Verified Solution • 2025–26 Session moderate

QJEEM25Ch42be: JEE Main 2025 Chemistry - Class 12 JEEM25Ch42be Solution: Based on the data given below : the strongest reducing agent is :

Based on the data given below :

ECr2O72−/Cr3+∘=1.33 Vamp;ECl2/Cl(−)∘=1.36 VEMnO4−/Mn2+0=1.51 Vamp;ECr3+/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+}}^0=1.51 \mathrm{~V} & \mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^{\circ}=-0.74 \mathrm{~V}\end{array}

the strongest reducing agent is :

Multiple Choice Options

  • A Cl−\mathrm{Cl}^{-}
  • B MnO4−\mathrm{MnO}_4^{-}
  • C Cr\mathrm{Cr}
  • D Mn2+\mathrm{Mn}^{2+}

Step-by-Step Verified Solution

For strongest reducing agent

Reduction potential should be lowest

Hence Cr is the strongest reducing agent.

Options (2)

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