Quantitative electrolysis | Applications of thermodynamics | AP Chemistry | Khan Academy

We already know that in an electrolytic cell, current or movement of electrons is used to drive a redox reaction. If we look at a generic reduction half-reaction, the stoichiometry of the half-reaction shows how many electrons are needed to reduce a generic metal ion. For example, if we say that our generic metal ion is M²⁺, it takes two electrons to turn M²⁺ into the solid metal.

So, one mole of M²⁺ ions requires two moles of electrons to form one mole of the metal. We could use this mole ratio of one mole of metal to two moles of electrons to figure out, say, how much of a metal is produced in an electrolytic reaction. This is an example of a quantitative electrolysis problem. Moles of electrons are related to the amount of charge that passes through the electrolytic cell because one mole of electrons carries 96,485 coulombs.

We could set this up as a constant called Faraday's constant. Faraday’s constant is symbolized by F, and it's equal to 96,485 coulombs per mole of electrons. Charge is related to current and time by this equation: I = Q/T, where I is the current, Q (capital Q) is the charge, and T is the time. Sometimes you see a lower case q for charge. Charge is measured in coulombs, and time is in seconds. One coulomb per second is equal to one ampere, so current is measured in amperes.

Next, let's look at a conversion chart that shows how to approach quantitative electrolysis problems. Let's start with this box on the far left, which says that charge is equal to current times time. This comes from our equation for current: I = Q/T.

So, solving for charge, Q is equal to I times T, or charge is equal to current times time. Charge is related to moles of electrons by Faraday's constant, so we can convert back and forth between the two using Faraday's constant. Moles of electrons are related to the moles of the substance that we're interested in the problem by the balanced half-reaction.

The balanced half-reaction allows us to convert back and forth between moles of electrons and moles of substance. From stoichiometry, we know that the moles of a substance are related to the grams of the substance by the molar mass. The molar mass allows us to convert back and forth between these quantities. So, it helps to look at a conversion chart when doing these sorts of problems.

However, if you don't remember the chart, you can always use dimensional analysis and units to figure out the answer to the question. Next, let's use our conversion chart to figure out a quantitative electrolysis problem. Let's say we have an aqueous solution of silver nitrate, and a constant current of 2.40 amperes is applied for 1,225 seconds. Our goal is to figure out the mass of silver that is formed.

As a quick reminder, an aqueous solution of silver nitrate contains Ag⁺ ions and NO₃⁻ anions in solution. The applied current is used to reduce the Ag⁺ ions to form solid silver. Since our goal is to figure out the mass of the solid silver that's formed, we're trying to find grams of substance. In the problem, we're given the current and the time.

So, we're starting over here in the first box with the current and the time. Let's think about the steps that we need to solve this problem. The first step is to find the charge, which we can get from the current and the time. The second step is to convert charge into moles of electrons. From moles of electrons, we can convert that into the moles of silver in this case. In the fourth step, we can convert moles of silver into grams of silver.

So, step one is to find the charge. Q, the charge, is equal to the current times the time. The current is equal to 2.40 amperes. Remember, one ampere is equal to one coulomb per second. So we plug in 2.40 coulombs per second, and the time is 1,225 seconds. When we plug that in, seconds cancel out, and we get that Q is equal to 2,940 coulombs.

That's the charge that's transferred in that time period. Now that we have the charge, in the second step, our goal is to find the moles of electrons. Charge is related to moles of electrons by Faraday's constant. Dividing the charge by Faraday's constant causes coulombs to cancel out, and we get 0.0305 moles of electrons.

Now that we have moles of electrons in step three, we can find the moles of silver. Looking at the balanced equation for our half-reaction showing the reduction of silver ions, one mole of silver ions requires one mole of electrons to form one mole of solid silver. So our mole ratio of silver to moles of electrons is 1:1, which gives us this conversion factor of one mole of silver per one mole of electrons.

If we multiply our moles of electrons by our conversion factor, moles of electrons cancels out, and that gives us 0.0305 moles of silver. In this case, our mole ratio was 1:1, but if we were reducing a different metal ion, our mole ratio might not be 1:1. So we have to be extra careful on this step to make sure we get the right mole ratio.

Since we have moles of silver, in our fourth and final step, we can convert moles of silver into grams of silver. So multiplying moles of silver by the molar mass of silver, which is 107.87 grams per mole, moles cancel, and this gives us 3.29 grams of silver. It's also possible to do all four steps at once in a dimensional analysis approach.

So looking at our units, if we multiply 1,225 seconds by 2.40 coulombs per second, seconds cancel out and give us coulombs. Multiplying that by one mole of electrons per 96,485 coulombs cancels out coulombs. Next, we have our conversion factor of one mole of silver per one mole of electrons, so moles of electrons cancel out. Finally, multiplying that by the molar mass of 107.87 grams per one mole of silver will cancel out moles of silver and give us a final answer of 3.29 grams.

So, we get the same answer no matter which approach we take.