CBSE Class 11-science Answered

ANODIC Half Reaction

H₂O₂(aq) ---> O ₂(g)

In order to balance the Oxidation Number put the Electrons exchanged on right hand side

H₂O₂(aq) ---> O ₂(g) + 2e^-

Since the reaction is carried out in basic medium, balance the charge and hydrogen atoms by adding OH   on left hand side and water on right hand side

H₂O₂(aq) +2 OH^- (aq)--> O ₂(g) + 2e ^-+ 2 H₂O(aq)
It results balanced.

CATHODIC Half .Reaction

Cl₂O₇(aq) ---> ClO₂^-(aq)

first balance the CHLORINE ATOMS

Cl₂O₇(aq) --->2ClO₂^-(aq)

Then balance the Oxidation Number by adding electrons on left hand side

Cl₂O₇(aq) +8 e  ---> 2ClO₂^-(aq)

Now add water molecules in order to balance the Oxygen Atoms
Cl₂O₇(aq) +8 e  ---> 2ClO₂^-(aq) + 3 H₂O(aq)

Now to balance the hydrogen atoms add OH ^- ions on right hand side and water on left hand side(since electrons are on left) and water on left

Cl₂O₇(aq) +8 e   + 3 H₂O(aq)---> 2ClO₂^-(aq) + 6 OH^-(aq)
It results balanced.

Now, add the Half-Reactions in order to remove the elctrons

H₂O₂(aq) +2 OH^- (aq)--> O ₂(g) + 2e ^-+ 2 H₂O(aq)
Cl₂O₇(aq) +8 e   + 3 H₂O(aq)---> 2ClO₂^-(aq) + 6 OH^-(aq)
Multiply the first equation by 4

4 H₂O₂(aq) +8 OH^- (aq)--> 4O ₂(g) + 8e ^-+ 8 H₂O(aq)

Adding this to other half reaction

Cl₂O₇(aq) +8 e   + 3 H₂O(aq)---> 2ClO₂^-(aq) + 6 OH^-(aq)

Net reaction is

4 H₂O₂(aq) + Cl₂O₇(aq) + 2 OH^-(aq) --->4O₂(g) + 2 ClO₂ ^-(aq) + 5 H₂O(aq)