J.R. S. answered 03/29/23
Ph.D. University Professor with 10+ years Tutoring Experience
From the table of standard reduction potentials:
Zn2+ + 2e- ==> Zn Eº = -0.76 V
Ag+ + e- ==> Ag Eº = +0.80 V
So Ag + will be reduced and Zn will be oxidized and the half reactions will be:
2Ag+(aq) +2e- ==> 2Ag(s) ... +0.80 (CATHODE)
Zn(s) ==> Zn2+(aq) ... -0.76 (ANODE)
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2Ag+(aq) + Zn(s) ==> 2Ag(s) + Zn2+(aq) .. overall reaction for the galvanic cell
Eºcell = cathode - anode = 0.80 - (-0.76) = 0.80 + 0.76
Eºcell = 1.56 V