OXIDATION AND REDUCTION:
A chemical change that takes place as a result of electron transfer is called a
or oxidation –reduction reaction.
redox
The chemical reactions in which the oxidation number of atoms get changed are termed
as redox reactions.
OXIDATION
It is a process which involves
REDUCTION
It is a process which involves
Gain
Gain
i)
ii)
gain of oxygen
gain of
electronegative
radical
iii)
iv)
loss of hydrogen
loss of electropositive
radical
loss of electron
i)
ii)
Loss
iii)
v)
gain of hydrogen
gain of
electropositive
radical
gain of electron
Loss
iv)
v)
loss of oxygen
loss of
electronegative
radical
ILLUSTRATION:
Gain and Loss of Oxygen: i) Burning of hydrogen in oxygen
2H2 + O2 →
2H2O
The hydrogen is oxidized and the oxygen is reduced
ii) The combination of nitrogen and oxygen which occurs at high
temperatures follows the same pattern.
N2 + O2 → 2 NO
Here, nitrogen is oxidize and oxygen is reduced
c) Combustion of methane,
CH4 + 2O2 → CO2 + 2H2O
Both carbon and hydrogen are oxidized (gain oxygen) and oxygen is
reduced.
OXIDIZING AND REDUCING AGENTS:
OXIDIZING AGENTS:
Substances that oxidize other substances are known as oxidizing agents. , oxidants or
oxidizers. Oxidants are usually chemical substances with elements in high oxidation
numbers (e.g., H2O2,MnO4-,CrO3, Cr2O72-,etc.) or highly electronegative substances that
can gain one or two extra electrons by oxidizing a substance (e.g.O2,F2,Cl2,Br2 etc.)
REDUCING AGENTS:
Substances that reduce other substances are known as reducing agents ,reductants, or
reducers. Reductants are electropositive elemental metals like (Li, Na, Mg, Fe, Zn,
Al)or hidirdies like NaBH4, LiAlH4 that are widely used in the reduction of carbony
compouds to alcohols.Also hydrogen gas, H2 with palladium, platinum,nickel as catqlyst
 The chemical which is oxidised is the reducing agent.
The chemical which is reduced is the oxidising agent.
 Oxidation and reduction always occur side by side that is simultaneously.
OXIDATION NUMBER OR OXIDATION STATE:
“Oxidation number or oxidation state of an element is the charge which it has or appears
to have in the combined state.”
The determination of oxidation number or oxidation state of an atom is based on
certain rules.
Rules for determining oxidation state:
1. All elements in their elemental state have zero oxidation number Diatomic gases such
as O2 and H2 are also in this category.
2. Hydrogen atom in combined state has +1 oxidation number except in case of hydrides
where its oxidation number is -1
3. Oxygen atom in combined state has -2 oxidation number except in peroxides where it
has -1 oxidation number.
4. Halogens have -1 oxidation number in their combined state. In case of interhalogens
the more electronegative atom will possess negative oxidation number whereas the other
will have positive oxidation number.
4 Alkali metals have +1 oxidation number in combined state and alkaline earth metals
have +2 oxidation number..
6. In radicals or small covalent molecules, the element with the greatest electronegativity
has its natural ion charge as its oxidation state.
7. The sum of oxidation numbers of all atoms in a neutral compound is taken as zero.
8. In case of complex ion the sum of oxidation numbers of all atoms present in is equal to
the net charge present on it.
OXIDATION AND REDUCTION IN TERMS OF OXIDATION STATE:
A raise (increase) of oxidation state is an oxidation whereas a lower (decrease)
of oxidation state is a reduction.
BALANCING OF A CHEMICAL EQUATION:
Balancing of a chemical equation:
Chemical equations can be balanced using oxidation numbers of the species
involved. The following steppes are used
1. Write the skeleton equation:
2. Assign oxidation numbers to all atoms
3. Make the number of atoms that change oxidation number the same on both
sides by inserting temporary coefficient.
4. Identify oxidized and reduced species
5. Determine the number of electrons lost (oxidation) and gained (reduction)
6. Balance the oxidation number by adding electrons to whatever side required.
7. For reactions in acidic or basic solution, add H+/H2O (acidic) or OH-/H2O
(basic) to balance the charges.
8. Multiply these by appropriate factors make electrons lost = electrons gained
9. Complete the equation by final inspection.
Balancing Equations using Oxidation Numbers
e.g.1 Balance the following reaction
Cu (s) + Ag+(aq) →
Cu2+(aq) +
Ag (s)
Following steps
Initial Ox. No.
0
+1
Final Ox. No.
+2
0
Loss of 2 electrons (Oxidation)
Cu(s)
+ Ag+(aq)
→
Cu2+(aq)
+ Ag(s)
Gain of 1 electron (reduction)
It is clear that the number of electrons lost by copper is not equal the number gained by
silver. Therefore multiply Ag by 2, giving us a total of two silvers, and multiply copper
by one.
