During the discharge of a lead storage battery, the density of sulphuric acid fell from 1.5 to 1.1 g/ml. Sulphuric acid of density 1.5 g/ml is 40% H2SO4, by weight, and that of density 1.1 g/ml is 10%

Question

During the discharge of a lead storage battery, the density of sulphuric acid fell from 1.5 to 1.1 g/ml. Sulphuric acid of density 1.5 g/ml is 40% H2SO4, by weight, and that of density 1.1 g/ml is 10% H2SO4, by weight. The battery holds 3.6 L of the acid and the volume remained practically constant during the discharge. Calculate the number of ampere-hours which the battery should have been used. The electrode reactions are: Pb + SO4^2– -> PbSO4 + 2e–; PbO2 + 4H^+ + SO4^2– + 2e– -> PbSO4 + 2H2O

Prashant Kotian · Accepted Answer

Correct answer: B. Initial mass of solution = 3600 mL × 1.5 g/mL = 5400 g. Initial H2SO4 = 0.40 × 5400 = 2160 g. Final mass of solution = 3600 mL × 1.1 g/mL = 3960 g. Final H2SO4 = 0.10 × 3960 = 396 g. Mass of H2SO4 consumed = 2160 - 396 = 1764 g. Moles of H2SO4 consumed = 1764 / 98 = 18 mol. The overall discharge reaction consumes 2 moles of H2SO4 per 2 Faradays (1 mole of reaction). Thus, 18 moles H2SO4 corresponds to 18 F of charge. Ampere-hours = 18 F × (96500 C/F) / (3600 s/h) = 18 × 26.805 = 482.5 Ah. Therefore, correct answer is B.

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