[Single-digit Integer] The electrode reactions for charging of a lead storage battery are: PbSO4 + 2e^- - Pb + SO4^2- and PbSO4 + 2 - + SO4^2- + 4H^+ + 2e^-. The electrolyte in the battery is an aqu

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[Single-digit Integer] The electrode reactions for charging of a lead storage battery are: PbSO4 + 2e^- -> Pb + SO4^2- and PbSO4 + 2$H_2O$ -> $PbO_2$ + SO4^2- + 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

Chemistry diagram for: [Single-digit Integer] The electrode reactions for charging of a lead storage battery are: PbSO4 + 2e^- -> Pb + SO4^2- a

Answer: 0002

💡 Solution & Explanation

Prashant Kotian · Accepted Answer

Correct answer: 0002. Step 1 - Calculation of the Initial Mass of H2SO4 Before charging, the specific gravity of the liquid is 1.10 g/mL (which equals its density), the mass percentage of H2SO4 is 16%, and the total volume of the liquid is 2 L = 2000 mL. The total initial mass of the electrolyte solution is: 31782 ext{Mass of initial solution} = ext{Volume} imes ext{Density}_{ ext{initial}} = 2000 ext{ mL} imes 1.10 ext{ g/mL} = 2200 ext{ g}31782 The initial mass of pure H2SO4 present in the solution is: 31782 ext{Mass of }\ce{H2SO4} ext{ (initial)} = rac{16}{100} imes 2200 ext{ g} = 352 ext{ g}31782 Step 2 - Calculation of the Final Mass of H2SO4 After charging, the specific gravity increases to 1.42 g/mL, the mass percentage of H2SO4 becomes 40%, and the volume remains constant at 2000 mL. The total final mass of the electrolyte solution is: 31782 ext{Mass of final solution} = ext{Volume} imes ext{Density}_{ ext{final}} = 2000 ext{ mL} imes 1.42 ext{ g/mL} = 2840 ext{ g}31782 The final mass of pure H2SO4 is: 31782 ext{Mass of }\ce{H2SO4} ext{ (final)} = rac{40}{100} imes 2840 ext{ g} = 1136 ext{ g}31782 Step 3 - Determining Moles of H2SO4 Produced The net increase in mass of H2SO4 during charging: 31782\Delta m_{\ce{H2SO4}} = 1136 ext{ g} - 352 ext{ g} = 784 ext{ g}31782 Using the molar mass of H2SO4 (98 g/mol): 31782n(\ce{H2SO4}) = rac{784 ext{ g}}{98 ext{ g/mol}} = 8 ext{ mol}31782 Step 4 - Relating Moles of H2SO4 to Moles of Electrons Transferred The net cell reaction during charging: 31782\ce{2PbSO4(s) + 2H2O(l) -> Pb(s) + PbO2(s) + 2H2SO4(aq)}31782 From stoichiometry: 2 mol H2SO4 produced per 2 mol electrons transferred. Therefore n(e-) = n(H2SO4) = 8 mol. Step 5 - Calculation of Average Charging Current Total electrical charge: 31782Q = n_{e^-} imes F = 8 ext{ mol} imes 96500 ext{ C/mol} = 772000 ext{ C}31782 Time in seconds: 31782t = rac{965}{9} ext{ h} imes 3600 ext{ s/h} = 386000 ext{ s}31782 Average current: 31782I = rac{Q}{t} = rac{772000 ext{ C}}{386000 ext{ s}} = oxed{2} ext{ A}31782

💡 Solution & Explanation

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