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Punjab Medical Punjab - MET Solved Paper-2006

  • question_answer
    Given the electrode potentials\[F{{e}^{3+}}+{{e}^{-}}\xrightarrow{{}}F{{e}^{2+}},\,\,E{}^\circ =0.771\,\,volt\]\[{{I}_{2}}+2{{e}^{-}}\xrightarrow{{}}2{{I}^{-}},\,\,E{}^\circ =0.536\,\,volt\] \[E_{cell}^{\text{o}}\] for the cell reaction\[2F{{e}^{3+}}+2{{I}^{-}}\xrightarrow{{}}2F{{e}^{2+}}+{{I}_{2}},\]is:

    A) \[1.006V\]                          

    B) \[0.503V\]

    C) \[0.235V\]                          

    D) \[-0.235V\]

    Correct Answer: C

    Solution :

    Cell reaction: \[2F{{e}^{3+}}+2{{I}^{-}}\to 2F{{e}^{2+}}+{{I}_{2}}\]\[F{{e}^{3+}}\]shows reduction while I shows oxidation (reduction occurs at cathode while oxidation takes place at anode) \[E_{cell}^{o}=E_{cathode}^{o}-E_{anode}^{o}\] Given, \[E_{cathode}^{o}(F{{e}^{3+}}/F{{e}^{2+}})=0.771V\] and\[E_{anode}^{\text{o}}({{I}^{-}}/{{I}_{2}})=0.536V\] \[\therefore \]\[E_{cell}^{o}=0.771-0.536\] \[=0.235V\]


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