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 find the number of coulombs of charge required for the electrolysis of water, we need to use Faraday's law of electrolysis, which states that the amount of substance produced or consumed during electrolysis is directly proportional to the amount of charge passed through the electrolyte.
The equation relating the amount of substance produced/moles (n) to the charge (Q) passed through the electrolyte is given by:
n = Q / (zF)
where:
- n is the amount of substance produced or consumed in moles,
- Q is the charge passed through the electrolyte in coulombs,
- z is the number of moles of electrons exchanged per mole of substance, and
- F is the Faraday constant (96485 C/mol).
In this case, we are provided with the volume of O2 produced (30.3 liters), which we need to convert to moles using the ideal gas law.
The ideal gas law equation is given by:
PV = nRT
where:
- P is the pressure in atm,
- V is the volume in liters,
- n is the amount of substance in moles,
- R is the ideal gas constant (0.0821 L.atm/mol.K),
- T is the temperature in Kelvin.
From the given information:
- P = 1 atm
- V = 30.3 liters
- T = 273 K
Using the ideal gas law, we can solve for the number of moles of O2 produced (n):
n = (P * V) / (R * T)
Let's substitute the values into the equation:
n = (1 atm * 30.3 L) / (0.0821 L.atm/mol.K * 273 K)
≈ 1.1646 moles
Now that we have the number of moles of O2 produced (n), we can calculate the charge (Q) using Faraday's law:
Q = n * z * F
The number of moles of electrons exchanged per mole of O2 is 4, as oxygen gas is produced by the combination of 4 electrons and 4 H+ ions during the electrolysis of water.
Substituting the values:
Q = 1.1646 moles * 4 moles * 96485 C/mol
≈ 447,241.256 C
Therefore, approximately 447,241.256 coulombs of charge were required for the electrolysis of water to produce 30.3 liters of O2 at 273K and 1 atm.