Balance the following equation by oxidation number method.
$Z n + H N O_{3} \to Z n (N O_{3})_{2} + N H_{4} N O_{3} + H_{2} O$

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Solution

Writing oxidation numbers of all the atoms,
$0 Z n + + 1 H + 5 N_{3} \to + 2 Z n (+ 5 N_{3})_{2} + - 3 N_{4} + 5 N - 2 N_{3} +_{2} - 2 O$
The oxidation numbers of $Z n$ and $N$ have changed.
$0 Z n \to + 2 Z n (N O_{3})_{2}$ .......(i)
$H + 5 N O_{3} \to - 3 N H_{4} N O_{3}$ ..........(ii)
Increase in Ox. no. of $Z n = 2$ units per $Z n$ atom
Decrease in Ox. no. of $N = 8$ units per $H N O_{3}$ molecule
Eq.(i) should be multiplied by $4$ and added to eq. (ii),
$4 Z n + H N O_{3} \to 4 Z n (N O_{3})_{2} + N H_{4} N O_{3}$ .
To balance nitrogen, $9$ molecules of $H N O_{3}$ should be added on LHS.
To balance hydrogen and oxygen, $3 H_{2} O$ molecules should be added on RHS. Hence, the balanced equation is
$4 Z n + 10 H N O_{3} \to 4 Z n (N O_{3})_{2} + N H_{4} N O_{3} + 3 H_{2} O$ .

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