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Worked solutions to textbook questions
1
Chapter 28 Electrolysis: driving chemical reactions by
electricity
E1.
Use the Internet or another reference to find out why phenol (and, if possible, phenol
sulfonic acid) is regarded as a hazardous material.
AE1.
High levels of exposure of humans to phenol is reported to cause effects such as
headaches, respiratory irritation, fatigue, liver damage, diarrhoea and haemolytic
anaemia. It causes irritation to the lungs of animals. Phenol remains in the soil for
about 2–5 days and in water for longer than 9 days.
Hazards of phenol sulfonic acid include being corrosive, causing severe burns to the
eyes and skin, damaging the liver and other organs, and causing pulmonary oedema
(accumulation of fluids in the lungs).
E2.
Before the steel plate enters the tin plating cell it is cleaned thoroughly. Why is this
cleaning process essential?
AE2.
In order for the tin plating to adhere to the metal, iron oxide and grease present on the
steel surface must first be removed.
E3.
Why does the presence of a strong acid in the electrolyte allow higher rates of tin
plating than a weak acid?
AE3.
There are more ions present in a solution of a strong acid compared with a solution of
a weak acid of the same concentration. As a consequence, a strong acid solution will
conduct electricity better and allow higher rates of plating.
Q1.
A student wishes to copper-plate her iron locker key.
a Design a suitable electroplating cell. Identify the anode and cathode, electrode
polarities and the electrolyte to be used.
b If the student connects the cell to the wrong terminals of the power supply, what
happens to the locker key?
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Worked solutions to textbook questions
A1.
a
b
If the student were to connect the iron key to the positive terminal of the power
supply, it would be oxidised. Copper metal would plate the negative electrode.
Q2.
Draw a diagram to show the migration of ions in a nickel-plating cell containing
nickel sulfate solution.
A2.
Q3.
Calculate the quantity of electric charge, in coulomb, represented by:
a a current of 6.00 A flowing for 25.0 minutes
b a current of 25 mA flowing for 2.0 days
A3.
a
b
Q = It
Q = 6.00 A  (25.0  60) s
= 9.00  103 C
Q = It
Q = (25  10–3) A  (2  24  60  60) s
= 4.3  103 C
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Worked solutions to textbook questions
3
Q4.
The electrolyte in an electrolytic cell used to chromium-plate bicycle frames contains
Cr3+ ions. If the cell operates at 30.0 A for 25.0 minutes, calculate:
a the quantity of charge that passes through the cell
b the amount, in mol, of electrons that pass through the cell, in mol
c the mass of metal deposited on the frames
A4.
a
b
c
Q = It
Q = 30.0 A  (25  60) s
= 4.50  104 C
Q
n(e–) =
F
4.50  10 4 C
n(e–) =
96 500 C mol 1
= 0.466 mol
Step 1 Write the equation.
Cr3+(aq) + 3e–  Cr(s)
Step 2 Use stoichiometry to calculate the amount of Cr.
n(Cr )
1
=

3
n (e )
0.4663
n(Cr)
=
3
= 0.1554 mol
Step 3 Calculate the mass of Cr.
m(Cr) = 0.1554 mol  52.00 g mol–1
= 8.08 g
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4
Q5.
A 0.300 g mass of silver is to be used to plate a sporting trophy. How long should the
trophy be left in a silver-plating cell using a current of 6.00 A?
A5.
Step 1
Step 2
Step 3
Step 4
Step 5
Write the equation.
Ag+(aq) + e–  Ag(s)
Calculate the amount of Ag.
0.300 g
n(Ag) =
107.87 g mol 1
= 0.002781 mol
Use stoichiometry to calculate the amount in mol of electrons.
1
n (e  )
=
n(Ag ) 1
So n(e) = 0.002781 mol
Calculate the quantity of electricity required.
Q
= n(e)  F
= 0.002781 mol  96 500 C mol–1
= 268 C
Q
Calculate the time required, using I = .
t
Q
t
=
I
268 C
=
6.00 A
= 44.7 s
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Worked solutions to textbook questions
5
Q6.
A solution used to gold-plate bathroom fittings contains Au3+ ions. What current
would be required to plate 5.00 g of gold onto a tap in 30.0 minutes?
A6.
Step 1
Step 2
Step 3
Step 4
Step 5
Write the equation.
Au3+(aq) + 3e–  Au(s)
Calculate the amount of Au.
5.00 g
n(Au) =
196.967 g mol 1
= 0.0254 mol
Use stoichiometry to calculate the amount of electrons.
3
n (e  )
=
1
n(Au )
n(e)
= 3  0.0254 mol
= 0.076 mol
Calculate the quantity of electricity required.
Q
= n(e)  F
= 0.076  96 500
= 7348.9 C
Q
Calculate the current, using I = .
t
7348.9 C
I
=
30.0  60 s
= 4.08 A
Q7.
Predict the products at each electrode during electrolysis (using unreactive electrodes)
of the following 1 M solutions. The nitrate ion is not involved in any of the reactions.
a copper(II) bromide
b sodium iodide
c lead(II) nitrate
d zinc chloride
e aluminium nitrate
A7.
a
b
c
d
e
Br2(aq); Cu(s)
I2(aq); H2(g) and OH–(aq)
O2(g) and H+(aq); Pb(s)
O2(g) and H+(aq) (Cl2(g) is formed in practice); Zn(s)
O2(g) and H+(aq); H2(g) and OH–(aq)
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Worked solutions to textbook questions
6
Q8.
From the industrial electrolytic cells described in the text, select one that uses a
molten electrolyte and another that uses an aqueous electrolyte.
a Write the half reactions that occur at each electrode.
b What materials are used for the anode and cathode in each cell?
c Explain the reasons for the choice of the electrode materials.
d Why is an aqueous electrolyte more energy efficient than a molten electrolyte?
e Explain why aqueous solutions are not employed in the cell using a molten
electrolyte.
f What factors may be important in deciding where to site an industrial cell of each
type?
A8.
Refer to the appropriate descriptions of each cell in the text to answer this question.
Chapter review
Q9.
a
b
Explain what is meant by a faraday of charge.
State the number of faradays of charge needed to produce 1 mole of:
i silver atoms from silver nitrate solution
ii zinc atoms from zinc nitrate solution
iii chlorine molecules from molten potassium chloride
iv hydrogen molecules from water
A9.
a
b
A faraday is the charge on 1 mole of electrons, 96 500 C.
i 1F
ii 2 F
iii 2 F
iv 2 F
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Worked solutions to textbook questions
7
Q10.
A car battery is charged at a current of 2.0 A for 9.00 hours. Calculate the mass of
lead deposited at the cathode if the electrode reaction is represented by the equation:
PbSO4(s) + 2e–  Pb(s) + SO42–(aq)
A10.
Step 1
Calculate the number of coulomb, using Q = It.
Q
= 2.0 A  (9.00  60  60) s
= 64 800 C
Step 2
Calculate the amount in mol of electrons, using n =
64 800 C
96 500 C mol 1
= 0.6715 mol
Use stoichiometry to calculate the amount of Pb.
n( Pb)
1
=

2
n (e )
0.6715
n(Pb)
=
2
= 0.3358 mol
m
Calculate the mass of Pb, using n =
.
M
m(Pb)= 0.3358 mol  207.2 g mol–1
= 69.57 g
= 70 g
n(e–) =
Step 3
Step 4
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Q
.
F
Worked solutions to textbook questions
8
Q11.
Sodium chloride solution is electrolysed in a diaphragm cell to manufacture chlorine.
The cell operates at 100 000 A. For one day’s operation, calculate:
a the volume of chlorine produced, measured at SLC
b the mass of sodium chloride consumed
A11.
a
Step 1
Step 2
Write a balanced equation.
2Cl–(aq)  Cl2(g) + 2e–
Calculate the number of coulomb, using Q = It.
Q = 100 000 A  (24  60  60) s
= 8.64  109 C
Step 3
Calculate the amount in mol of electrons, using n =
8.64  10 9 C
96 500 C mol 1
= 8.953  104 mol
Use stoichiometry to calculate the amount of Cl2.
n(Cl 2 )
1
=

2
n (e )
n(e–)
Step 4
=
8.953  10 4
=
2
= 4.477  104 mol
n(Cl2)
Step 5
b
Step 1
Step 2
V
.
Vm
V(Cl2) = (4.477  104) mol  24.5 mol L–1
= 1.10  106 L
Calculate the number of coulomb, using Q = It.
Q = 100 000 A  (24  60  60) s
= 8.64  109 C
Calculate the volume of Cl2, using n =
Calculate the amount in mol of electrons, using n =
8.64  10 9 C
96 500 C mol 1
= 8.953  104 mol
Use stoichiometry to calculate the amount of NaCl.
n( NaCl)
1
=

