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NEET Chemistry Electrochemistry / विद्युत् रसायन Question Bank Self Evaluation Test - Electrochemistry

  • question_answer
    On the basis of the information available from the reaction \[\frac{4}{3}Al+{{O}_{2}}\to \frac{2}{3}A{{l}_{2}}{{O}_{3}},\Delta G\] \[=-827\text{ }kJ\text{ }mo{{l}^{-1}}\] of \[{{O}_{2}}\] the minimum e.m.f required to carry out an electrolysis of \[A{{l}_{2}}{{O}_{3}}\] is \[\left( F=96500C\text{ }mo{{l}^{-1}} \right)\]

    A) 8.56 V              

    B) 2.14 V

    C) 4.28 V              

    D) 6.42 V

    Correct Answer: B

    Solution :

    [b] \[\Delta G=-nEF\] For 1 mole of Al, n = 3 \[\therefore \] for \[\frac{4}{3}\] mole of Al, \[n=3\times \frac{4}{3}=4\] According to question, \[-827\times 1000=-4\times E\times 96500\]             \[E=\frac{827\times 1000}{4\times 96500}=2.14V\] \[\therefore \] minimum e.m.f required = 2.14 V


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