The standard reduction potential for \(Cu^{2+} / Cu\) is + 0.34. Calculate the reduction potential at pH = 14 for the above couple is:
\(K_{sp}\space of \space Cu(OH)_2\space is \space 1.0\times10^{-19}\).
The standard reduction potential for \(Cu^{2+} / Cu\) is + 0.34. Calculate the reduction potential at pH = 14 for the above couple is:
\(K_{sp}\space of \space Cu(OH)_2\space is \space 1.0\times10^{-19}\).
- -0.22 V
- + 0.22 V
- -0.44 V
- +0.44V
The Correct Option is A
Solution and Explanation
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
\(Cu^{2+}+2e^{-} ? Cu\)
\(E=E^{\circ}-\frac{0.059}{2} \, log\, \frac{1}{\left[Cu^{2+}\right]}\)
\(=0.34-\frac{0.059}{2} \, log \, \frac{1}{10^{-19}}=-0.22 V\)
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Concepts Used:
Nernst Equation
This equation relates the equilibrium cell potential (also called the Nernst potential) to its concentration gradient across a membrane. If there is a concentration gradient for the ion across the membrane, an electric potential will form, and if selective ion channels exist the ion can cross the membrane.
