Question

For the reaction, $A{g}_{\left(\text{aq}\right)}^{+}+C{l}_{\left(\text{aq}\right)}^{-}\rightleftharpoons AgC{l}_{\left(\text{s}\right)}$
Given:
$\begin{array}{cc}\text{Species}& \Delta G_f^{\circ }\left(\text{kJ/mol}\right)\\ A{g}_{\left(\text{aq}\right)}^{+}& +77\\ C{l}_{\left(\text{aq}\right)}^{-}& -129\\ AgC{l}_{\left(\text{s}\right)}& -109\end{array}$
Calculate $E_{\text{cell}}^{\circ }$ at $298\text{K}$.
Also find the solubility product of $AgCl$.

• A. $0.59\text{V}$ and ${\text{K}}_{\text{sp}}={10}^{-10}$
• B. $0.79\text{V}$ and ${\text{K}}_{\text{sp}}=3\times {10}^{-5}$
• C. $0.36\text{V}$ and ${\text{K}}_{\text{sp}}={10}^{-14}$
• D. $-0.59\text{V}$ and ${\text{K}}_{\text{sp}}={10}^{-10}$

Answer 1

The correct option is A $0.59\text{V}$ and ${\text{K}}_{\text{sp}}={10}^{-10}$

$A{g}^{+1}\left(aq\right)+C{l}^{-1}\rightleftharpoons AgCl\left(s\right)$

$\Delta G_r^{o}xn=\Delta G_f^o\left(products\right)-\Delta G_f^o\left(reactants\right)$

$=-109+129-77$

$=-55kJmo{l}^{-1}$

$Ag\left(s\right)\to A{g}^{+}+e^{-}\left(aq\right)$ ----1

$\frac{1}{2}C{l}_2\left(1atm\right)+e^{-}\to C{l}^{-}\left(aq\right)$ ----2

Adding 1 and 2

$Ag\left(s\right)+\frac{1}{2}C{l}_2\left(g\right)\to A{g}^{+}\left(aq\right)+C{l}^{-}\left(aq\right)$

$n=1$

$\Delta G_{rxn}^o=-55$

For the reverse reaction,

$\Delta G_{rxn}^o=55$

$-nF\Delta E_{rxn}^o=55$

$-1\times 96500\times \Delta E^o=55$

$\Delta E^o=-0.59$

$\Delta E^o=\frac{0.059}{1}log\left(Ks{p}_{AgCl}\right)$

$-0.59=0.059log\left(Ks{p}_{AgCl}\right)$

$log\left(Ks{p}_{AgCl}\right)=-10$

$Ks{p}_{AgCl}={10}^{-10}$