Suppose a car's motor draws 231 Ampere of current for 6.3 sec to start the car. What mass of Pb is oxidized in the lead-acid battery to supply this much electricity?

231 A x 6.3 sec = about 1,500 Coulombs but you need to do it more accurately.

96,485 coulombs will oxidize 1/2 mol Pb which is 207.2/2 = 103.6 g Pb. The amount of Pb deposited will be

103.6 x (1,500/95,485) = ? g Pb.

Thanks!

To determine the mass of Pb (lead) oxidized in the lead-acid battery, we need to use Faraday's law to calculate the amount of charge passed through the circuit.

Faraday's law states that the amount of charge (Q) passing through an electrolytic cell is directly proportional to the amount of substance transformed at the electrode.

The equation for calculating the charge passed (Q) is:

Q = I * t

Where:
Q = charge passed (in Coulombs)
I = current (in Amperes)
t = time (in seconds)

In this case, the current (I) is 231 Amperes, and the time (t) is 6.3 seconds. Thus, we can calculate the charge (Q) as follows:

Q = 231 A * 6.3 s
Q = 1452.3 Coulombs

Since one mole of electrons represents a charge of approximately 96,500 Coulombs (Faraday's constant), we can calculate the number of moles of Pb oxidized as follows:

moles of Pb = Q / Faraday's constant

moles of Pb = 1452.3 C / 96500 C/mol
moles of Pb ≈ 0.015 moles

To determine the mass, we need to know the molar mass of Pb. The molar mass of lead (Pb) is approximately 207.2 grams/mol.

mass of Pb = moles of Pb * molar mass of Pb
mass of Pb = 0.015 mol * 207.2 g/mol
mass of Pb ≈ 3.1 grams

Therefore, approximately 3.1 grams of lead (Pb) would be oxidized in the lead-acid battery to supply the necessary electricity for the car's motor to start.