Download REDOX EQUILIBRIA SL - chemistryatdulwich

Topic 9: REDOX EQUILIBRIA SL
9. 1 Oxidation and reduction
9.1.1
9.1.2
9.1.3
9.1.4
Define oxidation and reduction in terms of electron loss and gain.
Deduce the oxidation number of an element in a compound.
State the names of compounds using oxidation numbers.
Deduce whether an element undergoes oxidation or reduction in reactions using oxidation numbers.
Oxidation and reduction is applied to both ionic and covalent interactions during which electrons are
transferred or lost/gained control over.
OILRIG :
Oxidation Is Loss Reduction Is Gain
 reduction = gain of 1 or more electron; an atom has gained the control of electron(s) from another
atom;
 in a covalent bond, it results in the atom having a higher electron density around it;
 during ionic bond formation, the atom becomes a negative ion;
 oxidation = loss of 1 or more electrons; an atom has lost control over 1 or more of its electrons;
 in a covalent bond, it results in a lower electron density around the atom;
 in ionic bond formation, the atom becomes a positive ion.
Recognising redox reactions:
oxidation = happens when oxidation number of element increases;
reduction = when oxidation number of element decreases;
The oxidation number or the oxidation state of an element in a compound indicates the number of electrons
over which an atom/ion of an element has gained or lost control during a reaction. Oxidation numbers are
used to keep track of how many electrons are lost or gained by each atom or ion. They are used to identify
oxidation and reduction reactions.
Rules for working out oxidation number:
1. The oxidation number of an atom in its element form or uncombined form is always 0.
For example the atoms in Na, He, N2 and S8 have oxidation numbers of 0.
2. The oxidation number of a monatomic ion equals the charge of the ion.
For example, the oxidation number of Na+ is +1; the oxidation number of S2- is -2 and of Al3+ is +3.
3. The usual oxidation number of hydrogen is +1 (when bonded with a non-metal) but it is -1 when
hydrogen is bonded with a metal (metal hydrides) e.g. as in CaH2.
IB chemistry topic 9
7 hours
1|Page
4. The oxidation number of oxygen in compounds is usually -2.
Exceptions include OF2, since F is more electronegative than O, and H2O2, due to the structure of the
peroxide particle which is [O-O]2-.
5. The oxidation number of a group 1 element in a compound is +1.
6. The oxidation number of a group 2 element in a compound is +2.
7. The oxidation number of a group 7 element in a compound is -1, except when that element is
combined with one having a higher electronegativity.
For example in ICl, the oxidation number of I is +1, the oxidation number of Cl is -1 in HCl, but the
oxidation number of Cl is +1 in HOCl.
8. The sum of the oxidation numbers in a neutral compound is 0.
Examples:
NaCl
(oxidation number of Na) + (oxidation number of Cl) = 0
(+1) + (-1) = 0
Na2O
2(Ox. No. of Na) + (Ox. No. of O) = 0
2(+ 1) + (-2) = 0
CuS
(Ox. No. of Cu) + (Ox. No of S) = 0
(+2) + (-2) = 0
CaBr2
(Ox. No. of Ca) + 2(Ox. No. of Br) = 0
(+2) + 2(-1) = 0
9. The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.
For example, the sum of the oxidation numbers for SO42- is -2 because S = +6 and O = -2
so +6 + 4(-2) = -2
10. When elements show more than one oxidation state the oxidation number is shown using a
Roman numeral when naming the compound.
Examples: Fe (II)Cl2 = iron (II) chloride, Fe(III)Cl3 iron (III) chloride; Cu2O = copper (I) oxide; CuO =
copper (II) oxide
Examples
Example What is the oxidation number of thallium in TiCl3?
Method Chlorine always has the oxidation number -1.
Therefore (Ox. No. of Ti) + 3( -1) = 0 and the oxidation number of thallium is +3.
Example What is the oxidation number of Cl in Cl2O7?
Method The exceptions to the rule that the oxidation number of Cl equals -1 are compounds with O and
F. Oxygen is the reference point with the oxidation number -2.
IB chemistry topic 9
7 hours
2|Page
Therefore 2 (Ox. No. of Cl) + 7(-2) = 0 and the oxidation number of chlorine is +7.
Example What is the oxidation number of Cr in Cr(CN)63- ?
Method The cyanide ion, CN-, has a charge of - 1.
Therefore (Ox. No. of Cr) + 6( -1) = - 3 and the oxidation number of chromium is + 3.
Use http://www.occc.edu/kmbailey/chem1115tutorials/oxidation_numbers.htm
Exercises
In exercises 1 to 3 the main emphasis is on the following elements and their oxidation states: Fe (II) and
(III); Mn (II) and (VII), Cr (III) and (VI) and Cu (I) and (II), halogens and their ions and S in its oxyacids.
1. Give the oxidation number of each of the elements in the following species:
a) H2O
b) K2O
c) AlCl3
j) CrO42-
k) S2O32-
l) PbO2
d) OF2
e) MnO2
f) MnO4-
g) ClO3-
h) CuO
i) Cr2O72-
m) K2C2O4
2.
Work out the oxidation states of nitrogen in: NH3, N2H4, N2, N2O, NO, NO2, NO3-.
3.
Give the oxidation numbers of the first element in each of the following compounds.
Remember the oxidation numbers of the elements in a compound add up to zero. Take oxidation
numbers for hydrogen (+1), oxygen (-2), fluorine (-1) and chlorine (-1) as reference points.
CuO, Cu2O, H2S, SO2, SO3, PbO, SF6, SCl2, TiCl4, V2O.
4. Say whether the underlined atom is oxidized, reduced or remains unchanged?
(a) PbCl2 + Cl2  PbCl4
(b) NaOH + HCl  NaCl + H2O
(c) 2IO3 + 5HSO3-  I2 + 5SO42- + 3H+ + H2O
(d) Cu + ½ O2  CuO
Oxidation states in names: some elements can have different oxidation states in different compounds.
The oxidation state is often indicated using Roman numerals or by the name, eg:




