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Balancing Redox Reactions

Table of Content

Oxidation Number / State Method For Balancing Redox Reactions

Half-Reaction or Ion-Electron Method For Balancing Redox Reactions

Related Resources

Oxidation Number / State Method For Balancing Redox Reactions

This method is based on the principle that the number of electrons lost in oxidation must be equal to the number of electrons gained in reduction. The steps to be followed are :

Write the equation (if it is not complete, then complete it) representing the chemical changes.

By knowing oxidation numbers of elements, identify which atom(s) is(are) undergoing oxidation and reduction. Write down separate equations for oxidation and reduction.

Add respective electrons on the right of oxidation reaction and on the left of reduction reaction. Care must be taken to ensure that the net charge on both the sides of the equation is same.

Multiply the oxidation and reduction reactions by suitable numbers to make the number of electrons lost in oxidation reactions equal to the number of electrons gained in reduction reactions.

Transfer the coefficient of the oxidizing and reducing agents and their products to the main equation.

By inspection, arrive at the co-efficients of the species not undergoing oxidation or reduction.

Solved Examples

Question 1

Balance the equation :

KMnO4 + H2SO4 + FeSO4 →  K2SO4 + MnSO4 + Fe2(SO4)3 + H2O

Solution:

Mn+7 →  Mn+2 ; Fe+2 →  Fe2+3

Mn+7 + 5e- →  Mn+2 (1)  2;          2Fe+2 →  Fe2+3 + 2e- (2)  5

2Mn+7 + 10e- →  2Mn+2;   10Fe2+2 →  5Fe2+3 + 10e-.

Therefore coefficient of KMnO4 is 2, that of MnSO4 is 2, that of FeSO4 is 10 and that of Fe2(SO4) is 5.

Therefore 2KMnO4 + 10FeSO4 →  2MnSO4 + 5Fe2(SO4)3

Coefficient of K2SO4 should be 1.

2KMnO4 + 10 FeSO4 → K2SO4 + 2MnSO4 + 5Fe2(SO4)3

To balance SO42- so that left hand side must have 8H2SO4, thus H2O also becomes 8H2O.

2KMnO4 + 10 FeSO4 + 8H2SO4 →  K2SO4 + 2Mn(SO4) + 5Fe2(SO4)3 + 8H2O

______________________________________________

Question 2:

Balance the equation :

MnO4- + C2O42–  + H+ →  CO2 + Mn+2 + H2O

Solution:

Mn+7 →  Mn+2 ;  C2 →  C+4

5e- + Mn+7 →  Mn+2     (1) ´2;

C2 →  2C+4 + 2e-   (2) ´5

10e- + 2Mn+7 →  2Mn+2 ;  5C2 →  10C+4 + 10e-

Therefore 2MnO4- + 5C2O4-2 →  10CO2- + 2Mn+2

There must be 8H2O on the right to balance O. Therefore there must be 10H+ to balance H.

2MnO4- + 5C2O4-2 + 16H+ →  10CO2 + 2Mn+2 + 8H2O

Half-Reaction or Ion-Electron Method For Balancing Redox Reactions

This method involves the following steps :

Divide the complete equation into two half reactions, one representing oxidation and the other reduction.

Balance the atoms in each half reaction separately according to the following steps:

First of all balance the atoms other than H and O.

In a reaction taking place in acidic or neutral medium, oxygen atoms are balanced by adding molecules of water to the side deficient in oxygen atoms while hydrogen atoms are balanced by adding H+ ions to the other side deficient in hydrogen atoms. On the other hand, in alkaline medium (OH-), every excess of oxygen atom on one side is balanced by adding one H2O to the same side and 2OH- to the other side. In case hydrogen is still unbalanced, then balance by adding one OH-, for every excess of H atom on the same side as the excess and one H2O on the other side.

Equalize the charge on both sides by adding a suitable number of electrons to the side deficient in negative charge.

Multiply the two half reactions by suitable integers so that the total number of electrons gained in one half reaction is equal to the number of electrons lost in the other half reaction.

Refer to the following Video For Balancing Redox Reactions

Add the two balanced half equations and cancel any term common to both sides.

Solved Example

Question 1:

Balance the redox reaction  Al + NO3- →  Al(OH)4- +NH3 in alkaline medium.

Solution:

Dividing the equation in two half reactions
NO3-  →NH3  (Oxidation Half Reaction)
Al + 4OH- →Al(OH)4- (Reduction Half Reaction)

All the atoms other than O & H are balanced (N and Al in this case which are already balanced).
NO 3- + H2O →NH3 + OH-
Al + 4OH- →Al(OH)4-

Then O and H are balanced using OH- and H2O
NO3- + 6H2O →NH3 + 9OH-
Al + 4 OH- → Al(OH) 4 -

Electrons need to be added in order to balance the charge.
8 e- + NO 3 - + 6 H 2 O →NH 3 + 9 OH -
Al + 4 OH- → Al(OH)4- + 3 e-

Loss and gain of electrons need to be made equal:
For this, we need to multiply first equation by 3 and second equation by 8

Adding both equations we get the balanced redox equation
3NO 3 - + 18H 2 O+ 8Al + 5 OH - → 8Al(OH) 4 - +3NH 3

____________________________________________________

Question 2:

Balance the equation

Cr2O7 2-(g) + SO2(aq) → Cr3+ + SO42-(aq) (in acidic medium)

Solution:

Dividing the equation in two half reactions
SO2 →SO4 2-  (Oxidation half reaction)
Cr2O7 2-→Cr3+  (Reduction half reaction)

Balance the atoms other than O and H
SO2 →SO42-
Cr2O7 2-→2Cr3+

O and H are balanced using H+ and H 2 O
SO2  + 2H2O →SO4 2-  +4H+
Cr2O7 2-+14H+→2Cr3+  +7H2O

Electrons need to be added to balance the charge.
SO2  + 2H2O →SO4 2-  +4H+ +2e- ...............................(2)
Cr2O7 2-+14H+  +6e- →2Cr3+  +7H  2O .........................(1)

Loss and gain of electrons need to be made equal. For this, we need to multiply equation (2) by 3

Adding both equations to get the balanced redox reaction
Cr2O7 2-+2H+    +3SO 2  →2Cr3+  +7H2O

Question 1: Which of the following equations represents a balanced redox reaction?

a. KMnO4 + H2SO4 + FeSO4 →  K2SO4 + MnSO4 + Fe2(SO4)3 + H2O
b. 2KMnO4 + 10 FeSO4 + 8H2SO4 →  K2SO4 + 2Mn(SO4) + 5Fe2(SO4)3 + 8H2O
c. Cr2O7 2-(g) + SO2(aq) → Cr3+ + SO42-(aq)
d. Al + NO3- →  Al(OH)4- +NH3

Question 2: Ion-electron method for balancing Redox reactions is also known as

a. Oxidation number method

b. State method

c. Half reaction method

d. Half life method

Question 3: In a reaction taking place in acidic or neutral medium, oxygen atoms are balanced by adding

a. H2O & H+

b. H2O & OH-

c. H+ & OH-

d. H2O only

Question 4: What is the number of electrons being transferred in the redox reaction 2KMnO4 + 10 FeSO4 + 8H2SO4 →  K2SO4 + 2Mn(SO4) + 5Fe2(SO4)3 + 8H2O?

a. Zero

b. One

c. Five

d. Ten

Q.1

Q.2

Q.3

Q.4

b

c

a

d

Related Resources

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Have a look at syllabus of IIT JEE

You can refer to oxidation number

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