In alkaline medium, the reduction of permanganate anion involves a gain of _______ electrons.

We need to determine how many electrons are gained during the reduction of the permanganate anion ($$MnO_4^-$$) in alkaline medium.

The product of reduction depends on the medium:

- In acidic medium: $$MnO_4^- \to Mn^{2+}$$ (gain of 5 electrons)

- In neutral/weakly alkaline medium: $$MnO_4^- \to MnO_2$$ (gain of 3 electrons)

- In strongly alkaline medium: $$MnO_4^- \to MnO_4^{2-}$$ (gain of 1 electron)

For the standard alkaline medium reduction, the product is $$MnO_2$$.

In $$MnO_4^-$$: Let the oxidation state of Mn be $$x$$. Then $$x + 4(-2) = -1$$, giving $$x = +7$$.

In $$MnO_2$$: Let the oxidation state of Mn be $$y$$. Then $$y + 2(-2) = 0$$, giving $$y = +4$$.

Change in oxidation state: $$+7 \to +4$$, a decrease of 3.

$$MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$$

We can verify this is balanced: Mn: 1 on each side; O: 4+2=6 on left, 2+4=6 on right; H: 4 on left, 4 on right; Charge: -1-3=-4 on left, -4 on right.

The permanganate anion gains 3 electrons in alkaline medium.