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(D) The balanced chemical equation for the reaction of $MnO_4^{-}$ with $I^{-}$ in acidic medium is: $2 \ MnO_4^{-} + 10 \ I^{-} + 16 \ H^{+} \longrig

Vedclass · Accepted Answer

$6$

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(D) The balanced chemical equation for the reaction of $MnO_4^{-}$ with $I^{-}$ in acidic medium is: $2 \ MnO_4^{-} + 10 \ I^{-} + 16 \ H^{+} \longrightarrow 2 \ Mn^{2+} + 5 \ I_2 + 8 \ H_2O$ The $n$-factor for $MnO_4^{-}$ (reduction from $+7$ to $+2$) is $5$. The $n$-factor for $I^{-}$ (oxidation from $-1$ to $0$ in $I_2$) is $1$ per $I^{-}$ ion. Using the law of equivalence: $n_1 	imes M_1 	imes V_1 = n_2 	imes M_2 	imes V_2$ $5 	imes 0.02 	imes V = 1 	imes 0.01 	imes 60$ $0.1 	imes V = 0.6$ $V = \frac{0.6}{0.1} = 6 \ mL$

The volume of $0.02 \ M$ acidified acidified permanganate solution required for complete reaction of $60 \ mL$ of $0.01 \ M \ I^{-}$ ion solution to form $I_2$ in $mL$ is

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