Question

At $298$K, the standard reduction potentials are $1.51$V for $Mn{O}_4^{-}$|$M{n}^{2+}$, $1.36$V for $C{l}_2$|$C{l}^{-}$, $1.07$V for $B{r}_2$|$B{r}^{-}$ and $0.54$V for $I_2$|$I^{-}$. At pH$=3$ permanganate is expected to oxidize: $\left(\frac{RT}{F}0.059V\right)$.

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

Answer 1

Answer: (B)

$C{l}^{-},B{r}^{-}$ and $I^{-}$

$E_{Mn{O}_4^{-}|M{n}^{2+}}^o=1.51;E_{C{l}_2|C{l}^{-}}^o=1.36;E_{B{r}_2|B{r}^{-}}^o=1.07;E_{I_2|I^{-}}^o=0.54$

$E_{cell}^o$ for $C{l}^{-}|C{l}_2||Mn{O}_4^{-}|M{n}^{2+}$

$E_{cell}^o=1.51-1.36$

$=0.15V$

$E_{cell}^o$ for $B{r}^{-}|B{r}_2||Mn{O}_4^{-}|M{n}^{2+}$

$E_{cell}^o=1.51-1.07$

$=0.44V$

$E_{cell}^o$ for $I^{-}|I_2||Mn{O}_4^{-}|M{n}^{2+}$

$E_{cell}^o=1.51-0.54$

$=0.97V$

Since $E_{cell}^o>0$ for all the three cases, therefore $Mn{O}_4^{-}$ can oxidize all three of $C{l}^{-},B{r}^{-},I^{-}$