1
n (e )
n(NaCl) = 8.953  104 mol
m
Calculate the mass of NaCl, using n =
.
M
m(NaCl) = 8.953  104 mol  58.5 g mol–1
= 5.237  106 g
= 5.24 tonnes
n(e–)
Step 3
Step 4
Q
.
F
=
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Q
.
F
Worked solutions to textbook questions
Q12.
Iron is plated with tin in an electrolytic cell containing K2Sn(OH)6 as the electrolyte.
A cell operates for 5.00 hours at a current of 25.0 A.
a What is the charge on the tin ions in the electrolyte?
b How many faradays of charge are required to produce 1.00 mole of tin?
c Calculate the mass of tin deposited during this period.
A12.
a
b
c
Write a balanced equation. Remember that K2SnF6 contains Sn4+ ions.
Sn4+(aq) + 4e–  Sn(s)
Step 2 A faraday is the charge on 1 mol electrons. So the number of faraday is
the same as the mole of electrons. Use stoichiometry to calculate the
amount of electrons.
4
n(e  )
=
1
n(Sn )
–
n(e ) = 4  1.00 mol
= 4.00 mol
So number of faraday = 4.00 F
Step 1 Calculate the number of coulomb, using Q = It.
Q
= 25.0 A  (5.00  60  60) s
= 450 000 C
Q
Step 2 Calculate the amount in mol of electrons, using n = .
F
450 000 C
n(e–) =
96 500 C mol 1
= 4.663 mol
Step 3 Use stoichiometry to calculate the amount of Sn.
n(Sn )
1
=

4
n (e )
4.663
n(Sn) =
4
= 1.166 mol
m
Step 4 Calculate the mass of tin, using n =
.
M
m(Sn) = 1.166 mol  118.69 g mol–1
= 139 g
Step 1
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Worked solutions to textbook questions
Q13.
A charge of 0.400 faraday was passed through 1.00 L of 1.00 M copper(II) sulfate
solution using carbon electrodes.
a Write equations for the reactions at each electrode.
b Calculate the concentration of the copper ions in solution after electrolysis.
A13.
2H2O(l)  4H+(aq) + O2(g) + 4e–
Cu2+(aq) + 2e–  Cu(s)
b Step 1 Calculate the amount in mol of electrons, remembering that 1 F is the
charge on to 1 mol electrons.
n(e–) = 0.400 mol
Step 2 Use stoichiometry to calculate the amount of Cu2+ deposited as Cu.
1
n(Cu 2 )
=

2
n (e )
0.400
n(Cu2+)
=
2
= 0.200 mol
Step 3 Calculate the amount of Cu2+ remaining.
n(Cu2+)remaining = n(Cu2+)initial – n(Cu2+)discharged
= (1.00 L  1.00 M) mol – 0.200 mol
= 0.800 mol
Step 4 Calculate the concentration of Cu2+ remaining.
0.800 mol
c(Cu2+) =
1.00 L
= 0.800 M
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Worked solutions to textbook questions
11
Q14.
Three electrolytic cells containing silver nitrate solution, copper(II) sulfate solution
and chromium(III) sulfate solution, respectively, were connected in series so that the
same amount of electric charge passed through each cell. Metal was deposited at the
cathode of each cell. If 10.0 g of silver was obtained from one cell, what mass of
metal would be obtained from each of the other cells?
A14.
Step 1
Write a balanced equation for each reaction.
Ag+(aq) + e–  Ag(s)
Cu2+(aq) + 2e–  Cu(s)
Cr3+(aq) + 3e–  Cr(s)
Step 2
Calculate the amount of Ag deposited, using n =
10.0 g
107.87 g mol 1
= 0.09270 mol
Use stoichiometry to calculate the amount in mol of electrons that passed
through all cells.
1
n (e  )
=
1
n(Ag )
–
n(e )
= 0.09270 mol
Use stoichiometry to calculate the amount of Cu and Cr deposited.
n (Cu )
1
=

2
n (e )
0.09270
n(Cu) =
2
= 0.04635 mol
n(Cr )
1
=

3
n (e )
0.09270
n(Cr) =
3
= 0.0309 mol
Calculate the masses of Cu and Cr deposited.
m(Cu) = 0.04635 mol  63.55 g mol–1
= 2.946 g
= 2.95 g
m(Cr) = 0.0309 mol  52.00 g mol–1
= 1.607 g
= 1.61 g
n(Ag) =
Step 3
Step 4
Step 5
m
.
M
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Worked solutions to textbook questions
12
Q15.
Electrolysis of concentrated sodium chloride solution produced 15.0 L of chlorine gas,
measured at SLC. What mass of silver would be produced if the same amount of
electricity were passed through a silver-plating cell?
A15.
Step 1
Write a balanced equation.
2Cl–(aq)  Cl2(g) + 2e–
Step 2
Calculate the amount of Cl2, using n =
15.0 L
24.5 mol L1
= 0.6122 mol
Use stoichiometry to calculate the amount in mol of electrons released.
n (e  )
2
=
1
n(Cl 2 )
–
n(e )
= 2  0.6122
= 1.224 mol
Write a balanced equation.
Ag+(aq) + e– Ag(s)
Use stoichiometry to calculate the amount of Ag deposited.
n ( Ag )
1
=

1
n (e )
n(Ag) = 1.224 mol
m
Calculate the mass of Ag, using n =
.
M
m(Ag) = 1.224 g  107.87 g mol–1
= 132 g
n
Step 3
Step 4
Step 5
Step 6
V
.
Vm
=
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13
Q16.
A nickel teapot, with a surface area of 0.0900 m2, is to be silver plated.
a Which electrode should the teapot be connected to?
b What is the polarity of this electrode?
c If the plating is to be 0.005 00 cm thick and the density of silver is 10.5 g cm–3,
how long should the pot be put in a silver-plating cell with a current of 0.500 A?
A16.
a
b
c
cathode
negative
Step 1 Calculate the volume of Ag required, using volume = area  thickness.
V(Ag) = (0.0900  10 000) m2  0.00500 cm
= 4.5 cm3
m
Step 2 Calculate the mass of Ag required, using density =
.
V
m(Ag) = 10.5 g cm–3  4.5 cm3
= 47.25 g
Step 3 Calculate the amount of Ag.
47.25 g
n(Ag) =
107.87 g mol 1
= 0.4380 mol
Step 4 Use stoichiometry to calculate the amount in mol of electrons required.
1
n (e  )
=
1
n(Ag )
= 0.4380 mol
Q
Step 5 Calculate the amount of charge required, using n(e–) = .
F
–1
Q
= 0.4380 mol  96 500 C mol
= 42 269 C
Step 6 Calculate the time required, using Q = It.
42 269 C
t =
0.500 A
= 84 539 s
= 23.5 h
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Worked solutions to textbook questions
Q17.
Electrolysis of a molten ionic compound with a current of 0.50 A for 30.0 minutes
yielded 0.700 g of a metallic element at the cathode. If the element has a relative
atomic mass of 150, calculate the charge on the metal ions.
A17.
Step 1
Step 2
Step 3
Step 4
Step 5
Calculate the charge which passed through circuit, using Q = It.
Q
= 0.50 A  (30.0  60) s
= 900 C
Q
Calculate the amount in mol of electrons, using n(e–) =
.
F
900 C
n(e–) =
96 500 C mol 1
= 0.0093264 mol
m
Calculate the amount of metal, using n =
.
M
0.700 g
n =
150 g mol 1
= 0.004667 mol
Calculate the amount of electrons per mole of metal.
0.0093264 mol
n (e  )
=
0.004667 mol
n(metal )
= 1.998
=2
Calculate the charge on metal ion.
Charge must be a whole number. So it is 2+.
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Worked solutions to textbook questions
15
Q18.
A student wishes to copper-plate a nickel medallion and sets up an electrolytic cell
using 250 mL of 1.00 M copper(II) sulfate solution and a large copper anode. The
student runs a current of 10.0 A through the cell for 20 minutes.
a What mass of copper will be plated onto the medallion?
b What will be the concentration of copper(II) sulfate remaining in solution?
The student replaces the copper anode with an inert graphite electrode and runs
the same current through the cell for a further 20 minutes.
c What gas will be produced at the anode?
d What will be the concentration of copper(II) sulfate remaining in solution?
A18.
a
Step 1
Step 2
Step 3
Write balanced half equations.
Cathode: Cu2+(aq) + 2e–  Cu(s)
Anode: Cu(s)  Cu2+(aq) + 2e–
Calculate the quantity of charge, using Q = It.
Q = 10.0 A  (20  60) s
= 12 000 C
Calculate the amount in mol of electrons, using n(e–) =
12 000 C
96 500 C mol 1
= 0.1244 mol
Step 4 Use stoichiometry to calculate the amount of Cu2+ deposited as Cu.
1
n(Cu 2 )
=

