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 62.0 is fed into a car battery for 23.0 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 1

Answer: The mass of lead deposited on the cathode of the battery is 1.523 g.

Explanation:

Given: Current = 62.0 A

Time = 23.0 sec

Formula used to calculate charge is as follows.

$Q = I \times t$

where,

Q = charge

I = current

t = time

Substitute the values into above formula as follows.

$Q = I \times t\\= 62.0 A \times 23.0 sec\\= 1426 C$

It is known that 1 mole of a substance tends to deposit a charge of 96500 C. Therefore, number of moles obtained by 1426 C of charge is as follows.

$Moles = \frac{1426 C}{96500 C/mol}\\= 0.0147 mol$

The oxidation state of Pb in $PbSO_{4}$ is 2. So, moles deposited by Pb is as follows.

$Moles of Pb = \frac{0.0147}{2}\\= 0.00735 mol$

It is known that molar mass of lead (Pb) is 207.2 g/mol. Now, mass of lead is calculated as follows.

$No. of moles = \frac{mass}{molar mass}\\ 0.00735 = \frac{mass}{207.2 g/mol}\\mass = 1.523 g$

Thus, we can conclude that the mass of lead deposited on the cathode of the battery is 1.523 g.