How to write an overall redox reaction in a neutral solution?

Question

The half reactions are:

reduction: MnO₄^2- + 2H₂0 = MnO_2(s) + 4OH^-

oxidation: SO₃^2- +H₂0 = SO₄^2- + 2H⁺ + 2e-

Write an overall redox reaction in a neutral solution.

Answer 1

The half reaction MnO₄^2- + 2H₂0 = MnO_2(s) + 4OH^- is not complete. Note it is not balanced for charge as there is a -2 on the left and a -4 on the right. So 2 electrons need to be added to the left side. Also, this does not appear to be a neutral solution because there is H₂SO₄ (acid) and excess OH^- (base). Nonetheless...

MnO₄^2- + 2H₂0 + 2e^- = MnO_2(s) + 4OH^-

SO₃^2- +H₂0 = SO₄^2- + 2H⁺ + 2e-

————————-———————

MnO₄^2- + SO₃^2- + 3H₂O => MnO₂ + SO₄^2- + 2H⁺ + 4OH^-

MnO₄^- + SO₃^2- + 3H₂O => MnO₂ + H₂SO₄ + 4OH^- or

MnO₄^2- + SO₃^2- + H₂O => MnO₂ + 2OH^- + SO₄^2-