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Ion Electron Method

Balance the f...

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.
$P_{4 (s)} + O H -_{(a q)} ⟶ P H_{3 (g)} + H P O_{2 (a q)}^{-}$

$N_{2} H_{4 (I)} + C I O_{3 (a q)}^{-} ⟶ N O_{(g)} + C I_{(g)}^{-}$

$C l_{2} O_{7 (g)} + H_{2} O_{2 (a q)} ⟶ C I O_{2 (a q)}^{-} + O_{2 (g)} + H_{(a q)}^{+}$

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Solution

The O.N. (oxidation number) of P decreases from 0 in $P_{4} t o - 3 i n P H_{3}$ and increases from 0 in $P_{4} t o + 2 i n H P O_{2}^{-}$ . Hence, $P_{4}$ acts both as an oxidizing agent and a reducing agent in this reaction.
Ion–electron method:
The oxidation half equation is:
$P_{4 (s)} ⟶ H P O_{2 (a q)}^{-}$
The P atom is balanced as:
${0 P}_{4 (s)} ⟶ 4 H 2 + P O_{2 (a q)}^{-}$
The O.N. is balanced by adding 8 electrons as:
$P_{4 (s)} ⟶ 4 H P O_{2 (a q)}^{-} + 8 e^{-}$
The charge is balanced by adding $12 O H^{-}$ as:
$P_{4 (s)} + 12 O H_{(a q)}^{-} ⟶ 4 H P O_{2 (a q)}^{-} + 8 e^{-}$
The H and O atoms are balanced by adding $4 H_{2} O$ as:
$P_{4 (s)} + 12 O H_{(a q)}^{-} ⟶ 4 H P O_{2 (a q)}^{-} + 4 H_{2} O_{(I)} + 8 e^{-} . . . . . . (i)$
The reduction half equation is:
$P_{4 (s)} ⟶ P H_{3 (g)}$
The P atom is balanced as
${0 P}_{4 (s)} ⟶ 4 - 3 P H_{3 (g)}$
The O.N. is balanced by adding 12 electrons as:
$P_{4 (s)} + 12 e^{-} ⟶ 4 P H_{3 (g)}$
The charge is balanced by adding $12 O H^{-}$ as:
$P_{4 (s)} + 12 e^{-} ⟶ 4 P H_{3 (g)} + 12 O H_{(a q)}^{-}$
The O and H atoms are balanced by adding $12 H_{2} O$ as:
$P_{4 (s)} + 12 H_{2} O_{(I)} + 12 e^{-} ⟶ 4 P H_{3 (g)} + 12 H O_{(a q)}^{-} (i i)$
By multiplying equation (i) with 3 and (ii) with 2 and then adding them, the balanced chemical equation can be obtained as:
$5 P_{4 (s)} + 12 H_{2} O_{(I)} + 12 H O_{(a q)}^{-} ⟶ 8 P H_{3 (g)} + 12 H P O_{2 (a q)}^{-}$

The oxidation number of N increases from – 2 in $N_{2} H_{4}$ to + 2 in NO and the oxidation number of Cl decreases from + 5 in $C I O_{3}^{-}$ to – 1 in $C l^{-}$ . Hence, in this reaction, $N_{2} H_{4}$ is the reducing agent and $C I O_{3}^{-}$ is the oxidizing agent.
Ion - electron method:
The oxidation half equation is:
$_{2} H_{4 (I)} ⟶ + 2 N O_{(g)}$
The N atoms are balanced as:
$N_{2} H_{4 (I)} ⟶ 2 N O_{(g)} + 8 e^{-}$
The charge is balanced by adding $8 O H^{-}$ ions as:
$N_{2} H_{4 (I)} + 8 O H_{(a q)}^{-} ⟶ 2 N O_{(g)} + 8 e^{-}$
The O atoms are balanced by adding $6 H_{2} O$ as:
$N_{2} H_{4 (I)} + 8 O H_{(a q)}^{-} ⟶ 2 N O_{(g)} + 6 H_{2} O_{(I)} + 8 e^{-} . . . . . . (i)$
The reduction half equation is:
$+ 5 C I O_{3 (a q)}^{-} ⟶ - 1 C I_{(a q)}^{-}$
The oxidation number is balanced by adding 6 electrons as:
$C I O_{3 (a q)}^{-} + 6 e^{-} ⟶ C l_{(a q)}^{-}$
The charge is balanced by adding $6 O H^{-}$ ions as:
$C l O_{3 (a q)}^{-} + 6 e^{-} ⟶ C l^{-} (a q) + 6 O H_{(a q)}^{-}$
The O atoms are balanced by adding $3 H_{2} O$ as:
$C I O_{3 (a q)}^{-} + 3 H_{2} O_{(I)} + 6 e^{-} ⟶ C l_{(a q)}^{-} + 6 O H_{(a q)}^{-} . . . . . . . (i i)$
The balanced equation can be obtained by multiplying equation (i) with 3 and equation (ii) with 4 and then adding them as:
$3 N_{2} H_{4 (I)} + 4 C l O_{3 (a q)}^{-} ⟶ 6 N O_{(g)} + 4 C l_{(a q)}^{-} + 6 H_{2} O_{(I)}$
Oxidation number method:
Total decrease in oxidation number of $N = 2 \times 4 = 8$
Total increase in oxidation number of $C l = 1 \times 6 = 6$
On multiplying $N_{2} H_{4}$ with 3 and $C I O_{3}^{-}$ with 4 to balance the increase and decrease in O.N.,
we get:
$3 N_{2} H_{4 (I)} + 4 C I O_{3 (a q)}^{-} ⟶ N O_{(g)} + C I_{(a q)}^{-}$
The N and Cl atoms are balanced as:
$3 N_{2} H_{4 (I)} + 4 C I O_{3 (a q)}^{-} ⟶ 6 N O_{(g)} + 4 C I_{(a q)}^{-}$
The O atoms are balanced by adding $6 H_{2} O$ as:
$3 N_{2} H_{4 (I)} + 4 C I O_{3 (a q)}^{-} ⟶ 6 N O_{(g)} + 4 C I_{(a q)}^{-} + 6 H_{2} O_{(I)}$
This is the required balanced equation.