FeO and Fe2O3 are called iron(II) oxide and iron(III) oxide. (Roman numerals are used; there is no
space between the numeral and the element.)
CuCl2 and Cu2O: copper(II) chloride and copper(I) oxide.
The following ions, ClO- , ClO2- , ClO3- , ClO4- are called chlorate(I), chlorate(III), chlorate(V) and
chlorate(VII).
sulphur: sulphuric acid, H2SO4, (S has oxidation state +6) and sulphurous acid, H2SO3, (S is +4);
 nitrogen: nitric acid, HNO3, (N has +5) and nitrous acid, HNO2, (N has +3);
IB chemistry topic 9
7 hours
3|Page
9.2 Redox equations
9.2.1. Deduce simple oxidation and reduction half-equations given the species involved in a redox reaction.
9.2.2 Deduce redox equations using half equations.
9.2.3 Define the terms oxidizing agent and reducing agent.
9.2.4 Identify the oxidizing and reducing agents in redox equations.
A redox react ion is a reaction during which an oxidation and reduction happen at the same time; as a
result an electron transfer takes place between the different reacting species in the reaction.
We can now distinguish two different species:
 oxidizing agent= specie which oxidizes another substance during a redox reaction; the oxidizing agent
itself is reduced i.e. it accepts electrons from the other substance;

reducing agent = specie which reduces another substance in a redox reaction; itself is oxidised i.e. it
gives electrons to the other species;
We use reduction and oxidation half equations to describe each type of reaction (oxidation or reduction) in
a redox reaction.
Reduction and oxidation half equation = stoichiometric equation showing the oxidation or reduction of a
specie; such an equation involves electron symbols.
Examples:
oxidation half equation (e- on product side !!!)
Fe2+ (aq)
 Fe3+ (aq) +
reduction half equation (e- on reactant side !!!)
e-
+
2e-

Cu (s)
Zn2+ (aq) +
2e-

Zn (s)
Cu2+ (aq)
2Cl- (l)  Cl2 (g) + 2e-
Such equations must balance in respect to the charges and the number of atoms or ions.
When presented with an equation you need to be able to:




