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Question

The emf (E) of the following cells are:
Ag/Ag+(1 M)Cu2+(1 M)|Cu;E=0.46 volt
Zn/Zn2+(1 M)Cu2+(1 M)|Cu;E=+1.10 volt
Calculate the emf of the cell:
Zn|Zn2+(1 M)Ag+(1 M)|Ag.

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Solution

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.

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