The oxidation number of Cl decreases from + 7 in $C l_{2} O_{7}$ to + 3 in $C I O_{2}^{-}$ and the oxidation number of O increases from -1 in $H_{2} O_{2}$ to zero in $O_{2}$ . Hence, in this reaction, $C l_{2} O_{7}$ is the
oxidizing agent and $H_{2} O_{2}$ is the reducing agent.
Ion–electron method:
The oxidation half equation is:
$H_{2}_{2 (a q)} ⟶ {0 O}_{2 (g)}$
The oxidation number is balanced by adding 2 electrons as:
$H_{2} O_{2 (a q)} + ⟶ O_{2 (g)} + 2 e^{-}$
The charge is balanced by adding $2 O H^{-}$ ions as:
$H_{2} O_{2 (a q)} + 2 O H_{(a q)}^{-} ⟶ O_{2 (g)} + 2 e^{-}$
The oxygen atoms are balanced by adding $2 H_{2} O$ as:
$H_{2} O_{2 (a q)} + 2 O H_{(a q)}^{-} ⟶ O_{2 (g)} + 2 H_{2} O_{(I)} + 2 e^{-} (i)$
The reduction half equation is:
$+ 7 C l_{2} O_{7 (g)} ⟶ + 3 C I O_{2 (a q)}^{-}$
The Cl atoms are balanced as:
$C l_{2} O_{2 (g)} ⟶ 2 C I O_{2 (a q)}^{-}$
The oxidation number is balanced by adding 8 electrons as:
$C l_{2} O_{7 (g)} + 8 e^{-} ⟶ 2 C I O_{2 (a q)}^{-}$
The charge is balanced by adding $6 O H^{-}$ as:
$C l_{2} O_{7 (g)} + 8 e^{-} ⟶ 2 C I O_{2 (a q)}^{-} + 6 O H_{(a q)}^{-}$
The oxygen atoms are balanced by adding $3 H_{2} O$ as:
$C l_{2} O_{7 (g)} + 3 H_{2} O_{I} 8 e^{-} ⟶ 2 C I O_{2 (a q)}^{-} + 6 O H_{(a q)}^{-} (i i)$
The balanced equation can be obtained by multiplying equation (i) with 4 and adding equation (ii) to it as:
$C l_{2} O_{7 (g)} + 4 H_{2} O_{2 (a q)} + 2 O H_{(a q)}^{-} ⟶ 2 C I O_{2 (a q)}^{-} + 4 O_{2 (a q)} + 5 H_{2} O_{(I)}$
Oxidation number method:
Total decrease in oxidation number of $C l_{2} O_{7} = 4 \times 2 = 8$
Total increase in oxidation number of $H_{2} O_{2} = 2 \times 1 = 2$
By multiplying $H_{2} O_{2}$ and $O_{2}$ with 4 to balance the increase and decrease in the oxidation number, we get:
$C l_{2} O_{7 (g)} + 4 H_{2} O_{2 (a q)} ⟶ C I O_{2 (a q)}^{-} + 4 O_{2 (g)}$
The Cl atoms are balanced as:
$C l_{2} O_{7 (g)} + 4 H_{2} O_{2 (a q)} ⟶ 2 C I O_{2 (a q)}^{-} + 4 O_{2 (g)}$
The O atoms are balanced by adding $3 H_{2} O$ as:
$C l_{2} O_{7 (g)} + 4 H_{2} O_{2 (a q)} ⟶ 2 C I O_{2 (a q)}^{-} + 4 O_{2 (g)} + 3 H_{2} O_{(I)}$
The H atoms are balanced by adding $2 O H^{-} a n d 2 H_{2} O$ as:
$C l_{2} O_{7 (g)} + 4 H_{2} O_{2 (a q)} + 2 O H_{(a q)}^{-} ⟶ 2 C I O_{2 (g)}^{-} + 4 O_{2 (g)} + 5 H_{2} O_{(I)}$
This is the required balanced equation.

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