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