Calculate the volume of H2 gas (25 Celsius and 760 mm Hg) that will collect at the cathode when an aqueous solution of Na2SO4 is electrolytes for 90 minutes with a 12amp current.
Coulombs = amp x seconds = 12 x 90 x 60 = approx 65,000
2H^+ + 2e ==> H2
96,500 coulombs will release 1/2 mol H2.
Now PV = nRT =? You have P = 1 atmosphere.
V = ?
n = 1/2
R you know
T in kelvin.
Solve for V in liters.
Post your work if you get stuck.
Hey ryan thats what I did but I thought it was right? I'll show you tomorrow in lab.
Student
Well, let's break it down.
First, we need to find the number of moles of hydrogen gas (H2) produced. To do that, we can use the formula Q = It, where Q is the total charge, I is the current, and t is the time.
Given that the current (I) is 12 amps and the time (t) is 90 minutes, we need to convert the time to seconds to match the units of the current. There are 60 seconds in a minute, so 90 minutes would be 90 * 60 = 5400 seconds.
Now, let's calculate the total charge (Q). Since one mole of electrons carries a charge of 96485 coulombs, we can use the equation Q = It, where I is the current and t is the time.
Q = (12 amps) * (5400 seconds)
Q = 64800 coulombs
Now that we have the total charge, we can find the number of moles of hydrogen gas using Faraday's Law. According to the equation 2 moles of electrons produce 1 mole of hydrogen gas, we have:
Number of moles of H2 = Q / (2 * 96485)
Number of moles of H2 = 64800 / (2 * 96485)
Number of moles of H2 ≈ 0.335 moles
Finally, we can use the ideal gas law to find the volume of hydrogen gas at a given temperature and pressure. However, since we don't know the volume, we can't calculate it precisely.
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To calculate the volume of H2 gas collected at the cathode during electrolysis, we need to take into account the stoichiometry of the reaction and Faraday's law of electrolysis.
The electrolysis of water can be represented by the following equation:
2H2O(l) -> 2H2(g) + O2(g)
According to Faraday's law, the amount of substance (in moles) produced at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the cell. One Faraday (F) of electricity is equivalent to 96,485 C (coulombs).
First, let's calculate the number of moles of electrons passed through the cell:
Q = I * t
Where:
Q = quantity of electricity passed (C)
I = current (A)
t = time (s)
Given that the current is 12 A and the time is 90 minutes (90 min = 5400 s), we can calculate Q:
Q = 12 A * 5400 s
Q = 64,800 C
Next, let's calculate the number of moles of H2 gas produced at the cathode. From the balanced equation, we know that 2 moles of electrons are required to produce 1 mole of H2 gas.
Moles of electrons = Q / (n * F)
Where:
n = number of electrons in the balanced equation (2 for H2 gas)
F = Faraday's constant (96,485 C/mol)
Moles of electrons = 64,800 C / (2 * 96485 C/mol)
Moles of electrons = 0.3356 mol
Since the balanced equation tells us that 2 moles of H2 gas are produced per 2 moles of electrons, the number of moles of H2 gas produced is also 0.3356 mol.
Finally, let's calculate the volume of H2 gas using the ideal gas law:
PV = nRT
Where:
P = pressure (in this case, 760 mm Hg)
V = volume (to be determined)
n = number of moles of H2 gas (0.3356 mol)
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (25°C = 273 + 25 = 298 K)
We can rearrange the equation to solve for V:
V = (nRT) / P
V = (0.3356 mol * 0.0821 L·atm/(mol·K) * 298 K) / (760 mm Hg * 1 atm/760 mm Hg)
V ≈ 0.0358 L
Therefore, the volume of H2 gas collected at the cathode during electrolysis is approximately 0.0358 liters.