Answer:
The answers to the question are as follows
First part
The mass of PbO2 (s) reduced at the cathode during the period is = 467.55_g
Second part
The electrical charge are transferred from Pb to PbO2 is 377186.86_C or 3.909 F
Explanation:
To solve this, we write the equation for the discharge of the lead acid battery as
H₂SO₄ → H⁺ + HSO₄⁻
Pb (s) + HSO⁻₄ → PbSO₄ + H⁺ + 2e⁻
at the cathode we have
PbO₂ + 3H⁺ + HSO⁻₄ + 2e⁻ → PbSO₄ + 2H₂O
Summing the two equation or the total equation for discharge is
Pb (s) + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O
From the above one mole of lead and one mole of PbO₂ are consumed simultaneously hence
Number of moles of lead contained in 405 g of Pb with molar mass = 207.2 g/mole = (405 g)/ (207.2 g/mole) = 1.95 mole of Pb
Hence number of moles of PbO₂ reduced at the cathode = 1.95 mole
mass of PbO₂ reduced at the cathode = (number of moles)×(molar mass)
= 1.95 mole × 239.2 g/mol = 467.55 g of Lead (IV) Oxide is reduced at the cathode
Part B
Each mole of Pb transfers 2e⁻ or 2 electrons, therefore 1.95 moles of Pb will transfer 2 × 1.95 = 3.909 moles of electrons transferred
Each electron carries a charge equal to -1.602 × 10⁻¹⁹ C or one mole of electrons carry a charge equal to 96,485.33 coulombs
hence 3.909 moles carries a charge = 3.909 × 96,485.33 coulombs =377186.86 Coulombs of electrical charge
or transferred electrical charge = 377186.86 C or 3.909 Faraday
