Question 7.8: The Clark cell Zn|Zn^2+, Hg2SO4|Hg is often employed as a st......

Before we commence, we note that the spontaneous cell reaction is

Z\mathrm{n}+\mathrm{Hg_{2}S O_{4}}+7\mathrm{H}_{2}O\longrightarrow\mathrm{ZnSO_{4}}\cdot7\mathrm{H}_{2}O+2\mathrm{Hg^{0}}

so the cell reaction is a two-electron process.
Next, we recall from Equation (7.15) that \Delta G_{\mathrm{(cell)}}=-n F\times e m f. Inserting values for the cell at 298 K gives

\Delta G_{\mathrm{(cell)}}=-2\times96\,485\,\,\mathrm{Cmol}^{-1}\times1.423\,\,\mathrm{V}

\Delta G_{\mathrm{(cell)}}=-275\mathrm{~kJmol}^{-1}

Then, from Equation (7.18), we recall that

\Delta S_{\mathrm{(cell)}}=n F\left({\frac{\mathrm{d}(e m\!f)}{\mathrm{d}T}}\right)_{p}

Inserting values:

\Delta S_{\mathrm{(cell)}}=2\times96485\mathrm{~Cmol}^{-1}\times(-1.2\times10^{-4}\mathrm{~VK}^{-1})

\Delta S_{\mathrm{(cell)}}=-23.2\mathrm{~JK}^{-1}\mathrm{mol}^{-1}

Finally, from Equation (4.21), we say that \Delta H_{\mathrm{(cell)}}=\Delta G_{\mathrm{(cell)}}+T\Delta S_{\mathrm{(cell)}}. We again insert values:

\Delta H_{\mathrm{(cell)}}=(-275\ \mathrm{kJmol}^{-1})+(298\ \mathrm{K}\times-23.2\ \mathrm{JK}^{-1}\ \mathrm{mol}^{-1})

\Delta G_{\mathrm{(system)}}=\Delta H_{\mathrm{(system)}}-T\,\Delta S_{\mathrm{(system)}}              (4.21)

so

\Delta H_{\mathrm{(cell)}}=-282{\mathrm{~kJmol}}^{-1}