Question

Give equivalent mass of KMnO4 in acidic and basic medium ?

Answer 1

Dear Student,


$$\begin{gathered} \mathrm{In}\mathrm{acidic}\mathrm{medium}: \\ {{\mathrm{MnO}}_4}^{-}+8{\mathrm{H}}^{+}+5{\mathrm{e}}^{-}\to {\mathrm{Mn}}^{2+}+4{\mathrm{H}}_2\mathrm{O} \\ \mathrm{So},\mathrm{n}=5 \\ \mathrm{Equivalent}\mathrm{weight}=\frac{\mathrm{Molecular}\mathrm{weight}}{\mathrm{n}} \\ =\frac{158}{5}=31.6 \\ \mathrm{In}\mathrm{basic}\mathrm{medium}: \\ {{\mathrm{MnO}}_4}^{-}+{\mathrm{e}}^{-}\to {{\mathrm{MnO}}_4}^{2-} \\ \mathrm{So},\mathrm{n}=1 \\ \mathrm{Equivalent}\mathrm{weight}=\frac{\mathrm{Molecular}\mathrm{weight}}{\mathrm{n}} \\ =\frac{158}{1}=158 \end{gathered}$$