Assume a cell with the following reaction
Cu₍ₛ₎ + 2Ag + (1 × 10⁻³M) → Cu²⁺(0.250M) + 2Ag₍ₛ₎
E_(cell)^Θ = 2.97 V
E cell for the above reaction is ______________ V. (Nearest integer)
[Given : log 2.5 = 0.3979, T = 298 K]

Solution

Chemistry ma’am · 4 steps

Write Q with products over reactants, square the silver ion, and the Nernst correction is tiny.

1. Electrons and Q. Copper goes from 0 to +2, so n = 2. Solids do not appear in Q, so Q = [Cu²⁺]/[Ag⁺]², with [Ag⁺] squared because of its coefficient 2.

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

2. Find log Q. Q = 0.250/(1 × 10⁻³)² = 2.5 × 10⁵. So log Q = log 2.5 + 5 = 5.3979.
3. Put in numbers. E_(cell) = 2.97 - 0.059/2 × 5.3979 = 2.97 - 0.159 = 2.81 V.
4. So. E_(cell) ≈ 2.81 V, which rounds to 3.

The idea to keep. In the Nernst equation, raise each ion's concentration to its coefficient in the balanced reaction, and leave solids out of Q. Then use n from the electrons actually transferred.

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