REDOX REACTIONS

Where there is oxidation, there is always reduction –

Chemistry is essentially a study of redox systems.

After studying this unit you will be

able to Chemistry deals with varieties of matter and change
of one kind of matter into the other. Transformation of
 identify redox reactions as a class
of reactions in which oxidation matter from one kind into another occurs through the
and reduction reactions occur various types of reactions. One important category of such
simultaneously; reactions is Redox Reactions. A number of phenomena,
 define the terms oxidation, both physical as well as biological, are concerned with
reduction, oxidant (oxidising redox reactions. These reactions find extensive use in
agent) and reductant (reducing pharmaceutical, biological, industrial, metallurgical and
agent); agricultural areas. The importance of these reactions is
 explain mechanism of redox apparent from the fact that burning of different types of
reactions by electron transfer fuels for obtaining energy for domestic, transport and
process; other commercial purposes, electrochemical processes
 use the concept of oxidation for extraction of highly reactive metals and non-metals,
number to identify oxidant and manufacturing of chemical compounds like caustic
reductant in a reaction;
soda, operation of dry and wet batteries and corrosion of
 classify redox reaction into metals fall within the purview of redox processes. Of late,
combination (synthesis),
environmental issues like Hydrogen Economy (use of
decomposition, displacement
and dis pr o p o r t i o n a t i o n liquid hydrogen as fuel) and development of ‘Ozone Hole’
reactions; have started figuring under redox phenomenon.
 suggest a comparative order
7.1 CLASSICAL IDEA OF REDOX REACTIONS –
among various reductants and
oxidants; OXIDATION AND REDUCTION REACTIONS
 balance chemical equations Originally, the term oxidation was used to describe the
using (i) oxidation number addition of oxygen to an element or a compound. Because
(ii) half reaction method; of the presence of dioxygen in the atmosphere (~20%),
 learn the concept of redox many elements combine with it and this is the principal
reactions in terms of electrode reason why they commonly occur on the earth in the
processes. form of their oxides. The following reactions represent
oxidation processes according to the limited definition of
oxidation:
2 Mg (s) + O2 (g)  2 MgO (s) (7.1)
S (s) + O2 (g)  SO2 (g) (7.2)
 In reactions (7.1) and (7.2), the elements broadened these days to include removal
magnesium and sulphur are oxidised on of oxygen/electronegative element from
account of addition of oxygen to them. a substance or addition of hydrogen/
Similarly, methane is oxidised owing to the electropositive element to a substance.
addition of oxygen to it. According to the definition given above,
CH4 (g) + 2O2 (g)  CO (g) + 2H2O (l) (7.3) the following are the examples of reduction
processes:
A careful examination of reaction (7.3) in
which hydrogen has been replaced by oxygen 2 HgO (s) 2 Hg (l) + O2 (g) (7.8)
prompted chemists to reinterpret oxidation (removal of oxygen from mercuric oxide )
in terms of removal of hydrogen from it and, 2 FeCl3 (aq) + H2 (g) 2 FeCl (aq) + 2 HCl(aq)
therefore, the scope of term oxidation was (7.9)
broadened to include the removal of hydrogen
(removal of electronegative element, chlorine
from a substance. The following illustration is
from ferric chloride)
another reaction where removal of hydrogen
can also be cited as an oxidation reaction. CH = CH (g) + H (g)  H C – CH (g) (7.10)
(addition of hydrogen)
2 H S(g) + O (g)  2 S (s) + 2 H O (l) (7.4)
2HgCl (aq) + SnCl (aq)  Hg Cl (s)+SnCl (aq)
2 2 2 2 4
As knowledge of chemists grew, it was
natural to extend the term oxidation for (7.11)
reactions similar to (7.1 to 7.4), which do (addition of mercury to mercuric chloride)
not involve oxygen but other electronegative In reaction (7.11) simultaneous oxidation
elements. The oxidation of magnesium with of stannous chloride to stannic chloride is
fluorine, chlorine and sulphur etc. occurs also occurring because of the addition of
according to the following reactions : electronegative element chlorine to it. It was
soon realised that oxidation and reduction
Mg (s) + F2 (g)  MgF (s) (7.5) always occur simultaneously (as will be
Mg (s) + Cl2 (g)  MgCl (s) (7.6) apparent by re-examining all the equations
given above), hence, the word “redox” was
Mg (s) + S (s)  MgS (s) (7.7) coined for this class of chemical reactions.
Incorporating the reactions (7.5 to
7.7) 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 [Fe(CN) ](aq) + H O (aq) 2K [Fe(CN) ](aq)
4 6 2 2 3 6
+ 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
REDOX REACTIONS 237

