The equivalent weight of KMnO_4 in an alkalin | vedclass

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Question

(B) In an alkaline medium,$KMnO_4$ acts as an oxidizing agent and is reduced to $MnO_4^{2-}$ (manganate ion).The chemical reaction is: $MnO_4^- + e^-

Vedclass · Accepted Answer

$158$

Vedclass · Answer

(B) In an alkaline medium,$KMnO_4$ acts as an oxidizing agent and is reduced to $MnO_4^{2-}$ (manganate ion).The chemical reaction is: $MnO_4^- + e^- ightarrow MnO_4^{2-}$.The change in oxidation state of $Mn$ is from $+7$ to $+6$,so the change in oxidation number is $1$.The equivalent weight is calculated as: $	ext{Equivalent weight} = \frac{	ext{Molar mass}}{	ext{n-factor}}$.The molar mass of $KMnO_4$ is $158 \ g/mol$.The n-factor is $1$.Therefore,$	ext{Equivalent weight} = \frac{158}{1} = 158$.

The equivalent weight of $KMnO_4$ in an alkaline medium is .......

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