Zn|Zn2+(1 M)∥Ag+(1 M)|Ag
Ecell=EOX(Zn/Zn2+)+Ered(Ag+/Ag)
With the help of the following two cells, the above equation can be obtained:
Ag|Ag+(1 M)∥Cu2+(1 M)|Cu;E∘=−0.46 volt
or Cu|Cu2+(1 M)∥Ag+(1 M)|Ag;E∘ will be +0.46 volt
or +0.46=EOX(Cu/Cu2+)+Ered(Ag+/Ag)....(i)
Zn|Zn2+(1 M)∥Cu2+|Cu;E∘=+1.10 volt
+1.10=EOx(Zn/Zn2+)+Ered(Cu2+/Cu)....(ii)
Adding eqs. (i) and (ii),
+1.56=E+Ered(Ag+/Ag)+EOX(Zn/Zn2+)+Ered(Cu2+/Cu)
Since, EOX(Cu/Cu2+)=−Ered(Cu2+/Cu)
So, +1.56=EOX(Zn/Zn2+)+Ered(Ag+/Ag)
Thus, the emf of the following cell is
Zn|Zn2+(1 M)∥Ag+(1 M)|Ag is +1.56 volt.