2
n (e )
0.1244
n(Cu2+) =
2
= 0.0622 mol
Step 5 Calculate the mass of Cu deposited.
m(Cu) = 0.0622 mol  63.5 g mol–1
= 3.95 g
2+
Cathode: Cu (aq) + 2e–  Cu(s)
Anode: Cu(s)  Cu2+(aq) + 2e–
Because the reverse reaction is occurring at each electrode [Cu] will remain
constant, 1.00 M.
Write balanced half equations after the change of electrode.
Cathode: Cu2+(aq) + 2e–  Cu(s)
Anode: 2H2O(l)  O2(g) + 4H+(aq) + 4e–
 O2(g) is produced
n(e–)
b
c
Q
.
F
=
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Worked solutions to textbook questions
d
Step 1
Calculate the quantity of charge, using Q = It.
Q = 10.0 A  (20  60) s
= 12 000 C
Step 2
Calculate the amount in mol of electrons, using n(e–) =
Step 4
Step 5
Q
.
F
12 000 C
96 500 C mol 1
= 0.1244 mol
Use stoichiometry to calculate the amount of Cu2+ deposited as Cu.
1
n(Cu 2 )
=

2
n (e )
0.1244
n(Cu2+)
=
2
= 0.0622 mol
Calculate the amount of Cu2+ ions remaining in solution.
n(Cu2+)remaining = n(Cu)initial – n(Cu)discharged
= (0.250  1.00) mol – 0.0622 mol
= 0.1878 mol
Calculate the concentration of Cu2+ ions remaining in solution.
0.1878 mol
[Cu2+remaining] =
0.250 L
= 0.7512 M
= 0.75 M
n(e–)
Step 3
16
=
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Worked solutions to textbook questions
17
Q19.
A number of important metals are produced by electrolysis of molten salts. In one
such process a current of 25 000 A produces 272 kg of metal in 24 hours. If the cation
in the salt has a 2+ charge, what is the metal?
A19.
Step 1
Step 2
Step 3
Step 4
Step 5
Step 6
Calculate the quantity of charge.
Q
= It
= 25 000  24  60  60
= 2.16  109 C
Calculate the amount in mol of electrons.
Q
n(e–) =
F
2.16 109
=
96 500
= 2.24  104 mol
Write a balanced equation.
M2+ + 2e–  M
i.e. mole ratio M2+ : e– = 1 : 2
Calculate the number of moles of M.
1
n(M) = n(e–)
2
1
=  2.24  104 mol
2
= 1.12  104 mol
Mass M produced = 2.72  105 g
m
Using n =
M
M
=
m
n
2.72 105
1.12 104
= 24.3 g mol–1
Inspection of a periodic table shows that this divalent metal must be
magnesium.
=
Step 7
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18
Q20.
Two electrolytic cells are connected in series as shown in Figure 28.25. A current of
5.00 A flows for 15.0 minutes.
Figure 28.25
Electrolytic cells in series.
a
b
c
d
e
Calculate the charge flowing through each cell.
Write half equations for the reactions occurring at each electrode (A–D) when the
current commences to flow.
Calculate the change in mass of electrode C after 15 minutes.
Calculate the volume of gas, measured at SLC, that is formed at electrode B after
15.0 minutes.
As the current flows through the cells, how does the reaction at electrode D
change?
A20.
a
b
Q = It
Q = 5.00 A  (15  60) s
= 4500 C
A: 2H2O(l)  O2(g) + 4H+(aq) + 4e–
B: 2H2O(l) + 2e–  H2(g) + 2OH–(aq)
C: Cu(s)  Cu2+(aq) + 2e–
D: 2H2O(l) + 2e–  H2(g) + 2OH–(aq)
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Worked solutions to textbook questions
c
Step 1
Write the half equation.
Cu(s)  Cu2+(aq) + 2e–
Step 2
Calculate the amount in mol of electrons, using n(e–) =
e
Q
.
F
4500 C
96 500 C mol 1
= 0.0466 mol
Step 3 Use stoichiometry to calculate the amount of Cu consumed.
n (Cu )
1
=

2
n (e )
0.0466
n(Cu) =
2
= 0.0233 mol
Step 4 Calculate the mass of Cu consumed.
m(Cu) = 0.0233 mol  63.5 g mol–1
= 1.48 g
Step 1 Write the half equation.
2H2O(l) + 2e–  H2(g) + 2OH–(aq)
Q
Step 2 Calculate the amount in mol of electrons, using n(e–) =
.
F
4500 C
n(e–)
=
96 500 C mol 1
= 0.0466 mol
Step 3 Use stoichiometry to calculate the amount of H2 produced.
n( H 2 )
1
=

2
n (e )
0.0466
n(H2) =
2
= 0.0233 mol
Step 4 Calculate the volume of H2 gas produced at SLC.
V(H2) = 0.0233 mol  24.5 L mol–1
= 0.570997 L
= 0.571 L
2+
–
Cu (aq) + 2e  Cu(s)
n(e–)
d
19
=
Q21.
For electrolytic and galvanic cells, compare:
a the polarity of the anode and cathode
b the direction of electron flow
c the energy transformation occurring in the cells
d the tendency of the cell reaction to occur spontaneously
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A21.
a
b
c
d
In galvanic cells the anode is negative and the cathode is positive; in electrolytic
cells the anode is positive and the cathode is negative.
In galvanic cells the direction of electron flow is determined by the cell reaction;
in electrolytic cells the direction of electron flow is determined by the external
power supply.
In galvanic cells chemical energy is converted into electrical energy; in
electrolytic cells electrical energy is converted into chemical energy.
Galvanic cell reactions occur spontaneously; electrolytic cell reactions are nonspontaneous.
Q22.
Sodium and chlorine are manufactured by passing direct current through molten
sodium chloride.
a Why would alternating current be unsuitable?
b Why is it necessary for the electrolyte to be molten?
A22.
a
b
If electrodes were connected to a source of alternating current, the polarity of the
electrodes would change (typically 50 times per second), with the result that little
reaction was observed at either electrode.
The electrolyte needs to be molten, rather than solid, because ions must be able to
move to create a current and allow reactions to occur at the electrodes.
Q23.
Using the electrochemical series, complete the table by predicting the initial products
of each of the following electrolysis experiments.
Experiment
Electrolyte
Electrodes
a
b
molten potassium iodide
copper(II) sulfate
solution
potassium bromide
solution
a mixture of copper(II)
nitrate and nickel(II)
nitrate solutions
lithium fluoride
platinum
platinum
c
d
e
Cathode
reaction
Anode
reaction
copper
carbon
platinum
A23.
Cathode reaction
a K+(l) + e–  K(l)
b Cu2+(aq) + 2e– Cu(s)
c 2H2O(l) + 2e–  H2(g) + 2OH–(aq)
d Cu2+(aq) + 2e– Cu(s) 

–
–
e 2H2O(l) + 2e  H2(g) + 2OH (aq)
Anode reaction
2I (l)  I2(g) + 2e–
2H2O(l) O2(g) + 4H+(aq) + 4e–
Cu(s) Cu2+(aq) + 2e–
2H2O(l) O2(g) + 4H+(aq) + 4e–
2H2O(l) O2(g) + 4H+(aq) + 4e–
–
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Q24.
Early attempts to produce aluminium by electrolysis of aqueous solutions of
aluminium compounds were unsuccessful. Use the electrochemical series to explain
why.
A24.
The electrochemical series shows that the oxidising strength of H2O is greater than
that of Al3+. If water is present in an electrolysis cell, it therefore reacts preferentially
at the cathode and electrolysis of aqueous aluminium salts does not yield aluminium
metal.
Q25.
A student attempts to electroplate magnesium using a cell made of a magnesium
anode, iron cathode and magnesium nitrate solution as electrolyte. The student is
dismayed to discover that no magnesium has plated onto the cathode, despite careful
checks of all the electrical connections in the electrolytic cell.
a Why didn’t magnesium plate onto the cathode?
b What reaction occurred at the iron cathode?
A25.
a
b
The electrolyte solution contains water. As water is a stronger oxidant than Mg2+,
water will be reduced in preference to the Mg2+.
2H2O(l) + 2e–  H2(g) + 2OH–(aq)
Q26.
Sodium and chlorine are produced by electrolysis of molten sodium chloride in the
Downs cell. A typical Downs cell runs at 600°C, 7 V and with a current of 30 kA. In a
24-hour period:
a what mass of sodium metal will be produced?
b what volume of chlorine gas (at SLC) will be produced?
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A26.
a
Step 1
Step 2
Step 3
Write half equations.
Na+(l) + e–  Na(l)
2Cl–(l)  Cl2(g) + 2e–
Calculate the quantity of charge, using Q = It.
Q = (30  103) A  (24  60  60) s
= 2.593  109 C
Calculate the amount in mol of electrons, using n(e–) =
Q
.
F
2.593  10 9 C
n(e )
=
96 500 C mol 1
= 26 860.1 mol
Use stoichiometry to calculate the amount of Na.
n ( Na )
1
=