identify if it is a redox equation by using oxidation numbers
if it is, identify which species are reduced and which one is oxidized
deduce the oxidation and reduction half equations
identify the reducing and oxidizing agent
Example
Fe2O3 (s) +
2Al (s)  Al2O3 (s) + 2Fe (s)
As iron’s oxidation number changes from +3 to 0;
therefore it has been reduced.
aluminium’s oxidation number changes from 0 to +3;
it has been oxidized.
Reduction half equation: Fe3+ + 3e  Fe3+
Oxidation half equation: Al  Al3+ +
IB chemistry topic 9
7 hours
e4|Page
Fe2O3 is the oxidizing agent as it causes the oxidation of aluminium. Aluminium is the best reducing agent
as it reduces Fe3+ to Fe.
Exercises
1. For each of the redox equations below:
 deduce the oxidation and reduction half equations
 deduce the oxidizing and reducing agent
(a) Mg (s) + Cl2 (g)  MgCl2 (s)
(b) 2NaCl (aq) + F2 (aq)  2NaF (aq) + CI2 (aq)
(c) CuSO4 (aq) + Zn (s)  ZnSO4 (aq) + Cu (s)
(d) Cu(s) + 2AgNO3 (aq) → Cu(NO3)2 (aq) + 2Ag(s)
(e) CuO (s) + H2 (g)  Cu (s) + H2O (l)
(f) 2Fe2+(aq) + I2(aq)  2Fe3+(aq) + 2I-(aq)
(g) Sn2+ (aq) + 2Fe3+ (aq)  Sn4+ (aq) + 2Fe2+ (aq)
(h) 3I2(aq) + 3OH - (aq)  IO3 - (aq) + 5I - (aq) + 3H+ (aq)
2. In the following reactions, identify which species have been odixized and which have been reduced.
Also
identify the reducing agent and the oxidizing agent. You could even write half-equations for each
reaction!!!
(a) Fe + S  FeS
(b) MnO2 + 4HCl  MnCl2 + Cl2 +
2H2O
- a little tricky
(c) 2FeCl2 + Cl2  2FeCl3
(d) CuO + H2SO4  CuSO4 + H2O
(e) PbO2 + 4HCl  PbCl2
+
Cl2
+
2H2O
(f) 4Fe(OH)2 + O2  2 H2O + 4Fe(OH)3
(g) Zn +
2V3+  Zn2+ +
2V2+
(h) 2Ca + O2  2CaO
(i) CuO + H2  Cu + H2O
(j) 2 AgNO3 + Cu  2 Ag + Cu(NO3)2
IB chemistry topic 9
7 hours
5|Page
Balancing redox equations
Example:
Step 1: Write the unbalanced equation for the reaction in ionic form.
 Fe3+ +
Fe2+ + Cr2O72-
Cr3+
Step 2: Separate the equation into two half-equations:
oxidation (loss of electrons) :
Fe2+
reduction (gain of electrons):
Cr2O72-
 Fe3+

Cr3+
Step 3: Balance the atoms other than O and H in each half-equation separately.
oxidation (loss of electrons) :
Fe2+
reduction (gain of electrons):
Cr2O72-
 Fe3+
(was already balanced)
2Cr

3+
Step 4: For reactions in an acid medium, add H2O to balance the O atoms and H+ to balance the H atoms
that have been added after the addition of water.
Cr2O72-

