The EMF for the cell: Ag(s) \| AgCl(s) \| KCl(0.2 M) \|\| KBr (0.001 M) \| AgBr(s) \| Ag(s) at 25°C is (Ksp(AgCl) = 2.0 × 10^–10; Ksp(AgBr) = 4.0 × 10^–13, 2.303RT/F = 0.06, log 2 = 0.3)

Question

Prashant Kotian · Accepted Answer

Correct answer: B. 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.

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