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KVPY Sample Paper KVPY Stream-SX Model Paper-1

  • question_answer
    The electrode potentials for \[C{{u}^{2+}}_{(aq)}+{{e}^{-}}\to C{{u}^{+}}_{(aq)}+\] and \[C{{u}^{+}}_{(aq)}{{e}^{-}}\to C{{u}_{(s)}}\]respectively. The value of \[E{{{}^\circ }_{C{{u}^{2+}}/Cu}}\] will be -

    A) 0.500 V            

    B) 0.325 V

    C) 0.650 V            

    D) 0.150 V

    Correct Answer: B

    Solution :

    [B]
    \[\frac{\begin{matrix}    C{{u}^{2+}}+1{{e}^{-}}\to C{{u}^{+}} & E_{1}^{0}=0.15\,v\,\,\Delta \,\,G_{1}^{0}=-\,{{n}_{1}}\,\,E_{1}^{0}\,\,F  \\    C{{u}^{2+}}+1{{e}^{-}}\to Cu & E_{2}^{0}=0.50\,v\,\,\Delta \,\,G_{2}^{0}=-\,{{n}_{2}}\,\,E_{2}^{0}\,\,F  \\ \end{matrix}}{\begin{matrix}    C{{u}^{2+}}+2{{e}^{-}}\to Cu & \Delta \,G{}^\circ   \\ \end{matrix}=\Delta G{{{}^\circ }_{1}}+\Delta G{{{}^\circ }_{2}}}\]
    \[(-\,1)n\,\,E{}^\circ F=(-\,1)\,{{n}_{1}}\,\,E_{1}^{0}F+(-\,1)\,{{n}_{2}}E_{2}^{0}F\]
    \[E{}^\circ =\frac{{{n}_{1}}E_{1}^{0}+{{n}_{2}}E_{2}^{0}}{n}=\frac{0.15\times 1+0.50\times 1}{2}\]
    \[=0.325V\]


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