Calculate the maximum work that can be obtained from the daniell call given below,
`Zn(s)|Zn^(2+)(aq)||Cu^(2+)(aq)|Cu(s)`. Given that `E_(Zn^(2+)//Zn)^(@)=-0.76V` and `E_(Cu^(2+)//Cu)^(@)=+0.34V`.

Explanation

The maximum work that can be obtained from a cell is equal to free change for the cell reaction .
`Zn (s) + Cu^(2+) (aq) to Cu (s) + Zn^(2+) (aq)`
`E_(cell)^(Theta) = E^(Theta) ("cathode") - E^(Theta) ("anode")`
`E_((Cu^(2+) |Cu))^(Theta) - E_(Zn^(2+) |Zn)^(Theta)`
`= 0.34 - (-0.76) = 1.10 V `
Now , `Delta G^(Theta) =- n F E_(cell)^(Theta)`
`E_(cell) ^(Theta) = 1.10 V , n = 2 F = 96500 C`
`therefore Delta G^(Theta) = - (2 mol) xx (96500 C mol^(-1)) xx (1.10 V)`
`=-212 , 300 CV " " (1 J = 1 C xx V)`
`= - 212, 300 J or - 212.3 kJ`.
Thus , 212.3 kJ is the maximum work that can be obtained by the operation of this type of cell .