Respuesta :
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
Explanation:
To calculate the charge deposited, we use the equation:
[tex]\text{Charge}=\text{Current}\times \text{Time (in seconds)}[/tex]
We are given:
Current supplied = 96.0 A
Time = 37.0 seconds
Putting values in above equation, we get:
[tex]\text{Charge}=96.0\times 37=3552C[/tex]
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 = [tex]\frac{1}{96500}\times 3552=0.0368moles[/tex] of electrons
The half reaction follows:
At cathode: [tex]Pb(aq.)+2e^-\rightarrow Pb(s)[/tex]
Number of electrons exchanged are 2
So, moles of lead deposited = [tex]\text{Number of moles of electrons}}{\text{Number of electrons}}=\frac{0.0368}{2}=0.0184mol[/tex]
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Molar mass of lead = 207.2 g/mol
Moles of lead = 0.0184 moles
Putting values in above equation, we get:
[tex]0.0184mol=\frac{\text{Mass of lead}}{207.2g/mol}\\\\\text{Mass of lead}=(0.0184mol\times 207.2g/mol)=3.81g[/tex]
Hence, the mass of lead deposited on the cathode of the batter is 3.81 grams
