Question

A silver wire dipped in $0.1\text{M}HCl$ solution saturated with $AgCl$ develops oxidation potential of $-0.208\text{V}$.
The standard electrode potential of $E_{A{g}^{+}/Ag}^{\circ }=-0.799V$, then $K_{sp}$ of $AgCl$ in pure water will be

• A. ${10}^{-11}$
• B. ${10}^{-10}$
• C. ${10}^{-20}$
• D. ${10}^{-21}$

Answer 1

The correct option is A ${10}^{-11}$

We know that
$E_{\text{cell}}=E_{\text{cell}}^{\circ }-\frac{0.0591}{n}\text{log}\left[A{g}^{+}\right]$
On substituting the values in the formula,
$-0.208V=-0.799V-\frac{0.0591}{1}\text{log}\left[A{g}^{+}\right]$
$\Rightarrow -\text{log}\left[A{g}^{+}\right]=\frac{-0.208+0.799}{0.0591}=10$
$\Rightarrow A{g}^{+}={10}^{-10}$

$\therefore AgCl\to A{g}^{+}+C{l}^{-}$
$HCl\to H^{+}+C{l}^{-}$
Here the concentration of $C{l}^{-}$ will be predominently from $HCl$.
So, $\left[C{l}^{-}\right]=0.1\text{M}$
$\therefore K_{sp}=\left(A{g}^{+}\right)\left(C{l}^{-}\right)$
$\Rightarrow K_{sp}={10}^{-10}\times 0.1$
$\Rightarrow K_{sp}={10}^{-11}$