The correct option is A Ecell=E∘−0.0296 log([Zn2+][Cu2+])
Let us first recall the general form of Nernst equation which is
Ecell=E∘−2.303RTnFlog([Cc][Dd][Aa][Bb])
From the given cell reaction, we can write
Reduction on the right:Cu2+(aq) + 2e−→Cu(s)
Oxidation on the left: Zn(s)→Zn2+(aq) + 2e−
Net reaction of cell is Zn(s)+Cu2+(aq) →Cu(s)+Zn2+(aq)
From here we see that, n = 2
We already know
R=8.314 JK−1mol−1
T=25∘C=298 K
Ecell=E∘−2.303RTnFlog([Zn2+][Cu(s)][Cu2+][Zn(s)])
Don't forget that we consider the concentration of solids as unity.
If we put these into the above equation, we end up with
E = E∘−0.0296 log([Zn2+][Cu2+])