Question

The standard electrode potentials $E{}_{I_2/I^{-}}^{\circ },E{}_{B{r}^{-}/B{r}_2}^{\circ }\text{and}E{}_{Fe/F{e}^{2+}}^{\circ }$ are respectively $+0.54V$, $-1.09V$ and $0.44V$.

On the basis of the above data which of the following process is nonspontaneous?

• A. $B{r}_2+2I^{-}\to 2B{r}^{-}+I_2$
• B. $Fe+B{r}_2\to F{e}^{2+}+2B{r}^{-}$
• C. $Fe+I_2\to F{e}^{2+}+2I^{-}$
• D. $I_2+2B{r}^{-}\to 2I^{-}+B{r}_2$

Answer 1

The correct option is D $I_2+2B{r}^{-}\to 2I^{-}+B{r}_2$

$B{r}_2+2I^{{}_{-}}⟶2B{r}^{-}+I_2$

$E_{cell}=E_{B{r}_2/B{r}^{-}}-E_{I_2/I^{-}}=1.36$ $V$ spontaneous

$Fe+B{r}_2⟶F{e}^{2+}+2B{r}^{-}$

$E_{cell}=E_{B{r}_2/B{r}^{-}}-E_{Fe/F{e}^{2+}}=0.65$ $V$ spontaneous

$Fe+I_2⟶F{e}^{2+}+2I^{-}$

$E_{cell}=E_{I_2/I^{-}}-E_{Fe/F{e}^{2+}}=0.1$ $V$ spontaneous

$I_2+2B{r}^{-}⟶2I^{-}+B{r}_2$

$E_{cell}=E_{I_2/I^{-}}-E_{B{r}_2/B{r}^{-}}=-1.36$ $V$ Non- spontaneous