Question

Based on data given below for the electrode potential, reducing power of $F{e}^{2+},Al$ and $B{r}^{-}$ will increase in the order

$B{r}_2(aq)+2e^{-}\to 2B{r}^{-}(aq);E^o=+1.08V$

$A{l}^{3+}(aq)+3e^{-}\to Al(s);E^o=-1.66V$

$F{e}^{3+}(aq)+e^{-}\to F{e}^{+2}(aq);E^o=+0.77V$

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

Answer 1

Answer: (A)

Reducing power, i.e. the tendency to lose electrons increases as the reduction potential decreases.