JEE Advanced 2020 — Paper 2 — Question 19

In slightly alkaline medium, $\mathrm{MnO}_{4}^{-}$reduces to $\mathrm{MnO}_{2}$ while $\mathrm{I}^{-}$is oxidized to $\mathrm{I}_{2}$.

$\left.\mathrm{MnO}_{4}^{-}+2 \mathrm{H}_{2} \mathrm{O}+3 \mathrm{e}^{-} \longrightarrow \mathrm{MnO}_{2}+4 \mathrm{OH}^{-}\right) \times 2$

$\left.2 \mathrm{I}^{-} \longrightarrow \mathrm{I}_{2}+2 \mathrm{e}^{-}\right) \times 3$

$2 \mathrm{MnO}_{4}^{-}+6 \mathrm{I}^{-}+4 \mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{MnO}_{2}+3 \mathrm{I}_{2}+8 \mathrm{OH}^{-}$

So, 2 mol of $\mathrm{KMnO}_{4}$ can produce 3 mole of $\mathrm{I}_{2}$

$\therefore 4 \mathrm{~mol}$ of $\mathrm{KMnO}_{4}$ can produce 6 mole of $\mathrm{I}_{2}$