In neutral medium KMnO4 is reduced to MnO2. Its equivalent mass in this medium is (K = 39, Mn = 55, O = 16):

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In neutral medium $KMnO X_{4}$ is reduced to $MnO X_{2}$ . Its equivalent mass in this medium is (K = 39, Mn = 55, O = 16):

Mn goes from +7 to +4, so n-factor = 3 and equivalent mass = 158/3 = 52.7. The 31.6 value belongs to acidic medium.