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MnO4^– + 8H⁺ + 5e^– → Mn²⁺ + 4H₂O.

So the n factor is 5

r u sure mentor sir ?

i thought this way.
this is the general equation.

3 K₂MnO₄ + 2 H₂O → 2 KMnO₄ + MnO₂ + 4 KOH

now, here Mn (+6) changes to Mn(+7) and Mn(+4). So, change in oxidation state is 1 and 2 respec.

so , equivalent weight = M / 1 + M / 2 = 3M / 2 = M / (2/3).

thus, the n-factor is 2/3.

ohh ya sorry i wrote it for KMNO₄