I need help balancing this reaction in a basic solution.

Question

I- + MnO4^2- → MnO2 + I2 

Answer 1

I'll try to give a step by step procedure for balancing this in basic medium.

I^- + MnO₄^- ==> MnO₂ + I₂ (note: I believe it should be MnO₄^- and not MnO₄^2-)

Do the half reactions separately and then add them together at the end.

I- ==> I₂ ... oxidation half reaction

2I^- ==> I₂ ... balances the Iodines

2I^- ==> I₂ + 2e- ... balances the charge and this is the final balanced equation for the oxidation 1/2 reaction

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

MnO₄^- ==> MnO₂ + 2H₂O ... balances the oxygens

MnO₄^- + 4H₂O ==> MnO₂ + 2H₂O + 4OH^- ... balances the hydrogens using basic solution

MnO₄^- + 4H₂O + 3e- ==> MnO₂ + 2H₂O + 4OH^- ... balances the charge

MnO₄^- + 2H₂O + 3e- ==> MnO₂ + 4OH^- ... balanced equation for the reduction 1/2 reaction

Since 3 electrons were transferred in the reduction reaction, and 2 electrons were transferred in the oxidation reaction, we must equalize the electrons by multiplying the reduction reaction by 2 and the oxidation reaction by 3. Thus, we have...

6I^- ==> 3I₂ + 6e- ... oxidation reaction

2MnO₄^- + 4H₂O + 6e- ==> 2MnO₂ + 8OH^- ... reduction reaction

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6I^- + 2MnO₄^- + 4H₂O ==> 3I₂ + 2MnO₂ + 8OH^- ... balanced redox reaction