Balancing Redox Reactions

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Oxidation Number / State Method For Balancing Redox Reactions
Half-Reaction or Ion-Electron Method For Balancing Redox Reactions
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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) $\times$ 2; 2Fe⁺² → Fe₂⁺³ + 2e^- (2) $\times$ 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 SO₄^2- so that left hand side must have 8H₂SO₄, thus H₂O also becomes 8H₂O.

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

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Question 2:

Balance the equation :

MnO₄^- + C₂O₄^2– + 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.

Watch this Video for more reference

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^- +NH₃in 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

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Question 2:

Balance the equation

Cr₂O₇ ^2-(g) + SO₂(aq) → Cr³⁺ + SO₄^2-(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 →SO₄^2-
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⁺ + SO₄^2-(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

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