The number of moles of KMnO_4 required to oxi | vedclass

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(D) The balanced redox reaction in acidic medium is: $2KMnO_4 + 10KI + 8H_2SO_4 \longrightarrow 2MnSO_4 + 5I_2 + 6K_2SO_4 + 8H_2O$. In this reaction,$

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$1/5$

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(D) The balanced redox reaction in acidic medium is: $2KMnO_4 + 10KI + 8H_2SO_4 \longrightarrow 2MnSO_4 + 5I_2 + 6K_2SO_4 + 8H_2O$. In this reaction,$KMnO_4$ acts as an oxidizing agent where the oxidation state of $Mn$ changes from $+7$ to $+2$,so the $n$-factor for $KMnO_4$ is $5$. For $KI$,the oxidation state of $I$ changes from $-1$ to $0$,so the $n$-factor for $KI$ is $1$. According to the law of equivalence,the number of equivalents of $KMnO_4$ must be equal to the number of equivalents of $KI$. $	ext{Equivalents of } KMnO_4 = 	ext{Equivalents of } KI$. $n_{KMnO_4} 	imes n	ext{-factor}_{KMnO_4} = 	ext{Equivalents of } KI$. $n_{KMnO_4} 	imes 5 = 1$. $n_{KMnO_4} = 1/5$.

The number of moles of $KMnO_4$ required to oxidize one equivalent of $KI$ in the presence of sulphuric acid is

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