrochemical series. Calcium atom loses
electrons more readily in cell reactions than in reactions with air, water and dilute acids.
53 A The position of nickel in the electrochemical series is higher than that of silver.
142
54 C A copper-silver chemical cell is shown below:
EJHJUBMNVMUJNFUFS
BTWPMUNFUFS
FMFDUSPOGMPX
DPQQFSTUSJQ
OFHBUJWFFMFDUSPEF
m
F
Fm
$V
$V
TJMWFSTUSJQ
QPTJUJWFFMFDUSPEF
DPQQFS**
TVMQIBUF
TPMVUJPO
In the copper-silver chemical cell, copper atoms lose electrons and form copper(II) ions. These ions then
go into the copper(II) sulphate solution.
Cu(s)
Cu2+(aq) + 2e–
Electrons given up by copper atoms flow along the conducting wires to the silver strip. Copper(II) ions
in the copper(II) sulphate solution near to the silver strip gain these electrons and form copper atoms.
As a result, a deposit of copper forms on the silver strip.
Cu2+(aq) + 2e–
Cu(s)
Copper is transferred from the copper strip to the silver strip.
The concentration of copper(II) ions in the electrolyte remains the same. The blue colour of the solution
remains unchanged.
55 D In a copper-zinc chemical cell using dilute sulphuric acid as the electrolyte, the zinc atoms lose electrons
to form zinc ions.
7
DPQQFSFMFDUSPEF
Fm
Fm
)
;O
FMFDUSPOGMPX
[JODFMFDUSPEF
EJMVUFTVMQIVSJDBDJE
Zinc ions are colourless. Hence there is NO change in the colour of the dilute sulphuric acid.
The copper electrode is the positive electrode. The copper atoms would NOT lose electrons to form
copper(II) ions.
56 A
57 D Electrons flow in the external circuit of the chemical cell, NOT through the salt bridge.
58 D In a Daniell cell, a glass container CANNOT be used to contain the zinc sulphate solution.
The glass container does NOT allow ions to move between the two solutions.
143
Unit 20 Oxidation and reduction
Fill in the blanks
1
reduction
2
oxidation
3
reducing
4
oxidizing
5
reducing
6
a) bromide
–
b) Br2(aq) + 2e
2Br–(aq)
c) yellow-brown
7
a) iron(III)
b) Fe2+(aq)
Fe3+(aq) + e–
c) yellow-brown
8
a) manganese(II)
b) MnO4–(aq) + 8H+(aq) + 5e–
Mn2+(aq) + 4H2O(l)
c) purple
9
a) chromium(III)
b) Cr2O72–(aq) + 14H+(aq) + 6e–
2Cr3+(aq) + 7H2O(l)
c) orange; green
10 a) yellow-brown; bromine
b) Cl2(aq) + 2Br–(aq)
2Cl–(aq) + Br2(aq)
11 a) brown; iodine
b) Cl2(aq) + 2I–(aq)
2Cl–(aq) + I2(aq)
12 a) chloride; hypochlorite
b) Cl2(g) + 2NaOH(aq)
NaCl(aq) + NaOCl(aq) + H2O(l)
13 a) oxidizing agent
b) acid
14 a) i) bromide
ii) NaBr(s) + H2SO4(l)
NaHSO4(s) + HBr(g)
b) i) bromine
ii) 2HBr(g) + H2SO4(l)
Br2(g) + SO2(g) + 2H2O(l)
15 a) orange; green
b) Cr2O72–(aq) + 3SO2(aq) + 2H+(aq)
144
2Cr3+(aq) + 3SO42–(aq) + H2O(l)
True or false
16 F
Not all chemical reactions are redox reactions. For example, neutralization reactions are not redox
reactions.
17 T
18 T The position of potassium in the electrochemical series is higher than that of sodium.
19 F
Dilute nitric acid is an oxidizing agent. It will compete with the potassium permanganate solution for
the reducing agent in reactions. Hence it is NOT used to acidify potassium permanganate solution.
Usually dilute sulphuric acid is used to acidify potassium permanganate solution.
20 T The ionic half-equation for the reduction of acidified potassium dichromate solution is
Cr2O72–(aq) + 14H+(aq) + 6e–
2Cr3+(aq) + 7H2O(l)
The oxidation of chromium in the Cr2O72– ion is +6 while that in the Cr3+ ion is +3.
Hence the oxidation number of chromium changes from +6 to +3.
21 T Acidified potassium permanganate solution is a strong oxidizing agent. When it reacts with iron(II)
sulphate solution, the permanganate ions will be reduced to manganese(II) ions. The permanganate
solution turns from purple to colourless while the iron(II) sulphate solution changes from pale green to
yellow-brown.
On the other hand, there is NO observable change when acidified potassium permanganate solution is
mixed with iron(III) sulphate solution.
Hence acidified potassium permanganate solution can be used to distinguish between iron(II) sulphate
solution and iron(III) sulphate solution.
22 F
In the reaction between aqueous chlorine and potassium iodide solution, chlorine is reduced to chloride
ions while iodide ions are oxidized to iodine. The resulting reaction mixture is brown in colour due to
the presence of iodine.
Cl2(aq) + 2e–
2l–(aq)
23 F
2Cl–(aq)
I2(aq) + 2e–
Both aqueous chlorine and acidified potassium permanganate solution are oxidizing agents. There is NO
reaction between them.
24 T
25 T When aqueous bromine is mixed with sodium sulphite solution, the yellow-brown aqueous bromine
becomes colourless.
This is because the yellow-brown bromine is reduced to colourless bromide ions.
On the other hand, there is NO observable change when aqueous bromine is mixed with sodium
sulphate solution.
Hence aqueous bromine can be used to distinguish between sodium sulphate solution and sodium
sulphite solution.
145
26 F
Magnesium reacts with dilute nitric acid to give nitrogen monoxide.
+
–
3Mg(s) + 2NO3 (aq) + 8H (aq)
2+
3Mg (aq) + 2NO(g) + 4H2O(l)
27 T The equation for the reaction between zinc and concentrated nitric acid is:
Zn(s) + 2NO3–(aq) + 4H+(aq)
Zn2+(aq) + 2NO2(g) + 2H2O(l)
The oxidation number of nitrogen in the NO3– ion is +5 while that in NO2 is +4.
Hence the oxidation number of nitrogen changes from +5 to +4.
28 F
Nitrogen monoxide gas is colourless. When this gas is exposed to air, it reacts with oxygen in the air to
give brown nitrogen dioxide gas.
2NO(g) + O2(g)
colourless
29 F
2NO2(g)
brown
Concentrated sulphuric acid liberates steamy fumes of hydrogen chloride when it reacts with sodium
chloride.
NaCl(s) + H2SO4(l)
NaHSO4(s) + HCl(g)
Hydrogen chloride is a weak reducing agent. It is NOT oxidized by concentrated sulphuric acid.
Hence NO chlorine (halogen) is produced in the reaction.
30 T Aqueous sulphur dioxide is a good reducing agent.
NO reaction occurs when sulphur dioxide gas is bubbled into potassium iodide solution.
Multiple choice questions
31 D Suppose the oxidation number of Cl in HClO4 is x.
(+1) + x + (–2) x 4 = 0
x = +7
32 C Sum of oxidation number of all atoms in the OH– ion = charge on the ion = –1
Suppose the oxidation number of Al in [Al(OH)4]– is x.
x + (–1) x 4 = –1
x = +3
33 C Suppose the oxidation number of Co in Co(NH3)6Cl3 is x.
x + 0 + (–1) x 3 = 0
x = +3
34 D
146
Option
Compound
Oxidation number of S in the compound
A
ZnS
–2
B
Na2S2O3
+2
C
NaHSO3
+4
D
H2S2O7
+6
∴ sulphur exhibits the highest oxidation number in H2S2O7.
35 C
Option
A
Substances
Oxidation number of underlined element
PbO2
+4
MnO4–
+7
Fe2O3
+3
CO2
+4
NO2
+4
CO32–
+4
B
C
–
D
OH
–2
Cr2O72–
+6
∴ in NO2 and CO32–, the oxidation number of the underlined element is the same.
36 C
Option
A
Compound
Oxidation number of chromium in the compound
Cr2O3
+3
K2CrO7
+6
Na2CrO4
+6
CrCl2
+2
Cr2O3
+3
K2Cr2O7
+6
Cr2O3
+3
CrCl3
+3
Cr(NO3)3
+3
Na2CrO4
+6
CrO3
+6
Cr(NO3)3
+3
B
C
D
∴ in Cr2O3, CrCl3 and Cr(NO3)3, the oxidation number of chromium is the same.
37 A
Option
Stage
A
1
B
2
C
3
D
4
Conversion
0
+4
S
SO2
+4
SO2
+6
SO3
+6
H2SO4
Change in oxidation number of sulphur
4
+6
2
SO3
+6
H2SO4
+6
(NH4)2SO4
0
0
∴ Stage 1 involves the largest change in oxidation number of sulphur.
147
38 C
Option
Conversion
+3
A
Cr
B
ClO–
3+
+1
+6
C
Cr2O72–
D
MnO4–
+7
∴ the conversion Cr2O72–
Change in oxidation number
+2
Cr
1 unit
2+
+3
2 units
ClO2–
+3
3+
3 units
Mn2+
5 units
Cr
+2
Cr
3+
involves an oxidation number change of 3 units.
39 D The unbalanced ionic half-equation is:
Br2
2BrO3–
Balance the ionic half-equation
with respect to the number of
atoms.
Balance the ionic half-equation
with respect to the number of
charges.
To balance the 6
side.
Br2 + 6H2O
To balance the 12
side.
Br2 + 6H2O
oxygen atoms in 2BrO3–, add 6H2O on the left-hand
2BrO3–
hydrogen atoms in 6H2O, add 12H+ on the right-hand
2BrO3– + 12H+
Charge on left-hand side = 0
Charge on right-hand side = 2 x (–1) + 12 x (+1)
= +10
–
∴ add 10e on the right-hand side to balance the charge.
The balanced ionic half-equation is:
2BrO3– + 12H+ + 10e–
Br2 + 6H2O
∴ x is 6, y is 12 and z is 10.
40 B Writing the redox equation using ionic half-equations
1 Write down the oxidizing
agent and the reducing
agent involved.
NO3– is the oxidizing agent and Fe2+ is the reducing agent.
2 a) Write an ionic halfequation for the
reduction process.
–
The unbalanced ionic half-equation for NO3 is:
NO
NO3–
i) Balance the ionic halfequation with respect
to the number of
atoms.
–
To balance the 3 oxygen atoms in NO3 , add 2H2O on the right-hand side.
–
NO + 2H2O
NO3
+
To balance the 4 hydrogen atoms in 2H2O, add 4H on the left-hand side.
+
–
NO + 2H2O
NO3 + 4H
ii) Balance the ionic halfequation with respect
to the number of
charges.
Charge on left-hand side = (–1) + 4 x (+1) = +3
Charge on right-hand side = 0
–
∴ add 3e on the left-hand side to balance the charge.
The balanced ionic half-equation for acidified NO3– is:
+
–
NO + 2H2O......(i)
NO3– + 4H + 3e
148
2 b) Write an ionic halfequation for the
oxidation process.
2+
The unbalanced ionic half-equation for Fe
2+
3+
Fe
Fe
is:
i) Balance the ionic halfequation with respect
to the number of
atoms.
The number of atoms is balanced.
ii) Balance the ionic halfequation with respect
to the number of
charges.
Charge on left-hand side = +2
Charge on right-hand side = +3
∴ add 1e– on the right-hand side to balance the charge.
2+
The balanced ionic half-equation for Fe
2+
3+
–
Fe + e ......(ii)
Fe
3 Make the number of
electrons gained in one ionic
half-equation equal to that
lost in the other.
(ii) x 3
4 Combine the two ionic halfequations and eliminate the
electrons.
(i)
NO3– + 4H+ + 3e–
2+
(ii) x 3 3Fe
is:
NO + 2H2O
3Fe3+ + 3e–
3Fe2+ + 4H+ + NO3–
3Fe3+ + 2H2O + NO
∴ x is 3, y is 4 and z is 2.
Writing the redox equation using oxidation number method
1 Write down the oxidizing
agent and the reducing
agent involved. Determine
their products.
2 Assign oxidation numbers to
all atoms.
3 Notice atoms which undergo
a change in oxidation
number. Determine the
number of electrons lost or
gained per formula unit.
2+
NO3– is the oxidizing agent and Fe
NO
NO3–
2+
3+
Fe
Fe
+5
+2
+2
NO3– + Fe2+
is the reducing agent.
+3
NO + Fe3+
reduction: gain of 3e– per NO3–
+5
NO3
–
+2
+
2+
Fe
+2
+3
NO
+
–
3+
Fe
2+
oxidation: loss of 1e per Fe
4 Insert an appropriate
coefficient before the
formula of each reagent
on the left-hand side of
the equation to make the
number of electrons gained
equal to that lost.
2+
–
NO3 + 3Fe
NO + Fe3+
5 Add appropriate coefficients
on the right to balance
the number of atoms
which have gained or lost
electrons.
2+
–
NO3 + 3Fe
NO + 3Fe3+
149
6 Balance the number of all
other atoms except O and
H.
The number of all other atoms except O is balanced.
7 Add H+ to the side deficient
in positive charges to make
the number of charges on
both sides equal.
Total charge on left-hand side = (–1) + 3 x (+2) = +5
Total charge on right-hand side = 3 x (+3) = +9
+
Add 4H to the left-hand side.
+
2+
–
NO + 3Fe3+
NO3 + 4H + 3Fe
8 Add H2O to the appropriate
side to balance the number
of O atoms.
3Fe2+ + 4H+ + NO3–
3+
3Fe
+ 2H2O + NO
∴ x is 3, y is 4 and z is 2.
41 B Number of H atoms on left-hand side = number of H atoms on right-hand side
i.e. 2x = 2 x 4 + 2
x= 5
Number of O atoms on left-hand side = number of O atoms on right-hand side
i.e. 2 x 4 + x + 3y = 2 x 4 + 2 + 4y
x – 2= y
∴ y= 3
42 C Writing the redox equation using ionic half-equations
1 Write down the oxidizing
agent and the reducing
agent involved.
AuCl4– is the oxidizing agent and Sn2+ is the reducing agent.
2 a) Write an ionic halfequation for the
reduction process.
–
The unbalanced ionic half-equation for AuCl4 is:
–
2Au
2AuCl4
i) Balance the ionic halfequation with respect
to the number of
atoms.
–
To balance the 8 chlorine atoms in 2AuCl4 on the left-hand side, add
–
8Cl on the right.
–
2AuCl4–
2Au + 8Cl
ii) Balance the ionic halfequation with respect
to the number of
charges.
Charge on left-hand side = 2 x (–1) = –2
Charge on right-hand side = 8 x (–1) = –8
–
∴ add 6e on the left-hand side to balance the charge.
–
The balanced ionic half-equation for AuCl4 is:
–
–
–
2Au + 8Cl ......(i)
2AuCl4 + 6e
150
2 b) Write an ionic halfequation for the
oxidation process.
2+
The unbalanced ionic half-equation for Sn
2+
4+
Sn
Sn
is:
i) Balance the ionic halfequation with respect
to the number of
atoms.
The number of atoms is balanced.
ii) Balance the ionic halfequation with respect
to the number of
charges.
Charge on left-hand side = +2
Charge on right-hand side = +4
∴ add 2e– on the right-hand side to balance the charge.
2+
The balanced ionic half-equation for Sn
2+
4+
–
Sn + 2e ......(ii)
Sn
3 Make the number of
electrons gained in one ionic
half-equation equal to that
lost in the other.
(ii) x 3
4 Combine the two ionic halfequations and eliminate the
electrons.
(i)
2AuCl4– + 6e–
(ii) x 3 3Sn2+
is:
2Au + 8Cl–
3Sn4+ + 6e–
2AuCl4– + 3Sn2+
3Sn4+ + 2Au + 8Cl–
∴ the value of x is 3.
Writing the redox equation using oxidation number method
1 Write down the oxidizing
agent and the reducing
agent involved. Determine
their products.
2 Assign oxidation numbers to
all atoms.
3 Notice atoms which undergo
a change in oxidation
number. Determine the
number of electrons lost or
gained per formula unit.
2+
AuCl4– is the oxidizing agent and Sn
–
Au + Cl
AuCl4–
2+
4+
Sn
Sn
+3
+2
0
AuCl4– + Sn2+
is the reducing agent.
+4
Au + Cl– + Sn4+
reduction: gain of 3e– per AuCl4–
+3
–
AuCl4
+2
+
2+
Sn
0
Au
+
Cl
–
–
+4
+
Sn
4+
2+
oxidation: loss of 2e per Sn
4 Insert an appropriate
coefficient before the
formula of each reagent
on the left-hand side of
the equation to make the
number of electrons gained
equal to that lost.
2+
–
2AuCl4 + 3Sn
Au + Cl– + Sn4+
5 Add appropriate coefficients
on the right to balance
the number of atoms
which have gained or lost
electrons.
2+
–
2AuCl4 + 3Sn
2Au + 8Cl– + 3Sn4+
∴ the value of x is 3.
151
43 D Writing the redox equation using ionic half-equations
1 Write down the oxidizing
agent and the reducing
agent involved.
H3AsO4 is the oxidizing agent and Zn is the reducing agent.
2 a) Write an ionic halfequation for the
reduction process.
The unbalanced ionic half-equation for H3AsO4 is:
H3AsO4
AsH3
i) Balance the ionic halfequation with respect
to the number of
atoms.
To balance the 4 oxygen atoms in H3AsO4, add 4H2O on the right-hand
side.
AsH3 + 4H2O
H3AsO4
+
To balance the 8 hydrogen atoms in 4H2O, add 8H on the left-hand side.
+
AsH3 + 4H2O
H3AsO4 + 8H
ii) Balance the ionic halfequation with respect
to the number of
charges.
Charge on left-hand side = 8 x (+1) = +8
Charge on right-hand side = 0
–
∴ add 8e on the left-hand side to balance the charge.
The balanced ionic half-equation for H3AsO4 is:
+
–
AsH3 + 4H2O......(i)
H3AsO4 + 8H + 8e
2 b) Write an ionic halfequation for the
oxidation process.
The unbalanced ionic half-equation for Zn is:
Zn2+
Zn
i) Balance the ionic halfequation with respect
to the number of
atoms.
The number of atoms is balanced.
ii) Balance the ionic halfequation with respect
to the number of
charges.
Charge on left-hand side = 0
Charge on right-hand side = +2
∴ add 2e– on the right-hand side to balance the charge.
The balanced ionic half-equation for Zn is:
2+
–
Zn + 2e ......(ii)
Zn
3 Make the number of
electrons gained in one ionic
half-equation equal to that
lost in the other.
(ii) x 4
4 Combine the two ionic halfequations and eliminate the
electrons.
(i)
H3AsO4 + 8H+ + 8e–
(ii) x 4 4Zn
∴ x is 4, y is 8 and z is 4.
152
H3AsO4 + 4Zn + 8H+
AsH3 + 4H2O
4Zn2+ + 8e–
AsH3 + 4Zn2+ + 4H2O
Writing the redox equation using oxidation number method
1 Write down the oxidizing
agent and the reducing
agent involved. Determine
their products.
2 Assign oxidation numbers to
all atoms.
3 Notice atoms which undergo
a change in oxidation
number. Determine the
number of electrons lost or
gained per formula unit.
H3AsO4 is the oxidizing agent and Zn is the reducing agent.
H3AsO4
AsH3
2+
Zn
Zn
+5
0
–3
+2
AsH3 + Zn2+
H3AsO4 + Zn
reduction: gain of 8e– per H3AsO4
+5
H3AsO4
0
+
Zn
–3
+2
AsH3
+
Zn2+
oxidation: loss of 2e– per Zn
4 Insert an appropriate
coefficient before the
formula of each reagent
on the left-hand side of
the equation to make the
number of electrons gained
equal to that lost.
H3AsO4 + 4Zn
AsH3 + Zn2+
5 Add appropriate coefficients
on the right to balance
the number of atoms
which have gained or lost
electrons.
H3AsO4 + 4Zn
AsH3 + 4Zn
6 Balance the number of all
other atoms except O and
H.
The number of all other atoms except O is balanced.
7 Add H+ to the side deficient
in positive charges to make
the number of charges on
both sides equal.
Total charge on left-hand side = 0
Total charge on right-hand side = 4 x (+2) = +8
+
Add 8H to the left-hand side.
+
AsH3 + 4Zn2+
H3AsO4 + 4Zn + 8H
8 Add H2O to the appropriate
side to balance the number
of O atoms.
H3AsO4 + 4Zn + 8H+
2+
AsH3 + 4Zn2+ + 4H2O
∴ x is 4, y is 8 and z is 4.
153
44 B Writing the redox equation using ionic half-equations
1 Write down the oxidizing
agent and the reducing
agent involved.
MnO4– is the oxidizing agent and H2O2 is the reducing agent.
2 a) Write an ionic halfequation for the
reduction process.
–
The unbalanced ionic half-equation for MnO4 is:
2+
MnO4–
Mn
i) Balance the ionic halfequation with respect
to the number of
atoms.
To balance the 4 oxygen atoms in MnO4–, add 4H2O on the right-hand
side.
Mn2+ + 4H2O
MnO4–
+
To balance the 8 hydrogen atoms in 4H2O, add 8H on the left-hand side.
+
2+
–
Mn + 4H2O
MnO4 + 8H
ii) Balance the ionic halfequation with respect
to the number of
charges.
Charge on left-hand side = (–1) + 8 x (+1) = +7
Charge on right-hand side = +2
–
∴ add 5e on the left-hand side to balance the charge.
–
The balanced ionic half-equation for MnO4 is:
+
–
2+
–
Mn + 4H2O......(i)
MnO4 + 8H + 5e
2 b) Write an ionic halfequation for the
oxidation process.
The unbalanced ionic half-equation for H2O2 is:
O2
H2O2
i) Balance the ionic halfequation with respect
to the number of
atoms.
To balance the hydrogen atoms, add 2H+ on the right-hand side.
O2 + 2H+
H2O2
ii) Balance the ionic halfequation with respect
to the number of
charges.
Charge on left-hand side = 0
Charge on right-hand side = 2 x (+1) = +2
–
∴ add 2e on the right-hand side to balance the charge.
The balanced ionic half-equation for H2O2 is:
+
–
H2O2
O2 + 2H + 2e ......(ii)
3 Make the number of
electrons gained in one ionic
half-equation equal to that
lost in the other.
(i) x 2
(ii) x 5
4 Combine the two ionic halfequations and eliminate the
electrons.
(ii) x 5 5H2O2
(i)
2MnO4– + 16H+ + 10e–
5H2O2 + 2MnO4– + 16H+
5O2 + 10H+ + 10e–
2Mn2+ + 8H2O
5O2 + 10H+ + 2Mn2+ + 8H2O
Since H+ appears on both sides of the equation, simplify the equation by
collecting like terms.
+
–
5O2 + 2Mn2+ + 8H2O
5H2O2 + 2MnO4 + 6H
∴ x is 2, y is 6 and z is 8.
154
Writing the redox equation using oxidation number method
1 Write down the oxidizing
agent and the reducing
agent involved. Determine
their products.
2 Assign oxidation numbers to
all atoms.
3 Notice atoms which undergo
a change in oxidation
number. Determine the
number of electrons lost or
gained per formula unit.
MnO4– is the oxidizing agent and H2O2 is the reducing agent.
–
2+
MnO4
Mn
H2O2
O2
+7
–1
+2
0
Mn2+ + O2
MnO4– + H2O2
reduction: gain of 5e– per MnO4–
+7
–
MnO4
–1
+
+2
Mn2+
H2O2
0
+
O2
oxidation: loss of 2e– per H2O2
4 Insert an appropriate
coefficient before the
formula of each reagent
on the left-hand side of
the equation to make the
number of electrons gained
equal to that lost.
–
2MnO4 + 5H2O2
Mn2+ + O2
5 Add appropriate coefficients
on the right to balance
the number of atoms
which have gained or lost
electrons.
–
2MnO4 + 5H2O2
2Mn
6 Balance the number of all
other atoms except O and
H.
The number of all other atoms except O and H is balanced.
7 Add H+ to the side deficient
in positive charges to make
the number of charges on
both sides equal.
Total charge on left-hand side = 2 x (–1) = –2
Total charge on right-hand side = 2 x (+2) = +4
+
Add 6H to the left-hand side.
+
5O2 + 2Mn2+
5H2O2 + 2MnO4– + 6H
8 Add H2O to the appropriate
side to balance the number
of O atoms.
5H2O2 + 2MnO4– + 6H+
2+
+ 5O2
5O2 + 2Mn2+ + 8H2O
∴ x is 2, y is 6 and z is 8.
155
45 D Writing the redox equation using ionic half-equations
1 Write down the oxidizing
agent and the reducing
agent involved.
IO3– is the oxidizing agent and SO2 is the reducing agent.
2 a) Write an ionic halfequation for the
reduction process.
–
The unbalanced ionic half-equation for IO3 is:
2IO3–
I2
i) Balance the ionic halfequation with respect
to the number of
atoms.
To balance the 6 oxygen atoms in 2IO3–, add 6H2O on the right-hand
side.
I2 + 6H2O
2IO3–
+
To balance the 12 hydrogen atoms in 6H2O, add 12H on the left-hand
side.
+
–
I2 + 6H2O
2IO3 + 12H
ii) Balance the ionic halfequation with respect
to the number of
charges.
Charge on left-hand side = 2 x (–1) + 12 x (+1) = +10
Charge on right-hand side = 0
–
∴ add 10e on the left-hand side to balance the charge.
–
The balanced ionic half-equation for IO3 is:
+
–
–
I2 + 6H2O......(i)
2IO3 + 12H + 10e
2 b) Write an ionic halfequation for the
oxidation process.
i) Balance the ionic halfequation with respect
to the number of
atoms.
ii) Balance the ionic halfequation with respect
to the number of
charges.
The unbalanced ionic half-equation for SO2 is:
SO42–
SO2
To balance the oxygen atoms, add 2H2O on the left-hand side.
SO42–
SO2 + 2H2O
To balance the 4 hydrogen atoms in 2H2O, add 4H+ on the right-hand
side.
SO42– + 4H+
SO2 + 2H2O
Charge on left-hand side = 0
Charge on right-hand side = (–2) + 4 x (+1) = +2
–
∴ add 2e on the right-hand side to balance the charge.
The balanced ionic half-equation for SO2 is:
+
–
SO42– + 4H + 2e ......(ii)
SO2 + 2H2O
3 Make the number of
electrons gained in one ionic
half-equation equal to that
lost in the other.
(ii) x 5
4 Combine the two ionic halfequations and eliminate the
electrons.
(i)
2IO3– + 12H+ + 10e–
(ii) x 5 5SO2 + 10H2O
2IO3– + 12H+ + 5SO2 + 10H2O
I2 + 6H2O
5SO42– + 20H+ + 10e–
I2 + 6H2O + 5SO42– + 20H+
Since H+ and H2O appear on both sides of the equation, simplify the
equation by collecting like terms.
+
–
I2 + 5SO42– + 8H
2IO3 + 5SO2 + 4H2O
∴ x is 5, y is 4 and z is 8.
156
Writing the redox equation using oxidation number method
1 Write down the oxidizing
agent and the reducing
agent involved. Determine
their products.
2 Assign oxidation numbers to
all atoms.
IO3– is the oxidizing agent and SO2 is the reducing agent.
–
2IO3
I2
SO2
SO42–
+5
+4
0
2IO3– + SO2
+6
I2 + SO42–
3 Notice atoms which undergo
a change in oxidation
number. Determine the
number of electrons lost or
gained per formula unit.
reduction: gain of 10e– per 2IO3–
4 Insert an appropriate
coefficient before the
formula of each reagent
on the left-hand side of
the equation to make the
number of electrons gained
equal to that lost.
