Balance the following equations by oxidation number method.
$[F e (C N)_{6}]^{3 -} + N_{2} H_{4} + O H^{-} \to [F e (C N)_{6}]^{4 -} + N_{2} + H_{2} O$

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Solution

Two half reactions
$[+ 3 F e (C N)_{6}]^{3 -} \to [+ 2 F e (C N)_{6}]^{4 -}$
(change of Ox. no. per $F e$ atom $= - 1$ )
$_{2} H_{4} \to_{2}$
(change in Ox. no. per $N$ atom $= + 2$ )
Total increase $= 2 \times (+ 2) = + 4$
$4 [F e (C N)_{6}]^{3 -} + N_{2} H_{4} \to 4 [F e (C N)_{6}]^{4 -} + N_{2}$
$[4 F e (C N)_{6}]^{3 -} + N_{2} H_{4} + 4 O H^{-} \to 4 [F e (C N)_{6}]^{4 -} + N_{2} + 4 H_{2} O$ .

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