Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.
(a) $\mathrm{P}_{4(s)}+\mathrm{OH}_{\text {(aq) }} \longrightarrow \mathrm{PH}_{3(g)}+\mathrm{HPO}_{2}^{-}$(aq)
(b) $\mathrm{N}_{2} \mathrm{H}_{4(())}+\mathrm{ClO}_{3(a q)}^{-} \longrightarrow \mathrm{NO}_{(g)}+\mathrm{Cl}_{(g)}^{-}$
(c) $\mathrm{Cl}_{2} \mathrm{O}_{7(g)}+\mathrm{H}_{2} \mathrm{O}_{2(a q)} \longrightarrow \mathrm{ClO}_{2(a q)}^{-}+\mathrm{O}_{2(g)}+\mathrm{H}_{(a q)}^{+}$
(a)
The O.N. (oxidation number) of $\mathrm{P}$ decreases from 0 in $\mathrm{P}_{4}$ to $-3$ in $\mathrm{PH}_{3}$ and increases from 0 in $\mathrm{P}_{4}$ to $+2$ in $\mathrm{HPO}_{2}^{-}$. Hence, $\mathrm{P}_{4}$ acts both as an oxidizing agent and a reducing agent in this reaction.
Ion–electron method:
The oxidation half equation is:
$\mathrm{P}_{4}(\mathrm{~s}) \rightarrow \mathrm{HPO}_{2}^{-}(\mathrm{aq})$
The P atom is balanced as:
$\mathrm{P}_{4}(\mathrm{~s}) \rightarrow 4 \mathrm{HPO}_{2}^{-}(\mathrm{aq})$
The O atom is balanced by adding 8 H2O molecules:
$\mathrm{P}_{4}(\mathrm{~s})+8 \mathrm{H}_{2} \mathrm{O} \rightarrow 4 \mathrm{HPO}_{2}^{-}(\mathrm{aq})$
The H atom is balanced by adding 12 H+ ions:
$\mathrm{P}_{4}(\mathrm{~s})+8 \mathrm{H}_{2} \mathrm{O} \rightarrow 4 \mathrm{HPO}_{2}^{-}(\mathrm{aq})+12 \mathrm{H}^{+}$
The charge is balanced by adding e– as:
$\mathrm{P}_{4}(\mathrm{~s})+8 \mathrm{H}_{2} \mathrm{O} \rightarrow 4 \mathrm{HPO}_{2}^{-}(\mathrm{aq})+12 \mathrm{H}^{+}+8 \mathrm{e}^{-} \quad \ldots$ (i)
The reduction half equation is:
$\mathrm{P}_{4(s)} \longrightarrow \mathrm{PH}_{3(g)}$
The P atom is balanced as:
$\mathrm{P}_{4}(\mathrm{~s}) \rightarrow 4 \mathrm{PH}_{3}(\mathrm{~g})$
The H is balanced by adding 12 H+ as:
$\mathrm{P}_{4}(\mathrm{~s})+12 \mathrm{H}^{+} \rightarrow 4 \mathrm{PH}_{3}(\mathrm{~g})$
The charge is balanced by adding 12e– as:
$\mathrm{P}_{4}(\mathrm{~s})+12 \mathrm{H}^{+}+12 \mathrm{e}^{-} \rightarrow 4 \mathrm{PH}_{3}(\mathrm{~g})$
By multiplying equation (i) with 3 and (ii) with 2 and then adding them, the balanced chemical equation can be obtained as:
$5 \mathrm{P}_{4}(\mathrm{~s})+24 \mathrm{H}_{2} \mathrm{O} \rightarrow 12 \mathrm{HPO}_{2}^{-}+8 \mathrm{PH}_{3}(\mathrm{~g})+12 \mathrm{H}^{+}$
As, the medium is basic, add 12OH– both sides as:
$5 \mathrm{P}_{4}(\mathrm{~s})+12 \mathrm{H}_{2} \mathrm{O}+12 \mathrm{OH}^{-} \rightarrow 12 \mathrm{HPO}_{2}^{-}+8 \mathrm{PH}_{3}(\mathrm{~g})$
This is the required balanced equation.
Oxidation number method:
Let, total no of P reduced = x
$\mathrm{P}_{4}(\mathrm{~s})+\mathrm{OH}^{-} \rightarrow x \mathrm{PH}_{3}(\mathrm{~g})+(4-x) \mathrm{HPO}_{2}^{-}$
Total decrease in oxidation number of P = x × 3 = 3x
Total increase in oxidation number of $P=(4-x) \times 2=8-2 x$
$\because 3 x=8-2 x$
$x=8 / 5$
From (i),
$5 \mathrm{P}_{4}(\mathrm{~s})+5 \mathrm{OH}^{-} \rightarrow 8 \mathrm{PH}_{3}(\mathrm{~g})+12 \mathrm{HPO}_{2}^{-}$
Since, reaction occures in basic medium, the charge is balanced by adding 7OH– on LHS as:
$5 \mathrm{P}_{4}(\mathrm{~s})+12 \mathrm{OH}^{-} \rightarrow 8 \mathrm{PH}_{3}(\mathrm{~g})+12 \mathrm{HPO}_{2}^{-}$
The $\mathrm{O}$ atoms are balanced by adding $12 \mathrm{H}_{2} \mathrm{O}$ as:
$5 \mathrm{P}_{4}(\mathrm{~s})+12 \mathrm{H}_{2} \mathrm{O}+12 \mathrm{OH}^{-} \rightarrow+12 \mathrm{HPO}_{2}^{-}+8 \mathrm{PH}_{3}(\mathrm{~g})$
This is the required balanced equation.
(b)
The oxidation number of $\mathrm{N}$ increases from $-2$ in $\mathrm{N}_{2} \mathrm{H}_{4}$ to $+2$ in $\mathrm{NO}$ and the oxidation number of $\mathrm{Cl}$ decreases from $+5$ in $\mathrm{ClO}_{3}^{-}$to $-1$ in $\mathrm{Cl}^{-}$. Hence, in this
reaction, $\mathrm{N}_{2} \mathrm{H}_{4}$ is the reducing agent and $\mathrm{ClO}_{3}^{-}$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 $\mathrm{N}_{2} \mathrm{H}_{4}$ with 3 and $\mathrm{ClO}_{3}^{-}$with 4 to balance the increase and decrease in $\mathrm{O} . \mathrm{N}$., 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.
(c)
The oxidation number of $\mathrm{Cl}$ decreases from $+7$ in $\mathrm{Cl}_{2} \mathrm{O}_{7}$ to $+3$ in $\mathrm{ClO}_{2}^{-}$and the oxidation number of $\mathrm{O}$ increases from $-1$ in $\mathrm{H}_{2} \mathrm{O}_{2}$ to zero in $\mathrm{O}_{2}$. Hence, in this
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 $\mathrm{Cl}_{2} \mathrm{O}_{7}=4 \times 2=8$
Total increase in oxidation number of $\mathrm{H}_{2} \mathrm{O}_{2}=2 \times 1=2$
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.