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JEE Main & Advanced JEE Main Paper Phase-I Held on 07-1-2020 Morning

  • question_answer
    Given that the standard potentials \[(E{}^\circ )\] of \[C{{u}^{2+}}\text{/}\,Cu\text{ }and\text{ }C{{u}^{+}}\text{/}\,Cu\text{ }are\text{ }0.34\text{ }V\]and\[0.522\text{ }V\]respectively, the \[E{}^\circ \] of \[C{{u}^{2}}^{+}/C{{u}^{+}}\]is:                                                   [JEE MAIN Held on 07-01-2020 Morning]

    A) +0.158 V

    B) -0.158 V

    C) -0.182 V

    D) 0.182 V

    Correct Answer: A

    Solution :

    [a] \[C{{u}^{2}}^{+}2{{e}^{-}}\to Cu,\Delta G_{1}^{{}^\circ }=-2F(0.34)...(i)\] \[C{{u}^{+}}+{{e}^{-}}\to Cu,\Delta G_{2}^{{}^\circ }=-F(0.522)...(ii)\] Subtract (ii) from (i) \[C{{u}^{2+}}+{{e}^{-}}\to C{{u}^{+}},\Delta G_{3}^{{}^\circ }=-F({{E}^{0}})\] \[\therefore \Delta G_{1}^{{}^\circ }-\Delta G_{2}^{{}^\circ }=\Delta G_{3}^{{}^\circ }\] \[\Rightarrow -FE{}^\circ =-2F(0.34)+F(0.522)\] \[\Rightarrow E{}^\circ =0.68-0.522=0.158\,\,V\]


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