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

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(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 HPO2- . 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: 

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(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 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 as: 

This is the required balanced equation. 

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(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 H2Ois 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 H2Oand 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.

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