Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2007

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

    A)  \[2.14V\]           

    B)  \[4.28V\]

    C)  6.42V                                   

    D)  \[8.56V\]

    Correct Answer: A

    Solution :

    \[\Delta G=-827\,kJ\,mo{{l}^{-1}}\] We know that  \[\Delta G=-nEF\{n=4\}\] \[\therefore \]  \[-827\times {{10}^{3}}J=-4\times E\times 96500\] \[\therefore \]  \[E=\frac{827\times {{10}^{3}}}{4\times 96500}\] \[\therefore \]  \[E=2.14V\]

adversite



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