Two students use same stock solution of ZnSO4 but different solutions of CuSO4. The EMF of one cell — Electrochemistry Chemistry Question

💡 Solution & Explanation

For the Daniel cell, E = E° - 0.03 * log([Zn^2+]/[Cu^2+]). Let Cell 1 have the higher EMF and [Cu^2+]_1 = 0.5 M. E1 - E2 = 0.03 * log([Cu^2+]_1 / [Cu^2+]_2) = 0.03 V. Thus, log([Cu^2+]_1 / [Cu^2+]_2) = 1, giving [Cu^2+]_1 / [Cu^2+]_2 = 10. Substituting [Cu^2+]_1 = 0.5 M, we get [Cu^2+]_2 = 0.5 / 10 = 0.05 M. Therefore, correct answer is A.