J.R. S. answered 07/23/21
Ph.D. University Professor with 10+ years Tutoring Experience
I'll do a few, but we need to look up the standard reduction potentials for each electrode. I'll leave that to you. I'm also assuming that the cell isn't the way it is written, but we have to figure out which is the cathode and which is the anode.
Cu-Zn
Cu2+ + 2e- => Cu(s) Eº = 0.34 V (cathode)
Zn2+ + 2e- => Zn(s) Eº = -0.76 V (anode)
Eºcell = 0.34 - (-0.76) = 1.1 V
Cu-Pb
Cu2+ + 2e- => Cu(s) Eº = 0.34 V (cathode)
Pb2+ + 2e- => Pb(s) Eº = -0.13 V (anode)
Eºcell = 0.34 - (-0.13) = 0.47 V
Zn-Pb
Zn2+ + 2e- => Zn(s) Eº = -0.76 V (anode)
Pb2+ + 2e- => Pb(s) Eº = -0.13 V (cathode)
Eºcell = -0.13 - (-0.76) = 0.63 V
You should try the rest.