Question

At 298 K, the standard reduce potentials are 1.51 V for $Mn{O}_4^{-}$, 1.36 V for $C{l}_2|C{l}^{-}$, 1.07 V for $B{r}_2|B{r}^{-}$, 0.54 V for $I_2|I^{-}$. At $pH=3$, permangnate is expected to oxidize:
$(\frac{RT}{F}=0.059)$

• A. $C{l}^{-}$ and $B{r}^{-}$
• B. $B{r}^{-}$ and $I^{-}$
• C. $I^{-}$ only
• D. $C{l}^{-},B{r}^{-}$ and $I^{-}$

Answer 1

The correct option is B $B{r}^{-}$ and $I^{-}$

For $Mn{O}_4^{-}$ at $pH=3$,

$Mn{O}_4^{-}+8H^{+}+5e^{-}⟶M{n}^{2+}+4H_{2}O$

Here, $\left[Mn{O}_4^{-}\right]=\left[M{n}^{2+}\right]=1M$ since in standard state so,

$E_{cell}=E_{cell}^o-\frac{0.059}{5}log\frac{1}{[H^{+}{]}^8}$

$E_{cell}=1.51-\frac{0.059\times 8}{5}\times pH$$=1.51-0.2832=1.23V$

The reduction potential for $Mn{O}_4^{-}$ is 1.23 V which is higher than the standard reduction potentials 1.07 V and 0.54 V for $B{r}_2|B{r}^{-}$ and $I_2|I^{-}$ respectively.

Hence, permanganate is expected to oxidize bromide and iodide ions.

Higher is the value of the reduction potential, higher is the oxidizing power. Thus, the oxidizing power of permanganate is higher than that of bromide and iodide.