Question

Equivalent mass of KMnO$_4$ is different in acidic and alkaline media. Why?

Hint:

Always calculate equivalent mass using: \[ \text{Eq. mass} = \frac{\text{Molar mass}}{\text{Number of electrons gained or lost}} \] For KMnO$_4$: 5 electrons in acidic medium, 3 electrons in alkaline medium.

Answer 1

The equivalent mass of a substance depends on the number of electrons gained or lost per mole of substance in a redox reaction. KMnO$_4$ shows different oxidation states of Mn in acidic and alkaline media, hence the number of electrons exchanged is different. In acidic medium:
\[ MnO_4^- + 8H^+ + 5e^- \longrightarrow Mn^{2+} + 4H_2O \] Manganese changes from +7 to +2 oxidation state (\(\Delta = 5e^-\)).
\[ \text{Equivalent mass} = \frac{\text{Molar mass of KMnO}_4}{5} \] In neutral/alkaline medium:
\[ MnO_4^- + 2H_2O + 3e^- \longrightarrow MnO_2 + 4OH^- \] Manganese changes from +7 to +4 oxidation state (\(\Delta = 3e^-\)).
\[ \text{Equivalent mass} = \frac{\text{Molar mass of KMnO}_4}{3} \] Therefore, the equivalent mass of KMnO$_4$ differs because the change in oxidation number of Mn (and hence electrons transferred) is different in acidic and alkaline medium.