Question

The standard potential of the following cell is $ 0.23 \,V$ at $15^\circ C$ and $0.21 \,V$ at $35^\circ C $
$Pt I H2(g ) | HCl(aq) | AgCl(s) | Ag(s)$
(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
$(Ag^+ (aq)/Ag (s)$ is $0.80 \,V$ at $25^\circ C$.

Answer 1

At anode $\frac{1}{2} H_2 \longrightarrow H^+ + e^- ; E^\circ = 0 $
$\underline{ \text{At cathode} AgCl(s)+e^- \longrightarrow Ag + Cl^- + ; E^\circ = ? }$
(i) Cell reaction $\frac{1}{2}H_2 + AgCl(s) \longrightarrow Ag + H^+ + Cl^- $
(ii) $\Delta G^\circ = - nE^\circ F=\Delta H^\circ -T\, \Delta S^\circ$
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 $
$\Rightarrow \Delta S^\circ = - \frac{96500 \times 0.02}{20} = -96.5\, J$
Substituting value of $\Delta S^\circ$ in (i)
$\Delta S^\circ = 288 \times (-96.5)-0.23 \times 96500 = - 49.987\, kJ$
(iii) At $25^\circ C$
$-E^\circ \times 96500 = - 49987-298(-96.5) $
$\Rightarrow E^\circ =0.22\, V$
$\Rightarrow AgCl(s)+e^- \longrightarrow Ag+Cl^-; E^\circ = 0.22\, V$
$Ag \longrightarrow Ag^+ +e^-; E^\circ = 0.80\, V$
Adding: $ AgCl(s) \longrightarrow Ag^+ +Cl^-; E^\circ = 0.58\, V$
$\Rightarrow E^\circ =0.0592\, \log\, K_{sp}$
$\Rightarrow \log\, K_{sp}=\frac{-0.58}{0.0592} = - 9.79$
$\Rightarrow K_{sp}= 1.6 \times 10 ^{-10} $