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JEE Main & Advanced Sample Paper JEE Main Sample Paper-40

  • question_answer
    The standard reduction potential for \[C{{u}^{2+}}/Cu\]is + 0.34. Calculate the reduction potential at pH = 14 for the above couple. \[({{K}_{sp}}Cu{{(OH)}_{2}}=1\times {{10}^{-19}})\]

    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 \[C{{u}^{2+}}+2{{e}^{-}}\xrightarrow[{}]{{}}Cu\] \[E={{E}^{o}}-\frac{0.059}{2}\log \frac{1}{[C{{u}^{2+}}]}\] \[=0.34-\frac{0.059}{2}\log \frac{1}{{{10}^{-19}}}=-0.22\,\text{V}\]


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