Question

The standard reduction potential for the following two reactions are given:
$AgCl+e^{-}\to Ag\left(s\right)+C{l}^{-2}(aq.);E^{\circ }=0.22V....\left(i\right)$
$A{g}^{+}(aq.)+e^{-}\to Ag\left(s\right)E^{\circ }=0.80V....\left(ii\right)$
The solubility product of $AgCl$ under standard condition will be:

• A. $1.613\times {10}^{-5}{M}^2$
• B. $1.535\times {10}^{-8}{M}^2$
• C. $3.213\times {10}^{-10}{M}^2$
• D. $1.535\times {10}^{-10}{M}^2$

Answer 1

Answer: (D)

$1.535\times {10}^{-10}{M}^2$

From eqs. (i) and (ii),
$AgCl\left(s\right)\rightleftharpoons A{g}^{+}+C{l}^{-};E^{\circ }=0.22-0.80=-0.58V$
$E_{cell}=E_{cell}^{\circ }-\frac{0.0591}{2}\log \left[A{g}^{+}\right]\left[C{l}^{-}\right]$
$0=-0.58-\frac{0.0591}{2}\log \left[A{g}^{+}\right]\left[C{l}^{-}\right]$
$K_{sp}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]=1.535\times {10}^{-10}{M}^2$.