For the given cell :
Cu(s) | Cu²⁺(C₁M) || Cu²⁺(C₂M) | Cu(s)
change in Gibbs energy (ΔG) is negative, if :

Solution

Chemistry ma’am · 3 steps

In a concentration cell, current flows only when the cathode side holds the stronger solution.

1. Read the cell. Both electrodes are copper, so E°_(cell) = 0. The left side is the anode: Cu → Cu²⁺(C₁) + 2e⁻. The right side is the cathode: Cu²⁺(C₂) + 2e⁻ → Cu.

   E_(cell) = 0 - 0.0591/2 log C₁/C₂ = 0.0591/2 log C₂/C₁

2. Link to Gibbs energy. Δ G = -nFE_(cell), so Δ G is negative only when E_(cell) is positive. That needs log C₂/C₁ > 0, that is, C₂ > C₁.

   Option	C₂/C₁	Sign of E_(cell)	Sign of Δ G
   (a) C₂ = √2 C₁	√2 > 1	positive	negative
   (b) C₂ = C₁/√2	1/√2 < 1	negative	positive
   (c) C₁ = 2C₂	1/2 < 1	negative	positive
   (d) C₁ = C₂	1	zero	zero

3. So. Only (a) makes the cathode solution more concentrated than the anode solution, so E_(cell) > 0 and Δ G < 0.

The idea to keep. For a concentration cell, E° = 0 and E = 0.0591/n log C_(cathode)/C_(anode). It is spontaneous only when the cathode (right) side is more concentrated.

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