Question

The standard potential of the following cell is $0.23V$ at $15^{\circ }C$ and $0.21V$ at $35^{\circ }C$
$PtIH2(g)|HCl(aq)|AgCl(s)|Ag(s)$
(i) Write the cell reaction
(ii) Calculate $\Delta H^{\circ }C$ and $\Delta S^{\circ }C$ for the cell reaction by assuming
that these quantities remain unchanged in the range $15^{\circ }C$ to $35^{\circ }C$.
(iii) Calculate the solubility of $AgCl$ in water at $25^{\circ }C$.
Given, the standard reduction potential of the
$(A{g}^{+}(aq)/Ag(s)$ is $0.80V$ at $25^{\circ }C$.

Answer 1

At anode $\frac{1}{2}{H}_2⟶H^{+}+e^{-};E^{\circ }=0$
$\underline{\text{At cathode}AgCl(s)+e^{-}⟶Ag+C{l}^{-}+;E^{\circ }=?}$
(i) Cell reaction $\frac{1}{2}{H}_2+AgCl(s)⟶Ag+H^{+}+C{l}^{-}$
(ii) $\Delta G^{\circ }=-n{E}^{\circ }F=\Delta H^{\circ }-T\Delta S^{\circ }$
At $15^{\circ }C=-0.23\times 96500=\Delta H^{\circ }-288\Delta S^{\circ }$...(i)
At $35^{\circ }C=-0.21\times 96500=\Delta H^{\circ }-308\Delta S^{\circ }$...(ii)
$\Rightarrow 96500(0.23-0.21)=-20\Delta S^{\circ }$
$\Rightarrow \Delta S^{\circ }=-\frac{96500\times 0.02}{20}=-96.5J$
Substituting value of $\Delta S^{\circ }$ in (i)
$\Delta S^{\circ }=288\times (-96.5)-0.23\times 96500=-49.987kJ$
(iii) At $25^{\circ }C$
$-E^{\circ }\times 96500=-49987-298(-96.5)$
$\Rightarrow E^{\circ }=0.22V$
$\Rightarrow AgCl(s)+e^{-}⟶Ag+C{l}^{-};E^{\circ }=0.22V$
$Ag⟶A{g}^{+}+e^{-};E^{\circ }=0.80V$
Adding: $AgCl(s)⟶A{g}^{+}+C{l}^{-};E^{\circ }=0.58V$
$\Rightarrow E^{\circ }=0.0592\log K_{sp}$
$\Rightarrow \log K_{sp}=\frac{-0.58}{0.0592}=-9.79$
$\Rightarrow K_{sp}=1.6\times 10^{-10}$