Question

In the chemical reaction between stoichiometric quantities of $KMnO _{4}$ and $KI$ in weakly basic solution, what is the number of moles of $I _{2}$ released for $4$ moles of $KMnO _{4}$ consumed?

Answer 1

In slightly alkaline medium, $MnO _{4}^{-}$reduces to $MnO _{2}$ while $I$ is oxidized to $I _{2}$.
$\left. MnO _{4}^{-}+2 H _{2} O +3 e ^{-} \longrightarrow MnO _{2}+4 OH ^{-}\right) \times 2$
$\left.2 I ^{-} \longrightarrow I _{2}+2 e ^{-}\right) \times 3$
$2 \,MnO _{4}^{-}+6 I ^{-}+4 H _{2} O \longrightarrow 2 MnO _{2}+3 I _{2}+8 OH ^{-}$
So, $2\, mol$ of $KMnO _{4}$ can produce $3\, mole$ of $I _{2}$
$\therefore 4 \,mol$ of $KMnO _{4}$ can produce $6$ mole of $I _{2}$