Electrochemistry 2 Question 40
39. The standard potential of the following cell is $0.23 \mathrm{~V}$ at $15^{\circ} \mathrm{C}$ and $0.21 \mathrm{~V}$ at $35^{\circ} \mathrm{C}$.
$$ \mathrm{Pt}\left|\mathrm{H}_{2}(g)\right| \mathrm{HCl}(a q)|\operatorname{AgCl}(s)| \mathrm{Ag}(s) $$
(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 $\mathrm{AgCl}$ in water at $25^{\circ} \mathrm{C}$. Given, the standard reduction potential of the $\left(\mathrm{Ag}^{+}(a q) / \mathrm{Ag}(s)\right.$ is $0.80 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$.
(2001, 10M)
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Solution:
- At anode $\quad \frac{1}{2} \mathrm{H}_{2} \longrightarrow \mathrm{H}^{+}+e^{-} ; E^{\circ}=0$
At cathode $\mathrm{AgCl}(s)+e^{-} \longrightarrow \mathrm{Ag}+\mathrm{Cl}^{-} ; E^{\circ}=$ ?
(i) Cell reaction : $\frac{1}{2} \mathrm{H}_{2}+\mathrm{AgCl}(s) \longrightarrow \mathrm{Ag}+\mathrm{H}^{+}+\mathrm{Cl}^{-}$
(ii) $\Delta G^{\circ}=-n E^{\circ} F=\Delta H^{\circ}-T \Delta S^{\circ}$
$$ \begin{aligned} & \text { At } 15^{\circ} \mathrm{C}:-0.23 \times 96500=\Delta H^{\circ}-288 \Delta S^{\circ} \ & \text { At } 35^{\circ} \mathrm{C}:-0.21 \times 96500=\Delta H^{\circ}-308 \Delta S^{\circ} \ & \Rightarrow \quad 96500(0.23-0.21)=-20 \Delta S^{\circ} \ & \Rightarrow \quad \Delta S^{\circ}=-\frac{96500 \times 0.02}{20}=-96.5 \mathrm{~J} \end{aligned} $$
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 $25^{\circ} \mathrm{C}$
$$ \begin{aligned} & -E^{\circ} \times 96500=-49987-298(-96.5) \ \Rightarrow \quad E^{\circ} & =0.22 \mathrm{~V} \ \Rightarrow \quad \mathrm{AgCl}(s)+e^{-} & \longrightarrow \mathrm{Ag}+\mathrm{Cl}^{-} ; E^{\circ}=0.22 \mathrm{~V} \ \mathrm{Ag} & \longrightarrow \mathrm{Ag}^{+}+e^{-} ; \quad E^{\circ}=-0.80 \mathrm{~V} \end{aligned} $$
Adding : $\mathrm{AgCl}(s) \longrightarrow \mathrm{Ag}^{+}+\mathrm{Cl}^{-} ; \quad E^{\circ}=-0.58 \mathrm{~V}$
$$ \begin{aligned} & \Rightarrow \quad E^{\circ}=0.0592 \log K_{\mathrm{sp}} \ & \Rightarrow \quad \log K_{\text {sp }}=\frac{-0.58}{0.0592}=-9.79 \ & \Rightarrow \quad K_{\mathrm{sp}}=1.6 \times 10^{-10} \end{aligned} $$
