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, we need to use Faraday's laws of electrolysis.
Faraday's laws state that the amount of substance produced or consumed during electrolysis is proportional to the amount of charge passed through the electrolytic cell.
To calculate the number of coulombs, we need to know the number of moles of O2 produced during the electrolysis of water.
To find the number of moles, we can use the ideal gas law:
PV = nRT
Where:
P = pressure (1 atm)
V = volume (30.3 liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (273 K)
Substituting the given values into the equation, we have:
(1 atm) * (30.3 liters) = n * (0.0821 L·atm/(mol·K)) * (273 K)
Simplifying the equation, we find that the number of moles of O2 produced is approximately 1.26 moles.
Next, we need to determine the number of electrons transferred during the electrolysis of water. The balanced reaction for the electrolysis of water is:
2 H2O(l) -> 2 H2(g) + O2(g)
From the balanced equation, we can see that for each O2 molecule produced, 4 electrons are transferred.
Therefore, for 1 mole of O2 produced, 4 moles of electrons are required.
Multiplying the number of moles of O2 produced (1.26 moles) by the number of moles of electrons per mole of O2 (4 moles), we find that approximately 5.04 moles of electrons are required.
Finally, to convert moles of electrons to coulombs, we can use Faraday's constant, which is defined as the charge of one mole of electrons (approximately 96,485 C/mol).
Multiplying the number of moles of electrons (5.04 moles) by Faraday's constant (96,485 C/mol), we find that approximately 485,554 coulombs of charge were required for the electrolysis of water to produce 30.3 liters of O2 at 273K and 1 atm.