Analytical Chemistry Redox Problem

Question

Calculate Eo and K for the following reaction:

5MnO₂(s) + 4H⁺ -> 2MnO₄^- + 3Mn²⁺ + 2H₂O

I'm struggling with balancing the half-reactions of this problem. Any help would be appreciated, thanks so much.

Answer 1

The overall reaction as given is already balanced. Not sure why you need to balance the half reactions. But since you asked ....

Oxidation half reaction:

MnO₂ ==> MnO₄^- .. unbalanced

MnO₂ + 2H₂O ==> MnO₄^- .. balanced for Mn and O

MnO₂ + 2H₂O ==> MnO₄^- + 4H⁺ .. balanced for Mn, O and H

MnO₂ + 2H₂O ==> MnO₄^- + 4H⁺ + 3e- .. balanced for Mn, O, H and charge = balanced equation

Reduction half reaction:

MnO₂ ==> Mn²⁺ .. unbalanced

MnO₂ ==> Mn²⁺ + 2H₂O .. balanced for Mn and O

MnO₂ + 4H⁺ ==> Mn²⁺ + 2H₂O .. balanced for Mn, O and H

MnO₂ + 4H⁺ + 2e- ==> Mn²⁺ + 2H₂O .. balanced for Mn, O, H and charge = balanced equation

Multiply oxidation reaction by 2 and multiply reduction reaction by 3 to equalize electrons, then add together:

2MnO₂ + 4H₂O ==> 2MnO₄^- + 8H⁺ + 6e-

3MnO₂ + 12H⁺ + 6e- ==> 3Mn²⁺ + 6H₂O

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2MnO₂ + 4H₂O + 3MnO₂ + 12H⁺ + 6e- ==> 2MnO₄^- + 8H⁺ + 6e- + 3Mn²⁺ + 6H₂O

5MnO₂ + 4H⁺ ==> 2MnO₄^- + 3Mn²⁺ + 2H2O .. BALANCED REDOX EQUATION