RAJASTHAN ­ PET Rajasthan PET Solved Paper-2012

  • question_answer If \[E_{F{{e}^{2+}}/Fe}^{o}=-0.441V\] and       \[E_{F{{e}^{3+}}/F{{e}^{2+}}}^{o}=0.771V\]the standard emf of the reaction \[Fe+3F{{e}^{3+}}\xrightarrow[{}]{{}}3F{{e}^{2+}}\]will be

    A)  0.330V      

    B)  1.653V

    C)  1.212V      

    D)  0.111 V

    Correct Answer: C

    Solution :

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

adversite


You need to login to perform this action.
You will be redirected in 3 sec spinner

Free
Videos