Balance the following equation by ion electron method:
$A s_{2} S_{3} + N O_{3}^{-} + H^{+} + H_{2} O \to H_{3} A s O_{4} + N O + S$ .

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Solution

Half reactions
$[A s_{2} S_{3} + 8 H_{2} O \to 2 H_{3} A s O_{4} + 3 S + 10 H^{+} + 10 e] \times 3$
$[N O^{-} + 4 H^{+} + 3 e \to N O + 2 H_{2} O] \times 10$
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$3 A s_{2} S_{3} + 4 H_{2} O + 10 N O_{3}^{-} + 10 H^{+} \to 6 H_{3} A s O_{4} + 9 S + 10 N O$
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H g S + C l^{-} + H^{+} + N O_{3}^{-} \to H g C l_{4}^{2 -} + S + N O + H_{2} O

C u + H N O_{3} \to C u (N O_{3})_{2} + N O + H_{2} O

Z n + N O_{3}^{-} + O H^{-} \to Z n O_{2}^{2 -} + N H_{3} + H_{2} O

C u + N O_{3}^{-} + 8 H^{+} \to C u (N O_{3})_{2} + N O + H_{2} O

N a_{2} H A s O_{3} + K B r O_{3} + H C l \to N a C l + K B r + H_{3} A s O_{4}

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