Redox Reactions And Electrochemistry Ques 686
Question: Given $ F{{e}^{3+}}(aq)+{{e}^{-}}\to F{{e}^{2+}}(aq);E{}^\circ =+0.77V $ $ A{{l}^{3+}}(aq)+3{{e}^{-}}\to Al(s);E{}^\circ =-1.66V $ $ Br_2(aq)+2{{e}^{-}}\to 2B{{r}^{-}}(aq);E{}^\circ =+1.09V $ Considering the electrode potentials, which of the following represents the correct order of reducing power-
Options:
A) $ F{{e}^{2+}}<Al<B{{r}^{-}} $
B) $ B{{r}^{-}}<F{{e}^{2+}}<Al $
C) $ Al<B{{r}^{-}}<F{{e}^{2+}} $
D) $ Al<F{{e}^{2+}}<B{{r}^{-}} $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Reducing character decreases down the series. Hence the correct order is $ Al<F{{e}^{2+}}<B{{r}^{-}} $
