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

$\begin{array}{l}{\mathrm{Cd}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}⟶{\mathrm{Cd}}_{\left(\mathrm{s}\right)};{\mathrm{E}}^0=-0.40\mathrm{V}\\ {\mathrm{Ag}}_{\left(\mathrm{aq}\right)}^{+}+{\mathrm{e}}^{-}⟶{\mathrm{Ag}}_{\left(\mathrm{s}\right)};{\mathrm{E}}^0=0.80\mathrm{V}\end{array}$
The standard free energy change for the reaction $2{\mathrm{Ag}}_{\left(\mathrm{aq}\right)}^{+}+{\mathrm{Cd}}_{\left(\mathrm{s}\right)}⟶2{\mathrm{Ag}}_{\left(\mathrm{s}\right)}+{\mathrm{Cd}}^{2+}\left(\mathrm{aq}\right)$ is given by

${\mathrm{Ag}}^{+}+{\mathrm{e}}^{-}\to {\mathrm{Ag}}_{\left(\mathrm{s}\right)} {\mathrm{E}}^{\mathrm{o}} = 0.80 \mathrm{V}, ∆{\mathrm{G}}^{\mathrm{o}} = 1\mathrm{F} (0.80)$
$$\begin{gathered} 2\mathrm{Ag}+2{\mathrm{e}}^{-}\to {\mathrm{Ag}}_{\left(\mathrm{s}\right)} ∆{\mathrm{G}}^{\mathrm{o}} = -2\mathrm{F}(0.80) \\ = -1.6 \mathrm{F} \end{gathered}$$
${{\mathrm{Cd}}^{2+}}_{\left(\mathrm{aq}\right)}+2{\mathrm{e}}^{-}\to {\mathrm{cd}}_{\left(\mathrm{s}\right)} {\mathrm{E}}^0 = -0.40 \mathrm{V}$
$\begin{array}{rcl}{\mathrm{Cd}}_{\left(\mathrm{s}\right)} & \to & {\mathrm{Cd}}^{2+}+2{\mathrm{e}}^{-} {\mathrm{E}}^{\mathrm{o}} = 0.40 \mathrm{V}\\ ∆{\mathrm{G}}^0 & =& -2\mathrm{F}(0.40)\\ & =& -0.8 \mathrm{F} \end{array}$
Total 
$\begin{array}{rcl}∆{\mathrm{G}}^{\mathrm{o}} & =& -2.4 \mathrm{F}\\ & =& -2.4\times 96,500 \mathrm{J}\\ & =& -231.6 \mathrm{KJ} \end{array}$