Given :
\(E^\circ_{\frac{1}{2}\mathrm{Cl_2/Cl^-}} = 1.36\ \mathrm{V},\ E^\circ_{\mathrm{Cr^{3+}/Cr}} = -0.74\ \mathrm{V}\)
\(E^\circ_{\mathrm{Cr_2O_7^{2-}/Cr^{3+}}} = 1.33\ \mathrm{V},\ E^\circ_{\mathrm{MnO_4^-/Mn^{2+}}} = 1.51\ \mathrm{V}\)
The correct order of reducing power of the species \((\mathrm{Cr, Cr^{3+}, Mn^{2+}}\) and \(\mathrm{Cl^-})\) will be :
\(E^\circ_{\frac{1}{2}\mathrm{Cl_2/Cl^-}} = 1.36\ \mathrm{V},\ E^\circ_{\mathrm{Cr^{3+}/Cr}} = -0.74\ \mathrm{V}\)
\(E^\circ_{\mathrm{Cr_2O_7^{2-}/Cr^{3+}}} = 1.33\ \mathrm{V},\ E^\circ_{\mathrm{MnO_4^-/Mn^{2+}}} = 1.51\ \mathrm{V}\)
The correct order of reducing power of the species \((\mathrm{Cr, Cr^{3+}, Mn^{2+}}\) and \(\mathrm{Cl^-})\) will be :
Show Hint
- \(\mathrm{Mn^{2+}} \lt \mathrm{Cl^-} \lt \mathrm{Cr^{3+}} \lt \mathrm{Cr}\)
- \(\mathrm{Mn^{2+}} \lt \mathrm{Cl^{3+}} \lt \mathrm{Cl^-} \lt \mathrm{Cr}\)
- \(\mathrm{Cr^{3+}} \lt \mathrm{Cl^-} \lt \mathrm{Mn^{2+}} \lt \mathrm{Cr}\)
- \(\mathrm{Cr^{3+}} \lt \mathrm{Cl^-} \lt \mathrm{Cr} \lt \mathrm{Mn^{2+}}\)
The Correct Option is A
Solution and Explanation
\(\mathrm{Cr \rightarrow Cr^{3+}}\): E°(oxidation) = +0.74 V.
\(\mathrm{Cr^{3+} \rightarrow Cr_2O_7^{2-}}\): E°(oxidation) = -1.33 V.
\(\mathrm{Cl^- \rightarrow \frac{1}{2}Cl_2}\): E°(oxidation) = -1.36 V.
\(\mathrm{Mn^{2+} \rightarrow MnO_4^-}\): E°(oxidation) = -1.51 V.
Larger (less negative) oxidation potential means stronger reducing agent: \(\mathrm{Cr} \gt \mathrm{Cr^{3+}} \gt \mathrm{Cl^-} \gt \mathrm{Mn^{2+}}\).
Hence \(\mathrm{Mn^{2+}} \lt \mathrm{Cl^-} \lt \mathrm{Cr^{3+}} \lt \mathrm{Cr}\).
Answer: (1) \(\mathrm{Mn^{2+}} \lt \mathrm{Cl^-} \lt \mathrm{Cr^{3+}} \lt \mathrm{Cr}\)
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