1
n (e )
n(Na) = 26 860.1 mol
Calculate the mass of Na.
m(Na) = 26 860.1 mol  23 g mol–1
= 618 kg
= 6.2  102 g
Use stoichiometry to calculate the amount of Cl2.
n(Cl 2 )
1
=

2
n (e )
26 860.1 mol
n(Cl2) =
2
= 13 430.05 mol
Calculate the volume of Cl2 at SLC.
V(Cl2) = 13 430.05 mol  24.5 L mol–1
= 329 036.225 L
= 3.3  105 L
–
Step 4
Step 5
b
Step 1
Step 2
Q27.
Account for the fact that:
a most chlorine is produced by electrolysis of sodium chloride solution, rather than
by electrolysis of the molten salt
b fluorine was first isolated from fluorine compounds by electrolysis
c the concentration of Sn2+ ions in a tin-plating cell (Figure 28.3) remains constant
as the cell operates
d although calcium chloride is present in the electrolyte in the Downs cell, calcium
metal is not formed at the cathode
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A27.
a
b
c
d
It is cheaper to obtain chlorine by electrolysis of concentrated sodium chloride
solution than from molten sodium chloride because electrical energy is required
in order to melt sodium chloride.
Fluorine is the strongest oxidant known. Since a stronger oxidant than fluorine
would be required to convert F– ions into fluorine, the element cannot be made by
direct reaction. However, it is generated at the anode during electrolysis of
molten metal fluorides.
As Sn2+ ions are reduced to tin metal at the cathode, the tin anode of the cell is
oxidised to form Sn2+ ions. Since one electrode reaction is the opposite of the
other, there is no overall change in the concentration of the electrolyte.
Na+ ions are stronger oxidants than Ca2+ ions, so sodium metal is formed at the
cathode in preference to calcium.
Q28.
Magnesium and zinc are both extracted from their compounds by electrolysis. Sketch
and label cells that you consider would be suitable for commercial extraction of these
metals.
A28.
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Q29.
An electrolytic cell for the extraction of aluminium operates at a current of 150 000 A.
In order to produce 1.00 tonne (106 g), calculate:
a how long the cell must operate
b the mass of carbon consumed at the anodes
c the volume of carbon dioxide gas produced, measured at STP
A29.
a
Step 1
Step 2
Step 3
Step 4
Step 5
Write balanced half equations.
Cathode: Al3+(in cryolite) + 3e–  Al(l)
Anode: C(s) + 2O2–(in cryolite)  CO2(g) + 4e–
m
Calculate the amount of Al, using n =
.
M
1.00  10 6 g
n(Al)
=
26.98 g mol 1
= 3.706  104 mol
Use stoichiometry to calculate the amount in mol of electrons.
3
n (e  )
=
1
n(Al)
–
n(e )
= 3  3.706  104 mol
= 1.112  105 mol
Q
Calculate the number of coulombs, using n =
.
F
Q= 1.112  105 mol  96 500 C mol–1
= 1.073  1010 C
Calculate the time for which current flowed, using Q = It.
1073  1010 C
t =
150 000 A
= 71 530 s
= 19.85 hours
= 19 h 51 min
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b
Step 1
Step 2
Use stoichiometry to calculate the amount in mol of electrons.
3
n (e  )
=
1
n(Al)
n(e–) = 3  3.706  104 mol
= 1.112  105 mol
Use stoichiometry to calculate the amount of C.
n (C)
1
=

