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

  • question_answer
    Three faradays of electricity are passed through molten \[A{{l}_{2}}{{O}_{3}}\], aqueous solution of \[CuS{{O}_{4}}\] and molten \[NaCl\] taken in different electrolytic cells. The amount of \[Al,\,Cu\] and \[Na\] deposited at the cathodes will be in the ratio of [BHU 1990]

    A)                 1 mole : 2 mole : 3 mole

    B)  3 mole : 2 mole : 1mole

    C)  1 mole : 1.5 mole : 3 mole

    D)  1.5 mole : 2 mole : 3 mole

    Correct Answer: C

    Solution :

               At cathode : \[A{{l}^{3+}}+3{{e}^{-}}\to Al\]            \[{{E}_{Al}}=\frac{\text{Atomic mass}}{3}\]            At cathode : \[C{{u}^{2+}}+2{{e}^{-}}\to Cu\]            \[{{E}_{Cu}}=\frac{\text{Atomic}\,\text{mass}}{2}\]            At cathode : \[N{{a}^{+}}+{{e}^{-}}\to Na\]            \[{{E}_{Na}}=\frac{\text{Atomic}\,\text{mass}}{1}\]            For the passage of 3 faraday;            mole atoms of \[Al\] deposited = 1            mole atoms of Cu deposited \[=\frac{1\times 3}{2}=1.5\]                                 mole atoms of Na deposited \[=1\times 3=3\].


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