The e.m.f. (`E^(0)` )of the following cells are
`Ag| Ag^(+)(1M) | | Cu^(2+)(1M) | Cu, E^(0) = -0.46V`,`Zn|Zn^(2+)(1M)| | Cu^(2+)(1M) | Cu : E^(0) = +1.10V`
Calculate the e.m.f. of the cell `Zn | Zn^(2+)(1M) | | Ag^(+)(1M) | Ag`

Explanation

`Zn|Zn^(2+) (1 M)||Ag^(+) (1 M)|Ag`
`E_(cell)=E_(o x(Zn//Zn^(2+)))+E_(red (Ag^(+)//Ag))`
With the help of the following two cells, the above equation can be obtained :
`Ag|Ag^(+) (1 M)||Cu^(2+) (1M)|Cu, E^(@)=-0.46" volt"`
or `Cu|Cu^(2+) (1 M)||Ag^(+) (1 M) |Ag, E^(@)" will be "+0.46" volt"`
or `+0.46=E_("ox"(Cu//Cu^(2+)))+R_("red "(Ag^(+)//Ag))` ...(i)
`Zn|Zn^(2+) (1 M)||Cu^(2+)|Cu, E^(@)=+1.10" volt"`
`+1.10=E_(o x (Zn//Zn^(2+)))+E_(red (Cu^(2+)//Cu))` ...(ii)
Adding eqs. (i) and (ii),
`+1.56=E=E_("red "(Ag^(+)//Ag))+E_(o x (Zn//Zn^(2+)))+E_("red "(Cu^(2+)//Cu))`
since, `E_(o x(Cu//Cu^(2+)))=-E_("red "(Cu^(2+)//Cu))`
So, `+1.56=E_(o x (Zn//Zn^(2+)))+E_("red "(Ag^(+)//Ag))`
Thus, the emf of the following cell is
`Zn|Zn^(2+)(1 M)||Ag^(+) (1 M)|Ag` is `+ 1.56` volt