Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox…

Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox titration. Some half cell reactions and their standard potentials are given below
$\mathrm{MnO}_{4}^{-}(\mathrm{aq})+8 \mathrm{H}^{+}(\mathrm{aq})+5 \mathrm{e}^{-} ightarrow \mathrm{Mn}^{+2}(\mathrm{aq})+4 \mathrm{H}_{2} \mathrm{O}(l) ; \mathrm{E}^{\circ}=1.51 \mathrm{~V}$
$\mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-}(\mathrm{aq})+14 \mathrm{H}^{+}(\mathrm{aq})+6 \mathrm{e}^{-} ightarrow 2 \mathrm{Cr}^{+3}(\mathrm{aq})+7 \mathrm{H}_{2} \mathrm{O}(l) ; \mathrm{E}^{\circ}=1.38 \mathrm{~V}$
$\begin{array}{l}
\mathrm{Fe}^{+3}(\mathrm{aq})+\mathrm{e}^{-} ightarrow \mathrm{Fe}^{+2} ; \mathrm{E}^{0}=0.77 \mathrm{~V} \\
\mathrm{Cl}_{2}(\mathrm{~g})+2 \mathrm{e}^{-} ightarrow 2 \mathrm{Cl}^{-}(\mathrm{aq}) ; \mathrm{E}^{0}=1.40 \mathrm{~V}
\end{array}$
Identify the only incorrect statement regarding the quantitative estimation of aqueous $\mathrm{Fe}\left(\mathrm{NO}_{3}ight)_{2}$
  1. $\mathrm{MnO}_{4}^{-}$can be used in aqueous $\mathrm{HCl}$
  2. $\mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-}$ can be used in aqueous $\mathrm{HCl}$
  3. $\mathrm{MnO}_{4}^{-}$can be used in aqueous $\mathrm{H}_{2} \mathrm{SO}_{4}$
  4. $\mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-}$ can be used in aqueous $\mathrm{H}_{2} \mathrm{SO}_{4}$

Solution

The reaction between $\mathrm{MnO}_{4}^{-}$ and $\mathrm{HCl}$ may be represented as follows:
$2 \mathrm{MnO}_{4}^{-}(\mathrm{aq})+16 \mathrm{H}^{+}+10 \mathrm{Cl}^{-} ightarrow$
$2 \mathrm{Mn}^{2+}(\mathrm{aq})+8 \mathrm{H}_{2} \mathrm{O}(l)+5 \mathrm{Cl}_{2}(\mathrm{~g})$
Thus, on the basis of this reaction following electrochemical cell will be represented
$\mathrm{Pt}, \mathrm{Cl}(\mathrm{g})(\mathrm{1atm})\left|\mathrm{Cl}^{-}(\mathrm{aq}) \| \mathrm{MnO}_{4}^{-}(\mathrm{g})ight| \mathrm{Mn}^{2+}(\mathrm{aq})$
Hence, $E_{\text {cell }}^{0}=E_{\text {cathode }}^{0}-E_{\text {anode }}^{0}$
From given data $E_{\text {cell }}^{0}=1.51-1.40=0.11 \mathrm{~V}$
$\mathrm{E}^{\circ}$ cell is positive, hence $\Delta \mathrm{G}^{\circ}$ is negative. Thus, above cell reaction is feasible but $\mathrm{MnO}_{4}^{-}$ion can oxidise, $\mathrm{Fe}^{2+}$ to $\mathrm{Fe}^{3+}$ and $\mathrm{Cl}^{-}$to $\mathrm{Cl}_{2}$ in aqueous medium also. Therefore, for quantitative estimation of aqueous $\mathrm{Fe}\left(\mathrm{NO}_{3}ight)_{2}$ it is not a suitable reagent.

Asked in: JEE-TOPICTESTS-CHEMISTRY

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