An electrolytic cell produces aluminum from Al2O3 at the rate of ten kilograms a day. Assuming a yield of 100%
a) how many moles of electrons must pass through the cell in one day?
b) how many amperes are passing through the cell?
c) how many moles of oxygen (O2) are being produced simultaneously?
I know I can figure out this problem easily, I'm just really struggling with writing the equation. Can anyone help me with this?
Use unit multipliers:
Start with what you want:
MolesElectrons= 3*molesAluminumatoms
= above* avagnumberatoms/moleAl
or
MolesElectrons=3*10kg/26.98 moles/day
then for amperes, change the above to moles electron per second.
Then convert moles of electrons to ampere (you know the charge on one electron, so you then know the charge/moleElectrons)
For oxygen, write the balanced equation, and use the coefficents to solve.
I don't quite understand how to write the balanced equation for oxygen. I know I have Al2O3 that ends up as Al (is it aluminium ion, or just Al?) and O2 gas, but I don't know what to do with electrons and what not.
Would I have something like this:
Al2O3 --> 2Al + 3/2O2
? But this doesn't seem right because no electrons are included in the equation.
I'm wondering the exact problem! I don't do how to write the balanced equation..
To write the balanced equation for the electrolysis of Al2O3, you need to consider the transfer of electrons involved in the process.
The overall reaction can be represented as follows:
2 Al2O3 (s) → 4 Al (s) + 3 O2 (g)
In this equation, Al2O3 is being converted into Al and O2. However, since this is an electrolytic cell, you need to account for the flow of electrons.
During electrolysis, aluminum is reduced at the cathode, where it gains electrons. Simultaneously, oxygen is oxidized at the anode, releasing electrons. The net balanced equation can be written as:
4 Al (s) + 3 O2- (aq) → 4 Al (s) + 3 O2 (g)
Here, the oxide ions (O2-) are supplying the necessary electrons for the reduction of aluminum. By balancing the equation, we can see that 2 moles of Al2O3 produce 4 moles of Al and 3 moles of O2.
To calculate the number of moles of O2 being produced, you already have the rate of aluminum production (10 kg) and assuming 100% yield, you know the molar ratio of Al to O2 is 4 to 3.
Therefore, the moles of O2 produced in one day would be:
MolesO2 = (10 kg Al * (1 mol Al / 26.98 g Al) * (3 mol O2 / 4 mol Al))
= 11.123 mol O2/day
This reflects the number of moles of O2 produced per day in the electrolysis of Al2O3.
The balanced equation for the electrolysis of Al2O3 can be written as follows:
2Al2O3 → 4Al + 3O2
In this equation, aluminum oxide (Al2O3) is being reduced to aluminum (Al) while oxygen gas (O2) is being produced at the positive electrode.
The equation doesn't explicitly include electrons because it is an overall representation of the reaction. Electrons are involved in the half-reactions that occur at each electrode.
At the negative electrode (cathode):
Al3+ + 3e- → Al
At the positive electrode (anode):
2O2- → O2 + 4e-
In these half-reactions, electrons are either gained or lost to form the desired products.
I hope this clarifies the balanced equation for the electrolysis of Al2O3 and how the electrons are involved in the reaction.