–
2IO3 + 5SO2
I2 + SO42–
5 Add appropriate coefficients
on the right to balance
the number of atoms
which have gained or lost
electrons.
–
2IO3 + 5SO2
I2 + 5SO42–
6 Balance the number of all
other atoms except O and
H.
The number of all other atoms except O is balanced.
7 Add H+ to the side deficient
in positive charges to make
the number of charges on
both sides equal.
Total charge on left-hand side = 2 x (–1) = –2
Total charge on right-hand side = 5 x (–2) = –10
+
Add 8H to the right-hand side.
–
I2 + 5SO42– + 8H+
2IO3 + 5SO2
8 Add H2O to the appropriate
side to balance the number
of O atoms.
2IO3– + 5SO2 + 4H2O
+5
–
2IO3
+
+4
0
S O2
I2
+6
+
SO42–
oxidation: loss of 2e– per SO2
I2 + 5SO42– + 8H+
∴ x is 5, y is 4 and z is 8.
46 A Combine the two ionic half-equations to obtain a redox equation:
(i)
Cr2O72–(aq) + 14H+(aq) + 6e–
(ii) x 3 3C2O42–(aq)
2Cr3+(aq) + 7H2O(l)
6CO2(g) + 6e–
Cr2O72–(aq) + 14H+(aq) + 3C2O42–(aq)
2Cr3+(aq) + 7H2O(l) + 6CO2(g)
According to the equation, 1 mole of Cr2O72–(aq) ions will react completely with one mole of C2O42–(aq)
3
ions.
157
47 B Combine the two ionic half-equations to obtain a redox equation:
+
–
2+
–
(i) x 4 4MnO4 (aq) + 32H (aq) + 20e
4Mn (aq) + 16H2O(l)
+
3–
(ii) x 5 5As2O3(s) + 25H2O(l)
10AsO4 (aq) + 50H (aq) + 20e
–
4MnO4–(aq) + 32H+(aq) + 5As2O3(s) + 25H2O(l)
4Mn2+(aq) + 16H2O(l) + 10AsO43–(aq) + 50H+(aq)
Since H+(aq) and H2O(l) appear on both sides of the equation, simplify the equation by collecting like
terms.
4Mn2+(aq) + 10AsO43–(aq) + 18H+(aq)
4MnO4–(aq) + 5As2O3(s) + 9H2O(l)
According to the equation, 4 moles of MnO4–(aq) ions will react completely with 5 moles of As2O3(s), i.e.
0.80 mole of MnO4–(aq) ions will react completely with 1 mole of As2O3(s).
48 B
Option
Reaction
–3
A
–3
NH3 + HNO3
+5
3NO2 + H2O
–3, +5
+2
+2, +4, +5
2HNO3 + NO
+5
+5
3Zn + 8HNO3
+2
3Zn(NO3)2 + 2NO + 4H2O
–3
D
+5
NH4NO3
+4
B
C
Oxidation number(s) exhibited by nitrogen
+5
–3
(NH4)2SO4
+2, +5
–3
–3
NH4HSO4 + NH3
∴ in the reaction in Option B, nitrogen exhibits three different oxidation numbers.
49 B
Option
Reaction
+4
A
+4
CaSO3 + H2SO3
–2
B
C
Oxidation number(s) exhibited by sulphur
+4
0
2H2S + SO2
+6
Cu + 2H2SO4
+6
–2, 0, +4
2H2O + 3S
+6
D
+4
Ca(HSO3)2
+4
+4
CuSO4 + 2H2O + SO2
+6
+4, +6
+6
2H2SO4 + SO3
+6
H2S2O7
∴ in the reaction in Option B, sulphur exhibits three different oxidation numbers.
50 C
Option
Conversion
+3
A
3+
Fe (aq)
+5
B
–
NO3 (aq)
+7
C
MnO4 (aq)
D
H2SO4(l)
–
+6
+2
Fe2+(aq)
+2
NO(g)
+2
Mn2+(aq)
+4
SO2(g)
Change in oxidation number of underlined element
1 unit
3 units
5 units
2 units
∴ the conversion in Option C involves the greatest change in oxidation number of the underlined
element.
158
51 B
Option
Conversion
Is the conversion an oxidation?
A
0
–2
S
H2S
B
VO2+
+4
+6
C
0
Cr2O72–
The oxidation number of Cr remains unchanged,
∴ the conversion is not an oxidation.
CH3CH2OH
CH3COOH loses oxygen in the conversion, ∴ the
conversion is a reduction.
+6
2–
CH3COOH
+2
The oxidation number of V increases from +4 to +5,
∴ the conversion is an oxidation.
+5
VO2+
CrO4
D
The oxidation number of S decreases from 0 to –2,
∴ the conversion is a reduction.
+2.5
–1
S4O62– + 2I–
52 B 2S2O32– + I2
The oxidation number of S increases from +2 to +2.5 while that of I decreases from 0 to –1.
Hence the equation represents a redox reaction.
+2
– 2 +1
+4 – 2
+2 +4 – 2
53 B Ca(OH)2 + S O2
+1 – 2
CaS O3 + H2O
Oxidation numbers of all elements remain unchanged in the reaction.
Hence this is NOT a redox reaction.
54 D
Option
Equation
Does the underlined substance become oxidized?
A
+1 – 2 +1
+1
KOH + HNO3
KNO3 + H2O
B
8NH3 + 3Cl2
C
Zn + FeSO4
0
Oxidation numbers of all elements in KOH remain
unchanged, ∴ KOH does NOT become oxidized.
+1
The oxidation number of Cl decreases from 0 to –1,
∴ Cl2 does NOT become oxidized.
–1
6NH4Cl + N2
+2
The oxidation number of Fe decreases from +2 to 0,
∴ FeSO4 does NOT become oxidized.
0
ZnSO4 + Fe
–1
D
–2
0
2HBr + H2SO4
SO2 + 2H2O + Br2
The oxidation number of Br increases from –1 to 0,
∴ HBr becomes oxidized.
55 A The position of calcium in the electrochemical series is the highest among calcium, magnesium, sodium
and zinc. Hence calcium is the strongest reducing agent.
56 D
Element
Atomic number
Name of element
W
10
neon
X
12
magnesium
Y
14
silicon
Z
17
chlorine
Chlorine is an oxidizing agent.
159
+2
0
57 C CO(g) + PdCl2(aq) + H2O(l)
CO2(g) + Pd(s) + 2HCl(aq)
The oxidation number of Pd decreases from +2 to 0, ∴ PbCl2 is reduced in the reaction.
0
+6
+2
58 B Co + SO42– + 4H+
+4
Co2+ + H2SO3 + H2O
The oxidation number of Co increases from 0 to +2, ∴ Co is oxidized.
The oxidation number of S decreases from +6 to +4, ∴ SO42– is reduced.
+7
+2
59 C 16H+(aq) + 2MnO4–(aq) + 5C2O42–(aq)
10CO2(g) + 2Mn2+(aq) + 8H2O(l)
The oxidation number of Mn decreases from +7 to +2, ∴ MnO4–(aq) is reduced.
+4
+6
60 B Br2(aq) + SO2(g) + Na2SO4(aq) + 2H2O(l)
2H2SO4(aq) + 2NaBr(aq)
The oxidation number of S increases from +4 to +6, ∴ SO2(g) is oxidized in the reaction.
61 A
Option
Equation
A
C + 2PbO
B
Br2 + 2KI
C
Zn + CuSO4
D
2Na + Cl2
Explanation
Carbon reduces lead(II) oxide to lead. It is a reducing agent.
CO2 + 2Pb
Bromine oxidizes iodide ions to iodine. It is an oxidizing
agent.
2KBr + I2
Copper(II) sulphate oxidizes zinc to zinc ions. It is an
oxidizing agent.
ZnSO4 + Cu
Chlorine oxidizes sodium to sodium ions. It is an oxidizing
agent.
2NaCl
+2
62 B Option A — Zn(s) + Ni2+(aq)
0
Zn2+(aq) + Ni(s)
Zinc causes the oxidation number of Ni to decrease from +2 to 0.
Hence zinc acts a reducing agent in the reaction.
Option B — Zinc can displace nickel from nickel(II) sulphate solution. Hence the position of zinc in the
electrochemical series is higher than that of nickel.
oxidizing power of
metal ion increasing
Zn2+(aq) + 2e–
Zn(s)
2+
–
Ni (aq) + 2e
Ni(s)
reducing power of
metal increasing
2+
2+
∴ Ni (aq) ion is a stronger oxidizing agent than Zn (aq) ion.
Option C — Zinc is more reactive than nickel.
∴ the position of nickel in the reactivity series is lower than that of zinc.
Option D — Zinc forms ions more readily than nickel does. When zinc and nickel are used as the
electrodes in a lemon cell, electrons flow from the zinc electrode to the nickel electrode
in the external circuit.
Hence zinc is the negative electrode, i.e. the anode.
160
63 D Option D — Sodium sulphite solution can decolorize acidified potassium permanganate solution.
This is because the purple permanganate ions are reduced to pale pink / colourless
manganese(II) ions.
5SO32–(aq) + 2MnO4–(aq) + 6H+(aq)
5SO42–(aq) + 2Mn2+(aq) + 3H2O(l)
64 B X–(aq) can reduce Y and Z while Z–(aq) can only reduce Y.
Hence the reducing power of X–(aq) is greater than that of Z–(aq).
Y–(aq) has NO reaction with both X and Z.
Hence the reducing power of Y–(aq) is the lowest.
∴ the order of reducing power of the ions is X– > Z– > Y–.
65 D From the equation:
2Cr2+(aq) + Co2+(aq)
2Cr3+(aq) + Co(s)
it can be deduced that Co2+(aq) is a stronger oxidizing agent than Cr3+(aq) as it can oxidize Cr2+(aq) to
Cr3+(aq).
From the equation:
Co(s) + Pb2+(aq)
Co2+(aq) + Pb(s)
it can be deduced that Pb2+(aq) is a stronger oxidizing agent than Co2+(aq) as it can oxidize Co(s) to
Co2+(aq).
From the equation:
Fe(s) + 2Cr3+(aq)
Fe2+(aq) + 2Cr2+(aq)
it can be deduced that Cr3+(aq) is a stronger oxidizing agent than Fe2+(aq) as it can oxidize Fe(s) to
Fe2+(aq).
Hence the order of oxidizing power of the species is Pb2+(aq) > Co2+(aq) > Cr3+(aq) > Fe2+(aq)
∴ Pb2+(aq) is the strongest oxidizing agent.
66 C The order of the species in the electrochemical series is shown below:
higher in electrochemical series
2+
–
Fe(s)
–
Cr (s)
Co (aq) + 2e
–
Co(s)
Pb2+(aq) + 2e–
Pb(s)
Fe (aq) + 2e
3+
oxidizing power
increasing
Cr (aq) + e
2+
2+
reducing power
increasing
lower in electrochemical series
161
Option C — The position of Pb2+(aq) in the electrochemical series is lower than that of Cr3+(aq).
2+
2+
2+
2+
3+
Hence when Cr (aq) and Pb (aq) are mixed, Pb (aq) can oxidize Cr (aq) to Cr (aq).
reduced form of half-equation
higher in
electrochemical
series
Cr3+(aq) + e–
oxidizing power
increasing
lower in
electrochemical
series
Cr2+(aq)
reducing power
increasing
oxidize
Pb2+(aq) + 2e–
Pb(s)
oxidized form of half-equation
67 C Ionic half-equation for Cr2O72–(aq):
Cr2O72–(aq) + 14H+(aq) + 6e–
3+
2Cr (aq) + 7H2O(l)......(i)
Ionic half-equation for H2S(aq):
S(s) + 2H+(aq) + 2e–......(ii)
H2S(aq)
The redox equation is obtained by combining the two ionic half-equations:
(i)
Cr2O72–(aq) + 14H+(aq) + 6e–
2Cr3+(aq) + 7H2O(l)
3S(s) + 6H+(aq) + 6e–
(ii) x 3 3H2S(aq)
Cr2O72–(aq) + (14 – 6)H+(aq) + 3H2S(aq)
= 8
2Cr3+(aq) + 7H2O(l) + 3S(s)
Option B is correct.
Option C — According to the ionic half-equation for Cr2O72–(aq), 1 mole of Cr2O72– ions requires 6
moles of electrons for reduction.
68 B Ionic half-equation for VO2+(aq):
VO2+(aq) + 2H+(aq) + e–
VO2+(aq) + H2O(l)......(i)
Ionic half-equation for I–(aq):
2I–(aq)
I2(aq) + 2e–......(ii)
The redox equation is obtained by combining the two ionic half-equations:
(i) x 2 2VO2+(aq) + 4H+(aq) + 2e–
(ii)
162
2VO2+(aq) + 2H2O(l)
2I–(aq)
I2(aq) + 2e–
2VO2+(aq) + 4H+(aq) + 2I–(aq)
2VO2+(aq) + 2H2O(l) + I2(aq)
Option A — Suppose the oxidation number of V in VO2+ ion is x.
x + (–2) x 2 = +1
x = +5
∴ the oxidation number of V in VO2+ ion is +5.
Options C and D — VO2+(aq) ion can oxidize I–(aq) ion.
Hence it is a stronger oxidizing agent than I2(aq).
reduced form of half-equation
higher in
electrochemical
series
I2(aq) + 2e–
oxidizing power
increasing
lower in
electrochemical
series
2I–(aq)
oxidize
VO2+(aq) + 2H+(aq) + e–
reducing power
increasing
VO2+(aq) + H2O(l)
oxidized form of half-equation
69 A
Option
Conversion
–
2–
reduction: gain of 6e per Cr2O7
A
+6
+3
2Cr3+(aq)
Cr2O72–(aq)
reduction: gain of 5e– per IO3–
B
+5
IO3–(aq)
1 0
I2(aq)
2
–
reduction: gain of 3e per MnO4–
C
+7
+4
MnO4–(aq)
MnO2(s)
reduction: gain of 2e– per PbO2
D
+4
PbO2(s)
+2
Pb2+(aq)
∴ the conversion in Option A involves the greatest number of moles of electrons.
163
70 B Option B — Aqueous chlorine oxidizes iodide ions to iodine.
–
–
Cl2(aq) + 2I (aq)
2Cl (aq) + I2(aq)
Option D — Silver chloride forms when sodium chloride solution is added to silver nitrate solution.
+
–
Ag (aq) + Cl (aq)
AgCl(s)
This is NOT a redox reaction.
71 D
Option
Equation for the process
2–
Ba2+(aq) + SO4 (aq)
A
+6
BaSO4(s)
Cr2O72–(aq) + H2O(l)
NOT a redox reaction
Mg(OH)2(s)
NOT a redox reaction
2+
–
Mg (aq) + 2OH (aq)
C
+3
–1
+2
2Fe3+(aq) + 2I–(aq)
D
NOT a redox reaction
+6
2CrO42–(aq) + 2H+(aq)
B
Does a redox reaction occur?
0
2Fe2+(aq) + I2(aq)
a redox reaction
72 A Option A — Chlorine can oxidize sulphurous acid.
Cl2(aq) + SO32–(aq) + H2O(l)
–
+
2Cl (aq) + SO42–(aq) + 2H (aq)
73 B Option B — Aqueous chlorine can oxidize potassium bromide solution to yellow-brown bromine.
Cl2(aq) + 2Br–(aq)
2Cl–(aq) + Br2(aq)
There is NO observable change when aqueous chlorine is added to potassium fluoride
solution.
Hence aqueous chlorine can be used to distinguish potassium bromide solution from
potassium fluoride solution.
74 C Option C — Chlorine has the highest oxidizing power among chlorine, bromine and iodine.
75 C Options A, B and D — Copper does NOT react with dilute ethanoic acid, dilute hydrochloric acid and
dilute sulphuric acid.
76 A Option A — When nitrogen monoxide is exposed to air, it reacts with oxygen in the air to give
nitrogen dioxide gas.
2NO(g) + O2(g)
2NO2(g)
Option B — Reaction 2 involves the following conversion:
+4
NO2(g)
+5
HNO3(aq)
The oxidation number of nitrogen increases from +4 to +5.
Hence nitrogen dioxide undergoes oxidation in Reaction 2.
Option C — The following equation represents the reaction that occurs when dilute nitric acid is added
to copper:
3Cu(s) + 2NO3–(aq) + 8H+(aq)
3Cu2+(aq) + 2NO(g) + 4H2O(l)
HNO3(aq) is converted to NO(g), NOT NO2(g), in the reaction.
164
Option D — The following equation represents the reaction that occurs when concentrated nitric acid
is added to carbon:
C(s) + 4HNO3(aq)
CO2(g) + 4NO2(g) + 2H2O(l)
HNO3(aq) is converted to NO2(g), NOT NO(g), in the reaction.
77 A Option A — Concentrated sulphuric acid liberates steamy fumes of hydrogen chloride when it reacts
with sodium chloride.
NaCl(s) + H2SO4(l)
NaHSO4(s) + HCl(g)
Hydrogen chloride is a weak reducing agent. It is NOT oxidized by concentrated sulphuric
acid.
Hence NO chlorine (halogen) is produced in the reaction.
Option B — When concentrated sulphuric acid is added to sodium iodide, hydrogen iodide is
produced.
NaI(s) + H2SO4(l)
NaHSO4(s) + HI(g)
Hydrogen iodide is oxidized by concentrated sulphuric acid to iodine.
8HI(g) + H2SO4(l)
4I2(s) + H2S(g) + 4H2O(l)
Option C — Aqueous chlorine oxidizes iodide ions to iodine.
Cl2(aq) + 2I–(aq)
2Cl–(aq) + I2(aq)
Option D — Aqueous chlorine oxidizes bromide ions to bromine.
–
Cl2(aq) + 2Br (aq)
2Cl–(aq) + Br2(aq)
78 D Sulphur dioxide is formed when heating sulphur with concentrated sulphuric acid. Sulphur dioxide
reduces yellow-brown iron(III) ions to pale green iron(II) ions.
SO32–(aq) + H2O(l) + 2Fe3+(aq)
SO42–(aq) + 2H+(aq) + 2Fe2+(aq)
79 A Option A — Aqueous sulphur dioxide is a reducing agent. It turns acidified potassium permanganate
solution from purple to pale pink / colourless.
5SO32–(aq) + 2MnO4–(aq) + 6H+(aq)
5SO42–(aq) + 2Mn2+(aq) + 3H2O(l)
Carbon dioxide gives a white precipitate with calcium hydroxide solution.
CO2(g) + Ca(OH)2(aq)
CaCO3(s) + H2O(l)
80 C Option C — Carbon reacts with concentrated sulphuric acid according to the following equation:
0
+6
C(s) + 2H2SO4(aq)
+4
+4
CO2(g) + 2SO2(g) + 2H2O(l)
The oxidation number of oxygen remains the same.
81 C Option C — Copper reacts with concentrated sulphuric acid according to the following equation:
Cu(s) + 2H2SO4(l)
CuSO4(aq) + SO2(g) + 2H2O(l)
SO2(g) is acidic and turns universal indicator red.
SO2(g) turns limewater milky.
SO2(g) + Ca(OH)2(aq)
CaSO3(s) + H2O(l)
165
82 A Ammonia reacts with dilute sulphuric acid or concentrated sulphuric acid to give ammonium sulphate.
83 A
Option
Conversion
+4
A
84 C
Na2SO3 + H2O
+4
The oxidation number of S in
SO2 increases from +4 to +6,
∴ SO2 undergoes oxidation, NOT
reduction.
+6
SO2 + Cl2 + 2H2O
H2SO4 + 2HCl
+4
+6
5SO2 + 2KMnO4 + 2H2O
+6
+6
K2SO4 + 2MnSO4 + 2H2SO4
Equation for the process
2–
2–
Cu(s) + 2H2SO4(l)
CuSO4(aq) + SO2(g) + 2H2O(l)
there is no reaction between potassium chloride and
aqueous bromine
D
Cl2(aq) + 2Br–(aq)
2Cl–(aq) + Br2(aq)
85 A
Option
Equation for the process
0
A
B
C
D
+4
Cl2(aq) + SO32–(aq) + H2O(l)
–1
+6
–
+
2Cl (aq) + SO42–(aq) + 2H (aq)
+4
SO2(aq) + H2O(l)
SO2(aq) + Ca(OH)2(aq)
The oxidation number of S in
SO2 increases from +4 to +6,
∴ SO2 undergoes oxidation, NOT
reduction.
Colour change
+
3SO3 (aq) + Cr2O7 (aq) + 8H (aq)
3SO42–(aq) + 2Cr3+(aq) + 4H2O(l)
B
C
The oxidation number of S in
SO2 remains unchanged, ∴ SO2
does NOT undergo reduction.
+4
SO2 + 2NaOH
Option
A
3S + 2H2O
+4
C
The oxidation number of S in
SO2 decreases from +4 to 0,
∴ SO2 undergoes reduction.
0
SO2 + 2H2S
B
D
–2
Does SO2 undergo reduction?
+4
H2SO3(aq)
CaSO3(s) + H2O(l)
from orange to green
a blue solution is formed
NO obvious colour change
from colourless to yellow-brown
Does SO2
act as a reducing agent?
SO2 causes the oxidation number
of Cl to decrease from 0 to –1,
∴ SO2 acts as a reducing agent.
It is NOT a redox equation.
It is NOT a redox equation.
There is NO reaction between sulphur dioxide and iron(II) sulphate solution.
86 A Option A — Sodium sulphite solution can turn acidified potassium permanganate solution from purple
to pale pink / colourless (due to the formation of maganese(II) ions).
There is NO observable change when sodium sulphate solution is mixed with acidified
potassium permanganate solution.
Hence acidified potassium permanganate solution can be used to distinguish between
sodium sulphite solution and sodium sulphate solution.
166
87 D
higher in electrochemical series
SO42–(aq) + 4H+(aq) + 2e–
oxidizing power
increasing
H2SO3(aq) + H2O(l)
Fe3+(aq) + e–
Fe2+(aq)
Br2(aq) + 2e–
2Br–(aq)
Cr2O72–(aq) + 14H+(aq) + 6e–
reducing power
increasing
2Cr3+(aq) + 7H2O(l)
lower in electrochemical series
Hence Cr2O72–(aq) is the most powerful oxidizing agent.
88 A (1)
reduced form of half-equation
higher in
electrochemical
series
2–
SO4 (aq) + 4H+(aq) + 2e–
oxidizing power
increasing
lower in
electrochemical
series
H2SO3(aq) + H2O(l)
oxidize
–
2Br–(aq)
Br2(aq) + 2e
oxidized form of half-equation
–
Br2(aq) can oxidize H2SO3(aq) and itself is reduced to Br (aq).
89 C
Reaction
(1)
(2)
(3)
Involve oxidation and reduction?
+4 – 2
+4 – 2
N2O4(g)
0
+2
Pb(s) + PbO2(s) + 2H2SO4(aq)
+4
NO oxidation and reduction involved
2NO2(g)
+4
–1
2PbSO4(s) + 2H2O(l)
+2
MnO2(s) + 4HCl(aq)
0
MnCl2(aq) + Cl2(g) + 2H2O(l)
oxidation and reduction are involved
oxidation and reduction are involved
90 D (1) Magnesium displaces iron from iron(II) sulphate solution.
Mg(s) + FeSO4(aq)
MgSO4(aq) + Fe(s)
The oxidation number of iron changes from +2 to 0.
(2) Iron reacts with dilute hydrochloric acid to give iron(II) chloride and hydrogen.
Fe(s) + 2HCl(aq)
FeCl2(aq) + H2(g)
The oxidation number of iron changes from 0 to +2.
(3) Carbon can extract iron from iron(III) oxide. The oxidation number of iron changes from +3 to 0.
167
91 D (2) The redox equation for the reaction between acidified potassium dichromate solution and iron(II)
sulphate solution:
2–
2+
+
3+
Cr2O7 (aq) + 6Fe (aq) + 14H (aq)
3+
2Cr (aq) + 6Fe (aq) + 7H2O(l)
(3) The redox reaction for the reaction between acidified potassium dichromate solution and potassium
sulphite solution:
Cr2O72–(aq) + 3SO32–(aq) + 8H+(aq)
2Cr3+(aq) + 3SO42–(aq) + 4H2O(l)
92 A As MnO4– ion is a stronger oxidizing agent than I2, it will oxidize the I– ion to I2.
Combining the two ionic half-equations, we get the ionic equation for the reaction:
(i) x 2 2MnO4–(aq) + 16H+(aq) + 10e–
2Mn2+(aq) + 8H2O(l)
(ii) x 5 10I–(aq)
5I2(aq) + 10e–
2MnO4–(aq) + 10I–(aq) + 16H+(aq)
2Mn2+(aq) + 5I2(aq) + 8H2O(l)
(1) The reaction mixture is brown in colour due to the presence of iodine.
(3) The oxidation of iodine changes from –1 to 0.
Hence iodide ions are oxidized by permanganate ions in the reaction.
2+
3+
93 D (1) Fe (aq) ions can decolorize acidified potassium permanganate solution but Fe (aq) ions cannot.
MnO4–(aq) + 5Fe2+(aq) + 8H+(aq)
Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
(2) Fe2+(aq) ions give a green precipitate with dilute aqueous ammonia.
Fe3+(aq) ions give a reddish brown precipitate with dilute aqueous ammonia.
Fe2+(aq) + 2OH–(aq)
3+
–
Fe (aq) + 3OH (aq)
Fe(OH)2(s)
Fe(OH)3(s)
3+
(3) Fe (aq) ions can react with SO2(g)
but Fe2+(aq) ions cannot.
SO32–(aq) + H2O(l) + 2Fe3+(aq)
SO42–(aq) + 2H+(aq) + 2Fe2+(aq)
The iron(III) sulphate solution changes from yellow-brown to pale green.
94 D (1) Oxidation number of H = +1
Suppose the oxidation number of O in H2O2 is x.
(+1) x 2 + 2x = 0
x = –1
∴ the oxidation number of O in H2O2 is –1.
(2) Ionic half-equation for MnO4–(aq):
MnO4–(aq) + 8H+(aq) + 5e–
Mn2+(aq) + 4H2O(l)......(i)
Ionic half-equation for H2O2(aq):
H2O2(aq)
O2(g) + 2H+(aq) + 2e–......(ii)
The redox equation is obtained by combining the two ionic half-equations:
168
(i) x 2 2MnO4–(aq) + 16H+(aq) + 10e–
2Mn2+(aq) + 8H2O(l)
(ii) x 5 5H2O2(aq)
5O2(g) + 10H (aq) + 10e
+
–
+
–
2+
2MnO4 (aq) + 5H2O2(aq) + (16 –10)H (aq)
= 6
2Mn (aq) + 8H2O(l) + 5O2(g)
(3) The oxidation of oxygen increases from –1 to 0.
Hence H2O2(aq) is oxidized by MnO4–(aq) ions in the reaction.
95 D (1) The formula of chromate is CrO42–.
Hence the chemical formula of silver chromate is Ag2CrO4.
(2) CrO42–(aq) ions are converted to Cr2O72–(aq) ions upon the addition of H2SO4(aq):
+6 – 2
+1
+6
2CrO42–(aq) + 2H+(aq)
–2
+1 – 2
Cr2O72–(aq) + H2O(l)
The oxidation numbers of all elements remain unchanged. It is NOT a redox reaction.
3+
2+
(3) Zinc causes the conversion of Cr (aq) ions to Cr (aq) ions.
The oxidation number of Cr decreases from +3 to +2, ∴ Cr3+(aq) ions are reduced.
Hence zinc acts as a reducing agent in the conversion.
96 B (2) Chlorine oxidizes iodide ions to iodine.
Cl2(aq) + 2I–(aq)
2Cl–(aq) + I2(aq)
The colour of ammonium iodide solution gradually changes to brown due to the presence of
iodine.
