For an electrochemical cell
Sn(s) | Sn²⁺ (aq,1M)||Pb²⁺ (aq,1M)|Pb(s)
the ratio [ Sn²⁺ ]/[ Pb²⁺ ] when this cell attains equilibrium is _________. (Round off to the nearest integer)
(Given E_(Sn²⁺|Sn)⁰ = - 0.14V,
E_(Pb²⁺|Pb)⁰ = - 0.13V, 2.303RT/F = 0.06)

Solution

Chemistry ma’am · 4 steps

At equilibrium the cell is dead, so put E_(cell) = 0 in the Nernst equation and solve for the ratio.

1. Standard cell potential. Sn is the anode and Pb is the cathode. So E°_(cell) = E°_(cathode) - E°_(anode) = -0.13 - (-0.14) = 0.01 V.
2. Cell reaction. The reaction is Sn + Pb²⁺ → Sn²⁺ + Pb, with n = 2 electrons transferred. Solids do not appear in Q, so Q = [Sn²⁺]/[Pb²⁺].

   0 = E°_(cell) - 0.06/n log[Sn²⁺]/[Pb²⁺]

3. Put in numbers. log[Sn²⁺]/[Pb²⁺] = (0.01 × 2)/0.06 = 1/3, so the ratio is 10^(1/3) = 2.154.
4. So. The ratio [Sn²⁺]/[Pb²⁺] at equilibrium is about 2.15, which rounds to 2.

The common slip: 1. Taking n = 1 instead of 2 gives log Q = 1/6, a ratio of about 1.47, which rounds to 1.

The idea to keep. Equilibrium means E_(cell) = 0, so log K = nE°/0.06. Always check n from the balanced reaction before dividing.

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