Question

$1$ mole of $Mn{O}_4^{2-}$ in neutral aqueous medium disproportionate to

• A. $\frac{2}{3}$ mole of $Mn{O}_4^{-}$and $\frac{1}{3}$ mole of $Mn{O}_2$
• B. $\frac{1}{3}$ mole of $Mn{O}_4^{-}$and $\frac{2}{3}$ mole of $Mn{O}_2$
• C. $\frac{1}{2}$ mole of $M{n}_2{O}_7$ and $\frac{1}{3}$ mole of $Mn{O}_2$
• D. $\frac{2}{3}$ mole of $M{n}_2{O}_7$ and $\frac{1}{3}$ mole of $Mn{O}_2$

Answer 1

Answer: (A)

The oxidation state of $Mn$ in $Mn{O}_4^{2-},Mn{O}_4^{-}$and $Mn{O}_2$ are $+6,+7$ and $+4$ respectively.
$M{n}^{6+}\to M{n}^{7+}+1e^{-}$
$M{n}^{6+}+2e^{-}\to M{n}^{4+}$
To balance the number of electrons, oxidation half reaction is multiplied with $2$ and added to reduction half reaction.
$\therefore 3Mn{O}_4^{2-}\to 2Mn{O}_4^{-}+Mn{O}_2$
$1$ mole of $Mn{O}_4^{2-}$ in neutral aqueous medium disproportionates to $\frac{2}{3}mole$ of $Mn{O}_4^{-}$and $\frac{1}{3}$ mole of $Mn{O}_2$