Arrange the following electrodes in decreasing order of the strength of the respective ions as reducing agent.
(A) \( Cu^{2+} + 2e^-→ Cu(s)\,\,\,\, E° = 0.34 V \)
(B) \( Sn^{2+} + 2e^- → Sn(s)\,\,\,\, E° = -0.14 V \)
(C) \( AgBr(s) + e^- → Ag(s) + Br^-\,\,\,\, E° =0.10 V \)
(D) \( Pb^{2+} + 2e^-→ Pb(s)\,\,\,\, E° = -0.13 V \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (B), (D), (C), (A)
The standard reduction potential \( E^{\circ} \) measures the tendency of an electrode to be reduced. The lower (more negative) the reduction potential, the more easily the reduced form on the right of the half-reaction gives up its electrons, and the stronger it is as a reducing agent. So the strength as reducing agents decreases as \( E^{\circ} \) increases.
Arrange the four electrodes by their \( E^{\circ} \) values:
(B) \( \ce{Sn^2+/Sn} \): \( E^{\circ} = -0.14\,\mathrm{V} \)
(D) \( \ce{Pb^2+/Pb} \): \( E^{\circ} = -0.13\,\mathrm{V} \)
(C) \( \ce{AgBr/Ag, Br-} \): \( E^{\circ} = +0.10\,\mathrm{V} \)
(A) \( \ce{Cu^2+/Cu} \): \( E^{\circ} = +0.34\,\mathrm{V} \)
Shortcut: The strongest reducing agent has the lowest \( E^{\circ} \) (B, −0.14 V) and the weakest the highest (A, +0.34 V). Only Option (1) starts with (B) and ends with (A).
Note: It seems that there is a mistake in the original NTA paper. The question speaks of the strength of the respective ions as reducing agents; the reducing agents are the metals (the reduced forms Sn, Pb, Ag and Cu), so it should read “strength of the respective metals as reducing agents”.