263REDOX REACTIONS
Chemistry deals with varieties of matter and change of one
kind of matter into the other. T ransformation of matter from
one kind into another occurs through the various types of
reactions. One important category of such reactions is
Redox Reactions. A number of phenomena, both physical
as well as biological, are concerned with redox reactions.
These reactions find extensive use in pharmaceutical,
biological, industrial, metallurgical and agricultural areas.
The importance of these reactions is apparent from the fact
that burning of different types of fuels for obtaining energy
for domestic, transport and other commercial purposes,
electrochemical processes for extraction of highly reactive
metals and non-metals, manufacturing of chemical
compounds like caustic soda, operation of dry and wet
batteries and corrosion of metals fall within the purview of
redox processes. Of late, environmental issues like
Hydrogen Economy (use of liquid hydrogen as fuel) and
development of ‘Ozone Hole’ have started figuring under
redox phenomenon.
8.1 CLASSICAL IDEA OF REDOX REACTIONS –
OXIDATION AND REDUCTION REACTIONS
Originally, the term oxidation was used to describe the
addition of oxygen to an element or a compound. Because
of the presence of dioxygen in the atmosphere (~20%),
many elements combine with it and this is the principal
reason why they commonly occur on the earth in the
form of their oxides. The following reactions represent
oxidation processes according to the limited definition of
oxidation:
2 Mg (s) + O
2
(g) → 2 MgO (s) (8.1)
S (s) + O
2
(g) → SO
2
(g) (8.2)
After studying this unit you will be
able to
••
••
• identify redox reactions as a class
of reactions in which oxidation
and reduction reactions occur
simultaneously;
••
••
• define the terms oxidation,
reduction, oxidant (oxidising
agent) and reductant (reducing
agent);
••
••
• explain mechanism of redox
reactions by electron transfer
process;
••
••
• use the concept of oxidation
number to identify oxidant and
reductant in a reaction;
••
••
• classify redox reaction into
combination (synthesis),
decomposition, displacement
and disproportionation
reactions;
••
••
• suggest a comparative order
among various reductants and
oxidants;
••
••
• balance chemical equations
using (i) oxidation number
(ii) half reaction method;
••
••
• learn the concept of redox
reactions in terms of electrode
processes.
UNIT 8
REDOX REACTIONS
Where there is oxidation, there is always reduction –
Chemistry is essentially a study of redox systems.
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264 CHEMISTRY
In reactions (8.1) and (8.2), the elements
magnesium and sulphur are oxidised on
account of addition of oxygen to them.
Similarly, methane is oxidised owing to the
addition of oxygen to it.
CH
4
(g) + 2O
2
(g) → CO
2
(g) + 2H
2
O (l) (8.3)
A careful examination of reaction (8.3) in
which hydrogen has been replaced by oxygen
prompted chemists to reinterpret oxidation in
terms of removal of hydrogen from it and,
therefore, the scope of term oxidation was
broadened to include the removal of hydrogen
from a substance. The following illustration is
another reaction where removal of hydrogen
can also be cited as an oxidation reaction.
2 H
2
S(g) + O
2
(g) → 2 S (s) + 2 H
2
O (l) (8.4)
As knowledge of chemists grew, it was
natural to extend the term oxidation for
reactions similar to (8.1 to 8.4), which do not
involve oxygen but other electronegative
elements. The oxidation of magnesium with
fluorine, chlorine and sulphur etc. occurs
according to the following reactions :
Mg (s) + F
2
(g) → MgF
2
(s) (8.5)
Mg (s) + Cl
2
(g) → MgCl
2
(s) (8.6)
Mg (s) + S (s) → MgS (s) (8.7)
Incorporating the reactions (8.5 to 8.7)
within the fold of oxidation reactions
encouraged chemists to consider not only the
removal of hydrogen as oxidation, but also the
removal of electropositive elements as
oxidation. Thus the reaction :
2K
4
[Fe(CN)
6
](aq) + H
2
O
2
(aq) →2K
3
[Fe(CN)
6
](aq)
+ 2 KOH (aq)
is interpreted as oxidation due to the removal
of electropositive element potassium from
potassium ferrocyanide before it changes to
potassium ferricyanide. To summarise, the
term “oxidation” is defined as the addition
of oxygen/electronegative element to a
substance or removal of hydrogen/
electropositive element from a substance.