(3) Chlorine oxidizes SO32–(aq) ions to SO42–(aq) ions.
2Cl–(aq) + SO42–(aq) + 2H+(aq)
Cl2(aq) + SO32–(aq) + H2O(l)
However, the sodium sulphite solution would NOT change to brown.
97 B (1) and (3) The atomic size and boiling point of the halogens are in the order:
chlorine < bromine < iodine
98 D (3) Stronger reducing agents lose electrons more readily. The reducing power of halide ions is in the
order of I– > Br– > Cl–.
0
–1
99 D (2) and (3) Cl2(g) + 2NaOH(aq)
+1
NaCl(aq) + NaOCl(aq) + H2O(l)
The oxidation number of the element chlorine is 0. The oxidation numbers of Cl in NaCl
and NaOCl are –1 and +1 respectively. Therefore chlorine is simultaneously reduced and
oxidized.
100 D A reaction in which the same species is simultaneously reduced and oxidized is called disproportionation.
+4
(1) 2OH–(aq) + 2NO2(g)
+5
+3
NO3–(aq) + NO2–(aq) + H2O(l)
The oxidation number of N in NO2 is +4. Oxidation numbers of N in NO3– and NO2– are +5 and +3
respectively.
Hence NO2 is simultaneously reduced and oxidized.
169
+1
(2) 2CuCl(aq)
+2
0
CuCl2(aq) + Cu(s)
The oxidation number of Cu in CuCl is +1. The oxidation numbers of Cu in CuCl2 and Cu are +2
and 0 respectively.
Hence CuCl is simultaneously reduced and oxidized.
0
(3) Cl2(g) + H2O(l)
–1
+1
HCl(aq) + HOCl(aq)
The oxidation number of chlorine is 0. The oxidation numbers of Cl in HCl and HOCl are –1 and
+1 respectively.
Hence Cl2 is simultaneously reduced and oxidized.
101 A Concentrated nitic acid and zinc react according to the following equation:
Zn(s) + 2NO3–(aq) + 4H+(aq)
Zn2+(aq) + 2NO2(g) + 2H2O(l)
(1) Nitrogen dioxide, a brown gas, is evolved.
(2) Ionic half-equation for the NO3–(aq) ion:
NO3–(aq) + 2H+(aq) + e–
NO2(g) + H2O(l)
Hence one mole of NO3–(aq) ions requires one mole of electrons for reduction.
(3) Zinc causes the oxidation number of nitrogen to decrease from +5 (in NO3–) to +4 (in NO2).
Hence zinc acts as a reducing agent in the reaction.
102 A (1) Copper reacts with dilute nitric acid according to the following equation:
3Cu(s) + 2NO3–(aq) + 8H+(aq)
3Cu2+(aq) + 2NO(g) + 4H2O(l)
NO(g) is produced.
(2) Copper reacts with concentrated sulphuric acid according to the following equation:
Cu(s) + 2H2SO4(l)
CuSO4(aq) + SO2(g) + 2H2O(l)
SO2(g) is produced.
(3) There is NO reaction between copper and concentrated ethanoic acid.
103 C (1) There is NO reaction between magnesium sulphate solution and acidified potassium permanganate
solution.
(2) Iron(II) sulphate solution can decolorize acidified potassium permanganate solution.
MnO4–(aq) + 5Fe2+(aq) + 8H+(aq)
Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
(3) Sodium sulphite solution can decolorize acidified potassium permanganate solution.
+
2–
–
5SO3 (aq) + 2MnO4 (aq) + 6H (aq)
5SO42–(aq) + 2Mn2+(aq) + 3H2O(l)
104 A (1) Iodine is reduced to iodide ion.
SO32–(aq) + H2O(l) + I2(aq)
SO42–(aq) + 2H+(aq) + 2I–(aq)
Brown I2(aq) changes to colourless I–(aq) ions in the reaction.
(3) Sulphur dioxide acts as a reducing agent. It is oxidized.
170
105 B Aqueous bromine and sodium sulphite solution react according to the following equation:
+
2–
SO3 (aq) + H2O(l) + Br2(aq)
–
SO42–(aq) + 2H (aq) + 2Br (aq)
(1) The yellow-brown aqueous bromine is reduced to bromide ions.
Hence the reaction mixture is colourless.
2–
(2) Ionic half-equation for SO3 (aq) ion:
SO32–(aq) + H2O(l)
SO42–(aq) + 2H+(aq) + 2e–
Two moles of electrons are required for the oxidation of 1 mole of SO32–(aq) ions.
(3) The oxidation number of Br decreases from 0 to –1 in the reaction.
Hence bromine is reduced in the reaction.
106 D (1) Sulphur dioxide and carbon dioxide dissolve in water to form sulphurous acid and carbonic acid
respectively.
SO2(g) + H2O(l)
H2SO3(aq)
CO2(g) + H2O(l)
H2CO3(aq)
(2) Both sulphur dioxide and carbon dioxide can turn limewater milky.
SO2(g) + Ca(OH)2(aq)
CaSO3(s) + H2O(l)
CO2(g) + Ca(OH)2(aq)
CaCO3(s) + H2O(l)
107 C (1) There is NO reaction between chlorine and potassium nitrate solution.
(2) Carbon dioxide and sodium hydroxide solution react according to the following equation:
CO2(g) + NaOH(aq)
NaHCO3(aq)
(3) Sulphur dioxide and iron(III) sulphate solution react according to the following equation:
SO32–(aq) + H2O(l) + 2Fe3+(aq)
SO42–(aq) + 2H+(aq) + 2Fe2+(aq)
108 B (1) When sulphur dioxide dissolves in water, it reacts to form sulphurous acid. Therefore the solution
conducts electricity better than water.
(2) Aqueous sulphur dioxide can reduce a solution containing iron(III) ions. The colour of the solution
changes from yellow-brown to pale green.
SO32–(aq) + H2O(l) + 2Fe3+(aq)
SO42–(aq) + 2H+(aq) + 2Fe2+(aq)
(3) Aqueous sulphur dioxide can reduce acidified potassium permanganate solution. Colourless / pale
pink manganese(II) ions are formed.
5SO32–(aq) + 2MnO4–(aq) + 6H+(aq)
5SO42–(aq) + 2Mn2+(aq) + 3H2O(l)
109 C (1) There is NO reaction between aqueous bromine and ammonium chloride solution.
(2) Aqueous bromine can oxidize iodide ions to iodine.
Br2(aq) + 2I–(aq)
2Br–(aq) + I2(aq)
(3) Sulphur dioxide can reduce bromine to bromide ions.
SO32–(aq) + H2O(l) + Br2(aq)
+
–
SO42–(aq) + 2H (aq) + 2Br (aq)
171
110 A (1) Potassium iodide solution and acidified potassium permanganate solution react according to the
following equation:
–
–
+
2+
2MnO4 (aq) + 10I (aq) + 16H (aq)
2Mn (aq) + 5I2(aq) + 8H2O(l)
(2) Potassium iodide solution and silver nitrate solution react according to the following equation:
+
–
Ag (aq) + I (aq)
AgI(s)
(3) There is NO reaction between potassium iodide solution and sodium sulphite solution.
111 B Potassium reacts with water according to the following equation:
2K(s) + 2H2O(l)
2KOH(aq) + H2(g)
Potassium causes the oxidation number of hydrogen to decrease from +1 in H2O to 0. Hence potassium
acts as a reducing agent.
112 D The thermal decomposition of calcium carbonate is NOT a redox reaction.
CaCO3(s)
CaO(s) + CO2(g)
113 D Lead is less reactive than iron.
Hence lead CANNOT displace iron from iron(II) nitrate solution.
114 A The position of sodium in the electrochemical series is higher than that of magnesium.
115 D Dilute nitric acid is an oxidizing agent. It will compete with the potassium permanganate solution for
the reducing agent in reactions.
Hence the common oxidizing agent potassium permanganate solution is usually acidified with dilute
sulphuric acid, NOT dilute nitric acid.
116 C Acidified potassium permanganate solution and iron(II) sulphate solution react according to the
following equation:
MnO4–(aq) + 5Fe2+(aq) + 8H+(aq)
Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
Purple permanganate ions are reduced to colourless maganese(II) ions in the reaction.
Pale green iron(II) ions are oxidized to yellow-brown iron(III) ions.
Hence the reaction mixture is NOT colourless.
117 C Acidified potassium permanganate solution and potassium iodide solution react according to the
following equation:
2MnO4–(aq) + 10I–(aq) + 16H+(aq)
2Mn2+(aq) + 5I2(aq) + 8H2O(l)
The reaction mixture is brown in colour due to the presence of iodine.
118 C The oxidation number of nitrogen may NOT be –3 in compounds.
e.g. the oxidation number of nitrogen in NO3– is +5;
the oxidation number of nitrogen in NO2 is +4.
172
119 D Both bromine and chlorine can oxidize iodide ions to iodine.
–
2Br (aq) + I2(aq)
–
2Cl (aq) + I2(aq)
Br2(aq) + 2I (aq)
Cl2(aq) + 2I (aq)
–
–
Chlorine is a stronger oxidizing agent than bromine.
120 C Aqueous bromine and sodium iodide solution react according to the following equation:
Br2(aq) + 2I–(aq)
2Br–(aq) + I2(aq)
The reaction mixture is brown in colour due to the presence of iodine.
121 A Zinc liberates nitrogen monoxide, NOT hydrogen, from 1 mol dm–3 nitric acid.
3Zn(s) + 2NO3–(aq) + 8H+(aq)
3Zn2+(aq) + 2NO(g) + 4H2O(l)
122 D Copper liberates nitrogen dioxide, NOT hydrogen, from concentrated nitric acid.
Cu(s) + 2NO3–(aq) + 4H+(aq)
Cu2+(aq) + 2NO2(g) + 2H2O(l)
Concentrated nitric acid is a common oxidizing agent.
123 C When concentrated sulphuric acid is added to sodium chloride, hydrogen chloride gas is formed.
NaCl(s) + H2SO4(l)
NaHSO4(s) + HCl(g)
Hydrogen chloride is a weak reducing agent. It is NOT oxidized by concentrated sulphuric acid.
Hence chlorine is NOT formed in the reaction.
124 D There is NO reaction between sulphur dioxide gas and iron(II) sulphate solution.
125 D The following reaction occurs when sulphur dioxide gas is passed into sodium hydroxide solution.
SO2(g) + NaOH(aq)
NaHSO3(aq)
This is NOT a redox reaction.
126 A Sodium sulphite solution and aqueous bromine react according to the following equation:
SO32–(aq) + H2O(l) + Br2(aq)
SO42–(aq) + 2H+(aq) + 2Br–(aq)
The aqueous bromine is changed from yellow-brown to colourless.
127 A Fe2+(aq) ions can decolorize acidified potassium permanganate solution but Fe3+(aq) ions cannot.
MnO4–(aq) + 5Fe2+(aq) + 8H+(aq)
Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
128 A Acidified potassium dichromate solution and sodium sulphite solution react according to the following
equation:
3SO32–(aq) + Cr2O72–(aq) + 8H+(aq)
3SO42–(aq) + 2Cr3+(aq) + 4H2O(l)
There is NO observable change when acidified potassium dichromate solution is mixed with sodium
sulphate solution.
173
Unit 21 Oxidation and reduction in chemical cells
Fill in the blanks
1
Zn2+(aq) + 2e–
a) i) Zn(s)
ii) 2NH4+(aq) + 2e–
2NH3(aq) + H2(g)
b) hydrogen
2
a) negative
b) positive
–
c) i) Pb(s) + SO42–(aq)
PbSO4(s) + 2e
+
–
ii) PbO2(s) + 4H (aq) + SO42–(aq) + 2e
PbSO4(s) + 2H2O(l)
d) i) negative
ii) positive
3
a) hydrogen
b) oxygen
c) concentrated potassium hydroxide solution
d) i) H2(g) + 2OH–(aq)
2H2O(l) + 2e–
ii) O2(g) + 2H2O(l) + 4e–
4
4OH–(aq)
a) methanol
b) hydrogen
c) hydrogen; oxygen
+
–
d) i) CH3OH(l) + H2O(l)
CO2(g) + 6H (aq) + 6e
3
ii)
O2(g) + 6H+(aq) + 6e–
3H2O(l)
2
True or false
5
F
When a zinc-carbon cell is supplying electricity, reduction occurs at the positive electrode:
2NH4+(aq) + 2e–
2NH3(aq) + H2(g)
Manganese(IV) oxide, an oxidizing agent, is used to remove the hydrogen.
2MnO2(s) + H2(g)
Mn2O3(s) + H2O(l)
Hence manganese(IV) oxide is reduced.
6
T
EJHJUBMNVMUJNFUFS
BTWPMUNFUFS
FMFDUSPOGMPX
m
m
[JODFMFDUSPEF
OFHBUJWFFMFDUSPEF
m
F
F
)
;O
174
DPQQFSFMFDUSPEF
QPTJUJWFFMFDUSPEF
EJMVUFTVMQIVSJDBDJE
In the above chemical cell, zinc atoms lose electrons and form zinc ions.
Hence the zinc electrode gradually dissolves.
7
T When a lead-acid accumulator discharges, the following chemical change occurs at the lead plates (the
anode):
PbSO4(s) + 2e–
Pb(s) + SO42–(aq)
∴ oxidation occurs at the anode.
8
F
In an alkaline fuel cell, only hydrogen acts as the fuel.
9
F
In a direct methanol fuel cell, methanol is oxidized in the presence of water at the negative electrode.
CO2(g) + 6H+(aq) + 6e–
CH3OH(l) + H2O(l)
10 T When a glass of wine is exposed to the air for some time, the wine would become sour due to the
formation of ethanoic acid from ethanol.
CH3CH2OH
ethanol
oxidation
CH3COOH
ethanoic acid
Multiple choice questions
11 D
EJHJUBMNVMUJNFUFS
BTWPMUNFUFS
FMFDUSPOGMPX
DPOEVDUJOHXJSF
NBHOFTJVN
Fm
Fm
TBMU
CSJEHF
NBHOFTJVN
OJUSBUFTPMVUJPO
.H
MFBE
MFBE**
OJUSBUF
TPMVUJPO
1C
Option A — Magnesium forms ions more readily than lead does.
Hence electrons flow from magnesium to lead in the external circuit.
Options B and D — Magnesium atoms lose electrons and form magnesium ions.
Hence the magnesium strip gradually dissolves.
Electrons given up by magnesium atoms flow along the conducting wires to the
lead strip. Lead(II) ions in the lead(II) nitrate solution near to the lead strip gain
these electrons and form lead atoms. As a result, a deposit of lead forms on the
lead strip.
2+
–
Pb (aq) + 2e
Pb(s)
Hence lead(II) ions are reduced to lead.
175
Option C — Magnesium atoms lose electrons and form magnesium ions.
2+
–
Mg (aq) + 2e
Mg(s)
Hence magnesium is oxidized.
12 B The position of X is higher than that of zinc in the electrochemical series.
Hence X forms ions more readily than zinc does.
Option A — Atoms of X lose electrons and form ions.
FMFDUSPOGMPX
7
DPOEVDUJOHXJSF
[JODTUSJQ
Fm
Fm
TUSJQNBEFPG9
TBMU
CSJEHF
;O40BR
;O
940BR
9
Hence electrons flow from the strip made of X to the zinc strip in the external circuit.
Option B — Atoms of X lose electrons and form ions.
X2+(aq) + 2e–
X(s)
Hence oxidation occurs at the strip made of X.
Option C — Zinc ions in the zinc sulphate solution near to the zinc strip gain electrons and form zinc
atoms. As a result, a deposit of zinc forms on the zinc strip.
Option D — Electrons flow in the external circuit, NOT through the salt bridge.
13 A Chemical change at the strip made of X, the anode:
X(s)
X2+(aq) + 2e–
Chemical change at the zinc strip, the cathode:
Zn2+(aq) + 2e–
Zn(s)
14 B It is observed that nickel is placed in a zinc sulphate solution, no reaction occurs.
It can be deduced that zinc is more reactive than nickel. The position of zinc in the electrochemical
series is higher than that of nickel.
Option A —
FMFDUSPOGMPX
7
OJDLFMFMFDUSPEF
Fm
Fm
TBMU
CSJEHF
OJDLFM**
TVMQIBUF
TPMVUJPO
/J
[JODFMFDUSPEF
[JODTVMQIBUF
TPMVUJPO
;O
Zinc forms ions more readily than nickel does.
Hence electrons flow from the zinc electrode to the nickel electrode in the external circuit.
176
Option B — Zinc atoms lose electrons and form zinc ions.
2+
–
Zn (aq) + 2e
Zn(s)
Hence oxidation occurs at the zinc electrode.
Option C — Reduction occurs at the nickel electrode.
2+
–
Ni (aq) + 2e
Ni(s)
Hence the nickel electrode is the cathode.
Option D —
LFZ
OJDLFM
FMFDUSPEF
TBMU
CSJEHF
[JOD
FMFDUSPEF
[JOD
TVMQIBUF
TPMVUJPO
OJDLFM**
TVMQIBUF
TPMVUJPO
;OJPOT
40mJPOT
QPTJUJWFJPOTJFDBUJPOT
GSPNUIFTBMUCSJEHF
OFHBUJWFJPOTJFBOJPOT
GSPNUIFTBMUCSJEHF
∴ anions in the salt bridge move towards the zinc half-cell.
15 D
7
FMFDUSPOGMPX
TUSJQNBEFPG9
TUSJQNBEFPG:
TBMU
CSJEHF
9/0
BR
:/0
BR
Electrons flow from X to Y in the external circuit. Hence X forms ions more readily than Y does.
Option A — In the above chemical cell, atoms of X lose electrons and form ions.
Hence the strip made of X gradually dissolves.
Option B — In the electrochemical series, the position of X is higher than that of Y.
Option C — Oxidation occurs at the strip made of X.
X2+(aq) + 2e–
X(s)
Hence the strip made of X is the anode.
Option D — Electrons given up by atoms of X flow along the conducting wires to the strip made of
Y. Y2+(aq) ions in Y(NO3)2(aq) near to the strip made of Y gain these electrons and form
atoms of Y.
Y2+(aq) + 2e–
Y(s)
2+
Hence Y (aq) ions are reduced to Y(s).
177
16 A
FMFDUSPOGMPX
[JODQMBUF
"
m
F
Fm
DPQQFSQMBUF
)
;O
NPMENm
TVMQIVSJDBDJE
Zinc forms ions more readily than copper does.
Hence electrons flow from the zinc plate to the copper plate in the external circuit.
Option A — Zinc atoms lose electrons and form zinc ions.
2+
Zn(s)
–
Zn (aq) + 2e
As a result, the zinc plate gradually dissolves.
Option B — Zinc ions are colourless.
Hence the sulphuric acid would NOT turn blue.
Option C — Electrons given up by zinc atoms flow along the conducting wires to the copper plate.
Hydrogen ions in dilute sulphuric acid near to the copper plate gain these electrons and
form hydrogen.
2H+(aq) + 2e–
H2(g)
Hydrogen ions in the acid are reduced to hydrogen gas.
Option D — Oxidation occurs at the zinc plate.
17 C X reacts with Z(NO3)2 solution according to the following equation:
X(s) + Z2+(aq)
X2+(aq) + Z(s)
It can be deduced that X is more reactive than Z. The position of X in the electrochemical series is
higher than that of Z.
The following chemical cell uses X and Z as the electrodes with sodium chloride solution as the
electrolyte.
EJHJUBMNVMUJNFUFS
BTWPMUNFUFS
FMFDUSPOGMPX
9OFHBUJWFFMFDUSPEF
;QPTJUJWFFMFDUSPEF
TPEJVNDIMPSJEFTPMVUJPO
Option A — X forms ions more readily than Z does.
Hence atoms of X lose electrons and form ions. As a result, the mass of X decreases.
Option B — Electrons flow from X to Z in the external circuit.
178
Option C — Oxidation occurs at X.
Hence X is the anode.
Option D — Z is the positive electrode.
Reduction occurs at Z.
18 C
EJHJUBMNVMUJNFUFS
BTWPMUNFUFS
FMFDUSPOGMPX
DPQQFSFMFDUSPEF:
DPQQFSFMFDUSPEF9
TBMU
CSJEHF
NPMENm$V40BR
#FBLFS"
NPMENm$V40BR
#FBLFS#Ş
In the set-up, electrons flow in such a direction that the concentration of Cu2+(aq) ions in each half-cell
becomes the same eventually, i.e. electrons flow from Y to X in the external circuit.
At electrode Y
Cu(s)
Cu2+(aq) + 2e–
At electrode X
Cu2+(aq) + 2e–
Cu(s)
Option A — The concentration of Cu2+(aq) ions in the solution in beaker A decreases.
Option B — The concentration of Cu2+(aq) ions in the solution in beaker B increases.
Hence the blue colour of the solution in beaker B becomes more intense.
Option D — Copper atoms of electrode Y lose electrons and form copper(II) ions.
As a result, the mass of electrode Y decreases.
19 A
EJHJUBMNVMUJNFUFS
BTWPMUNFUFS
FMFDUSPOGMPX
OJDLFM
FMFDUSPEF
m
NPMEN /J BR
Fm
Fm
/J
1E
DBUJPONPWFNFOU
QBMMBEJVN
FMFDUSPEF
NPMENm
1EBR
BOJPONPWFNFOU
QPSPVTCBSSJFS
Cations (i.e. positive ions) move from the nickel half-cell to the palladium half-cell.
179
It can be deduced that nickel atoms lose electrons and form nickel(II) ions.
Ni(s)
2+
–
Ni (aq) + 2e
Option A — Nickel forms ions more readily than palladium does.
Hence the position of nickel in the electrochemical series is higher than that of palladium.
Option B — Electrons flow from the nickel electrode to the palladium electrode in the external circuit.
Option C — Nickel is oxidized to form nickel(II) ions.
Option D — Electrons flow in the external circuit, NOT through the porous barrier.
20 B In the zinc-carbon cell, the carbon rod is the positive electrode; the outer zinc case is the negative
electrode.
Oxidation occurs at the negative electrode, i.e. the anode:
Zn(s)
Zn2+(aq) + 2e–
Reduction occurs at the positive electrode, i.e. the cathode:
2NH4+(aq) + 2e–
2NH3(aq) + H2(g)
21 D Manganese(IV) oxide, an oxidizing agent, is used to remove the hydrogen produced at the cathode.
2MnO2(s) + H2(g)
Mn2O3(s) + H2O(l)
22 A Hydrogen is produced and collected on the surface of the positive electrode. Since hydrogen is a poor
conductor of electricity, the accumulation of hydrogen at the positive electrode may hinder further
reactions. The voltage drops as a result.
23 B Options A and B — At the aluminium can:
Al(s) + 4OH–(aq)
[Al(OH)4]–(aq) + 3e–
The oxidation number of Al increases from 0 to +3.
Oxidation occurs at the aluminium can.
Hence electrons flow the aluminium can to the carbon rod in the external circuit.
Option C — Oxidation occurs at the aluminium can.
Hence the aluminium can is the anode.
Option D — At the carbon rod:
OCl–(aq) + H2O(l) + 2e–
Cl–(aq) + 2OH–(aq)
The oxdiation number of Cl decreases from +1 to –1.
24 A Option A — In the left beaker, iron(II) ions are oxidized.
Fe2+(aq)
Fe3+(aq) + e–
In the right beaker, permanganate ions are reduced.
MnO4–(aq) + 8H+(aq) + 5e–
180
Mn2+(aq) + 4H2O(l)
25 C Option A — In the left beaker, the iron(II) sulphate solution changes from pale green to yellow-brown
gradually. This is because the iron(II) ions (Fe2+(aq)) lose electrons and change to iron(III)
ions (Fe3+(aq)).
Option B — In the right beaker, the purple colour of the acidified potassium permanganate solution
fades gradually. This is because the permanganate ions (MnO4–(aq)) gain electrons and
change to manganese(II) ions (Mn2+(aq)).
Option C —
EJHJUBMNVMUJNFUFS
BTWPMUNFUFS
FMFDUSPOGMPX
DBSCPOFMFDUSPEF9
DBSCPOFMFDUSPEF:
TBMU
CSJEHF
JSPO**
TVMQIBUFTPMVUJPO
BDJEJGJFEQPUBTTJVN
QFSNBOHBOBUFTPMVUJPO
Electrons flow from electrode X to electrode Y in the external circuit.
Option D — Reduction occurs at electrode Y.
Hence electrode Y is the cathode, NOT the anode.
26 B Option B — In the left beaker, iodide ions are oxidized.
2I–(aq)
I2(aq) + 2e–
In the right beaker, iron(III) ions are reduced.
Fe3+(aq) + e–
Fe2+(aq)
27 B Option A —
EJHJUBMNVMUJNFUFS
BTWPMUNFUFS
FMFDUSPOGMPX
DBSCPOFMFDUSPEF:
DBSCPOFMFDUSPEF9
,*BR
XJUITPNF
TUBSDITPMVUJPO
TBMU
CSJEHF
'F40
BR
Electrons flow from electrode X to electrode Y in the external circuit.
Option B — The iodine produced in the left beaker gives a blue colour when mixed with starch.
Option C — In the right beaker, the iron(III) sulphate solution changes from yellow-brown to pale
green gradually. This is because the iron(III) ions (Fe3+(aq)) gain electrons and change to
iron(II) ions (Fe2+(aq)).
Option D — Oxidation occurs at electrode X.
Hence electrode X is the anode, NOT the cathode.
181
28 B Br2(aq) is a stronger oxidizing agent than Fe3+(aq) ion.
2+
3+
It can be deduced that Br2(aq) would oxidize Fe (aq) ion to Fe (aq) ion.
reduced form of half-equation
higher in
electrochemical
series
3+
Fe (aq) + e
oxidizing power
increasing
lower in
electrochemical
series
–
2+
Fe (aq)
oxidize
–
Br2(aq) + 2e
2Br–(aq)
oxidized form of half-equation
The following chemical changes would occur:
Fe2+(aq)
3+
–
Fe (aq) + e
Br2(aq) + 2e–
2Br–(aq)
Option B — Oxidation occurs at the anode. Hence the following chemical change would occur at the
anode:
Fe2+(aq)
29 C
3+
–
Fe (aq) + e
EJHJUBMNVMUJNFUFS
BTWPMUNFUFS
FMFDUSPOGMPX
DBSCPOFMFDUSPEF:
DBSCPOFMFDUSPEF9
NJYUVSFPG,#SBR
BOE#SBR
TBMU
CSJEHF
NJYUVSFPG/B40BR
BOE/B40BR
Electrons flow from electrode Y to electrode X in the external circuit.
It can be deduced that oxidation occurs at electrode Y while reduction occurs at electrode X.
Option C — Electrode Y is the anode as oxidation occurs here.
Electrode X is the cathode as reduction occurs here.
182
Chemical change that occurs at the anode (i.e. electrode Y):
2–
SO3 (aq)
SO42–(aq)
Chemical change that occurs at the cathode (i.e. electrode X):
2Br–(aq)
Br2(aq)
30 D
EJHJUBMNVMUJNFUFS
BTWPMUNFUFS
DBSCPO
FMFDUSPEF9
TBMU
CSJEHF
NJYUVSFPG
'F$MBR
BOE
'F$MBR
DBSCPO
FMFDUSPEF:
NJYUVSFPG
$V$MBR
BOE
$V$MBR
It is known that Fe3+(aq) ion is a stronger oxidizing agent than Cu2+(aq) ion.
It can be deduced that Fe3+(aq) ion would oxidize Cu+(aq) ion to Cu2+(aq) ion.
reduced form of half-equation
higher in
electrochemical
series
Cu2+(aq) + e–
oxidizing power
increasing
lower in
electrochemical
series
Cu+(aq)
oxidize
3+
–
Fe (aq) + e
Fe2+(aq)
oxidized form of half-equation
The following chemical changes would occur:
Fe3+(aq) + e–
Cu+(aq)
Fe2+(aq)
Cu2+(aq) + e–
Options A and B — Oxidation occurs at electrode Y.
