2. Solved Papers
  3. J & K CET Medical

J & K CET Medical J & K - CET Medical Solved Paper-2011

  • 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 three different electrolytic cells. Then the mole ratio of\[Al,\text{ }Cu\]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 o\[Al\] \[\because \] \[C{{u}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Cu\] \[\therefore \] 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.


You need to login to perform this action.
You will be redirected in 3 sec spinner