A) 0.330 V
B) 1.653 V
C) 1.212 V
D) 0.111 V
Correct Answer: C
Solution :
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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