If the oxidation of oxalic acid by acidic `MnO_(4)^(c-)` solution is carried out in a reversible cell, then what is the electrode reaction and equilibirum constant of the cell reaction.
Given `:`
`E^(c-)._((MnO_(4)^(c-)|Mn^(2+)))=1.51V`
`E^(c-)._((CO_(2)|C_(2)O_(4)^(2-)))-0.49V`

Explanation

To solve the problem, we need to determine the electrode reactions and the equilibrium constant for the oxidation of oxalic acid by acidic permanganate solution in a reversible cell.

### Step 1: Identify the half-reactions

1. **Oxidation of Oxalic Acid (C2O4^2- to CO2)**:
The half-reaction for the oxidation of oxalic acid (C2O4^2-) to carbon dioxide (CO2) involves the loss of electrons:
\[
C_2O_4^{2-} \rightarrow 2 CO_2 + 2 e^-
...