In this video we will learn how to use the Faraday constant to calculate the mass or volume of substance liberated during electrolysis.

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### Video Transcript

In this video, we will learn what the Faraday constant is and how it can be used to determine the mass of a substance produced or the volume of a substance liberated during electrolysis.

Michael Faraday was an English-born scientist and inventor who was especially interested in chemistry and physics. He had little formal schooling but managed to get a job in the laboratory of an inventor. Over time, he began to be involved in the experiments himself and made major science discoveries and inventions. He discovered the main principles of electromagnetic induction and electrolysis. So great were his contributions to science that a special constant is named after him, the Faraday constant. What is this constant?

In an electrolytic cell, we know that the potential supplied by the battery drives a nonspontaneous chemical reaction. Negatively charged electrons flow out the negative terminal. We can say therefore that positive charge or conventional current 𝐼 flows out the positive terminal in the opposite direction. Positive ions in the electrolyte are attracted to the negatively charged electrode, the cathode. And negatively charged ions are attracted towards the positively charged electrode, the anode. Metal ions tend to plate or deposit onto the cathode as they gain electrons. As a result, the mass of the cathode increases, and this can be measured to determine how much metal has plated on.

Depending on the nonmetal, nonmetals often form gas bubbles at the anode, as they release electrons into this positively charged electrode. Depending on the reaction conditions, for example, the current flowing, and the time that has passed, a certain volume of gas will be liberated. The mass of a metal deposited or the volume of a gas released can be calculated using Faraday’s constant.

Now that we know the usefulness of the Faraday constant, let’s derive it. You may have seen this equation in physics: 𝑄 equals 𝐼𝑡, where 𝑄 is the quantity of charge transferred in a circuit, for example, an electrolytic cell. Charge is measured in coulombs, symbol capital C. 𝐼 is the conventional current that is flowing, measured in amperes, symbol capital A, and 𝑡, the length of time that the current flows for, measured in seconds, lowercase s.

One coulomb is the magnitude of charge of approximately 6.24 times 10 to the 18 electrons. We know that electrons are negatively charged. And in fact their charge is negative 1.602 times 10 to the negative 19 coulombs. And we know that protons are positively charged by the same magnitude. This value is called 𝑒, the elementary charge. We know that, in reality, it is the negatively charged electrons that flow in a circuit. But by convention, current 𝐼 is positive. That is why the quantity of charge transferred in a circuit 𝑄 is equal to the magnitude of charge transferred. So one coulomb is equal to the size or magnitude of charge on approximately 6.24 times 10 to the 18 electrons. We are not looking at the nature of the charge, whether it is positive or negative.

So let’s refine this definition by removing electrons over here. And we will write 𝑒, which is elementary charge or protons. This may all seem like semantics, and in some ways it is. We know it is electrons that flow in a circuit, but we tend to think of the flow of positive charge. And so we have refined our definition to suit our mental model. And so we can define an ampere as the flow of approximately 6.24 times 10 to the 18 elementary charges or electrons per second.

However, in chemistry, it is more convenient to talk about the number of moles of charge transferred. If we take the elementary charge, which is the magnitude only of an electron charge, and multiply it by Avogadro’s constant, symbol 𝐿, sometimes written as symbol capital 𝑁 subscript A, which is 6.022 times 10 to the 23 entities in one mole or per mole, and if we include all the decimals, which I have not shown here, we get a value of 96,485.3321 coulombs per mole, which we can round off to 96,500 coulombs per mole. This value is the Faraday constant, symbol 𝐹. So 𝐹 is equal to the elementary charge multiplied by Avogadro’s constant. 𝐹 is equal to 𝑒𝐿 or 𝐹 equals 𝐿𝑒 as you might more commonly see it. And sometimes we see this as 𝐹 equals 𝑁 subscript A 𝑒. We have finally derived the Faraday constant, but what does this value mean?

Faraday’s constant is the charge per one mole of elementary charge. In other words, for every one mole of elementary charge transferred in an electrolytic circuit, the amount of charge transferred is 96,500 coulombs. Incidentally, this value over here is one Faraday or one 𝐹. We must be careful. Don’t get confused! Capital F written behind a number as a unit is the amount of charge in faradays. This capital 𝐹 here is the Faraday constant whose unit is coulombs per mole. Let’s practice understanding the value of the Faraday constant.

If one mole of potassium ions in an electrolyte are reduced in electrolysis to potassium metal, according to this half equation, we can see that one mole of electrons are required to convert one mole of the ions to one mole of the metal. According to Faraday’s constant, one mole of electrons transferred is equivalent to 96,500 coulombs of charge transferred, which we could write like this.

Here is another example. If two moles of electrons are required to generate one mole of pure copper from one mole of copper two plus ions according to this half equation, then two faradays of charge are needed for two moles of electrons transferred. So we have determined that 193,000 coulombs of charge are needed to produce one mole of pure copper from one mole of copper ions, usually from a molten copper electrolyte.

There is one more handy equation we can derive from this information. 𝑄 equals 𝑛𝐹, where 𝑄 is charge transferred, 𝑛 is the number of moles of charge transferred, and 𝐹 Faraday’s constant. This equation makes sense, since a value in moles multiplied by Faraday’s constant, whose unit is coulombs per mole, would give an answer in coulombs since the moles would cancel.

We now know what the Faraday constant is and how it is derived. And we know that one faraday of charge is transferred in a circuit when one mole of electrons is transferred. But we still need to answer the question, how do we use the Faraday constant to determine the mass of a metal plated or the volume of gas liberated during electrolysis?

Taking the copper chloride electrolytic cell, in this case the electrolyte is aqueous and not molten. With the copper reduction half equation we had earlier, we could calculate the mass of copper deposited as follows. If we know the current flowing and the time that has elapsed, we can determine the quantity of charge transferred during this time using our key equation 𝑄 equals 𝐼𝑡. Then, knowing Faraday’s constant, we can determine the number of moles of electrons transferred using the key equation 𝑄 equals number of moles times Faraday’s constant.

Next, we can take the number of moles of electrons transferred and relate it to the moles of copper produced using the stoichiometric coefficients from the balanced equation. We can then use the number of moles of copper produced with its molar mass to determine the mass of copper deposited during electrolysis. In a similar way, we can work out the volume of chlorine gas liberated. Knowing the stoichiometric coefficients of the chlorine oxidation half reaction and knowing the number of moles of electrons transferred, we can then calculate the number of moles of chlorine gas produced. If the reaction occurs under a set of known standard conditions, we can use the key equation number of moles is equal to volume over molar volume. Then, using the appropriate known molar volume for those conditions, we can determine the volume of chlorine gas liberated. Let’s practice this type of calculation.

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.

Now, let’s summarize what we’ve learnt about the Faraday constant. We learnt the equation 𝑄 equals 𝐼𝑡, where 𝑄 is the charge transferred in coulombs, 𝐼 the current in amperes, and 𝑡 the time in seconds. This equation is useful in electrolysis to determine the charge transferred from a power source to an electrolyte. We learnt about the Faraday constant, symbol 𝐹, which is the amount of charge per mole of elementary charge. And this value is 96,500 coulombs per mole. We know that it is negatively charged electrons which flow in an electrolytic circuit but that by convention we talk about the flow of positive charge or elementary charge. Finally, we learnt that we can use the Faraday constant to calculate the mass of a substance plated or deposited or the volume of a gas generated during electrolysis.