Question

The standard reduction potential for two reactions is given below.
$AgC{l}_{\left(s\right)}+e^{-}\to A{g}_{\left(s\right)}+C{l}_{\left(aq\right)}^{-};E^{\circ }=0.22V$
$A{g}_{\left(aq\right)}^{+}+e^{-}\to A{g}_{\left(s\right)};E^{\circ }=0.08V$
The solubility product of $AgCl$ under standard conditions of temperature is given by

• A. $1.6\times 10^{-5}$
• B. $1.5\times 10^{-8}$
• C. $1.5\times 10^{-10}$
• D. $3.2\times 10^{-8}$

Answer 1

Answer: (C)

$AgCl\rightleftharpoons A{g}^{+}+C{l}^{-}{K}_{sp}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$
$\frac{AgC{l}_{\left(s\right)}+e^{-}\to A{g}_{\left(s\right)}+C{l}_{\left(aq\right)}^{-};E^{\circ }=0.22VA{g}_{\left(aq\right)}^{+}+e^{-}\to A{g}_{\left(s\right)};E^{\circ }=0.08V}{AgCl\to A{g}^{+}+C{l}^{-};E_{cell}^{\circ }=0.22+\left(-0.80\right)=-0.58V}$
Applying $E_{cell}^{\circ }=\frac{0.0591}{n}$ lof $K_{sp}$
log $K_{sp}=-\frac{0.58\times 1}{0.0591}=-9.8138$
Taking antilog $K_{sp}=1.5\times 10^{-10}$