What is the standard cell potential E for this half reaction Pd^+2 (aq) +2e- ---> Pd(s)?

Question

Standard reduction potential

Cu(s) + pd^+2 (aq) -> Cu^+2 (aq) + pd(s) E nut = 0.650 V

CU^+2 (aq) + 2e- -> Cu (s) E nut = 0.337 V

Answer 1

Cu(s) + Pd^2+(aq) --> Cu^2+(aq) + Pd(s)........Eo = 0.650 V

Rewrite to fit the balanced redox reaction above....

Cu(s) --> Cu^2+(aq) + 2e-............Eo= -0.337 V

We want

Pd^2+(aq) + 2e- --> Pd(s)

Since total E of the electrochemical cell is 0.650V and the E(anode) = -0.337, we can simply solve for E(cathode), which would be the Eo for Pd^2+(aq) + 2e- --> Pd(s)

So, knowing that

E(total) = E(cathode) - E(anode)

Plug everything in

0.650 = E(cathode) - (-0.337)

E(cathode) = 0.313

Thus,

Pd^2+(aq) + 2e- --> Pd(s) has Eo = 0.313V

[Answer: see above]