Given

 $\begin{array}{l}{\mathrm{Fe}}^{3+}\left(\mathrm{aq}\right)+{\mathrm{e}}^{-}\to {\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{E}^{\circ }=+0.77\mathrm{V}\\ {\mathrm{Al}}^{3+}\left(\mathrm{aq}\right)+3{\mathrm{e}}^{-}\to \mathrm{Al}\left(\mathrm{s}\right)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{E}^{\circ }=-1.66\mathrm{V}\\ {\mathrm{Br}}_2\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\to {\mathrm{Br}}^{-}\left(\mathrm{aq}\right)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{E}^0=+1.09\mathrm{V}\end{array}$

Considering the electrode potentials, which of the following represents the correct order of reducing power?

More negative the standard potential, least the reduction tendency of the ion/molecule. The corresponding atom/ ion has largest oxidation tendency and thus is a stronger reducing agent. Hence, the order of reducing power is

 $\mathrm{Al}\left(\mathrm{s}\right)>{\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right)>\mathrm{Br}\left(\mathrm{aq}\right)$, i.e., ${\mathrm{Br}}^{-}\left(\mathrm{aq}\right)<{\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right)<\mathrm{Al}\left(\mathrm{s}\right)$