Calculate the standard Gibbs energy for the cell :
`Zn(s)|Zn^(2+)(aq)||Cd^(2+)(aq)|Cd(s)`
`E_((Zn^(2+)//Zn))^(@)=-0.76V,E_((Cd^(2+)//Cd))^(@)=-0.403V,F=96500C`.

Explanation

Cell representation is,
`Zn(s)|Zn^(2+)(aq)||Cd^(2+)(aq)|Cd(s)`
Cell reactions are
Oxidation: `Zn(s) to Zn^(2+)(aq)+Ze^(-)`
Reduction:
`Cd^(2+)(aq)+2e^(-) to Zn^(2+)(aq)+Cd(s)`
Here, n=2
and F=96500 C `mol^(-1)`
`EMF^(@)=E_(R)^(@)-E_(L)^(@)`
`=E_(Cd^(2+)//Cd)^(@)-E_(Zn^(2+)//Zn)^(@)`
`=[-0.403-(-0*76)]V`
`=(0.76-0.403)V=0.357V`
`DeltaG^(@)=-nE^(@)F`
`=-2xx(0.357V)xx(96500C" "mol^(-1))`
`=-68901J" "mol^(-1)(1Cxx1V=1J)`
`=-68.9kJ" "mol^(-1)`