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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2011

  • question_answer
        The half-cell reactions for the corrosion are \[2{{H}^{+}}+\frac{1}{2}{{O}_{2}}+2{{e}^{-}}\xrightarrow[{}]{{}}{{H}_{2}}O;\]      \[E{}^\circ =1.23\,V\] \[F{{e}^{2+}}+2{{e}^{-}}\xrightarrow[{}]{{}}Fe(s);\]                         \[E{}^\circ =-0.44\,V\] Find the\[\Delta G{}^\circ \](in kJ) for the overall reaction.

    A)  \[-76\]                                

    B)  \[-322\]

    C)  \[-161\]                              

    D)  \[-152\]

    Correct Answer: B

    Solution :

                    \[E{}^\circ =E_{{{H}^{+}}/H}^{o}+E_{Fe/F{{e}^{2+}}}^{o}\] \[=1.23+0.44=1.67\,V\]                 \[\Delta G{}^\circ =-nFE{}^\circ \]                 \[=-2\times 96500\times 1.67\]                 \[=-322310\,J\]                 \[=-322.31\,kJ\]


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