Answer:
The mass is [tex]mass = 3.813 g[/tex]
Explanation:
From the question we are told that
The current supplied is [tex]I = 96.06 A[/tex]
The time taken is t = 37.0 sec
Generally the charge been deposited on the cathode is mathematically represented as
[tex]Q = It[/tex]
Substituting values
[tex]Q = 96.0 * 37[/tex]
[tex]= 3552C[/tex]
Generally 1 mole of the lead deposited contains [tex]96500C[/tex]
Then [tex]3552C[/tex] would contain how many moles
Not let say x is the number of moles deposited then
Using this mathematical relation
[tex]1 mole ---------> 96500C \\x mole ----------> 3552C[/tex]
[tex]x = \frac{3552}{96500}[/tex]
[tex]= 0.0368 \ moles[/tex]
From the the question we are told that lead is been oxidized to lead 2 oxide which implies that the two electron would be accepted by the lead cation
Hence the number of moles that would be deposited at the cathode is
[tex]n = \frac{x}{2} = \frac{0.036808}{2} = 0.018404 \ moles[/tex]
Now the molar mass of lead is constant with a value [tex]M= 207.2 g/mol[/tex]
Generally mass is mathematically represented as
[tex]mass = n * M[/tex]
For lead
Substituting values
[tex]mass =0.018404 * 207.2[/tex]
[tex]mass = 3.813 g[/tex]