14H+
+ Cr2O72-
Cr3+ +

7H2O
(as there are 7 O atoms on the left side)
2Cr3+ +
7H2O
(to balance the 14 H atoms on the right side)
Step 5: Add electrons to one side of each half-reaction to balance the charges. If necessary, equalize the
number of electrons in the two half-equations by multiplying one or both half-reactions by appropriate
coefficients.
Fe2+  Fe3+ + e14H+ + Cr O 2- + 6e-  2Cr3+ + 7H O (the number of electrons makes sense
2
7
2
as Cr oxidation number changes from +6 to +3
per atom- there is 2)
To equalize the number of electrons in each half-equations, the oxidation equation needs to be
multiplied by 6.
6Fe2+  6Fe3+ + 6eStep 6: Add the two half-equations together and balance the final equation by inspection. The electrons on
both sides must balance.
The two half-equations are added to give
14H+
IB chemistry topic 9
+ Cr2O72- +
6Fe2+ +
6e- 
2Cr3+ +
7 hours
7H2O + 6Fe3+ +
6e6|Page
The electrons on both sides cancel, and we are left with a balanced net ionic equation:
Step 7: Check that the equation has the same types and numbers of atoms and the same charges on both
sides of the equation.
A final check shows that the resulting equation is ‘atomically’ and ‘electrically’ balanced.
Exercise
Complete and balance the following equations (all in acidic solutions): In each redox reaction, identify
the reducing agent and the oxidising agent and state how the oxidation numbers change for each of these
reagents.
a. Cu (s) + NO3- (aq)  Cu2+ (aq) + NO2 (g)
b. Mn2+ (aq) + NaBiO3 (s)  Bi3+ (aq) + MnO4- (aq) + Na+ (aq)
c. Cr2O72- (aq) + CH3OH (aq)  HCOOH (aq) + Cr3+ (aq)
d. MnO4- (aq) + Cl- (aq)  Mn2+ (aq) + Cl2 (g)
e. Cr2O72- (aq) + NO2- (aq)  NO3- (aq) + Cr3+ (aq)
f.
I- + NO3-
 IO3- + NO2
g. H2O2 (aq) + Fe2+ (aq)  H2O (l) + Fe3+ (aq)
h. Cr2O72- (aq) + C2O42- (aq)  CO2 (g) + Cr3+ (aq)
i.
MnO4- (aq) + Fe2+ (aq)  Mn2+ (aq) + Fe3+ (aq)
j.
MnO4- (aq) + SO32- (aq)  Mn2+ (aq) + SO42- (aq)
k. H2O2 + I- + H+ (aq)  H2O + I2
9. 3 Reactivity
9.3.1 Deduce a reactivity series based on the chemical behaviour of a group of oxidizing and reducing agents.
9.3.2 Deduce the feasibility of a redox reaction from a given reactivity series.
When metals react they lose electrons or are oxidized, when non-metals react they gain electrons or are
reduced. Therefore the reactivity of a metal or non-metal is about how easily it is oxidized or reduced or
how strong a reducing or oxidizing agent it is.
The strength of an oxidising or reducing agent can be found by using it in displacement reactions with other
oxidising or reducing agents e.g. like what you have done with the halogens or metals in group 1 in topic 3
on periodicity.
Displacement reactions involving elements, metals or non-metals, are called single replacement reactions
and these are all redox reactions as the reactant element is either oxidised (if it is a metal) or reduced (if it
is a halogen).
Metals are reducing agents: displacing other metal ions
IB chemistry topic 9
7 hours
7|Page
Metal displacement reactions involve placing one metal in a salt solution (e.g. a sulphate or nitrate) of
another metal.
The metal which displaces the metal in the compound is the best reducing agent as it reduces the metal
ion in the compound: it gives its electrons to the metal ion in the compound forcing it to come out of the
solution as an atom. The reducing agent itself becomes an ion.
Example:
CuCl2 (aq) +
 MgCl2 (aq)
Mg (s)
+
Cu (s)
This reaction is feasible as magnesium is the strongest reducing agent and reduces/displaces copper ions,
Cu2+, according to the following half-equations; the magnesium itself becomes a positive ion:
Half equation:


Cu2+ (aq) + 2e-  Cu (s)
= reduction
Mg (s)
 Mg2+ + 2e- (aq) = oxidation
We state that magnesium is more reactive than copper; it loses its electrons more readily, it oxidizes more
readily.
In this example, the magnesium is the reducing agent and copper (II) chloride is the oxidizing agent.
However, magnesium is a weaker reducing agent than sodium and can therefore never cause the
reduction of sodium ions so the following reaction is impossible:
2NaCl (aq) +
Mg (s)
 MgCl2 (aq)
+
2Na (s)
The above reaction is not feasible because magnesium is not a strong enough reducing agent to cause the
reduction of sodium ions.
However, if sodium atoms are placed in contact with magnesium ions than the following reaction will occur
as sodium is the strongest reducing agent – this reaction now is feasible :
MgCl2 (aq) +
 2NaCl (aq)
2Na (s)
+
Mg (s)
By carrying out a series of displacement reactions, which follow the pattern shown below, between metal
atoms and metal ions, a series of reactivity can be deduced with the strongest reducing agent at the top as
the most reactive metal.
XCl (aq) + Y (s)
 NaY (aq) + X
If the reaction above is feasible than Y is more reactive than X as it is a better reducing agent. It loses its
electrons more readily.
You are already familiar with some parts of this series which is called the reactivity series:
K
Na
Li
Ca
Mg
Al
Zn
strongest reducing agent;
its atoms will displace any ions
from elements more to the right
H
Cu
Ag
Au
weakest reducing agent
its ions will be displaced by any atom
from elements more to the left
K(s)  K+ (aq) + e-
IB chemistry topic 9
Fe
Au+ (aq) + e-  Au (s)
7 hours
8|Page
As the potassium atom, K, is a strong reducing agent, its ion, K+, is therefore a very weak oxidising
agent. Potassium atoms will displace/reduce ions of less reactive metals.
Also, in the series above Au+ (gold ion) is the strongest oxidising agent as Au (gold atom) is the
weakest reducing agent.
Non-metals
As non-metals usually gain electrons during reactions they are usually oxidising agents. Just like with
metals, single displacement reactions can also be used to place non-metals into a reactivity series – those
non-metals with the greatest tendency to accept electrons are the most reactive so the most reactive will be
the best oxidising agent as shown by the equation below:
When fluorine is added to a solution of potassium chloride, chloride ions are displaced by fluorine atoms.
2KCl (aq)
reducing agent
+

