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JEE Main & Advanced Chemistry Electrochemistry / विद्युत् रसायन Question Bank Faraday's law of electrolysis

  • question_answer
    Charge required to liberate 11.5 g sodium is [AIIMS 1992; DCE 2002]

    A)                 0.5 F      

    B)                 0.1 F

    C)                 1.5 F      

    D)                 96500 coulombs

    Correct Answer: A

    Solution :

               \[N{{a}^{+}}+{{e}^{-}}\to Na\]                    Charge (in F) = moles of e?  used = moles of Na  deposited                                 \[=\frac{11.5}{23}\ gm=0.5\]Faraday.


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