What volume of H2(g) and O2(g) is produced by electrolyzing water at a current of 4.00 A for 12.0 minutes (assuming ideal conditions)?
Volume of Hydrogen gas is 0.3343 dm3 and that of oxygen gas is 0.167.15 dm3
Refer to the problem just below that I worked for you (the Cu one).
To determine the volume of H2(g) and O2(g) produced by electrolyzing water, we need to use two key concepts: Faraday's Law of Electrolysis and the Ideal Gas Law.
1. Faraday's Law of Electrolysis states that the amount of substance produced or consumed during electrolysis is directly proportional to the quantity of electric charge passed through the electrolytic cell. The equation can be expressed as:
moles of substance = (electric charge passed) / (Faraday's constant)
In this case, we want to find the moles of H2(g) and O2(g) produced.
2. The Ideal Gas Law relates the volume of a gas to the number of moles, temperature, and pressure. The equation can be expressed as:
V = (n * R * T) / P
where:
V is the volume of gas,
n is the number of moles of gas,
R is the ideal gas constant (0.0821 L·atm/mol·K),
T is the temperature in Kelvin, and
P is the pressure.
Now, let's solve the problem step by step:
Step 1: Calculate the electric charge passed through the electrolytic cell.
Given: Current = 4.00 A and Time = 12.0 minutes
We need to convert the time from minutes to seconds:
Time = 12.0 minutes * 60 seconds/minute = 720 seconds
Electric charge (Q) = Current * Time
Q = 4.00 A * 720 s = 2880 C
Step 2: Calculate the moles of H2(g) and O2(g) produced using Faraday's Law of Electrolysis.
Since the electrolysis of water produces H2 and O2 in a 2:1 molar ratio, for every mole of H2 produced, 0.5 moles of O2 are produced.
Moles of H2(g) = (Electric charge passed) / (Faraday's constant)
Faraday's constant = 96,485 C/mol (charge required to deposit one mole of electrons)
Moles of H2(g) = 2880 C / (96,485 C/mol) = 0.0298 mol
Moles of O2(g) = 0.5 * Moles of H2(g) = 0.5 * 0.0298 mol = 0.0149 mol
Step 3: Calculate the volume of H2(g) and O2(g) using the Ideal Gas Law.
Given: Temperature (assumed to be at standard conditions) = 273.15 K (0 °C) and Pressure = 1 atm
V = (n * R * T) / P
For H2(g):
V(H2) = (0.0298 mol * 0.0821 L·atm/mol·K * 273.15 K) / 1 atm = 0.661 L
For O2(g):
V(O2) = (0.0149 mol * 0.0821 L·atm/mol·K * 273.15 K) / 1 atm = 0.330 L
Therefore, the volume of H2(g) produced is approximately 0.661 L and the volume of O2(g) produced is approximately 0.330 L.