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Karnataka Board PUCPUC Science Class 11

Balance the Following Equations in the 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) - Chemistry

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Question

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

\[\ce{N2H4(l) + ClO^-_3 (aq) → NO(g) + Cl–(g)}\]

Very Long Answer
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Solution

The oxidation number of N increases from – 2 in N2H4 to + 2 in NO and the oxidation number of Cl decreases from + 5 in `"ClO"_3^-` to - 1 in Cl. Hence, in this reaction, N2H4 is the reducing agent and `"ClO"_3^-` is the oxidizing agent.

Ion–electron method:

The oxidation half equation is:

\[\ce{^{-2}N2H_{4(l)} -> ^{+2}NO_{(g)}}\]

The N atoms are balanced as:

\[\ce{N_2H_{4(l)} -> 2NO_{(g)}}\]

The oxidation number is balanced by adding 8 electrons as:

\[\ce{N_2H_{4(l)} -> 2NO_{(g)} + 8e-}\]

The charge is balanced by adding 8 OHions as:

\[\ce{N_2H_{4(l)} -> 2NO_{(g)} + 8e-}\]

The O atoms are balanced by adding 6H2O as:

\[\ce{N_2H_{4(l)} + 8OH-_{(aq)} -> 2NO_{(g)} + 6H_2O_{(l)} + 8e-}\]  .....(i)

The reduction half equation is:

\[\ce{^{+5}ClO^-_{3(aq)} -> ^{-1}Cl^-_{(aq)}}\]

The oxidation number is balanced by adding 6 electrons as:

\[\ce{ClO^-_3_{(aq)} + 6e- -> Cl-_{(aq)}}\]

The charge is balanced by adding 6OH– ions as:

\[\ce{ClO^-_3_{(aq)} + 6e-  -> Cl-_{(aq)} + 6OH-_{(aq)}}\]

The O atoms are balanced by adding 3H2O as:

\[\ce{ClO^-_3_{(aq)} + 3H_2O_{(l)} + 6e- -> Cl-_{(aq)} + 6OH-_{(aq)}}\] .....(ii)

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

\[\ce{3N_2H_{4(l)} + 4ClO^-_3_{(aq)} -> 6NO_{(g)} + 4Cl-_{(aq)} + 6H_2O_{(l)}}\]

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:

\[\ce{3N_2H_{4(l)} + 4ClO^-_3_{(aq)} -> NO_{(g)} + Cl-_{(aq)}}\]

The N and Cl atoms are balanced as:

\[\ce{3N_2H_{4(l)} + 4ClO^-_3_{(aq)} -> 6NO_{(g)} + 4l-_{(aq)}}\]

The O atoms are balanced by adding 6H2O as:

\[\ce{3N_2H_{4(l)} + 4ClO^-_3_{(aq)} -> 6NO_{(g)} + 4Cl-_{(aq)} + 6H_2O_{(l)}}\]

This is the required balanced equation.

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