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, \[E_{cell}^{o}=E_{cathode}^{o}-E_{anode}^{o}\] \[=E_{F{{e}^{3+}}/F{{e}^{2+}}}^{o}-E_{F{{e}^{2+}}/Fe}^{o}\] \[=(+0.771)-(-0.441)=+1.212V\]
You need to login to perform this action.
You will be redirected in
3 sec
