Question

The standard electrode potentials ${\text{E}}_{{\text{I}}_2/1^{-}}^{\circ },{\text{E}}_{\text{B}{\text{r}}^{-}/\text{B}{\text{r}}_2}^{\circ }$ and ${\text{E}}_{\text{Fe/F}{\text{e}}^{2+}}^{\circ }$ are respectively $+0.54\text{V,}-1.09\text{V}$ and 0.44 V. On the basis of the above data which of the following process is nonspontaneous?

• A. $\text{B}{\text{r}}_2+2{\text{I}}^{-}\text{2B}{\text{r}}^{-}+{\text{I}}_2$
• B. $\text{Fe}+\text{B}{\text{r}}_2\text{F}{\text{e}}^{2+}+2\text{B}{\text{r}}^{-}$
• C. $\text{Fe}+{\text{I}}_2\text{F}{\text{e}}^{2+}+2{\text{I}}^{-}$
• D. ${\text{I}}_2+2\text{B}{\text{r}}^{-}2{\text{I}}^{-}+\text{B}{\text{r}}_2$

Answer 1

Answer: (D)

${\text{I}}_2+2\text{B}{\text{r}}^{-}2{\text{I}}^{-}+\text{B}{\text{r}}_2$

the one which is having higher oxidation potential will undergo oxidation,

in D, $I^{-}$ is having lower reduction potential in comparison to $B{r}^{-}$, so has to undergo oxidation but in option D, it is getting oxidized.

hence, it is a nonspontaneous reaction.