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NEET AIPMT SOLVED PAPER MAINS 2012

  • question_answer
    The Gibbs' energy for the decomposition of\[A{{l}_{2}}{{O}_{3}}\] at \[{{500}^{o}}C\] is as follows \[\frac{2}{3}A{{l}_{2}}{{O}_{3}}\xrightarrow[{}]{{}}\frac{4}{3}Al+{{O}_{2}};\]\[{{\Delta }_{r}}G=+\,960\,kJ\,mo{{l}^{-1}}\] The potential difference needed for the electrolytic reduction of aluminium oxide (\[A{{l}_{2}}{{O}_{3}}\]) at \[{{500}^{o}}C\] is at least

    A)  4.5 V             

    B)  3.0 V    

    C)  2.5 V                    

    D)  5.0 V

    Correct Answer: C

    Solution :

    The half cell reaction are                 at anode                 \[2{{O}^{2-}}+4{{e}^{-}}\xrightarrow{{}}{{O}_{2}}]\times 3\]                 At cathode                 \[A{{l}^{3+}}\xrightarrow{{}}Al+3{{e}^{-}}]4\]                 Net reaction                 \[4Al+6{{O}^{2-}}\xrightarrow{{}}3{{O}_{2}}+4Al\]                 or \[\frac{4}{3}Al+2{{O}^{2-}}\xrightarrow{{}}{{O}_{2}}+\frac{4}{3}Al\]                 \[\therefore \]  \[n=\frac{12}{3}=4\]                                 \[\Delta {{G}^{o}}=-nF{{E}^{o}}\]                 Here,     \[\Delta {{G}^{o}}=+960\,kJ\,mo{{l}^{-1}}\]                 \[=960\times 1000\,J\,mo{{l}^{-1}}\]                 \[n=4\]                 \[F=96500\,coulomb\]                 \[\therefore \]\[960\times 1000=-4\times 96500\times {{E}^{o}}\]                 \[{{E}^{o}}=-\frac{960000}{4\times 96500}\]                 \[=-2.48\,V\]


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