$\mathrm{Pt}\left|{\mathrm{Cl}}_2\left({\mathrm{P}}_1\right)\right|\mathrm{HCl}(0.1\mathrm{M})\left|\mathrm{Pt}\right|{\mathrm{Cl}}_2\left({\mathrm{P}}_2\right)$; cell reaction will be spontaneous if: 

$$\begin{gathered} {\mathrm{E}}_{\mathrm{cell}}={\mathrm{E}}_{\mathrm{cell}}^0-\frac{\mathrm{RT}}{1\mathrm{F}}\ln {\left[\frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}\right]}^{\frac{1}{2}} \\ {\mathrm{E}}_{\mathrm{cell}}^0={\mathrm{E}}_{\mathrm{C}}^0-{\mathrm{E}}_{\mathrm{a}}^0=0-0=0 \\ {\mathrm{E}}_{\mathrm{cell}}=\frac{\mathrm{RT}}{\mathrm{F}}\ln {\left[\frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}\right]}^{\frac{1}{2}} \\ {\mathrm{E}}_{\mathrm{cell}}=\frac{\mathrm{RT}}{2\mathrm{F}}\ln \left[\frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}\right] \\ {\mathrm{P}}_1>{\mathrm{P}}_2\Rightarrow {\mathrm{E}}_{\mathrm{cell}}=+\mathrm{ve} \\ \Rightarrow ∆\mathrm{G}=-\mathrm{ve} \\ \mathrm{Spontaneous} \mathrm{cell} \mathrm{reaction}. \end{gathered}$$

OR 

$$\begin{gathered} ∆\mathrm{G} = -2.303\mathrm{RTlogK} \\ ∆\mathrm{G} = -2.303\mathrm{RTlog}\frac{{\mathrm{P}}_1}{{\mathrm{P}}_2} \\ \mathrm{If} {\mathrm{P}}_1>{\mathrm{P}}_2 ∆\mathrm{G} = -\mathrm{ve} \\ \mathrm{Spontaneous}. \end{gathered}$$