A) -0.22V
B) + 0.22V
C) -0.44V
D) +0.44V
Correct Answer: A
Solution :
[a] When pH= 14 \[[{{H}^{+}}]={{10}^{-4}}\]and \[[O{{H}^{+}}]=1\]1 M \[{{K}_{sp}}=[C{{u}^{2+}}]{{[O{{H}^{-}}]}^{2}}={{10}^{-19}}\] \[\therefore \]\[[C{{u}^{2+}}]=\frac{{{10}^{-19}}}{{{[O{{H}^{-}}]}^{2}}}={{10}^{-19}}\] The half cell reaction \[C{{u}^{2+}}+2{{e}^{-}}\xrightarrow[{}]{{}}Cu\] \[E={{E}^{o}}-\frac{0.059}{2}\log \frac{1}{[C{{u}^{2+}}]}\] \[=0.34-\frac{0.059}{2}\log \frac{1}{{{10}^{-19}}}=-0.22\,\text{V}\]You need to login to perform this action.
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