Question

The standard reduction potential for $C{u}^{2+}/Cu$ is $+0.34$. Calculate the reduction potential at $pH=14$ for the above couple. $\left(K_{\text{sp }}Cu(OH{)}_2=1\times 10^{-19}\right)$

• A. $+0.22V$
• B. $-0.44V$
• C. $-0.22V$
• D. $+0.44V$

Answer 1

Answer: (C)

When $pH=14,\left[H^{+}\right]=10^{-4}$ and $\left[O{H}^{-}\right]=1M$
$K_{sp}=\left[C{u}^{2+}\right]{\left[O{H}^{-}\right]}^2=10^{-19}$
$\therefore \left[C{u}^{2+}\right]=\frac{10^{-19}}{{\left[O{H}^{-}\right]}^2}=10^{-19}$
The half-cell reaction
$C{u}^{2+}+2e^{-}\to Cu$
$E=E^{\circ }-\frac{0.059}{2}\log \frac{1}{\left[C{u}^{2+}\right]}$
$=0.34-\frac{0.59}{2}\log \frac{1}{10^{-19}}=-0.22V$