Question

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. 0.330 V
• B. 1.653 V
• C. 1.212 V
• D. 0.111 V

Answer 1

Answer: (C)

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 $