Question

Calculate the gram equivalent weights of $KMn{O}_4$ with different conditions.

Answer 1

In acidic medium, $M{n}^{7+}⟶M{n}^{2+}$

Hence, net gain of electrons $=5$

$Mn{O}_4+8H^{+}+5e^{-}⟶M{n}^{2+}+4H_{2}O$

$\therefore$ Equivalent weight (Eq. wt.) $=\frac{158}{5}=31.6g$

In alkaline medium,

$\underset{–}{\text{case}1}:$ $M{n}^{7+}+M{n}^{4+}$ $Mn{O}_4+2H_{2}O+3e^{-}⟶Mn{O}_2+4^{-}OH$

$\therefore$ Equivalent weight $=\frac{158}{3}=52.67g$

$\underset{–}{\text{case}2}:$ Highly alkaline, $M{n}^{7+}⟶M{n}^{6+}$

$\therefore$ Equivalent weight $\frac{158}{1}=158g.$