CHAPTER
09 Redox Reactions
1. Definition
Chemical reactions involve the transfer of electrons from one chemical substance to another. These electrons–transfer reactions are
termed as oxidation-reduction or redox-reactions.
2. Oxidation & Reduction
Oxidation Reduction
Oxidation is a process which involves; the addition of oxygen, Reduction is just reverse of oxidation. Reduction is a process
removal of hydrogen, the addition of non-metal, removal of which involves; removal of oxygen, the addition of hydrogen,
metal, Increase in +ve valency, loss of electrons and increase in removal of non-metal, addition of metal, a decrease in +ve
oxidation number. valency, a gain of electrons and decrease in oxidation number.
(1) Addition of oxygen : 2Mg + O2 → 2MgO (1) Removal of oxygen : CuO + C → Cu + CO
(2) Removal of hydrogen : H2S+Cl2 → 2HCl + S (2) Addition of hydrogen : Cl2 + H2 → 2HCl
(3) Addition of Non-metal : Fe + S → FeS (3) Removal of non-metal:
(4) Removal of metal: 2KI+H2O2 → 2KOH+I2 2HgCl2 + SnCl2 → Hg2Cl2 + SnCl4
(5) Increase in +ve valency : Fe2+ →Fe3+ + e− [Reduction of mercuric chloride]
(6) Loss of electrons: MnO42− → MnO4− + e− (4) Addition of metal : HgCl2 + Hg → Hg2Cl2
(7) Increase in oxidation number Mg 0 → Mg 2+ + 2e − (From 0 (5) A decrease in +ve valency:
to +2) [Fe (CN)6 ]3− → [Fe(CN)6 ]4−
(6) A gain of electrons: Zn2+ (aq) + 2e − → Zn(S)
(7) A decrease in oxidation number: Cl20 → 2Cl − − 2e−
(From 0 to –1)
3. Oxidation number or Oxidation state
3.1 Definition
Charge on an atom produced by donating or accepting electrons is called oxidation number or oxidation state. It is the
number of effective charges on an atom.
3.2 Valency and Oxidation Number
Oxidation number Valency
O.N. is the charge (real or imaginary) present on the atom It is the combining capacity of the element. No plus or
of the element when it is in combination. It may have plus minus sign is attached to it.
or minus sign.
O.N. of an element may have different values. It depends Valency of an element is usually fixed.
on the nature of a compound in which it is present.
O.N. of the element may be a whole number or fractional. Valency is always a whole number.
O.N. of the element may be zero. Valency of the element is never zero except for noble gases.
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Group Outer shell configuration Common oxidation numbers (states)except zero in a free state
1
IA ns +1
II A ns2 +2
III A ns2np1 +3, +1
IV A ns2np2 +4,+3,+2,+1, – 1, – 2, – 3, – 4
VA ns2np3 +5,+3,+1, –1, – 3
2 4
VI A ns np +6,+4,+2,– 2
VII A ns2np5 +7,+5,+3, +1, – 1
3.3 Oxidation Number and Nomenclature
(1) When an element forms two types of mono-atomic cations, the suffix – ous is used for the cation with lower oxidation
state and the suffix – ic is used for the cation with higher oxidation state.
(2) Albert Stock proposed a new system known as Stock system. In this system, Roman numeral written in parentheses
immediately after the name of the element indicates the oxidation states
3.4 Rules for the Determination of Oxidation Number of an Atom
(1) If there is a covalent bond between two same atoms then oxidation numbers of these two atoms will be zero
(2) If the covalent bond is between two different atoms then electrons are counted towards a more electronegative atom.
Thus oxidation number of a more electronegative atom is negative and the oxidation number of a less electronegative
atom is positive
(3) If there is a coordinate bond between two atoms then oxidation number of donor atom will be + 2 and of acceptor
atom will be – 2.
(4) The oxidation number of all the atoms of different elements in their respective elementary states is taken to be zero.
(5) The oxidation number of a mono-atomic ion is the same as the charge on it.
