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Based on the data given below, the correct order of reducing power is :
Fe³⁺_(aq)+ e $\to$ Fe²⁺ _(aq); E° = 0.77V
Al ³⁺ _(aq) + 3e $\to$ Al _(s); E° = –1.66V
Br_2(aq)+ 2e $\to$ 2Br^–_(aq) ; E° = + 1.08V

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Al < Br^–< Fe²⁺

Al < Fe²⁺ < Br^–

Br^– < Fe²⁺< Al

Fe²⁺ < Al < Br^–

answer is A.

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Detailed Solution

It is known that,

$E_{O P}^{0} = - E_{R P}^{0}$

Hence, the order of oxiding potential is Br⁻<Fe²⁺<Al.
Higher the oxiding potential, higher is the tendency to loose electron and higher is the reducing nature so order is

Br⁻<Fe²⁺<Al.

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