Hence electrons flow from electrode Y to electrode X in the external circuit.
Option C — Cu+(aq) ions are oxidized at electrode Y.
Option D — Reduction occurs at electrode X.
Hence electrode X is the cathode.
183
31 B
EJHJUBMNVMUJNFUFS
BTWPMUNFUFS
1U
DPCBMUTUSJQ
TBMU
CSJEHF
m
NPMEN $P/0
BR
NPMENm
,.O0BR
BOE
NPMENm)BR
JPOT
2+
In the left beaker, cobalt atoms lose electrons and form Co (aq) ions. Oxidation occurs at the cobalt
strip.
In the right beaker, permanganate ions are reduced.
MnO4–(aq) + 8H+(aq) + 5e–
Mn2+(aq) + 4H2O(l)
Option B — The cobalt strip is the anode.
The following chemical change occurs at the anode:
Co(s)
Co2+(aq) + 2e–
32 D Option D — During discharge, reduction occurs at the lead plates coated with lead(IV) oxide, i.e. the
cathode.
PbO2(s) + 4H+(aq) + SO42–(aq) + 2e–
PbSO4(s) + 2H2O(l)
Oxidation occurs at the lead plates, i.e. the anode.
Pb(s) + SO42–(aq)
PbSO4(aq) + 2e–
As the lead plates coated with lead(IV) oxide are the positive electrode, thus Z is the
positive terminal.
33 A
34 A Option A — During charging, the flow of electrons is reversed as compared to the discharging process.
The overall reaction for the charging process is the reverse of the discharging process.
At X (negative electrode of the external power source):
PbSO4(s) + 2e–
Pb(s) + SO42–(aq)
At Y (positive electrode of the external power source):
PbSO4(s) + 2H2O(l)
184
PbO2(s) + 4H+(aq) + SO42–(aq) + 2e–
35 D Option D —
MPBE
0H
)H
DPODFOUSBUFE
/B0)BR
0H
)H
)0H
FMFDUSPEF:
FMFDUSPEF9
At electrode X, the anode:
H2(g) + 2OH–(aq)
2H2O(l) + 2e–
At electrode Y, the cathode:
O2(g) + 2H2O(l) + 4e–
36 D Option D —
–
4OH (aq)
$0
NFUIBOPM
OFHBUJWFFMFDUSPEF
FMFDUSPO
GMPX
QPMZNFSJDNFNCSBOFBTFMFDUSPMZUF
)
QPTJUJWFFMFDUSPEF
PYZHFOBJS
BJS
37 D Option D — A direct methanol fuel cell uses methanol as the fuel and air as an oxidant.
At the anode:
CH3OH(l) + H2O(l)
38 C
CO2(g) + 6H+(aq) + 6e–
EJHJUBMNVMUJNFUFS
BTWPMUNFUFS
DBSCPO
FMFDUSPEF
NJYUVSFPG
,#SBR
BOE#SBR
TBMU
CSJEHF
DBSCPO
FMFDUSPEF
NJYUVSFPG
/B40BR
BOE
/B40BR
SO32–(aq) ion is a stronger reducing agent than Br–(aq) ion.
It can be deduced that SO32–(aq) ion would reduce Br2(aq) to Br–(aq) ion, i.e. Br2(aq) would oxidize SO32–(aq)
ion to SO42–(aq) ion.
185
reduced form of half-equation
higher in
electrochemical
series
SO42–(aq) + 2H+(aq) + 2e–
oxidize
oxidizing power
increasing
lower in
electrochemical
series
2–
SO3 (aq) + H2O(l)
Br2(aq) + 2e–
reducing power
increasing
2Br–(aq)
oxidized form of half-equation
The following chemical changes would occur:
Br2(aq) + 2e–
2Br–(aq)
SO42–(aq) + 2H+(aq) + 2e–
SO32–(aq) + H2O(l)
3+
2+
39 C It is know that Fe (aq) ion is a stronger oxidizing agent than Cu (aq) ion.
It can be deduced that Fe3+(aq) ion would oxidize Cu+(aq) ion to Cu2+(aq) ion.
reduced form of half-equation
higher in
electrochemical
series
Cu2+(aq) + e–
oxidizing power
increasing
lower in
electrochemical
series
oxidize
3+
–
Fe (aq) + e
oxidized form of half-equation
The following chemical changes would occur:
Fe3+(aq) + e–
186
Cu+(aq)
Fe2+(aq)
2+
–
Cu (aq) + e
Cu+(aq)
Fe2+(aq)
(1) Iron(III) ions gain electrons to form iron(II) ions.
(2)
FMFDUSPOGMPX
7
DBSCPO
FMFDUSPEF9
DBSCPO
FMFDUSPEF:
TBMU
CSJEHF
NJYUVSFPG
$V$MBR
BOE
$V$MBR
NJYUVSFPG
'F$MBR
BOE
'F$MBR
Electrons flow from electrode X to electrode Y in the external circuit.
(3) At electrode X, copper(I) ions undergo oxidation to form copper(II) ions.
40 A
7
TJMWFSSPE
Fm
FMFDUSPOGMPX
Fm
DPQQFSSPE
$V
DPQQFS**
TVMQIBUF
TPMVUJPO
$V
(1) In the above chemical cell, copper atoms lose electrons and form copper(II) ions. These ions then
go into the copper(II) sulphate solution.
Cu(s)
2+
–
Cu (aq) + 2e
As a result, the copper rod gradually dissolves.
(2) Electrons given up by copper atoms flow along the conducting wires to the silver rod. Copper(II)
ions in the copper(II) sulphate solution near to the silver rod gain these electrons and form copper
atoms. As a result, a deposit of copper forms on the silver rod.
Cu2+(aq) + 2e–
Cu(s)
Copper(II) ions are reduced, NOT oxidized, at the silver rod.
(3) Copper is transferred from the copper rod to the silver rod.
The concentration of copper(II) ions in the electrolyte remains the same. The colour intensity of the
solution remains unchanged.
41 D
NVMUJNFUFS
FMFDUSPOGMPX
JSPOSPE
WJOFHBS
DPQQFSDBO
(1) In the above chemical cell, iron atoms lose electrons and form iron(II) ions.
As a result, the iron rod gradually dissolves.
187
(2) Electrons given up by iron atoms flow along the conducting wire to the copper can. Hydrogen ions
in the vinegar gain these electrons and form hydrogen.
+
–
2H (aq) + 2e
H2(g)
Hence hydrogen ions in the vinegar are reduced to hydrogen gas.
(3) The iron rod would react with the vinegar.
Hence gas bubbles are given off at the surface of the iron rod immersed in the vinegar.
42 B The set-up is a short circuit cell. Electrons flow from the zinc plate to the copper plate.
(1) The mass of the zinc plate would decrease.
(2) Hydrogen ions in the dilute sulphuric acid gain the electrons and form hydrogen gas.
Hence gas bubbles are formed on the surface of the copper plate.
(3) The sulphuric acid would NOT turn blue as NO blue copper(II) ions are formed.
43 C (1) In the left beaker, the permanganate ions are reduced.
+
–
–
MnO4 (aq) + 8H (aq) + 5e
Mn2+(aq) + 4H2O(l)
The oxidation number of potassium remains unchanged.
∴ potassium ion is NOT reduced.
(2) and (3)
FMFDUSPOGMPX
7
QMBUJOVN
FMFDUSPEF9
m
NPMEN BDJEJGJFE
,.O0BR
TBMU
CSJEHF
QMBUJOVN
FMFDUSPEF:
NPMENm
/B40BR
At electrode Y, the sulphite ions are oxidized.
SO32–(aq) + H2O(l)
+
SO42–(aq) + 2H (aq) + 2e
–
Hence electrons flow from electrode Y to electrode X in the external circuit.
44 B (2) Electrode Y is made of porous nickel coated with nickel oxide.
45 A The following diagram shows the migration of ions in a lithium ion secondary cell during discharging.
QPTJUJWFFMFDUSPEF
DBUIPEF
FMFDUSPMZUF
EJTDIBSHJOH
-J
188
OFHBUJWFFMFDUSPEF
BOPEF
(2) Electrons flow from the negative electrode (i.e. the anode) to the positive electrode (i.e. the
cathode) in the external circuit.
(3) Lithium cobalt oxide or lithium manganese oxide forms the positive electrode, i.e. the cathode.
46 C Oxidation always occurs at the anode.
In the copper-magnesium chemical cell, electrons flow from the anode (i.e. the negative electrode) to
the cathode (i.e. the positive electrode) in the external circuit.
Hence a current flows from the cathode to the anode in the external circuit.
47 A The position of zinc in the electrochemical series is higher than that of silver.
Hence zinc forms ions more readily than silver does.
Zinc atoms lose electrons and form zinc ions, i.e. oxidation occurs at the zinc electrode.
48 A A copper-iron chemical cell is shown below:
EJHJUBMNVMUJNFUFS
BTWPMUNFUFS
FMFDUSPOGMPX
m
m
JSPOFMFDUSPEF
OFHBUJWFFMFDUSPEF
Fm
F
)
'F
DPQQFSFMFDUSPEF
QPTJUJWFFMFDUSPEF
WJOFHBS
In the above chemical cell, electrons given up by iron atoms flow along the conducting wires to the
copper electrode. Hydrogen ions in the vinegar near to the copper electrode gain these electrons and
form hydrogen gas. The hydrogen ions are reduced.
2H+(aq) + 2e–
H2(g)
Hence colourless gas bubbles are given off from the copper electrode.
49 D In a zinc-carbon cell, reduction occurs at the positive electrode:
2NH4+(aq) + 2e–
2NH3(aq) + H2(g)
Manganese(IV) oxide, an oxidizing agent, is used to remove the hydrogen produced at the positive
electrode.
Manganese(IV) oxide is reduced in the process.
50 B There is a slow direct reaction between the zinc electrode and ammonium ions. After some time, the
zinc case becomes too thin and the paste leaks out.
51 B The negative electrode is the anode. Oxidation occurs here.
Pb(s) + SO42–(aq)
PbSO4(s) + 2e–
The positive electrode is the cathode. Reduction occurs here.
PbO2(s) + 4H+(aq) + SO42–(aq) + 2e–
PbSO4(s) + 2H2O(l)
189
52 D The following diagram illustrates the charging process of a lead-acid cell.
FYUFSOBM
QPXFSTPVSDF
FMFDUSPOGMPX
MFBEQMBUFDPBUFE
XJUIMFBE*7
PYJEF
MFBEQMBUF
TVMQIVSJDBDJE
FMFDUSPMZUF
MFBE**
TVMQIBUF
The lead plate is connected to the negative terminal of the external power source.
During the charging process, the following chemical change occurs at the electrode connected to the
positive terminal of the external power source.
PbO2(s) + 4H+(aq) + SO42–(aq) + 2e–
PbSO4(s) + 2H2O(l)
Oxidation, NOT reduction, occurs here.
53 C In an alkaline fuel cell, only hydrogen functions as the fuel.
Unit 22 Electrolysis
Fill in the blanks
1
electrolysis
2
electrolytic cell
3
a) anode; positive
b) cathode; negative
4
a) anode
b) cathode
5
a) electrochemical
b) concentration
c) electrodes
6
a) i) chloride; hydroxide
ii) sodium; hydrogen
b) i) hydroxide; oxygen
ii) hydrogen; hydrogen
c) i) 4OH–(aq)
O2(g) + 2H2O(l) + 4e–
ii) 2H+(aq) + 2e–
d) i) hydrogen; acidic
ii) hydroxide; alkaline
190
H2(g)
7
a) i) chloride; chlorine
ii) hydrogen; hydrogen
b) i) 2Cl–(aq)
Cl2(g) + 2e–
ii) 2H+(aq) + 2e–
H2(g)
c) sodium; hydroxide; sodium hydroxide
8
a) i) sulphate; hydroxide
ii) copper(II); hydrogen
b) i) hydroxide; oxygen
ii) copper(II); copper
c) i) 4OH–(aq)
O2(g) + 2H2O(l) + 4e–
ii) Cu2+(aq) + 2e–
Cu(s)
d) less; decreases
9
a) i) copper; copper(II)
ii) copper(II); copper
b) i) Cu(s)
2+
2+
–
Cu (aq) + 2e
–
ii) Cu (aq) + 2e
Cu(s)
d) remains the same; remains the same / does not change
10 electroplating
True or false
11 F
In an electrolytic cell, reduction takes place at the cathode.
12 F
The type of electrode used for electrolysis would affect the products obtained.
13 F
During the electrolysis of dilute sulphuric acid using platinum electrodes, hydrogen ions and hydroxide
ions are discharged. The net effect is that water is decomposed. The number of hydrogen ions and
sulphate ions from the sulphuric acid remains the same. The concentration of sulphuric acid increases
at the end as water is consumed in the electrolysis.
14 F
During the electrolysis of 0.1 mol dm–3 copper(II) chloride solution, chloride ions and hydroxide ions are
attracted to the anode.
A hydroxide ion is higher than a chloride ion in the electrochemical series. That means, a hydroxide ion
is a stronger reducing agent than a chloride ion. Therefore hydroxide ions are preferentially discharged
to form oxygen gas at the anode.
15 T During the electrolysis of concentrated sodium chloride solution using carbon electrodes, hydrogen ions
and chloride ions are consumed. Sodium ions and hydroxide ions remain in the solution. Eventually, the
solution becomes sodium hydroxide solution.
191
16 T During the electrolysis of concentrated sodium chloride solution using a carbon anode and a mercury
cathode, chloride ions are preferentially discharged at the anode.
This is because the concentration of chloride ions in the solution is much higher than that of hydroxide
ions.
17 T During the electrolysis of dilute copper(II) sulphate solution using a copper anode and a carbon
cathode, copper is a stronger reducing agent than hydroxide ions and sulphate ions. The copper anode
dissolves to form copper(II) ions (oxidized).
Cu(s)
Cu2+(aq) + 2e–
Copper(II) ions are preferentially discharged (reduced) to form a deposit of copper on the cathode.
Cu2+(aq) + 2e–
Cu(s)
The net effect is the transfer of copper from the anode to the cathode.
The concentration of copper(II) ions in the electrolyte remains the same.
Hence the colour intensity of the copper(II) sulphate solution remains unchanged.
18 T
19 T During the refining process, the copper is gradually transferred from the anode to the cathode. The
concentration of copper(II) ions in the electrolyte drops gradually. This is because at the anode, iron and
zinc dissolve as ions readily while at the cathode, copper(II) ions are always preferentially discharged.
20 F
In an electroplating process, the object to be plated is made the cathode. The plating metal is made
the anode.
Multiple choice questions
21 B There are three kinds of ions in acidified water:
Cation
Anion
From sulphuric acid
H+(aq)
SO42–(aq)
From water
H+(aq)
OH (aq)
–
The sulphate ions and hydroxide ions are attracted to the anode. The hydrogen ions are attracted to
the cathode.
At the anode
A hydroxide ion is a stronger reducing agent than a sulphate ion. Therefore hydroxide ions are
preferentially discharged (oxidized) to form oxygen gas.
At the cathode
Hydrogen ions are discharged (reduced) to form hydrogen gas.
192
22 D Option A — When hydrogen ions and hydroxide ions are discharged, more water molecules dissociate.
The net effect is that water is decomposed.
The numbers of hydrogen ions and sulphate ions from the sulphuric acid remain the
same. The concentration of sulphuric acid increases at the end as water is consumed in
the electrolysis.
Option B — The gas collected above electrode X is oxygen. It can relight a glowing splint.
Option C — The gas collected above electrode Y is hydrogen. It burns with a ‘pop’ sound.
Option D — The overall cell reaction is:
2H2O(l)
2H2(g) + O2(g)
The volume ratio of gases collected above the electrodes is 2 : 1.
23 D In 1 mol dm–3 potassium sulphate solution, there are four kinds of ions:
Cation
Anion
From potassium sulphate
K+(aq)
SO42–(aq)
From water
H+(aq)
OH (aq)
–
Sulphate ions and hydroxide ions are attracted to the anode.
At the anode
A hydroxide ion is a stronger reducing agent than a sulphate ion. Therefore hydroxide ions are
preferentially discharged (oxidized) to form oxygen gas.
4OH–(aq)
O2(g) + 2H2O(l) + 4e–
24 C At the copper electrode (i.e. the cathode)
A hydrogen ion is a stronger oxidizing agent than a potassium ion. Therefore hydrogen ions are
preferentially discharged to form hydrogen gas.
Hence the mass of the copper electrode remains unchanged.
Cations move towards the copper electrode, i.e. to the right.
25 B In the dilute sodium nitrate solution, there are four kinds of ions:
Cation
Anion
From sodium nitrate
Na+(aq)
NO3–(aq)
From water
H+(aq)
OH–(aq)
The nitrate ions and hydroxide ions are attracted to the anode. The sodium ions and hydrogen ions are
attracted to the cathode.
At the anode (electrode X)
A hydroxide ion is a stronger reducing agent than a nitrate ion. Therefore hydroxide ions are
preferentially discharged (oxidized) to form oxygen gas.
At the cathode (electrode Y)
A hydrogen ion is a stronger oxidizing agent than a sodium ion. Therefore hydrogen ions are
preferentially discharged (reduced) to form hydrogen gas.
193
26 C Water ionizes continuously to replace the hydroxide ions discharged at the anode (electrode X).
H2O(l)
+
–
H (aq) + OH (aq)
Thus there is an excess of hydrogen ions near to the anode (electrode X). The solution there becomes
acidic.
At the same time, water dissociates continuously to replace the hydrogen ions discharged at the
cathode (electrode Y). Thus there is an excess of hydroxide ions near to the cathode (electrode Y). The
solution there becomes alkaline.
The solution near electrode X will turn red while the solution near electrode Y will turn blue.
27 A The concentration of sodium nitrate solution increases as water is consumed in the electrolysis.
28 D There are four kinds of ions in 2 mol dm–3 sodium chloride solution:
Cation
Anion
From sodium chloride
Na+(aq)
Cl–(aq)
From water
H+(aq)
OH–(aq)
The chloride ions and hydroxide ions are attracted to the anode. The sodium ions and hydrogen ions
are attracted to the cathode.
Option A — At the cathode
A hydrogen ion is a stronger oxidizing agent than a sodium ion. Therefore hydrogen ions
are preferentially discharged (reduced) to form hydrogen gas.
Options B and C — At the anode
The concentration of chloride ions in the solution is much greater than that of
hydroxide ions. Therefore chlorine ions are preferentially discharged (oxidized) to
form chlorine gas.
Option D — Water dissociates continuously to replace the hydrogen ions discharged at the cathode.
Thus there is an excess of hydroxide ions near to the cathode. The solution there becomes
alkaline.
Hence the phenolphthalein turns pink.
29 A Cations at lower positions of the electrochemical series are more readily discharged because they are
stronger oxidizing agents (i.e. they gain electrons more readily). Hence Ag+(aq) is the most readily
discharged.
.HBR
JODSFBTJOH
)BR
FBTFPG
EJTDIBSHFBU
$V BR
"HBR
194
DBUIPEF
30 B Anions at higher positions of the electrochemical series are more readily discharged because they are
stronger reducing agents (i.e. they lose electrons more readily). Hence OH–(aq) is the most readily
discharged.
0)mBR
JODSFBTJOH
$MmBR
FBTFPG
EJTDIBSHFBU
m
/0 BR
BOPEF
40mBR
31 D In the aqueous Rb2SO4 solution, there are four kinds of ions:
Cation
Anion
+
From Rb2SO4
Rb (aq)
SO42–(aq)
From water
H+(aq)
OH–(aq)
The sulphate ions and hydroxide ions are attracted to the anode. The rubidium ions and hydrogen ions
are attracted to the cathode.
At the anode
A hydroxide ion is a stronger reducing agent than a sulphate ion. Therefore hydroxide ions are
preferentially discharged (oxidized) to form oxygen gas.
At the cathode
A hydrogen ion is a stronger oxidizing agent than a rubidium ion. Therefore hydrogen ions are
preferentially discharged (reduced) to form hydrogen gas.
Water dissociates continuously to replace the hydroxide ions discharged at the anode.
H2O(l)
H+(aq) + OH–(aq)
Thus there is an excess of hydrogen ions near the anode. The solution there becomes acidic. The pH of
the solution around the anode decreases.
At the same time, water dissociates continuously to replace the hydrogen ions discharged at the
cathode. Thus there is an excess of hydroxide ions near to the cathode. The solution there becomes
alkaline. The pH of the solution around the cathode increases.
32 A In 1 mol dm–3 nickel(II) iodide solution, there are four kinds of ions:
Cation
Anion
From nickel(II) iodide
Ni2+(aq)
I–(aq)
From water
H+(aq)
OH–(aq)
The iodide ions and hydroxide ions are attracted to the anode. The nickel(II) ions and hydrogen ions
are attracted to the cathode.
195
At the anode
The concentration of iodide ions in the solution is much greater than that of hydroxide ions. Therefore
iodide ions are preferentially discharged (oxidized) to form iodine.
At the cathode
Nickel(II) ions are discharged (reduced) to form nickel.
(Aqueous solutions of salts of metals below zinc in the electrochemical series tend to liberate the
metal at the cathode. Refer to the electroplating of nickel on objects. This can be explained by the
overvoltage effect, which is not taught in the curriulum.)
33 C In concentrated sodium chloride solution, there are four kinds of ions:
Cation
Anion
From sodium chloride
Na+(aq)
Cl–(aq)
From water
H+(aq)
OH (aq)
–
The chloride ions and hydroxide ions are attracted to the anode. The sodium ions and hydrogen ions
are attracted to the cathode.
At the anode (electrode X)
The concentration of chloride ions in the solution is much greater than that of hydroxide ions.
Therefore chloride ions are preferentially discharged (oxidized) to form chlorine gas.
At the cathode (electrode Y)
A hydrogen ion is a stronger oxidizing agent than a sodium ion. Therefore hydrogen ions are
preferentially discharged (reduced) to form hydrogen gas.
34 C In dilute sodium iodide solution, there are four kinds of ions:
Cation
+
Anion
–
From sodium iodide
Na (aq)
I (aq)
From water
H+(aq)
OH–(aq)
The iodide ions and hydroxide ions are attracted to the anode. The sodium ions and hydrogen ions are
attracted to the cathode.
At the anode
The concentration of iodide ions in the solution is much greater than that of hydroxide ions. Therefore
iodide ions are preferentially discharged (oxidized) to form iodine.
35 D In dilute copper(II) sulphate solution, there are four kinds of ions
From copper(II) sulphate
From water
Cation
Anion
Cu2+(aq)
SO42–(aq)
H+(aq)
OH–(aq)
The copper(II) ions and hydrogen ions are attracted to the cathode
196
At the cathode
A copper(II) ion is a stronger oxidizing agent than a hydrogen ion. Hence copper(II) ions are
preferentially discharged (reduced) to form a deposit of copper on the cathode.
36 C Option A — The blue colour of the solution becomes less intense because the concentration of
copper(II) ions in the solution decreases.
Options B, C and D — Copper(II) ions and hydroxide ions are consumed in the electrolysis. Hydrogen
ions and sulphate ions remain in the solution. Thus the solution eventually
becomes sulphuric acid.
37 B In dilute copper(II) chloride solution, there are four kinds of ions:
From copper(II) chloride
From water
Cation
Anion
Cu2+(aq)
Cl–(aq)
H+(aq)
OH–(aq)
The chloride ions and hydroxide ions are attracted to the anode. The copper(II) ions and hydrogen ions
are attracted to the cathode.
At the anode
Copper is a stronger reducing agent than hydroxide ions and chloride ions. The copper anode dissolves
to form copper(II) ions (oxidized).
$VBR
SFEVDJOH
m
0) BR
QPXFS
JODSFBTJOH
m
$M BR
At the cathode
A copper(II) ion is a stronger oxidizing agent than a hydrogen ion. Therefore copper(II) ions are
preferentially discharged (reduced) to form a deposit of copper on the cathode.
38 C The net effect is the transfer of copper from the anode to the cathode. The rate at which copper
deposits on the cathode is equal to the rate at which the copper anode dissolves.
increase in mass of cathode = decrease in mass of anode
The concentration of copper(II) ions in the electrolyte remains the same.
197
39 A In copper(II) sulphate solution, there are four kinds of ions:
From copper(II) sulphate
From water
Cation
Anion
2+
Cu (aq)
SO42–(aq)
H+(aq)
OH–(aq)
The sulphate ions and hydroxide ions are attracted to the anode. The copper(II) ions and hydrogen ions
are attracted to the cathode.
At the anode
Copper is a stronger reducing agent than hydroxide ions and sulphate ions. The copper anode dissolves
to form copper(II) ions (oxidized).
$VBR
SFEVDJOH
m
0) BR
QPXFS
JODSFBTJOH
m
40 BR
At the cathode
A copper(II) ion is a stronger oxidizing agent than a hydrogen ion. Therefore copper(II) ions are
preferentially discharged (reduced) to form a deposit of copper on the cathode.
Option A — The net effect is the transfer of copper from the anode to the cathode. The rate at which
copper deposits on the cathode is equal to the rate at which the copper anode dissolves.
increase in mass of cathode = decrease in mass of anode
The concentration of copper(II) ions in the electrolyte remains the same.
Hence the blue colour of the solution remains unchanged.
Option B — The pH of the copper(II) sulphate solution remains the same.
40 C In copper(II) sulphate solution, there are four kinds of ions:
From copper(II) sulphate
From water
Cation
Anion
Cu2+(aq)
SO42–(aq)
H+(aq)
OH (aq)
–
The sulphate ions and hydroxide ions are attracted to the anode. The copper(II) ions and hydrogen ions
are attracted to the cathode.
At the anode
A hydroxide ion is a stronger reducing agent than a sulphate ion. Therefore hydroxide ions are
preferentially discharged (oxidized) to form oxygen gas.
198
At the cathode
A copper(II) ion is a stronger oxidizing agent than a hydrogen ion. Therefore copper(II) ions are
preferentially discharged (reduced) to form a deposit of copper on the cathode.
Options A and B — Changes in the solution
Copper(II) ions and hydroxide ions are consumed in the electrolysis. Hydrogen ions
and sulphate ions remain in the solution. Thus the solution eventually becomes
sulphuric acid.
Hence the pH and the colour of the solution would change.
Option C — The mass of the anode remains unchanged.
Option D — The mass of the cathode increases.
41 C Option C — During the electrolysis of dilute sodium bromide solution using carbon electrodes,
hydrogen ions and bromide ions are consumed, sodium ions and hydroxide ions remain in
the solution. The solution eventually becomes sodium hydroxide solution.
42 A
Option
A
B
C
D
43 A
Electrolyte
0.1 mol dm
–3
–3
0.1 mol dm
Anode
Cathode
Major product at anode
CuCl2
carbon
carbon
oxygen
CuCl2
copper
copper
copper(II) ions
–3
CaCl2
carbon
carbon
chlorine
–3
CaCl2
copper
copper
copper(II) ions
3 mol dm
3 mol dm
Option
Electrolyte
Major product at cathode
A
1 mol dm–3 silver nitrate solution
silver
B
1 mol dm–3 potassium hydroxide solution
hydrogen
C
2 mol dm–3 calcium chloride solution
hydrogen
D
2 mol dm–3 magnesium sulphate solution
hydrogen
44 B In concentrated sodium chloride solution, there are four kinds of ions:
Cation
Anion
From sodium chloride
Na+(aq)
Cl–(aq)
From water
H+(aq)
OH–(aq)
At the anode
The concentration of chloride ions in the solution is much greater than that of hydroxide ions. Chloride
ions are preferentially discharged (oxidized) to form chlorine gas.
