J.R. S. answered 09/11/21
Ph.D. University Professor with 10+ years Tutoring Experience
Cu2+ + 2e- ==> Cu(s) ... Eo = 0.34 V
Fe2+ + 2e- ==> Ag(s) ... Eo = -0.41 V
Since 0.34 is greater than -0.41, reduction will be the Cu reaction and oxidation will be the Fe reaction.
a) Anode = Cu and cathode is Fe
Eo = 0.34 + 0.41 = 0.75 V
b) Cu2+ + 2e- ==> Cu(s) Eo = 0.34 V
Ag+ + e- ==> Ag(s) Eo = 0.80 V
Eo = 0.80 - 0.34 = 0.46 V (your standard reduction potentials may vary slightly from the ones I used)