The document provides information on redox reactions including: - Defining oxidation and reduction in terms of electron loss and gain. - Explaining how to deduce the oxidation number of elements in compounds and use oxidation numbers to name compounds. - Stating that redox reactions involve the transfer of electrons between chemical species. - Describing how to write half reactions to show electron transfer and combine them to give overall redox equations. - Explaining how to assign oxidation numbers and use them to determine if an element is oxidized or reduced in a reaction.

2. Objectives
 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.

3. Introduction
 Redox reactions are reactions that involve the transfer of
electrons between chemical species.
 Electron transfer from one place to another is a flow of
electrons and a flow of electrons is an electric current.
 This study of redox reactions is therefore the combined
study of chemistry and electricity.
 In fact, an alternative title for this chapter could well be
‘electrochemistry’. The applications of electrochemistry
include;
 the use of redox reactions inside batteries to generate an
electric potential difference (voltage).
 predicting whether chosen substances will take part in a
redox reaction.

4. Some terms you need to know
 The term redox is a contraction of the words
reduction and oxidation.
 Oxidation and reduction take place together;
simultaneously.
 One substance is reduced while the other is oxidised.
 A substance that reduces another substance is called
a reducing agent, whilst the substance itself will be
oxidised.
 A substance that oxidises another substance is called
an oxidising agent, whilst the substance itself will be
reduced.

5. The Oxidation of Magnesium
 An example of a redox reaction is magnesium burning in
oxygen to produce magnesium oxide.
2Mg(s) + O2(g) → 2MgO(s)
 Magnesium is a metal, oxygen is a diatomic covalent gas,
and magnesium oxide is an ionic compound containing Mg2+
and O2- ions.
 A simple definition of redox reactions considers
oxidation as the gain of oxygen atoms and reduction as
the loss of oxygen atoms. Applying this definition to the
burning of magnesium:
 Magnesium is oxidised by oxygen
 Oxygen is reduced by magnesium

6. Half equations
 The balanced chemical reaction does not show
electron transfer. Splitting the equation into two
half-equations reveals more clearly what is
happening in the course of the reaction.
 One half equation shows the loss of electrons and
the other shows the gain of electrons.
 Magnesium loses electrons:
2Mg(s) → 2Mg2+(s) + 4e-
 Oxygen gains electrons
O2(g) +4e- → 2O2-(s)

7. Half Equations
 The half reaction described by the half equation does
not necessarily actually occur; the electrons may not
become free in the way suggested.
 Half equations are simply a useful way of thinking
about the two component parts of a redox reaction.
 A half reaction must balance with respect to charge
as well as with respect to the number of atoms.
 Notice how the sum of the charges on the left equals
the sum of the charges on the right.
2Mg(s) → 2Mg2+(s) + 4e-
O2(g) + 4e- → 2O2-(s)

8. Electron Transfer
 We can consider the reaction in terms of the
underlying electron transfer:
 Oxidation is the loss of electrons
 Reduction is the gain of electrons
 Oxidation is caused by the oxidant (or oxidising
agent): the oxidant (oxygen O2) is itself reduced in
the process (to O2-)
 Reduction is caused by the reductant (Magnesium
Mg) is itself oxidised in the process (to Mg2+)

9. Redox in the absence of
oxygen
 The example given so far shows that the oxidation
of magnesium may be interpreted in two ways:
either as a gain of oxygen or as a loss of
electrons.
 The idea of redox as electron transfer allows
many reactions not involving oxygen to be
classified as redox reactions.

10.  The reaction between aqueous copper (II)
sulphate and zinc metal is a redox reaction:
Zn(s) + CuSO4(aq) → ZnSO4(aq) +Cu(s)
 Remember that the formulae CuSO4(aq) and
ZnSO4(aq) refer to ionic substances dissolved in
water:
 CuSO4(aq) is therefore present as Cu2+
(aq) and SO4
2-
(aq) ions.
 ZnSO4(aq) is present as Zn2+
(aq) and SO4
2-
(aq) ions.
 The sulphate ion SO4
2-
(aq) is a spectator ion; it
does not take part in the chemical reaction and so
we may disregard it.

11.  The full balanced chemical equation is not very
informative. However, the two half-equations reveal the
redox nature of the reaction.
 The half-equation for zinc shows that the metal atoms
are oxidised as the result of electron loss:
Zn(s) → Zn2+
(aq) + 2e-
 The half-equation for copper shows that the aqueous
copper (II) ions are reduced as the result of electron
gain:
Cu2+
(aq) + 2e- → Cu(s)
 The balanced equation for the reduction of copper (II)
ions by zinc results from adding together the two half-
equations.
 In this example the combination is straightforward
because both half-equations involve two electrons (note
that we have not included the sulphate spectator ions
Zn(s) + Cu2+
(aq) → Zn2+
(aq) + Cu(s)

12. Using Half Equations – an
example
 Chlorine gas oxidises iodide ions in solution to iodine,
and in the process is reduced itself to chloride ions.
Write the ionic equation for the reaction.
 All you need to do is to write the two half equations,
and then combine them.
Chlorine gas is Cl2; chloride ions are Cl-. Write those
down on each side of the equation.
Cl2 → Cl-

