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

  • question_answer
    The e.m.f. of a cell whose half cells are given below is \[M{{g}^{2+}}+2{{e}^{-}}\to Mg(s)\ E{}^\circ =-2.37\ V\]\[C{{u}^{2+}}+2{{e}^{-}}\to Cu(s)\ E{}^\circ =+0.34\ V\]          [Pb.CET 2001]

    A)                 + 1.36 V               

    B)                 + 2.71 V

    C)                 + 2.17 V               

    D)                 ? 3.01 V

    Correct Answer: B

    Solution :

               \[E_{Cu}^{0}>E_{Mg}^{0}\]hence Cu acts as cathode and Mg acts as anode.                                 \[E_{Cell}^{0}=E_{Cu}^{0}-E_{Mg}^{0}\] \[=(0.34)-(-2.37)=+2.71\ V\].


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