4
n (e )
n(C)
Step 3
c
Step 1
Step 2
25
1.112  10 5
4
= 2.78  104 mol
=
m
.
M
m(C)
= (2.78  104 ) mol  12 g mol–1
= 3.334  105 g
= 333 kg
Use stoichiometry to calculate the amount of CO2.
n(CO 2 ) 1
=
1
n ( C)
n(CO2) = 2.78  104 mol
V
Calculate the volume of CO2, using n =
.
Vm
V(CO2) = 2.78  104 mol  22.4 mol L–1
= 6.23  105 L
Calculate the mass of C, using n =
Q30.
Commercial operation of an aluminium smelter depends upon the availability of
cheap electric power.
a Calculate the electric charge required to produce 1.00 tonne of aluminium from
alumina (Al2O3) by electrolysis.
b If the cell operates at 5.00 V, calculate the energy needed to make 1.00 tonne of
aluminium. (1 volt = 1 joule per coulomb)
c Calculate the cost of the electricity required to make 1.00 tonne of aluminium if
the smelter purchased electricity at the domestic rate of 12 cents per kilowatt
hour. (1 kilowatt hour = 3 600 000 J)
d Apart from the cost of electricity, name three other important costs involved in
the production of aluminium.
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A30.
a
Step 1
Write a balanced equation.
Al3+(in cryolite) + 3e–  Al(l)
Step 2
Calculate the amount of Al, using n =
1.00  10 6 g
26.98 g mol 1
= 37 064 mol
Step 3 Use stoichiometry to calculate the amount in mol of electrons.
3
n (e  )
=
1
n(Al)
–
n(e ) = 3  37 064 mol
= 111 193 mol
Q
Step 4 Calculate the electric charge required, using n(e–) = .
F
Q = 111 193 mol  96 500 C mol–1
= 1.073  1010 C
= 1.07  1010 C
Energy (J) = V  C
= 5.00 V  (1.07  1010) C
= 5.36  1010 J
Step 1 Calculate the number of kilowatt hours required.
5.36  1010 J
Number of kWh =
3 600 000 J
= 1.489  104 kWh
Step 2 Calculate the cost of electricity.
Cost
= 1.489  104  $0.12
= $1787
= $1780
Three other important costs in the production of aluminium are the capital cost of
the plant, the cost of producing the alumina and the cost of manufacturing the
carbon anodes.
n(Al)
b
c
d
m
.
M
=
Q31.
Chemists are investigating the possibility of replacing the carbon anodes in the
electrolytic cell for aluminium extraction with anodes made of an unreactive material.
a Write an equation for the anode reaction using unreactive electrodes.
b What would be the advantages of using unreactive anodes?
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A31.
a
b
2O2– (in cryolite)  O2(g) + 4e–
Unreactive electrodes would not be consumed in the anode reaction; frequent
replacement of the anodes would thus be avoided, reducing labour and materials.
The manufacture of carbon anodes is a major activity in an aluminium smelter
and its elimination could substantially reduce the cost of aluminium production.
Emissions of carbon dioxide gas from the smelter would also be reduced since
oxygen would be produced at the anode rather than carbon dioxide. However,
higher potentials would be involved in such cells and the energy consumption
would be greater.
Q32.
The Portland aluminium smelter in Victoria produces about 1000 tonnes of aluminium
a day. What volume of carbon dioxide gas (measured at SLC) is produced by the
smelter each day?
A32.
Step 1
Step 2
Step 3
Step 4
Write a balanced equation.
2Al2O3(in cryolite) + 3C(s)  4Al(l) + 3CO2(g)
m
Calculate the amount of Al, using n =
.
M
1.00  10 6 g
n(Al) =
26.98 g mol 1
= 37 064 mol
Use stoichiometry to calculate the amount of CO2.
n(CO 2 )
3
=
4
n(Al)
3  37 064
n(CO2)
=
mol
4
= 27 798 mol
Calculate the volume of CO2 at SLC.
V(CO2) = 27 798 mol  24.5 L mol–1
= 681 057.125 L
= 6.811  105 L
Q33.
Impure copper, called ‘blister copper’ is produced by the reduction of copper ore in a
blast furnace. Blister copper can be purified by the electrolytic process called
electrorefining. Use the description of this process in the text to help you answer the
following questions.
a To which electrode is the blister copper attached?
b What is the nature of the other electrode?
c What is the composition of the electrolyte?
d Write the half equations for the reactions involving copper that occur at the two
electrodes.
e What happens to impurities in the blister copper that are:
i more reactive than copper, such as zinc?
ii less reactive than copper, such as gold?
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A33.
a
b
c
d
e
anode
a thin sheet of pure copper
sulfuric acid
anode Cu(s) Cu2+(aq) + 2e–; cathode Cu2+(aq) + 2e– Cu(s)
i More reactive impurities are oxidised and remain in solution as ions.
ii Less reactive impurities fall to the bottom of the cell and are collected.
Q34.
Construct a concept map that includes the terms: electrolysis, electrolytic cell,
chemical energy, electrical energy, anode, cathode, reductant, oxidant, reduction,
oxidation, electrolyte and non-spontaneous reactions.
A34.
Q35.
Under the title ‘Using electrical energy to produce a useful chemical’, write one or
two paragraphs that explain to a Year 11 Chemistry student how a chemical of your
choice can be produced industrially by electrolysis.
A35.
You might choose to select one of the following chemicals: sodium, chlorine, sodium
hydroxide, aluminium or copper. Refer to the appropriate descriptions of each
electrolysis cell in the text to answer this question.
Q36.
When constructing a galvanic cell in the laboratory, why are two half cells usually used
whereas the reactants of an electrolysis cell are often placed in a single container?
A36.
The reaction in a galvanic cell is spontaneous and if the reactants in the cell were in
the one container and in contact with each other, the reaction between them could
occur directly, releasing energy as heat rather than as electricity.
In electrolysis cells, the reaction is non-spontaneous so that both electrode reactions
can occur within the same container. The products of the electrolysis reaction should
not be allowed to come into contact with each other, however, or a reaction may
occur.
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Unit 4 Area of Study 2 review
Multiple-choice questions
Q1.
Which one of the following lists includes only renewable energy sources?
A ethanol, coal, natural gas
B animal waste, the sun, crude oil
C uranium, ‘hot rocks’, hydrogen
D hot underground springs, biochemical fuels, wind
A1.
D. Non-renewable energy sources are being used up at a faster rate than they can be
replaced and include fossil fuels and uranium. Renewable energy sources are
continually being replaced by natural processes, e.g. wind, hot underground springs,
and photosynthesis in plants from which biochemical fuels such as ethanol (made
from sugar cane) or bio diesel (from vegetable oils) are derived.
Q2.
Two gaseous fuels that have been investigated as possible alternative energy sources
are hydrogen gas, H2, and biogas, which consists mostly of methane, CH4. The table
shows the energy released by the combustion of 1 mol of hydrogen and methane.
Fuel
Hydrogen
Methane
Equation for combustion reaction
Energy released per mole
of fuel (kJ mol–1)
O2(g)  H2O(l)
286
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
890
H2(g) +
1
2
Consider the following situations that may apply to the use of the fuels.
I Limiting greenhouse gas emissions is very important.
II Equal masses of the two gases are to be stored.
III Equal volumes of the gases are to be stored (at the same temperature and
pressure).
Hydrogen gas would be the preferred fuel in situations:
A I, II and III.
B I and II.
C II and III.
D I only.
A2.
B. Only methane releases the greenhouse gas CO2, increased levels of which in the
atmosphere are thought to cause global warming. 2 g of hydrogen releases 286 kJ,
therefore 1 g releases 143 kJ of energy. However, 16 g of methane releases 890 kJ,
therefore 1 g releases 56 kJ. Therefore, hydrogen releases more energy per unit mass
than methane; equal volumes of the two gases contain equal amounts in mol and, on
this basis, methane releases more energy than hydrogen.
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Q3.
Some eminent politicians and scientists have recently proposed that we should
consider utilising our reserves of uranium to generate electricity for our cities, instead
of the current situation where most electricity is supplied by coal-fired power stations.
An argument in favour of greater reliance on nuclear power, rather than coal, could be
based on:
A whether the energy source is renewable or non-renewable
B the relative amounts of greenhouse gases produced by the power stations
C the safe disposal of waste materials from the power plant
D the number of energy conversions involved between the energy source and the
electrical energy that is generated
A3.
B. Burning coal releases carbon dioxide gas into the atmosphere, increasing levels of
which are thought to be causing global warming. Sulfur dioxide can also be released
when coal is burned. Nuclear power stations don’t generate any air pollutants.
However the storage or radioactive waste is a contentious issue. Both fuels are nonrenewable and a similar number of energy conversions are involved in the two types
of power station.
The following information relates to Questions 4 and 5.
As a response to rising petrol prices the Australian Government in 2006 offered car
owners a $2000 incentive to convert their cars from petrol to LPG. Propane is one of
the gases in the mixture sold as LPG. The thermochemical equation for the complete
combustion of propane is:
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l); ΔH = –2220 kJ mol–1
Q4.
Which energy profile diagram best represents the energy changes that take place
during this reaction?
A
B
C
D
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A4.
C. As this is an exothermic reaction the energy of the reactants will be less than
energy of the products. There is 2220 kJ mol–1 difference between energy of the
reactants and products. Bond breaking always requires an input of energy, i.e. energy
is needed to break the bonds. Energy is always released when new bonds are formed.
This exothermic reaction involves a net release of energy, so more energy is released
during bond formation than was used up to break bonds of the reactants.
