Given $$ \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+}(\mathrm{aq}) ;…

Given $$ \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+}(\mathrm{aq}) ; \mathrm{E}^0=+0.77 \mathrm{~V} $$ $$ \begin{aligned} &\mathrm{Al}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Al}(\mathrm{s}) ; \mathrm{E}^0=-1.66 \mathrm{~V} \\ &\mathrm{Br}_2(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-} ; \mathrm{E}^0=+1.09 \mathrm{~V} \end{aligned} $$ Considering the electrode potentials, which of the following represents the correct order of reducing power?
  1. $\mathrm{Fe}^{2+} < \mathrm{A} l < \mathrm{Br}^{-}$
  2. $\mathrm{Br}^{-} < \mathrm{Fe}^{2+} < \mathrm{Al}$
  3. $\mathrm{Al} < \mathrm{Br}^{-} < \mathrm{Fe}^{2+}$
  4. $\mathrm{Al} < \mathrm{Fe}^{2+} < \mathrm{Br}^{-}$

Solution

Reducing character decreases down the series. Hence the correct order is $ \mathrm{Al} < \mathrm{Fe}^{2+} < \mathrm{Br}^{-} $

Asked in: JEE Main 2014 (11 Apr Online)

Practice more Electrochemistry questions on Aicharya