F2 (g)
2KF (aq)
+
Cl2 (g)
oxidising agent
In the above example, the following half equations can be deduced:
oxidation:
2Cl- (aq)  Cl2 (g) + 2e-
reduction:
F2 (g) + 2e-  2F- (aq)
redox:
2Cl- (aq) + F2 (g)  Cl2 (g) + 2F- (aq)
From experimental evidence from displacement reactions involving halogens the following series, which
you again are familiar with, has been deduced:
F2
Cl2
Br2
I2
At2
Fluorine atoms strongest
oxidizing agent
Astatine atoms weakest
oxidizing agent
Fluoride ions, F-,
weakest reducing agents
Astatide ions, At-,
strongest reducing agent
The oxidizing agents to the left in the series gain electrons from ions from the halogens more to the right.
Describe the experiments you would do to confirm experimentally which halogen is the strongest oxidising
agent. Describe observations for each tests. Write equations for those combinations in which reactions
take place.
IB chemistry topic 9
7 hours
9|Page
All common reducing and oxidizing reagents are in the table below which shows part of the electrochemical
series.
element
Lithium
Rubidium
Potassium
Sodium
Calcium
Magnesium
Aluminium
Zinc
Iron
Lead
HYDROGEN
Copper
Iodine
Silver
Bromine
Chlorine
Fluorine
comment
Strongest reducing agent, greatest tendency to lose electrons and
form positive ions in solutions.
Elements above hydrogen should displace the hydrogen ion from
acids. Elements near the top can also displace hydrogen from water.
Strongest oxidizing agent. Greatest tendency to gain electrons and to
form negative ions in solution.
Prediction of feasibility of redox reactions
These reactivity series can also be used to predict the feasibility of redox reactions:

Atoms of weaker reducing agents can never displace ions from stronger reducing agents.

