| \[F{{e}^{3+}}(aq)+{{e}^{-}}\to F{{e}^{2+}}(aq);E{}^\circ =+0.77V\] |
| \[A{{l}^{3+}}(aq)+3{{e}^{-}}\to Al(s);E{}^\circ =-1.66V\] |
| \[B{{r}_{2}}(aq)+2{{e}^{-}}\to 2B{{r}^{-}};E{}^\circ =+1.09V\] |
| Considering the electrode potentials, which of the following represents the correct order of reducing power? |
A) \[F{{e}^{2+}}<Al<B{{r}^{-}}\]
B) \[B{{r}^{-}}<F{{e}^{2+}}<Al\]
C) \[Al<B{{r}^{-}}<F{{e}^{2+}}\]
D) \[Al<F{{e}^{2+}}<B{{r}^{-}}\]
Correct Answer: D
Solution :
[d] Reducing character decreases down the series. Hence the correct order is \[Al<F{{e}^{2+}}<B{{r}^{-}}\]
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