Given Fe3+(aq)+e−→Fe2+(aq) E∘=+0.77VAl3+(aq)+3e−→Al(s) E∘=−1.66VBr2(aq)+2e−→Br−(aq) E0=+1.09VConsidering the electrode potentials, which of the following represents the correct order of reducing power?

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Given

$\begin{array}{ll} {Fe}^{3 +} (aq) + e^{-} \to {Fe}^{2 +} (aq) E^{\circ} = + 0.77 V \\ {Al}^{3 +} (aq) + 3 e^{-} \to Al (s) E^{\circ} = - 1.66 V \\ {Br}_{2} (aq) + 2 e^{-} \to {Br}^{-} (aq) E^{0} = + 1.09 V \end{array}$

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

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

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

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

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

answer is B.

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Detailed Solution

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

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