1 Cu(s) + 2Ag+(aq) → 1 Cu2+(aq) + 2Ag(s)
It is not necessary to put the "1" in front of copper.
No. of electrons lost by one Cu atom =2
No. of electrons gained by two Ag atoms= 2
Thus the no. of electrons gained = No. of the electrons gained
Therefore the balanced equation is
Cu(s) + 2Ag+(aq) → 1 Cu2+(aq) + 2Ag(s)
 The multipliers to balance electrons become the balancing coefficients in
the equation. Thus, we should put a "1" in front of copper and a "2" in
front of silver. However, it is not necessary to put the "1" in front of
copper.
e.g.2
Balance the following reaction:
KIO3 + KI + H2SO4 -> K2SO4 + H2O + I2
K+ + IO3- + K+ + I- +2H+ + SO42- → 2K+ + SO42- + H2O + I2
However the net ionic reaction may be written as
IO3- + I- + H+ → H2O + I2
Find the oxidtion number of each atom in the equation
Decrease in ox. no. ( reduction)
Ox. No.
+ 5 -2
-1
+1
IO3-
+ I-
+ H+
+1 -2
→
H2O
0
+
I2
Increase in ox. no. ( Oxidation)
Decrease in oxidation no.= 5 by I in IO3Increase in oxidation no =1 by I in INow multiply I- by 5 and IO3- by 1, we get
IO3-
+ 5I-
+ H+
→
H2O
+
I2
There are six atoms of iodine on the left, so we need three I2 molecules to balance iodine:
1IO3-
+ 5I-
H+
+
→
H2O
+
3I2
Fianally, balancing oxygen, hydrogen and water:
IO3-
+
5I-
+
6H+
→ 3H2O
+
3I2
This is the balanced equation.
Balance the following equation:
MnO4-(aq) + Cl-(aq) + H+(aq) → Mn2+(aq) + Cl2(aq) + H2O(l)
Assign oxidation numbers (o.n.) to atoms.
Increase in ox. no. (Oxidation)
Ox.no.
+7 -2
-1
+1
+2
0
+1 -2
MnO4-(aq) + Cl-(aq) + H+(aq) → Mn2+(aq) + Cl2(aq) + H2O(l)
Decrease in ox. no. (Reduction)
No. of electrons gained = 5 by Mn in MnO4No. of electrons lost
= 1 by Cl in Cl-
Multiply MnO4- by 1 and Cl- by 5 , we have
MnO4-(aq) + 5Cl-(aq) + H+(aq) → Mn2+(aq) + Cl2(aq) + H2O(l)
Balance chlorine atoms on both sides
MnO4-(aq) + 10Cl-(aq) + H+(aq) → Mn2+(aq) + 5Cl2(aq) + H2O(l)
Balance O and H atoms on both sides
MnO4-(aq) + 10Cl-(aq) + 8 H+(aq) → Mn2+(aq) + 5Cl2(aq) + 4H2O(l)
To make
eletrone gained = electrons lot
Multiply MnO4- and Mn2+ each by 2 ,thus we have finally the required balanced
equation
2 MnO4-(aq) + 10Cl-(aq) + 8 H+(aq) → 2Mn2+(aq) + 5Cl2(aq) + 4H2O(l)
Alternate methood half equation method
Balance the following equation:
MnO4-(aq) + Cl-(aq) + H+(aq) → Mn2+(aq) + Cl2(aq) + H2O(l)
Assign oxidation numbers (o.n.) to atoms.
Increase in ox. no. (Oxidation)
Ox.no.
+7 -2
-1
+1
+2
0
+1 -2
MnO4-(aq) + Cl-(aq) + H+(aq) → Mn2+(aq) + Cl2(aq) + H2O(l)
Decrease in ox. no. (Reduction)
No. of electrons gained = 5 by Mn in MnO4-
No. of electrons lost
= 1 by Cl in Cl-
Write two half equations i.e. one for oxidation half equation and other
reduction half equation:
Oxidation half reaction
Reduction half reaction
→
Cl-
→
MnO4-
Cl2
Mn2+
Balancig atoms other than O and H
2Cl-
→
MnO4-
Cl2
→
Mn2+
Balancing oxidation no by adding to Balancing oxidation no by adding to
whatever side required
whatever side required
2Cl-
→
MnO4- + 5e-
Cl2 + 2e-
→
Mn2+
+
Balancing of charge by adding H+ ions Balancing of charge by adding H ions
to whaever side required ( being acidic to whaever side required ( being acidic
medium)
medium)
2Cl-
→
MnO4- + 5e- + 8H+ →
Cl2 + 2e-
Mn2+
Balancing of O atoms on both sides by Balancing of O atoms on both sides by
adding H2O to whatever side required.
adding H2O to whatever side required.
