### Video Transcript

The industrial extraction of
aluminum using Hall–Héroult cells requires a very large electrical current of 120
kiloamps. How much aluminum is produced per
hour, taking the molar mass of aluminum to be 27.0 grams per mole and one faraday of
charge as 9.65 times 10 to the four coulombs? Give your answer in kilograms to
two decimal places.

The extraction of aluminum is an
electrolytic process. Aluminum ions in the molten phase
are reduced to liquid aluminum metal. For every one mole of aluminum
ions, three moles of electrons are needed, and this produces one mole of aluminum
metal. We are asked how much aluminum is
produced per hour. In other words, what mass of
aluminum is produced?

We are given the molar mass of
aluminum. We can see that we first need to
calculate the number of moles of aluminum produced. And using this value and the molar
mass, we can then determine the mass. We are not given the number of
moles of aluminum, but we are given other data. We are given the current that flows
in kiloamps and the time that has elapsed, which is one hour. We are also given the value of one
faraday of charge.

A useful equation in electrolysis
is 𝑄 equals 𝐼𝑡, where 𝑄 is the charge transferred in coulombs, 𝐼 the current in
amperes, and 𝑡 the time in seconds. We need to convert the current to
amperes and the time to seconds. Kiloamps can be converted to amps
by multiplying by this conversion factor. Kiloamps cancel, and we get 120,000
amperes. Then, we can convert time in hours
to minutes using this conversion factor and minutes to seconds using this conversion
factor. Hours cancel, minutes cancel, and
we get an answer of 3,600 seconds, which is equivalent to one hour.

Now we can work out the charge
transferred. Using our key equation 𝑄 equals
𝐼𝑡, we can substitute in the current in amperes and the time in seconds. Solving, we get the charge
transferred in one hour, which is 432,000,000 coulombs. The next step is to convert the
amount of charge transferred to a number of moles of electrons transferred. There are various ways to do
this. One way is to use the key equation
𝑄 equals 𝑛𝐹, where 𝑄 is the charge transferred in coulombs, 𝑛 the number of
moles of electrons transferred, and 𝐹 the Faraday constant.

Now, we are given one faraday of
charge as 9.65 times 10 to the four coulombs. The Faraday constant, however, is
this same value of charge in coulombs but per one mole of electrons transferred. The Faraday constant is, therefore,
the charge on one mole of elementary charge. So we can rearrange our key
equation to give 𝑛 is equal to 𝑄 divided by 𝐹. Substituting the amount of charge
that flowed in one hour divided by the charge on one mole of elementary charge,
which is equivalent to the charge on one mole of electrons, we can cancel the
coulomb units. We get the number of moles of
charge transferred in an hour, which is 4,476.68 moles.

Let’s clear some space. The next step is to convert the
moles of charge transferred to the moles of aluminum produced. From the balanced half equation, we
know that three moles of charge produces one mole of aluminum. But we don’t have three moles of
electrons or charge. We have 4,476.68 moles of
charge. And this will produce 𝑥 moles of
aluminum. We can then solve this ratio by
taking the moles of charge that we have dividing it by three. And we can solve for the moles of
aluminum produced, 1,492.23 moles, which is this value here.

Finally, we can use the key
equation number of moles is equal to mass divided by molar mass to determine the
mass of aluminum produced. We can rearrange the key equation
and multiply moles by molar mass, substituting in the mole value that we have just
calculated, 1,492.23 moles, multiplied by the given molar mass of 27.0 grams per
mole. We calculate an answer of 40,290.21
grams of aluminum produced in one hour in this Hall–Héroult cell.

However, we were asked to give an
answer in kilograms. So we can multiply the mass of
aluminum in grams by this conversion factor. Grams cancel out, giving an answer
in kilograms, which we must remember to round off to two decimal places, which is
then 40.29 kilograms of aluminum produced in one hour.