Q5.
When 100 g of propane undergoes complete combustion:
A 5.04  103 kJ of energy is absorbed
B 5.04  103 kJ of energy is released
C 3.92  101 kJ of energy is absorbed
D 3.92  101 kJ of energy is released
A5.
B. This exothermic reaction (with a negative ΔH value) involves a net release of
energy.
Step 1 The amount of propane that reacted is calculated.
100
mass
n(C3H8) =
=
= 2.26 mol
M
44.11
Step 2 The amount of energy released by 100 g of propane is calculated.
= 2.26  2220 = 5.024  103 kJ
Q6.
Which one of the following sources of power provides the highest quantity of energy
per gram of fuel?
A coal
B natural gas
C nuclear fission
D biochemical fuels
A6.
C. 235U has a significantly energy per kilogram (energy density) than fossil or
biochemical fuels. The energy density of 235U is 9 × 107 MJ kg–1, compressed natural
gas 54 MJ kg–1, coal 24 MJ kg–1 and a biochemical fuel such as ethanol 23 MJ kg–1.
Q7.
A temperature rise of 2.0°C occurred when 1.50 × 10–3 mol of ethane gas burnt in a
bomb calorimeter.
2C2H6(g) + 7O2(g) —> 4CO2(g) + 6H2O(l); H = –3120 kJ mol–1
The calibration constant of the calorimeter, in J oC–1, is
A 1170
B 3120
C 2340
D 4680
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A7.
A. 2 mole C2H6 generates 3120 kJ of energy.
1 mole generates 1560 kJ.
Energy produced by 1.50 × 10–3 mol = 1.50 × 10–3 × 1560 × 103 = 2340 J
Calibration factor = energy/T = 2340/2 = 1170 J oC–1
Q8.
The following equations involve ions of the transition metal vanadium. The half cell
potential E0(V) is shown for each equation.
E0(V)
VO2+(aq) + 2H+(aq) + e– VO2+(aq) + H2O(l)
+1.00
VO2+(aq) + 2H+(aq) + e–  V3+(aq) + H2O(l)
+0.36
3+
–
2+
V (aq) + e  V (aq) –0.25
Of the ions listed, the strongest oxidant and the strongest reductant are:
Strongest oxidant
A
B
C
D
VO2+
VO2+
V3+
V2+
Strongest
Reductant
V2+
VO2+
VO2+
VO2+
A8.
A. An oxidant gains electrons and is shown on the left side of the half equation in the
electrochemical series. The strongest oxidant is on the left in the half-cell reaction that
has the highest E0. A reductant loses electrons and is shown on the right side of the
half equation. The conjugate redox pair containing the strongest reductant has the
lowest E0.
Q9.
Use the electrochemical series to determine which one of the following would not be
expected to occur to an appreciable extent.
A 2H+(aq) + Fe(s)  Fe2+(aq) + H2(g)
B 2Ag+(aq) + Ni(s)  2Ag(s) + Ni2+(aq)
C Br2(aq) + 2Fe2+(aq)  2Br–(aq) + 2Fe3+(aq)
D 2I–(aq) + Pb2+(aq)  I2(s) + Pb(s)
A9.
D. A reaction will only occur if the oxidant of the conjugate redox pair with the higher
E0 is added to the reductant of the conjugate redox pair having the lower E0. In D, a
reaction will not occur because the E0 of the half reaction involving the reductant I– is
higher than the E0 of the half reaction involving the oxidant, Pb2+.
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Q10.
Which of the following best describes the features of an anode in a galvanic cell?
polarity electrode reaction
A positive oxidation
B positive reduction
C negative oxidation
D negative reduction
A10.
C. The oxidation reaction at the anode produces electron giving the electrode a
negative polarity.
Q11.
An electrochemical cell was made by dipping a copper rod into a solution of 1 M
CuSO4 in one beaker and dipping a nickel rod into a solution of 1 M NiSO4 in another
beaker. The metals were connected with wire and the two solutions were connected
by a piece of paper towel that had been soaked in a potassium nitrate solution. The
cell is shown in the diagram.
The solution in beaker I was initially coloured blue, owing to the presence of Cu2+
ions. The solution in beaker II was initially coloured green because of the presence of
Ni2+ ions. After the galvanic cell has been discharging for a period of time, it might be
possible to detect the following changes.
A The green colour in beaker II has faded and the mass of the copper electrode has
increased.
B The blue colour in beaker I has faded and the mass of the nickel electrode has
increased.
C The green colour in beaker II has faded and the mass of the copper electrode has
decreased.
D The blue colour in beaker I has faded and the mass of the nickel electrode has
decreased.
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A11.
D. The electrode half reactions are predicted as:
Cu2+(aq) + 2e–  Cu(s)
thus causing this electrode to increase in mass (more Cu formed) and the solution’s
colour to fade (less Cu2+ is present); and
Ni(s)  Ni2+(aq) + 2e–
this causing the electrode to decrease in mass (less Ni) and the solution colour to
deepen (more Ni2+(aq)).
The following information relates to Questions 12 and 13.
A car manufacturer has developed a prototype of an electric car that uses hydrogen
fuel cells to generate electricity.
Q12.
A reaction that might occur at the cathode of a fuel cell using hydrogen and oxygen as
reactants is
A 2H2(g) + O2(g)  2H2O(g)
B O2(g) + 2H2O(g) + 4e–  4OH–(aq)
C H2(g)  2H+(aq) + 2e–
D H2(g) + 2OH–(aq)  2H2O(g) + 2e–
A12.
B. At the cathode, oxygen is reduced to hydroxide ions in the presence of water.
Q13.
Consider the following features of the fuel cell powered car.
I Two half cell reactions occur spontaneously at two separated locations (the
electrodes).
II Oxygen behaves as an oxidant in the reaction that supplies energy.
III Hydrogen gas stores 143 kJ g–1 of fuel.
The fuel cell powered car differs from a conventional petrol-driven car with respect
to:
A I, II and III
B I and II
C I and III
D II and III
A13.
C. A fuel cell uses an electrochemical process, whereas petrol reacts directly with
oxygen in the internal combustion engine. Oxygen oxidises both fuels; petrol has a
different energy density from that of hydrogen, approximately 47 kJ g–1.
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The following information relates to Questions 14 and 15.
The diagram shows an electroplating cell that was devised by a student to place a
silver coating onto a key. An electric current is passed through a solution of 1 M silver
nitrate, AgNO3, using an inert carbon electrode as the anode.
Q14.
The reaction that occurs at the electrode connected to the negative terminal of the
external power supply is:
A Ag+(aq) + e–  Ag(s)
B 2H2O(l) + 2e–  H2(g) + 2OH–(aq)
C 2Cl–(aq)  Cl2(g) + 2e–
D 2H2O(l)  O2(g) + 4H+(aq) + 4e–
A14.
A. Electrons are supplied to the negative electrode to allow reduction (electron gain)
to occur; Ag+ is a stronger oxidant than H2O so it is reduced.
Q15.
If 0.055 g silver is deposited on the surface of the key when a current is passed
through the cell for 7.0 minutes, what current is required?
A 0.12 A
B 0.75 A
C 7.0 A
D 49 A
A15.
A. Cathode reaction: Ag+(aq) + e–  Ag(s)
m
n(Ag) =
M
0.055
=
= 5.1  10–4 mol
107.87
Since n(e–) = n(Ag+), 5.1  10–4 F of charge is required
Q = 5.1  10–4  96 500 = 49 C
Q
I =
t
49 C
=
= 0.12 A
7.0  60 s
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Q16.
Three beakers contain solutions of 1.0 M chromium(III) nitrate, 1.0 M copper(II)
nitrate and 1.0 M silver nitrate. Each solution has 0.60 F of electric charge passed
through it, using carbon electrodes and a power pack. In each beaker a different metal
is deposited onto the cathode. In decreasing order, the amount of metal (in moles)
deposited onto the cathode in each beaker will be:
A Cr > Cu > Ag
B Ag > Cu > Cr
C Cr > Ag > Cu
D the same in all three beakers
A16.
B. 0.60 F is carried by 0.60 mol of electrons.
Cr3+ + 3e–  Cr; therefore, 0.20 mol Cr is produced
Cu2+ + 2e–  Cu; therefore, 0.30 mol Cu is produced
Ag+ + e–  Ag; therefore 0.60 mol Ag is produced
Q17.
A series of experiments is described below.
I Zinc granules are added to a solution of tin(II) chloride
II Copper turnings are added to tin(II) chloride solution
III Electrolysis of molten tin(II) chloride
IV Electrolysis of an aqueous solution of tin(II) chloride
The experiments that would result in the production of metallic tin, Sn, are:
A all of them
B I, III and IV
C I and II
D only II
A17.
B. Tin metal can be produced by a direct reaction with a stronger reductant than Sn,
such as Zn (but not Cu), or by electrolysis of an aqueous solution of a tin(II) salt or
the molten salt. In all cases the reduction involves Sn2+ + 2e–  Sn.
Short-answer questions
Q18.
a
During the 20th century and continuing to the present day, atmospheric carbon
dioxide concentrations have increased significantly due mainly to the burning of
fossil fuels.
i Describe two undesirable atmospheric pollutants, other than carbon dioxide
emissions, produced by the burning of fossil fuels.
ii What is thought to be the environmental impact of increased levels of
atmospheric carbon dioxide?
iii Identify one fossil fuel and briefly discuss the sustainability of its continued
use into the future. Your response should include a statement about the extent
of the fossil fuel’s reserves and an outline one advantage and one
disadvantage of its continued use.
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b
c
d
37
The main use of coal is in the generation of electricity. Electricity can also be
generated using non-fossil fuel alternatives. Name one of theses alternatives and
identify one advantage (other than less CO2 emission) and one disadvantage
compared to burning fossil fuels for electricity generation.
Some countries use nuclear fission instead of fossil fuels for electricity
production.
i Explain what the term ‘nuclear fission’ means.
ii What energy transformation occurs when 235U undergoes fission?
There has been an increased awareness of the use of biochemical fuels as an
alternative to fossil fuels.
i How does a biochemical fuel differ from a fossil fuel?
ii Give two examples of biochemical fuels and describe briefly how they are
produced.
iii Why are biochemical fuels considered by some to be carbon dioxide neutral?
A18.
a
b
c
d
i
Carbon monoxide, carbon (soot), oxides of sulfur and nitrogen, sulfuric acid,
ozone
ii Increased levels of atmospheric carbon dioxide have been linked to global
warming
iii Estimated global fossil fuel reserves and their expected lifetimes are given in
