Question

Given the cell:$Cd\left(s\right)|Cd\left(OH{)}_2\right(s\left)|NaOH\right(aq,0.01M\left)|H_2\right(g,1bar\left)|Pt\right(s)$ with $E_{cell}=0.0V$. If $E_{C{d}^{2+}|Cd}^o=-0.39V$, then $K_{sp}$ of $Cd(OH{)}_2$ is:

• A. $0.1$
• B. ${10}^{-13}$
• C. ${10}^{-8}$
• D. ${10}^{-15}$

Answer 1

Answer: (D)

${10}^{-15}$

$E_{cell}=E_{H^{+}|H_2}-E_{O{H}^{-}|Cd(OJ{)}_2|Cd}$
$=E_{H^{+}|H_2}-E_{C{d}^{2+}|Cd}$
or $E_{cell}=E^{\circ }-\frac{0.06}{2}log\frac{\left[C{d}^{2+}\right]}{[H^{+}{]}^2}$
$\because$ $E_{cell}=0$,
$\therefore$ $E_{cell}^{\circ }=0.03log\times \frac{\left[C{d}^{2+}\right][O{H}^{-}{]}^2}{K_w^2}$
$log\frac{K_{sp}}{K_w^2}=\frac{0.39}{0.03}=13$
$K_{sp}={10}^{13}\times ({10}^{14}{)}^2={10}^{-15}$ {Option D}