Question

The standard reduction potential of a silver chloride electrode is $0.2V$ and that of a silver electrode is $0.79V$. The maximum amount of $AgCl$ that can be dissolved in ${10}^{6}L$ of a $0.1MAgN{O}_3$ solution is:

• A. $0.5\text{mmol}$
• B. $1.0\text{mmol}$
• C. $2.0\text{mmol}$
• D. $2.5\text{mmol}$

Answer 1

The correct option is B $1.0\text{mmol}$

$AgCl+e^{-}\to Ag+C{l}^{-}{E}^o=0.2VAg\to A{g}^{+}+e^{-}{E}^o=-0.79V$
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$AgCl\to A{g}^{+}+C{l}^{-}{E}^o=-0.59$

$E^o=\frac{0.059}{n}\log k\Rightarrow -0.59=\frac{0.059}{1}\log k_{sp}\Rightarrow k_{sp}={10}^{-10}$
Now solubility of $AgCl$ in $0.1MAgN{O}_3$
$S(S+0.1)={10}^{–10}$
$\approx 0.1S={10}^{-10}$
$\Rightarrow S={10}^{–9}mol/L$
Therefore $1$ mole of $AgCl$ dissolves in ${10}^{9}L$ solution.
Hence $1$ mmol of $AgCl$ can dissolved in ${10}^{6}L$ solution.