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

The Correct Option is A

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 
\(Cu^{2+}+2e^{-} ? Cu\)
\(E=E^{\circ}-\frac{0.059}{2} \, log\, \frac{1}{\left[Cu^{2+}\right]}\)
\(=0.34-\frac{0.059}{2} \, log \, \frac{1}{10^{-19}}=-0.22 V\)