KMnO_4 acts as an oxidising agent in the neut | vedclass

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(B) In a neutral medium,the reduction reaction of $KMnO_4$ is: $MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$. Here,the oxidation state of $Mn$ ch

Vedclass · Accepted Answer

$M / 3$

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(B) In a neutral medium,the reduction reaction of $KMnO_4$ is: $MnO_4^- + 2H_2O + 3e^- ightarrow MnO_2 + 4OH^-$. Here,the oxidation state of $Mn$ changes from $+7$ in $KMnO_4$ to $+4$ in $MnO_2$. The change in oxidation state is $7 - 4 = 3$. Thus,the number of electrons involved ($n$-factor) is $3$. The equivalent weight is calculated as: $	ext{Equivalent weight} = \frac{	ext{Molecular weight}}{n	ext{-factor}} = \frac{M}{3}$.

$KMnO_4$ acts as an oxidising agent in the neutral medium and gets reduced to $MnO_2$. What is the equivalent weight of $KMnO_4$ in a neutral medium?

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