Suppose that a fully charged lead-acid battery contains 1.50 L of 5.00 M H2SO4.

Question

Suppose that a fully charged lead-acid battery contains 1.50 L of 5.00 M H2SO4. What will be the concentration of H2SO4 in the battery after 2.10 A of current is drawn from the battery for 7.50 batteries?

Explanation / Answer

The electrode reactions that occur during discharge are: Cathode: PbO2(s) +HSO4- (aq) + 2e- ..........> PbSO4 (s) + 2H2O(l) Anode: Pb (s) + HSO4- (aq) .............> PbSO4 (s) + H+(aq) + 2e- Overall: PbO2(s) + Pb(s) + 2HSO4- (aq) +2H+ (aq) .........> 2PbSO4 (s) + 2H2O(l) Number of moles of electrons transfered from anode to cathodeto produce a current of 2.10A for 7.5 hours are = amperes * seconds * (1mol e- /96485C) = (2.10A)(7.5h (3600s / h))(1mol e- /96485C) = 0.587mol Total moles of H+ ions present initially =(1.5L)(5.0M)(2) = 15 moles Form the anode and cathode reactions it is clear that for each2moles of electron transfer 2moles of H+ consuming. So for 1.31moles of electron transfer (0.587mol e-)(2 mol H+/ 2 mol e-) = 0.587moles of H+ cosumed. Hence, moles of H+ remains after discharging = 15 mol -0.587mol = 14.413mol H+ Therefore, concentration of H2SO4 = ( 14.413 molH+ / 2) / (1.5L) = 4.80mol / L = 4.80M So concentration after discharging is 4.80M

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