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NEET AIPMT SOLVED PAPER SCREENING 2011

  • question_answer
    The electrode potentials for \[C{{u}^{2+}}(aq)+{{e}^{-}}\xrightarrow[{}]{{}}C{{u}^{+}}(aq)\]and \[C{{u}^{+}}(aq)+{{e}^{-}}\xrightarrow[{}]{{}}Cu(s)\] Are + 0.15 V and + 0.50 v respectively. The value of \[{{E}^{o}}_{C{{u}^{2+}}/Cu}\]will be

    A)  0.325 V         

    B)         0.650 V

    C)         0.150 V         

    D)         0.500 V

    Correct Answer: A

    Solution :

    \[C{{u}^{2+}}+{{e}^{-}}\xrightarrow[{}]{{}}C{{u}^{+}};\] \[E_{1}^{o}=015V;\Delta G_{1}^{o}=-{{n}_{1}}E_{1}^{o}F\] \[C{{u}^{+}}+{{e}^{-}}\xrightarrow[{}]{{}}Cu;\] \[\frac{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,E_{2}^{o}=0.50V;\Delta G_{2}^{o}=-{{n}_{2}}E_{2}^{o}F}{C{{u}^{2+}}+2{{e}^{-}}\xrightarrow[{}]{{}}Cu;{{E}^{o}}=?;\Delta {{G}^{o}}=-n{{E}^{o}}F}\] \[\Delta {{G}^{o}}=\Delta G_{1}^{o}+\Delta G_{2}^{o}\] or\[-2{{E}^{o}}F=-1F\times 0.15+(-1F\times 0.50)\] or\[-2{{E}^{o}}F=-0.15F-0.50F\] or\[-2F{{E}^{o}}=-F(0.15+0.50)\] \[\therefore \]\[{{E}^{o}}=\frac{0.65}{2}=0.325V\]


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