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

Explanation

Step I. Calculate 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) = 0.36V`
`[Zn^(+)(aq)] = 0.0004M, [Cd^(2+)(aq)] = 0.2M,n = 2`
`E = (0.36) -((0.0591V))/(2)log.(0.004)/(0.2)`
`= 0.36 -((0.0591V))/(2) xx (-2.69990)`
`= 0.36 V + 0.08 = 0.44 V`
Step II. Calculate of `DeltaG`:
`DeltaG =- nFE_(cell)`
`:. DeltaG =- (2mol) xx (96500 C mol^(-1)) xx (0.44V)`.
`=- 84920 CV =- 84920 J`