2Cl-
→
Cl2 + 2e- ---(1)
MnO4- + 5e- + 8H+ → Mn2++ 4H2O
--- (2)
In order to balance the the number of electrons gained and the number of electrons
lost, multiply equation (i) by 5 nd equation by 2 and add the two resultant equations:
2Cl-
→
Cl2 + 2e-
---(1) x 5
MnO4- + 5e- + 8H+ → Mn2++ 4H2O
---(2) x 2
----------------------------------------------------------------------------------------2 MnO4- + 10 Cl- + 16 H+ → 5 Cl2 + 2 Mn2+ + 8 H2O
------------------------------------------------------------------------------------EQUIVALENTS IN REDOX PROCESSES:
Equivalent weight : Equivalent of an element is the weight of the element that combines
with 1 g hydrogen, 8 g oxygen or 35.5 g chlorine.
In ither words; Equivalent weight of a substance is the number of parts by weight of it
(element or compound) that will combine with or replace, directly or indirectly, 1.008
parts by weight of hydrogen, 8.00 parts by weight of oxygen, or 35.5 parts by weight of
chlorine(or the equivalent weight of any other element or compound).
 For all elements, the atomic weight is equal to the equivalent weight times a small
whole number, called the valence of the element. An element can have more than one
valence and therefore more than one equivalent weight.
or
Equivalent weight is the amount of an element that reacts, or is involved in reaction
with, 1 mole of electrons.
or
The equivalent weight of an element or radical is equal to its atomic weight or formula
weight divided by the valence it assumes in compounds.
The unit of equivalent weight is the atomic mass unit i.e. the amount of a substance
in grams numerically equal to the equivalent weight is called a gram equivalent. e.g.
i) Hydrogen has atomic weight 1.008 and always assumes valence 1 in compounds, so its
equivalent weight is 1.008
ii). Oxygen has an atomic weight of 15.9994 and always assumes valence 2 in
compounds, so its equivalent weight is 7.9997.
iii) The sulphate radical (SO 4 2-) has formula weight 96.0636 and always has valence 2 in
compounds, so its equivalent weight is 48.0318.
iv) Some elements exhibit more than one valence in forming compounds and thus have
more than one equivalent weight. Iron (atomic weight 55.845) has an equivalent weight
of 27.9225 in ferrous compounds (valence 2) and 18.615 in ferric compounds (valence 3).
or
Equivalent weight: (For oxidation-reduction and electrolysis)
The equivalent weight is the weight associated with the loss or gain of 6.02  1023
electrons (Avogadro’s number) or 96500 coulombs of electric charge.
This is also the molecular weight divided by the number of electrons lost or gained.
Numerical-1
i) How many equivalents of KMnO4 will be required to react with 60 g of FeSO4 in the
reaction?
5 Fe2+ + MnO4- + 8 H+ ⇄ 5 Fe3+ + Mn2+
+ 4 H2O
Ans.
Following the relationship:
Equivalents of KmnO4 =
Now,
Equivalents of FeSO4
Formula weght of FeSO4
Equivalent weight of FeSO4 = --------------------------------------Oxidation number change of Fe
Equivalent weight of FeSO4
151.9
= ----------1
Equivalent weight of FeSO4
151.9 g FeSO4
= ---------------------------1 equivalent FeSO4
60 g FeSO4
∴ Equivalents FeSO4
=
------------------------------151.9 g/equialent FeSO4
= 0.40 equivalent FeSO4
Hence, 0.40 equivalent of KMnO4 is required.
 [substances react in the ratio of their equivalent weights]
Numerical- 2.
Find the conentration of 10-3 N KMnO4 in gL-1 in the following process.
2 MnO4- + 10 Cl- + 16 H+
Ans.
⇄
2 Mn2+ + 5 Cl2
+ 8 H2O
A 10-3 N KMnO4 solution contains 10-3 g equivalent of KMnO4 per litre.
Change in oxidation number of Mn in the given reaction = 5 , (+7 to +2)
Molecular weight
∴ Eq.Weight of KMnO4 = -------------------Ox.No.Change
158
= -----------5
Now, 1 g equivalent
∴
KMnO4
10-3 g equivalent of KMnO4
=
= 31.6
31.6 g
= 31.6 x 10-3 g
= 0.0316 g L-1
Numerical.3
Calculate the mass of KClO3 required for the preparation of 400 cm3 of
0.20 N KClO3 to be used in the following reaction.
ClO3- + 3 H2SO4 ⇄ Cl- + 3 SO42- + 6 H+
Change in oxidation number of Cl in the given reaction = 6 ( +5 to -1)
Molecular weighr
∴
Eq. weight of KClO3 = -- ---------------------Ox. No. Change
122.5
-------6
Now, 1 g eq.wt (1N) of KClO3 = 20.4 gL-1
=
∴
= 20.4 gL-1
0.2N KClO3 = 20.4 x .2 = 4.08 gL-1
i.e. 1000 ml of 0.2N KClO3 = 4.08 g
∴ 400 ml of 0.2N KClO3 = (4.08x400)/1000 = 1.632 g