JEE Main · 2022 · Shift-IImediumEC-010

The correct order of reduction potentials of the following pairs is | Label | Half-cell | |---|---| | A | Cl2/Cl- | | B…

Electrochemistry · Class 12 · JEE Main Previous Year Question

Question

The correct order of reduction potentials of the following pairs is

| Label | Half-cell | |---|---| | A | Cl2/Cl\mathrm{Cl_2/Cl^-} | | B | I2/I\mathrm{I_2/I^-} | | C | Ag+/Ag\mathrm{Ag^+/Ag} | | D | Na+/Na\mathrm{Na^+/Na} | | E | Li+/Li\mathrm{Li^+/Li} |

Choose the correct answer from the options given below.

Options
  1. a

    A > B > C > E > D

  2. b

    A > C > B > E > D

  3. c

    A > B > C > D > E

  4. d

    A > C > B > D > E

Correct Answerd

A > C > B > D > E

Detailed Solution

Strategy: Reduction potential (E°) values reflect the electrochemical series. Non-metals and noble metals have high positive potentials, while active metals (like Group 1) have highly negative potentials.

Step 1: Recall standard values

  • \ceCl2/Cl\ce{Cl2/Cl^-}: +1.36 V\approx +1.36\ \mathrm{V} (A)
  • \ceAg+/Ag\ce{Ag^+/Ag}: +0.80 V\approx +0.80\ \mathrm{V} (C)
  • \ceI2/I\ce{I2/I^-}: +0.54 V\approx +0.54\ \mathrm{V} (B)
  • \ceNa+/Na\ce{Na^+/Na}: 2.71 V\approx -2.71\ \mathrm{V} (D)
  • \ceLi+/Li\ce{Li^+/Li}: 3.05 V\approx -3.05\ \mathrm{V} (E)

Step 2: Compare and Order Higher values are: \ceCl2>Ag+>I2\ce{Cl2 > Ag^+ > I2}. Lowest values (active metals): \ceNa+>Li+\ce{Na^+ > Li^+}. Full order: A > C > B > D > E.

Answer: (d) A > C > B > D > E\boxed{\text{Answer: (d) A > C > B > D > E}}

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