13. Using Half Equations – an
example
 Chlorine gas oxidises iodide ions in solution to iodine,
and in the process is reduced itself to chloride ions.
Write the ionic equation for the reaction.
 All you need to do is to write the two half equations,
and then combine them.
Now balance the atoms
Cl2 → Cl-
Cl2 → 2Cl-

14. Using Half Equations – an
example
 Chlorine gas oxidises iodide ions in solution to iodine,
and in the process is reduced itself to chloride ions.
Write the ionic equation for the reaction.
 All you need to do is to write the two half equations,
and then combine them.
Finally, the charges need to be balanced by adding
electrons where necessary. In this case, you need two on
the left hand side:
Cl2 → 2Cl-
Cl2 + 2e-→ 2Cl-

15. Using Half Equations – an
example
 Chlorine gas oxidises iodide ions in solution to iodine,
and in the process is reduced itself to chloride ions.
Write the ionic equation for the reaction.
 All you need to do is to write the two half equations,
and then combine them.
Repeating this process (first balancing the atoms and
then the charges) for the other half equation involving
the iodine gives you:
2I- → I2 + 2e-

16. Using Half Equations – an
example
 Chlorine gas oxidises iodide ions in solution to iodine,
and in the process is reduced itself to chloride ions.
Write the ionic equation for the reaction.
 All you need to do is to write the two half equations,
and then combine them.
These two equations can now be literally added
together:
2I- → I2 + 2e-
Cl2 + 2e-→ 2Cl-
Cl2 + 2I-→ 2Cl- + I2
---------------------------------
Because there are two
electrons on both sides,
they cancel out, and we are
left with the ionic equation

17. Rules for constructing half
equations
 You are only allowed to write certain things into
half equations:
 The substance you start from and what it is
oxidised or reduced to.
 Hydrogen ions, H+ (unless the solution is alkaline, in
which case these are replaced by hydroxide ions,
OH-)
 Water
 Electrons

18. Example
 An acidic solution of hydrogen peroxide, H2O2,
oxidises iron (II) ions to iron (III) ions. The
hydrogen peroxide is reduced to water. Write the
ionic equation for the reaction.
Start with hydrogen peroxide reducing to water:
H2O2 → H2O

19. Example
 An acidic solution of hydrogen peroxide, H2O2,
oxidises iron (II) ions to iron (III) ions. The
hydrogen peroxide is reduced to water. Write the
ionic equation for the reaction.
Balance the oxygens:
H2O2 → H2O
H2O2 → 2H2O

20. Example
 An acidic solution of hydrogen peroxide, H2O2,
oxidises iron (II) ions to iron (III) ions. The
hydrogen peroxide is reduced to water. Write the
ionic equation for the reaction.
There aren’t enough hydrogen on the left hand side, add
two hydrogen:
H2O2 → 2H2O
2H+ + H2O2 → 2H2O

21. Example
 An acidic solution of hydrogen peroxide, H2O2,
oxidises iron (II) ions to iron (III) ions. The
hydrogen peroxide is reduced to water. Write the
ionic equation for the reaction.
The atoms balance, but not the charges. There are two
positive charges on the left, but zero charge on the
right. Add two electrons to the left:
2H+ + H2O2 → 2H2O
2H+ + H2O2 + 2e- → 2H2O

22. Example
 An acidic solution of hydrogen peroxide, H2O2,
oxidises iron (II) ions to iron (III) ions. The
hydrogen peroxide is reduced to water. Write the
ionic equation for the reaction.
Now for the iron, this is much easier;
Fe2+ → Fe3+

23. Example
 An acidic solution of hydrogen peroxide, H2O2,
oxidises iron (II) ions to iron (III) ions. The
hydrogen peroxide is reduced to water. Write the
ionic equation for the reaction.
Everything is OK except the charges. Add an electron to
supply a negative charge to the right, and cut 3+ down to
2+
Fe2+ → Fe3+
Fe2+ → Fe3+ + e-

24. Example
 An acidic solution of hydrogen peroxide, H2O2,
oxidises iron (II) ions to iron (III) ions. The
hydrogen peroxide is reduced to water. Write the
ionic equation for the reaction.
This leaves
Fe2+ → Fe3+ + e-
2H+ + H2O2 + 2e- → 2H2O

25. Example
 An acidic solution of hydrogen peroxide, H2O2,
oxidises iron (II) ions to iron (III) ions. The
hydrogen peroxide is reduced to water. Write the
ionic equation for the reaction.
Warning; you can’t just add these equations together –
the electrons will not cancel out, we must multiply the
second equation by two.
Fe2+ → Fe3+ + e-
2H+ + H2O2 + 2e- → 2H2O
x2 2Fe2+ →2 Fe3+ + 2e-

26. Example
 An acidic solution of hydrogen peroxide, H2O2,
oxidises iron (II) ions to iron (III) ions. The
hydrogen peroxide is reduced to water. Write the
ionic equation for the reaction.
Now the electrons balance and can cancel out:
2H+ + H2O2 + 2e- → 2H2O
2Fe2+ →2 Fe3+ + 2e-
2H+ + H2O2 + 2Fe2+ → 2H2O + 2Fe3+

27. Quick questions
 For each of the following redox reactions identify
the (i) oxidizing agent, (ii) the reducing agent,
(iii) the species which is oxidised and (iv) the
species which is reduced.
 2Ca(s) + O2(g) → 2CaO(s)
 CuO(s) + H2(g) → Cu(s) + H2O(l)
 2AgNO3(aq) + Cu(s) → 2Ag(s) + Cu(NO3)2(aq)
 For each of the reactions above write half
equations to show (i) electron loss during
oxidation and (ii) electron gain during reduction.

