JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
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 :
- A Cr
- B \(\mathrm{Cl}^{-}\)
- C \(\mathrm{MnO}_4^{-}\)
- D \(\mathrm{Mn}^{2+}\)
Answer & Solution
Correct Answer
(A) Cr
Step-by-step Solution
Detailed explanation
\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…
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