Given : $$ \begin{gathered} \mathrm{E}_{\frac{1}{2} \mathrm{Cl}_2 / \mathrm{Cl}^{-}}^{\mathrm{o}}=1.36…

Given : $$ \begin{gathered} \mathrm{E}_{\frac{1}{2} \mathrm{Cl}_2 / \mathrm{Cl}^{-}}^{\mathrm{o}}=1.36 \mathrm{~V}, \mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^{\mathrm{o}}=-0.74 \mathrm{~V} \\ \mathrm{E}_{\mathrm{Cr}_2 \mathrm{O}_7^{2-} / \mathrm{Cr}^{3+}}^{\mathrm{o}}=1.33 \mathrm{~V}, \mathrm{E}_{\mathrm{MnO}_4^{-} / \mathrm{Mn}^{2+}}^{\mathrm{o}}=1.51 \mathrm{~V} \end{gathered} $$ The correct order of reducing power of the species $\left(\mathrm{Cr}, \mathrm{Cr}^{3+}, \mathrm{Mn}^{2+}\right.$ and $\left.\mathrm{Cl}^{-}\right)$will be:
  1. $\mathrm{Mn}^{2+} < \mathrm{Cl}^{-} < \mathrm{Cr}^{3+} < \mathrm{Cr}$
  2. $\mathrm{Mn}^{2+} < \mathrm{Cl}^{3+} < \mathrm{Cl}^{-} < \mathrm{Cr}$
  3. $\mathrm{Cr}^{3+} < \mathrm{Cl}^{-} < \mathrm{Mn}^{2+} < \mathrm{Cr}$
  4. $\mathrm{Cr}^{3+} < \mathrm{Cl}^{-} < \mathrm{Cr} < \mathrm{Mn}^{2+}$

Solution

Lower the value of reduction potential higher will be reducing power hence the correct order will be $ \mathrm{Mn}^{2+} < \mathrm{Cl}^{-} < \mathrm{Cr}^{3+} < \mathrm{Cr} $

Asked in: JEE Main 2013 (23 Apr Online)

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