Question

The standard reduction potential of a $AgCl/Ag$ electrode is $0.2V$ and that of a silver electrode ($A{g}^{+}/Ag$) is $0.79V.$ The maximum amount of $AgCl$ that can dissolve in $10$${}^6$ L of a $0.1M$ $Ag$NO${}_3$ solution is:

• A. $0.5mmol$
• B. $1.0mmol$
• C. $2.0mmol$
• D. $2.5mmol$

Answer 1

The correct option is B $1.0mmol$

$AgCl+1e^{-}\to Ag+C{l}^{-};(E^0=0.2V)$

$Ag\to A{g}^{+}+1e^{-};(E^0=-0.79V)$
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$AgCl\stackrel{1e^{-}}{\to }A{g}^{+}+C{l}^{-};E^0=-0.59V$

$E^0=\frac{0.059}{n}logK\Rightarrow -0.59=\frac{0.059}{1}log{K}_{sp}$

$\Rightarrow K_{sp}={10}^{-10}$

Now solubility of $AgCl$ in $0.1M$ $AgNO$${}_3$

$K_{sp}=S(S+0.1)={10}^{-10}$ $\Rightarrow$ $S={10}^{-9}mol/L$

Hence 1 milli mole dissolves in $10$${}^6$L solution

Hence in $10$${}^6$ L amount that dissolves is $1mmol$.