Question

Calculate Ksp of Ag₂(C₂O₄) at 25 degrees C, (via standard Reduction potentials)

Answer 1

2CO2(g) + 2e- <-> C2O42- Eo = -0.49 V
Ag+ + 1e- ---> Ag Eo = 0.8 V
The reaction with higher reduction potential occurs at cathode:
At anode: 2CO2(g) + 2e- <-> C2O42- Eo = -0.49 V
At cathode: 2Ag+ + 2e- ---> Ag Eo = 0.8 V
Eo cell = Eo cathode -Eo anode = 0.8 - (-0.49) = 1.29 V

Use:
Eo cell = (0.059/n) log Kc
Here, number of electron transfered, n = 2
Kc here is Ksp because reaction taking place is:
Ag2(C2O4) <-----> 2Ag+   + C2O4 2-

Putting values:
Eo cell = (0.059/n) log Kc
1.29 = (0.059/2) log Ksp
Ksp = 5.36*10^43

This answer is a bit weird but still i think this is the process you should follow