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

  • question_answer
    A current of strength 2.5 amp was passed through \[CuS{{O}_{4}}\] solution for 6 minutes 26 seconds. The amount of copper deposited is                 (Atomic weight of \[Cu=63.5\])                 (1 faraday = 96500 coulombs) [EAMCET 1989; MP PET 1994]

    A)                 0.3175 g              

    B)                 3.175 g

    C)                 0.635 g 

    D)                 6.35 g

    Correct Answer: A

    Solution :

               \[Q=2.5\times 386=96500\,C\]            \[2F(2\times 96500C)\] deposited \[Cu=63.5\,g\]                                 \Hence 965 C will deposited;  \[Cu=0.3175\,\,gm\].


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