JEE Main 2022 · 25 June · evening · Single correct · +4 / −1

The correct order of reduction potentials of the following pairs is
A. Cl₂/Cl⁻
B. I²/I⁻
C. Ag⁺/Ag
D. Na⁺/Na
E. Li⁺/Li
Choose the correct answer from the options given below.

  1. (a) A > C > B > D > E
  2. (b) A > B > C > D > E
  3. (c) A > C > B > E > D
  4. (d) A > B > C > E > D
Answer and solution

Answer (a) A > C > B > D > E

Solution

Chemistry ma’am · 4 steps

Line up the five E° values you know; just watch iodine against silver and lithium against sodium.

  1. What the number means. A higher reduction potential means the oxidised form takes electrons more readily. Strong oxidising agents like Cl₂ sit at the top, and strong reducing metals like Li sit at the bottom.
    PairE° (V)Rank
    A. Cl₂/Cl⁻+1.361
    C. Ag⁺/Ag+0.802
    B. I₂/I⁻+0.543
    D. Na⁺/Na-2.714
    E. Li⁺/Li-3.055
  2. Silver beats iodine. I₂ is a much weaker oxidising agent than Cl₂. At +0.54 V it falls below Ag⁺/Ag at +0.80 V.
  3. Lithium is lowest. The very small Li⁺ ion has a very large hydration enthalpy. That makes Li the strongest reducing agent in water, so its E° is the most negative of all, below sodium.
  4. So. The order is A > C > B > D > E, which is option (a).

The idea to keep. Keep a few anchor E° values ready: I₂ below Ag, and Li below Na, because hydration of the tiny Li⁺ makes lithium the strongest reducer in water.

Answer from the NTA final answer key. Solution by PrepWiser.

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Asked about 170 times in JEE Main since 2020. This question is from 2022.