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NEET AIPMT SOLVED PAPER 2003

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

    A)                 6.42 V                    

    B)                 8.56 V                    

    C)                 2.14 V    

    D)                 4.28 V

    Correct Answer: C

    Solution :

                    \[\frac{4}{3}Al+{{O}_{2}}\to \frac{2}{3}A{{l}_{2}}{{O}_{3}},\Delta G=-827\,k\,J\,mo{{l}^{-1}}\]                 As we know that                 \[\because \,\Delta G=-nEf(value\,of\,n=4)\]                 \[-827\times {{10}^{3}}\,J=-4\times E\times 96500\]                 \[E=\frac{827\times {{10}^{3}}}{4\times 96500}\]


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