Arrange the following electrodes in increasing order of strength as an oxidizing agent.
(A) \( Ni^{2+} + 2e^- → Ni(s)\,\,\,\,\, E° = -0.25 V \)
(B) \( Cu^{2+} + 2e^- → Cu(s)\,\,\,\,\, E° = 0.34 V \)
(C) \( Br_2 + 2e^- → 2Br^-\,\,\,\,\, E° = 1.09 V \)
(D) \( O_2(g) + 2H^+ + 2e^- → H_2O_2\,\,\,\,\, E° = 0.68 V \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A), (B), (D), (C)
A standard reduction potential, \( E^{\circ} \), measures the tendency of the species on the left of the half-reaction to gain electrons. An oxidising agent is a species that gains electrons, so the higher the reduction potential, the stronger the oxidising agent.
(A) \( \ce{Ni^2+} \): \( E^{\circ} = -0.25\,\mathrm{V} \)
(B) \( \ce{Cu^2+} \): \( E^{\circ} = +0.34\,\mathrm{V} \)
(C) \( \ce{Br2} \): \( E^{\circ} = +1.09\,\mathrm{V} \)
(D) \( \ce{O2} \) (in acid, to \( \ce{H2O2} \)): \( E^{\circ} = +0.68\,\mathrm{V} \)
Shortcut: The weakest oxidising agent has the lowest \( E^{\circ} \), (A), and the strongest the highest, (C). Only Option (4) starts with (A) and ends with (C).