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Chemistry · 2013

JEE Main · 23 April 2013, Shift 1 · Q53

Given: E^0_1/2 Cl_2 / Cl^- =1.36 V, E_Cr^3+ / Cr^0=-0.74 V, E^0_Cr_2 O_7^2- / Cr^3+ =1.33 V, E^0_MnO_4^- / Mn^2+ =1.51 V The correct order of…

Given : $$\begin{aligned} & \mathrm{E}^0_{\frac{1}{2} \mathrm{Cl}_2 / \mathrm{Cl}^{-}}=1.36 \mathrm{~V}, \mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^0=-0.74 \mathrm{~V}, \\ & \mathrm{E}^0_{\mathrm{Cr}_2 \mathrm{O}_7^{2-} / \mathrm{Cr}^{3+}}=1.33 \mathrm{~V}, \mathrm{E}^0_{\mathrm{MnO}_4^{-} / \mathrm{Mn}^{2+}}=1.51 \mathrm{~V} \end{aligned} $$ The correct order of reducing power of the species ( $\displaystyle \mathrm{Cr}, \mathrm{Cr}^{3+}, \mathrm{Mn}^{2+}$ and $\displaystyle \mathrm{Cl}^{-}$) will be :
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JEE Main 2013 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.