Electrochemistry Result Question 43
43. The standard reduction potential for $\mathrm{Cu}^{2+} / \mathrm{Cu}$ is +0.34 V . Calculate the reduction potential at $\mathrm{pH}=14$ for the above couple. $K_{\mathrm{sp}}$ of $\mathrm{Cu}(\mathrm{OH})_2$ is $1.0 \times 10^{-19}$.
(1996, 3M)
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Answer:
Correct Answer: 43. $(-0.222 \mathrm{V})$
Solution:
- $\mathrm{pH}=14$
$\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 $
${\left[\mathrm{Cu}^{2+}\right] } =\frac{10^{-19}}{\left[\mathrm{OH}^{-}\right]^2}=10^{-19}$
For reaction : $\mathrm{Cu}^{2+}+2 e^{-} \longrightarrow \mathrm{Cu} ; E^{\circ}=0.34 \mathrm{V}$
$ E =E^{\circ}-\frac{0.0592}{2} \log \frac{1}{\left[\mathrm{Cu}^{2+}\right]} $
$=0.34-\frac{0.0592}{2} \log 10^{19}=-0.222 \mathrm{V}$
BETA