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 (g) + 2e–  2 Cl–(g)
Each of the above steps is called a half
reaction, which explicitly shows involvement
7.2 REDOX REACTIONS IN TERMS OF of electrons. Sum of the half reactions gives
ELECTRON TRANSFER REACTIONS the overall reaction :
We have already learnt that the reactions
2 Na(s) + Cl2 (g)  2 Na+ Cl– (s) or 2 NaCl (s)
2Na(s) + Cl 2(g)  2NaCl (s) (7.12)
4Na(s) + O (g)  2Na O(s) (7.13) Reactions 7.12 to 7.14 suggest that half
2 2 reactions that involve loss of electrons are
2Na(s) + S(s)  Na S(s) (7.14) called oxidation reactions. Similarly, the
are redox reactions because in each of these half reactions that involve gain of electrons
reactions sodium is oxidised due to the addition are called reduction reactions. It may not
of either oxygen or more electronegative be out of context to mention here that the
element to sodium. Simultaneously, chlorine, new way of defining oxidation and reduction
oxygen and sulphur are reduced because to has been achieved only by establishing a
each of these, the electropositive element correlation between the behaviour of species
sodium has been added. From our knowledge as per the classical idea and their interplay
of chemical bonding we also know that sodium in electron-transfer change. In reactions (7.12
chloride, sodium oxide and sodium sulphide to 7.14) sodium, which is oxidised, acts as
are ionic compounds and perhaps better a reducing agent because it donates electron
written as Na+Cl– (s), (Na+) 2O2–(s), and (Na+) 2 to each of the elements interacting with it and
S2–(s). Development of charges on the species thus helps in reducing them. Chlorine, oxygen
produced suggests us to rewrite the reactions and sulphur are reduced and act as oxidising
(7.12 to 7.14) in the following manner : 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).
 At this stage we may investigate the state
of equilibrium for the reaction represented by
equation (7.15). 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 Cu2+ ions
by passing H2S 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
7.2.1 Competitive Electron Transfer
sensitive test; yet the amount of Cu 2+ formed
Reactions
cannot be detected. We thus conclude that
Place a strip of metallic zinc in an aqueous
the state of equilibrium for the reaction (7.15)
solution of copper nitrate as shown in Fig.
greatly favours the products over the reactants.
7.1, for about one hour. You may notice
that the strip becomes coated with reddish Let us extend electron transfer reaction
metallic copper and the blue colour of the now to copper metal and silver nitrate solution
solution disappears. Formation of Zn2+ ions in water and arrange a set-up as shown in
among the products can easily be judged Fig. 7.2. The solution develops blue colour
when the blue colour of the solution due to due to the formation of Cu2+ ions on account
Cu2+ has disappeared. If hydrogen sulphide of the reaction:
gas is passed through the colourless solution
containing Zn2+ ions, appearance of white zinc
sulphide, ZnS can be seen on making the
solution alkaline with ammonia.
The reaction between metallic zinc and the (7.16)
aqueous solution of copper nitrate is :
Here, Cu(s) is oxidised to Cu2+(aq) and
Zn(s) + Cu2+ (aq)  Zn2+ (aq) + Cu(s) (7.15)
Ag+(aq) is reduced to Ag(s). Equilibrium greatly
In reaction (7.15), zinc has lost electrons favours the products Cu2+ (aq) and Ag(s).
to form Zn2+ and, therefore, zinc is oxidised.
By way of contrast, let us also compare
Evidently, now if zinc is oxidised, releasing the reaction of metallic cobalt placed in nickel
electrons, something must be reduced, sulphate solution. The reaction that occurs
accepting the electrons lost by zinc. Copper here is :
ion is reduced by gaining electrons from the zinc.
Reaction (7.15) may be rewritten as :
At equilibrium, chemical tests reveal that both However, as we shall see later, the charge
Ni2+(aq) and Co2+(aq) are present at moderate transfer is only partial and is perhaps better
concentrations. In this case, neither the described as an electron shift rather than a
reactants [Co(s) and Ni2+(aq)] nor the products complete loss of electron by H and gain by
[Co2+(aq) and Ni (s)] are greatly favoured. O. What has been said here with respect
This competition for release of electrons to equation (7.18) may be true for a good
incidently reminds us of the competition for number of other reactions involving covalent
release of protons among acids. The similarity compounds. Two such examples of this class
suggests that we might develop a table in of the reactions are:
which metals and their ions are listed on the H (s) + Cl (g)  2HCl(g) (7.19)
basis of their tendency to release electrons and,
just as we do in the case of acids to indicate CH (g) + 4Cl (g)  CCl (l) + 4HCl(g) (7.20)
the strength of the acids. As a matter of fact 4 2 4

