Redox Reactions And Electrochemistry Ques 593
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^{-}};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^{-}} $