Table 23.3.
Refer to Chapter 24 (Table 24.3 Some advantages and disadvantages of energy
sources available in Australia).
i Nuclear fission is the splitting of the nucleus of an atom
ii Nuclear binding energy or nuclear mass  heat and kinetic energy
Fossil fuels are derived from the remains of plants and animals that died millions
of years ago. Biochemical fuels are manufactured from living plant materials, e.g.
ethanol from sugar cane, biodiesel from vegetable oils, biogas from the decay of
plant and animal matter.
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Q19.
The heat of combustion of brown coal was measured in a bomb calorimeter that had
been calibrated by passing a current of 3.00 amperes at 5.00 volts through the
electrical heater for 2.00 minutes. The temperature rose by 1.1°C.
When a 0.50 g sample of freshly crushed brown coal was completed burned in the
calorimeter, the temperature rose by 3.0°C.
a Calculate the value of the calorimeter constant.
b Determine the heat of combustion of brown coal in kJ g–1
c Another sample of brown coal was heated at 100°C for two hours. A sample,
weighing 0.5 g, was then burned completely in the calorimeter. Would this
sample release more or less energy per gram than the freshly crushed sample?
Explain your answer.
d A brown coal fired power station produces 1.7 tonnes of carbon dioxide gas for
ever megawatt of electricity generated. A black coal fired power station releases
0.9 tonnes of carbon dioxide into the atmosphere per megawatt of electricity.
Explain the difference in the amount of carbon dioxide released to produce the
same amount of electricity. Assume that both types of power stations operate at
the same level of efficiency.
A19.
a
b
c
d
Energy supplied during calibration = VIt
= 5.00 V  3.00 A  (2.00  60) s
= 1800 J
Since this caused the temperature to rise by 1.1C, the energy needed for a 1C
1800
temperature change =
= 1636 J °C–1
1.1
When the coal burned, the energy released = calibration constant  temperature
change
= 1636J °C–1  3.0°C
= 4909.1 J
= 4.91 kJ
0.500 g of coal supplies 4.91 kJ of energy.
1.00
Therefore, 1 g of coal will give 4.91 
= 9.82 kJ
0..50
i.e. the heat of combustion is 9.82 kJ g–1.
Brown coal contains between 30% and 70% water. Heating the coal prior to
combustion in the calorimeter drives off some of this moisture. Since the mass of
the pre-dried and dried samples combusted are the same the dried sample will
contain a greater proportion of carbon and will produce more energy per gram.
Carbon dioxide is a combustion product of both black and brown coal. The
energy content of brown coal is about a third of the energy content of the black
coal used in power stations. Black coal has a much higher carbon content than
brown coal and its water content is significantly less. To produce one megawatt
of electricity less black coal needs to be burned and consequently less carbon
dioxide will be released into the atmosphere.
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Q20.
One of the boilers at a power station burns 75.6 kg of brown coal every second. Each
kilogram of brown coal produces 9.82 MJ of energy. The electrical output of the
generator connected to this boiler is 300 MJ s–1.
a Calculate the amount of energy produced by burning 75.6 kg of coal each second.
b Give two reasons why the answer to part a is much greater than the output of the
generator.
c Over three-quarters of the electricity generated in Victoria comes from brown
coal, a non-renewable energy source. Explain what is meant by the term nonrenewable energy source.
d Describe briefly how one renewable energy source can be used to generate
electricity.
A20.
a
b
c
d
1 kg coal supplies 9.82 MJ of energy. Therefore, 75.6 kg of coal will supply
9.82  75.6 = 742 MJ of energy, i.e. 742 MJ s–1.
Coal-fired power stations are only about 40% efficient at converting coal’s
chemical energy into electrical energy. Heat is lost from the chimney stacks at
power station with the waste gases and from the cooling towers in the power
plant.
A non-renewable energy source is one that is being used up at a faster rate than it
can be replaced.
Wind power: kinetic energy of moving air is converted to mechanical energy of
spinning turbines that are connected to electricity generators.
Q21.
When a Ni–Cad cell is converting chemical energy to electrical energy (discharge),
the electrode reactions are best described as follows:
positive electrode: NiOOH(s) + H2O(l) + e–  Ni(OH)2(s) + OH–(aq)
negative electrode: Cd(s) + 2OH–(aq)  Cd(OH)2(s) + 2e–
a Which metal, nickel or cadmium, is at the anode of this cell as it is discharging?
b Give the formula of a salt that might be expected to be found in the electrolyte
paste of a Ni–Cad cell.
Ni–Cad cells are superior to alkaline cells as they can be recharged.
c Explain why cells such as the Ni–Cad can be recharged.
d Write equations for the half reactions and the overall cell reaction occurring when
a Ni–Cad cell is recharged.
e Describe the energy transformation that occurs when a Ni–Cad cell is recharged.
A21.
a
b
c
d
e
Cadmium, Cd, since this is the electrode at which oxidation takes place.
KOH (or NaOH)
These cells can be recharged because the products of the discharge half reactions
remain in contact with the electrodes in a convertible form.
Positive electrode: Ni(OH)2(s) + OH–(aq)  NiOOH(s) + H2O(l) + e–
Negative electrode: Cd(OH)2(s)+ 2e–  Cd(s) + 2OH–(aq)
Overall: 2 Ni(OH)2(s) + Cd(OH)2(s)  2NiOOH(s) + 2H2O(l) + Cd(s)
electrical energy  chemical energy
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Q22.
Zinc–air cells have been developed for a range of different uses, including electric
vehicles, heart pacemakers and laptop computers. They offer a favourable alternative
to other batteries because of their non-toxic contents and high energy density. In the
cells, zinc metal is oxidised at the negative electrode and oxygen gas is reduced to
water at the positive electrode. A zinc–air cell used for a pacemaker operates with a
steady current of 3.5  10–5 A.
a Write a half equation for the half reaction at the negative electrode.
b If the cell will continue operating until 1.5 g of zinc in the negative electrode has
been used up, determine how long the pacemaker will last before it needs
replacing.
A22.
a
b
Zn(s)  Zn2+(aq) + 2e–
m
1 .5
n(Zn) =
=
= 0.023 mol
M
65.38
Therefore, n(e–) released from anode = 2  0.023 = 0.046 mol
Q = n(e–)  F = 0.046  96 500 = 4.4  103 C
Q
4400 C
t =
=
0.000035 A
I
= 1.3  108 s
= 4.0 years
Q23.
An ‘alcometer’ is a small, hand-held device that can be used to give reliable readings
of blood alcohol levels when a person blows air from their lungs into the instrument.
It relies on the equilibrium established between ethanol dissolved in the blood and
gaseous ethanol in the lungs:
C2H5OH(aq) (in blood)  C2H5OH(g) (in lungs)
which has an equilibrium constant, K = 4.35 10–4 at 37C.
The alcometer is a fuel cell and air from the lungs is directed onto the anode of the
cell where ethanol, C2H5OH, is oxidised to carbon dioxide, CO2. It consists of a
porous glass plate containing a phosphoric acid electrolyte and coated with layers of
platinum metal on the top and bottom. The fuel cell’s voltage depends on the
concentration of ethanol in the air blown onto the anode, so the instrument can be
calibrated to switch on a red light if blood alcohol concentration is greater than
0.05%, or a green light for zero blood alcohol concentration.
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The net reaction that occurs in this fuel cell is given by the equation:
C2H5OH(g) + 3O2(g)  2CO2(g) + 3H2O(l)
a By assigning oxidation numbers to carbon in both ethanol and carbon dioxide,
verify that ethanol is oxidised in the fuel cell reaction.
b The half reaction that occurs at the cathode is:
O2(g) + 4H+(aq) + 4e–  2H2O(l)
What is the equation for the half reaction that occurs at the anode in the
alcometer?
c On the diagram, place an arrow to show the direction of electron movement
through the external circuit.
d Give two different functions of the layers of platinum metal in the fuel cell.
A23.
a
b
Ethanol: the CH3 carbon is in the –3 oxidation state and the CH2OH carbon is in –1
oxidation state (attributing zero effect to the C–C bond)
Carbon dioxide: carbon is in the +4 oxidation state. Carbon is oxidised.
C2H5OH(g) + 3H2O(l)  2CO2(g) + 12H+(aq) + 12e–
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c
d
They are the sites of the oxidation and reduction half reactions; they behave as
catalysts to increase the rate of the electrode half reactions.
Q24.
An experiment was carried out as follows to compare the effectiveness of two
different methods of generating energy from liquid methanol (CH3OH) as a source of
fuel.
a The heat of combustion of methanol was determined in bomb calorimeter which
had a calibration factor of 8.26 kJ °C–1. When 2.0 g of methanol was completely
burned in excess oxygen the temperature rose from 20.0°C to 25.5°C.
i How much energy is released when 2.0 g of methanol is burned in excess
oxygen?
ii How much energy is released when 1.0 mole of methanol is burned in excess
oxygen?
iii What is the value of H for the reaction:
2CH3OH(l) + 3O2(g)  CO2(g) + 4H2O(l)
b The methanol fuel cell generates electricity by consuming methanol and oxygen.
The equations for the reactions occurring at the electrodes are:
CH3OH (l) + H2O(l)  CO2(g) + 6H+(aq) + 6e– Equation 1
O2(g) + 4H+(aq) + 4e–  2H2O(l)
Equation 2
i Which equation represents the reaction occurring at the anode?
ii Write an equation that represents the overall cell reaction.
iii Calculate the electric current generated by the methanol fuel cell when 0.64 g
of methanol is consumed in 10 minutes.
iv How much energy in kJ is delivered per mole of methanol from a fuel cell
that generates 1.26 V?
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43
Methanol can also be used as a fuel for an internal combustion engine driving an
electrical generator.
i Give one reason why the fuel cell would produce electricity more efficiently
