Given are the standard electrode potentials:
A. I2/I− = +0.54 V
B. Fe3+/Fe2+ = +0.77 V
C. Cu+/Cu = +0.52 V
D. Ag+/Ag = +0.80 V
Arrange these redox couples in the increasing order of oxidizing strength.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → C, A, B, D
A standard electrode potential (E°) is a reduction potential. It measures how readily the oxidised form of a couple (\( \ce{I2} \), \( \ce{Fe^{3+}} \), \( \ce{Cu^{+}} \), \( \ce{Ag^{+}} \)) accepts electrons. The more easily a species is reduced, the stronger it is as an oxidising agent. So:
The more positive the E° value, the stronger the oxidising agent.
Arranging the four couples by E°, from least to most positive:
(C) \( \ce{Cu^{+}} \)/Cu = +0.52 V: \( \ce{Cu^{+}} \) is the weakest oxidising agent here
(A) \( \ce{I2} \)/\( \ce{I^{-}} \) = +0.54 V
(B) \( \ce{Fe^{3+}} \)/\( \ce{Fe^{2+}} \) = +0.77 V
(D) \( \ce{Ag^{+}} \)/Ag = +0.80 V: \( \ce{Ag^{+}} \) is the strongest oxidising agent here
Shortcut: C has the lowest E° and D the highest. Only Option (3) starts with C and ends with D.
NCERT (2026-27 edition):
Measurement of Electrode Potential – …
| Reaction (Oxidised form + ne– → Reduced form) | Eo/V |
|---|---|
| Ag+ + e– → Ag(s) | 0.80 |
| Fe3+ + e– → Fe2+ | 0.77 |
| I2 + 2e– → 2I– | 0.54 |
| Cu+ + e– → Cu(s) | 0.52 |