MnO4– + SO32– + H2O → MnO2 + SO42– + OH–

Question

If the reaction occurs in an acidic solution where water and H+ ions are added to the half-reactions to balance the overall reaction, how many electrons are transferred in the balanced reduction half-reaction?

Answer 1

In an acidic solution the MnO4^- goes to Mn^2+ which is a gain of 5 electrons per mole MnO4^-. S in sulfite is 4+ and it changes to 6+ in sulfate which is a loss of 2 electrons/mol. After multiplying the Mn by 2 and the sulfite by 5 the total electons changes = 10.
So the reduction half (the MnO4^-) gains 5e/mol or 10e/2 mols in the balanced equation.