Atoms of weaker oxidising agents can never displace ions from stronger oxidizing agents.
Exercises on predicting the feasibility of the following reactions.
Decide which ones of the following reactions are feasible; if the reaction is feasible, write the halfequations, an ionic equation and the overall equation.
1. Cu (s) +
2. I2 (s) +
HCl (aq) 
NaBr (aq) 
2. Mg (s) +
CuSO4 (aq) 
3. Cl2 (g) +
KBr (aq) 
4. At2 (s) +
MgCl2 (aq) 
6. Cu (s) +
IB chemistry topic 9
Page
Fe2+ (aq) 
7 hours
10 |
7. I2 (s) +
2 Cl- (aq) 
9. 4 Voltaic or electrochemical cells
9.4.1 Explain how a redox reaction is used to produce electricity in a voltaic cell.
9.4.2 State that oxidation occurs at the negative electrode (anode) and reduction occurs at the positive electrode (cathode).
Spontaneous redox reactions can be used to produce an electric current by using difference in reactivity or
oxidising ability between metals.
A voltaic or electrochemical cell is made by connecting two half-cells using an external circuit and a salt
bridge as shown below. The external circuit allows electrons to be transferred from one half cell to the
other.
A half cell is a piece of a metal immersed in an aqueous salt solution of that metal, usually a nitrate or a
sulphate.
A salt bridge is a piece of filter paper soaked in a salt solution, usually potassium nitrate, and it allows the
flow of ions between both half cells; this is necessary to ensure each half cell remains electrically neutral.
(Image from http://www.saskschools.ca/curr_content/chem30_05/6_redox/redox2_2.htm)
In the half cell of the more reactive metal (better reducing agent), the metal atoms oxidise releasing
electrons. This makes this half cell the negative electrode as a current always flows from the negative to
positive electrode. Because oxidation occurs it is also called the anode.
In the electrochemical cell above, zinc is more reactive than copper. As a result oxidation takes place in
the zinc half cell and it becomes the negative electrode/anode.
In the half cell of the less reactive metal, the metal ions gain the electrons lost by the oxidation in the other
half cell and as a result these ions are reduced. This makes this half cell the positive electrode. Because
reduction occurs it is also called the cathode.
In the electrochemical cell above, reduction occurs in the copper half cell and as a result it becomes the
positive electrode/cathode.
Equations of reactions taking place in the zinc-copper voltaic cell:
IB chemistry topic 9
Page
7 hours
11 |
At the cathode
Zn (s)  Zn2+ (aq) + 2e-
oxidation
At the anode
Cu2+ (aq) + 2e-  Cu (s)
reduction
Zn (s) + Cu2+ (aq)
Overall
 Zn2+ (aq) + Cu (s)
redox
See animation on http://www.saskschools.ca/curr_content/chem30_05/6_redox/redox2_2.htm
Reactivity difference of metals
The greater the difference in reactivity between the metals, the greater the voltage produced.
The redox reaction in an electrochemical cell is a spontaneous reaction.
Other examples of voltaic cells
For each of the voltaic cells, deduce the half equations at each electrode and identify the reducing and
oxidising agent.
(a) iron and magnesium
(b) iron and copper
(c) Pb (s) +
PbO2 (s) + 2H2SO4 (aq)  2PbSO4 (aq) + 2H2O (l)
9.5 Electrolysis
9.5.1 Describe, using a diagram, the essential components of an electrolytic cell.
9.5.2. State that oxidation occurs at the positive electrode (anode) and reduction occurs at the negative electrode
(cathode).
9.5.3. Describe how current is conducted in an electrolytic cell.
9.5.4. Deduce the products of the electrolysis of a molten salt.
Electrolysis reaction is a non-spontaneous redox reaction (endothermic) which is made to happen using
electrical energy from an outside source.
To be able to carry out an electrolysis reaction an electrolytic cell is needed. An electrolytic cell is a piece
of apparatus which converts chemical energy into electrical energy (opposite to an electrochemical
reaction). It needs an external source of electrical energy, two electrodes and an electrolyte.
An electrolyte is a liquid containing ions which allows a current to be conducted through it.
During an electrolysis reaction…..
At the cathode/negative electrode
At the anode/positive electrode
Reduction of positive metal ions to atoms using the
electrons supplied by the external source
Oxidation of negative ions releases electrons which
return to the external source
IB chemistry topic 9
Page
7 hours
12 |
(from http://www.ausetute.com.au/elecysis.html)
The diagram below shows an electrolytic cell used to electrolyse molten sodium chloride.
2Na+ (l) + 2e-  2Na (s)
2Cl- (l)  Cl2 (g)+ 2e-
During electrolysis, the current in the electrolytic cell is carried by:
 the movement of ions in the electrolyte
 by electrons in the wires and electrodes
The flow of electrons is achieved in the following way:
Electrons from the negative terminal of the power pack accumulate in the negative electrode
(=cathode). The negatively charged field attracts the positive salt ions (cations) which accept the
free electrons available at the cathode. This removal keeps the flow of electrons going in that part
of the circuit. At the cathode a reduction of positive metal ions occurs.
At the positive electrode (=anode) there is a shortage of electrons creating a positively charged
field which attracts the anions in the solution. These negative ions move towards the anode and
become oxidised; they release their electrons to the anode. The released electrons are attracted to
the source/power pack, maintaining the flow of electrons in that part of the circuit
IB chemistry topic 9
Page
7 hours
13 |
External source of electric current
positive terminal;
source of electric
current
negative terminal;
source of electric
current
electrons
electrons
anode – oxidation of negative ions
electrons are added to circuit
cathode – reduction of positive ions;
electrons are removed from circuit
negative ions
positive ions
at anode, negative ions are attracted and lose their
electrons – the negative ions are oxidised
at the cathode the positive ions are attracted and
gain electrons from the cathode – they become
reduced
Products of electrolysis of molten salts
General half-equations at each electrode:

