The electrode reactions for charging of a lead storage battery are: ; . The electrolyte in the batte — Electrochemistry Chemistry Question
Question
The electrode reactions for charging of a lead storage battery are: $PbSO_4 + 2e \rightarrow Pb + SO_4^{2-}$ ; $PbSO_4 + 2H_2O \rightarrow PbO_2 + SO_4^{2-} + 4H^+ + 2e$. The electrolyte in the battery is an aqueous solution of sulphuric acid. Before charging, the specific gravity of the liquid was found to be 1.10 (16% $H_2SO_4$ by wt.). After charging for 965/9 h, the specific gravity of the liquid was found to be 1.42 (40% $H_2SO_4$ by weight). If the battery contained 2 L of the liquid and the volume remains constant during charge, the average current (in A) used for charging the battery is
💡 Solution & Explanation
Initial solution mass = $2000 \text{ mL} \times 1.10 \text{ g/mL} = 2200$ g. Initial $H_2SO_4 = 0.16 \times 2200 = 352$ g. Let $x$ be the Faradays passed. Charging produces $x$ moles (or $98x$ grams) of $H_2SO_4$ and consumes $x$ moles (or $18x$ grams) of $H_2O$. Final solution mass = $2200 + 98x - 18x = 2200 + 80x$. Final $H_2SO_4$ mass = $352 + 98x$. We are given final concentration is 40%: $(352 + 98x) / (2200 + 80x) = 0.40$. Solving: $352 + 98x = 880 + 32x \Rightarrow 66x = 528 \Rightarrow x = 8$ F. Total charge $Q = 8 \times 96500$ C. Time $t = (965 / 9) \text{ h} = (965 / 9) \times 3600 \text{ s} = 386000$ s. Current $I = Q / t = (8 \times 96500) / 386000 = 772000 / 386000 = 2$ A. Therefore, correct answer is 2.