NEET Sample Paper NEET Sample Test Paper-48

  • question_answer
    What will be the emf of the given cell: \[\left| {{\operatorname{PH}}_{2}}\left( {{P}_{1}} \right){{H}^{+}}\left( aq. \right) \right.\left| {{H}_{2}}\left( {{P}_{2}} \right)Pt \right|\]

    A) \[\frac{RT}{F}In\frac{{{P}_{1}}}{{{P}_{2}}}\]            

    B) \[\frac{RT}{2F}In\frac{{{P}_{1}}}{{{P}_{2}}}\]

    C) \[\frac{RT}{2F}In\frac{{{P}_{2}}}{{{P}_{1}}}\]          

    D) None

    Correct Answer: D

    Solution :

    \[\frac{RT}{2F}In\frac{{{P}_{1}}}{{{P}_{2}}}\] \[\operatorname{R}.H.S\,\,\,2{{H}^{+}}+2{{e}^{-}}\rightleftharpoons {{H}_{2}}\left( {{P}_{2}} \right)\] \[\operatorname{L}.H.S\,\,{{H}_{2}}\left( {{P}_{1}} \right)\rightleftharpoons 2{{H}^{+}}+2{{e}^{-}}\] \[\operatorname{Overall} reaction {{H}_{2}}\left( {{P}_{1}} \right)\rightleftharpoons {{H}_{2}}\left( {{P}_{2}} \right)\] \[\operatorname{E}={{E}^{o}}-\frac{RT}{nF}\,In\frac{{{P}_{2}}}{{{P}_{1}}}\] \[=O-\frac{RT}{nF}\,In\frac{{{P}_{2}}}{{{P}_{1}}}\] \[=\frac{RT}{nF}\ \,In\,\frac{{{P}_{2}}}{{{P}_{1}}}\]


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