For the following redox reaction: MnO₄⁻ (aq) + Ni (s) → MnO₂ (s) + Ni²⁺ (aq). Show how you would balance the reaction if it were in an acidic solution.

Question

For the following redox reaction: MnO₄⁻ (aq) + Ni (s) → MnO₂ (s) + Ni²⁺ (aq). Show how you would balance the reaction if it were in an acidic solution. 

Answer 1

MnO₄⁻ (aq) + Ni (s) → MnO₂ (s) + Ni²⁺ (aq)

I'll do the balancing step by step so you can hopefully see how to approach these problems.

MnO₄^- ==> MnO₂ ... oxidation half reaction

MnO₄^- ==> MnO₂ + 2H₂O ... balanced for oxygen

MnO₄^- + 4H⁺ ==> MnO₂ + 2H₂O ... balanced for hydrogen (in acid solution)

MnO₄^- + 4H⁺ + 3e- ==> MnO₂ + 2H₂O ... balanced for charge and is the final balanced oxidation reaction

Ni(s) ==> Ni²⁺ ... reduction half reaction

Ni(s) ==> Ni²⁺ + 2e- ... balanced for charge and is the final balanced reduction reaction

Next we must equalized the electrons in both reactions, so multiply oxidation reaction by 2 and multiply reduction reaction by 3 to get 6 electrons transferred.

MnO₄^- + 4H⁺ + 3e- ==> MnO₂ + 2H₂O [x2]

2MnO₄^- + 8H⁺ + 6e- ==> 2MnO₂ + 4H₂O

Ni(s) ==> Ni²⁺ + 2e- [x3]

3Ni(s) ==> 3Ni2+ + 6e-

add the two equations together to get the final balanced equation:

2MnO₄^- + 3Ni(s) + 8H⁺ ==> 2MnO₂ + 3Ni²⁺ + 4H₂O ... balanced redox equation in acid solution