When the reaction MnO2(s) <=> Mn2+(aq) + MnO4-(aq) is balanced in acidic solution, what is the coefficient of H+?

Question

I'm having trouble understanding how to solve these type of problems, is there a simple way to solve this?

Answer 1

Yes, there is a "simple" way to solve these, depending on what you call "simple". I think I've posted step by step instructions on how to balance redox in acid media, but I'll do it again here. I hope it makes sense.

This problem is a disproportionation reaction where Mn is both oxidized and reduced, but the process is the same.

(1). Do the oxidation and reduction half reactions separately

(2). Balance the mass

(3). Balance the oxygens by adding H2O

(4). Balance the hydrogens by adding H+

(5). Balance the charge by adding electrons

(6). After balancing both ox and red half reactions, equalize the electrons being transferred

(7). Add the two reactions together and combine/cancel like terms to get final equation.

(1). MnO₂ ==> Mn²⁺ ... reduction half reaction b/c Mn goes from 4+ to 2+

(2). MnO₂ ==> Mn²⁺ ... balanced for mass of Mn

(3). MnO₂ ==> Mn²⁺ + 2H₂O ... balanced for oxygen

(4). MnO₂ + 4H⁺ ==> Mn²⁺ + 2H₂O ... balanced for hydrogens

(5). MnO₂ + 4H⁺ +2e- ==> Mn²⁺ + 2H₂O ... balanced for charge

(1). MnO₂ ==> MnO₄^- ... oxidation half reaction b/c Mn goes from 4+ to 7+

(2). MnO₂ ==> MnO₄^- ... balanced for mass of Mn

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

(4). MnO₂ + 2H₂O ==> MnO₄^- + 4H⁺ ... balancded for hydrogen

(5). MnO₂ + 2H₂O ==> MnO₄^- + 4H⁺ + 3e- ... balanced for charge

Step 6 is applied to both equations since the reduction reaction has 2 electrons and the oxidation reaction has 3 electrons, so multiply reduction by 3 and oxidation by 2. You obtain the following:

3MnO₂ + 12H⁺ +6e- ==> 3Mn²⁺ + 6H₂O ... reduction reaction

2MnO₂ + 4H₂O ==> 2MnO₄^- + 8H⁺ + 6e- ... oxidation reaction

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(7). 3MnO₂ + 12H⁺ +6e- + 2MnO₂ + 4H₂O ==> 3Mn²⁺ + 6H₂O + 2MnO₄^- + 8H⁺ + 6e-

5MnO₂ + 4H⁺ ==> 3Mn²⁺ + 2MnO₄^- + 2H₂O ... balanced redox (balanced for mass and charge)

Let me know if this made sense and if it helps you.