Question

$Mn{O}_4^{2-}$ changes to $Mn{O}_2$ into $Mn{O}_4^{-}$ in acidic medium. Calculate equivalent weight of $Mn{O}_4^{2-}$.

• A. $1M$
• B. $\frac{1}{2}M$
• C. $\frac{3}{2}M$
• D. $3M$

Answer 1

Answer: (C)

$\frac{3}{2}M$

$3M{O}_4^{2-}+2H_{2}O\to M{\underset{+2}{n}}^{2+}+M\underset{+4}{n}{O}_2+4H^{+}$, It is a disproportionation reaction. In disproportionation reaction, equivalent weight of the species undergoing disproportionation =Eq. wt. (in oxidation) + Eq. wt. (in reduction)
$M\underset{+6}{n}{O}_4^{2-}\to M\underset{+4}{n}{O}_{2}Eq.wt.(redn.)=\frac{mol.wt}{2}$,
$M\underset{+6}{n}{O}_4^{2-}\to M\underset{+7}{n}{O}_4^{-}Eq.wt.(oxidn.)=\frac{mol.wt}{1}$,
$Eq.wt.ofMn{O}_4^{2-}=\frac{M}{2}+\frac{M}{1}=\frac{3}{2}M$