Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.
N2H4(l) + ClO–3(aq) → NO(g) + Cl–(g)
The changes in the oxidation state taking place in the reaction are as shown below : -
Hence, we can infer easily that the oxidation number of N increases from – 2 in N2H4 to + 2 in NO and the oxidation number of Cl decreases from + 5 in ClO3- to – 1 in Cl– . Hence, in this reaction N2H4 is the reducing agent and ClO3- is the oxidizing agent.
Solving by Ion–electron method:
The oxidation half equation is:-
The N atoms are balanced as:-
N2H4(l)→ 2NO(g)
The oxidation number is balanced by adding 8 electrons as:-
N2H4(l)→ 2NO(g) + 8e–
The charge is balanced by adding 8 OH- ions as:-
The O atoms are balanced by adding 6H2O as:-
The reduction half equation is:-
The oxidation number is balanced by adding 6 electrons as:-
The charge is balanced by adding 6OH- ions as:-
The O atoms are balanced by adding 3H2O as:-
Multiplying the oxidation half reaction by 3 and then adding it to the reduction half reaction multiplied by 4, to get the net balanced reaction equation as:-
Oxidation number method:
Total decrease in oxidation number of N = 2 × 4 = 8 Total increase in oxidation number of Cl = 1 × 6 = 6
On multiplying N2H4 with 3 and ClO3- with 4 to balance the increase and decrease in O.N., we get:
The N and Cl atoms are balanced as:-
The O atoms are balanced by adding 6H2O to the required balanced equation:-