we have already made certain comparisons. In order to keep track of electron shifts
By comparison we have come to know that in chemical reactions involving formation
zinc releases electrons to copper and copper of covalent compounds, a more practical
releases electrons to silver and, therefore, method of using oxidation number has
the electron releasing tendency of the metals been developed. In this method, it is always
is in the order: Zn>Cu>Ag. We would love to assumed that there is a complete transfer
make our list more vast and design a metal of electron from a less electronegative atom
activity series or electrochemical series. to a more electonegative atom. For example,
The competition for electrons between various we rewrite equations (7.18 to 7.20) to show
metals helps us to design a class of cells, charge on each of the atoms forming part of
named as Galvanic cells in which the chemical the reaction :
reactions become the source of electrical 0 0 +1 –2
energy. We would study more about these 2H (g) + O (g)  2H O (l) (7.21)
2 2 2
cells in Class XII. 0 0 +1 –1
H (s) + Cl (g)  2HCl(g) (7.22)
7.3 OXIDATION NUMBER
–4+1 0 +4 –1 +1 –1
A less obvious example of electron transfer is
realised when hydrogen combines with oxygen CH (g) + 4Cl (g)  CCl (l) +4HCl(g) (7.23)
to form water by the reaction: It may be emphasised that the assumption
2H 2(g) + O 2 (g)  2H 2O (l) (7.18) of electron transfer is made for book-keeping
Though not simple in its approach, yet purpose only and it will become obvious at
we can visualise the H atom as going from a a later stage in this unit that it leads to the
neutral (zero) state in H2 to a positive state in simple description of redox reactions.
H 2O, the O atom goes from a zero state in O 2 Oxidation number denotes the oxidation
to a dinegative state in H2O. It is assumed that state of an element in a compound
there is an electron transfer from H to O and ascertained according to a set of rules
consequently H2 is oxidised and O2 is reduced. formulated on the basis that electron pair
in a covalent bond belongs entirely to more bonding state of oxygen but this number
electronegative element. would now be a positive figure only.
It is not always possible to remember or 4. The oxidation number of hydrogen is +1,
make out easily in a compound/ion, which except when it is bonded to metals in binary
element is more electronegative than the compounds (that is compounds containing
other. Therefore, a set of rules has been two elements). For example, in LiH, NaH,
formulated to determine the oxidation and CaH2, its oxidation number is –1.
number of an element in a compound/ion. 5. In all its compounds, fluorine has an
If two or more than two atoms of an element oxidation number of –1. Other halogens (Cl,
are present in the molecule/ion such as Br, and I) also have an oxidation number
Na S O /Cr O2–, the oxidation number of the of –1, when they occur as halide ions in
atom of that element will then be the average their compounds. Chlorine, bromine and
of the oxidation number of all the atoms of iodine when combined with oxygen, for
that element. We may at this stage, state the example in oxoacids and oxoanions, have
rules for the calculation of oxidation number. positive oxidation numbers.
These rules are: 6. The algebraic sum of the oxidation number
of all the atoms in a compound must be
1. In elements, in the free or the uncombined zero. In polyatomic ion, the algebraic sum
state, each atom bears an oxidation of all the oxidation numbers of atoms of
number of zero. Evidently each atom in the ion must equal the charge on the ion.
H2, O2, Cl2, O3, P4, S8, Na, Mg, Al has the Thus, the sum of oxidation number of three
oxidation number zero. oxygen atoms and one carbon atom in the
2. For ions composed of only one atom, the carbonate ion, (CO )2– must equal –2.
oxidation number is equal to the charge By the application of above rules, we can
on the ion. Thus Na+ ion has an oxidation find out the oxidation number of the desired
number of +1, Mg2+ ion, +2, Fe3+ ion, +3, element in a molecule or in an ion. It is clear
Cl– ion, –1, O2– ion, –2; and so on. In their that the metallic elements have positive
compounds all alkali metals have oxidation oxidation number and nonmetallic elements
number of +1, and all alkaline earth metals have positive or negative oxidation number.
have an oxidation number of +2. Aluminium The atoms of transition elements usually
is regarded to have an oxidation number of display several positive oxidation states. The
+3 in all its compounds. highest oxidation number of a representative
3. The oxidation number of oxygen in most element is the group number for the first
compounds is –2. However, we come two groups and the group number minus 10
across two kinds of exceptions here. (following the long form of periodic table) for
One arises in the case of peroxides and the other groups. Thus, it implies that the
superoxides, the compounds of oxygen in highest value of oxidation number exhibited
which oxygen atoms are directly linked to by an atom of an element generally increases
each other. While in peroxides (e.g., H2O2, across the period in the periodic table. In the
Na2O2), each oxygen atom is assigned an third period, the highest value of oxidation
oxidation number of –1, in superoxides number changes from 1 to 7 as indicated
(e.g., KO2, RbO2) each oxygen atom is below in the compounds of the elements.
assigned an oxidation number of –(½). A term that is often used interchangeably
The second exception appears rarely, i.e. with the oxidation number is the oxidation
when oxygen is bonded to fluorine. In state. Thus in CO2, the oxidation state of
such compounds e.g., oxygen difluoride carbon is +4, that is also its oxidation number
(OF2) and dioxygen difluoride (O2F2), the and similarly the oxidation state as well
oxygen is assigned an oxidation number as oxidation number of oxygen is – 2. This
of +2 and +1, respectively. The number implies that the oxidation number denotes the
assigned to oxygen will depend upon the oxidation state of an element in a compound.
The oxidation number/state of a metal in a The idea of oxidation number has been
compound is sometimes presented according invariably applied to define oxidation,
to the notation given by German chemist, reduction, oxidising agent (oxidant), reducing
Alfred Stock. It is popularly known as Stock agent (reductant) and the redox reaction. To
notation. According to this, the oxidation summarise, we may say that:
number is expressed by putting a Roman Oxidation: An increase in the oxidation
numeral representing the oxidation number number of the element in the given substance.
in parenthesis after the symbol of the metal in
the molecular formula. Thus aurous chloride Reduction : A decrease in the oxidation
and auric chloride are written as Au(I)Cl and number of the element in the given substance.
Au(III)Cl3. Similarly, stannous chloride and Oxidising agent: A reagent which can
stannic chloride are written as Sn(II)Cl2 and increase the oxidation number of an element
Sn(IV)Cl4. This change in oxidation number in a given substance. These reagents are
implies change in oxidation state, which in called as oxidants also.
turn helps to identify whether the species
Reducing agent: A reagent which lowers the
is present in oxidised form or reduced form.
oxidation number of an element in a given
Thus, Hg2(I)Cl2 is the reduced form of Hg(II) Cl2.
substance. These reagents are also called as
reductants.
Redox reactions: Reactions which involve
change in oxidation number of the interacting
species.









 here that all decomposition reactions are not
redox reactions. For example, decomposition
of calcium carbonate is not a redox reaction.
+2 +4 –2 +2 –2 +4 –2
CaCO3 (s) CaO(s) + CO2(g)
3. Displacement reactions
In a displacement reaction, an ion (or an
7.3.1 Types of Redox Reactions atom) in a compound is replaced by an ion
(or an atom) of another element. It may be
1. Combination reactions
denoted as:
A combination reaction may be denoted in
the manner: X + YZ  XZ + Y
A+B  C D isplace ment reactions fit into two
Either A and B or both A and B must be in categories: metal displacement and non-metal
the elemental form for such a reaction to be displacement.
a redox reaction. All combustion reactions, (a) Metal displacement: A metal in a
which make use of elemental dioxygen, as well compound can be displaced by another metal
as other reactions involving elements other in the uncombined state. We have already
than dioxygen, are redox reactions. Some discussed about this class of the reactions
important examples of this category are: under section 7.2.1. Metal displacement
0 0 +4 –2 reactio ns find many appl i ca tio ns in
C(s) + O2 (g) CO2(g) (7.24) metallurgical processes in which pure metals
are obtained from their compounds in ores. A
few such examples are:
+2 +6 –2 0 0 +2 +6 –2
CuSO (aq) + Zn (s)  Cu(s) + ZnSO (aq)
4 4
(7.29)
2. Decomposition reactions +5 –2 0 0 +2 –2

Decomposition reactions are the opposite V2O5 (s) + 5Ca (s) 2V (s) + 5CaO (s)
of combination reactions. Precisely, a (7.30)
decomposition reaction leads to the breakdown +4 –1 0 0 +2 –1
of a compound into two or more components TiCl4 (l) + 2Mg (s) Ti (s) + 2 MgCl2 (s)
at least one of which must be in the elemental
(7.31)
state. Examples of this class of reactions are:
+1 –2 0 0 +3 –2 0 +3 –2 0

