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JEE Main & Advanced Chemistry Electrochemistry / विद्युत् रसायन Question Bank Electrode potential Ecell Nernst equation and ECS

  • question_answer
    EMF of a cell whose half cells are given below is                 \[M{{g}^{2+}}+2{{e}^{-}}\to Mg(s);\,\,E=-2.37\,V\]                 \[C{{u}^{2+}}+2{{e}^{-}}\to Cu(s);\,\,\,E=+0.33\,V\] [EAMCET 1987; MP PET 1994; Pb. PMT 2000]

    A)                 ? 2.03 V               

    B)                 1.36 V

    C)                 2.7 V     

    D)                 2.03 V

    Correct Answer: C

    Solution :

               \[E_{\text{cell}}^{o}=E_{\text{cathode}}^{o}-{{E}_{\text{anode}}}\];         \[E_{\text{cell}}^{o}=0.34-(-\,2.37)\]                                 \[E_{\text{cell}}^{o}=2.71\,\,V\].


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