The number of electrons involved in the reduction of permanganate to manganese dioxide in acidic medium is ______.

We need to find the number of electrons involved in the reduction of permanganate (MnO$$_4^-$$) to manganese dioxide (MnO$$_2$$). In MnO$$_4^-$$, let Mn have oxidation state $$x$$. Then $$x + 4(-2) = -1$$, so $$x = +7$$, and in MnO$$_2$$, $$x + 2(-2) = 0$$, so $$x = +4$$.

The change in oxidation state from +7 to +4 is a decrease of 3, so 3 electrons are involved.

The balanced half-reaction is $$ MnO_4^- + 4H^+ + 3e^- \to MnO_2 + 2H_2O $$. Verification: charge on left = $$-1 + 4 - 3 = 0$$, charge on right = 0. Oxygen atoms: left = 4, right = 2 + 2 = 4. Hydrogen atoms: left = 4, right = 4.

The number of electrons is 3.