2H2O (l) 2H2 (g) + O2(g) (7.26) Cr2O3 (s) + 2 Al (s) Al2O3 (s) + 2Cr(s)
(7.32)
+1 –1 0 0
In each case, the reducing metal is a
2NaH (s) 2Na (s) + H2(g) (7.27)
better reducing agent than the one that is
+1 +5 –2 +1 –1 0 being reduced which evidently shows more
2KClO3 (s) 2KCl (s) + 3O2(g) (7.28) capability to lose electrons as compared to
It may carefully be noted that there is no the one that is reduced.
change in the oxidation number of hydrogen (b) Non-metal displacement: The non-
in methane under combination reactions metal displacement redox reactions include
and that of potassium in potassium chlorate hydrogen displacement and a rarely occurring
in reaction (7.28). This may also be noted reaction involving oxygen displacement.
 All alkali metals and some alkaline earth Cu>Ag. Like metals, activity series also exists
metals (Ca, Sr, and Ba) which are very good for the halogens. The power of these elements
reductants, will displace hydrogen from cold as oxidising agents decreases as we move
water. down from fluorine to iodine in group 17 of the
0 +1 –2 +1 –2 +1 0
periodic table. This implies that fluorine is so
2Na(s) + 2H O(l)  2NaOH(aq) + H (g) reactive that it can replace chloride, bromide
2 2
(7.33) and iodide ions in solution. In fact, fluorine is
0 +1 –2 +2 –2 +1 0 so reactive that it attacks water and displaces
Ca(s) + 2H O(l)  Ca(OH) (aq) + H2(g) the oxygen of water :
(7.34) +1 –2 0 +1 –1 0

Less active metals such as magnesium and 2H O (l) + 2F (g)  4HF(aq) + O (g) (7.40)
iron react with steam to produce dihydrogen gas: It is for this reason that the displacement
0 +1 –2 +2 –2 +1 0
reactions of chlorine, bromine and iodine
using fluorine are not generally carried out in
Mg(s) + 2H2O(l) Mg(OH)2(s) + H2(g) aqueous solution. On the other hand, chlorine
(7.35)
can displace bromide and iodide ions in an
0 +1 –2 +3 –2 0 aqueous solution as shown below:
2Fe(s) + 3H2O(l) Fe2O3(s) + 3H2(g) (7.36) 0 +1 –1 +1 –1 0
Many metals, including those which do not Cl (g) + 2KBr (aq)  2 KCl (aq) + Br (l)
react with cold water, are capable of displacing (7.41)
hydrogen from acids. Dihydrogen from acids 0 +1–1 +1 –1 0
may even be produced by such metals which Cl (g) + 2KI (aq)  2 KCl (aq) + I (s)
do not react with steam. Cadmium and tin are (7.42)
the examples of such metals. A few examples As Br2 and I2 are coloured and dissolve in CCl4,
for the displacement of hydrogen from acids can easily be identified from the colour of the
are: solution. The above reactions can be written
0 +1 –1 +2 –1 0
in ionic form as:
Zn(s) + 2HCl(aq)  ZnCl (aq) + H2 (g) 0 –1 –1 0
(7.37) Cl2 (g) + 2Br – (aq)  2Cl – (aq) + Br (l) (7.41a)
0 +1 –1 +2 –1 0 0 –1 –1 0
Mg (s) + 2HCl (aq)  MgCl (aq) + H (g) Cl2 (g) + 2I – (aq)  2Cl – (aq) + I (s) (7.42b)
(7.38) Reactions (7.41) and (7.42) form the basis
0 +1 –1 +2 –1 0
of identifying Br– and I– in the laboratory
Fe(s) + 2HCl(aq)  FeCl (aq) + H (g) through the test popularly known as ‘Layer
(7.39) Test’. It may not be out of place to mention
Reactions (7.37 to 7.39) are used to here that bromine likewise can displace iodide
prepare dihydrogen gas in the laboratory. ion in solution:
Here, the reactivity of metals is reflected in 0 –1 –1 0
the rate of hydrogen gas evolution, which is
Br2 (l) + 2I (aq)  2Br (aq) + I (s)
– –
(7.43)
the slowest for the least active metal Fe, and
the fastest for the most reactive metal, Mg. The halogen displacement reactions have
Very less active metals, which may occur in a direct industrial application. The recovery
the native state such as silver (Ag), and gold of halogens from their halides requires an
(Au) do not react even with hydrochloric acid. oxidation process, which is represented by:
2X–  X + 2e– (7.44)
In section (7.2.1) we have already discussed
that the metals – zinc (Zn), copper (Cu) and here X denotes a halogen element. Whereas
silver (Ag) through tendency to lose electrons chemical means are available to oxidise Cl–,
show their reducing activity in the order Zn> Br– and I–, as fluorine is the strongest oxidising
agent; there is no way to convert F– ions to F fluorine shows deviation from this behaviour
by chemical means. The only way to achieve when it reacts with alkali. The reaction that
F from F– is to oxidise electrolytically, the takes place in the case of fluorine is as follows:
details of which you will study at a later stage. 2 F (g) + 2OH–(aq)  2 F–(aq) + OF (g) + H O(l)
2 2 2
4. Disproportionation reactions (7.49)
Disproportionation reactions are a special type (It is to be noted with care that fluorine in
of redox reactions. In a disproportionation reaction (7.49) will undoubtedly attack water
reaction an element in one oxidation state to produce some oxygen also). This departure
is simultaneously oxidised and reduced. shown by fluorine is not surprising for us as
One of the reacting substances in a we know the limitation of fluorine that, being
disproportionation reaction always contains the most electronegative element, it cannot
an element that can exist in at least three exhibit any positive oxidation state. This
oxidation states. The element in the form means that among halogens, fluorine does not
of reacting substance is in the intermediate show a disproportionation tendency.
oxidation state; and both higher and lower
oxidation states of that element are formed in
the reaction. The decomposition of hydrogen
peroxide is a familiar example of the reaction,
where oxygen experiences disproportionation.
+1 –1 +1 –2 0
2H O (aq)  2H O(l) + O (g) (7.45)
Here the oxygen of peroxide, which is present
in –1 state, is converted to zero oxidation state
in O2 and decreases to –2 oxidation state in
H2O.
Phosphorous, sulphur and chlorine
undergo disproportionation in the alkaline
medium as shown below :
0 –3 +1
P (s) + 3OH–(aq)+ 3H O(l)  PH (g) + 3H PO –
4 2 3 2 2
(aq)
(7.46) 
0 –2 +2
S (s) + 12 OH– (aq)  4S2– (aq) + 2S O 2–(aq)
8 2 3
+ 6H2O(l)
(7.47) 
0 +1 –1
Cl2 (g) + 2 OH– (aq)  ClO– (aq) + Cl– (aq) +
H2O (l)
(7.48)
The reaction (7.48) describes the
formation of household bleaching agents.
The hypochlorite ion (ClO –) formed in the
reaction oxidises the colour-bearing stains
of the substances to colourless compounds.
It is of interest to mention here that whereas
bromine and iodine follow the same trend
as exhibited by chlorine in reaction (7.48),

