If 30.3 liters of O2 at 273K and 1 atm are produced by the electrolysis of water, how many coulombs of charge were required?
To determine the number of coulombs of charge required for the electrolysis of water to produce 30.3 liters of O2 at 273K and 1 atm, we first need to understand the relationship between the amount of substance and charge.
The concept of Faraday's law can help us find the answer. Faraday's law states that the amount of chemical change in an electrolysis reaction is directly proportional to the quantity of electricity that passes through the cell. The proportionality constant is known as the Faraday constant (F) and is equal to 96485 coulombs per mole of electrons.
Now, let's calculate the number of moles of O2 produced using the ideal gas law equation:
PV = nRT
Where:
P = Pressure = 1 atm
V = Volume = 30.3 liters
n = Number of moles (to be determined)
R = Ideal gas constant = 0.0821 L·atm/(mol·K)
T = Temperature = 273K
Rearranging the equation to solve for n:
n = PV / RT
Substituting the given values:
n = (1 atm) * (30.3 liters) / (0.0821 L·atm/(mol·K) * 273K)
Simplifying the calculation:
n = 1.131 moles of O2
Since the electrolysis of water involves the production of O2 by the reduction of water molecules, we can conclude that for each mole of O2 produced, four moles of electrons are required.
Therefore, the total number of moles of electrons required for the given amount of O2 can be calculated as:
moles of electrons = 1.131 moles of O2 * 4 moles of electrons/mole of O2
moles of electrons = 4.524 moles of electrons
Finally, we can calculate the number of coulombs of charge required using Faraday's law:
coulombs of charge = moles of electrons * Faraday constant
coulombs of charge = 4.524 moles of electrons * 96485 coulombs/mole of electrons
coulombs of charge ≈ 436,594 coulombs
Therefore, approximately 436,594 coulombs of charge were required for the electrolysis of water to produce 30.3 liters of O2 at 273K and 1 atm.