Chemistry - Redox Reactions Question with Solution | TestHub

The oxidation of ferrous ion by ${\mathrm{KMnO}}_4$ takes place in acidic medium as per following reaction $\begin{array}{l}2{\mathrm{KMnO}}_4+8{\mathrm{H}}_2{\mathrm{SO}}_4+10{\mathrm{FeSO}}_4\\ \to {\mathrm{K}}_2{\mathrm{SO}}_4+2{\mathrm{MnSO}}_4+8{\mathrm{H}}_2\mathrm{O}+5{\mathrm{Fe}}_2{\left({\mathrm{SO}}_4\right)}_3\\ \therefore \text{ Eq. mass of }{\mathrm{KMnO}}_4\\ \frac{\text{ Molecular mass }}{\text{ changein oxidation number }}\\ =\frac{\text{ Molecular mass }}{5}=\frac{158}{5}=31.6\end{array}$