2024-25
 The Paradox of Fractional Oxidation Number
Sometimes, we come across with certain compounds in which the oxidation number of a particular
element in the compound is in fraction. Examples are:
C3O2 [where oxidation number of carbon is (4/3)],
Br3O8 [where oxidation number of bromine is (16/3)]
and Na2S4O6 (where oxidation number of sulphur is 2.5).
We know that the idea of fractional oxidation number is unconvincing to us, because electrons
are never shared/transferred in fraction. Actually this fractional oxidation state is the average
oxidation state of the element under examination and the structural parameters reveal that the
element for whom fractional oxidation state is realised is present in different oxidation states.
Structure of the species C O , Br O and S O 2– reveal the following bonding situations:
3 2 3 8 4 6

+2 0 +2
O = C = C*= C = O
Structure of C3O2
(carbon suboxide)

Structure of Br O (tribromooctaoxide) Structure of S O2– (tetrathionate ion)

3 8 4 6
The element marked with asterisk in each species is exhibiting the different oxidation
state (oxidation number) from rest of the atoms of the same element in each of the species.
This reveals that in C3O2, two carbon atoms are present in +2 oxidation state each, whereas
the third one is present in zero oxidation state and the average is 4/3. However, the realistic
picture is +2 for two terminal carbons and zero for the middle carbon. Likewise in Br3O8, each
of the two terminal bromine atoms are present in +6 oxidation state and the middle bromine
is present in +4 oxidation state. Once again the average, that is different from reality, is
2–
16/3. In the same fashion, in the species S O
4 6
, each of the two extreme sulphurs exhibits oxidation
state of +5 and the two middle sulphurs as zero. The average of four oxidation numbers of sulphurs
of the S O 2– is 2.5, whereas the reality being + 5,0,0 and +5 oxidation number respectively for
4 6
each sulphur.
We may thus, in general, conclude that the idea of fractional oxidation state should be taken
with care and the reality is revealed by the structures only. Further, whenever we come across
with fractional oxidation state of any particular element in any species, we must understand that
this is the average oxidation number only. In reality (revealed by structures only), the element in
that particular species is present in more than one whole number oxidation states. Fe 3O4, Mn3O4,
Pb3O4 are some of the other examples of the compounds, which are mixed oxides, where we come
across with fractional oxidation states of the metal atom. However, the oxidation states may be in
fraction as in O+ and O – where it is +½ and –½ respectively.
2 2
 (a) Oxidation Number Method: In writing
equations for oxidation-reduction reactions,
just as for other reactions, the compositions and
formulas must be known for the substances
that react and for the products that are
formed. The oxidation number method is now
best illustrated in the following steps:
Step 1: Write the correct formula for each
reactant and product.
Step 2: Identify atoms which undergo
change in oxidation number in the reaction
by assigning the oxidation number to all
elements in the reaction.
Step 3: Calculate the increase or decrease
in the oxidation number per atom and for
the entire molecule/ion in which it occurs. If
these are not equal then multiply by suitable
number so that these become equal. (If you
realise that two substances are reduced and
nothing is oxidised or vice-versa, something
is wrong. Either the formulas of reactants or
products are wrong or the oxidation numbers
have not been assigned properly).
Step 4: Ascertain the involvement of ions if
the reaction is taking place in water, add H+ or
OH– ions to the expression on the appropriate
side so that the total ionic charges of reactants
and products are equal. If the reaction is
carried out in acidic solution, use H + ions in
the equation; if in basic solution, use OH– ions.
Step 5 : Make the numbers of hydrogen
atoms in the expression on the two sides
equal by adding water (H2O) molecules to the
reactants or products. Now, also check the
number of oxygen atoms. If there are the same
number of oxygen atoms in the reactants and
products, the equation then represents the
7.3.2 Balancing of Redox Reactions balanced redox reaction.
Two methods are used to balance chemical Let us now explain the steps involved in
equations for redox processes. One of these the method with the help of a few problems
methods is based on the change in the given below:
oxidation number of reducing agent and the
oxidising agent and the other method is based
on splitting the redox reaction into two half
reactions — one involving oxidation and the
other involving reduction. Both these methods
are in use and the choice of their use rests
with the individual using them.
(b) Half Reaction Method: In this method,
the two half equations are balanced separately
and then added together to give balanced
equation.
Suppose we are to balance the equation
showing the oxidation of Fe2+ ions to Fe3+ ions
by dichromate ions (Cr 2O 7)2– in acidic medium,
wherein, Cr O 2– ions are reduced to Cr3+ ions.
The following steps are involved in this task.
Step 1: Produce unbalanced equation for the
reaction in ionic form :
Fe2+(aq) + Cr O2– (aq)  Fe3+ (aq) + Cr3+(aq)
2 7
(7.50)
Step 2: Separate the equation into half- Step 7: Verify that the equation contains the
reactions: same type and number of atoms and the same
+2 +3 charges on both sides of the equation. This
Oxidation half : Fe2+ (aq)  Fe3+(aq) (7.51) last check reveals that the equation is fully
+6 –2 +3 balanced with respect to number of atoms
Reduction half : Cr O2–(aq)  Cr3+(aq) and the charges.
(7.52) For the reaction in a basic medium, first
Step 3: Balance the atoms other than O and balance the atoms as is done in acidic medium.
H in each half reaction individually. Here the Then for each H+ ion, add an equal number of
oxidation half reaction is already balanced OH– ions to both sides of the equation. Where
with respect to Fe atoms. For the reduction H+ and OH– appear on the same side of the
half reaction, we multiply the Cr3+ by 2 to equation, combine these to give H 2O.
balance Cr atoms.
Cr O2–(aq)  2 Cr3+(aq) (7.53)
2 7