2Cl–(aq)
Cl2(g) + 2e–
199
At the cathode
A sodium ion is a weaker oxidizing agent than a hydrogen ion. However, if mercury is used as the
cathode, sodium ions are preferentially discharged (reduced) to form sodium metal. The metal formed
dissolves in the mercury cathode to form an alloy called sodium amalgam.
Na+(aq) + e– + Hg(l)
Na/Hg(l)
sodium amalgam
The sodium amalgam then moves towards the water. The sodium reacts with water to form sodium
hydroxide and hydrogen.
2Na/Hg(l) + 2H2O(l)
2NaOH(aq) + H2(g) + 2Hg(l)
45 B Sodium ions and chloride ions are consumed in the electrolysis. Thus the sodium chloride solution
becomes more and more dilute.
46 C Chemical changes that occur at the cathodes of the cells:
Ag+(aq) + e–
Ag(s)
Pd2+(aq) + 2e–
Pd(s)
For the same number of moles of electrons,
the number of moles of Ag : Pd is 2 : 1.
The relative atomic masses of Ag and Pd are approximately the same.
Thus the mass of Ag produced is approximately twice that of Pd.
47 B The following diagrama shows the set-up used for the electroplating:
TJMWFSSPE
BOPEF
m
"H
"H
EJTDUPCFQMBUFE
DBUIPEF
TPMVUJPODPOUBJOJOH
"HBR
JPOT
Option B — The disc is connected to the negative terminal of the battery.
Chemical change that occurs at the disc:
+
–
Ag (aq) + e
Ag(s)
∴ reduction occurs at the disc.
200
48 C
DPQQFSFMFDUSPEF
EJMVUFDPQQFS**
TVMQIBUFTPMVUJPO
[JODPCKFDU
Option C — At the anode (zinc object)
Zinc in the object gives up electrons to form zinc ions.
Hence the zinc object dissolves.
At the cathode (copper electrode)
A copper(II) ion is a stronger oxidizing agent than a zinc ion. Therefore copper(II) ions are
preferentially discharged to form copper.
Hence copper deposits on the copper electrode.
49 A The following diagram shows the set-up used for the purification of impure copper:
JNQVSFDPQQFS
DPOUBJOJOH[JOD
QVSFDPQQFS
SPE
DPQQFS**
TVMQIBUFTPMVUJPO
The impure copper is the anode while the pure copper rod is the cathode.
Option A — At the anode (impure copper)
Zinc forms ions more readily than copper. When the cell is operating at the correct
voltage, zinc in the anode gives up electrons first. Then copper gives up electrons to form
copper(II) ions.
Hence the mass of the anode decreases.
At the cathode (pure copper rod)
A copper(II) ion is a stronger oxidizing agent than a zinc ion. Therefore copper(II) ions are
preferentially discharged to form copper.
Overall cell reaction
Cu(s)
(anode)
Cu(s)
(cathode)
During the refining process, the copper is gradually transferred from the anode to the
cathode. The concentration of copper(II) ions in the electrolyte drops gradually. This is
because at the anode, zinc dissolves as ions readily while at the cathode, copper(II) ions
are always preferentially discharged.
Hence the colour intensity of the copper(II) sulphate solution decreases.
201
50 B
9
"
CBUUFSZ
:
#
QMBUJOVN
FMFDUSPEFT
DPQQFS
FMFDUSPEFT
$
EJMVUFDPQQFS**
TVMQIBUFTPMVUJPO
%
EJMVUFTPEJVN
OJUSBUFTPMVUJPO
Option B — On completing the circuit, the mass of electrode B increases.
This is because copper(II) ions are discharged (reduced) to form copper at electrode B.
Thus electrode A is the anode while electrode B is the cathode.
X is the positive terminal of the battery.
51 A Options A and B — At the anode (electrode A)
Copper is a stronger reducing agent than hydroxide and sulphate ions. The copper
anode (electrode A) dissolves to form copper(II) ions (oxidized).
The net effect is the transfer of copper from electrode A to electrode B. The rate
at which copper deposits on electrode B is equal to the rate at which electrode A
dissolves.
increase in mass of cathode = decrease in mass of anode
The concentration of copper(II) ions in the copper(II) sulphate solution remains the
same.
Hence the colour of the copper(II) sulphate solution remains unchanged.
Options C and D — Electrode C is the anode while electrode D is the cathode.
At electrode C, hydroxides ions are preferentially discharged to form oxygen gas.
At electrode D, hydrogen ions are preferentially discharged to form hydrogen gas.
52 C P and Q are two different metals. When a current flows in the external circuit of the set-up for some
time, copper is found to deposit on the carbon electrode R.
"
FMFDUSPEF
NBEFPG1
DBSCPO
FMFDUSPEF4
DPQQFS**
TVMQIBUF
TPMVUJPO
FMFDUSPEF
NBEFPG2
TFUVQ9
202
DBSCPO
FMFDUSPEF3
TFUVQ:
In this case, set-up X functions as a chemical cell while set-up Y is an electrolytic cell.
As copper is found to deposit on the carbon electrode R, it can be deduced that electrode R is the
cathode of the electrolytic cell, set-up Y.
Electrode S is the anode of set-up Y.
Electrons flow from the electrode made of Q to electrode R. Oxidation occurs at the electrode made of
Q, i.e. this is the anode of set-up X.
53 B (2)
"
FMFDUSPOGMPX
FMFDUSPEF
NBEFPG1
DBSCPO
FMFDUSPEF4
DPQQFS**
TVMQIBUF
TPMVUJPO
FMFDUSPEF
NBEFPG2
TFUVQ9
DBSCPO
FMFDUSPEF3
TFUVQ:
∴ electrons flow from the electrode made of Q to carbon electrode R in the external circuit.
(3) In set-up X, oxidation occurs at the electrode made of Q.
It can be deduced that Q forms ions more readily than P does.
Thus Q occupies a higher position in the electrochemical series than P.
54 B In dilute sodium bromide solution, there are four kinds of ions:
Cation
Anion
From sodium bromide
Na+(aq)
Br–(aq)
From water
H+(aq)
OH (aq)
–
The bromide ions and hydroxide ions are attracted to the anode. The sodium ions and hydrogen ions
are attracted to the cathode.
(1) and (2) At the anode
The concentration of bromide ions in the solution is much greater than that of hydroxide
ions. Therefore bromide ions are preferentially discharged (oxidized) to form bromine.
Hence a yellow-brown colour develops near the anode.
(3) At the cathode
A hydrogen ion is a stronger oxidizing agent than a sodium ion. Therefore hydrogen ions are
preferentially discharged (reduced) to form hydrogen gas.
Hydrogen ions and bromide ions are consumed in the electrolysis. Sodium ions and hydroxide ions
remain in the solution. Eventually, the solution becomes sodium hydroxide solution.
Hence the pH of the solution in the electrolytic cell increases.
203
55 B There are four kinds of ions in nickel(II) sulphate solution:
From nickel(II) sulphuric
From water
Cation
Anion
2+
Ni (aq)
SO42–(aq)
H+(aq)
OH–(aq)
The sulphate ions and hydroxide ions are attracted to the anode. The nickel(II) ions and hydrogen ions
are attracted to the cathode.
At the anode
A hydroxide ion is a stronger reducing agent than a sulphate ion. Therefore hydroxide ions are
preferentially discharged (oxidized) to form oxygen gas.
At the cathode
Nickel(II) ions are discharged (reduced) to form nickel.
(Aqueous solutions of salts of metals below zinc in the electrochemical series tend to liberate the
metal at the cathode. Refer to the electroplating of nickel on objects. This can be explained by the
overvoltage effect, which is not taught in the curriulum.)
(1) A deposit of nickel forms on the cathode.
(2) Nickel(II) ions and hydroxide ions are consumed in the electrolysis. Hydrogen ions and sulphate ions
remain in the solution. Eventually, the solution becomes sulphuric acid.
Hence the pH of the solution decreases.
(3) Nickel(II) ions are discharged at the cathode.
Hence the concentration of nickel(II) ions in the solution decreases.
56 D
Solution
Product at cathode
(1)
1 mol dm–3 silver nitrate solution
(2)
2 mol dm–3 potassium nitrate solution
hydrogen
(3)
3 mol dm–3 calcium nitrate solution
hydrogen
silver
57 D (1) Copper(II) ions are discharged (reduced) to form a deposit of copper on the cathode.
(2) Nickel(II) ions are discharged (reduced) to form a deposit of nickel on the cathode.
(3) A sodium ion is a weaker oxidizing agent than a hydrogen ion. However, if mercury is used as the
cathode, sodium ions are preferentially discharged (reduced) to form sodium metal.
58 A
204
Solution
Product at cathode
Product at anode
(1)
concentrated sodium sulphate soltuion
hydrogen
oxygen
(2)
concentrated sodium iodide solution
hydrogen
iodine
(3)
dilute copper(II) sulphate solution
copper
oxygen
59 D (1)
EJHJUBMNVMUJNFUFS
BTWPMUNFUFS
FMFDUSPOGMPX
DPQQFS
NBHOFTJVN
[JOD
QPUBUP"
DPQQFS
QPUBUP#
In the above set-up, two competing potato cells are connected together. The more powerful cell (i.e.
cell A) drives the weaker cell (i.e. cell B) into electrolysis.
Electrons flow from the magnesium strip to the zinc strip across the conducting wire between the
magnesiums trip and the zinc strip.
(2) Magnesium in potato A loses electrons and forms magnesium ions. Hence oxidation occurs at the
electrode.
Thus magnesium is the anode in potato A.
60 B (2) Iron is extracted by carbon reduction.
61 A
62 D During the electrolysis of concentrated sodium bromide solution using carbon electrodes, bromine is
produced at the anode. Bromine, rather than oxygen, is produced due to the concentration effect.
63 C During the electrolysis of 2 mol dm–3 nickel(II) sulphate solution using carbon electrodes, hydroxide ions
are discharged at the anode.
This is because a hydroxide ion is a stronger reducing agent than a sulphate ion. Hydroxide ions are
preferentially discharged (oxidized) to form oxygen gas.
64 C In the electrolysis of 2 mol dm–3 copper(II) chloride solution using carbon electrodes, chlorine is liberated
at the anode.
This is because the concentration of chloride ions in the solution is much greater than that of
hydroxide ions. Chloride ions are preferentially discharged (oxidized) to form chlorine.
65 B Reasons for using carbon anodes include:
• carbon is cheap;
• carbon is inert;
• carbon is a good conductor of electricity.
66 D Carbon anodes are used in the electrolysis of brine.
Chlorine gas can attack platinum.
67 C During the electroplating of nickel on a copper object, a nickel rod is made the anode.
68 A
205
Part B
Topic-based exercise
Multiple choice questions
1
C Option A — Zinc acts as the negative electrode of an alkaline manganese cell.
Oxidation occurs at zinc.
Option B — Maganese(IV) oxide acts the positive electrode, i.e. the cathode.
Option C — Zinc acts as the negative electrode, i.e. the anode.
Oxidation occurs at the anode.
Option D — Potassium hydroxide acts as the electrolyte.
2
D Option A — A silver oxide cell is NOT rechargeable.
Option B — Potassium hydroxide acts as the electrolyte.
Option C — Its anode (i.e. negative electrode) is made of zinc.
Option D — Its cathode (i.e. positive electrode) is made of silver oxide.
B Option B — The energy density of a nickel metal hydride cell is higher than that of a lead-acid
accumulator.
3FMBUJWFFOFSHZEFOTJUZ
3
/J.)DFMM
-FBEBDJE
BDDVNVMBUPS
Option C — Potassium hydroxide acts as the electrolyte.
Option D — Its cathode (i.e. positive electrode) is made of nickel(II) hydroxide.
206
4
D Option D — During discharge, lithium ions move from the anode to the cathode through the
electrolyte.
QPTJUJWFFMFDUSPEF
DBUIPEF
FMFDUSPMZUF
OFHBUJWFFMFDUSPEF
BOPEF
EJTDIBSHJOH
-J
5
D Option D — At electrode Y:
0
Zn + 2OH–
+2
ZnO + H2O + 2e–
The oxidation number of Zn increases from 0 to +2.
Hence oxidation occurs at electrode Y.
Electrode Y is the anode as oxidation occurs here.
6
A Option B — At electrode X:
+4
+2
NiO2(s) + 2H2O(l) + 2e–
Ni(OH)2(s) + 2OH–(aq)
The oxidation number of Ni decreases from +4 to +2.
Hence reduction, NOT oxidation, occurs at X.
Option C — Electrode Y is made of cadmium.
At electrode Y:
0
–
Cd(s) + 2OH (aq)
+2
Cd(OH)2(s) + 2e–
The oxidation number of cadmium increases from 0 to +2.
Hence oxidation occurs at the cadmium electrode.
The anode, NOT cathode, is made of cadmium.
Option D — At electrode Y, cadmium atoms lose electrons. Electrons given up by cadmium atoms flow
along the conducting wires to electrode X.
Hence electrons flow from Y to X in the external circuit when the cell operates.
7
B Electrons flow from P, Q and S to copper when tested. Hence P, Q and S form ions more readily than
copper does.
The Q / copper couple gives the highest voltage while the P / copper couple gives the lowest voltage.
Therefore the difference in the tendency to form ions between Q and copper is the greatest while that
between P and copper is the smallest.
Hence the descending order of tendency of the three metals to form ions is: Q > S > P.
Electrons flow from copper to R. Hence copper forms ions more readily than R does. Thus R has the
least tendency to form ions among the four metals.
207
8
B
Element
Number of protons in atom
Name of element
W
6
carbon
X
8
oxygen
Y
11
sodium
Z
18
argon
∴ element X (oxygen) is likely to be an oxidizing agent.
9
B A current flows from X to Y in the external circuit. Hence electrons flow from Y to X in the external
circuit.
"
FMFDUSPOGMPX
NFUBM:
NFUBM9
EJMVUFTVMQIVSJDBDJE
Option A — The electrode made of X is the positive electrode, i.e. the cathode.
Option B — Atoms of metal Y lose electrons and form ions.
Hence the electrode made of Y gradually dissolves.
Option C — Oxidation occurs at the electrode made of Y.
Option D — Y forms ions more readily than X does.
Hence the position of Y in the electrochemical series is higher than that of X.
10 D
Half-cells used
Positive
electrode
Negative
electrode
Reducing power of metal
I and IV
P
Cu
0.46
Cu forms ions more readily than P does, ∴
the reducing power of Cu is higher than
that of P.
II and IV
Cu
Q
0.57
Q forms ions more readily than Cu does,
∴ the reducing power of Q is higher than
that of Cu.
III and IV
Cu
R
1.10
R forms ions more readily than Cu does,
∴ the reducing power of R is higher than
that of Cu.
II and III
Q
R
0.53
R forms ions more readily than Q does,
∴ the reducing power of R is higher than
that of Q.
∴ the order of increasing reducing power is
P < Cu < Q < R
208
Voltage (V)
11 C The order of the metals in the electrochemical series is shown below:
R+(aq) + e–
R(s)
Q+(aq) + e–
Q(s)
+
–
Cu (aq) + e
SFEVDJOH
QPXFS
JODSFBTJOH
Cu(s)
Hence R is more reactive than Q and thus it can displace Q from a solution of a compound of Q.
12 B Option B — The overall cell reaction for the electrolysis of sea water:
2H+(aq) + 2Cl–(aq)
H2(g) + Cl2(g)
Hence a redox reaction occurs.
13 C Na3PO4 consists of Na+ ions and a PO43– ion.
Suppose the oxidation number of P in the PO43– ion is x.
x + (–2) x 4 = –3
x = +5
14 C (NH4)2Cr2O7 consists of NH4+ ions and a Cr2O72– ion.
Suppose the oxidation number of Cr in Cr2O72– is x.
2x + (–2) x 7 = –2
x = +6
15 C
Option
Species
Oxidation number of Mo in species
A
MoCl5
+5
B
Mo2S3
+3
C
MoO42–
+6
D
Mo6Cl12
+2
∴ the oxidation number of Mo in MoO42– is the highest.
209
16 A
17 B
Species
Oxidation number of N in species
N2
0
Li3N
–3
NO2–
+3
Option
Equation
+6
A
–2
+4
3CaSO4 + CaS
+6
B
4CaO + 4SO2
+6
+6
H2SO4 + SO3
H2S2O7
+6
+6
C
Cu + 2H2SO4
D
Na2S2O3 + 2HCl
+4
CuSO4 + 2H2O + SO2
+2
+4
0
2NaCl + SO2 + H2O + S
∴ in the reaction in Option B, sulphur exhibits the same oxidation numbers in all the species involved.
18 D
Option
Conversion
–
2–
oxidation: loss of 2e per C2O4
A
+3
+4
C2O42–(aq)
2CO2(g)
reduction: gain of 5e– per MnO4–
B
+7
+2
Mn2+(aq)
MnO4–(aq)
oxidation: loss of 2e– per SO32–
C
+4
SO32–(aq)
+6
SO42–(aq)
reduction: gain of 1e– per VO2+
D
+5
VO2+(aq)
+4
VO2+(aq)
∴ the chemical change in Option D involves the smallest number of moles of electrons.
19 C
Option
Conversion
+3
A
Cr2O3
B
TiCl4
0
Cr
0
+4
Ti
+4
+4
C
TiO2
D
H2SO4
+6
210
A reduction occurs?
TiCl4
+4
SO2
The oxidation number of Cr decreases from +3 to 0,
∴ a reduction occurs.
The oxidation number of Ti decreases from +4 to 0,
∴ a reduction occurs.
The oxidation number of Ti remains unchanged,
∴ the conversion is NOT a reduction.
The oxidation number of S decreases from +6 to +4,
∴ a reduction occurs.
20 A
Option
Conversion
–1
+7
Br2
MnO4
C
SO3
D
VO2+
The oxidation number of Mn decreases from +7 to +4,
∴ a reduction occurs.
+4
–
B
The oxidation number of Br increases from –1 to 0,
∴ an oxidation occurs.
0
2Br–
A
An oxidation occurs?
MnO2
The oxidation number of S remains unchanged,
∴ the conversion is NOT an oxidation.
+6
+6
H2S2O7
The oxidation number of V decreases from +5 to +4,
∴ a reduction occurs.
+4
+5
VO2+
21 D Option D — Aqueous bromine and sulphur dioxide gas react according to the following equation:
SO42–(aq) + 2H+(aq) + 2Br–(aq)
SO32–(aq) + H2O(l) + Br2(aq)
Yellow-brown aqueous bromine changes to colourless bromide ions in the reaction.
22 D
Option
Substance that can decolorize acidified
potassium permanganate solution
Equation for the reaction involved
A
concentrated hydrochloric acid
2KMnO4(aq) + 16HCl(aq)
2KCl(aq) + 2MnCl2(aq) + 8H2O(l) + 5Cl2(g)
B
iron(II) chloride solution
2+
+
–
MnO4 (aq) + 5Fe (aq) + 8H (aq)
2+
3+
Mn (aq) + 5Fe (aq) + 4H2O(l)
C
potassium sulphite solution
+
–
2–
2MnO4 (aq) + 5SO3 (aq) + 6H (aq)
2+
2Mn (aq) + 5SO42–(aq) + 3H2O(l)
D
There is NO reaction between sodium sulphate solution and acidified potassium permanganate
solution.
23 D
Option
+2
The oxidation number of Fe increases
from +2 to +3, ∴ the underlined
substance undergoes oxidation.
+3
2FeCl2 + Cl2
A
2FeCl3
+2
+2 +4 – 2
B
MgCO3 + 2HCl
C
Fe2(SO4)3 + 2KI
–1
+2
D
Does the underlined substance undergo
reduction?
Equation
2NH3 + 3CuO
+4 – 2
–2
MgCl2 + CO2 + H2O
0
2FeSO4 + K2SO4 + I2
0
3Cu + N2 + 3H2O
The oxidation number of all elements
in MgCO3 remain unchanged, ∴ NO
reduction occurs.
The oxidation number of I increases from
–1 to 0, ∴ the underlined substance
undergoes oxidation.
The oxidation number of Cu decreases
from +2 to 0, ∴ the underlined substance
undergoes reduction.
211
24 D
Option
A
0
–2
Fe + S
FeS
H2SO4 + CuO
C
NaOCl + SO2 + H2O
0
–2
CuSO4 + H2O
B
ZnSO4 + Mg
The oxidation number of S decreases
from 0 to –2, ∴ the underlined substance
becomes reduced.
+2
+2 –2
D
Does the underlined substance become
oxidized?
Equation
NaCl + H2SO4
+2
Zn + MgSO4
The oxidation number of all elements in
CuO remain unchanged, ∴ the underlined
substance does NOT become oxidized.
NaOCl loses oxygen in the process, ∴ the
underlined substance becomes reduced.
The oxidation number of Mg increases
from 0 to +2, ∴ the underlined substance
becomes oxidized.
25 C Writing the redox equation using ionic half-equations
1 Write down the oxidizing
agent and the reducing
agent involved.
HNO3 is the oxidizing agent and I2 is the reducing agent.
2 a) Write an ionic halfequation for the
reduction process.
The unbalanced ionic half-equation for HNO3 is:
NO
HNO3
i) Balance the ionic halfequation with respect
to the number of
atoms.
ii) Balance the ionic halfequation with respect
to the number of
charges.
To balance the 3 oxygen atoms in HNO3, add 2H2O on the right-hand
side.
NO + 2H2O
HNO3
+
To balance the 4 hydrogen atoms in 2H2O, add 3H on the left-hand side.
+
NO + 2H2O
HNO3 + 3H
Charge on left-hand side = 3 x (+1) = +3
Charge on right-hand side = 0
–
∴ add 3e on the left-hand side to balance the charge.
The balanced ionic half-equation for HNO3 is:
+
–
NO + 2H2O......(i)
HNO3 + 3H + 3e
2 b) Write an ionic halfequation for the
oxidation process.
The unbalanced ionic half-equation for I2 is:
I2
2HIO3
i) Balance the ionic halfequation with respect
to the number of
atoms.
To balance the oxygen atoms, add 6H2O on the left-hand side.
2HIO3
I2 + 6H2O
To balance the 12 hydrogen atoms in 6H2O, add 10H+ on the right-hand
side.
2HIO3 + 10H+
I2 + 6H2O
ii) Balance the ionic halfequation with respect
to the number of
charges.
Charge on left-hand side = 0
Charge on right-hand side = 10 x (+1) = +10
–
∴ add 10e on the right-hand side to balance the charge.
The balanced ionic half-equation for I2 is:
2HIO3 + 10H+ + 10e– ......(ii)
I2 + 6H2O
212
3 Make the number of
electrons gained in one ionic
half-equation equal to that
lost in the other.
(i) x 10
(ii) x 3
4 Combine the two ionic halfequations and eliminate the
electrons.
+
–
(i) x 10 10HNO3 + 30H + 30e
(ii) x 3 3I2 + 18H2O
10NO + 20H2O
6HIO3 + 30H+ + 30e–
3I2 + 10HNO3 + 18H2O
20H2O + 10NO + 6HIO3
Since H2O appears on both sides of the equation, simplify the equation
by collecting like terms.
3I2 + 10HNO3
2H2O + 10NO + 6HIO3
∴ x is 10, y is 2 and z is 6.
Writing the redox equation using oxidation number method
1 Write down the oxidizing
agent and the reducing
agent involved. Determine
their products.
2 Assign oxidation numbers to
all atoms.
3 Notice atoms which undergo
a change in oxidation
number. Determine the
number of electrons lost or
gained per formula unit.
HNO3 is the oxidizing agent and I2 is the reducing agent.
NO
HNO3
HIO3
I2
+5
0
+2
HNO3 + I2
+5
NO + HIO3
reduction: gain of 3e– per HNO3
+5
HNO3
+
0
+2
I2
NO
+5
+
HIO3
oxidation: loss of 10e– per I2
4 Insert an appropriate
coefficient before the
formula of each reagent
on the left-hand side of
the equation to make the
number of electrons gained
equal to that lost.
10HNO3 + 3I2
NO + HIO3
5 Add appropriate coefficients
on the right to balance
the number of atoms
which have gained or lost
electrons.
10HNO3 + 3I2
10NO + 6HIO3
6 Balance the number of all
other atoms except O and
H.
The number of all other atoms except O and H is balanced.
7 Add H2O to the appropriate
side to balance the number
of O atoms.
3I2 + 10HNO3
2H2O + 10NO + 6HIO3
∴ x is 10, y is 2 and z is 6.
213
26 C Writing the redox equation using ionic half-equations
1 Write down the oxidizing
agent and the reducing
agent involved.
O2(g) is the oxidizing agent and Fe2+(aq) is the reducing agent.
2 a) Write an ionic halfequation for the
reduction process.
The unbalanced ionic half-equation for O2(g) is:
O2(g)
2H2O(l)
i) Balance the ionic halfequation with respect
to the number of
atoms.
ii) Balance the ionic halfequation with respect
to the number of
charges.
2 b) Write an ionic halfequation for the
oxidation process.
+
To balance the 4 hydrogen atoms in 2H2O, add 4H on the left-hand side.
+
2H2O(l)
O2(g) + 4H (aq)
Charge on left-hand side = 4 x (+1) = +4
Charge on right-hand side = 0
–
∴ add 4e on the left-hand side to balance the charge.
The balanced ionic half-equation for O2(g) is:
+
–
2H2O(l)......(i)
O2(g) + 4H (aq) + 4e
2+
The unbalanced ionic half-equation for Fe (aq) is:
2+
3+
Fe (aq)
Fe (aq)
i) Balance the ionic halfequation with respect
to the number of
atoms.
The number of atoms is balanced.
ii) Balance the ionic halfequation with respect
to the number of
charges.
Charge on left-hand side = +2
Charge on right-hand side = +3
–
∴ add 1e on the right-hand side to balance the charge.
2+
The balanced ionic half-equation for Fe (aq) is:
2+
3+
–
Fe (aq) + e ......(ii)
Fe (aq)
3 Make the number of
electrons gained in one ionic
half-equation equal to that
lost in the other.
(ii) x 4
4 Combine the two ionic halfequations and eliminate the
electrons.
(i)
(ii) x 4
∴ x is 4, y is 4 and z is 2.
214
O2(g) + 4H+(aq) + 4e–
2+
4Fe (aq)
2H2O(l)
4Fe3+(aq) + 4e–
4Fe2+(aq) + O2(g) + 4H+(aq)
4Fe3+(aq) + 2H2O(l)
Writing the redox equation using oxidation number method
1 Write down the oxidizing
agent and the reducing
agent involved. Determine
their products.
2 Assign oxidation numbers to
all atoms.
3 Notice atoms which undergo
a change in oxidation
number. Determine the
number of electrons lost or
gained per formula unit.
2+
O2(g) is the oxidizing agent and Fe (aq) is the reducing agent.
O2(g)
H2O(l)
2+
3+
Fe (aq)
Fe (aq)
0
+2
–2
O2(g) + Fe2+(aq)
+3
H2O(l) + Fe3+(aq)
reduction: gain of 4e– per O2
0
O2(g)
+2
+
Fe2+(aq)
–2
H2O(l)
+3
+
Fe3+(aq)
oxidation: loss of 1e– per Fe2+
4 Insert an appropriate
coefficient before the
formula of each reagent
on the left-hand side of
the equation to make the
number of electrons gained
equal to that lost.
2+
O2(g) + 4Fe (aq)
H2O(l) + Fe3+(aq)
5 Add appropriate coefficients
on the right to balance
the number of atoms
which have gained or lost
electrons.
2+
O2(g) + 4Fe (aq)
2H2O(l) + 4Fe3+(aq)
6 Balance the number of all
other atoms except O and
H.
The number of all other atoms except H is balanced.
7 Add H+ to the side deficient
in positive charges to make
the number of charges on
both sides equal.
