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.$N_2H_4(l) + ClO_3^-(aq) \rightarrow NO(g) + Cl^-(g)$
✓
Answer
Oxidation number method:Total increase in O.N. of N = 2 × 4 = 8
Total decreases in O.N. of Cl = 1 × 6 = 6
Therefore, to balance increase/ decrease in O.N. multiply $N_2H_4$ by 3 and $ClO_3^-$
by 4, we have,
$3\text{N}_2\text{H}_4(\text{l})+4\text{ClO}_3^-(\text{aq})\rightarrow\text{NO(g)}+\text{Cl}^-(\text{aq})$
To balance N and Cl atoms, multiply NO by 6 and $Cl^-$ by 4, we have,
$3\text{N}_2\text{H}_4(\text{l})+4\text{ClO}_3^-(\text{aq})\rightarrow6\text{NO(g)}+4\text{Cl}^-(\text{aq})$
Balance O atoms by adding $6H_2O$,
$3\text{N}_2\text{H}_4(\text{l})+4\text{ClO}_3^-(\text{aq})\rightarrow6\text{NO(g)}+4\text{Cl}^-(\text{aq})+6\text{H}_2\text{O}(\text{l})\ ....(\text{i})$
H atoms get automatically balanced and thus eq. (i) represents the correct balanced equation.
Ion electron method:
Oxidation half reaction: $\stackrel{{-2}}{\ \ \ \ \ \ \ \ \ \ \ \hbox{N}_2\text{H}_4(\text{l})}\rightarrow\stackrel{{+2}}{\ \ \ \ \ \ \hbox{NO(g)}}$
Balance N atoms, $\text{N}_2\text{H}_4(\text{l})\rightarrow2\text{NO(g)}$
Balance O.N. by adding electrons,
$\text{N}_2\text{H}_4(\text{l})\rightarrow2\text{NO(g)}+8\text{e}^-$
Balance charge by adding $8OH^-$ ions,
$\text{N}_2\text{H}_4(\text{l})+8\text{OH}^{-}(\text{aq})\rightarrow2\text{NO(g)}+8\text{e}^-$
Balance O atoms by adding $6H_2O$,
$\text{N}_2\text{H}_4(\text{l})+8\text{OH}^{-}(\text{aq})\rightarrow2\text{NO(g)}+6\text{H}_2\text{O(l)}+8\text{e}^-\ ....(\text{ii})$
Thus, eq. (ii) represents the correct balanced oxidation half equation.
Reduction half reaction
$\stackrel{{+5}}{\ \ \ \ \ \ \ \ \ \ \ \hbox{ClO}^-_3(\text{aq})}\rightarrow\stackrel{{-1}}{\ \ \ \ \ \ \ \ \hbox{Cl}^{-}{(\text{aq})}}$
Balance O.N. by adding electrons,
$\text{ClO}_3^-(\text{aq})+6\text{e}^-\rightarrow\text{Cl}^-(\text{aq})$
Balance charge by adding $OH^-$ ions,
$\text{ClO}_3^-(\text{aq})+6\text{e}^-\rightarrow\text{Cl}^-(\text{aq})+6\text{OH}^-(\text{aq})$
Balance O atoms by adding $3H_2O$,
$\text{ClO}_3^-(\text{aq})+3\text{H}_2\text{O(l)}+6\text{e}^-\rightarrow\text{Cl}^-(\text{aq})+6\text{OH}^-(\text{aq})\ .....(\text{iii})$
Thus, eq. (iii) represents the correct balanced reduction half equation.
To cancel out electrons gained and lost, multiply eq. (ii) by 3 and eq. (iii) by 4 and add, we have,
$3\text{N}_2\text{H}_4(\text{l})+4\text{ClO}_3^-(\text{aq})\rightarrow6\text{NO(g)}+4\text{Cl}^{-}(\text{aq})+6\text{H}_2\text{O(l)}\ .....(\text{iv})$
Thus, eq. (iv) represents the correct balanced equation.
Total decreases in O.N. of Cl = 1 × 6 = 6
Therefore, to balance increase/ decrease in O.N. multiply $N_2H_4$ by 3 and $ClO_3^-$
by 4, we have,
$3\text{N}_2\text{H}_4(\text{l})+4\text{ClO}_3^-(\text{aq})\rightarrow\text{NO(g)}+\text{Cl}^-(\text{aq})$
To balance N and Cl atoms, multiply NO by 6 and $Cl^-$ by 4, we have,
$3\text{N}_2\text{H}_4(\text{l})+4\text{ClO}_3^-(\text{aq})\rightarrow6\text{NO(g)}+4\text{Cl}^-(\text{aq})$
Balance O atoms by adding $6H_2O$,
$3\text{N}_2\text{H}_4(\text{l})+4\text{ClO}_3^-(\text{aq})\rightarrow6\text{NO(g)}+4\text{Cl}^-(\text{aq})+6\text{H}_2\text{O}(\text{l})\ ....(\text{i})$
H atoms get automatically balanced and thus eq. (i) represents the correct balanced equation.
Ion electron method:
Oxidation half reaction: $\stackrel{{-2}}{\ \ \ \ \ \ \ \ \ \ \ \hbox{N}_2\text{H}_4(\text{l})}\rightarrow\stackrel{{+2}}{\ \ \ \ \ \ \hbox{NO(g)}}$
Balance N atoms, $\text{N}_2\text{H}_4(\text{l})\rightarrow2\text{NO(g)}$
Balance O.N. by adding electrons,
$\text{N}_2\text{H}_4(\text{l})\rightarrow2\text{NO(g)}+8\text{e}^-$
Balance charge by adding $8OH^-$ ions,
$\text{N}_2\text{H}_4(\text{l})+8\text{OH}^{-}(\text{aq})\rightarrow2\text{NO(g)}+8\text{e}^-$
Balance O atoms by adding $6H_2O$,
$\text{N}_2\text{H}_4(\text{l})+8\text{OH}^{-}(\text{aq})\rightarrow2\text{NO(g)}+6\text{H}_2\text{O(l)}+8\text{e}^-\ ....(\text{ii})$
Thus, eq. (ii) represents the correct balanced oxidation half equation.
Reduction half reaction
$\stackrel{{+5}}{\ \ \ \ \ \ \ \ \ \ \ \hbox{ClO}^-_3(\text{aq})}\rightarrow\stackrel{{-1}}{\ \ \ \ \ \ \ \ \hbox{Cl}^{-}{(\text{aq})}}$
Balance O.N. by adding electrons,
$\text{ClO}_3^-(\text{aq})+6\text{e}^-\rightarrow\text{Cl}^-(\text{aq})$
Balance charge by adding $OH^-$ ions,
$\text{ClO}_3^-(\text{aq})+6\text{e}^-\rightarrow\text{Cl}^-(\text{aq})+6\text{OH}^-(\text{aq})$
Balance O atoms by adding $3H_2O$,
$\text{ClO}_3^-(\text{aq})+3\text{H}_2\text{O(l)}+6\text{e}^-\rightarrow\text{Cl}^-(\text{aq})+6\text{OH}^-(\text{aq})\ .....(\text{iii})$
Thus, eq. (iii) represents the correct balanced reduction half equation.
To cancel out electrons gained and lost, multiply eq. (ii) by 3 and eq. (iii) by 4 and add, we have,
$3\text{N}_2\text{H}_4(\text{l})+4\text{ClO}_3^-(\text{aq})\rightarrow6\text{NO(g)}+4\text{Cl}^{-}(\text{aq})+6\text{H}_2\text{O(l)}\ .....(\text{iv})$
Thus, eq. (iv) represents the correct balanced equation.
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