A solution of H2SO4 was electrolyzed using the inert platinum electrodes.
A) Write the balanced half-reactions for the anode and cathode in this cell.
B) How many coulombs passed through the cell in 90 minutes at 8.0 amperes?
C) How many moles of electrons pass through the cell while the current flows?
D) What number of moles of gas was produced at the cathode after the cell operated for 90 minutes at 8.0 amperes?
E) What total volume of gas, at STP could be produced by this cell if it operated for 100 minutes at 20.0 amperes?
A) The balanced half-reactions for the anode and cathode in this cell can be determined as follows:
At the anode (oxidation):
2H2O(l) -> O2(g) + 4H+(aq) + 4e-
At the cathode (reduction):
H+(aq) + 4e- -> H2(g)
B) To calculate the total charge passed through the cell in coulombs, you can use the formula:
Charge (in coulombs) = Current (in amperes) x Time (in seconds)
Converting 90 minutes to seconds:
90 minutes x 60 seconds/minute = 5400 seconds
Since the current is 8.0 amperes, we can calculate the charge passed through the cell:
Charge = 8.0 A x 5400 s = 43200 Coulombs
Therefore, 43200 Coulombs passed through the cell in 90 minutes at 8.0 amperes.
C) The number of moles of electrons passing through the cell can be determined using Faraday's Law:
1 mole of electrons = 1 Faraday (F) = 96485 Coulombs
To calculate the moles of electrons passing through the cell, divide the total charge passed by the value of 1 Faraday:
Moles of electrons = Charge (in Coulombs) / 96485 Coulombs
Using the value obtained from part B (43200 Coulombs), we can calculate the moles of electrons:
Moles of electrons = 43200 C / 96485 C/mol = 0.4474 moles
Therefore, approximately 0.4474 moles of electrons pass through the cell while the current flows.
D) The number of moles of gas produced at the cathode can be determined using the stoichiometry of the balanced half-reaction at the cathode:
From the balanced half-reaction:
H+(aq) + 4e- -> H2(g)
The stoichiometric ratio is 1:1, meaning that for every mole of H+(aq) reduced at the cathode, 1 mole of H2(g) is produced.
Since 2H+ ions are required to produce 1 mole of H2 gas, the number of moles of gas produced is half the moles of electrons, as determined in part C.
Number of moles of gas = Moles of electrons / 2 = 0.4474 moles / 2 = 0.2237 moles
Therefore, approximately 0.2237 moles of gas were produced at the cathode after the cell operated for 90 minutes at 8.0 amperes.
E) To calculate the total volume of gas produced at STP (Standard Temperature and Pressure), we need to use the Ideal Gas Law which states:
PV = nRT
Where:
P = Pressure (at STP, 1 atm)
V = Volume (unknown, we need to determine)
n = moles of gas (to be calculated)
R = Ideal Gas Constant (0.0821 L·atm/mol·K)
T = Temperature (at STP, 273 K)
Using the value of 0.2237 moles from part D, and substituting the values into the equation, we get:
(1 atm)(V) = (0.2237 moles)(0.0821 L·atm/mol·K)(273 K)
Solving for V:
V = (0.2237 moles)(0.0821 L·atm/mol·K)(273 K) / (1 atm)
V = 4.63 L
Therefore, the total volume of gas, at STP, that could be produced by this cell if it operated for 100 minutes at 20.0 amperes is 4.63 liters.
A) To write the balanced half-reactions at the anode and cathode, let's first identify the species involved and their oxidation states.
At the cathode:
- The reduction half-reaction occurs, meaning that reduction is happening at the cathode.
- So, we look for species that can be reduced in the solution, H2SO4.
- In H2SO4, hydrogen atoms (H+) are at their lowest oxidation state (+1).
- Therefore, reduction of hydrogen ions to hydrogen gas (H2) is the most likely half-reaction at the cathode.
H+ (aq) + e- -> 1/2 H2 (g)
At the anode:
- The oxidation half-reaction occurs, meaning that oxidation is happening at the anode.
- In this case, we have sulfuric acid (H2SO4) in the solution.
- The element sulfur (S) is at its highest oxidation state (+6) in H2SO4.
- Therefore, oxidation of sulfur to sulfuric acid is the most likely half-reaction at the anode.
1/2 S (s) -> SO4^2- (aq) + 2e-
B) To determine the number of coulombs passing through the cell in 90 minutes at 8.0 amperes, we can use the formula:
Charge (coulombs) = Current (amperes) x Time (seconds)
First, we need to convert the given time of 90 minutes to seconds:
90 minutes x 60 seconds/minute = 5400 seconds
Now, we can calculate the charge:
Charge = 8.0 amperes x 5400 seconds = 43200 coulombs
Therefore, 43200 coulombs passed through the cell.
C) The number of moles of electrons passing through the cell while the current flows can be calculated using Faraday's Law. Faraday's constant (F) is the charge of one mole of electrons, which is 96485 coulombs per mole.
Number of moles of electrons = Charge (coulombs) / Faraday's constant
Using the charge value from part B:
Number of moles of electrons = 43200 coulombs / 96485 coulombs per mole
D) To determine the number of moles of gas produced at the cathode after the cell operated for 90 minutes at 8.0 amperes, we need to look at the balanced half-reaction at the cathode:
H+ (aq) + e- -> 1/2 H2 (g)
The stoichiometry indicates that for every 2 electrons, 1 mole of hydrogen gas is produced.
So, the number of moles of gas produced = (Number of moles of electrons from part C) / 2
E) To calculate the total volume of gas, at STP, that could be produced by this cell operating for 100 minutes at 20.0 amperes, we need to consider the balanced half-reaction at the cathode and standard conditions.
Assuming STP is 1 atmosphere pressure and 273.15 K temperature, we can use the ideal gas law, PV = nRT, to calculate the volume of hydrogen gas produced.
First, we need to calculate the number of moles of hydrogen gas produced using the same approach as part D.
Once we have the number of moles of hydrogen gas, we can use the ideal gas law:
V = (nRT) / P
where:
n = number of moles of hydrogen gas
R = gas constant (0.0821 L•atm/(mol•K))
T = temperature (in Kelvin)
P = pressure (1 atm)
Substituting the appropriate values, we can calculate the volume of gas produced.