Electrode potential data are given below:
$\mathrm{Fe}^{+3}(\mathrm{aq})+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{+2}(\mathrm{aq}) ; \mathrm{E}^{\circ}=+0.77 \mathrm{~V}$
$\mathrm{Al}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}_{(\mathrm{s})} ; \mathrm{E}^{\circ}=-1.66 \mathrm{~V}$
$\mathrm{Br}_2(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Br}^{-}(\mathrm{aq}) ; \mathrm{E}^{\circ}=+1.08 \mathrm{~V}$
Based on the data, the reducing power of $\mathrm{Fe}^{2+}$, $\mathrm{Al}$ and $\mathrm{Br}^{-}$will increase in the order

• Correct answer
  A

  $\mathrm{Br}^{-} < \mathrm{Fe}^{2+} < \mathrm{Al}$
• B

  $\mathrm{Fe}^{2+} < \mathrm{Al} < \mathrm{Br}^{-}$
• C

  $\mathrm{Al} < \mathrm{Br}^{-} < \mathrm{Fe}^{2+}$
• D

  $\mathrm{Al} < \mathrm{Fe}^{2+} < \mathrm{Br}^{-}$

Hence, the correct answer is option: 'A'.