(6) The oxidation number of hydrogen is + 1 when combined with non-metals and is –1 when combined with active
metals called metal hydrides
(7) The oxidation number of oxygen is – 2 in most of its compounds, except in peroxides like H2O2 , BaO2 etc. it is -1 and
in superoxides like KO2 it is -1/2.
(8) In compounds formed by the union of metals with non-metals, the metal atoms will have positive oxidation numbers
and the non-metals will have negative oxidation numbers.
(9) The oxidation number of alkali metals is always +1 and those of alkaline earth metals is + 2.
(10) The oxidation number of halogens is always –1 in metal halides such as KF, AlCl3, MgBr2, CdI2. etc.
(11) In interhalogen compounds of Cl, Br, and I; the more electronegative of the two halogens gets the oxidation number of
–1. For example, in BrCl3, the oxidation number of Cl is –1 while that of Br is +3.
(12) For a neutral molecule, the sum of the oxidation numbers of all the atoms is equal to zero
(13) In the case of representative elements, the highest oxidation number of an element is the same as its group number
while highest negative oxidation number is equal to (8 – Group number) with a negative sign with a few exceptions.
(14) Transition metals exhibit a large number of oxidation states due to the involvement of (n –1) d electron besides ns
electron.
(15) The oxidation number of a metal in carbonyl complex is always zero. E.g. Ni has zero oxidation state in Ni ( CO)4 .
3.5 Procedure for Calculation of Oxidation Number
(1) Write down the formula of the given molecule/ion leaving some space between the atoms.
(2) Write oxidation number on the top of each atom. In case of the atom whose oxidation number has to be calculated
write x.
(3) Beneath the formula, write down the total oxidation numbers of each element. For this purpose, multiply the oxidation
numbers of each atom with the number of atoms of that kind in the molecule/ion. Write the product in a bracket.
(4) Equate the sum of the oxidation numbers to zero for neutral molecule and equal to charge for the ion.
(5) Solve for the value of x.
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3.6 Exceptional Cases of Evaluation of Oxidation Numbers
Type I :
Oxidation number of S in H2SO5 Oxidation number of Cr in CrO5
By the usual method; The chemical structure of CrO5 is
H2SO5 2 1 + x + 5 (−2) = 0 or x = + 8 O O
Peroxide
Cr Peroxide
But this cannot be true as maximum oxidation number linkage
linkage
for S cannot exceed + 6. Since S has only 6 electrons O O
in its valence shell. This exceptional value is due to the O
fact that two oxygen atoms in H2SO5 shows peroxide
Therefore, the evaluation of O.N. of Cr should be made as
linkage. Therefore the evaluation of O.N. of sulphur
here should be made as follows, follows
Peroxide linkage x + 1 × (– 2) + 4 (–1) = 0 (for Cr) (for O) (for O–O) or x – 2 –
O
H O S O O H 4 = 0 or x = + 6.
O
2×(+1)+ x + 3 × (–2) + 2 × (–1) (for H) (for S) (for
O) (for O–O In peroxide linkage) or 2 + x – 6 – 2 = 0
or x = + 6.
Type II :
Oxidation number of carbon in H – N = C
→ Oxidation number of carbon in HC N is +2
The contribution of the coordinate bond is neglected since the In HC N , N is more electronegative than carbon,
bond is directed from a more electronegative N atom (donor) each bond gives an oxidation number of –1 to N. There
to a less electronegative carbon atom (acceptor). are three covalent bonds, the oxidation number of N in
HC N is taken as – 3
→
Therefore the oxidation number of N in HN = C remains – 3 as
Now HC N
it has three covalent bonds.
+1 + x – 3 = 0 x = + 2
1 × (+ 1) + 1 × (– 3) + x = 0 (for H) (for N) (for C)
or 1 + x – 3 = 0 or x = + 2.
