Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox…
$\mathrm{MnO}_{4}^{-}(\mathrm{aq} .)+8 \mathrm{H}^{+}(\mathrm{aq} .)+5 \mathrm{e}^{-} ightarrow \mathrm{Mn}^{2+}(\mathrm{aq} .)+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$
$\mathrm{E}^{\circ}=1.51 \mathrm{~V}$
$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$ (aq.) $+14 \mathrm{H}^{+}$ (aq.) $+6 \mathrm{e}^{-} ightarrow 2 \mathrm{Cr}^{3+}$ (aq.) $+7 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$
$\mathrm{E}^{\circ}=1.38 \mathrm{~V}$
$\mathrm{Fe}^{3+}$ (aq.) $+\mathrm{e}^{-} ightarrow \mathrm{Fe}^{2+}$ (aq.)
$\mathrm{E}^{\circ}=0.77 \mathrm{~V}$
$\mathrm{Cl}_{2}(\mathrm{~g})+2 \mathrm{e}^{-} ightarrow 2 \mathrm{Cl}^{-}$ (aq.)
$\mathrm{E}^{\circ}=1.40 \mathrm{~V}$
Identify the only incorrect statement regarding the quantitative estimation of aqueous $\mathrm{Fe}\left(\mathrm{NO}_{3}ight)_{2}$
- $\mathrm{MnO}_{4}^{-}$ can be used in aqueous $\mathrm{HCl}$
- $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$ can be used in aqueous $\mathrm{HCl}$
- $\mathrm{MnO}_{4}^{-}$ can be used in aqueous $\mathrm{H}_{2} \mathrm{SO}_{4}$
- $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$ can be used in aqueous $\mathrm{H}_{2} \mathrm{SO}_{4}$
Solution
$\begin{aligned}
2 \mathrm{MnO}_{4}^{-}+16 \mathrm{H}^{+}+10 \mathrm{Cl}^{-} \longrightarrow \\
2 \mathrm{Mn}^{2+}+8 \mathrm{H}_{2} \mathrm{O}+5 \mathrm{Cl}_{2} \uparrow
\end{aligned}$
The cell corresponding to this reaction is as follows:
$\mathrm{Pt}, \mathrm{Cl}_{2}(1 \mathrm{~atm})\left|\mathrm{Cl}^{-} \| \mathrm{MnO}_{4}^{-}, \mathrm{Mn}^{2+}, \mathrm{H}^{+}ight| \mathrm{Pt}$
$\mathrm{E}_{\mathrm{cell}}^{\mathrm{o}}=1.51-1.40=0.11 \mathrm{~V}$
$\mathrm{E}_{\text {cell }}^{\mathrm{o}}$ being $+\mathrm{ve}, \Delta \mathrm{G}^{\circ}$ will be -ve and hence the above
reaction is feasible. $\mathrm{MnO}_{4}^{-}$ will not only oxidise $\mathrm{Fe}^{2+}$
ion but also $\mathrm{Cl}^{-}$ ion simultaneously.
Asked in: JEE-TOPICTESTS-CHEMISTRY