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JEE Main 202524 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
24 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Electrochemistry

JEE Main 2025 · 24 January · Evening Shift · Chemistry · Electrochemistry

Based on the data given below : array ll E_ Cr_2 O_7^ 2- / Cr^ 3+ ^...

Based on the data given below : ECr2O72−/Cr3+∘=1.33 VECl2/Cl(−)∘=1.36 VEMnO4−/Mn2+∘=1.51 VECr3+/Cr∘=−0.74 V\begin{array}{ll}\mathrm{E}_{\mathrm{Cr}_2 \mathrm{O}_7^{2-} / \mathrm{Cr}^{3+}}^{\circ}=1.33 \mathrm{~V} & \mathrm{E}_{\mathrm{Cl}_2 / \mathrm{Cl}^{(-)}}^{\circ}=1.36 \mathrm{~V} \\ \mathrm{E}_{\mathrm{MnO}_4^{-} / \mathrm{Mn}^{2+}}^{\circ}=1.51 \mathrm{~V} & \mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^{\circ}=-0.74 \mathrm{~V}\end{array} the strongest reducing agent is :

Options

  1. ACr
  2. BCl−\mathrm{Cl}^{-}
  3. CMnO4−\mathrm{MnO}_4^{-}
  4. DMn2+\mathrm{Mn}^{2+}

Correct answer

A. Cr

Step-by-step solution

ECr2Of4∘/ca80=1.33 VC2/Cr∘=1.36 VEMnO4/Mm2∘=1.51 VECr2/Cr0=−0.74 V\begin{aligned} & \mathrm{E}_{\mathrm{Cr}_2 \mathrm{Of}_4^{\circ} / \mathrm{ca}^8}^0=1.33 \mathrm{~V}_{\mathrm{C}_2 / \mathrm{Cr}}^{\circ}=1.36 \mathrm{~V} \\ & \mathrm{E}_{\mathrm{MnO}_4 / \mathrm{Mm}^2}^{\circ}=1.51 \mathrm{~V} \mathrm{E}_{\mathrm{Cr}^2 / \mathrm{Cr}}^0=-0.74 \mathrm{~V} \end{aligned} The species which has the most negative value of standard reduction potential will be the strongest reducing agent. Since Cr3+/Cr\mathrm{Cr}^{3+} / \mathrm{Cr} has SRP value of −0.74 V,Cr-0.74 \mathrm{~V}, \mathrm{Cr} is the strongest reducing agent.
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