Type III :
Oxidation number of S in Na2S2O3 Oxidation Oxidation number of N in Oxidation number
number of NH4NO3 of Fe in Fe3O4
chlorine in
CaOCl2
The oxidation number of sulphur is In bleaching No doubt NH4NO3 has two In Fe3O4, Fe atoms are
evaluated from concepts of chemical powder, nitrogen atoms but one N has in two different
bonding. The chemical structure of Na2S2O3 Ca(OCl)Cl, the negative oxidation number oxidation states. Fe3O4
is two Cl atoms (attached to H) and the other can be considered as
are in different has positive oxidation number an equimolar mixture
oxidation of FeO [iron (II) oxide]
(attached to O). Hence the
states i.e., one and Fe2O3 [iron (III)
evaluation should be made
Cl– having oxide]. Thus in one
oxidation separately for NH4+ and molecule of Fe3O4,
Due to the presence of a coordinate bond number of –1 two Fe atoms are in +
NO3− NH4+
between two sulphur atoms, the acceptor and the other 3 oxidation state and
sulphur atom has oxidation number of – 2 as OCl– having x + 4 × (+1) = +1 or one Fe atom is in + 2
whereas the other S atom gets oxidation oxidation oxidation state.
number of number of +1 x = – 3 NO3−
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+ 6. x+3 (– 2)= –1 or x = + 5.
2 × (+1) + 3 × (–2) + x × 1 + 1 × (– 2)
=0
or + 2 – 6 + x – 2 = 0 or x = + 6
Thus two sulphur atoms in Na2S2O3 have
oxidation number of – 2 and +6.
4. Redox - Reactions
4.1 Definition
An overall reaction in which oxidation and reduction take place simultaneously is called redox or oxidation-reduction
reaction. These reactions involve the transfer of electrons from one atom to another. Thus every redox reaction is made up of
two half-reactions; one-half reaction represents the oxidation and the other half-reaction represents the reduction.
4.2 Types of Redox Reaction
(1) Direct Redox Reaction : The reactions in which oxidation and reduction take place in the same vessel are called
direct redox reactions.
(2) Indirect Redox Reaction : The reactions in which oxidation and reduction take place in different vessels are called
indirect redox reactions. Indirect redox reactions are the basis of electrochemical cells.
(3) Intermolecular Redox Reactions : In which one substance is oxidized while the other is reduced.
(4) Intramolecular Redox Reactions : In which one element of a compound is oxidized while the other is reduced.
(5) Disproportionation : One and the same substance may act simultaneously as an oxidizing agent and as a reducing
agent with the result that a part of it gets oxidized to a higher state and rest of it is reduced to a lower state of oxidation.
Such a reaction, in which a substance undergoes simultaneous oxidation and reduction is called disproportionation and
the substance is said to disproportionate.
(6) Spectator Ions : In ionic equations, the ions which do not undergo any change and are equal in number in both
reactants and products are termed as spectator ions and are not included in the final balanced equations. Example,
Zn + 2H+ + 2Cl − → Zn2+ + H2 + 2Cl − (Ionic equation), Zn + 2H + → Zn2+ + H2 (Final ionic equation)
In the above example, the CI– ions are the spectator ions and hence are not included in the final ionic balanced
equation.
5. Balancing of Redox Reactions
Oxidation Number Method Ion-Electron Method (half reaction method)
(1) Write the skeleton equation (if not given, frame it) (1) Write down the redox reaction in ionic form.
representing the chemical change.
(2) Split the redox reaction into two half-reactions, one for
(2) Assign oxidation numbers to the atoms in the equation oxidation and other for reduction.
and find out which atoms are undergoing oxidation and
reduction. Write separate equations for the atoms (3) Balance each half-reaction for the number of atoms of
undergoing oxidation and reduction. each element. For this purpose,
(3) Find the change in oxidation number in each equation. (i) Balance the atoms other than H and O for each half-
Make the change equal in both the equations by reaction using simple multiples.
multiplying with suitable integers. Add both the
(ii) Add water molecules to the side deficient in oxygen
equations.
and H+ to the side deficient in hydrogen. This is
(4) Complete the balancing by inspection. First balance done in acidic or neutral solutions.
those substances which have undergone a change in
(iii) In alkaline solution, for each excess of oxygen, add
oxidation number and then other atoms except –
hydrogen and oxygen. Finally, balance hydrogen and one water molecule to the same side and 2OH ions
oxygen by putting H2O molecules wherever needed. to the other side. If hydrogen is still unbalanced, add
–
one OH ion for each excess hydrogen on the same
side and one water molecule to the other side.
