What would be the equivalent weight of the reductant in the given reaction?
${[F e {(C N)}_{6}]}^{- 3} + H_{2} O_{2} + 2 {O h}^{-} \to 2 {[F e {C N}_{6}]}^{4 -} + {2 H}_{2} O + O_{2}$
$[G i v e n : F e = 56, C = 12, N = 14, O = 16, H = 1]$

17

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212

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34

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32

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Solution

The correct option is B $34$
${[F e {(C N)}_{6}]}^{- 3} + H_{2} O_{2} + 2 {O H}^{-} ⟶ 2 {[F e {(C N)}_{6}]}^{4 -} + 2 H_{2} O + O_{2}$
Here, $H_{2} O_{2}$ is a reductant
$∴$ Oxidation state of $O$ in $H_{2} O_{2} = - 1$
Oxidation state of $O$ in $O_{2} = 0$
$∴$ Change in oxidation state $= + 1$
$∴$ Equivalent weight $= \frac{M o l . w t o f H_{2} O_{2}}{1}$
$= 34$

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