Balance the following redox reaction if it occurs in acidic solution.

What are the coefficients in front of H2C2O4 and H2O in the balanced reaction?

MnO4 (aq) + H2C2O4 (aq) → Mn2 + (aq) + CO2 (g)

6H⁺ + 2MnO₄⁻ + 5H₂C₂O₄ → 2Mn²⁺ + 10CO₂ + 8H₂O

Explanation:

MnO₄⁻ (aq) + H₂C₂O₄ (aq) → Mn²⁺ (aq) + CO₂ (g)

First of all, determinate the half reaction (reduction/oxidation)

Reduction - Oxidation state decrease

Oxidation - Oxidation state increase

MnO₄⁻ → Mn²⁺

Manganese acts with + 7 in permanganate, and it's get reduced to Mn²⁺

C₂O₄⁻² → CO₂

Carbon changes from + 3 to + 4. It's been oxidized.

8H⁺ + MnO₄⁻ + 5e⁻ → Mn²⁺ + 4H₂O

We add 4 water in products side because we have 4 oxygen in reactant side (acidic medium). Aftewards, we balance the protons. Mn to change from + 7 to + 2, has gained 5 moles of electrons.

C₂O₄⁻² → 2CO₂ + 2e⁻

We can balance the carbon with 2, so now the half reaction is balanced in charges and atoms. Carbon has released 2 mol of e⁻ (3 + → 4+)

(8H⁺ + MnO₄⁻ + 5e⁻ → Mn²⁺ + 4H₂O).2

(C₂O₄⁻² → 2CO₂ + 2e⁻).5

Let's multiply the reaction x2 and x5, to balance the electrons

16H⁺ + 2MnO₄⁻ + 10e⁻ → 2Mn²⁺ + 8H₂O

5C₂O₄⁻² → 10CO₂ + 10e⁻

We sum both reaction, so we can quit the electrons

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

Finally the balanced redox reaction is this:

6H⁺ + 2MnO₄⁻ + 5H₂C₂O₄ → 2Mn²⁺ + 10CO₂ + 8H₂O