Electrochemistry 2 Question 44
43. The standard reduction potential for $\mathrm{Cu}^{2+} / \mathrm{Cu}$ is $+0.34 \mathrm{~V}$. Calculate the reduction potential at $\mathrm{pH}=14$ for the above couple. $K_{\text {sp }}$ of $\mathrm{Cu}(\mathrm{OH})_{2}$ is $1.0 \times 10^{-19}$.
(1996, 3M)
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Solution:
- $\mathrm{pH}=14$
$$ \begin{array}{rlrl} \Rightarrow & \mathrm{pOH} & =0 \ \Rightarrow & & {\left[\mathrm{OH}^{-}\right]} & =1.0 \mathrm{M} \ & & K_{\mathrm{sp}}=10^{-19}=\left[\mathrm{Cu}^{2+}\right]\left[\mathrm{OH}^{-}\right]^{2} \ \Rightarrow & & {\left[\mathrm{Cu}^{2+}\right]=} & \frac{10^{-19}}{\left[\mathrm{OH}^{-}\right]^{2}}=10^{-19} \end{array} $$
For reaction : $\mathrm{Cu}^{2+}+2 e^{-} \longrightarrow \mathrm{Cu} ; E^{\circ}=0.34 \mathrm{~V}$
$$ \begin{aligned} E & =E^{\circ}-\frac{0.0592}{2} \log \frac{1}{\left[\mathrm{Cu}^{2+}\right]} \ & =0.34-\frac{0.0592}{2} \log 10^{19}=-0.222 \mathrm{~V} \end{aligned} $$
