Question

The standard potential of the following cell is $0.23V$ at ${15}^{\circ }C$ and $0.21V$ at ${25}^{\circ }C$
$Pt{H}_2\left(g\right)|HCl(aq.)||AgCl\left(s\right)|Ag\left(s\right)$
(i) Write cell reaction.
(ii) Calculate $△H^{\circ }$ and $△S^{\circ }$ 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 $A{g}^{+}/Ag$ couple is $0.80volt$ at ${25}^{\circ }C$.

Answer 1

(i) Electrode process:
$\frac{1}{2}{H}_2\to H^{+}+e^{-}$ (Anode)
$\underset{–}{AgCl+e^{-}\to Ag+C{l}^{-}}$ (Cathode)
$\underset{–}{\frac{1}{2}{H}_2+AgCl\rightleftharpoons H^{+}+Ag+C{l}^{-}}$
(ii) We know that, $△G^{\circ }=△H^{\circ }-T△S^{\circ }$
$-22195=△H^{\circ }-288\times △S^{\circ }$
$-20265=△H^{\circ }-308\times △S^{\circ }$
On solving, $△S^{\circ }=-96.5J,△H^{\circ }=49.987kJ$
(iii) $E=E^{\circ }-\frac{0.0591}{n}{\log }_{10}Q$
At equilibrium, $E=0,Q=K=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$
$0=(0.8-0.22)+\frac{0.0591}{1}\log K_{sp}$
$\frac{(-0.8+0.22)}{0.0591}=\log K_{sp}$
$K_{sp}=1.47\times {10}^{-10}$
Solubility, $S=\sqrt{K_{sp}}$
$=\sqrt{1.47\times {10}^{-10}}=1.21\times {10}^{-5}M$.