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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2012

  • question_answer
    For the following cell with hydrogen electrodes at two different pressures\[{{p}_{1}}\]and\[{{p}_{2}},\]emf is given by \[\underset{{{p}_{1}}}{\mathop{pt({{H}_{2}})}}\,|\underset{1M}{\mathop{{{H}^{+}}(aq)}}\,|\underset{{{p}_{2}}}{\mathop{Pt({{H}_{2}})}}\,\]

    A)  \[\frac{RT}{F}{{\log }_{e}}\frac{{{p}_{1}}}{{{p}_{2}}}\]

    B)  \[\frac{RT}{2F}{{\log }_{e}}\frac{{{p}_{1}}}{{{p}_{2}}}\]

    C)  \[\frac{RT}{F}{{\log }_{e}}\frac{{{p}_{2}}}{{{p}_{1}}}\]

    D)  \[\frac{RT}{2F}{{\log }_{e}}\frac{{{p}_{2}}}{{{p}_{1}}}\]

    Correct Answer: B

    Solution :

     LHS half-cell \[\underset{{{p}_{1}}}{\mathop{{{H}_{2}}(g)}}\,\xrightarrow[{}]{{}}2{{H}^{+}}(1M)+2{{e}^{-}}\] RHS half-cell \[\underline{2{{H}^{+}}(1M)+2{{e}^{-}}\xrightarrow[{}]{{}}\underset{{{p}_{2}}}{\mathop{{{H}_{2}}(g)}}\,}\] \[\underline{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underset{{{p}_{1}}}{\mathop{{{H}_{2}}(g)}}\,\xrightarrow{{}}\underset{{{p}_{2}}}{\mathop{{{H}_{2}}(g)}}\,}\] \[K=\frac{{{p}_{2}}}{{{p}_{1}}},\] \[E_{cell}^{o}=0.00V,n=2\] \[{{E}_{cell}}=E_{cell}^{o}-\frac{RT}{nF}{{\log }_{e}}K\] \[=0-\frac{RT}{2F}{{\log }_{e}}\frac{{{p}_{2}}}{{{p}_{1}}}\] \[{{E}_{cell}}=\frac{RT}{2F}{{\log }_{e}}\frac{{{p}_{1}}}{{{p}_{2}}}\]


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