A current of 3.78 A is passed through a Pb(NO3)2 solution for 1.40 hours. How much lead is plated out of the solution?
To find the amount of lead plated out of the solution, we need to use the concept of Faraday's laws of electrolysis, which states that the amount of substance deposited or liberated during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.
The formula to calculate the amount of substance deposited is:
Amount of substance = (Current × Time) / Faraday's constant
In this case, we need to find the amount of lead plated out of the solution. The chemical formula of lead (II) nitrate is Pb(NO3)2.
First, we need to find the number of moles of electrons passed through the solution. The balanced chemical equation for the electrolysis of Pb(NO3)2 is:
Pb(NO3)2 -> Pb + 2NO3- + 2e-
We can see that 2 moles of electrons are needed to deposit 1 mole of lead from the solution.
The Faraday's constant is the charge of 1 mole of electrons, which is approximately 96,500 Coulombs.
Now, let's calculate the amount of lead plated out:
Amount of substance = (Current × Time) / Faraday's constant
Current = 3.78 A (Amperes)
Time = 1.40 hours = 1.40 × 3600 seconds (convert hours to seconds)
Amount of substance = (3.78 × 1.40 × 3600) / 96500
By plugging in the values into the formula and performing the calculations, we can determine the amount of lead plated out of the solution.
To calculate the amount of lead plated out of the solution, we need to use Faraday's law of electrolysis. According to Faraday's law, the amount of substance produced or consumed in an electrochemical reaction is directly proportional to the quantity of electricity passed through the solution.
The formula to calculate the amount of substance plated or consumed is:
Amount of substance = (Current × Time) / (Faraday's Constant × Number of Electrons)
Given:
Current (I) = 3.78 A
Time (t) = 1.40 hours
Charge on one mole of electrons (Faraday's Constant) = 96,485 C/mol (Coulombs per mole)
Number of electrons transferred for one mole of lead ions = 2
Let's calculate the amount of lead plated out of the solution using the formula:
Amount of substance = (Current × Time) / (Faraday's Constant × Number of Electrons)
Amount of substance = (3.78 A × 1.40 hours) / (96,485 C/mol × 2)
First, we need to convert time from hours to seconds:
Time (t) = 1.40 hours × 3600 seconds/hour = 5040 seconds
Amount of substance = (3.78 A × 5040 s) / (96,485 C/mol × 2)
Now, let's calculate the amount of lead plated out of the solution:
Amount of substance = (3.78 A × 5040 s) / (96,485 C/mol × 2)
= 19,003.2 C / 192,970 C/mol
= 0.0983 mol
Therefore, approximately 0.0983 moles of lead will be plated out of the solution.
coulombs = amperes x seconds
C = 3.78A x 1.4 hr x (60 min/hr) x (60 s/min) = ?C. Let's estimate it at 19,000 C.
96,485 C will deposit 207.2/2 grams Pb = 96,485 C will deposit 103.6 g Pb.
103.6g Pb x (19,000/96,485) = ?g Pb deposited.