Question

Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent. 

Answer 1

(a) The O.N. (oxidation number) of P decreases from 0 in P₄ to – 3 in PH₃ and increases 

from 0 in P₄ to + 2 in HPO^₂- . Hence, P₄ acts both as an oxidizing agent and a reducing agent in this reaction. Ion–electron method:

The oxidation half equation is: 

The P atom is balanced as: 

The O.N. is balanced by adding 8 electrons as: 

 

The charge is balanced by adding 12OH^– as: 

The H and O atoms are balanced by adding 4H₂O as:

The reduction half equation is: 

The P atom is balanced as 

The O.N. is balanced by adding 12 electrons as: 

The charge is balanced by adding 12OH^– as: 

The O and H atoms are balanced by adding 12H₂O as: 

By multiplying equation (i) with 3 and (ii) with 2 and then adding them, the balanced chemical equation can be obtained as: 

Answer 2

(b) 

The oxidation number of N increases from – 2 in N₂H₄ to + 2 in NO and the oxidation number of Cl decreases from + 5 in to – 1 in Cl^– . Hence, in this reaction, N₂H₄ is the reducing agent and is the oxidizing agent. 

Ion–electron method: 

The oxidation half equation is: 

The N atoms are balanced as: 

The oxidation number is balanced by adding 8 electrons as: 

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: 

The balanced equation can be obtained by multiplying equation (i) with 3 and equation (ii) with 4 and then adding them 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 ClO^3- 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 as: 

This is the required balanced equation. 

Answer 3

(c) 

The oxidation number of Cl decreases from + 7 in Cl₂O₇ to + 3 in ClO^3-  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.