Question

A lead-acid battery uses a redox reaction in which lead(0) and lead(IV) are both converted to lead(II). This reaction is facilitated by the presence of sulfuric acid, H2SO4, as shown by the reaction

Pb+PbO2+2H2SO4→2PbSO4+2H2O

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 3.30 A of current is drawn from the battery for 7.50 hours?

Express your answer with the appropriate units.

Answer 1

Pb+PbO2+2H2SO4→2PbSO4+2H2O

fully charged lead-acid battery contains 1.50 L of 5.00 M H2SO4

(7.50 hours) (60 minutes / hour) (60 seconds / min) = 27,000seconds
3.30A = 3.30 Coulombs / sec
(3.30 Coulombs / sec) (27,000 seconds) = 89,100 Coulombs
an anode is the site for oxidation, (loss of electrons).
a lead anode will lose mass, by releasing Pb+2 ions into the solution,
as it loses 2 electrons per lead atom.
using Faraday's constant, F : (Charge of 1 mole of electrons = 96500 Coulombs)
find the number of moles of electrons lost
(89,100 Coulombs) (1 mole of electrons / 96500 Coulombs) = 0.923316 moles of electrons
find the moles of lead which lost those electrons:
(0.923316 moles of electrons) (1 mole lead / 2 moles of e- lost) = 0.461658 moles of lead
an equal amount of PbO2 was reduced to Pb+2 ions by taking the same electrons that the anode lost
                 = an additional 0.461658 moles of Pb+2
for a total of 0.923316 moles of Pb+2 ion released to the solution
by the reaction:
1 Pb (anode) + 1 PbO2 (cathode) + 2 H2SO4→2 PbSO4 + 2 H2O
an equal amount of H2SO4 leaves the solution as the Pb+2 forms the precipitate PbSO4
so 0.923316 moles of H2SO4 are removed
(1.50 L) (5.00 mol / L H2SO4) = 7.5 moles H2SO4 was present
(7.5 moles H2SO4 was present) - (0.923316 moles removed) = 6.576684 moles remain

(6.576684 moles remain) / (1.50 L) = 4.384456 Molar H2SO4 remains