Question

The reduction potential at $pH=14$ for the $C{u}^{2+}/Cu$ couples is [ Given , $E_{C{u}^{2+}}^{\circ }/Cu=0.34V;K_{sp}Cu(OH{)}_2=1\times 10^{-19}$

• A. 0.34 V
• B. - 0.34V
• C. 0.22 V
• D. -0.22V

Answer 1

Answer: (D)

Given, $pH=14$;
$\therefore POH=0$
and $\left[O{H}^{-}\right]=1M$
$\left[C{u}^{2+}\right]{\left[O{H}^{-}\right]}^2=K_{\text{sp }}=1\times 10^{-19}$
$\left[C{u}^{2+}\right]=1\times 10^{-19}M$
For the half reaction, $C{u}^{2+}+2e^{-}⟶Cu$
$E_{C{u}^{2+}/cu}=E_{C{u}^{2+}}^{\circ }/Cu-\frac{0.0591}{2}\log \frac{1}{\left[C{u}^{2+}\right]}$
$=0.34-\frac{0.0591}{2}\log 10^{19}=-0.22V$