Question

The standard reduction potential for $Cu^{2+} / Cu$ is + 0.34. Calculate the reduction potential at pH = 14 for the above couple is:
$K_{sp}\space of \space Cu(OH)_2\space is \space 1.0\times10^{-19}$.

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

Answer 1

Answer: (A)

-0.22 V

Answer 2

The correct answer is A:-0.22V
When pH =$14 \left[H^{+}\right] =10^{-14}$
and $\left[OH^{-}\right]=1 M$
$K_{sp} =\left[Cu^{2}\right]\left[OH^{-}\right]^{2} =10^{-19}$
$\therefore\quad\left[Cu^{2+}\right]=\frac{10^{-19}}{\left[OH^{-}\right]^{2}} =10^{-19}$
The half cell reaction
$Cu^{2+}+2e^{-} ? Cu$
$E=E^{\circ}-\frac{0.059}{2} \, log\, \frac{1}{\left[Cu^{2+}\right]}$
$=0.34-\frac{0.059}{2} \, log \, \frac{1}{10^{-19}}=-0.22 V$