A dilute aqueous solution of Na2SO4 is electrolyzed between Pt electrodes for 3.70 h with a current of 2.77 A .
Part A
What volume of gas, saturated with water vapor at 25∘C and at a total pressure of 742 mmHg , would be collected at the anode? The vapor pressure of water at 25∘C is 23.8 mmHg.
My answer is Incorrect. Anyone know what went wrong?
1 Coulomb = 1 amp / second, so 2.77 A = 37395 Coulombs
total = O2 pressure + Water vapor pressure
O2 Pressure = 742 - 23.8 = 718.2 mm Hg
Mole fraction of water vapor = 23.8/742 = 0.0321
V=nRT/P = (0.1)(0.082057)(298.15K)/(742/760) = 2.39 L
Thanks a lot;)
Get number or moles of electrons yung Q=It=nF then get number of moles of SO2. Use PV=nRT to get volume.
Your approach to calculating the volume of gas collected at the anode seems correct. However, it looks like there is a small error in your calculation of the mole fraction of water vapor.
To calculate the mole fraction of water vapor, you need to consider both the water vapor pressure (23.8 mmHg) and the total pressure (742 mmHg). The mole fraction of water vapor is given by:
mole fraction of water vapor = water vapor pressure / total pressure
mole fraction of water vapor = 23.8 / 742 = 0.0320
Note that I rounded the value to four decimal places based on the given significant figures. This slight change in the mole fraction value may lead to a different final answer for the volume of gas collected at the anode.
Your calculations are correct up until the calculation of the mole fraction of water vapor (0.0321). However, the calculation for the volume of gas collected at the anode seems to be incorrect.
To find the volume of gas collected at the anode, you need to use the ideal gas law equation:
PV = nRT
Where:
P = total pressure (742 mmHg)
V = volume of gas collected
n = number of moles of gas
R = ideal gas constant (0.0821 L atm/mol K)
T = temperature (25 °C = 298.15 K)
First, let's calculate the number of moles of water vapor initially present in the system using the mole fraction:
moles of water vapor = mole fraction of water vapor * total moles of gas
moles of water vapor = 0.0321 * 0.1 moles (assuming 0.1 moles of gas were produced during electrolysis, as you did not provide this information explicitly)
Next, let's calculate the number of moles of oxygen gas produced during electrolysis. Since the reaction is 4e- + 4H2O -> 2H2 + 4OH-, for every 1 mole of gas produced, 4 moles of electrons are transferred. So, the number of moles of oxygen gas produced is:
moles of oxygen gas = (number of moles of electrons transferred during electrolysis) / 4
Given that 1 Coulomb = 1 amp / second and 2.77 A were used for 3.70 h, the number of moles of oxygen gas produced is:
moles of oxygen gas = (2.77 A * 3.70 h * 3600 s/h) / (1 C/e^-) / 4 = 12.11 moles (approximately)
Now, let's calculate the total number of moles of gas:
total moles of gas = moles of water vapor + moles of oxygen gas
total moles of gas = 0.0321 moles + 12.11 moles = 12.1421 moles (approximately)
Finally, we can substitute these values into the ideal gas law equation to find the volume of gas collected:
V = (nRT) / P
V = (12.1421 moles * 0.0821 L atm/mol K * 298.15 K) / 742 mmHg
V = 2.05 L (approximately)
Therefore, the corrected volume of gas collected at the anode should be approximately 2.05 L, not 2.39 L as you originally calculated.
Hope this explanation helps! Let me know if you have any further questions.
To start the coulombs is not right.
2.77 A x 1 hr x (60 min/hr) x (60 sec/min) = ? and I quit reading there.