The cell potential of the cell $\mathrm{Pt}\mid {\mathrm{H}}_2\left(\mathrm{g}, 0.25 \mathrm{atm}\left|\mathrm{HCl}\right(\mathrm{aq}, c\left)\right|{\mathrm{H}}_2(\mathrm{g}, 0.5 \mathrm{atm})\mid \mathrm{Pt}\right.$ at 25 °C  is.

The cell reaction is 

Reduction at right-hand electrode $2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\to {\left({\mathrm{H}}_2\left(\mathrm{g}\right)\right)}_{\mathrm{R}}$

Oxidation at left-hand electrode  $\frac{{\left({\mathrm{H}}_2\left(\mathrm{g}\right)\right)}_{\mathrm{L}}\to 2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}}{{\left({\mathrm{H}}_2\left(\mathrm{g}\right)\right)}_{\mathrm{L}}\to {\left({\mathrm{H}}_2\left(\mathrm{g}\right)\right)}_{\mathrm{R}}}$

The cell potential is

$E_{\text{cell }}=-\frac{R\mathrm{T}}{2F}\ln \left(\frac{{\left(p_{{\mathrm{H}}_2}\right)}_{\mathrm{R}}}{{\left(p_{{\mathrm{H}}_2}\right)}_{\mathrm{L}}}\right)=-\frac{0.059\mathrm{V}}{2}\log \left(\frac{0.5}{0.25}\right)=-\frac{0.059\mathrm{V}}{2}\left(0.30\right)=-0.0089\mathrm{V}$