The EMF for the cell: Ag(s) \| AgCl(s) \| KCl(0.2 M) \|\| KBr (0.001 M) \| AgBr(s) \| Ag(s) at 25°C — Electrochemistry Chemistry Question

💡 Solution & Explanation

Anode [Ag^+] = Ksp(AgCl) / [Cl^-] = 2e-10 / 0.2 = 10^-9 M. Cathode [Ag^+] = Ksp(AgBr) / [Br^-] = 4e-13 / 0.001 = 4e-10 M. This is a simple Ag concentration cell, so E_cell = 0.06 * log([Ag^+]_cathode / [Ag^+]_anode) = 0.06 * log(4e-10 / 1e-9) = 0.06 * log(0.4). log(0.4) = log(4) - 1 = 0.6 - 1 = -0.4. E_cell = 0.06 * (-0.4) = -0.024 V. Therefore, correct answer is B.