JEE Main 2025 · 23 January · evening · Single correct · +4 / −1

Standard electrode potentials for a few half cells are mentioned below :
E_(Cu²⁺ / Cu)°=0.34 V, E_(Zn²⁺ / Zn)°=-0.76 V ; E_(Ag⁺ / Ag)°=0.80 V, E_(Mg²⁺ / Mg)°=-2.37 V
Which one of the following cells gives the most negative value of Δ G° ?

  1. (a) Zn|Zn²⁺(1 M) ∥ Ag⁺(1 M)| Ag
  2. (b) Ag|Ag⁺(1 M)||Mg²⁺(1 M)| Mg
  3. (c) Zn|Zn²⁺(1 M)||Mg²⁺(1 M)| Mg
  4. (d) Cu|Cu²⁺(1 M)||Ag⁺(1 M)| Ag
Answer and solution

Answer (a) Zn|Zn²⁺(1 M) ∥ Ag⁺(1 M)| Ag

Solution

Chemistry ma’am · 3 steps

Read each cell as written, left as anode and right as cathode, then pick the largest positive E°.

  1. Link the two. The most negative Δ G° needs the largest positive nE°_(cell). A cell with a negative E°_(cell) has a positive Δ G°, so it is not spontaneous as written.

    Δ G° = -nFE°_(cell), E°_(cell) = E°_(right) - E°_(left)

  2. Use the notation. In cell notation the left electrode is the anode and the right one is the cathode. Each cell here transfers n = 2 electrons, so only E°_(cell) changes from one option to the next.
    Cell (left to right)E°_(cell)Δ G° = -2FE°_(cell)
    (a) Zn to Ag0.80 - (-0.76) = +1.56 V-3.12F, most negative
    (b) Ag to Mg-2.37 - 0.80 = -3.17 V+6.34F, positive
    (c) Zn to Mg-2.37 - (-0.76) = -1.61 V+3.22F, positive
    (d) Cu to Ag0.80 - 0.34 = +0.46 V-0.92F
  3. So. Cell (a) has the largest positive E°_(cell). Its Δ G° = -2 × 96500 × 1.56 ≈ -301 kJ mol⁻¹, the most negative of the four. The answer is (a).

The idea to keep. Δ G° = -nFE° follows the cell as written. Take E°_(right) - E°_(left), and do not quietly flip a cell that is written backwards.

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

Study Electrochemistry

Asked about 170 times in JEE Main since 2020. This question is from 2025.