1.Given the standard potentials: EFe3+/Fe2+∘=+0.77E^\circ_{Fe^{3+}/Fe^{2+}} = +0.77EFe3+/Fe2+∘=+0.77 V, EBr2/Br−∘=+1.09E^\circ_{Br_2/Br^-} = +1.09EBr2/Br−∘=+1.09 V, EAl3+/Al∘=−1.66E^\circ_{Al^{3+}/Al} = -1.66EAl3+/Al∘=−1.66 V. The correct order of reducing power is:a.Fe2+^{2+}2+ >>> Al >>> Br−^-−b.Al <<< Br−^-− <<< Fe2+^{2+}2+c.Al <<< Br <<< Fe2+^{2+}2+d.Br−^-− <<< Fe2+^{2+}2+ <<< AlLogin to continueOnly logged in users canattempt or see the solution.
