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JEE Main & Advanced Chemistry Electrochemistry / विद्युत् रसायन Question Bank Electrolytes and Electrolysis

  • question_answer
    The passage of current liberates \[{{H}_{2}}\] at cathode and \[C{{l}_{2}}\] at anode. The solution is        [EAMCET 1979,87]

    A)                 Copper chloride in water             

    B)                 \[NaCl\] in water

    C)                 \[{{H}_{2}}S{{O}_{4}}\]

    D)                 Water

    Correct Answer: B

    Solution :

               Since discharge potential of water is greater than that of sodium so water is reduced at cathode instead of \[N{{a}^{+}}\]            Cathode:  \[{{H}_{2}}O+{{e}^{-}}\to \frac{1}{2}{{H}_{2}}+O{{H}^{-}}\]                                 Anode:    \[C{{l}^{-}}\to \frac{1}{2}C{{l}_{2}}+{{e}^{-}}\].


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