Equivalent weight of KMnO_4 acting as an oxid | vedclass

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(D) The equivalent weight of an oxidizing agent is calculated as: $	ext{Equivalent weight} = \frac{	ext{Molecular weight}}{	ext{n-factor}}$.In an a

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$\frac{1}{5} 	imes$ Molecular weight of $KMnO_4$

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(D) The equivalent weight of an oxidizing agent is calculated as: $	ext{Equivalent weight} = \frac{	ext{Molecular weight}}{	ext{n-factor}}$.In an acidic medium,$KMnO_4$ undergoes the following reduction reaction: $MnO_4^- + 8H^+ + 5e^- 	o Mn^{2+} + 4H_2O$.Here,the change in oxidation state of $Mn$ is from $+7$ to $+2$,which involves a transfer of $5$ electrons.Therefore,the n-factor is $5$.Thus,the equivalent weight of $KMnO_4$ is $\frac{1}{5} 	imes 	ext{Molecular weight}$.

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

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