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

  • question_answer
    When 0.04 faraday of electricity is passed through a solution of \[CaS{{O}_{4}}\], then the weight of \[C{{a}^{2+}}\] metal deposited at the cathode is            [BHU 1996]

    A)                 0.2 gm 

    B)                 0.4 gm

    C)                 0.6 gm 

    D)                 0.8 gm

    Correct Answer: D

    Solution :

               \[C{{a}^{++}}+2{{e}^{-}}\to Ca\]            \[{{E}_{Ca}}=\frac{40}{2}=20\]                                 \[{{W}_{Ca}}={{E}_{Ca}}\]× No. of faradays \[=20\times 0.04=0.8\,gm\].


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