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NEET Chemistry NEET PYQ-Electrochemistry

  • question_answer
    If \[E_{F{{e}^{2+}}/Fe}^{o}=-0.441\,V\] and \[E_{F{{e}^{3+}}/F{{e}^{2+}}}^{o}=0.771\,V,\] the standard emf of the reaction:   \[Fe+2F{{e}^{3+}}\to ~3F{{e}^{2+}}\]will be: [AIPMT (S) 2006]

    A)       0.330 V

    B) 1.653 V

    C) 1.212 V

    D) 0.111 V

    Correct Answer: C

    Solution :

    [c] Given that  \[E_{F{{e}^{2+}}/Fe}^{o}=-0.441\,V\]
                So, \[Fe\to F{{e}^{2+}}+2{{e}^{-}},{{E}^{o}}=+0.441\,V....(i)\]
                and  \[E_{F{{e}^{3+}}/F{{e}^{2+}}}^{o}=0.771\,V\]
                \[So,F{{e}^{3+}}+{{e}^{-}}\to F{{e}^{2+}},{{E}^{o}}=0.771\,V...(ii)\]
                Cell reaction
                \[\begin{align}   & (i)\,Fe\to F{{e}^{2+}}+2{{e}^{-}},{{E}^{o}}=0.441\,V \\  & (ii)\,2F{{e}^{3+}}+2{{e}^{-}}\to 2F{{e}^{2+}},{{E}^{o}}=+0.771\,V \\  & \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\  & Fe+2F{{e}^{3+}}\xrightarrow[{}]{{}}3F{{e}^{2+}},\,\,\,\,\,\,\,\,E_{cell}^{o}=1.212\,V \\ \end{align}\]
                or
                So, on the basis of cell reaction following half-cell reactions are written
                At anode :
                (1) \[Fe\to F{{e}^{2+}}+2{{e}^{-}}\] (oxidation)
                At  cathode :
                (2) \[2F{{e}^{3+}}+2{{e}^{-}}\to 2F{{e}^{2+}}\] (reduction)
                So,  \[E_{cell}^{o}=E_{cathode}^{o}-E_{anode}^{o}\]
                   \[=E_{F{{e}^{3+}}/F{{e}^{2+}}}^{o}-E_{F{{e}^{2+}}/Fe}^{o}\]
                \[=(+0.771)-(-0.441)=+1.212\,V\]


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