Balance the equation in acidic solution

Question

MnO₂^-+ C₂O₄^- ------ Mn²⁺+ CO₂

Answer 1

I am going to use C₂O₄^2- as this is the oxalate ion.

1) Write half reactions for elements oxidized and reduced (balance those elements - Mn and C)

MnO₄^- → Mn²⁺

C₂O₄^2- → 2CO₂ (C's balanced)

2) Use H₂O to balance Os

MnO₄^- → Mn²⁺ + 4H₂O

C₂O₄^2- → 2CO₂ (Os balanced already)

3) Add H⁺ to balance Hs

8H⁺ + MnO₄^- → Mn²⁺ + 4H₂O

C₂O₄^2- → 2CO₂ (No Hs)

4) Add e^- to balance charge

5e^- + 8H⁺ + MnO₄^- → Mn²⁺ + 4H₂O

C₂O₄^2- → 2CO₂ + 2e^-

5) Combine equations so as to cancel e^- in the red and ox 1/2 reactions (2 x rxn 1 + 5 x rxn 2)

16H⁺ + 5C₂O₄^2- + 2MnO₄^- → 2Mn²⁺ + 8H₂O + 10CO₂

This is correct in acidic medium. For basic, you can solve for as if acidic and add OH^- to both sides and turn H⁺ into H₂O, There is another method using Oxidation States that I prefer where you calculate the electrons needed in the reactions to satisfy the change of oxidation state, use the charged species (H⁺ or OH^-) to balance charge, use H₂O to balance O or H and check with the other element balance. This method is symmetric and has a built-in check.