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

  • question_answer
    An electrolytic cell contains a solution of \[A{{g}_{2}}S{{O}_{4}}\] and have platinum electrodes. A current is passed until 1.6 gm of \[{{O}_{2}}\] has been liberated at anode. The amount of silver deposited at cathode would be [CPMT 1971]

    A)                 107.88 gm          

    B)                 1.6 gm

    C)                 0.8 gm 

    D)                 21.60 gm

    Correct Answer: D

    Solution :

               At cathode: \[A{{g}^{+}}+{{e}^{-}}\to Ag\]            At Anode: \[2O{{H}^{-}}\to {{H}_{2}}O+\frac{1}{2}{{O}_{2}}+2{{e}^{-}}\]            \[{{E}_{Ag}}=\frac{108}{1}=108;\,\,{{E}_{{{O}_{2}}}}=\frac{\frac{1}{2}\times 32}{2}=8\]                 \[\frac{{{W}_{Ag}}}{{{E}_{Ag}}}=\frac{{{W}_{{{O}_{2}}}}}{{{E}_{{{O}_{2}}}}}\]; \[{{W}_{Ag}}=\frac{1.6\times 108}{8}=21.6\,\,gm\].


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