Equivalent weight of KMnO_4 acting as an oxid | vedclass

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Question

(D) In an acidic medium,$KMnO_4$ acts as an oxidizing agent and undergoes reduction as follows:$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$Here

Vedclass · Accepted Answer

One-fifth of its molecular weight

Vedclass · Answer

(D) In an acidic medium,$KMnO_4$ acts as an oxidizing agent and undergoes reduction as follows:$MnO_4^- + 8H^+ + 5e^- ightarrow Mn^{2+} + 4H_2O$Here,the oxidation state of $Mn$ changes from $+7$ to $+2$,which corresponds to a change of $5$ electrons per molecule of $KMnO_4$.Equivalent weight is calculated as:$	ext{Equivalent weight} = \frac{	ext{Molecular weight}}{	ext{n-factor}}$Since the n-factor is $5$,the equivalent weight is $\frac{1}{5}$ of its molecular weight.Therefore,the correct option is $D$.

Equivalent weight of $KMnO_4$ acting as an oxidant in acidic medium is

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