Dear Student
The EMF of the Cell, E0 = 0.46 V
The cell reaction is:
Cu (s) + 2Ag+ (aq) ----> Cu2+ (aq) + 2Ag (s)
The half-cell reactions are:
2Ag+ + 2e- ---> 2Ag (s)
Cu (s) ------> Cu2+ (aq) + 2e-
Standard Reduction Potential of Ag+/ Ag = - 0.80 V
[Note: The value of EoAg+/Ag should be +0.80 V, and because they have said reduction potential, then value becomes - 0.80 V]
The question has a printing mistake.
Therefore, Eocell = Eo right - Eo left
= EoAg+/Ag - EoCu2+/Cu
Thus, 0.46 V = - 0.80 V - EoCu2+/Cu
Hence, EoCu2+/Cu = (-0.80 - 0.46) V = -1.26 V
Regards