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:

(B), (D), (C), (A)

The correct answer is Option (1) → (B), (D), (C), (A)

For reducing strength of ions, we consider their tendency to get oxidized.
An ion will be a stronger reducing agent if the corresponding standard reduction potential (E°) is more negative.

Given half-cell reactions and E° values:

• (A) Cu²⁺/Cu → +0.34 V
• (B) Sn²⁺/Sn → −0.14 V
• (C) AgBr/Ag → +0.10 V
• (D) Pb²⁺/Pb → −0.13 V

More negative E° ⇒ stronger reducing agent (ion)

Decreasing order of reducing strength:

$\text{Sn}^{2+} > \text{Pb}^{2+} > \text{AgBr} > \text{Cu}^{2+}$

✅ Correct answer: (B), (D), (C), (A)