The question is incomplete, here is the complete question:
When a lead acid car battery is recharged by the alternator, it acts essentially as an electrolytic cell in which solid lead(II) sulfate PbSO₄ is reduced to lead at the cathode and oxidized to solid lead(II) oxide PbO at the anode.
Suppose a current of 96.0 A is fed into a car battery for 37.0 seconds. Calculate the mass of lead deposited on the cathode of the battery. Round your answer to 3 significant digits. Also, be sure your answer contains a unit symbol.
Answer: The mass of lead deposited on the cathode of the batter is 3.81 grams
Step-by-step explanation:
To calculate the charge deposited, we use the equation:

We are given:
Current supplied = 96.0 A
Time = 37.0 seconds
Putting values in above equation, we get:

We know that:
96500 C of charge deposition is contained in 1 mole of electrons
So, 3552 C of charge deposition will be contained in =
of electrons
The half reaction follows:
At cathode:

Number of electrons exchanged are 2
So, moles of lead deposited =

To calculate the number of moles, we use the equation:
Molar mass of lead = 207.2 g/mol
Moles of lead = 0.0184 moles
Putting values in above equation, we get:

Hence, the mass of lead deposited on the cathode of the batter is 3.81 grams