**(c) **

The oxidation number of Cl decreases from + 7 in Cl_{2}O_{7} to + 3 in ClO^{3-} and the oxidation number of O increases from – 1 in H_{2}O_{2} to zero in O_{2}. Hence, in this reaction, Cl_{2}O_{7} is the

oxidizing agent and H_{2}O_{2 }is the reducing agent.

**Ion–electron method: **

The oxidation half equation is:

The oxidation number is balanced by adding 2 electrons as:

The charge is balanced by adding 2OH^{–} ions as:

The oxygen atoms are balanced by adding 2H_{2}O as:

The reduction half equation is:

The Cl atoms are balanced as:

The oxidation number is balanced by adding 8 electrons as:

The charge is balanced by adding 6OH^{–} as:

The oxygen atoms are balanced by adding 3H_{2}O as:

The balanced equation can be obtained by multiplying equation (i) with 4 and adding equation (ii) to it as:

**Oxidation number method: **

Total decrease in oxidation number of Cl_{2}O_{7} = 4 × 2 = 8

Total increase in oxidation number of H_{2}O_{2} = 2 × 1 = 2

By multiplying H_{2}O_{2 }and O_{2} with 4 to balance the increase and decrease in the oxidation number, we get:

The Cl atoms are balanced as:

The O atoms are balanced by adding 3H_{2}O as:

The H atoms are balanced by adding 2OH^{–} and 2H_{2}O as:

This is the required balanced equation.