At $298 \mathrm{~K}$, the standard reduction potentials are $1.51 \mathrm{~V}$ for $\mathrm{MnO}_{4}^{-}…

At $298 \mathrm{~K}$, the standard reduction potentials are $1.51 \mathrm{~V}$ for $\mathrm{MnO}_{4}^{-} \mid \mathrm{Mn}^{2+}, 1.36 \mathrm{~V}$ for $\mathrm{Cl}_{2} \mid \mathrm{Cl}^{-}, 1.07\mathrm{~V}$ for $\mathrm{Br}_{2} \mid \mathrm{Br}^{-}$, and $0.54 \mathrm{~V}$ for $\mathrm{I}_{2} \mid \mathrm{I}^{-}$. At $\mathrm{pH}=3$, permanganate is expected to oxidize $\left(\frac{\mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V}ight)$
  1. $\mathrm{Cl}^{-}, \mathrm{Br}^{-}$ and $\mathrm{I}^{-}$
  2. $\mathrm{Br}^{-}$ and $\mathrm{I}^{-}$
  3. $\mathrm{Cl}^{-}$ and $\mathrm{Br}^{-}$
  4. $\mathrm{I}^{-}$ only

Solution

$\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}$
$\mathrm{E}=1.51-\frac{0.059}{5} \log \frac{\left[\mathrm{Mn}^{2+}ight]}{\left[\mathrm{MnO}_{4}^{-}ight]\left[\mathrm{H}^{+}ight]^{8}}$
Taking $\mathrm{Mn}^{2+}$ and $\mathrm{MnO}_{4}^{-}$ in standard state i.e. $1 \mathrm{M}$,
$\mathrm{E}=1.51-\frac{0.059}{5} \times 8 \log \frac{1}{\left[\mathrm{H}^{+}ight]}$
$=1.51-\frac{0.059}{5} \times 8 \times 3=1.2268 \mathrm{~V}$
Hence at this $\mathrm{pH}, \mathrm{MnO}_{4}^{-}$ will oxidise only
$\mathrm{Br}^{-}$ and $\mathrm{I}^{-}$ as $\mathrm{SRP}$ of $\mathrm{Cl}_{2} / \mathrm{Cl}^{-}$ is $1.36 \mathrm{~V}$ which
is greater than that for $\mathrm{MnO}_{4}^{-} / \mathrm{Mn}^{2+}$. ,

Asked in: JEE-TOPICTESTS-CHEMISTRY

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