Answer:
1. C has been oxidized in the reaction.
2. Mn has been reduced in the reaction.
3/4 The reducing agent is the C and the oxidizing agent is the Mn.
The reaction in an acidic solution is:
16H⁺ + 2MnO₄⁻ + 5C₂O₄⁻² → 10CO₂ + 2Mn²⁺ + 8H₂O
10 moles of electrons are been transferred.
Step-by-step explanation:
MnO₄⁻ (aq) + H₂C₂O₄ (aq) → Mn²⁺ (aq) + CO₂ (g)
In the permanganate, Mn acts with +7 as oxidation number, and in product side, we have Mn2+. It has decrease the oxidation number, so it has been reduced. This is the oxidizing agent.
In the oxalic acid, carbon has +3 as oxidation number, and in CO2, we see that acts with +4 so, it has increase it. This element is the reducing agent and has been oxidated.
8H⁺ + MnO₄⁻ + 5e⁻ → Mn²⁺ + 4H₂O
It is completed on the opposite side where there are oxygen, with as many water as there are oxygen, and on the opposite side complete with protons to balance the H
C₂O₄⁻² → 2CO₂ + 2e⁻
In oxalate, the carbon that acted with +3 gained an electron to oxidize to +4, but since there are 2 carbons, it gained 2 electrons. The oxygen is balanced by adding a 2 in stoichiometry
The halfs reaction have to be multipplied .2 (reduction) and .5 (oxidation) to balance the electrons in the main equation.
(8H⁺ + MnO₄⁻ + 5e⁻ → Mn²⁺ + 4H₂O) . 2
(C₂O₄⁻² → 2CO₂ + 2e⁻ ) . 5
16H⁺ + 2MnO₄⁻ + 10e⁻ + 5C₂O₄⁻² → 10CO₂ + 10e⁻ + 2Mn²⁺ + 8H₂O
16H⁺ + 2MnO₄⁻ + 5C₂O₄⁻² → 10CO₂ + 2Mn²⁺ + 8H₂O