In the beginning, reduction was
considered as removal of oxygen from a
compound. However, the term reduction has
been broadened these days to include removal
of oxygen/electronegative element from a
substance or addition of hydrogen/
electropositive element to a substance.
According to the definition given above, the
following are the examples of reduction
processes:
2 HgO (s)
2 Hg (l) + O
2
(g) (8.8)
(removal of oxygen from mercuric oxide )
2 FeCl
3
(aq) + H
2
(g) →2 FeCl
2
(aq) + 2 HCl(aq)
(8.9)
(removal of electronegative element, chlorine
from ferric chloride)
CH
2
= CH
2
(g) + H
2
(g) → H
3
C – CH
3
(g) (8.10)
(addition of hydrogen)
2HgCl
2
(aq) + SnCl
2
(aq) → Hg
2
Cl
2
(s)+SnCl
4
(aq)
(8.11)
(addition of mercury to mercuric chloride)
In reaction (8.11) simultaneous oxidation
of stannous chloride to stannic chloride is also
occurring because of the addition of
electronegative element chlorine to it. It was
soon realised that oxidation and reduction
always occur simultaneously (as will be
apparent by re-examining all the equations
given above), hence, the word “redox” was
coined for this class of chemical reactions.
Problem 8.1
In the reactions given below, identify the
species undergoing oxidation and
reduction:
(i) H
2
S (g) + Cl
2
(g) → 2 HCl (g) + S (s)
(ii) 3Fe
3
O
4
(s) + 8 Al (s) → 9 Fe (s)
+ 4Al
2
O
3
(s)
(iii) 2 Na (s) + H
2
(g) → 2 NaH (s)
Solution
(i) H
2
S is oxidised because a more
electronegative element, chlorine is added
to hydrogen (or a more electropositive
element, hydrogen has been removed
from S). Chlorine is reduced due to
addition of hydrogen to it.
(ii) Aluminium is oxidised because
oxygen is added to it. Ferrous ferric oxide
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265REDOX REACTIONS
(Fe
3
O
4
) is reduced because oxygen has
been removed from it.
(iii) With the careful application of the
concept of electronegativity only we may
infer that sodium is oxidised and
hydrogen is reduced.
Reaction (iii) chosen here prompts us to
think in terms of another way to define
redox reactions.
8.2 REDOX REACTIONS IN TERMS OF
ELECTRON TRANSFER REACTIONS
We have already learnt that the reactions
2Na(s) + Cl
2
(g) → 2NaCl (s) (8.12)
4Na(s) + O
2
(g) → 2Na
2
O(s) (8.13)
2Na(s) + S(s) → Na
2
S(s) (8.14)
are redox reactions because in each of these
reactions sodium is oxidised due to the
addition of either oxygen or more
electronegative element to sodium.
Simultaneously, chlorine, oxygen and sulphur
are reduced because to each of these, the
electropositive element sodium has been
added. From our knowledge of chemical
bonding we also know that sodium chloride,
sodium oxide and sodium sulphide are ionic
compounds and perhaps better written as
Na
+
Cl
–
(s), (Na
+
)
2
O
2–
(s), and (Na
+
)
2
S
2–
(s).
Development of charges on the species
produced suggests us to rewrite the reactions
(8.12 to 8.14) in the following manner :
For convenience, each of the above
processes can be considered as two separate
steps, one involving the loss of electrons and
the other the gain of electrons. As an
illustration, we may further elaborate one of
these, say, the formation of sodium chloride.
