Q.Write the balanced half-reaction for the reduction of $\text{MnO}_4^-$ in neutral or faintly alkaline medium, name the coloured product formed, and state its n-factor.

In neutral or faintly alkaline medium, permanganate is not reduced all the way to $\text{Mn}^{2+}$ as it is in strongly acidic solution; instead, it is reduced only as far as the $+4$ state, precipitating as brown, insoluble manganese dioxide:

Checking the balance. Mn: 1 each side. O: left, $4 + 2 = 6$; right, $2$ (in $\text{MnO}_2$) $+ 4$ (in $4\text{OH}^-$) $= 6$. H: left, $2 \times 2 = 4$; right, $4 \times 1 = 4$. Charge: left, $(-1) + 0 + (-3) = -4$; right, $0 + 4(-1) = -4$ -- balanced.

Since manganese goes from the $+7$ state (in $\text{MnO}_4^-$) to the $+4$ state (in $\text{MnO}_2$), this is a three-electron change, so the n-factor of $\text{KMnO}_4$ in neutral/faintly alkaline medium is 3 -- distinctly different from its n-factor of 5 in acidic medium (where $\text{Mn}^{7+}$ is reduced all the way to $\text{Mn}^{2+}$). …

Write the reduction half-reaction for MnO4- to MnO2 in neutral/alkaline medium, verify the atom/charge balance, identify the +7-to-+4 oxi …

Using the acidic-medium half-reaction (n=5, product Mn2+) when the neutral/alkaline medium is specified (n=3, product MnO2) -- the two are not interchangeable; forgett …