Calculate the standard free energy change for the reaction, `Zn+ Cu^(2+) rarr Cu + Zn^(2+)`
Given that standard free energy (G°) for `Zn2+(aq), Cu2+(aq), are -147.2 kJ mol-1 and 65.0 kJ mol-1` respectively.

Explanation

At anode `Zn^(2+)(aq) + 2e^(-) rarr Zn(s), E^(@)= -0.76V`
At cathode : `Cu^(2+) (aq) + 2e^(-) rarr Cu(s), E^(@) = 0.34V`
`E_("cell")^(@) = E_("cathode")^(@) - E_("anode")^(@)= 0.34 -(-0.76)= 1.1V`
Since two mols of electrons are associated with the reaction, `Delta G^(@)= -nFE_("cell")^(@)= -2 mol xx 96500 C mol^(-1) xx 1.1V= -212300J= -212.3kJ`
Thus the maximum work aviable from the cell is 212.3kJ