Step 4: For reactions occurring in acidic

medium, add H O to balance O atoms and H+
to balance H atoms.
Thus, we get :
Cr O2– (aq) + 14H+ (aq)  2 Cr3+(aq) + 7H O (l)
2 7 2

(7.54)
Step 5: Add electrons to one side of the half
reaction to balance the charges. If need be,
make the number of electrons equal in the
two half reactions by multiplying one or both
half reactions by appropriate number.
The oxidation half reaction is thus
rewritten to balance the charge:
Fe2+ (aq)  Fe3+ (aq) + e– (7.55)
Now in the reduction half reaction there
are net twelve positive charges on the left hand
side and only six positive charges on the right
hand side. Therefore, we add six electrons on
the left side.
Cr O2– (aq) + 14H+ (aq) + 6e–  2Cr3+(aq) +
2 7

7H2O (l) (7.56)

To equalise the number of electrons in both
the half reactions, we multiply the oxidation
half reaction by 6 and write as :
6Fe2+ (aq)  6Fe3+(aq) + 6e– (7.57)
Step 6: We add the two half reactions to
achieve the overall reaction and cancel the
electrons on each side. This gives the net ionic
equation as :
6Fe2+(aq) + Cr O2–(aq) + 14H+(aq)  6 Fe3+(aq) +
2 7

2Cr 3+(aq) + 7H O(l) (7.58)

 (ii) If there is no dramatic auto-colour change
(as with MnO– titration), there are indicators
which are oxidised immediately after the
last bit of the reactant is consumed,
producing a dramatic colour change. The
best example is afforded by Cr O2–, which
2 7
is not a self-indicator, but oxidises the
indicator substance diphenylamine just
after the equivalence point to produce an
intense blue colour, thus signalling the
end point.
(iii) There is yet another method which is
interesting and quite common. Its use is
restricted to those reagents which are able
to oxidise I– ions, say, for example, Cu(II):
2Cu2+(aq) + 4I–(aq)  Cu I (s) + I (aq) (7.59)
2 2 2
This method relies on the facts that
iodine itself gives an intense blue colour with
starch and has a very specific reaction with
thiosulphate ions (S 2O 2–
3
), which too is a redox
reaction:
I (aq) + 2 S O2–(aq)2I–(aq) + S O 2–(aq) (7.60)
2 2 3 4 6
I2, though insoluble in water, remains in
solution containing KI as KI3.
On addition of starch after the liberation
7.3.3 Redox Reactions as the Basis for of iodine from the reaction of Cu2+ ions on
Titrations iodide ions, an intense blue colour appears.
This colour disappears as soon as the iodine
In acid-base systems we come across with a
titration method for finding out the strength is consumed by the thiosulphate ions. Thus,
of one solution against the other using a the end-point can easily be tracked and the
pH sensitive indicator. Similarly, in redox rest is the stoichiometric calculation only.
systems, the titration method can be adopted 7.3.4 Limitations of Concept of Oxidation
to determine the strength of a reductant/ Number
oxidant using a redox sensitive indicator.
The usage of indicators in redox titration is As you have observed in the above discussion,
illustrated below: the concept of redox processes has been
evolving with time. This process of evolution
(i) In one situation, the reagent itself is
intensely coloured, e.g., permanganate is continuing. In fact, in recent past the
ion, MnO–4. Here MnO 4– acts as the self oxidation process is visualised as a decrease
in electron density and reduction process as
indicator. The visible end point in this case
is achieved after the last of the reductant an increase in electron density around the
(Fe2+ or C O 2–) is oxidised and the first atom(s) involved in the reaction.
2 4
lasting tinge of pink colour appears at 7.4 REDOX REACTIONS AND ELECTRODE
MnO – concentration as low as 10–6 mol dm–3 PROCESSES
(10–6 mol L–1). This ensures a minimal
‘overshoot’ in colour beyond the equivalence The experiment corresponding to reaction
point, the point where the reductant and (7.15), can also be observed if zinc rod is
the oxidant are equal in terms of their dipped in copper sulphate solution. The
mole stoichiometry. redox reaction takes place and during the
reaction, zinc is oxidised to zinc ions and are represented as Zn2+/Zn and Cu2+/Cu.
copper ions are reduced to metallic copper In both cases, oxidised form is put before
due to direct transfer of electrons from zinc the reduced form. Now we put the beaker
to copper ion. During this reaction heat is containing copper sulphate solution and the
also evolved. Now we modify the experiment beaker containing zinc sulphate solution
in such a manner that for the same redox side by side (Fig. 7.3). We connect solutions
reaction transfer of electrons takes place in two beakers by a salt bridge (a U-tube
indirectly. This necessitates the separation containing a solution of potassium chloride
of zinc metal from copper sulphate solution. or ammonium nitrate usually solidified by
We take copper sulphate solution in a beaker boiling with agar agar and later cooling to a
and put a copper strip or rod in it. We also jelly like substance). This provides an electric
take zinc sulphate solution in another beaker contact between the two solutions without
and put a zinc rod or strip in it. Now reaction allowing them to mix with each other. The
takes place in either of the beakers and at the zinc and copper rods are connected by a
interface of the metal and its salt solution in metallic wire with a provision for an ammeter
each beaker both the reduced and oxidized and a switch. The set-up as shown in Fig.7.3
forms of the same species are present. These is known as Daniell cell. When the switch is
represent the species in the reduction and in the off position, no reaction takes place in
oxidation half reactions. A redox couple is either of the beakers and no current flows
defined as having together the oxidised and through the metallic wire. As soon as the
reduced forms of a substance taking part in switch is in the on position, we make the
an oxidation or reduction half reaction. following observations:
This is represented by separating the 1. The transfer of electrons now does not
oxidised form from the reduced form by take place directly from Zn to Cu2+ but
a vertical line or a slash representing an through the metallic wire connecting the
interface (e.g. solid/solution). For example two rods as is apparent from the arrow
in this experiment the two redox couples which indicates the flow of current.
2. The electricity from solution in one beaker
to solution in the other beaker flows by
Fig.7.3 The set-up for Daniell cell. Electrons
produced at the anode due to oxidation the migration of ions through the salt
of Zn travel through the external circuit bridge. We know that the flow of current
to the cathode where these reduce the is possible only if there is a potential
copper ions. The circuit is completed difference between the copper and zinc
inside the cell by the migration of ions rods known as electrodes here.
through the salt bridge. It may be noted
that the direction of current is opposite to
The potential associated with each
the direction of electron flow. electrode is known as electrode potential.
If the concentration of each species taking
part in the electrode reaction is unity (if any
gas appears in the electrode reaction, it is
confined to 1 atmospheric pressure) and
further the reaction is carried out at 298K,
then the potential of each electrode is said
to be the Standard Electrode Potential. By
convention, the standard electrode potential
(Eg) of hydrogen electrode is 0.00 volts. The
electrode potential value for each electrode
process is a measure of the relative tendency
of the active species in the process to remain
in the oxidised/reduced form. A negative Eg
means that the redox couple is a stronger
REDOX REACTIONS 251

