Calculate the equivalent weight of potassium permanganate $\left({\mathrm{KMnO}}_4\right)$ in  alkaline medium, by oxidation number change method (at. wt, $\mathrm{Mn}=55,\mathrm{K}=39$ $\mathrm{O}=16)$.

Mol. wt. of $\begin{array}{r}{\mathrm{KMnO}}_4=39+55+(4\times 16)\end{array}$

                                  $\begin{array}{r}=158\mathrm{g}{\mathrm{mol}}^{-}\end{array}$

(i) In neutral medium : ${\mathrm{Mn}}^{7+}+3{\mathrm{e}}^{-}⟶{\mathrm{Mn}}^{4+}$

$\therefore \text{ Eq. wt. of }{\mathrm{KMnO}}_4$,

                                    $\begin{array}{l}=\frac{\text{ Mol. Wt. of }{\mathrm{KMnO}}_4}{\text{ no. of electrons gained by }}\\ \text{ one molecule of }{\mathrm{KMnO}}_4\\ =\frac{158}{3}=52.67\text{ Ans. }\end{array}$

(ii) In acidic medium : 

                                $\begin{array}{l}{\mathrm{Mn}}^{7+}+5{\mathrm{e}}^{-}⟶{\mathrm{Mn}}^{2+}\\ \therefore \text{ Eq. wt. of }{\mathrm{KMnO}}_4=\frac{\text{ Mol. wt. of }{\mathrm{KMnO}}_4}{\text{ no. of electrons gained by }}\\ \text{ one molecule of }{\mathrm{KMnO}}_4\\ =\frac{158}{5}=31.6\text{ Ans. }\end{array}$

(iii) In alkaline medium, ${\mathrm{Mn}}^{7+}+1{\mathrm{e}}^{-}⟶{\mathrm{Mn}}^{6+}$

$\begin{array}{l}\therefore \text{ Eq. wt. of }{\mathrm{KMnO}}_4=\frac{\text{ Mol. wt. of }{\mathrm{KMnO}}_4}{\text{ no. of electrons gained by }}\\ \text{ one molecule of }{\mathrm{KMnO}}_4\\ =\frac{158}{1}=158\text{ Ans. }\end{array}$