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

It is known that,

$E_{OP}^0=-E_{RP}^0$ 

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.