Which of these shows the correct balanced half-reaction for the following? The reaction takes place in acidic solution.

C2O42–(aq) → CO2(g)

To balance a half-reaction, you need to follow a few steps. In this case, you are given the unbalanced half-reaction as follows:

C2O42-(aq) → CO2(g)

To balance this half-reaction in acidic solution, you can follow these steps:

Step 1: Separate the equation into two half-reactions: oxidation and reduction.
In this case, we can identify that carbon in C2O42-(aq) is going from an oxidation state of +3 to +4. Hence, it is being oxidized. On the other hand, oxygen in CO2(g) is going from an oxidation state of -2 to 0, indicating it is being reduced.

Step 2: Balance the atoms other than oxygen and hydrogen.
Starting with the oxidation half-reaction:
C2O42-(aq) → 2CO2(g)

Step 3: Balance the oxygen atoms by adding water molecules to the side that needs more oxygen.
Since there are four oxygens on the left side and only two on the right side, we need to add two H2O molecules to the right side:
C2O42-(aq) → 2CO2(g) + 2H2O(l)

Step 4: Balance the hydrogen atoms by adding H+ ions to the side that needs more hydrogen.
As we are working with an acidic solution, we need to balance the hydrogen atoms. Since there are four hydrogens on the right side and none on the left side, we need to add 4 H+ ions to the left side:
C2O42-(aq) + 4H+ → 2CO2(g) + 2H2O(l)

Step 5: Balance the charge by adding electrons (e-) to the appropriate side.
In this case, we can see that the left side has a total charge of -6 (2-, considering two C2O42-), while the right side has a charge of 0. To balance the charges, we need to add 6 electrons (e-) to the left side:
C2O42-(aq) + 4H+ + 6e- → 2CO2(g) + 2H2O(l)

Now, you have balanced the oxidation half-reaction.

Remember that in a redox reaction, there is always a reduction half-reaction that accompanies the oxidation half-reaction. In this case, the reduction half-reaction is the formation of CO2(g) from C2O42-(aq).

To obtain the balanced reduction half-reaction, you can flip the oxidation half-reaction and reverse the sign of the electrons:

CO2(g) + 2H2O(l) → C2O42-(aq) + 4H+ + 6e-

Finally, you have the balanced half-reactions:

Oxidation: C2O42-(aq) + 4H+ + 6e- → 2CO2(g) + 2H2O(l)
Reduction: CO2(g) + 2H2O(l) → C2O42-(aq) + 4H+ + 6e-