reduction of metal ion at cathode: M+

oxidation of non-metal ion at anode: X-
+ e-  M
 X + e-
Examples of half equations at electrodes in different electrolytic cells
electrolyte
half - equations
2K+ + 2e-  2K
reaction at cathode
molten KCl
2Cl-
reaction at anode
2K+ + 2Cl-
overall equation - redox
2I-
reaction at anode
overall equation – redox
IB chemistry topic 9
Page
 I2 + 2e-
Mg2+ + 2I-
 I2 + Mg
2Al3+ + 6e-  2Al
reaction at cathode
molten Al2O3
 Cl2 + 2K
Mg2+ + 2e-  Mg
reaction at cathode
molten MgI2
 Cl2 + 2e-
3O2-
reaction at anode
7 hours
 1½ O2 + 6e14 |
2Al3+ + 3O2-
overall equation - redox
 O2 + 2Al
Overall pattern: metal at cathode and non-metal at anode!!!!!
Write both half-equations and overall equation for the electrolysis of the following salts:
a) molten sodium fluoride
b) molten calcium oxide
c) molten copper chloride
Electrochemical reactions V electrolysis
Electrochemical reactions are spontaneous redox reactions carried out in voltaic cells. They take place
whenever a more reactive metal is placed in contact by an electrical wire with a less reactive metal dipped
in a solution of one of its salts.
Electrolysis is the decomposition of compounds, which can ionise or dissociate using electricity, as
electrolysis reactions are non-spontaneous.
energy conversions
electrochemical reaction
(voltaic cell)
electrolysis
(electrolytic cell)
converts chemical energy into
electrical energy
spontaneous redox reaction
converts electrical energy into
chemical energy
non-spontaneous redox reaction
voltaic cells: two separate
solutions and a salt bridge
electrolytic cell + power supply;
one solution and no salt bridge
oxidation
reduction
reduction
oxidation
spontaneity
apparatus
reaction at negative electrode
reaction at positive electrode
IB questions
1. (M07)
IB chemistry topic 9
Page
7 hours
15 |
2. (N07)
3. (N07)
4. (N07)
IB chemistry topic 9
Page
7 hours
16 |
5. (N06) Which are examples of reduction?
I. Fe3+ becomes Fe2+
II. Cl- becomes Cl2
III. CrO3 becomes Cr3+
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
6. (N06) Which statement is correct for the electrolysis of a molten salt?
A. Positive ions move toward the positive electrode.
B. A gas is produced at the negative electrode.
C. Only electrons move in the electrolyte.
D. Both positive and negative ions move toward electrodes.
7. (N06) Which statement about the following reaction is correct?
2 Br- (aq) + Cl2 (aq)  Br2 (aq) + 2 Cl- (aq)
A. Br- (aq) is reduced and gains electrons.
-
C. B r (aq) is oxidized and loses electrons.
B. Cl2 (aq) is reduced and loses electrons.
D. Cl2 (aq) is oxidized and gains electrons.
8. (M05) What happens when molten sodium chloride is electrolysed in an electrolytic cell?
A.
B.
C.
D.
Chlorine is produced at the positive electrode.
Sodium ions lose electrons at the negative electrode.
Electrons flow through the liquid from the negative electrode to the positive electrode.
Oxidation occurs at the negative electrode and reduction at the positive electrode.
9. (N05) Which equations represent reactions that occur at room temperature?
IB chemistry topic 9
Page
7 hours
17 |
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
10. (N05) Which equation represents a redox reaction?
11. (N05) The following information is given about reactions involving the metals X, Y and Z and solutions
of their sulfates.
When the metals are listed in decreasing order of reactivity (most reactive first), what is the correct
order?
A. Z > Y > X
B. X > Y > Z
C. Y > X > Z
D. Y > Z > X
12. (N02) In the reaction
A.
B.
C.
D.
Br2 is only oxidised.
Br2 is only reduced.
Br2 is neither oxidised nor reduced.
Br2 is both oxidised and reduced.
13. (N02) Consider the following statements regarding electrolysis of molten lead(II) bromide.
I. Oxidation takes place at the anode where lead ions gain electrons.
II Reduction takes place at the cathode where lead ions gain electrons.
IB chemistry topic 9
Page
7 hours
18 |