reducing agent than the H+/H couple. A information from them. The values of standard
positive Eg means that the redox couple is a electrode potentials for some selected electrode
weaker reducing agent than the H+/H couple. processes (reduction reactions) are given in
The standard electrode potentials are very Table 7.1. You will learn more about electrode
important and we can get a lot of other useful reactions and cells in Class XII.
.
Redox reactions form an important class of reactions in which oxidation and reduction
occur simultaneously. Three tier conceptualisation viz, classical, electronic and oxidation
number, which is usually available in the texts, has been presented in detail. Oxidation,
reduction, oxidising agent (oxidant) and reducing agent (reductant) have been viewed
according to each conceptualisation. Oxidation numbers are assigned in accordance with a
consistent set of rules. Oxidation number and ion-electron method both are useful means in
writing equations for the redox reactions. Redox reactions are classified into four categories:
combination, decomposition displacement and disproportionation reactions. The concept of
redox couple and electrode processes is introduced here. The redox reactions find wide
applications in the study of electrode processes and cells.

7.1 Assign oxidation number to the underlined elements in each of the following species:
(a) NaH2PO4 (b) NaHSO4 (c) H4P2O7 (d) K2MnO4
(e) CaO2 (f) NaBH4 (g) H2S2O7 (h) KAl(SO4)2.12 H2O
7.2 What are the oxidation number of the underlined elements in each of the following
and how do you rationalise your results ?
(a) KI3 (b) H2S4O6 (c) Fe3O4 (d) CH3CH2OH (e) CH3COOH
7.3 Justify that the following reactions are redox reactions:
(a) CuO(s) + H (g)  Cu(s) + H O(g)
(b) Fe O (s) + 3CO(g)  2Fe(s) + 3CO (g)
(c) 4BCl (g) + 3LiAlH (s)  2B H (g) + 3LiCl(s) + 3 AlCl (s)
(d) 2K(s) + F (g)  2K+F– (s)
(e) 4 NH (g) + 5 O (g)  4NO(g) + 6H O(g)
7.4 Fluorine reacts with ice and results in the change:
H O(s) + F (g)  HF(g) + HOF(g)
Justify that this reaction is a redox reaction.
7.5 Calculate the oxidation number of sulphur, chromium and nitrogen in H SO , Cr O 2–
2 5 2 7
and NO–. Suggest structure of these compounds. Count for the fallacy.
7.6 Write formulas for the following compounds:
(a) Mercury(II) chloride (b) Nickel(II) sulphate
(c) Tin(IV) oxide (d) Thallium(I) sulphate
(e) Iron(III) sulphate ( f ) Chromium(III) oxide
7.7 Suggest a list of the substances where carbon can exhibit oxidation states from –4
to +4 and nitrogen from –3 to +5.
7.8 While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing
agents in their reactions, ozone and nitric acid act only as oxidants. Why ?
7.9 Consider the reactions:
(a) 6 CO (g) + 6H O(l)  C H O (aq) + 6O (g)
2 2 6 12 6 2
REDOX REACTIONS 253

(b) O (g) + H O (l)  H O(l) + 2O (g)