than a generator driven by an internal combustion engine.
ii Give one reason why fuel cells are not used to generate electricity on a large
scale.
A24.
a
b
c
energy = calibration factor × temperature change = 8.26 × 5.5 = 45.43 kJ
2 g produces 45.43kJ.
2
mole = 45.43
32
32
1 mole = 45.43 ×
kJ = 726.88 kJ = 727 kJ (three significant figures)
2
iii The equation as written involves two moles. H = 2 × 727 = 1454 kJ mol–1
i Oxidation occurs at the anode. The oxidation state of carbon increases from
–2 in CH3OH to + 4 in CO2, i.e. carbon has been oxidised. The anode
reaction is:
CH3OH(l) + H2O(l)  CO2(g) + 6H+(aq) + 6e–
ii The half equations can be combined when both have the same number of
electrons. There will be equal numbers of electrons when equation 1 is
multiplied by 2 and equation 2 is multiplied by 3.
2CH3OH(l) + 2H2O(l)  CO2(g) + 12H+(aq) + 12e–
3O2(g) + 12H+(aq) + 12e–  6H2O(l)
2CH3OH (l) + 2H2O(l) + 3O2(g) + 12H+(aq) + 12e–  CO2(g) + 12H+(aq) +
12e– + 6H2O(l)
This simplifies to:
2CH3OH (l) + 3O2(g)  CO2(g) + 4H2O(l)
iii 1 mole CH3OH generates 6 mole of electrons = 6 × 96500 C
0.64
0.64 g CH3OH =
= 0.02 mol.
32
0.02 mol CH3OH = 0.02 × 6 × 96500 = 11580 C
Q 11580
current, I =
=
= 19.3 A
10 x 60
t
iv Since 6 electrons are involved in the half equation, the charge of 6 mol of
electrons is
6 × 96500C. The energy produced when the cell generates 1.26 V is:
1.26 × 6 × 96500 = 730 kJ mol–1
In a fuel cell, the energy conversions are chemical  electrical energy.
In a combustion engine, heat is lost at each step in the energy conversions:
chemical  kinetic  mechanical  electrical.
i
ii
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Q25.
a
Complete the half equations for the reactions predicted when each of the
following aqueous solutions undergo electrolysis, using the electrodes described
in the table. The process was carried out in a U-tube, as shown in the diagram.
Solution
Electrode materials
Positive electrode
KI(aq)
PbCl2(aq)
AlCl3(aq)
b
carbon
carbon
copper
Negative
electrode
carbon
carbon
copper
Half equation for the reaction at
the
Positive
Negative
electrode
electrode
For one of the solutions described in the table, explain how you might verify that
the predicted products were being formed at each electrode.
A25.
a
KI(aq)
PbCl2(aq)
AlCl3(aq)
Positive
2I–(aq)  I2(s) + 2e–
2H2O(l)  O2(g) + 4H+(aq) + 4e–
Cu(s)  Cu2+(aq) + 2e–
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Negative
2H2O(l) + 2e–  H2(g) + 2OH–(aq)
Pb2+(aq) + 2e–  Pb(s)
2H2O(l) + 2e–  H2(g) + 2OH–(aq)
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b
KI(aq)
PbCl2(aq)
AlCl3(aq)
Positive
Look for the yellow–brown
colour of iodine in the solution.
Negative
Use an acid-base indicator to test for
OH–, look for bubbles of H2 gas,
collect and test gas with burning
string ‘pop’ confirms H2 gas.
Use an acid-base indicator to test
for presence of H+(aq), look for
gas bubbles, collect gas and test
with glowing splint, if this
ignites confirms presence of O2.
Blue colour will appear in
solution as Cu electrode is
consumed.
A grey deposit of lead will form on
the electrode.
Use an acid-base indicator to test for
OH–, look for bubbles of H2 gas,
collect and test gas with burning
string ‘pop’ confirms H2 gas.
Q26.
A 100 mL solution containing a mixture of magnesium chloride, MgCl2, and tin(II)
chloride, SnCl2, is prepared by dissolving 0.025 mol of each salt in water. Platinum
electrodes are placed in the solution and a small current is passed through it, as shown
in the diagram.
a
b
c
Write half equations for the electrode reactions occurring just after the
electrolysis is started
i at the anode
ii at the cathode
After electrolysis has been occurring for a considerable period of time all of the
metal that was first plated on the cathode will have been used up and a new
electrode reaction will occur at the cathode. Write a half equation for the next
electrode reaction that occurs at the cathode.
What major difference would occur in the cell if the solution that was used had
been a saturated solution of tin(II) chloride?
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A26.
a
b
c
i
In dilute solutions water, the stronger reductant will be oxidised at the anode
rather than Cl(aq).
2H2O(l)  O2(g) + 4H+(aq) + 4e–
ii Sn2+(aq) is a stronger oxidant than either water or Mg2+(aq) and will be
reduced
Sn2+(aq) + 2e–  Sn(s)
Once all the Sn2+(aq) has been reduced, water, being a stronger oxidant than
Mg2+(aq), will then be reduced.
2H2O(l) + 2e–  H2(g) + 2OH–(aq)
Chlorine gas would be released at the positive electrode (anode) due to the very
high concentration of chloride ions in the solution.
2Cl–(aq)  Cl2(g) + 2e–
Q27.
In an electrolysis experiment, a student is provided with a solution of nickel(II)
nitrate, Ni(NO3)2. The electrodes to be used were a carbon rod, as the positive
electrode, and a metal spatula, as the negative electrode.
a Will the nickel coating appear on the carbon rod or the metal spatula during the
experiment described?
b Write equations for the half reactions that occur at each electrode.
c A current of 2.5 A was passed through this nickel-plating cell for 15 minutes.
Calculate the mass of nickel that could be plated on the cathode.
A27.
a
b
c
On the metal spatula: electrons are supplied to the negative electrode, where
reduction of Ni2+ to Ni will occur.
Spatula (negative): Ni2+(aq) + 2e–  Ni(s)
Carbon rod (positive): 2H2O(l)  O2(g) + 4H+(aq) + 4e–
Q = It
= 2.5 A  (15  60) s = 2250 C
Q
2250
n(e–) delivered to negative electrode =
=
= 0.023 mol
F
96 500
1
n(Ni) formed =
 n(e–) = 0.012 mol
2
m(Ni)
=nM
= 0.012  58.69 = 0.68 g
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Q28.
Lithium metal is prepared by electrolysis of a molten mixture of lithium chloride and
potassium chloride.
a Write a half equation for the reaction occurring at the cathode in this cell.
b Write a half equation for the reaction occurring at the anode in the cell.
c Why is it not possible to produce lithium metal by the electrolysis of an aqueous
solution of lithium chloride?
d Suggest why a mixture of lithium chloride and potassium chloride is used, rather
than pure lithium chloride, in this electrolytic process.
e Assuming that lithium reacts in a similar way to sodium, describe two precautions
that would need to be taken in this preparation.
A28.
a
b
c
d
e
Li+(l) + e–  Li(l)
2Cl–(l)  Cl2(g) + 2e–
Li+ is a weaker oxidant than H2O (Li+/Li has a lower E0); therefore the negative
electrode reaction is:
2H2O(l) + 2e–  H2(g) + 2OH–(aq)
rather than Li+(l) + e–  Li(l).
Addition of KCl lowers the melting temperature of the electrolyte, thus lowering
the energy used in the process and lowering costs.
The products of the half reactions would need to be kept separated from one
another (as in the Downs cell) and water would need to be excluded from the
entire process as it is a more powerful oxidant than Li
Q29.
This question involves both electrochemical cells and electrolytic cells.
a Compare the energy transformation occurring in an electrochemical cell with that
occurring in an electrolytic cell.
b Explain why the negative electrode of an electrochemical cell is the site of
oxidation, whereas the negative electrode of an electrolytic cell is the site of
reduction.
A29.
a
b
Electrochemical cell: chemical energy  electrical energy
Electrolytic cell: electrical energy  chemical energy
In the electrochemical cell, electrons produced at the site of oxidation make that
electrode develop a negative polarity, i.e. the half reaction causes the negative
polarity of the electrode. However, in the electrolytic cell, the negative polarity of
one electrode is imposed on it by the external power source, which ‘forces’
(negatively charged) electrons towards this electrode and allows reduction to
occur there.
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Q30.
Some industrially important chemicals are produced by the electrolysis of a molten
electrolyte while others are produced by electrolysis of an aqueous solution.
a Name one chemical produced by the electrolysis of:
i a molten electrolyte
ii an aqueous solution
b For each chemical that you have selected:
i Write an equation for the reaction occurring at the anode
ii Write an equation for the cathode reaction.
iii Describe one feature of the electrolytic cell used to manufacture of each
selected chemical.
c Explain why some chemicals can be produced by the electrolysis of a molten
electrolyte but not from the electrolysis of an aqueous solution.
A30.
a
b
c
i sodium, aluminium
ii chlorine, sodium hydroxide, copper
Equations depend on chemicals selected in part a (Refer to Chapter 28.)
In the electrolysis of aqueous solutions the possibility that water may be oxidised
at the anode or reduced at the cathode needs to be considered. Water is a stronger
oxidant than either sodium or aluminium and will be reduced at the cathode in
preference to the metal ions. These reactive metals are produced by the
electrolysis of a molten salt.
Q31.
Write equations to represent the anode and cathode reactions that occur when an
electric current is passed through:
a a dilute solution of sodium chloride
b a concentrated solution of sodium chloride
c molten sodium chloride
A31.
a
b
c
Anode. Both the reductants, Cl– and H2O need to be considered. Water is the
stronger reductant and would be expected to oxidised in preference to Cl–(aq).
2H2O(l)  O2(g) + 4H+(aq) + 4e–
Cathode. Water is the stronger oxidant and would be expected to be reduced in
preference to Na+(aq).
2H2O(l) + 2e–  H2(g) + 2OH–(aq)
Anode: The electrochemical series can be used as a guide to predict reactions at
concentrations of 1 M. In the case of a concentrated sodium chloride solution, Cl–
(aq) is oxidised rather than water.
2Cl–(aq)  Cl2(g) + 2e–
Cathode: Water is the stronger oxidant and would be expected to be reduced in
preference to Na+(aq).
2H2O(l) + 2e–  H2(g) + 2OH–(aq)
Anode: Chloride ions are oxidised.
2Cl–(aq)  Cl2(g) + 2e–
Cathode: Sodium ions are reduced.
Na+(l) + e–  Na(l)
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