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Given: E⁢∘Cl2/Cl− = 1.36 V, E⁢∘Cr3+/Cr = −0.74 V E⁢∘Cr2⁢O2-7/Cr3+ = 1.33 V, E⁢∘MnO−4/Mn2+ = 1.51 V Among the following, the strongest reducing agent is: - Chemistry (Theory)

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Question

Given: \[\ce{E^{\circ}_{Cl_2/Cl^-}}\] = 1.36 V, \[\ce{E^{\circ}_{Cr^{3+}/Cr}}\] = −0.74 V

\[\ce{E^{\circ}_{Cr_2O^{2-}_7/Cr^{3+}}}\] = 1.33 V, \[\ce{E^{\circ}_{MnO^-_4/Mn^{2+}}}\] = 1.51 V

Among the following, the strongest reducing agent is:

Options

  • Cr3+

  • Cl

  • Cr

  • Mn2+

MCQ
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Solution

Cr

Explanation:

A strong reducing agent has a low (more negative) standard reduction potential, meaning it readily loses electrons (gets oxidised).

From the data:

\[\ce{E^{\circ}_{Cr^{2+}/Cr}}\] = −0.74 V

Cr is oxidised to Cr3+.

All other values are more positive.

Thus, Cr has the most negative E° value, making it the strongest reducing agent among the options.

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Chapter 3: Electrochemistry - OBJECTIVE (MULTIPLE CHOICE) TYPE QUESTIONS [Page 201]

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Nootan Chemistry Part 1 and 2 [English] Class 12 ISC
Chapter 3 Electrochemistry
OBJECTIVE (MULTIPLE CHOICE) TYPE QUESTIONS | Q 69. | Page 201
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