28. Oxidation numbers
 Each element in any chemical species may be
assigned an oxidation number. The oxidation
number of an element is the number of electrons
that need to be added to the element to make a
neutral atom.
 E.g. The iron ion Fe2+ requires the addition of two
electrons to make a neutral atom. The oxidation
number for iron in the Fe2+ ion form is therefore
+2.
 The oxidation state of this ion is written as
iron(II) or Fe(II); it is expressed in Roman
numerals and describes the extent of oxidation
of the species.

29. Assigning oxidation numbers
 Uncombined elements always have oxidation number = 0
 Some elements always have the same oxidation numbers in
their compounds:
 Group I metals always +I
 Group II metals always +II
 Al always +III
 H, +I, except in compounds with Group 1 or 2 metals (metal
hydrides) where it is –I
 F, always –I
 O, -II, except in peroxides and compounds with F, where it is –I
 Cl, -I except in compounds with F and O, where it has positive
values.
 The sum of all the oxidation numbers in a compound = 0
 The sum of the oxidation numbers of a complex ion = the
charge of the ion

30. Quick questions
 What is the oxidation number of the metal M in
each of the following:
 MCl
 M2O3
 MSO4
 M

31. Oxidation of Iron (II)
 The Fe3+ ion has an oxidation number of +3. The
oxidation state of this ion is written as iron (III) or
as Fe(III) , indicating that it is a more highly oxidised
species than Fe2+
 When an iron (II) ion,(Fe2+)reacts to form an iron
(III) ion, (Fe3+), an electron is lost thus the Fe(II) is
oxidised:
Fe2+
(aq) → Fe3+
(aq) + e-
 When it is oxidised, the oxidation number change for
iron is from +2 to +3; the oxidation state change for
iron is from Fe(II) to Fe(III)
 Oxidation involves an increase in oxidation number.

32. Reduction of Iron (III)
 In the reverse process, reduction , an iron (III)
ion gains one electron to become an iron (II) ion:
Fe3+
(aq) + e- → Fe2+
(aq)
 When it is reduced, the oxidation number change
for iron is from 3+ to 2+; the oxidation state
change for iron is from Fe(III) to Fe(II).
 Reduction involves a decrease in oxidation number.

33. Reactivity Series
 As some substances are clearly better at
oxidizing than others we can place them in order
of their oxidizing ability.
 Experimentally this series was arrived at initially
by seeing which metal could displace the ions of
another metal in aqueous solution.
 The most reactive metals are those that give up
electrons most readily, so the reactivity series is
normally written as a reduction series with the
best reducing agents at the top of the table

34. The Reactivity series
Activity:
 Carry out
Experiment 9.1
or watch the
video by clicking
on the test-tube.
 Record your
results in an
appropriate table.
 Write down the
half equations for
each of the
reactions.
 Arrange the half
equations in order
of reactivity.

35. K(s) e- + K+
(aq)
Na(s) e- + Na+
(aq)
Li(s) e- + Li+
(aq)
Ca(s) 2e- + Ca2+
(aq)
Mg(s) 2e- + Mg2+
(aq)
Al(s) 3e- + Al3+
(aq)
Zn(s) 2e- + Zn2+
(aq)
Fe(s) 2e- + Fe2+
(aq)
Pb(s) 2e- + Pb2+
(aq)
½ H2(s) e- + 2H+
(aq)
Cu(s) 2e- + Cu2+
(aq)
Ag(s) e- + Ag+
(aq)
The Reactivity Series
Increasing
reducing
ability

36. Electrical energy from a redox
reaction
 Activity: Carry
out
Experiment
9.2 or watch
the video.

37. Voltaic cells
 A simple voltaic cell is a device for obtaining electrical
energy from a spontaneous chemical reaction.
 The oxidation reaction and the reduction reaction, as
represented by their two half equations, are separated,
and each is carried out experimentally in a separate
electrochemical half cell.
The simplest half cell consists of a metal
in contact with a solution of its own ions
e.g. A zinc half cell consists of a strip of
zinc metal placed in an aqueous solution of
zinc sulphate.
An equilibrium will be established between the metal
and the ions in solution Zn

38. Some electrochemical cells
 Activity:
Carry out
Experiment
9.3 or watch
the video.
 Make a note
of the
voltage
changes and
compare the
values.
 Make a note
of the
effect of
heating and
dilution on
e.m.f. Values.