Electrochemistry Result Question 39
39. The standard potential of the following cell is 0.23 V at $15^{\circ} \mathrm{C}$ and 0.21 V at $35^{\circ} \mathrm{C}$.
$ \text { Pt }\left|\mathrm{H}_2(g)\right| \mathrm{HCl}(a q)|\mathrm{AgCl}(s)| \mathrm{Ag}(s) $
(i) Write the cell reaction.
(ii) Calculate $\Delta H^{\circ}$ and $\Delta S^{\circ}$ for the cell reaction by assuming that these quantities remain unchanged in the range $15^{\circ} \mathrm{C}$ to $35^{\circ} \mathrm{C}$.
(iii) Calculate the solubility of AgCl in water at $25^{\circ} \mathrm{C}$.
Given, the standard reduction potential of the
$\left(\mathrm{Ag}^{+}(a q) / \mathrm{Ag}(s)\right.$ is $0.80 V$ at $25^{\circ} \mathrm{C}$.
(2001, 10M)
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Solution:
(ii) $\Delta G^{\circ}=-n E^{\circ} F=\Delta H^{\circ}-T \Delta S^{\circ}$
At $15^{\circ} \mathrm{C}:-0.23 \times 96500=\Delta H^{\circ}-288 \Delta S^{\circ}$ ….(i)
At $35^{\circ} \mathrm{C}:-0.21 \times 96500=\Delta H^{\circ}-308 \Delta S^{\circ}$ ….(ii)
$\Rightarrow 96500(0.23-0.21)=-20 \Delta S^{\circ} $
$\Rightarrow \Delta S^{\circ}=-\frac{96500 \times 0.02}{20}=-96.5 $ $ \mathrm{J}$
Substituting value of $\Delta S^{\circ}$ in (i)
$ \Delta H^{\circ}=288 \times(-96.5)-0.23 \times 96500=-49.987 $ $\mathrm{kJ} $
(iii) At $25^{\circ} \mathrm{C}$
$ -E^{\circ} \times 96500=-49987-298(-96.5) $
$ \Rightarrow \quad E^{\circ}=0.22 \mathrm{~V} $
$ \Rightarrow \quad \mathrm{AgCl}(s)+e^{-} \longrightarrow \mathrm{Ag}+\mathrm{Cl}^{-} ;\quad E^{\circ}=0.22 \mathrm{~V} $
$ \mathrm{Ag} \longrightarrow \mathrm{Ag}^{+}+e^{-} ; \quad E^{\circ}=-0.80 \mathrm{~V} $
$ \text { Adding : } \mathrm{AgCl}(s) \longrightarrow \mathrm{Ag}^{+}+\mathrm{Cl}^{-} ; \quad E^{\circ}=-0.58 \mathrm{~V} $
$ \Rightarrow \quad E^{\circ}=0.0592 \log K_{\mathrm{sp}} $
$ \Rightarrow \quad \log K_{\mathrm{sp}}=\frac{-0.58}{0.0592}=-9.79 $
$ \Rightarrow \quad K_{\text {sp }}=1.6 \times 10^{-10} $
BETA
