For the cell
Cu(s) | Cu²⁺ (aq) (0.1 M) || Ag⁺(aq) (0.01 M) | Ag(s)
the cell potential E₁ = 0.3095 V
For the cell
Cu(s) | Cu²⁺ (aq) (0.01 M) || Ag⁺(aq) (0.001 M) | Ag(s)
the cell potential = ____________ × 10⁻² V. (Round off the nearest integer).
[Use : 2.303RT/F = 0.059]

Solution

Chemistry ma’am · 4 steps

Same electrodes, so E° is the same: get it from the first cell, then reuse it in the second.

1. Cell reaction. Copper is oxidised at the anode and silver ions are reduced at the cathode: Cu + 2Ag⁺ → Cu²⁺ + 2Ag. Two electrons are transferred, so n = 2, and [Ag⁺] appears squared in the reaction quotient.

   E = E°_(cell) - 0.059/2 log[Cu²⁺]/[Ag⁺]²

2. First cell. Q = 0.1/(0.01)² = 10³, so log Q = 3. Then 0.3095 = E° - 0.0295 × 3 = E° - 0.0885, which gives E°_(cell) = 0.398 V.
3. Second cell. Q = 0.01/(0.001)² = 10⁴, so log Q = 4. Then E₂ = 0.398 - 0.0295 × 4 = 0.398 - 0.118 = 0.280 V.
4. So. E₂ = 0.28 V = 28 × 10⁻² V, so the answer is 28.

The common slip: 31. Forgetting to square [Ag⁺] makes Q = 10 for both cells, so E₂ comes out equal to E₁ and you get 31, not 28.

The idea to keep. When two cells share the same electrodes, E° does not change. Find it from the given cell, then only Q changes. Always raise each ion's concentration to the power of its coefficient.

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