III Oxidation takes place at the anode where bromide ions lose electrons.
IV. Reduction takes place at the cathode where bromide ions lose electrons.
Which of the above statements are correct?
A. I and II only
B. I and IV only
C. II and III only
D. II and IV only
14. (M02) Which of the following changes represents a reduction reaction?
15. (M02) During the electrolysis of a molten salt, the cation moves toward the .. I.. and undergoes .II..
I
A.
B.
C.
D.
II
negative electrode
negative electrode
positive electrode
positive electrode
reduction
oxidation
oxidation
reduction
12. (N01) What is the oxidation number of phosphorus in NaH2PO4?
A. +3
B. -3
D. –5
C. + 5
13. (N01) Which product is formed at the cathode (negative electrode) when molten MgCl2 is electrolysed?
A. Mg2+
B. Cl-
C. Mg
D. Cl2
14. (N00)
2AgNO3 (aq) + Zn (s)  2Ag(s) + Zn(NO3)2 (aq)
Zn(NO3 )2 (aq) + Co (s)  no reaction
2AgNO3 (aq) + Co (s)  2Ag(s) + Co(NO3)2 (aq)
IB chemistry topic 9
Page
7 hours
19 |
Using the above information, the order of increasing reactivity of the metals is
A. Ag Zn  Co
B. Co Ag  Zn
C. Co Zn  Ag
D. Ag  Co  Zn
15. (N00) The oxidation number of sulfur in HS2O5- is
A. –1
B. +3
C. +4
D. +5
PAPER 2
1. (M08/TZ1)
IB chemistry topic 9
Page
7 hours
20 |
2. (N06)
(a) For each of the following reactions in aqueous solution, state one observation that would
be made, and deduce the equation.
(i)
The reaction between chlorine and sodium iodide.
[2]
(ii)
The reaction between silver ions and chloride ions.
[2]
(b) Deduce whether or not each of the reactions in (a) is a redox reaction, giving a reason in
each case.
[4]
(c) (i) Draw a diagram of apparatus that could be used to electrolyse molten potassiumbromide. Label
the diagram to show the polarity of each electrode and the product formed.
[3]
(ii) Describe the two different ways in which electricity is conducted in the apparatus.
[2]
(iii) Write an equation to show the formation of the product at each electrode.
[2]
3. (M04) Consider the following redox equation.
IB chemistry topic 9
Page
7 hours
21 |
5Fe2+(aq) +MnO 4 (aq) +8H(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O(l)
(i)
Determine the oxidation numbers for Fe and Mn in the reactants and in the products.
(2)
(ii)
Based on your answer to (i), deduce which substance is oxidized.
(1)
(iii)
The compounds CH3OH and CH3O contain carbon atoms with different oxidation
numbers. Deduce the oxidation numbers and state the kind of chemical change needed
to make CH2O from CH3OH.
(3)
4. (M04) A part of the reactivity series of metals, in order of decreasing reactivity, is shown below.
magnesium
zinc
iron
lead
copper
silver
If a piece of copper metal were placed in separate solutions of silver nitrate and zinc nitrate
(i)
determine which solution would undergo reaction.
(1)
(ii)
identify the type of chemical change taking place in the copper and write the
half-equation for this change.
(2)
(iii)
state, giving a reason, what visible change would take place in the solutions.
(2)
5. (M04)
(i) Solid sodium chloride does not conduct electricity but molten sodium chloride does. Explain this
difference, and outline what happens in an electrolytic cell during the electrolysis of molten
sodium chloride using carbon electrodes.
(4)
(ii) State the products formed and give equations showing the reactions at each electrode.
(4)
(iii) State what practical use is made of this process.
(1)
4. For the following reaction
2Cu+  Cu + Cu2+
(a) state the oxidation number of each species.
IB chemistry topic 9
Page
[1]
7 hours
22 |
(b) write a balanced half-reaction for the oxidation process.
[1]
(c) write a balanced half-reaction for the reduction process.
[1]
IB chemistry topic 9
Page
7 hours
23 |
Mark scheme
PAPER 1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
PAPER 2
1. (N06)
IB chemistry topic 9
Page
7 hours
24 |
2.
IB chemistry topic 9
Page
7 hours
25 |
3.
IB chemistry topic 9
Page
7 hours
26 |
IB chemistry topic 9
Page
7 hours
27 |