In the reaction 2[Fe(CN)_6]^{3-} + H_2O_2 + 2 | vedclass

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(A) $1$. Identify the oxidation and reduction half-reactions.$2$. In the given reaction,$H_2O_2$ is oxidized to $O_2$ $(H_2O_2 \rightarrow O_2 + 2H^+

Vedclass · Accepted Answer

$17$

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(A) $1$. Identify the oxidation and reduction half-reactions.$2$. In the given reaction,$H_2O_2$ is oxidized to $O_2$ $(H_2O_2 ightarrow O_2 + 2H^+ + 2e^-)$.$3$. Since $H_2O_2$ loses electrons,it acts as the reductant.$4$. The change in oxidation state of oxygen in $H_2O_2$ $(-1)$ to $O_2$ $(0)$ involves a change of $1$ unit per oxygen atom. For two oxygen atoms in $H_2O_2$,the total change in oxidation number is $2$.$5$. The equivalent weight of a substance is calculated as $	ext{Equivalent weight} = \frac{	ext{Molar mass}}{	ext{n-factor}}$.$6$. Molar mass of $H_2O_2 = (2 	imes 1) + (2 	imes 16) = 34 \ g/mol$.$7$. The n-factor for $H_2O_2$ in this reaction is $2$.$8$. Therefore,$	ext{Equivalent weight} = \frac{34}{2} = 17$.

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

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