(c)
The oxidation number of Cl decreases from + 7 in Cl2O7 to + 3 in ClO3- and the oxidation number of O increases from – 1 in H2O2 to zero in O2. Hence, in this reaction, Cl2O7 is the
oxidizing agent and H2O2 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 2H2O 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 3H2O 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 Cl2O7 = 4 × 2 = 8
Total increase in oxidation number of H2O2 = 2 × 1 = 2
By multiplying H2O2 and O2 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 3H2O as:
The H atoms are balanced by adding 2OH– and 2H2O as:
This is the required balanced equation.