Calculate the Ksp of lead iodide from the following standard electrode potentials, at 25°C: 2 e- + PbI₂(S) → Pb(s) + 2 I-(aq) E° = -0.365 V
2 e- + Pb²⁺(aq) → Pb(s) E° = -0.126 V

Question

Calculate the Ksp of lead iodide from the following standard electrode potentials, at 25°C: 2 e- + PbI₂(S) → Pb(s) + 2 I-(aq) E° = -0.365 V 2 e- + Pb²⁺(aq) → Pb(s) E° = -0.126 V

Quizplus · Accepted Answer

The Ksp can be calculated using the Nernst equation and the given standard electrode potentials. The reaction for PbI2 dissolving is PbI2(s) ⇌ Pb2+(aq) + 2I-(aq). The cell potential for this reaction is E°cell = E°(Pb2+/Pb) - E°(PbI2/Pb). Substituting the given values, E°cell = -0.126 V - (-0.365 V) = 0.239 V. Using the relation ΔG° = -nFE°cell and ΔG° = -RTlnK, we find Ksp = 8 × 10^-9.

Calculate the Ksp of Lead Iodide from the Following Standard