JEE Main & Advanced Sample Paper JEE Main - Mock Test - 41

  • question_answer
    Given (i) \[C{{u}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Cu,E{}^\circ =0.337\,V\] (ii) \[C{{u}^{2+}}+{{e}^{-}}\xrightarrow{{}}C{{u}^{+}},E{}^\circ =0.153\,V\] Electrode potential, \[E{}^\circ \]for the reaction, \[C{{u}^{+}}{{e}^{-}}\xrightarrow{{}}Cu\], will be

    A) 0.90 V           

    B)        0.30 V

    C) 0.38 V            

    D)        0.52 V

    Correct Answer: D

    Solution :

    [d] \[\begin{align}   & \underline{\begin{align}   & C{{u}^{+2}}+2{{e}^{-}}\to Cu\,\,\,\,\Delta G_{i}^{{}^\circ } \\  & C{{u}^{+2}}+{{e}^{-}}\to C{{u}^{+}}~~\Delta G_{2}^{{}^\circ } \\ \end{align}} \\  & C{{u}^{+}}+{{e}^{-}}\to Cu\,\,\,\Delta G_{3}^{{}^\circ } \\ \end{align}\] \[\Delta G_{3}^{{}^\circ }=\Delta G_{1}^{{}^\circ }-\Delta G_{2}^{{}^\circ }\] \[-1\times F\times E=-2F\times 0.337-\left( -1\times F\times 0.153 \right)\] \[\Rightarrow E=0.521\,V\]


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