Calculate the cell e.m.f. and `DeltaG` for the cell reaction at `25^(@)C` for the cell:
`Zn_((s))|Zn^(2+)(0.004M)||Cd^(2+)(0.2M)|Cd_((s))`
`E^(@)` values at `25^(@)C,Zn^(2+)//Zn=-0.763V`
`Cd^(+2)//Cd=-0.403V`
`F=96,500,R=8.314JK^(-1)mo"le"^(-1)`.

Explanation

At anode, `Zn rarr Zn^(2+)+2e^(-)`
At cathode,
`ul(Cd^(2+) + 2e^(-) rarr Cd)`
`ul(Zn +Cd^(2+) rarr Zn^(2+)+ Cd)`
`E= E^(@)- (0.059)/(2) "log" ([Zn^(2+)])/([Cd^(2+)])`
`=0.36- (0.059)/(2)"log" (0.04)/(0.2)`
`=0.36 + (0.059)/(2) xx 1.699`
=0.41V
`Delta G^(@)= -nEF`
`= -2(0.41) (96500)`
`= -79130J mol^(-1)`
`= -79.130 kJ mol^(-1)`