First, we write the half equations for the reduction of the chemical species present:
CuāŗĀ² + 2e ā Cu; E° = 0.34 V
NiāŗĀ² + 2e ā Ni;Ā E° =Ā - 0.23 V
In order to determine the potential of the cell, we find the difference between the two values. For this:
E(cell) = 0.34 - (-0.23)
E(cell) = 0.57 V
The second option is correct. (The difference in values is due to different values in literature, and it is negligible)