JEE Main & Advanced Chemistry Electrochemistry / विद्युत् रसायन Question Bank Topic Test - Electrochemistry

  • question_answer
    Acidulated water is electrolysed by 1 A current for 16 min and 5s using inert electrodes, The volume of gases liberated at STP will be

    A) 22.4 mL

    B) 112 mL

    C) 168 mL

    D) 336 mL

    Correct Answer: C

    Solution :

    [c] At cathode
    \[{{H}_{2}}O+2{{e}^{-}}\xrightarrow{\,}\,{{H}_{2}}+2O{{H}^{\odot -}}\]
    At Anode
    \[{{H}_{2}}O\xrightarrow{\,}\frac{1}{2}\,{{O}_{2}}+2{{H}^{\oplus }}+2{{e}^{-}}\]
    i.e., on passing 2 mol of current, 1 mol \[{{H}_{2}}(g)\] and \[\frac{1}{2}\,mol\,\,{{O}_{2}}(g)\] is liberated at cathode and anode respectively.
    From \[Q=it\]
    \[=\,1\times (16\times 60+5)=965\,\,C\]
    the amount of gases liberated by passing 965 C of electricity \[=\frac{1.5\times 965}{2\times 96500}\text{mol}\]
    = 0.0075 mol
    and value of gases liberated
    \[=0.0075\times 22400\,mL=168\,mL\]

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