Question

The standard reduction potentials for two reactions are given below:
$\text{AgCl}\left(\text{s}\right)\text{+}{\text{e}}^{\text{-}}\to \text{Ag}\left(\text{s}\right)+\text{C}{\text{l}}^{\text{-}}\left(\text{aq}\right)$ ${\text{E}}^{\circ }=\text{0}\text{.22V}$
$\text{A}{\text{g}}^{+}\left(aq\right)+e^{-}\to Ag\left(s\right)$ ${\text{E}}^{\circ }=\text{0}\text{.80V}$
The solubility product of $AgCl$ under standard conditions of temperature (298 K) is given by:

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

Answer 1

Answer: (D)

$1.5\times {10}^{-10}$

Solution:- (D) $1.5\times {10}^{-10}$

${AgCl}_{\left(s\right)}+e^{-}⟶{Ag}_{\left(s\right)}+{{Cl}^{-}}_{(aq.)}{E}^{\circ }=0.22V.....\left(1\right)$

${{Ag}^{+}}_{(aq.)}+e^{-}⟶{Ag}_{\left(s\right)}{E}^{\circ }=0.80V$

${Ag}_{\left(s\right)}⟶{{Ag}^{+}}_{(aq.)}+e^{-}.....E^{\circ }=-0.80V.....\left(2\right)$

Adding equation $\left(1\right)\&\left(2\right)$, we get

${AgCl}_{\left(s\right)}⟶{{Ag}^{+}}_{(aq.)}+{{Cl}^{-}}_{(aq.)}{E}^{\circ }=-0.58V$

From Nernst equation,

$E_{cell}={E^{\circ }}_{cell}-\frac{RT}{nF}\ln K$

$0=-0.58-\frac{8.314\times 298\times 2.303}{1\times 96500}\log K$

$\Rightarrow \log K=\frac{0.58}{0.0591}=-9.81$

$\Rightarrow K=1.5\times {10}^{-10}$

Hence the solubility product of $AgCl$ is $1.5\times {10}^{-10}$.