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(5) The final balanced equation should be checked to (4) Add electrons to the side deficient in electrons as to
ensure that there are as many atoms of each element on equalize the charge on both sides.
the right as there are on the left.
(5) Multiply one or both the half reactions by a suitable
(6) In ionic equations, the net charges on both sides of the number so that the number of electrons becomes equal in
equation must be exactly the same. Use H+ ion/ions in both the equations.
acidic reactions and OH– ion/ions in basic reactions to
balance the charge and number of hydrogen and (6) Add the two balanced half-reactions and cancel any term
oxygen atoms common to both sides.
6. Oxidising and Reducing Agents
6.1 Definition
The substance (atom, ion or molecule) that gains electrons and is thereby reduced to a low valency state is called an oxidant
or oxidizing agent, while the substance that loses electrons and is thereby oxidized to a higher valency state is called a
reductant or reducing agent.
6.2 Important Oxidising Agents
(1) Molecules made up of electronegative elements. Example: O2, O3, and X2 (halogens).
(2) Compounds containing an element which is in the highest oxidation state. Example : KMnO4 , K2Cr2O7 , Na2Cr2O7
(3) Oxides of elements MgO, CuO, CrO3 , CO2 , P4 O10 , etc.
(4) Fluorine is the strongest oxidizing agent.
6.3 Important Reducing Agents
(1) All metals e.g. Na, Zn, Fe, Al, etc.
(2) A few non-metals e.g. C, H2, S etc.
(3) Hydracids: HCl, HBr, HI, H2S etc.
(4) Metallic hydrides e.g. NaH, LiH etc.
(5) Organic compounds like HCOOH and (COOH)2 and their salts, aldehydes, alkanes etc.
(6) Lithium is the strongest reducing agent in solution.
(7) Cesium is the strongest reducing agent in absence of water. Other reducing agents are Na2S2O3 and KI.
(8) Hypo prefix indicates that the central atom of the compound has the minimum oxidation state so it will act as a
reducing agent. Eg : H3 PO2
6.4 Equivalent Weight Of Oxidising And Reducing Agents
(1) Equivalent weight of a substance is equal to molecular weight divided by a number of electrons lost or gained by one
molecule of the substance in a redox reaction.
Molecular weight Molecular weight
(2) Eq. wt. of O. A. = =
No.of electrons gained by one molecule Change in O. N. per mole
Molecular weight Molecular weight
(3) Eq. wt. of R. A. = =
No.of electrons lost by one molecule Change in O. N. per mole
7. Autoxidation
7.1 Autoxidation
(1) Turpentine and numerous other olefinic compounds, phosphorus and certain metals like Zn and Pb can absorb oxygen
from the air in presence of water. The water is oxidised to hydrogen peroxide. This phenomenon of formation of H2O2
by the oxidation of H2O is known as autoxidation.
(2) The substance which can activate the oxygen is called activator. The activator is supposed to first combine with oxygen
to form an addition compound, which acts as an autoxidator and reacts with water or some other acceptor so as to
oxidise the latter.
7.2 Induced Oxidation
(1) The concept of autoxidation helps to explain the phenomenon of induced oxidation. Na2SO3 is oxidised by air but
Na3AsO3 is not oxidised by air. If a mixture of both is taken, it is observed both are oxidised
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8. Molecular and Ionic Equations
Molecular Equations Ionic Equations
A molecular equation is an equation in which the formulas of the This equation is called an ionic equation, an
compounds are written as though all substances exist as molecules. equation in which dissolved ionic compounds are
However, there is a better way to show what is happening in this reaction. shown as free ions.
All of the aqueous compounds should be written as ions because they are
present in the water as separated ions because of their dissociation. Na+(aq) + Cl–(aq) + Ag+(aq) → Na+(aq) + No3–
(aq) + AgCl(s)
NaCl(aq) + AgNO3(aq) → NaNO3(aq) + AgCl(s)
