Q7 / 75 · JEE Main 2025 (24 Jan Shift 2) · Chemistry
Based on the data given below :
$\begin{array}{ll}\mathrm{E}_{\mathrm{Cr}_2 \mathrm{O}_7^{2-} / \mathrm{Cr}^{3+}}^{\circ}=1.33 \mathrm{~V} & \mathrm{E}_{\mathrm{Cl}_2 / \mathrm{Cl}^{(-)}}^{\circ}=1.36 \mathrm{~V} \\ \mathrm{E}_{\mathrm{MnO}_4^{-} / \mathrm{Mn}^{2+}}^{\circ}=1.51 \mathrm{~V} & \mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^{\circ}=-0.74 \mathrm{~V}\end{array}$
the strongest reducing agent is :
- Correct answerACr
- B$\mathrm{Cl}^{-}$
- C$\mathrm{MnO}_4^{-}$
- D$\mathrm{Mn}^{2+}$
Solution
$\begin{aligned}
& \mathrm{E}_{\mathrm{Cr}_2 \mathrm{Of}_4^{\circ} / \mathrm{ca}^8}^0=1.33 \mathrm{~V}_{\mathrm{C}_2 / \mathrm{Cr}}^{\circ}=1.36 \mathrm{~V} \\
& \mathrm{E}_{\mathrm{MnO}_4 / \mathrm{Mm}^2}^{\circ}=1.51 \mathrm{~V} \mathrm{E}_{\mathrm{Cr}^2 / \mathrm{Cr}}^0=-0.74 \mathrm{~V}
\end{aligned}$
The species which has the most negative value of standard reduction potential will be the strongest reducing agent. Since $\mathrm{Cr}^{3+} / \mathrm{Cr}$ has SRP value of $-0.74 \mathrm{~V}, \mathrm{Cr}$ is the strongest reducing agent.
Hence, the correct answer is option: 'A'.