In the following oxidising reaction of KMnO4, the equivalent weight o…

The correct answer is Option (3) → M/5

The equivalent weight of an oxidising agent is its molar mass divided by the number of electrons gained by one formula unit (its n-factor):

    \( \displaystyle \text{Equivalent weight}=\frac{\text{Molar mass}}{\text{electrons gained per formula unit}} \)

This number equals the change in oxidation state of the atom that is reduced. It depends on the reaction, not only on the compound.

In this reaction, manganese changes from +7 in MnO₄⁻ (x + 4(−2) = −1, so x = +7) to +2 in Mn²⁺ (acidic medium):

    \( \displaystyle \ce{MnO4^- + 8H^+ + 5e^- -> Mn^2+ + 4H2O} \)

Each \( \ce{MnO4^{-}} \) gains 5 electrons. The sulphite ion is oxidised, sulphur going from +4 in \( \ce{SO3^{2-}} \) (x + 3(−2) = −2) to +6 in \( \ce{SO4^{2-}} \) (x + 4(−2) = −2):

    \( \displaystyle \ce{SO3^2- + H2O -> SO4^2- + 2H^+ + 2e^-} \)

Multiplying the reduction by 2 and the oxidation by 5 balances the electrons (10 on each side) and gives the equation of the question:

    \( \displaystyle \ce{5SO3^2- + 2MnO4^- + 6H^+ -> 2Mn^2+ + 3H2O + 5SO4^2-} \)

    \( \displaystyle \text{Equivalent weight of }\ce{KMnO4}=\frac{M}{5} \)

So the equivalent weight of \( \ce{KMnO4} \) is M/5.