During a period of discharge of a lead-acid battery, 405 g of Pb from the anode is converted into PbSO4 (s).

What mass of PbO2 (s) is reduced at the cathode during this same period?

and

How many coulombs of electrical charge are transferred from Pb to PbO2?

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