Total charge on left-hand side = 4 x (+2) = +8
Total charge on right-hand side = 4 x (+3) = +12
+
Add 4H to the left-hand side.
2+
+
4Fe3+(aq) + 2H2O(l)
4Fe (aq) + O2(g) + 4H (aq)
∴ x is 4, y is 4 and z is 2.
27 A Combine the two ionic half-equations to obtain a redox equation:
(i) x 2 2MnO4–(aq) + 16H+(aq) + 10e–
2Mn2+(aq) + 8H2O(l)
(ii) x 5 5Sn2+(aq)
5Sn4+(aq) + 10e–
2MnO4–(aq) + 16H+(aq) + 5Sn2+(aq)
2Mn2+(aq) + 8H2O(l) + 5Sn4+(aq)
According to the equation, 2 (i.e. 0.4) mole of MnO4–(aq) ions will react completely with one mole of
5
Sn2+(aq) ions.
215
28 B
7
FMFDUSPOGMPX
DPCBMUTUSJQ
[JODTUSJQ
TBMU
CSJEHF
$P/0
BR
;O/0
BR
The position of zinc in the electrochemical cell is higher than that of cobalt, i.e. zinc forms ions more
readily than cobalt does.
Option A — Electrons flow from the zinc strip to the cobalt strip in the external circuit.
Option B — Zinc atoms lose electrons and form zinc ions. These ions then go into the zinc nitrate
solution.
Hence the zinc strip gradually dissolves.
Electrons given up by zinc atoms flow along the conducting wires to the cobalt strip.
Cobalt(II) ions in the cobalt(II) nitrate solution near to the cobalt strip gain these electrons
and form cobalt atoms. As a result, a deposit of cobalt forms on the cobalt strip.
Option C — The zinc strip is the negative electrode. Oxidation occurs here.
Option D — The cobalt strip is the positive electrode. Reduction occurs here.
29 C
7
FMFDUSPOGMPX
OJDLFM
FMFDUSPEF9
OJDLFM
FMFDUSPEF:
DPODFOUSBUFE
OJDLFM**
TVMQIBUF
TPMVUJPO
EJMVUFOJDLFM**
TVMQIBUFTPMVUJPO
HMBTTXPPMQMVH
In the set-up, electrons flow in such a direction that the concentration of Ni2+(aq) ions on both sides
becomes the same eventually, i.e. electrons flow from Y to X in the external circuit.
At electrode Y
Ni(s)
Ni2+(aq) + 2e–
At electrode X
Ni2+(aq) + 2e–
Ni(s)
Option A — Electrons flow in the external circuit, NOT between the two solutions.
Option B — Reduction occurs at X.
Option D — The mass of Y decreases while the mass of X increases.
216
30 C
Option
Reaction with aqueous chlorine
Equation for reaction involved
A
ammonium iodide solution
B
iron(II) chloride solution
Cl2(aq) + 2Fe2+(aq)
D
zinc bromide solution
Cl2(aq) + 2Br–(aq)
Cl2(aq) + 2I–(aq)
2Cl–(aq) + l2(aq)
–
3+
2Cl (aq) + 2Fe (aq)
–
2Cl (aq) + Br2(aq)
31 D Option A — Reaction 1 can be brought about by adding dilute nitric acid to magnesium.
3Mg(s) + 2NO3–(aq) + 8H+(aq)
3Mg2+(aq) + 2NO(g) + 4H2O(l)
Option B — Reaction 2 can be brought about by allowing concentrated nitric acid to decompose
under sunlight.
4HNO3(aq)
+4
Option C — NO2(g)
4NO2(g) + O2(g) + 2H2O(l)
Reaction 3
+5
HNO3(aq)
The oxidation number of nitrogen increases from +4 to +5 in Reaction 3.
Option D — Nitrogen monoxide is converted to nitrogen dioxide when exposed to air.
2NO(g) + O2(g)
2NO2(g)
32 B We can represent the reaction between copper in the alloy and concentrated nitric acid by the
following equation:
Cu(s) + 2NO3–(aq) + 4H+(aq)
Cu2+(aq) + 2NO2(g) + 2H2O(l)
We can represent the reaction between zinc in the alloy and concentrated nitric acid by the following
equation:
Zn(s) + 2NO3–(aq) + 4H+(aq)
Zn2+(aq) + 2NO2(g) + 2H2O(l)
The resulting solution contains Cu2+(aq) ions and Zn2+(aq) ions. It is pale blue in colour.
33 D
Option
A
B
Reaction
+4
+6
SO2 + 2H2O + Cl2
+4
H2SO4 + 2HCl
+4
SO2 + 2NaOH
+4
C
D
Does SO2 act as an oxidizing agent?
2SO2 + O2
+4
SO2 + 2H2S
Na2SO3 + H2O
The oxidation number of S increases from +4 to +6,
∴ SO2 acts as a reducing agent.
The oxidation number of S remains unchanged,
∴ SO2 does NOT act as an oxidizing agent.
The oxidation number of S increases from +4 to +6,
∴ SO2 acts as a reducing agent.
+6
2SO3
0
2H2O + 3S
The oxidation number of S decreases from +4 to 0,
∴ SO2 acts as an oxidizing agent.
34 D Option A — Sulphur dioxide can be prepared by heating copper turnings with concentrated sulphuric
acid.
Cu(s) + 2H2SO4(l)
CuSO4(aq) + SO2(g) + 2H2O(l)
Option B — Sulphur dioxide can turn limewater milky.
SO2(g) + Ca(OH)2(aq)
CaSO3(s) + H2O(l)
217
Option C — Sulphur dioxide can be absorbed by sodium hydroxide solution.
SO2(g) + NaOH(aq)
NaHSO3(aq)
Option D — There is NO reaction between sulphur dioxide and potassium iodide solution.
35 B Option A — Concentrated sulphuric acid liberates steamy fumes of hydrogen chloride when it reacts
with sodium chloride.
NaCl(s) + H2SO4(l)
NaHSO4(s) + HCl(g)
Hydrogen chloride is a weak reducing agent. It is NOT oxidized by concentrated sulphuric
acid.
Hence NO chlorine (halogen) is produced in the reaction.
Option B — When concentrated sulphuric acid is added to sodium bromide, hydrogen bromide is given
off.
NaBr(s) + H2SO4(l)
NaHSO4(s) + HBr(aq)
Hydrogen bromide is more easily oxidized by concentrated sulphuric acid.
2HBr(g) + H2SO4(l)
Br2(g) + SO2(g) + 2H2O(l)
Options C and D — There is NO reaction between sulphur dioxide gas and potassium bromide solution
/ potassium iodide solution.
36 B From the equation:
Cd(s) + Ga3+(aq)
no reaction
it can be deduced that the position of Cd in the electrochemical series is lower than that of Ga.
From the equation:
Hg2+(aq) + Pd(s)
Pd2+(aq) + Hg(s)
it can be deduced that the position of Pd in the electrochemical series is higher than that of Hg.
From the equation:
2+
3Pd (aq) + 2Ga(s)
3+
2Ga (aq) + 3Pd(s)
it can be deduced that the position of Ga in the electrochemical series is higher than that of Pd.
Hence the position of Ga in the electrochemical series is the highest,
∴ Ga is the strongest reducing agent.
37 B Y can oxidize both X–(aq) and Z–(aq). Hence the oxidizing power of Y is the strongest.
X can oxidize Z–(aq). Hence the oxidizing power of X is higher than that of Z.
∴ the order of decreasing oxidizing power is Y > X > Z.
218
38 A
7
DBSCPO
FMFDUSPEF:
DBSCPO
FMFDUSPEF9
TBMU
CSJEHF
NPMENm
'F40
BR
NPMENm,MBR
XJUITPNFTUBSDI
TPMVUJPO
In the left beaker, the solution changes from yellow-brown to green gradually. This is because the
iron(III) ions (Fe3+(aq)) gain electrons and change to iron(II) ions (Fe2+(aq)). The iron(III) ions are reduced.
Fe3+(aq) + e–
2+
Fe (aq)
In the right beaker, the iodide ions (I–(aq)) lose electrons and change to iodine molecules (I2(aq)). The
iodide ions are oxidized.
2I–(aq)
I2(aq) + 2e–
Option A — Electrons flow from potassium iodide solution (i.e. X) to iron(III) sulphate solution (i.e. Y)
in the external circuit.
Hence a current flows from Y to X in the external circuit.
Option B — Reduction occurs at Y.
Fe3+(aq) + e–
2+
Fe (aq)
Option C — Iodine give a blue colour when mixed with starch.
39 C Br2(aq) is a stronger oxidizing agent than Fe3+(aq) ion.
It can be deduced that Br2(aq) would oxidize Fe2+(aq) ions to Fe3+(aq) ions.
reduced form of half-equation
higher in
electrochemical
series
3+
Fe (aq) + e
oxidizing power
increasing
lower in
electrochemical
series
–
2+
Fe (aq)
oxidize
–
Br2(aq) + 2e
–
2Br (aq)
oxidized form of half-equation
The following chemical change would occur:
Fe2+(aq)
Br2(aq) + 2e–
Fe3+(aq) + e–
2Br–(aq)
219
Option A — Reduction occurs at X.
Hence X is the cathode.
Option B — Oxidation occurs at Y.
Option C —
EJHJUBMNVMUJNFUFS
BTWPMUNFUFS
FMFDUSPOGMPX
DBSCPO
FMFDUSPEF9
DBSCPO
FMFDUSPEF:
TBMU
CSJEHF
NJYUVSFPG
,#SBR
BOE#SBR
NJYUVSFPG
'F40BR
BOE
'F40
BR
Electrons flow from Y to X in the external circuit.
40 D Option C — The overall reaction during the discharging process is:
Pb(s) + PbO2(s) + 4H+(aq) + 2SO42–(aq)
2PbSO4(s) + 2H2O(l)
As sulphuric acid is consumed in the process, thus the pH of the electrolyte increases.
Option D — The following diagram shows the discharging process of a lead-acid cell:
MPBE
FMFDUSPOGMPX
MFBEQMBUFDPBUFE
XJUIMFBE*7
PYJEF
FMFDUSPEF:
MFBEQMBUF
FMFDUSPEF9
TVMQIVSJDBDJE
FMFDUSPMZUF
Electrons flow from electrode X to electrode Y in the external circuit.
41 C The following diagram shows the charging process of a lead-acid cell:
FYUFSOBM
QPXFSTPVSDF
FMFDUSPOGMPX
MFBEQMBUF
MFBE**
TVMQIBUF
220
MFBEQMBUFDPBUFE
XJUIMFBE*7
PYJEF
TVMQIVSJDBDJE
FMFDUSPMZUF
During charging, lead plates and lead plates coated with lead(IV) oxide are connected to the negative
and positive terminals of an external power source respectively. The flow of electrons is reversed as
compared to the discharging process. The overall reaction for the charging process is the reverse of the
discharging process.
The chemcial change that occurs at the electrode connected to the positive terminal of the external
power source is:
PbSO4(s) + 2H2O(l)
PbO2(s) + 4H+(aq) + 2SO42–(aq) + 2e–
Oxidation of PbSO4 occurs here.
42 D Electrode Y is the negative electrode while electrode X is the positive electrode.
Hence electrons flow from Y to X in the external circuit.
The H2PO4–(aq) ions move towards Y, the negative electrode.
43 D In dilute sodium nitrate solution, there are four kinds of ions:
Cation
+
Anion
From sodium nitrate
Na (aq)
NO3–(aq)
From water
H+(aq)
OH–(aq)
The nitrate ions and hydroxide ions are attracted to the anode. The sodium ions and hydrogen ions are
attracted to the cathode.
Option A — At the anode (electrode X)
A hydroxide ion is a stronger reducing agent than a nitrate ion. Therefore hydroxide ions
are preferentially discharged (oxidized) to form oxygen gas.
Option B — At the cathode (electrode Y)
A hydrogen ion is a stronger oxidizing agent than a sodium ion. Therefore hydrogen ions
are discharged (reduced) to form hydrogen gas.
Option C — Water dissociates continuously to replace the hydroxide ions discharged at the anode
(electrode X). Thus there is an excess of hydrogen ions near to electrode X and the
solution there becomes acidic. The universal indicator turns red after some time.
Option D — The sodium nitrate solution becomes more concentrated as water is decomposed in the
process.
44 A The following diagram shows the set-up used to electroplate silver on a copper obejct:
TJMWFSBOPEF
DPQQFSPCKFDU
DBUIPEF
TPMVUJPODPOUBJOJOH
"HJPOT
221
45 B Let the oxidation number of M in compound X be +n. The chemical formula of X is MCln.
M + n Cl2
The overall cell reaction is MCln
2
The number of M atoms produced equals the number of chlorine molecules produced.
i.e. n = 1
2
n = 2
∴ the oxidation number of M in compound X is +2.
46 C Writing the redox equation using ionic half-equations
1 Write down the oxidizing
agent and the reducing
agent involved.
IO3–(aq) is the oxidizing agent and HSO3–(aq) is the reducing agent.
2 a) Write an ionic halfequation for the
reduction process.
–
The unbalanced ionic half-equation for IO3 (aq) is:
–
I2(aq)
2IO3 (aq)
i) Balance the ionic halfequation with respect
to the number of
atoms.
–
To balance the 6 oxygen atoms in 2IO3 , add 6H2O on the right-hand
side.
–
I2(aq) + 6H2O(l)
2IO3 (aq)
+
To balance the 12 hydrogen atoms in 6H2O, add 12H on the left-hand
side.
+
–
I2(aq) + 6H2O(l)
2IO3 (aq) + 12H (aq)
ii) Balance the ionic halfequation with respect
to the number of
charges.
Charge on left-hand side = (–2) + 12 x (+1) = +10
Charge on right-hand side = 0
–
∴ add 10e on the left-hand side to balance the charge.
–
The balanced ionic half-equation for IO3 (aq) is:
+
–
–
I2(aq) + 6H2O(l)......(i)
2IO3 (aq) + 12H (aq) + 10e
2 b) Write an ionic halfequation for the
oxidation process.
–
The unbalanced ionic half-equation for HSO3 (aq) is:
–
2–
SO4 (aq)
HSO3 (aq)
i) Balance the ionic halfequation with respect
to the number of
atoms.
To balance the oxygen atoms, add H2O on the left-hand side.
SO42–(aq)
HSO3–(aq) + H2O(l)
+
To balance the 3 hydrogen atoms on the left-hand side, add 3H on the
right-hand side.
–
SO42–(aq) + 3H+(aq)
HSO3 (aq) + H2O(l)
ii) Balance the ionic halfequation with respect
to the number of
charges.
Charge on left-hand side = –1
Charge on right-hand side = (–2) + 3 x (+1) = +1
–
∴ add 2e on the right-hand side to balance the charge.
–
The balanced ionic half-equation for HSO3 (aq) is:
+
–
HSO3–(aq) + H2O(l)
SO42–(aq) + 3H (aq)+ 2e ......(ii)
3 Make the number of
electrons gained in one ionic
half-equation equal to that
lost in the other.
222
(ii) x 5
4 Combine the two ionic halfequations and eliminate the
electrons.
(i)
2IO3–(aq) + 12H+(aq) + 10e–
(ii) x 5 5HSO3–(aq) + 5H2O(l)
I2(aq) + 6H2O(l)
5SO42–(aq) + 15H+(aq) + 10e–
2IO3–(aq) + 12H+(aq) + 5HSO3–(aq) + 5H2O(l)
+
I2(aq) + 6H2O(l) + 5SO42–(aq)+ 15H (aq)
Since H+(aq) and H2O(l) appear on both sides of the equation, simplify the
equation by collecting like terms.
–
–
5SO42–(aq) + I2(aq) + 3H+(aq) + H2O(l)
2IO3 (aq) + 5HSO3 (aq)
∴ x is 2, y is 5 and z is 5.
Writing the redox equation using oxidation number method
1 Write down the oxidizing
agent and the reducing
agent involved. Determine
their products.
IO3–(aq) is the oxidizing agent and HSO3–(aq) is the reducing agent.
I2(aq)
IO3–(aq)
SO42–(aq)
HSO3–(aq)
2 Assign oxidation numbers to
all atoms.
+5
3 Notice atoms which undergo
a change in oxidation
number. Determine the
number of electrons lost or
gained per formula unit.
reduction: gain of 5e– per IO3–
4 Insert an appropriate
coefficient before the
formula of each reagent
on the left-hand side of
the equation to make the
number of electrons gained
equal to that lost.
–
–
2IO3 (aq) + 5HSO3 (aq)
I2(aq) + SO42–(aq)
5 Add appropriate coefficients
on the right to balance
the number of atoms
which have gained or lost
electrons.
–
–
2IO3 (aq) + 5HSO3 (aq)
I2(aq) + 5SO42–(aq)
6 Balance the number of all
other atoms except O and
H.
The number of all other atoms except O and H is balanced.
7 Add H+ to the side deficient
in positive charges to make
the number of charges on
both sides equal.
Total charge on left-hand side = 2 x (–1) + 5 x (–1) = –7
Total charge on right-hand side = 5 x (–2) = –10
+
Add 3H to the right-hand side.
–
–
I2(aq) + 5SO42–(aq) + 3H+(aq)
2IO3 (aq) + 5HSO3 (aq)
8 Add H2O to the appropriate
side to balance the number
of O atoms.
2IO3–(aq) + 5HSO3–(aq)
+4
IO3–(aq) + HSO3–(aq)
+5
–
IO3 (aq)
+4
+
HSO3–(aq)
0
+6
I2(aq) + SO42–(aq)
0
I2(aq)
+6
+
SO42–(aq)
oxidation: loss of 2e– per HSO3–
5SO42–(aq) + I2(aq) + 3H+(aq) + H2O(l)
∴ x is 2, y is 5 and z is 5.
223
47 A (1) Oxidation number of O = –2
–
Suppose the oxidation number of I in IO3 is x.
x + (–2) x 3 = –1
x = +5
∴ the oxidation number of I in IO3– is +5.
(2) IO3– ion is reduced by HSO3–(aq) ion in the reaction.
(3) One of the ionic half-equations of this reaction is
2IO3–(aq) + 12H+(aq) + 10e–
48 B
Option
49 A (1)
I2(aq) + 6H2O(l).
Reaction
+6
Role of reagent
FeSO4 / H+
A
Cr2O72–(aq)
B
Cr3+(aq)
C
Cr3+(aq)
D
[Cr(OH)6]3–(aq)
+3
Zn / H+
+3
NaOH
Zn is a reducing agent; it causes the oxidation
number of Cr to decrease from +3 to +2.
+2
Cr2+(aq)
+3
[Cr(OH)6]3–(aq)
H2O2
+3
FeSO4 is a reducing agent; it causes the oxidation
number of Cr to decrease from +6 to +3.
+3
Cr3+(aq)
+6
CrO42–(aq)
NaOH is NOT an oxidizing agent as the oxidation
number of Cr remains unchanged.
H2O2 is an oxidizing agent; it causes the oxidation
number of Cr to increase from +3 to +6.
Species
Oxidation number of Cr in species
Cr2O72–(aq)
+6
Cr3+(aq)
+3
Cr2+(aq)
+2
[Cr(OH)6]3–(aq)
+3
CrO42–(aq)
+6
∴ the oxidation numbers of chromium shown in the reaction scheme are +2, +3 and +6.
(2)
Reaction
Conversion
1
Cr2O72–(aq)
orange
4
[Cr(OH)6]3–(aq)
green
Cr3+(aq)
green
CrO42–(aq)
yellow
∴ Reactions 1 and 4 involve a colour change.
(3)
+3
Cr3+(aq)
Reaction 3
+3
[Cr(OH)6]3–(aq)
∴ Reaction 3 does NOT involve a change in the oxidation number of chromium.
50 B (2) The position of fluorine in the electrochemical series is lower than that of chlorine.
Hence fluorine is a stronger oxidizing agent than chlorine.
224
–1
+4
–2
51 B (1) H2O2 + Na2SO3
+6
H2O + Na2SO4
The oxidation number of O decreases from –1 to –2 while that of S increases from +4 to +6.
Therefore it is a redox reaction.
– 3 +1
+1 +5 –2
– 3 +1 +5 –2
(2) NH3 + HNO3
NH4NO3
This is NOT a redox reaction because oxidation numbers of all elements remain unchanged in the
reaction.
+5 – 2
+4
(3) 4HNO3
0
2H2O + 4NO2 + O2
The oxidation number of N decreases from +5 to +4 while that of O increases from –2 to 0.
Therefore it is a redox reaction.
52 C (1) Oxidation number of O = –2
Suppose the oxidation number of S in S2O32– is x.
2x + (–2) x 3 = –2
x = +2
∴ the oxidation number of S in S2O32– is +2.
0
–1
S4O62–(aq) + 2I–(aq)
(2) 2S2O32–(aq) + I2(aq)
The oxidation number of iodine decreases from 0 to –1.
2–
I2(aq) is reduced by S2O3 (aq) ions in the reaction.
(3) S2O32–(aq) ion causes the reduction of iodine.
Hence S2O32–(aq) ion acts as a reducing agent in the reaction.
53 D A reaction in which the same species is simultaneously reduced and oxidized is called disproportionation.
+1
0
(1) Cu2O(s) + 2H+(aq)
+2
Cu(s) + Cu2+(aq) + H2O(l)
The oxidation of Cu in Cu2O is +1. The oxidation numbers of Cu in Cu and Cu2+ are 0 and +2
respectively.
Hence Cu2O is simultaneously reduced and oxidized.
0
–1
(2) 3Br2(l) + 6KOH(aq)
+5
5KBr(aq) + KBrO3(aq) + 3H2O(l)
The oxidation number of bromine is 0. The oxidation numbers of Br in KBr and KBrO3 are –1 and
+5 respectively.
Hence Br2 is simultaneously reduced and oxidizied.
+6
(3) 3MnO42–(aq) + 4H+(aq)
+7
+4
2MnO4–(aq) + MnO2(s) + 2H2O(l)
The oxidation number of Mn in MnO42– is +6. The oxidation numbers of Mn in MnO4– and MnO2
are +7 and +4 respectively.
Hence MnO42– is simultaneously reduced and oxidized.
54 B (1) Aqueous chlorine oxidizes bromide ions to bromine (which is yellow-brown in colour).
Cl2(aq) + 2e–
2Br–(aq)
2Cl–(aq)
Br2(aq) + 2e–
225
(2) Aqueous chlorine oxidizes sulphite ions to sulphate ions (which is colourless).
–
–
Cl2(aq) + 2e
2Cl (aq)
2–
+
2–
SO3 (aq) + H2O(l)
–
SO4 (aq) + 2H (aq) + 2e
(3) Aqueous chlorine oxidizes iodide ions to iodine (which is brown in colour).
Cl2(aq) + 2e–
2I–(aq)
2Cl–(aq)
I2(aq) + 2e–
55 C (2) Aqueous bromine oxidizes iodide ions to iodine.
(3) Aqueous sulphur dioxide reduces bromine to bromide ions.
56 A (1) When aqueous chlorine is added to sodium bromide solution, the solution becomes yellow-brown
due to the formation of bromine. Adding an organic solvent to the reaction mixture gives an
orange organic layer.
When aqueous chlorine is added to sodium iodide solution, the solution becomes brown due to the
formation of iodine. Adding an organic solvent to the reaction mixture gives a purple organic layer.
57 A (1) Sodium sulphite solution can turn aqueous bromine from yellow-brown to colourless (due to the
formation of bromide ions).
There is NO observable change when sodium sulphate solution is mixed with aqueous bromine.
Hence aqueous bromine can be used to distinguish between sodium sulphite solution and sodium
sulphate solution.
(2) Both sodium sulphite solution and sodium sulphate solution give a white precipitate with barium
nitrate solution.
Ba2+(aq) + SO32–(aq)
BaSO3(s)
Ba2+(aq) + SO42–(aq)
BaSO4(s)
58 D (1) Dilute nitric acid acts as an oxidizing agent when it reacts with zinc. Nitrogen monoxide, NOT
hydrogen, is produced.
(2) and (3) Dilute nitric acid acts as an acid in both cases.
59 B Dilute nitric acid and copper react according to the following equation:
3Cu(s) + 2NO3–(aq) + 8H+(aq)
3Cu2+(aq) + 2NO(g) + 4H2O(l)
(1) Nitrogen monoxide gas is evolved.
(2) The ionic half-equation for the reduction of nitrate ions is:
NO3–(aq) + 4H+(aq) + 3e–
NO(g) + 2H2O(l)
∴ one mole of NO3–(aq) ions requires three moles of electrons for reduction.
(3) A redox reaction is invovled.
60 A (1) Dilute hydrochloric acid reacts with iron powder only.
(2) Dilute nitric acid reacts with both copper powder and iron powder.
(3) Concentrated nitric acid reacts with both copper powder and iron powder.
226
61 A (1) Iron reacts with dilute nitric acid to give nitrogen monoxide gas. The gas gives a brown gas when
mixed with air.
Iron reacts with dilute sulphuric acid to give hydrogen gas.
Hence iron can be used to distinguish between dilute nitric acid and dilute sulphuric acid.
(2) Lead(II) nitrate solution gives a white precipitate (lead(II) sulphate) with dilute sulphuric acid, but
NOT dilute nitric acid.
Pb2+(aq) + SO42–(aq)
PbSO4(s)
Hence lead(II) nitrate solution can be used to distinguish between dilute nitric acid and dilute
sulphuric acid.
(3) Solid potassium carbonate reacts with both acids to give a colourless solution and carbon dioxide
gas.
CO32–(aq) + 2H+(aq)
H2O(l) + CO2(g)
Solid potassium carbonate dissolves in water to give mobile CO32–(aq) ions.
Hence solid potassium carbonate CANNOT be used to distinguish between dilute nitric acid and
dilute sulphuric acid.
62 B (2) Oxygen has NO characteristic odour.
63 B (1) When mixed with sodium sulphite solution, iodine solution changes from brown to colourless (due
to the formation of iodide ions).
SO42–(aq) + 2H+(aq) + 2I–(aq)
SO32–(aq) + H2O(l) + I2(aq)
(2) There is NO reaction between sodium sulphite solution and chromium(III) sulphate solution.
(3) When mixed with sodium sulphite solution, acidified potassium permanganate solution changes
from purple to colourless / pale pink (due to the formation of manganese(II) ions).
5SO32–(aq) + 2MnO4–(aq) + 6H+(aq)
5SO42–(aq) + 2Mn2+(aq) + 3H2O(l)
64 C (1) Bromine CANNOT displace chlorine from sodium chloride solution.
(2) Sulphur dioxide and calcium hydroxide solution give a white precipitate when mixed.
SO2(g) + Ca(OH)2(aq)
CaSO3(s) + H2O(l)
(3) Zinc nitrate solution gives a white precipitate with dilute sodium hydroxide solution. The precipitate
dissolves in excess dilute sodium hydroxide solution.
Zn2+(aq) + 2OH–(aq)
Zn(OH)2(s)
–
[Zn(OH)4]2–(aq)
Zn(OH)2(s) + 2OH (aq)
65 A (1) Copper and concentrated sulphuric acid react to give sulphur dioxide.
Cu(s) + 2H2SO4(l)
CuSO4(aq) + SO2(g) + 2H2O(l)
(2) Magnesium and concentrated nitric acid react to give nitrogen dioxide, NOT nitrogen monoxide.
+
–
Mg(s) + 2NO3 (aq) + 4H (aq)
Mg2+(aq) + 2NO2(g) + 2H2O(l)
227
(3) When concentrated sulphuric acid is added to sodium iodide, hydrogen iodide is produced.
NaI(s) + H2SO4(l)
NaHSO4(s) + HI(g)
Concentrated sulphuric acid then oxidizes hydrogen iodide to iodine.
