Given:
Fe$$^{3+}$$(aq) + e$$^-$$ → Fe$$^{2+}$$(aq); E$$^0$$ = +0.77 V
Al$$^{3+}$$(aq) + 3e$$^-$$ → Al(s); E$$^0$$ = $$-1.66$$ V
Br$$_2$$(aq) + 2e$$^-$$ → 2Br$$^-$$; E$$^0$$ = +1.09 V
Considering the electrode potentials, which of the following represents the correct order of reducing power?

A more positive reduction potential indicates a greater tendency to gain electrons (stronger oxidizing agent). Consequently, its conjugate (the reduced form) is a weaker reducing agent. 

A more negative reduction potential indicates a lower tendency to gain electrons. Consequently, the reduced form is a stronger reducing agent.

Since lower $$E^0$$ corresponds to higher reducing power, the order of reducing power is:

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