Copper sulphate solution (250 ml) was electrolysed using a platinum anode and a copper cathode. A constant current of 2 mA was passed for 19.3 min. It was found that after electrolysis the absorbance

Question

Copper sulphate solution (250 ml) was electrolysed using a platinum anode and a copper cathode. A constant current of 2 mA was passed for 19.3 min. It was found that after electrolysis the absorbance of the solution was reduced to 50% of its original value. Calculate the concentration of copper sulphate in the solution to begin with.

Prashant Kotian · Accepted Answer

Correct answer: A. Charge Q = 2 × 10^–3 A × 19.3 × 60 s = 2.316 C. Faradays = 2.316 / 96500 = 2.4 × 10^–5 F. Since Cu^2+ + 2e^– -> Cu, moles of Cu^2+ reduced = 2.4 × 10^–5 / 2 = 1.2 × 10^–5 mol. Absorbance is proportional to concentration. A 50% reduction in absorbance means 50% of the initial Cu^2+ was electrolyzed. Initial moles = 2 × 1.2 × 10^–5 = 2.4 × 10^–5 mol. Initial concentration = 2.4 × 10^–5 mol / 0.25 L = 9.6 × 10^–5 M. Therefore, correct answer is A.

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