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

Cell is `Zn (s) |Zn^(2+) (0.0004M) || Cd^(2+) (0.2 M) | Cd(s)`
`Zn (s) + Cd^(2+) (aq) to Zn^(2+) (aq) + Cd(s)`
`E_(cell)^(Theta) = E_((Cd^(2+) |Cd))^(Theta) - E_((Zn^(2+) |Zn))^(Theta)`
`= - 0.403 - (- 0.763) = 0.36 V`
According to Nernst equation ,
`E = E^(Theta) - (0.059)/(2) "log" ([Zn^(2+)])/([Cd^(2+)])`
`[Zn^(2+)] = 0.0004 M , [Cd^(2+)] = 0.2` M
`E = 0.36 - (0.059)/(2) "log" (0.0004)/(0.2)`
=` 0.36 - (0.059)/(2) "log" 2 xx 10^(-3)`
= `0.36 - (0.059)/(2) xx (-2.6990)`
`= 0.36 + 0.08 = 0.44 ` V
Cell emf = 0.44 V
Cell emf = 0.44 V
Now `Delta G = - nFE`
`n = 2 mol , F = 96500 C , E = 0.44` V
`Delta G = - 2 xx 96500 xx 0.44 CV`
`= -84920 CV = -84.920 kJ`.