The standard reduction potentials for the two half-cell reactions are given below :

${\mathrm{Cd}}_{\left(\mathrm{aq}\right)}^{2+}+2{\mathrm{e}}^{-}\to {\mathrm{Cd}}_{\left(\mathrm{s}\right)};{\mathrm{E}}^0=-0.40\mathrm{V}$

${\mathrm{Ag}}_{\left(\mathrm{aq}\right)}^{+}+{\mathrm{e}}^{-}\to {\mathrm{Ag}}_{\left(\mathrm{s}\right)};{\mathrm{E}}^0=0.80\mathrm{V}$

The standard free energy change for the reaction

$2{\mathrm{Ag}}_{\left(\mathrm{aq}\right)}^{+}+{\mathrm{Cd}}_{\left(\mathrm{s}\right)}\to 2{\mathrm{Ag}}_{\left(\mathrm{s}\right)}+{\mathrm{Cd}}^{2+}$ given by

From given data ${{\mathrm{E}}^0}_{{\mathrm{Cd}}^{+2}/\mathrm{Cd}}$ is  less than   ${{\mathrm{E}}^0}_{{\mathrm{Ag}}^{+}/\mathrm{Ag}}$. So,

$$\begin{gathered} {{\mathrm{E}}^0}_{\mathrm{Cell}}={{\mathrm{E}}^0}_{\mathrm{Cathode}}-{{\mathrm{E}}^0}_{\mathrm{Anode}} \\ {\mathrm{E}}_{\mathrm{cell}}^0={\mathrm{E}}_{\mathrm{Ag}}^0-{\mathrm{E}}_{\mathrm{Cd}}^0 \\ {\mathrm{E}}_{\mathrm{cell}}^0=0.80 \mathrm{V}-(-0.40 \mathrm{V})=1.20 \mathrm{V} \end{gathered}$$

Maximum work done by galvanic cell is equal to decrease in its Gibbs free energy

(${\mathrm{\Delta G}}^0$ ) is  thermodynamic quantity obtained from ${\mathrm{E}}^0$ values. 

$\therefore$ ${\mathrm{\Delta G}}^0=-{\mathrm{nFE}}_{\mathrm{cell}}^0$                          n= no of electrons involved

$∆{\mathrm{G}}^{\circ }=-2\times 96500\times 1.2=-231.6 \mathrm{kJ}$