Calculate the standard Gibbs energy for the cell :
`Zn(s)+Zn^(2+)(aq)||Cu^(2+)(aq)|Cu(s)`
`E_((Zn^(2+)//Zn)^(@))=-0.76V,E_((Cu^(2+)//Cu))^(@)=0.34V`, `F=96500 C`.

Explanation

`Zn(s)|Zn^(2+)(aq)||Cu^(2+)(aq)|Cu(s)`
`EMF^(@)=E_(R)^(@)-E_(L)^(@)`
`=E_(Cu^(2+)//Cu)^(@)-E_(Zn^(2+)//Z)^(@)`
`=0.34V-(-0.76V)`
`=(0.34+0.76)V=1.10V`
Cell reaction are,
Oxidation : `Zn(s) to Zn^(2+)(aq)+Ze^(-)`
Reduction : `Cu^(2+)(aq)+2e^(-) to Cu(s)`
Overall :
`Zn(s)+Cu^(2+)(aq) to Cu(s)+Zn^(2+)(aq)`
Thus, n=2
`DeltaG=-nE^(@)F`
`=-2xx1.10Vxx96500C" "mol^(-1)`
`=-212300C" "V" "mol^(-1)`
`(1Cxx1V=1J)`
`=-212*3kJ" "mol^(-1)`