8HI(g) + H2SO4(l)
4I2(s) + H2S(g) + 4H2O(l)
∴ hydrogen iodide is NOT prepared from sodium iodide and concentrated sulphuric acid.
To prepare hydrogen iodide, treat sodium iodide with concentrated phosphoric acid.
KI(s) + H3PO4(l)
66 C
KH2PO4(s) + HI(g)
EJHJUBMNVMUJNFUFS
BTWPMUNFUFS
m
NFUBM9
NFUBM:
TBMU
CSJEHF
TPMVUJPOPG
DIMPSJEFPG9
TPMVUJPOPG
DIMPSJEFPG:
The digital mulitimeter gives a positive reading.
It can be deduced that the metal connected to the positive terminal of the voltmeter is the positive
electrode while the metal connected to the negative terminal of the voltmeter is the negative electrode.
(1) X forms ions more readily than Y does.
Hence the position of X in the electrochemical series is higher than that of Y.
(2) Atoms of metal X lose electrons and form ions, i.e. oxidation occurs at metal X.
Hence metal X is the anode.
(3) Electrons flow from X to Y in the external circuit, i.e. a current flows from Y to X in the external
circuit.
67 A (1) X can displace Y from a solution containing Y2+(aq) ions.
It can be deduced that X is more reactive than Y.
(2) Y2+(aq) ion causes the oxidation number of X to increase from 0 to +2.
Hence Y2+(aq) ion acts as an oxidizing agent in the reaction.
(3) The following chemical cell uses X and Y as electrodes.
EJHJUBMNVMUJNFUFS
BTWPMUNFUFS
FMFDUSPOGMPX
m
m
NFUBM9BOPEF
Fm
F
NFUBM:DBUIPEF
FMFDUSPMZUF
228
9
In the above chemical cell, atoms of X lose electrons and form ions, i.e. oxidation occurs at X.
Hence X acts as the anode.
68 A
EJHJUBMNVMUJNFUFS
BTWPMUNFUFS
DBSCPOSPE
DBSCPOSPE
NJYUVSFPG
4O BR
BOE
4OBR
JPOT
TBMU
CSJEHF
NJYUVSFPG
#SmBR
JPOT
BOE#SBR
4+
Br2(aq) is a stronger oxidizing agent than Sn (aq) ion.
It can be deduced that Br2(aq) would oxidize Sn2+(aq) ion to Sn4+(aq) ion.
reduced form of half-equation
higher in
electrochemical
series
Sn4+(aq) + 2e–
oxidizing power
increasing
lower in
electrochemical
series
Sn2+(aq)
oxidize
–
Br2(aq) + 2e
2Br–(aq)
oxidized form of half-equation
The following chemical changes would occur:
(1) Sn2+(aq)
(2) Br2(aq) + 2e–
Sn4+(aq) + 2e–
2Br–(aq)
229
69 C
7
DPQQFS
FMFDUSPEF9
DBSCPO
FMFDUSPEF:
TBMU
CSJEHF
NPMENm
$V40BR
NPMENm)40BR
NPMENm,.O0BR
NPMENm.O40BR
(1) In the right beaker, the permanganate ions are reduced.
MnO4–(aq) + 8H+(aq) + 5e–
Mn2+(aq) + 4H2O(l)
(3) In the left beaker, atoms of copper are oxidized.
Cu2+(aq) + 2e–
Cu(s)
70 D Sodium is a strong reducing agent.
Hence sodium has a high tendency to lose electrons.
At electrode B
Na+(l) + e–
Na(l)
(1) Electrons flow from electrode B to electrode A in the external circuit, i.e. a current flows from
electrode A to electrode B in the external circuit.
0
+3
71 C (1) at X: Al(s) + 3OH–(aq)
Al(OH3)(s) + 3e–
The oxidation number of Al increases from 0 to +3.
Hence oxidation occurs at X.
0
–2
(2) at Y: O2(g) + 2H2O(l) + 4e–
4OH–(aq)
The oxidation number of O decreases from 0 to –2.
Hence reduction occurs at Y.
So, Y is the cathode.
(3) Electrons flow from the anode to the cathode, i.e. from X to Y, in the external circuit.
72 A at the aluminium can:
0
Al(s) + 4OH–(aq)
+3
–
–
[Al(OH)4] (aq) + 3e
at the carbon rod:
+1
OCl–(aq) + H2O(l) + 2e–
–1
Cl–(aq) + 2OH–(aq)
(1) The aluminium can acts as the anode.
Electrons flow from the anode to the cathode, i.e. from the aluminium can to the carbon rod, in
the external circuit.
230
(2) The oxidation number of Al increases from 0 to +3.
Hence oxidation occurs at the aluminium can.
So, the aluminium can acts as the anode.
(3) The oxidation number of Cl decreases from +1 to –1.
–
Hence OCl (aq) ions are reduced.
73 D
74 C In 2 mol dm–3 sodium chloride solution, there are four kinds of ions:
Cation
+
Anion
–
From sodium chloride
Na (aq)
Cl (aq)
From water
H+(aq)
OH–(aq)
The chloride ions and hydroxide ions are attracted to the anode. The sodium ions and hydrogen ions
are attracted to the cathode.
(1) At the cathode
A hydrogen ion is a stronger oxidizing agent than a sodium ion. Therefore hydrogen ions are
preferentially discharged (reduced) to form hydrogen gas.
(2) At the anode
The concentration of chloride ions in the solution is much greater than that of hydroxide ions.
Therefore chloride ions are preferentially discharged (oxidized) to form chlorine.
(3) Hydrogen ions and chloride ions are consumed in the electrolysis. Sodium ions and hydroxide ions
remain in the solution. Eventually, the solution becomes sodium hydroxide solution.
Hence the pH of the solution in the electrolytic cell increases.
75 C (1) The plating metal, nickel, is the anode.
(2) Reduction occurs at the iron bolt:
Ni2+(aq) + 2e–
Ni(s)
(3) Overall cell reaction
Ni(s)
nickel electrode
Ni(s)
iron bolt
Changes in the solution
The net effect is the transfer of nickel from the nickel electrode to the iron bolt. The rate at which
nickel deposits on the iron bolt is equal to the rate at which the nickel electrode dissolves.
increase in mass of iron bolt = decrease in mass of nickel electrode
The concentration of Ni2+(aq) ions in the solution remains unchanged.
231
76 D
Solution
Product at cathode
–3
copper(II) chloride solution
–3
sodium sulphate solution
hydrogen
3 mol dm–3 calcium chloride solution
hydrogen
(1)
1 mol dm
(2)
2 mol dm
(3)
77 A
Solution
copper
Product at anode
(1)
1 mol dm–3 nickel(II) sulphate solution
oxygen
(2)
2 mol dm–3 sodium hydroxide solution
oxygen
(3)
–3
3 mol dm
78 A
magnesium chloride solution
Solution
chlorine
Product at anode
Product at cathode
(1)
dilute copper(II) chloride solution
chlorine
copper
(2)
molten calcium chloride
chlorine
calcium
(3)
dilute sodium hydroxide solution
oxygen
hydrogen
79 D
80 A (1) Yellow-brown bromine is produced during the electrolysis of concentrated magnesium bromide
solution.
(2) Brown nitrogen dioxide gas is produced when copper is added to concentrated nitric acid.
(3) There is NO reaction between sulphur dioxide gas and sodium iodide solution.
81 D The size of silver oxide cells are too small for portable CD players.
Silver oxide cells are NOT rechargeable.
82 C A lithium ion cell is not very robust and cannot deliver high discharge currents. Loading the cell with
excess discharge current may overheat the pack and lead to explosion.
83 C Alkaline manganese cells are NOT rechargeable.
84 B In a zinc-carbon cell, reduction occurs at the positive electrode:
2NH4+(aq) + 2e–
2NH3(aq) + H2(g)
Ammonium ions are consumed.
Oxidation always occurs at the anode.
85 A The position of nickel in the electrochemical series is higher than that of silver.
Hence nickel releases electrons more readily than silver does.
In a nickel-silver chemical cell using sodium chloride solution as the electrolyte, the nickel atoms lose
electrons and form ions. Hence oxidation occurs at the nickel electrode.
So, the nickel electrode is the anode.
232
86 A A simple copper-zinc chemical cell using dilute sulphuric acid as the electrolyte is shown below:
EJHJUBMNVMUJNFUFS
BTWPMUNFUFS
FMFDUSPOGMPX
m
m
[JODFMFDUSPEF
OFHBUJWFFMFDUSPEF
m
F
F
DPQQFSFMFDUSPEF
QPTJUJWFFMFDUSPEF
)
;O
EJMVUFTVMQIVSJDBDJE
In the copper-zinc chemical cell, zinc atoms lose electrons and form zinc ions.
Electrons given up by zinc atoms flow along the conducting wires to the copper electrode. Hydrogen
ions in the solution near to the copper electrode gain these electrons and form hydrogen.
Hence colourless gas bubbles are given off from the copper electrode.
87 B Sodium and water react according to the following equation:
0
+1
2Na(s) + 2H2O(l)
+1
0
2NaOH(aq) + H2(g)
The oxidation number of Na increases from 0 to +1 while that of H decreases from +1 to 0. Therefore
it is a redox reaction.
88 C There is NO reaction between aqueous chlorine and acidified potassium permanganate solution.
89 A Aqueous chlorine can oxidize sodium bromide solution.
Cl2(aq) + 2Br–(aq)
2Cl–(aq) + Br2(aq)
90 C When ammonium iodide solution is mixed with aqueous bromine, iodine is formed.
Br2(aq) + 2I–(aq)
2Br–(aq) + I2(aq)
The resulting solution is brown in colour due to the presence of iodine.
91 C In the reaction between dilute nitric acid and iron(II) carbonate, dilute nitric acid acts as an acid.
FeCO3(s) + 2HNO3(aq)
Fe(NO3)2(aq) + H2O(l) + CO2(g)
92 C When concentrated sulphuric acid is added to sodium bromide / sodium iodide, hydrogen bromide /
hydrogen iodide is given off.
NaBr(s) + H2SO4(l)
NaI(s) + H2SO4(l)
NaHSO4(s) + HBr(g)
NaHSO4(s) + HI(g)
Concentrated sulphuric acid oxidizes hydrogen bromide to bromine, or hydrogen iodide to iodine.
2HBr(g) + H2SO4(l)
8HI(g) + H2SO4(l)
Br2(g) + SO2(g) + 2H2O(l)
4I2(s) + H2S(g) + 4H2O(l)
233
93 D Both concentrated nitric acid and concentrated sulphuric acid act as an acid when reacting with sodium
carbonate powder.
Hence sodium carbonate powder CANNOT be used to distinguish between concentrated nitric acid and
concentrated sulphuric acid.
94 D When sulphur dioxide gas is passed into an iodine solution prepared by dissolving iodine in potassium
iodide solution, the colour of the solution changes from brown to colourless.
This is because the sulphite ion reduces the brown iodine to colourless iodide ions.
I2(aq) + SO32–(aq) + H2O(l)
2I–(aq) + SO42–(aq) + 2H+(aq)
95 A Aqueous bromine can be reduced by sodium sulphite solution to colourless bromide ions.
Br2(aq) + SO32–(aq) + H2O(l)
2Br–(aq) + SO42–(aq) + 2H+(aq)
96 C In direct methanol fuel cell, only methanol acts as the fuel.
97 D Chlorine gas can attack platinum. Therefore platinum electrodes are NOT used in the electrolysis of
dilute or concentrated sodium chloride solution.
98 D A hydroxide ion is a stronger reducing agent than a chloride ion.
Hence during the electrolysis of 0.1 mol dm–3 CuCl2(aq) using platinum electrodes, hydroxide ions are
preferentially discharged to form oxygen gas at the anode.
99 C Iron and zinc in impure copper form ions more readily than copper. During the refining of copper using
electrolysis, iron and zinc in the anode give up electrons first. Then copper gives up electrons to form
copper(II) ions. Impurities such as silver, gold and platinum settle at the bottom of the container.
A copper(II) ion is a stronger oxidizing agent than zinc ion and iron(II) ion. Therefore copper(II) ions are
discharged to form copper at the cathode.
During the refining process, the copper is gradually transferred from the anode the cathode. The
concentration of copper(II) ions in the electrolyte drops gradually. This is because at the anode, iron and
zinc dissolve as ions readily while at the cathode, copper(II) ions are always preferentially discharged.
100 D During the electroplating of nickel on an iron object, a nickel electrode is made the anode while the
iron obejct is made the cathode.
At the anode
Ni(s)
Ni2+(aq) + 2e–
At the cathode
Ni2+(aq) + 2e–
Ni(s)
Hence during the electroplating process, oxidation occurs at the anode while reduction occurs at the
cathode.
234
Short questions
101
Species
Oxidation number of the underlined element
KMnO4
+7
ClO
–
4
+7
Co(NH3)4Cl2
+2
H2S2O7
+6
NH4NO3
–3
Cr2O72–
+6
NH4VO3
+5
[Pb(OH)4]2–
+2
Na2S2O3
+2
H2O2
–1
102 a) This is not a redox reaction
(1)
because oxidation numbers of all elements remain unchanged in the reaction.
+2
–2 +1
+2
Cu2+ + 2OH–
Cu(OH)2
b) The oxidation number of P increases from +3 to +5 while that of Cl decreases from 0 to –1.
Therefore it is a redox reaction.
+3
0
(1)
P Cl5
c) The oxidation number of S decreases from +4 to 0 while that of Mg increases from 0 to +2.
Therefore it is a redox reaction.
0
+2
SO2 + 2Mg
(1)
2MgO + S
(1)
because oxidation numbers of all elements remain unchanged in the reaction.
+1 +1 +4 –2
+1
2NaHCO3
+4 –2
+1 –2
(1)
+4 –2
Na2CO3 + H2O + CO2
e) The oxidation number of Zn increases from 0 to +2 while that of Ag decreases from +1 to 0.
Therefore it is a redox reaction.
+1
+2
Zn + 2AgNO3
0
Zn(NO3)2 + 2Ag
Therefore it is a redox reaction.
+2
2NH3 + 3CuO
(1)
(1)
f) The oxidation number of N increases from –3 to 0 while that of Cu decreases from +2 to 0.
–3
(1)
0
d) This is not a redox reaction
0
(1)
+5 –1
PCl3 + Cl2
+4
(1)
–2 +1
0
(1)
(1)
0
3Cu + N2 + 3H2O
235
g) This is not a redox reaction
(1)
because oxidation numbers of all elements remain unchanged in the reaction.
+6 –2
+1
+6
2CrO42– + 2H+
–2
(1)
+1 –2
Cr2O72– + H2O
h) The oxidation number of Cl decreases from +5 to –1 while that of O increases from –2 to 0.
Therefore it is a redox reaction.
+5 –2
–1
2KClO3
(1)
(1)
0
2KCl + 3O2
i) The oxidation number of N decreases from +5 to +4 while that of O increases from –2 to 0.
Therefore it is a redox reaction.
+5 –2
+4
4HNO3
(1)
(1)
0
2H2O + 4NO2 + O2
j) The oxidation number of N decreases from +4 to +3 and increases from +4 to +5.
Therefore it is a redox reaction.
+4
+3
N2O4 + H2O
(1)
(1)
+5
HNO2 + HNO3
103 a) The oxidation number of Cu decreases from +2 to 0.
Therefore CuSO4 is reduced.
+2
(1)
(1)
0
Mg + CuSO4
MgSO4 + Cu
b) The oxidation numbers of all elements remain unchanged in the reaction. This is not a redox reaction.
(1)
Therefore CrO42– is not reduced.
+6 –2
+1
2CrO42– + 2H+
+6
–2
(1)
+1 –2
Cr2O72– + H2O
c) The oxidation numbers of all elements remain unchanged in the reaction. This is not a redox reaction.
(1)
Therefore CaCO3 is not reduced.
+2 +4 –2
+1 –1
CaCO3 + 2HCl
+2
–1
+4 –2
(1)
+1
–2
CaCl2 + CO2 + H2O
d) The oxidation number of Fe decreases from +3 to +2.
Therefore Fe2(SO4)3 is reduced.
(1)
+2
+3
Fe2(SO4)3 + H2S
2FeSO4 + S + H2SO4
e) The oxidation number of Br increases from –1 to 0.
Therefore KBr is oxidized, not reduced.
–1
2KBr + Cl2
236
(1)
0
2KCl + Br2
(1)
(1)
f) The oxidation number of S decreases from +6 to +4.
Therefore H2SO4 is reduced.
+6
C + 2H2SO4
(1)
(1)
+4
CO2 + 2SO2 + 2H2O
g) The oxidation number of N decreases from +5 to +4.
Therefore NO3– is reduced.
+5
(1)
(1)
+4
Mg + 2NO3– + 4H+
Mg2+ + 2NO2 + 2H2O
h) The oxidation number of V decreases from +5 to +4.
Therefore VO2+ is reduced.
+5
(1)
(1)
+4
2VO2+ + 4H+ + 2I–
2VO2+ + 2H2O + I2
104 a) The yellow-brown aqueous bromine becomes colourless.
This is because the yellow-brown bromine is reduced to colourless bromide ions.
Br2(aq) + SO32–(aq) + H2O(l)
2Br–(aq) + SO42–(aq) + 2H+(aq)
b) A brown colour develops.
(1)
(1)
(1)
(1)
This is due to the formation of iodine in the reaction.
(1)
Cl2(aq) + 2I–(aq)
(1)
2Cl–(aq) + I2(aq)
c) A brown colour develops.
(1)
This is due to the formation of iodine in the reaction.
(1)
2I–(aq) + 2Fe3+(aq)
I2(aq) + 2Fe2+(aq)
(1)
d) The purple colour of acidified potassium permanganate solution fades.
(1)
This is because the purple permanganate ions are reduced to pale pink / colourless manganese(II) ions.
(1)
5SO32–(aq) + 2MnO4–(aq) + 6H+(aq)
5SO42–(aq) + 2Mn2+(aq) + 3H2O(l)
e) The silver dissolves. / A brown gas is given off.
(1)
(1)
This is because the nitrate ions are reduced to brown nitrogen dioxide gas.
(1)
Ag(s) + NO3–(aq) + 2H+(aq)
(1)
Ag+(aq) + NO2(g) + H2O(l)
f) The zinc dissolves. / A colourless gas is given off. This gas gives a brown gas when mixed with air. (1)
This is because the nitrate ions are reduced to colourless nitrogen monoxide gas. This gas gives brown
nitrogen dioxide gas when mixed with air.
(1)
3Zn(s) + 2NO3–(aq) + 8H+(aq)
3Zn2+(aq) + 2NO(g) + 4H2O(l)
(1)
237
g) Colourless gas bubbles are given off.
(1)
This is because concentrated sulphuric acid oxidizes carbon. Carbon dioxide gas and sulphur dioxide gas
are formed.
(1)
C(s) + 2H2SO4(l)
CO2(g) + 2SO2(g) + 2H2O(l)
(1)
105 a) Any one of the following:
• Add concentrated sulphuric acid to each solid separately.
Concentrated sulphuric acid gives reddish brown fumes when it reacts with sodium bromide.
(1)
(1)
Concentrated sulphuric acid gives purple fumes that condense to a black solid when it reacts with
sodium iodide.
(1)
• Dissolve each solid in water. Then add aqueous chlorine to each solution separately.
Solution of sodium bromide becomes yellow-brown in colour (bromine is formed).
(1)
(1)
Add an organic solvent to the reaction mixture.
The bromine formed dissolves in the solvent and an orange layer appears.
Solution of sodium iodide becomes brown in colour (iodine is formed).
(1)
Add an organic solvent to the reaction mixture.
The iodine formed dissolves in the solvent and a purple layer appears.
b) Add copper to each acid separately.
(1)
Concentrated nitric acid gives a brown gas with copper.
(1)
Concentrated sulphuric acid gives a colourless gas with copper.
(1)
c) Any one of the following:
• Mix aqueous bromine with each solution separately.
Sodium sulphite solution turns the yellow-brown aqueous bromine to colourless.
(1)
Sodium sulphate solution shows no observable change.
(1)
• Mix acidified potassium permanganate solution with each solution separately.
238
(1)
(1)
Sodium sulphite solution turns the purple permanganate solution to colourless.
(1)
Sodium sulphate solution shows no observable change.
(1)
106
Material of
Product at
Solution
Change in the solution
anode
cathode
anode
cathode
platinum
platinum
oxygen
hydrogen
becomes more concentrated
b) Concentrated hydrochloric
acid
carbon
carbon
chlorine
hydrogen
becomes more dilute
c) Very dilute sodium chloride
carbon
carbon
oxygen
hydrogen
becomes more concentrated
d) Dilute sodium chloride
carbon
carbon
chlorine
hydrogen
becomes sodium hydroxide
solution
e) Dilute magnesium sulphate
carbon
carbon
oxygen
hydrogen
becomes more concentrated
f) Concentrated sodium
chloride
carbon
mercury
chlorine
sodium
becomes more dilute
g) Dilute copper(II) sulphate
carbon
carbon
oxygen
copper
becomes sulphuric acid
h) Dilute copper(II) sulphate
copper
carbon
copper(II)
ions
copper
remains unchanged
i) Dilute copper(II) sulphate
copper
copper
copper(II)
ions
copper
remains unchanged
j) Dilute copper(II) chloride
carbon
carbon
chlorine
copper
becomes more dilute
a) Dilute sulphuric acid
Structured questions
107 a) i) Manganese(IV) oxide
ii) Zinc
b) Potassium hydroxide
(1)
(1)
(1)
c) Any two of the following:
Voltage falls slowly over discharge (1) / Able to supply a steady current (1) / Able to supply a large current
(1) / Long shelf life (1) / Long service life (1) / Leak proof (1)
d) More expensive
(1)
e) Any one of the following:
Portable CD players (1) / Motorized toys (1) / Flash guns (1) / Appliances with moderate and continuous
drains (1)
239
108 a) i) The size of an alkaline manganese cell is too large.
(1)
ii) A toy car with an electric motor requires a large current. A zinc-carbon cell is used when small currents
are needed.
(1)
iii) The size of a silver oxide cell is small and cannot supply enough electrical energy for torches.
b) i) A: Zn + 2OH–
ZnO + H2O + 2e–
B: Ag2O + H2O + 2e–
2Ag + 2OH–
ii) Electrode B is the cathode.
This is because reduction occurs at this electrode.
109 a) Hydrogen
(1)
(1)
(1)
(1)
(1)
(1)
b) Z, Y, zinc, X
(1)
Z is the most reactive because only Z reacts with the water in zinc sulphate solution to give hydrogen.
(1)
Y is more reactive than zinc because it displaces zinc from zinc sulphate solution.
(1)
X is less reactive than zinc because it cannot displace zinc from zinc sulphate solution. Therefore X is the
least reactive.
(1)
c) A metal higher than zinc in the reactivity series is suitable as it forms ions more readily than zinc. This
matches the direction of current flow in the external circuit of the potato cell.
(1)
Z and zinc are furthest apart in the reactivity series (or electrochemical series). The difference between
the tendencies for Z and zinc to form ions is the greatest.
(1)
Hence Z should be used to build a potato cell with the maximum voltage.
110 a) i) Magnesium and copper
(1)
ii) They are furthest apart in the electrochemical series.
b) i) (1) Zn(s)
Zn2+(aq) + 2e–
(2) 2H+(aq) + 2e–
(1)
H2(g)
ii) Zinc loses electrons more readily than copper.
So electrons flow from the zinc strip to the copper strip.
(1)
(1)
(1)
(1)
(1)
c) i) Any two of the following:
240
• the volume of acid used / the depth of immersion of the metal strips
(1)
• the separation between the metal strips
(1)
• temperature
(1)
• the size of the metal strips
(1)
ii) The electrical conductivity of the sulphuric acid increases with its concentration.
(1)
∴ the current produced by the cell increases.
111 a) i) Nickel
(1)
The oxidation number of nickel increases from 0 to +2.
ii) Copper(II) ions
(1)
The oxidation number of copper decreases from +2 to 0.
iii) Ni(s) + Cu2+(aq)
(1)
Ni2+(aq) + Cu(s)
b) i) From the nickel electrode to the copper electrode
because nickel forms ions more readily than copper.
ii) Nickel
(1)
(1)
(1)
(1)
(1)
Oxidation occurs at the nickel electrode.
iii) The mass increases.
(1)
(1)
Copper(II) ions in the copper(II) sulphate solution near to the copper electrode gain electrons and form
copper atoms. As a result, a deposit of copper forms on the copper electrode.
(1)
Cu2+(aq) + 2e–
Cu(s)
(1)
iv) Sodium carbonate reacts with both solutions.
c) i) Decreases
(1)
(1)
Nickel atoms lose electrons and form nickel(II) ions.
(1)
ii) Nickel(II) ions near to the copper electrode gain electrons and form nickel atoms.
The net effect is the transfer of nickel from the nickel electrode to the copper electrode.
(1)
The concentration of nickel(II) ions in the nickel(II) sulphate remains unchanged.
(1)
Hence the colour intensity of the solution remains unchanged.
(1)
112 a) No reaction would occur.
(1)
The position of tin in the electrochemical series is higher than that of copper, i.e. tin is more reactive
than copper.
(1)
Hence copper cannot displace tin from tin(II) chloride solution.
b) A reaction would occur.
(1)
Permanganate ion is a stronger oxidizing agent than silver ion.
Hence acidified potassium permanganate solution would oxidize silver to silver ions.
(1)
MnO4–(aq) + 5Ag(s) + 8H+(aq)
(1)
Mn2+(aq) + 5Ag+(aq) + 4H2O(l)
241
c) A reaction would occur.
(1)
Silver ion is a stronger oxidizing agent than hydrogen ion.
Hence silver nitrate solution can oxidize hydrogen gas to hydrogen ions.
(1)
H2(g) + 2Ag+(aq)
(1)
2H+(aq) + 2Ag(s)
d) A reaction would occur.
(1)
Aqueous bromine is a stronger oxidizing agent than tin(II) ion.
Hence aqueous bromine can oxidize tin to tin(II) ions.
Br2(aq) + Sn(s)
2Br–(aq) + Sn2+(aq)
(1)
(1)
113 a) –1
(1)
b) i) H2O2(aq)
O2(g) + 2H+(aq) + 2e–
(1)
ii) The colour of the acidified potassium permanganate solution changes from purple to colourless. (1)
5H2O2(aq) + 2MnO4–(aq) + 6H+(aq)
c) i) H2O2(aq) + 2H+(aq) + 2e–
5O2(g) + 2Mn2+(aq) + 8H2O(l)
2H2O(l)
(1)
ii) The colour of the iron(II) sulphate solution changes from pale green to yellow-brown.
H2O2(aq) + 2Fe2+(aq) + 2H+(aq)
(1)
2H2O(l) + 2Fe3+(aq)
(1)
(1)
d) The oxidation number of oxygen in H2O2 is –1.
The oxidation number of oxygen in O2 is 0.
(0.5)
The oxidation number of oxygen in H2O is –2.
(0.5)
Oxygen is simultaneously reduced and oxidized.
(1)
Hence this is a disproportionation reaction.
114 a) MnO4–(aq) + 5Fe2+(aq) + 8H+(aq)
Sn2+(aq) + 2Fe3+(aq)
Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
Sn4+(aq) + 2Fe2+(aq)
5SO32–(aq) + 2MnO4–(aq) + 6H+(aq)
5SO42–(aq) + 2Mn2+(aq) + 3H2O(l)
(1)
(1)
(1)
b) i) As revealed in the given experimental results, oxidizing power is in the order:
242
MnO4–(aq) ion > Fe3+(aq) ion > Sn4+(aq) ion
(1)
∴ acidified KMnO4(aq) can oxidize Sn2+(aq) ions.
