Question

Based on the date given below, the correct order of reducing power is:
$F{e}^{3+}\left(aq\right)+e\to F{e}^{2+}\left(aq\right);E^o=+0.77V$
$A{l}^{3+}\left(aq\right)+3e\to Al\left(s\right);E^o=-1.66V$
$B{r}_2\left(aq\right)+2e\to 2B^{-}\left(aq\right);E^o=+1.08V$

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

Answer 1

The correct option is A $B{r}^{-}<F{e}^{2+}<Al$

As we know,

$E_{OP}^o=-E_{RP}^o$ so order of oxiding potential is $B{r}^{-}<F{e}^{2+}<Al$.

Higher the oxiding potential, higher is the tendency to loose electron and higher is the reducing nature so order is $B{r}^{-}<F{e}^{2+}<Al$.