Question

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

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

Answer 1

Answer: (D)

-0.22V

Answer 2

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