Question

Given, $F{e}^{3+}\left(aq\right)+e^{-}\to F{e}^{2+}\left(aq\right);E^{\circ }=+0.77V$
$A{l}^{3+}\left(aq\right)+3e^{-}\to Al\left(s\right);E^{\circ }=-1.66V$
$B{r}_2\left(aq\right)+2e^{-}\to 2B{r}^{-};E^{\circ }=+1.09V$

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

• A. $Al<F{e}^2<B{r}^{-}$
• B. $F{e}^2<Al<B{r}^{-}$
• C. $B{r}^{-}<F{e}^2<Al$
• D. $Al<B{r}^{-}<F{e}^{2+}$

Answer 1

Answer: (C)

A negative $E^o$ means that the redox couple is a stronger reducing agent. Hence, we can say that Aluminium is strongest reducing agent.