Calculate the standard Gibbs energy for the cell :
`Zn(s)|Zn^(2+)(aq)||Sn^(2+)(aq)|Sn(s)`
`E_((Zn^(2+)//Zn))^(@)=-0.76V,E_((Sn^(2+)//Sn))^(@)`= `-0.16V`,`F=96500C`.

Explanation

Cell representation is,
`Zn(s)|Zn^(2+)(aq)||Sn(aq)|Sn(s)`
`EMF^(@)=E_(R)^(@)-E_(L)^(@)`
`=E_(Sn^(2+)//Sn)^(@)-E_(Zn^(2+)//Zn)^(@)`
`=-0.16V-(-0.76V)`
`=(0.76V-0.16)V=0.60V `
Cell reaction are,
Oxidation : `Zn(s) to Zn^(2+)(aq)+Ze^(-)`
Reduction : `Sn^(2+)(aq)+Ze^(-) to Sn(s)`
Overall : `Zn(s)+Sn^(2+)(aq) to Zn^(2+)(aq) + Sn(s)`
Here, `n=2 and F=96500C" "mol^(-1)`
`DeltaG=-nE^(@)F`
`=-2xx(0.60V)xx(96500C" "mol^(-1))`
`=-1.2xx96500C" V " "mol^(-1)`
`=-115800J" "mol^(-1)(1Cxx1V=1J)`
`=-115.88J" "mol^(-1)`