The solubility of AgCl in water at 25°C is (2.303 RT/F = 0.058) — Electrochemistry Chemistry Question

💡 Solution & Explanation

The cell potential E°_cell at 25°C is exactly E°_AgCl/Ag since the anode is the standard hydrogen electrode (E° = 0 V). Interpolating the given values gives E°_cell = 0.22 V. We also know E°_AgCl/Ag = E°_Ag+/Ag + 0.058 × log(Ksp). Thus, 0.22 = 0.80 + 0.058 × log(Ksp). Solving for Ksp: log(Ksp) = (0.22 - 0.80) / 0.058 = -0.58 / 0.058 = -10. Ksp = 10^-10. Solubility = √(Ksp) = 10^-5 M. Therefore, correct answer is C.