500 mL of 1.2 M KI solution is mixed with 5 | vedclass

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(C) In basic medium, the reaction between $KMnO_4$ and $KI$ is: $2MnO_4^{-} + I^{-} + H_2O \rightarrow 2MnO_2 + I_2 + 2OH^{-}$.Here, $n$-factor for $K

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$3$

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(C) In basic medium, the reaction between $KMnO_4$ and $KI$ is: $2MnO_4^{-} + I^{-} + H_2O ightarrow 2MnO_2 + I_2 + 2OH^{-}$.Here, $n$-factor for $KMnO_4$ is $3$ (change in oxidation state of $Mn$ from $+7$ to $+4$).Equivalents of $KMnO_4 = 0.2 	imes 0.5 	imes 3 = 0.3 \ 	ext{eq}$.Since $I_2$ is produced from $KMnO_4$, the equivalents of $I_2$ produced are equal to the equivalents of $KMnO_4$ reacted, which is $0.3 \ 	ext{eq}$.The titration reaction is $I_2 + 2S_2O_3^{2-} ightarrow 2I^{-} + S_4O_6^{2-}$.Equivalents of $I_2 = 	ext{Equivalents of } Na_2S_2O_3$.$0.3 = 0.1 	imes V 	imes 1$.$V = 3 \ L$.

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