### Video Transcript

The reduction of aluminum ions is described by the ionic half equation Al³⁺ liquid plus 3e⁻ react to form Al solid. What charge must be delivered to reduce 200 kilograms of aluminum ions to aluminum metal? The charge of a single electron is 1.602 times 10 to the minus 19 coulombs.

The reduction of aluminum ions means the addition of electrons. Three electrons are required to reduce one Al³⁺ ion to aluminum solid. While aluminum ions need not be liquid in order to be reduced, it is most commonly the case in industry. What the question is asking is, how much charge needs to be delivered to 200 kilograms of aluminum ions to turn them all into aluminum metal?

The charge in question is delivered by electrons. And the charge of an individual electron is given in the question 1.602 times 10 to the minus 19 coulombs. The first thing we need to do to answer this question is calculate the number of moles of Al³⁺ in 200 kilograms. Then we need to calculate the number of moles of electrons required to reduce the aluminum ions. Finally, we need to work out the total charge of all those electrons.

To work out the number of moles of aluminum 3+, we need to know the atomic mass of aluminum. If you look at this up in our periodic table, you should see the atomic mass of aluminum is 26.982𝑢. One 𝑢 is equivalent to one gram per mole. So the molar mass of aluminum is equal to 26.982 grams per mole. Aluminum 3+ does not weigh exactly the same as an atom of aluminum. But electrons have a very small mass. So the molar mass of aluminum 3+ is approximately 26.982 grams per mole. The number of moles of Al³⁺ is equal to the mass we have divided by its molar mass.

The units for molar mass are grams per mole. But we’ve been given the mass of aluminum ions in kilograms. There are 1000 grams in a kilogram. So we can multiply 200 kilograms by 1000 grams per kilogram which is equal to 200000 grams. Another way of thinking about this is remembering that kilo means 10 to the power of three or 1000. So 200 kilograms is equal to 200 times 1000 grams. Either way, we get the value 200000 grams. Now, we can substitute this value into our equation. A number of moles of Al³⁺ is equal to 200000 grams divided by 26.982 grams per mole. This is equal to 7412.35 moles of Al³⁺. Let’s save that value and move on to the next step, calculating the number of moles of electrons.

In the equation, it takes three electrons to reduce each Al³⁺ ion. Since the ratio of Al³⁺ to e⁻ is one to three, we need to multiply the number of moles of Al³⁺ by three to get the number of moles of electrons. So 7412.35 moles of Al³⁺ multiplied by three moles of electrons per mole of Al³⁺ is equal to 22237.0 moles of electrons. Here, we’re just cancelling the unit moles of Al³⁺, so that we end up with the proper units of moles of electrons. Now, we can store away the answer from part two and move on to part three calculating the charge of the electrons.

The question has provided the charge of a single electron. So if we knew the number of electrons, we could multiply them together to get the charge of the electrons. However, we only have the number of electrons in moles. Avogadro’s constant, given the symbol N A, tells us how many of something there are in a mole. It’s equal to 6.022 times 10 to the 23 per mole. So we can calculate the number of electrons by taking the number of moles of electrons 22237.0 and multiplying it by Avogadro’s constant 6.022 times 10 to the 23 per mole. This gives us, in standard form, 1.33911 times 10 to the 28 electrons.

So the combined charge of the electrons is equal to the number electrons 1.33911 times 10 to the 28 multiplied by the individual charge of an electron 1.602 times 10 to the minus 19 coulombs. This is equal to 2.14526 times 10 to the nine coulombs. But we’re not quite finished. We need to round our answer. The value in our question with the smallest number of significant figures is 200, meaning three significant figures. So we should round our answer to the same precision. That means our answer is 2.15 times 10 to the nine coulombs which can otherwise be written as 2.15 gigacoulombs.

Another way of doing this question would have been to use Faraday’s constant which tells you how many coulombs there are per mole of electrons. If you multiply the number of moles of electrons by Faraday’s constant, you would have got the same answer. But that’s hardly surprising because Faraday’s constant is equal to Avogadro’s constant multiplied by the electron charge. The way I did it is just as valid because I multiplied through by the Avogadro’s constant first and then by the charge of a single electron. But doing it with Faraday’s constant would have been a little quicker. Whichever way you want to do it, the value should be 2.15 times 10 to the nine coulombs.