Question

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.

Answer 1

`-DeltaG^(@)=nxxFxxE^(@)=2xx96500xx1.1=212.3kJ`
`DeltaG^(@)=-212.3kJ" "mol^(-1)`
`DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)`
`DeltaS^(@)=(DeltaH^(@)-DeltaG^(@))/(T)=(-216.7-(-212.3))/(298)`
`=-0.01476" kJ "K^(-1)mol^(-1)`
`=-14.76" J "K^(-1)mol^(-1)`.