Two students use same stock solution of ZnSO4 but different solutions of CuSO4. The EMF of one cell is 0.03 V higher than the other. The concentration of CuSO4 in the cell with higher EMF value is 0.5

Question

Two students use same stock solution of ZnSO4 but different solutions of CuSO4. The EMF of one cell is 0.03 V higher than the other. The concentration of CuSO4 in the cell with higher EMF value is 0.5 M. The concentration of CuSO4 in the other cell is (2.303RT/F = 0.06)

Prashant Kotian · Accepted Answer

Correct answer: A. 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.

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