(1)
2MnO4–(aq) + 5Sn2+(aq) + 16H+(aq)
(1)
2Mn2+(aq) + 5Sn4+(aq) + 8H2O(l)
ii) The experimental results only reveal the following:
Oxidizing power: MnO4–(aq) ion > Fe3+(aq) ion, and MnO4–(aq) ion > SO42–(aq) ion
No comparison of oxidizing power between Fe3+(aq) ion and SO42–(aq) ion can be deduced from the
experimental results.
(1)
∴ it is not possible to predict whether a reaction would occur.
(1)
115 a) The oxidation number of chlorine changes from 0 to +1 in NaOCl.
(0.5)
The oxidation number of chlorine changes from 0 to –1 in NaCl.
(0.5)
Chlorine is simultaneously reduced and oxidized / chlorine undergoes disproportionation.
b) i) NaOCl(aq) + 2HCl(aq)
Cl2(g) + NaCl(aq) + H2O(l)
(1)
(1)
ii) (1) No observable change
(1)
(2) The iron(II) sulphate solution changes from pale green to yellow-brown.
2Fe2+(aq) + Cl2(aq)
(1)
2Fe3+(aq) + 2Cl–(aq)
(1)
(3) The potassium bromide solution changes from colourless to yellow-brown.
(1)
Cl2(g) + 2Br–(aq)
2Cl–(aq) + Br2(aq) / Cl2(g) + 2KBr(aq)
2KCl(aq) + Br2(aq)
(1)
iii) Any two of the following:
• Save chemicals / reduce the cost of chemicals used.
(1)
• Reduce the chemical waste produced / disposal of chemicals after experiment is easier.
(1)
• Shorten the time for conducting the experiment.
(1)
• Require less work space for carrying out an experiment.
(1)
• Less dangerous / easier to handle.
(1)
116 a) i) By exposing nitrogen monoxide to air.
(1)
ii) The colourless nitrogen monoxide gas changes to brown nitrogen dioxide gas.
(1)
iii) 2NO(g) + O2(g)
(1)
2NO2(g)
b) i) By adding concentrated nitric acid to magnesium / zinc / copper.
(1)
ii) Any one of the following:
The magnesium / zinc / copper dissolves.
(1)
A brown gas is given off.
(1)
iii) Mg(s) + 2NO3–(aq) + 4H+(aq)
Mg2+(aq) + 2NO2(g) + 2H2O(l) /
Zn(s) + 2NO3–(aq) + 4H+(aq)
Zn2+(aq) + 2NO2(g) + 2H2O(l) /
Cu(s) + 2NO3–(aq) + 4H+(aq)
Cu2+(aq) + 2NO2(g) + 2H2O(l)
(1)
243
c) i) By adding dilute nitric acid to magnesium / zinc / copper.
ii) 3Mg(s) + 2NO3–(aq) + 8H+(aq)
3Mg2+(aq) + 2NO(g) + 4H2O(l) /
(1)
(1)
3Zn(s) + 2NO3–(aq) + 8H+(aq)
3Zn2+(aq) + 2NO(g) + 4H2O(l) /
(1)
3Cu(s) + 2NO3–(aq) + 8H+(aq)
3Cu2+(aq) + 2NO(g) + 4H2O(l)
(1)
117 a) i) Corrosive property
(1)
ii) Oxidizing hazard warning symbol
(1)
b) i) Copper
(1)
ii) Cu(s) + 2NO3–(aq) + 4H+(aq)
Cu2+(aq) + 2NO2(g) + 2H2O(l)
(1)
iii) By filtration
(1)
iv) Copper(II) nitrate solution
(1)
v) Cu2+(aq) + 2OH–(aq)
Cu(OH)2(s)
(1)
118 a) i) Any two of the following:
• Gas bubbles were given off.
(1)
• The liquid turned blue.
(1)
• The copper turnings dissolved.
(1)
ii) Cu(s) + 2H2SO4(l)
CuSO4(aq) + SO2(g) + 2H2O(l)
b) The blue litmus solution turned red
(1)
(1)
because sulphur dioxide dissolved in water to form an acidic solution.
(1)
c) Sodium hydroxide solution was used to absorb the excess sulphur dioxide.
(1)
Potential hazard — if sodium hydroxide solution was not used, toxic sulphur dioxide gas might escape
into the air and attack human’s respiratory system.
(1)
d) i) Set-up 1 was not suitable because sulphur dioxide was soluble in water.
Set-up 2 was not suitable because sulphur dioxide was denser than air.
(1)
(1)
ii)
HBTKBS
PS
HBTTZSJOHF
(1)
244
119 a) C(s) + 2H2SO4(l)
CO2(g) + 2SO2(g) + 2H2O(l)
(1)
b) Oxidizing property
(1)
c) i) The purple colour of the acidified potassium permanganate solution fades gradually.
(1)
Sulphite ions are formed when sulphur dioxide produced in test tube A dissolves in water. Sulphite
ions reduce the permanganate ions to manganese(II) ions.
(1)
+
ii) 5SO32–(aq) + 2MnO4–(aq) + 6H (aq)
5SO42–(aq) + 2Mn2+(aq) + 3H2O(l)
d) The carbon dioxide gas produced in test tube A turns the limewater milky.
(1)
(1)
This is because the gas reacts with calcium hydroxide solution to form insoluble white calcium
carbonate.
(1)
e) Any one of the following:
Conduct the experiment in fume cupboard
(1)
because the sulphur dioxide given off in the reaction is toxic.
(1)
Wear safety glasses and protective gloves
(1)
because concentrated sulphuric acid is corrosive.
(1)
120 a) i) –2
(1)
ii) Sodium iodide
iii) H2SO4(l) + 8H+(aq) + 8e–
(1)
H2S(g) + 4H2O(l)
(1)
b) i) +4
(1)
ii) Sodium bromide
iii) 2NaBr(s) + 3H2SO4(l)
(1)
2NaHSO4(s) + SO2(g) + Br2(g) + 2H2O(l)
(1)
c) i) Hydrogen fluoride / hydrogen chloride
(1)
ii) Sodium fluoride / sodium chloride
(1)
iii) NaF(s) + H2SO4(l)
121 a) i) Zn(s) + H2SO4(aq)
NaHSO4(s) + HF(g) or NaCl(s) + H2SO4(l)
ZnSO4(aq) + H2(g)
ii) This shows the acidic property of dilute sulphuric acid.
b) i) Zn(s) + 2H2SO4(l)
ZnSO4(aq) + SO2(g) + 2H2O(l)
ii) This shows the oxidizing property of concentrated sulphuric acid.
NaHSO4(s) + HCl(g)
(1)
(1)
(1)
(1)
(1)
c)
(1)
245
d) Any two of the following:
• Wear safety glasses / protective gloves.
(1)
• Never add water to the acid.
(1)
• In case any acid is spilt on the skin or clothes, wash the affected area with plenty of water.
(1)
e) Dilute concentrated sulphuric acid by adding it slowly to a large amount of water while stirring.
(1)
122 a) Any two of the following:
• Save chemicals / reduce the cost of chemicals used.
(1)
• Reduce the chemical waste produced / disposal of chemicals after experiment is easier.
(1)
• Shorten the time for conducting the experiment.
(1)
• Require less work space for carrying out an experiment.
(1)
• Less dangerous / easier to handle.
(1)
b) i) The pH paper turns red.
(1)
ii) This shows the acidic property of sulphur dioxide.
(1)
c) i) The aqueous bromine turns from yellow-brown to colourless.
(1)
ii) This shows the reducing property of sulphur dioxide.
iii) SO32–(aq) + H2O(l) + Br2(aq)
(1)
SO42–(aq) + 2H+(aq) + 2Br–(aq)
d) i) The iron(III) sulphate solution turns from yellow-brown to pale green.
(1)
(1)
ii) This shows the reducing property of sulphur dioxide.
(1)
iii) SO32–(aq) + H2O(l) + 2Fe3+(aq)
(1)
SO42–(aq) + 2H+(aq) + 2Fe2+(aq)
e) Test the gas with a piece of filter paper soaked with acidified potassium dichromate solution.
(1)
The gas turns the paper from orange to green.
Cr2O72–(aq) + 3SO2(aq) + 2H+(aq)
2Cr3+(aq) + 3SO42–(aq) + H2O(l)
123 a) It is used as a salt bridge / to complete the circuit.
b)
(1)
(1)
7
DBSCPOSPE
NPMENm'F40
BR
DBSCPOSPE
TBMU
CSJEHF
m
NPMEN ,MBR
(1 mark for correct set-up; 1 mark for correct labels of the solutions and voltmeter; 1 mark for correct
label of the salt bridge; 0 mark if the set-up is not workable.)
(3)
246
c) Oxidizing agent: iron(III) sulphate
(1)
Reducing agent: potassium iodide
(1)
d) i) Iron(III) sulphate solution changes from yellow-brown to pale green.
(1)
Potassium iodide solution changes from colourless to brown.
(1)
ii) 2Fe3+(aq) + 2I–(aq)
2Fe2+(aq) + I2(aq)
e) Chemical energy is converted into electrical energy in a chemical cell,
while chemical energy is converted into heat when the solutions are mixed directly.
124 a) i) MnO4–(aq) + 8H+(aq) + 5e–
Mn2+(aq) + 4H2O(l)
ii) The purple colour of the mixture fades
because purple permanganate ions are reduced to pale pink / colourless manganese(II) ions.
b) i) Sn2+(aq)
Sn4+(aq) + 2e–
ii) An oxidation occurs
(1)
(1)
(1)
(1)
(1)
(1)
(1)
(1)
because the oxidation number of tin increases from +2 to +4.
(1)
c) From electrode Y to electrode X
(1)
d) • It completes the circuit by allowing ions to move from one half-cell to the other.
(1)
• It provides ions that can move into the half-cells to prevent the build-up of excess positively or negatively
charged ions in the solutions which would cause the reaction to stop.
(1)
125 a) i) Br2(aq) + 2e–
2Br–(aq)
ii) The colour of the mixture fades
because the amount of yellow-brown bromine in the mixture decreases.
b) i) Fe2+(aq)
Fe3+(aq) + e–
ii) An oxidation occurs
because the oxidation number of iron increases from +2 to +3.
(1)
(1)
(1)
(1)
(1)
(1)
c) From electrode Y to electrode X
(1)
d) No
(1)
Sodium sulphite solution can react with Br2(aq) / Fe2(SO4)3(aq).
(1)
247
126 a) Positive terminal
(1)
b) At the cathode: PbO2(s) + 4H+(aq) + SO42–(aq) + 2e–
At the anode: Pb(s) + SO42–(aq)
PbSO4(s) + 2H2O(l)
PbSO4(s) + 2e–
c) i) Negative terminal
(1)
(1)
(1)
The overall reaction for the recharging process is the reverse of the discharging process.
(1)
Hence reduction should take place at the lead plate during the recharging process,
(1)
i.e. it should be made the cathode (connected to the negative terminal of the external power
source).
ii) PbSO4(s) + 2e–
Pb(s) + SO42–(aq)
(1)
127 a) i) From the cadmium strip to the nickel strip
ii) Cd(s)
(1)
Cd2+(aq) + 2e–
(1)
iii) It should be a soluble ionic compound.
(1)
It must not react with the two solutions in the half-cells.
b) i) +4
(1)
(1)
ii) X is the cathode.
(1)
Reduction occurs at X.
(1)
iii) (1) Electrode Y
(1)
The overall reaction for the recharging process is the reverse of the discharging process.
(1)
Hence reduction should occur at Y during the recharging process.
(1)
Hence Y should be made the cathode (connected to the negative terminal of the external power
source).
(2) Cd(OH)2(s) + 2e–
Cd(s) + 2OH–(aq)
(1)
iv) Nickel metal hydride cells offer more energy than nickel-cadmium cells.
Nickel metal hydride cells contain much less toxic cadmium than nickel-cadmium cells.
128 a) Cathodic reaction: O2(g) + 2H2O(l) + 4e–
Anodic reaction: H2(g) + 2OH–(aq)
b) From electrode X to electrode Y
4OH–(aq)
2H2O(l) + 2e–
(1)
(1)
(1)
(1)
(1)
c) Any two of the following:
248
• Fuel cells do not emit air pollutants.
(1)
• Fuel cells have high fuel efficiency.
(1)
• Fuel cells can operate continuously if the flow of hydrogen and oxygen can be maintained (they do
not run down or require charging).
(1)
129 a) The magnesium strip.
(1)
Oxidation occurs at it.
(1)
b) At the magnesium strip: Mg(s)
Mg2+(aq) + 2e–
(1)
Cu(s)
(1)
At the copper strip: Cu2+(aq) + 2e–
c) The multimeter reading would decrease.
(1)
When the zinc and copper strips in lemon A are interchanged, the set-up consists of two competing
lemon cells.
(1)
The total voltage of the set-up is less than the sum of the voltages of the two lemon cells.
130 a) i) The liquid around pencil A turned red.
(1)
ii) Around pencil A, hydroxide ions were preferentially discharged to form oxygen gas.
4OH–(aq)
(1)
O2(g) + 2H2O(l) + 4e–
(1)
(1)
Water dissociated continuously to replace the hydroxide ions discharged. Thus there was an excess of
hydrogen ions near pencil A. The solution there became acidic.
(1)
The universal indicator turned red.
b) i) The liquid around pencil B turned blue.
(1)
ii) Around pencil B, hydrogen ions were preferentially discharged to form hydrogen gas.
2H+(aq) + 2e–
H2(g)
(1)
(1)
Water dissociated continuously to replace the hydrogen ions discharged. Thus there was an excess of
hydroxide ions near pencil B. The solution there became alkaline.
(1)
The universal indicator turned blue.
c)
PYZHFO
PYZHFO
EJMVUFTPEJVN
TVMQIBUFTPMVUJPO
QMBUJOVN
FMFDUSPEF
BOPEF
IZESPHFO
IZESPHFO
QMBUJOVN
FMFDUSPEF
DBUIPEF
PS
EJMVUFTPEJVN
TVMQIBUFTPMVUJPO
DBSCPO
FMFDUSPEF
BOPEF
DBSCPO
FMFDUSPEF
DBUIPEF
(1 mark for correct labels of the electrodes; 1 mark for showing gases collected above the electrodes; 1
mark for correct labels of hydrogen and oxygen; 0 mark if the set-up is not workable.)
(3)
249
131 a) Oxygen
b) The gas relights a glowing splint.
c) i) Use concentrated sodium chloride solution instead of very dilute sodium chloride solution.
(1)
The concentration of chloride ions in the solution is much higher than that of hydroxide ions.
(1)
Therefore chloride ions are preferentially discharged to form chlorine gas.
(1)
ii) Sodium hydroxide solution
(1)
iii) Any one of the following:
• Manufacture of bleach
(1)
• Manufacture of soaps / detergents
(1)
d) i) Zinc loses electrons to form zinc ions.
(1)
Ammonium ions gain electrons to form ammonia and hydrogen gas.
(1)
Manganese(IV) oxide is an oxidizing agent used to remove the hydrogen gas.
(1)
When the cell is connected to an external circuit, electrons will flow from the zinc case to the carbon
rod in the external circuit.
(1)
ii) • If a current is drawn from the cell rapidly, the gaseous product cannot be removed fast enough.
The voltage drops as a result.
(1)
• There is a slow direct reaction between the zinc electrode and ammonium ions. After some time,
the zinc case becomes too thin and the paste leaks out.
(1)
iii) It may lead to the explosion if the dry cell is disposed of in fire.
132 a) i) 2Na(s) + 2H2O(l)
2NaOH(aq) + H2(g)
(1)
(1)
ii) Sodium undergoes oxidation
because the oxidation number of sodium increases from 0 to +1.
(1)
Water undergoes reduction
because the oxidation number of hydrogen decreases from +1 to 0.
(1)
b) Any one of the following:
• Experiment I is preferred
because a sodium hydroxide solution of high purity can be obtained.
(1)
• Experiment II is preferred
because the reaction in Experiment I is very vigorous and difficult to control.
(1)
c) i) The concentration of chloride ions in concentrated sodium chloride solution is much higher than that
of hydroxide ions.
(1)
250
ii) A hydrogen ion is a stronger oxidizing agent than a sodium ion, i.e. it is more readily reduced.
(1)
iii) Hydrogen ions and chloride ions are consumed in the electrolysis. Sodium ions and hydroxide ions
remain in the solution.
(1)
iv) At the anode: 2Cl–(aq)
Cl2(g) + 2e–
At the cathode: 2H+(aq) + 2e–
d)
(1)
H2(g)
DIMPSJOF
(1)
IZESPHFO
DPODFOUSBUFETPEJVN
DIMPSJEFTPMVUJPO
DBSCPOFMFDUSPEF
BOPEF
DBSCPOFMFDUSPEF
DBUIPEF
(1 mark for correct labels of the electrodes; 1 mark for showing the collection of gases above the
electrodes; 1 mark for correct labels of chlorine and hydrogen; 0 mark if the set-up is not workable.)
(3)
133 a) At electrode X:
The chloride ions and hydroxide ions in the copper(II) chloride solution move towards electrode X (the
anode) in the electrolysis process.
The concentration of chloride ions in the solution is much higher than that of hydroxide ions.
(1)
Therefore chloride ions are preferentially discharged to form chlorine gas.
(1)
At electrode Y:
The copper(II) ions and hydrogen ions are attracted to the electrode Y.
A copper(II) ion is a stronger oxidizing agent than a hydrogen ion.
(1)
Therefore copper(II) ions are preferentially discharged to form a deposit of copper on electrode Y.
(1)
b) At electrode X: 2Cl–(aq)
At electrode Y: Cu2+(aq) + 2e–
Cl2(g) + 2e
–
(1)
Cu(s)
(1)
c) The blue colour of the electrolyte becomes less intense
because the concentration of copper(II) ions in the electrolyte decreases.
d) The copper(II) chloride solution would remain unchanged
because the net effect is the transfer of copper from the anode to the cathode.
(1)
(1)
(1)
(1)
251
134 a) The ammeter is an instrument used to measure the electric current passing through the circuit.
The rheostat is used to vary the resistance in the circuit and regulate the current.
b) i) Electrode A dissolves.
(1)
(1)
(1)
A reddish brown deposit forms on electrode B.
Cu2+(aq) + 2e–
ii) At electrode A: Cu(s)
At electrode B: Cu2+(aq) + 2e–
(1)
(1)
Cu(s)
(1)
iii) The blue colour of the dilute copper(II) sulphate solution remains unchanged.
(1)
Copper is transferred from A to B. The concentration of copper(II) ions in the copper(II) sulphate
solution remains the same.
(1)
c) i) 2H–(l)
H2(g) + 2e–
(1)
ii) Let the oxidation number of M in compound X be +n. The chemical formula of X is MHn.
The overall cell reaction is MHn
M +
n
H
2 2
(1)
The number of M atoms produced equals the number of hydrogen molecules produced,
i.e.
n
= 1
2
n= 2
∴ the oxidation number of M in compound X is +2.
135 a) From the electrode made of metal M to electrode B / from electrode A to the copper electrode.
Silver ions receive electrons and are reduced at electrode B.
b) As a chemical cell / source of electricity
(1)
(1)
(1)
(1)
c) Zinc (or any metal higher than copper in the electrochemical series, except for potassium and sodium)
(1)
Metal M is the negative electrode of the chemical cell;
(1)
it loses electrons more readily than copper and is thus higher than copper in the electrochemical series.
(1)
d) i) Bubbles of gas (oxygen) are given off.
4OH–(aq)
(1)
2H2O(l) + O2(g) + 4e–
(1)
ii) Metal M (or zinc) dissolves.
M(s)
Mn+(aq) + ne– or Zn(s)
(1)
Zn2+(aq) + 2e–
iii) A reddish-brown deposit forms on the electrode.
Cu2+(aq) + 2e–
252
Cu(s)
(1)
(1)
(1)
136 a) Chromium
(1)
b) To make it a conductor of electricity.
(1)
c) To improve its appearance.
(1)
d) Negative terminal
(1)
e) i) Cr(s)
Cr3+(aq) + 3e–
ii) Cr3+(aq) + 3e–
(1)
Cr(s)
(1)
f) Cr(NO3)3
(1)
It is because to deposit 1 mole of Cr from 1 mole of Cr(NO3)4, 4 moles of electrons are required
[Cr4+(aq) + 4e–
Cr(s)] while it requires only 3 moles of electrons to deposit 1 mole of Cr from 1
mole of Cr(NO3)3 [Cr3+(aq) + 3e–
Cr(s)].
(1)
g) Knobs / water taps / any other suitable items
(1)
h) i) Heavy metal ions are toxic / may poison water life.
(1)
ii) To recover the chromium metal.
iii) 3SO32–(aq) + Cr2O72–(aq) + 8H+(aq)
(1)
3SO42–(aq) + 2Cr3+(aq) + 4H2O(l)
iv) (1) Sodium hydroxide solution
(2) Cr3+(aq) + 3OH–(aq)
137 a) i) Cu2S(s) + O2(g)
(1)
(1)
Cr(OH)3(s)
2Cu(s) + SO2(g)
ii) The sulphide ion undergoes oxidation
(1)
(1)
(0.5)
because the oxidation number of sulphur increases from –2 to +4.
(0.5)
The copper(I) ion undergoes reduction
(0.5)
because the oxidation number of copper decreases from +1 to 0.
(0.5)
The oxygen undergoes reduction
(0.5)
because the oxidation number of oxygen decreases from 0 to –2.
(0.5)
b) i) Anode
(1)
ii) Iron and zinc form ions more readily than copper. Iron and zinc in the anode give up electrons first.
(1)
The copper gives up electrons to form copper(II) ions.
(1)
Impurities such as silver and gold settle at the bottom of the container.
(1)
Copper(II) ions are preferentially discharged to form copper at the cathode.
(1)
iii) The concentration of copper(II) ions in the copper(II) sulphate solution drops gradually.
(1)
This is because at the anode iron and zinc dissolves as ions readily while at the cathode copper(II) ions
are always preferentially discharged.
(1)
253
138 a) i) Cl2(g) + 2Br–(aq)
2Cl–(aq) + Br2(g)
ii) Chlorine is the oxidizing agent
(1)
because it causes the oxidation number of bromine to increase from –1 to 0.
b) Br2(g) + H2O(l)
HBr(aq) + HOBr(aq)
c) i) Br2(g) + SO2(g) + 2H2O(g)
(1)
2HBr(g) + H2SO4(g)
ii) A redox reaction is involved
(1)
(1)
(1)
(1)
because the oxidation number of Br decreases from 0 to –1 while that of S increases from +4 to
+6.
(1)
d) Glass is hard to corrode.
(1)
e) Any one of the following:
• Wearing protective suits
(1)
• Wearing eye protection
(1)
f)
PS
(1)
139 a) In China, driving distances are relatively short and speeds are frequently low because of traffic jams. (1)
So the limitations of electric cars are less of a problem.
b) The negative electrode consists of lithium atoms lying between graphite sheets.
The positive electrode is made of lithium cobalt oxide.
(1)
(1)
(1)
During discharge, lithium atoms at the negative electrode give out electrons to form lithium ions. The
lithium ions migrate across the electrolyte to the positive electrode.
(1)
Electrons flow from the negative electrode to the positive electrode in the external circuit.
(1)
c) Any one of the following:
• Charging the cell with a high current
(1)
• Loading the cell with excess discharge current
(1)
d) Any two of the following:
254
• There are not enough recharging stations in convenient locations for electric cars.
(1)
• Electric cars are more expensive than petrol-driven cars.
(1)
• There are not enough maintenance centers for providing services to electric cars.
(1)
• Any other sensible suggestion
(1)
e) Electricity is required for the charging of lithium ion cells.
China gets much of its electricity from coal, which produces more soot and more greenhouse gases than
other fuels.
(1)
140 a) Cathodic reaction: O2(g) + 2H2O(l) + 4e–
Anodic reaction: Al(s) + 3OH–(aq)
4OH–(aq)
(1)
Al(OH)3(s) + 3e–
(1)
b) • Does not emit air pollutants
(1)
• Can operate continuously if replaceable cartridges are available (it does not run down or require
recharging using an external power source)
(1)
c) Hydrogen is explosive. Aluminium is easier to handle.
(1)
d) Bauxite consists mainly of aluminium oxide. Aluminium is extracted from bauxite using electrolysis of the
molten ore.
(1)
The electrolysis is carried out in a steel tank.
The tank is lined with a graphite cathode while huge blocks of graphite anodes hang in the middle of
the tank.
(1)
During the electrolysis, aluminium ions are attracted to the cathode where they undergo reduction to
form aluminium.
(1)
Oxide ions are attracted to the anode where they undergo oxidation to form oxygen gas.
141 Place a few cm3 of sodium bromide solution in a test tube and then add aqueous chlorine to it.
The solution changes from colourless to yellow-brown.
(1)
(1)
(1)
In this experiment, bromide ions are oxidized to yellow-brown bromine by chlorine and chlorine is reduced
to colourless chloride ions. Therefore the oxidizing power of chlorine is stronger than that of bromine. (1)
Now place a few cm3 of aqueous bromine in a test tube and then add sodium sulphite solution to it.
(1)
The yellow-brown solution becomes colourless.
(1)
In this experiment, sulphite ions are oxidized to sulphate ions by bromine and bromine is reduced to colourless
bromide ions. Therefore the oxidizing power of bromine is stronger than that of sulphate ion.
(1)
(3 marks for organization and presentation)
255
142 Conversion of chemical energy into electrical energy
Any one of the following:
• Daniell cell
A Daniell cell consists of a zinc strip (the negative electrode) immersed in zinc sulphate solution contained
in a porous pot and a copper container (the positive electrode) containing copper(II) sulphate solution.
(1)
Zinc atoms give up electrons to form zinc ions.
(1)
The copper(II) ions in the copper(II) sulphate solution gain electrons from the external circuit to form
copper metal.
(1)
• Zinc-carbon cell
A zinc-carbon cell contains a carbon rod surrounded by a mixture of powdered carbon and manganese(IV)
oxide. The carbon rod is the positive electrode. The outer zinc case is the negative electrode. The electrolyte
is a moist paste of ammonium chloride.
(1)
Zinc atoms give up electrons to form zinc ions.
(1)
Ammonium ions gain electrons from the external circuit to form ammonia and hydrogen.
(1)
• Any chemical cell upon discharge
Set-up of the cell
(1)
Chemical changes that occur at the electrodes
(2)
Conversion of electrical energy into chemical energy
Any one of the following:
• Electrolysis of acidified water
During the electrolysis of acidified water using platinum electrodes, sulphate ions and hydroxide ions are
attracted to the anode. Hydrogen ions are attracted to the cathode.
(1)
Hydroxide ions are preferentially discharged (oxidized) to form oxygen gas at the anode.
(1)
Hydrogen ions are discharged (reduced) to form hydrogen gas at the cathode.
(1)
• Electrolysis of concentrated sodium chloride solution
During the electrolysis of concentrated sodium chloride solution using carbon electrodes, chloride ions
and hydroxide ions are attracted to the anode. Sodium ions and hydrogen ions are attracted to the
cathode.
(1)
Chloride ions are preferentially discharged (oxidized) to form chlorine gas at the anode.
(1)
Hydrogen ions are discharged (reduced) to form hydrogen gas at the cathode.
(1)
• Electrolysis of other species
The electrodes used and the movement of anions and cations.
256
(1)
The ions preferentially discharged at the anode.
(1)
The ions preferentially discharged at the cathode.
(1)
(3 marks for organization and presentation)
143 First dissolve the copper(II) sulphate crystals in the beaker of distilled water.
(1)
Then connect the positive terminal of the d.c. power supply to the copper plate
(1)
and the negative terminal of the d.c. power supply to the metal key.
(1)
Dip both the copper plate and the key into the copper(II) sulphate solution.
(1)
Pass electricity through the circuit for some time. The copper plate becomes thinner.
(1)
A layer of shiny metal deposits onto the key.
(1)
(3 marks for organization and presentation)
257