2 Na(s) → 2 Na
+
(g)
+ 2e
–
Cl
2
(g) + 2e
–
→ 2 Cl
–
(g)
Each of the above steps is called a half
reaction, which explicitly shows involvement
of electrons. Sum of the half reactions gives
the overall reaction :
2 Na(s) + Cl
2
(g) → 2 Na
+
Cl
–
(s) or 2 NaCl (s)
Reactions 8.12 to 8.14 suggest that half
reactions that involve loss of electrons are
called oxidation reactions. Similarly, the
half reactions that involve gain of electrons
are called reduction reactions. It may not
be out of context to mention here that the new
way of defining oxidation and reduction has
been achieved only by establishing a
correlation between the behaviour of species
as per the classical idea and their interplay in
electron-transfer change. In reactions (8.12 to
8.14) sodium, which is oxidised, acts as
a reducing agent because it donates electron
to each of the elements interacting with it and
thus helps in reducing them. Chlorine, oxygen
and sulphur are reduced and act as oxidising
agents because these accept electrons from
sodium. To summarise, we may mention that
Oxidation: Loss of electron(s) by any species.
Reduction: Gain of electron(s) by any species.
Oxidising agent : Acceptor of electron(s).
Reducing agent : Donor of electron(s).
Problem 8.2 Justify that the reaction :
2 Na(s) + H
2
(g) → 2 NaH (s) is a redox
change.
Solution
Since in the above reaction the compound
formed is an ionic compound, which may
also be represented as Na
+
H
–
(s), this
suggests that one half reaction in this
process is :
2 Na (s) → 2 Na
+
(g) + 2e
–
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266 CHEMISTRY
and the other half reaction is:
H
2
(g) + 2e
–
→ 2 H
–
(g)
This splitting of the reaction under
examination into two half reactions
automatically reveals that here sodium is
oxidised and hydrogen is reduced,
therefore, the complete reaction is a redox
change.
8.2.1 Competitive Electron Transfer
Reactions
Place a strip of metallic zinc in an aqueous
solution of copper nitrate as shown in Fig. 8.1,
for about one hour. You may notice that the
strip becomes coated with reddish metallic
copper and the blue colour of the solution
disappears. Formation of Zn
2+
ions among the
products can easily be judged when the blue
colour of the solution due to Cu
2+
has
disappeared. If hydrogen sulphide gas is
passed through the colourless solution
containing Zn
2+
ions, appearance of white zinc
sulphide, ZnS can be seen on making the
solution alkaline with ammonia.
The reaction between metallic zinc and the
aqueous solution of copper nitrate is :
Zn(s) + Cu
2+
(aq) → Zn
2+
(aq) + Cu(s) (8.15)
In reaction (8.15), zinc has lost electrons
to form Zn
2+
and, therefore, zinc is oxidised.
Evidently, now if zinc is oxidised, releasing
electrons, something must be reduced,
accepting the electrons lost by zinc. Copper
ion is reduced by gaining electrons from the zinc.
Reaction (8.15) may be rewritten as :
At this stage we may investigate the state
of equilibrium for the reaction represented by
equation (8.15). For this purpose, let us place
a strip of metallic copper in a zinc sulphate
solution. No visible reaction is noticed and
attempt to detect the presence of Cu
2+
ions by
passing H
2
S gas through the solution to
produce the black colour of cupric sulphide,
CuS, does not succeed. Cupric sulphide has
such a low solubility that this is an extremely
sensitive test; yet the amount of Cu
2+
formed
cannot be detected. We thus conclude that the
state of equilibrium for the reaction (8.15)
greatly favours the products over the reactants.
Let us extend electron transfer reaction now
to copper metal and silver nitrate solution in
water and arrange a set-up as shown in
Fig. 8.2. The solution develops blue colour due
to the formation of Cu
2+
ions on account of the
reaction:
Fig. 8.1 Redox reaction between zinc and aqueous solution of copper nitrate occurring in a beaker.
