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  4. The reduction potential at pH = 14 for the Cu2+ /Cu couples is [ Given , E°Cu2+/Cu = 0.34 V; Ksp Cu(OH)2 = 1× 10-19

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Q. The reduction potential at $pH = 14$ for the $Cu^{2+} /Cu$ couples is [ Given , $E^{\circ}_{Cu^{2+}}/Cu = 0.34 \,V; K_{sp} Cu(OH)_2 = 1\times 10^{-19}$

AIIMSAIIMS 2012

Solution:

Given, $pH =14$;
$\therefore \,\,POH =0 $
and $\left[ OH ^{-}\right] =1 M $
$\left[ Cu ^{2+}\right]\left[ OH ^{-}\right]^{2} =K_{\text {sp }}=1 \times 10^{-19} $
$\left[ Cu ^{2+}\right] =1 \times 10^{-19} M$
For the half reaction, $Cu ^{2+}+2 e^{-} \longrightarrow Cu$
$E_{ Cu ^{2+}/cu} =E_{ Cu ^{2+}}^{\circ}/Cu-\frac{0.0591}{2} \log \frac{1}{\left[ Cu ^{2+}\right]} $
$=0.34-\frac{0.0591}{2} \log 10^{19}=-0.22 \,V $