A current of 3.05 A is passed through a Pb(NO3)2 solution. How long, in hours, would this current have to be applied to plate out 9.10 g of lead?

Answer 1

Pb²⁺(aq) + 2e- ==> Pb(s)

9.10 g Pb x 1 mol / 207 g = 0.0440 mols Pb

moles of electrons = 0.0440 mols Pb x 2 mols e- / mol Pb = 0.088 mols electrons

Coulombs = 0.088 mols e- x 96485 C / mol = 8491 C

3.05 amps = 3.05 C / sec, so...

8491 C x 1 sec / 3.05 amp = 2784 seconds

Converting to hours ..

2784 sec x 1 min / 60 sec x 1 hour / 60 min = 0.773 hours

Answer 2

We can use Faraday's laws of electrolysis to solve this problem. Faraday's first law states that the amount of substance produced at an electrode is directly proportional to the amount of electricity passed through the electrolyte. The constant of proportionality is called the Faraday constant (F), which is equal to 96,485 C/mol of electrons.

We can use this law to find the amount of electrons (n) required to plate out 9.10 g of lead. The molar mass of lead is 207.2 g/mol, so the number of moles of lead is:

n = (9.10 g)/(207.2 g/mol) = 0.044 moles

The number of electrons required to reduce each Pb2+ ion to Pb is 2, so the total number of electrons required is:

q = 2n = 0.088 C

The current is given as 3.05 A, so the time required to pass this amount of charge is:

t = q/I = 0.088 C / 3.05 A = 0.0289 hours

Therefore, the current would have to be applied for approximately 0.0289 hours, or 1.73 minutes, to plate out 9.10 g of lead.