What is the equivalent weight of the reductan | vedclass

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(A) In the given reaction: $2[Fe(CN)_6]^{3-} + H_2O_2 + 2OH^{-} \rightarrow 2[Fe(CN)_6]^{4-} + 2H_2O + O_2$.$H_2O_2$ is acting as a reducing agent bec

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$17$

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(A) In the given reaction: $2[Fe(CN)_6]^{3-} + H_2O_2 + 2OH^{-} ightarrow 2[Fe(CN)_6]^{4-} + 2H_2O + O_2$.$H_2O_2$ is acting as a reducing agent because it is oxidized to $O_2$.The oxidation half-reaction is: $H_2O_2 + 2OH^{-} ightarrow O_2 + 2H_2O + 2e^{-}$.The change in oxidation state of oxygen in $H_2O_2$ $(-1)$ to $O_2$ $(0)$ involves a loss of $2$ electrons per molecule of $H_2O_2$.Thus,the $n$-factor for $H_2O_2$ is $2$.The molar mass of $H_2O_2 = (2 	imes 1) + (2 	imes 16) = 34 \ g/mol$.The equivalent weight = $\frac{	ext{Molar Mass}}{n	ext{-factor}} = \frac{34}{2} = 17$.

What is the equivalent weight of the reductant in the reaction $2[Fe(CN)_6]^{3-} + H_2O_2 + 2OH^{-} \rightarrow 2[Fe(CN)_6]^{4-} + 2H_2O + O_2$? (Given: $Fe = 56, C = 12, N = 14, O = 16, H = 1$)

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