(8.16)
Here, Cu(s) is oxidised to Cu
2+
(aq) and
Ag
+
(aq) is reduced to Ag(s). Equilibrium greatly
favours the products Cu
2+
(aq) and Ag(s).
By way of contrast, let us also compare the
reaction of metallic cobalt placed in nickel
sulphate solution. The reaction that occurs
here is :
(8.17)
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267REDOX REACTIONS
Fig. 8.2 Redox reaction between copper and aqueous solution of silver nitrate occurring in a beaker.
At equilibrium, chemical tests reveal that both
Ni
2+
(aq) and Co
2+
(aq)
are present at moderate
concentrations. In this case, neither the
reactants [Co(s) and Ni
2+
(aq)] nor the products
[Co
2+
(aq) and Ni (s)] are greatly favoured.
This competition for release of electrons
incidently reminds us of the competition for
release of protons among acids. The similarity
suggests that we might develop a table in
which metals and their ions are listed on the
basis of their tendency to release electrons just
as we do in the case of acids to indicate the
strength of the acids. As a matter of fact we
have already made certain comparisons. By
comparison we have come to know that zinc
releases electrons to copper and copper
releases electrons to silver and, therefore, the
electron releasing tendency of the metals is in
the order: Zn>Cu>Ag. We would love to make
our list more vast and design a metal activity
series or electrochemical series. The
competition for electrons between various
metals helps us to design a class of cells,
named as Galvanic cells in which the chemical
reactions become the source of electrical
energy. We would study more about these cells
in Class XII.
8.3 OXIDATION NUMBER
A less obvious example of electron transfer is
realised when hydrogen combines with oxygen
to form water by the reaction:
2H
2
(g) + O
2
(g) → 2H
2
O (l) (8.18)
Though not simple in its approach, yet we
can visualise the H atom as going from a
neutral (zero) state in H
2
to a positive state in
H
2
O, the O atom goes from a zero state in O
2
to a dinegative state in H
2
O. It is assumed that
there is an electron transfer from H to O and
consequently H
2
is oxidised and O
2
is reduced.
However, as we shall see later, the charge
transfer is only partial and is perhaps better
described as an electron shift rather than a
complete loss of electron by H and gain by O.
What has been said here with respect to
equation (8.18) may be true for a good number
of other reactions involving covalent
compounds. Two such examples of this class
of the reactions are:
H
2
(s) + Cl
2
(g) → 2HCl(g) (8.19)
and,
CH
4
(g) + 4Cl
2
(g) → CCl
4
(l) + 4HCl(g) (8.20)
In order to keep track of electron shifts in
chemical reactions involving formation of
covalent compounds, a more practical method
of using oxidation number has been
developed. In this method, it is always
assumed that there is a complete transfer of
electron from a less electronegative atom to a
more electonegative atom. For example, we
rewrite equations (8.18 to 8.20) to show
charge on each of the atoms forming part of
the reaction :
0 0 +1 –2
2H
2
(g) + O
2
(g) → 2H
2
O (l) (8.21)
0 0 +1 –1
H
2
(s) + Cl
2
(g) → 2HCl(g) (8.22)
–4+1 0 +4 –1 +1 –1
CH
4
(g) + 4Cl
2
(g) → CCl
4
(l) +4HCl(g) (8.23)
It may be emphasised that the assumption
of electron transfer is made for book-keeping
purpose only and it will become obvious at a
later stage in this unit that it leads to the simple
description of redox reactions.
Oxidation number denotes the
oxidation state of an element in a
compound ascertained according to a set
of rules formulated on the basis that
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268 CHEMISTRY
electron pair in a covalent bond belongs
entirely to more electronegative element.
It is not always possible to remember or
make out easily in a compound/ion, which
element is more electronegative than the other.
Therefore, a set of rules has been formulated
to determine the oxidation number of an
element in a compound/ion. If two or more
than two atoms of an element are present in
the molecule/ion such as Na
2
S
2
O
3
/Cr
2
O
7
2–
, the
oxidation number of the atom of that element