Why it is more appropriate to write these reactions as :
(a) 6CO (g) + 12H O(l)  C H O (aq) + 6H O(l) + 6O (g)
(b) O (g) + H O (l)  H O(l) + O (g) + O (g)
Also suggest a technique to investigate the path of the above (a) and (b) redox
reactions.
7.10 The compound AgF2 is unstable compound. However, if formed, the compound acts
as a very strong oxidising agent. Why ?
7.11 Whenever a reaction between an oxidising agent and a reducing agent is carried out,
a compound of lower oxidation state is formed if the reducing agent is in excess and
a compound of higher oxidation state is formed if the oxidising agent is in excess.
Justify this statement giving three illustrations.
7.12 How do you count for the following observations ?
(a) Though alkaline potassium permanganate and acidic potassium permanganate
both are used as oxidants, yet in the manufacture of benzoic acid from toluene
we use alcoholic potassium permanganate as an oxidant. Why ? Write a balanced
redox equation for the reaction.
(b) When concentrated sulphuric acid is added to an inorganic mixture containing
chloride, we get colourless pungent smelling gas HCl, but if the mixture contains
bromide then we get red vapour of bromine. Why ?
7.13 Identify the substance oxidised reduced, oxidising agent and reducing agent for
each of the following reactions:
(a) 2AgBr (s) + C6H6O2(aq) 2Ag(s) + 2HBr (aq) + C6H4O2(aq)
(b) HCHO(l) + 2[Ag (NH ) ]+(aq) + 3OH–(aq) 2Ag(s) + HCOO–(aq) + 4NH (aq)
+ 2H2O(l)
(c) HCHO (l) + 2 Cu2+(aq) + 5 OH–(aq) Cu2O(s) + HCOO–(aq) + 3H O(l)
(d) N H (l) + 2H O (l)  N (g) + 4H O(l)
(e) Pb(s) + PbO (s) + 2H SO (aq)  2PbSO (s) + 2H O(l)
7.14 Consider the reactions :
2 S O2– (aq) + I (s)  S O2–(aq) + 2I–(aq)
2 3 2 4 6
S O2–(aq) + 2Br (l) + 5 H O(l)  2SO2–(aq) + 4Br–(aq) + 10H+(aq)
2 3 2 2 4
Why does the same reductant, thiosulphate react differently with iodine and
bromine ?
7.15 Justify giving reactions that among halogens, fluorine is the best oxidant and among
hydrohalic compounds, hydroiodic acid is the best reductant.
7.16 Why does the following reaction occur ?
XeO 4– (aq) + 2F– (aq) + 6H+(aq)  XeO (g)+ F (g) + 3H O(l)
What conclusion about the compound Na XeO (of which XeO4– is a part) can be
4 6 6
drawn from the reaction.
7.17 Consider the reactions:
(a) H3PO2(aq) + 4 AgNO3(aq) + 2 H2O(l)  H3PO4(aq) + 4Ag(s) + 4HNO3(aq)
(b) H3PO2(aq) + 2CuSO4(aq) + 2 H2O(l)  H3PO4(aq) + 2Cu(s) + H2SO4(aq)
(c) C6H5CHO(l) + 2[Ag (NH3)2]+(aq) + 3OH–(aq)  C6H5COO–(aq) + 2Ag(s) +
4NH3 (aq) + 2 H2O(l)
(d) C6H5CHO(l) + 2Cu2+(aq) + 5OH–(aq)  No change observed.
254 CHEMISTRY

What inference do you draw about the behaviour of Ag+ and Cu2+ from these
reactions ?
7.18 Balance the following redox reactions by ion – electron method :
(a) MnO – (aq) + I– (aq)  MnO (s) + I (s) (in basic medium)
(b) MnO – (aq) + SO (g)  Mn2+ (aq) + HSO –
(aq) (in acidic solution)
(c) H O (aq) + Fe2+ (aq)  Fe3+ (aq) + H O (l) (in acidic solution)
(d) Cr O 2– + SO (g)  Cr3+ (aq) + SO 2–
(aq) (in acidic solution)
2 7 2 4

7.19 Balance the following equations in basic medium by ion-electron method and
oxidation number methods and identify the oxidising agent and the reducing agent.
(a) P (s) + OH–(aq)  PH (g) + HPO– (aq)
4 3 2

(b) N H (l) + ClO–(aq)  NO(g) + Cl–(g)

2 4 3

(c) Cl O (g) + H O (aq)  ClO–(aq) + O (g) + H+

2 7 2 2 2 2
7.20 What sorts of informations can you draw from the following reaction ?
(CN)2(g) + 2OH–(aq)  CN–(aq) + CNO–(aq) + H2O(l)
7.21 The Mn3+ ion is unstable in solution and undergoes disproportionation to give Mn2+,
MnO2, and H+ ion. Write a balanced ionic equation for the reaction.
7.22 Consider the elements :
Cs, Ne, I and F
(a) Identify the element that exhibits only negative oxidation state.
(b) Identify the element that exhibits only postive oxidation state.
(c) Identify the element that exhibits both positive and negative oxidation states.
(d) Identify the element which exhibits neither the negative nor does the positive
oxidation state.
7.23 Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess
of chlorine is removed by treating with sulphur dioxide. Present a balanced equation
for this redox change taking place in water.
7.24 Refer to the periodic table given in your book and now answer the following questions:
(a) Select the possible non metals that can show disproportionation reaction.
(b) Select three metals that can show disproportionation reaction.
7.25 In Ostwald’s process for the manufacture of nitric acid, the first step involves the
oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is
the maximum weight of nitric oxide that can be obtained starting only with 10.00
g. of ammonia and 20.00 g of oxygen ?
7.26 Using the standard electrode potentials given in the Table 8.1, predict if the reaction
between the following is feasible:
(a) Fe3+(aq) and I–(aq)
(b) Ag+(aq) and Cu(s)
(c) Fe3+ (aq) and Cu(s)
(d) Ag(s) and Fe3+(aq)
(e) Br2(aq) and Fe2+(aq).
REDOX REACTIONS 255

7.27 Predict the products of electrolysis in each of the following:

(i) An aqueous solution of AgNO3 with silver electrodes
(ii) An aqueous solution AgNO3 with platinum electrodes
(iii) A dilute solution of H2SO4 with platinum electrodes
(iv) An aqueous solution of CuCl2 with platinum electrodes.
7.28 Arrange the following metals in the order in which they displace each other from
the solution of their salts.
Al, Cu, Fe, Mg and Zn.
7.29 Given the standard electrode potentials,
K+/K = –2.93V, Ag+/Ag = 0.80V,
Hg2+/Hg = 0.79V
Mg2+/Mg = –2.37V. Cr3+/Cr = –0.74V
arrange these metals in their increasing order of reducing power.
7.30 Depict the galvanic cell in which the reaction Zn(s) + 2Ag+(aq)  Zn2+(aq) +2Ag(s)
takes place, Further show:
(i) which of the electrode is negatively charged,
(ii) the carriers of the current in the cell, and
(iii) individual reaction at each electrode.