J.R. S. answered 04/27/20
Ph.D. University Professor with 10+ years Tutoring Experience
a) V2+ + 2e- ==> V(s) Eº = -1.18 V
U3+ + 3e- ===> U(s) Eº = -1.79 V
Reduction at cathode is V2+ + 2e- ==> V(s) Eº = -1.18 V
Oxidation at anode is U(s) ===> U3+ + 3e- Eº = -1.79
Eºcell = cathode - anode = -1.18 - (-1.79)
Eºcell = 0.61 V
b) Sn4+ + 2e- ===> Sn2+ Eº = +0.15 V
Pb4+ + 2e- ===> Pb2+ Eº = +1.67 V
Reduction at cathode is Pb4+ + 2e- ===> Pb2+ Eº = +1.67 V
Oxidation at anode is Pb2+ ===> Pb4+ + 2e- Eº = +0.15 V
Eºcell = cathode - anode = 1.67 - 0.15
Eºcell = 1.52 V