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J & K CET Engineering J and K - CET Engineering Solved Paper-2011

  • question_answer
    Three Faradays of electricity are passed through molten\[\text{A}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}\text{,}\]aqueous solution of \[\text{CuS}{{\text{O}}_{\text{4}}}\]and molten NaCI taken in three different electrolytic cells. Then the mole ratio of Al, Cu and Na deposited on the cathode will be                     

    A)  3:4:6        

    B)  2:1:6

    C)  3: 2: 1         

    D)  2: 3 : 6

    Correct Answer: D

    Solution :

     \[\because \]\[A{{l}^{3+}}+3{{e}^{-}}\xrightarrow{{}}Al\] \[\therefore \]3F of electricity will deposite 1 mole of Al. \[\because \] \[C{{u}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Cu\] \[\therefore \]3F of electricity will deposite 1.5 mole of Cu. \[\because \]  \[N{{a}^{+}}+{{e}^{-}}\xrightarrow{{}}Na\] \[\therefore \]3F of electricity will deposite 3 mole of Na. Hence, the mole ratio of Al, Cu and Na deposited on the cathode will be 1 : 1.5 : 3 or 2:3:6.


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