1. what element is being reduced in this redox reaction?
2. what element is being oxidized in the reaction?
3. what is the reducing agent?
4. what is the oxidizing agent?
5. balance the reaction in an acidic solution and indicate how many moles of electrons are being transferred.

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.

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

sorry i was late, but i don't know what assignment your talking about, i'm in flvs as well, im in earth space science. but u are behind pace bc every class should be in the 3's or 4'

your answer would be “dry”