Question

$1$ mole of $MnO_{4}^{2-} $ in neutral aqueous medium disproportionate to

• A. $\frac{2}{3}$ mole of $MnO _{4}^{-}$and $\frac{1}{3}$ mole of $MnO _{2}$
• B. $\frac{1}{3}$ mole of $MnO _{4}^{-}$and $\frac{2}{3}$ mole of $MnO _{2}$
• C. $\frac{1}{2}$ mole of $Mn _{2} O _{7}$ and $\frac{1}{3}$ mole of $MnO _{2}$
• D. $\frac{2}{3}$ mole of $Mn _{2} O _{7}$ and $\frac{1}{3}$ mole of $MnO _{2}$

Answer 1

Answer: (A)

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