The emf of the cell reaction
`Zn(s) +Cu^(2+) (aq) rarr Zn^(2+) (aQ) +Cu(s)`
is `1.1V`. Calculate the free enegry change for the reaction. If the enthalpy of the reaction is `-216.7 kJ mol^(-1)`, calculate the entropy change for the reaction.

Explanation

`-DeltaG^(Theta) = n xxF xx E^(Theta) = 2 xx 96500 xx 1.1 = 212.3 kJ`
`DeltaG^(Theta) =- 212.3 kJ mol^(-1)`
`DeltaG^(Theta) = DeltaH^(Theta) - T DeltaS^(Theta)`
`DeltaS^(Theta) = (DeltaH^(Theta) -DeltaG^(Theta))/(T) = (-216.7-(-212.3))/(298)`
`=- 0.01476 kJ mol^(-1) =- 14.76 J K^(-1) mol^(-1)`