Calculate the cell e.m.f. and `Delta`G for the cell reaction at 298 K for the cell.
`Zn(s)|Zn^(2+)(0.0004 M)|| Cd^(2+)(0.2 M)| Cd(s)`
Given `E_(Zn^(2+)//Zn)^(@)=-0.763 V,E_(cd^(2+)//cd)^(@)=-0.403 V " at "298 K, F=96500 C " mol"^(-1)`.

Explanation

Step I. Calculation of cell e.m.f.
According to Nernst equation : `E=E^(@)-(0.0591)/(n)"log"([Zn^(2+)(aq)])/([Cd^(2+)(aq)])`
`E_(cell)^(@)=E_((Cd^(2+)//Cd))^(@)-E_((Zn^(2+)//Zn))^(@)`
=(-0.403)-(-0.763)=0.36 V
`[Zn^(2+)(aq)=0.0004 M,[Cd^(2+)(aq)]`= 0.2 M, n=2
`:. " " E=(0.36 V)-(0.0591V)/(2)"log"(0.004)/(0.2)`
`0.36V-(0.0591V)/(2)xx(-2.6999)`
=0.36 V+0.08 V=0.44 V
Step II. Calculate of `DeltaG," " DeltaG=-nFE_(cell)`
`E_(cell)=0.44 V, n=2 " mol ",F`
=96500 C `mol^(-1)`
`:. " " DeltaG=-(2" mol")xx(96500 " C mol"^(-1))xx(0.44 V)`
`=-84920 " CV"=-84920 J`