Respuesta :
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.
[tex]Q = I \times t[/tex]
where,
Q = charge
I = current
t = time
Substitute the values into above formula as follows.
[tex]Q = I \times t\\= 62.0 A \times 23.0 sec\\= 1426 C[/tex]
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.
[tex]Moles = \frac{1426 C}{96500 C/mol}\\= 0.0147 mol[/tex]
The oxidation state of Pb in [tex]PbSO_{4}[/tex] is 2. So, moles deposited by Pb is as follows.
[tex]Moles of Pb = \frac{0.0147}{2}\\= 0.00735 mol[/tex]
It is known that molar mass of lead (Pb) is 207.2 g/mol. Now, mass of lead is calculated as follows.
[tex]No. of moles = \frac{mass}{molar mass}\\ 0.00735 = \frac{mass}{207.2 g/mol}\\mass = 1.523 g[/tex]
Thus, we can conclude that the mass of lead deposited on the cathode of the battery is 1.523 g.
