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)

Redox reaction equations involve loss and gain of electrons

To balance the redox reaction, we must first break the whole reaction into half equations;

Oxidation half equation;

5C2O4^2-(aq) + 10e + 8H^+> 10CO2(g) + 4H2O(l)

Reduction half equation;

2MnO4^- (aq) + 10e + 8H^+> 2Mn^2+(aq) + 4H2O(l)

Having written the two half equation, we can now combine, excluding the number of electrons lost or gained to obtain;

5C2O4^2-(aq) + 2MnO4^- (aq) + 16H^+> 10CO2(g) + 2Mn^2+(aq) + 8H2O(l)

The coefficient of H2C2O4 is 5 and that of water is 8

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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

the answer to your question is c