overall equations.

Question

According to the half reactions posted above, what will be my over all equation?

reduction: MnO4^- -> Mn^2+

oxidation: SO3^2- -> SO4^2-

using the half reaction rule, show all of the work please, and explain why so i can understand better.

Answer 1

First, you must balance the half reactions.

We'll do the reduction half reaction first:

MnO₄^- ==> Mn²⁺

MnO₄^- ==> Mn²⁺ + 4H2O ... balanced for O

MnO₄^- + 8H⁺ ==> Mn²⁺ + 4H2O ... balanced for H

MnO₄^- + 8H⁺ + 5e- ==> Mn²⁺ + 4H2O ... balance for charge, O, and H and Mn (BALANCED REACTION)

Next, we'll do the oxidation half reaction:

SO₃^2- ==> SO₄^2-

SO₃^2- + H2O ==> SO₄^2- ... balanced for O

SO₃^2- + H2O ==> SO₄^2- + 2H⁺ ... balanced for H

SO₃^2- + H2O ==> SO₄^2- + 2H⁺ + 2e- ... balanced for charge, O, H, and S (BALANCED REACTION)

MnO₄^- + 8H⁺ + 5e- ==> Mn²⁺ + 4H2O ... balanced reduction

SO₃^2- + H2O ==> SO₄^2- + 2H⁺ + 2e- ... balance oxidation

To equalize electrons, we must multiply reduction reaction by 2 and oxidation reaction by 5 to get...

2MnO₄^- + 16H⁺ + 10e- ==> 2Mn²⁺ + 8H2O ... balanced reduction

5SO₃^2- + 5H2O ==> 5SO₄^2- + 10H⁺ + 10e- ... balanced oxidation

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2MnO₄^- + 6H⁺ + 5SO₃^2- => 2Mn²⁺ + 3H2O + 5SO₄^2- combined reactions = BALANCED EQUATION