9.10: Electrolysis and Faraday’s Law
You made it! The last section of AP Chemistry. Before jumping into this final section of unit 9, congratulate yourself for making it this far! You’ve learned everything from acids and bases to equilibrium to thermodynamics and even more. In this final section we’ll take a look at nonspontaneous redox reactions and how we can use electrolytic cells to make them happen. In this section we’ll learn how electrolytic cells work, how we can differentiate them from galvanic cells, and how to use Faraday’s Law to make calculations surrounding electrolysis.
Review of Electrolytic Cells
As we’ve stated before, an electrolytic cell is a cell in which power from an external source (usually a battery) is used to spur nonspontaneous redox reactions. This means that a species that would normally be oxidized is instead reduced and vice versa. Here’s an example of an electrolytic cell with Zinc and Copper:
Image From Abigail Giordano
In this image, we can see that electrons are being pumped out of Cu to form Cu2+ and transferred to Zn2+ to form Zinc metal (If you’re confused as to how we know that, look at the direction of the flow of electrons). The redox reaction that is occurring is: Cu + Zn2+ → Cu2+ + Zn. Let’s use a table of reduction potentials to calculate the cell potential:
First let’s write out our half reactions:
Cu → Cu2+ + 2e- (E = -0.34V)
Zn2+ + 2e- → Zn (E = -0.76V)
Adding our Es we find that E°cell = -1.10V
Therefore we must have power to make this redox reaction occur. Note that on the battery in the cell diagram we see that it must be a battery with a voltage of over 1.10V. This is because the voltage must be enough to overcome the non-spontaneity of the redox reaction we want to occur (and also the spontaneousness of the reverse reaction which has an E value of +1.10V). When we connect this battery, the voltage pulls electrons (remember voltage is electromotive force) in the direction of the non-spontaneous reaction. Therefore, the mass of the copper electrode will decrease and the mass of the zinc electrode will increase.
Comparing Electrolytic and Galvanic Cells
There are some important differences between a galvanic cell and an electrolytic cell that you’ll need to know and recognize for the AP exam. Let’s look at a galvanic cell and an electrolytic cell side by side:
Image From LibreTexts
On the left we see the galvanic cell we know and love for the reaction between cadmium and Cu2+. Let’s do a quick review of what we see in this galvanic cell. We have an anode where oxidation occurs and a cathode where reduction occurs and electrons travelling through a wire from the anode to the cathode. As they travel, a voltmeter measures the electromotive force in volts that the reaction releases. Finally, a salt bridge with a neutral salt connects the two half cells to maintain neutrality. For this cell, the copper electrode will grow and the cadmium electrode will shrink.
On the right we have the same two metals, cadmium and copper, but this time in an electrolytic cell. Note that the first major difference is the direction electrons travel. In the galvanic cell electrons travelled from the cadmium to the copper electrode (and eventually to the Cu2+ to create Cu metal). In the electrolytic cell the opposite is occurring. Copper metal is being oxidized into Cu2+ and Cd2+ is being reduced into Cd metal which is the opposite reaction. Instead of a voltmeter measuring the voltage, a power supply (similar to a battery) is used to produce an electromotive force of at least 0.74V. This is because the reaction occurring in the electrolytic cell is nonspontaneous meaning it requires the external source of energy that the power supply supplies. Because our reaction is separated into half cells (unlike the electrolytic cell from before), a salt bridge is again required for the same reason. In this cell, the copper electrode will shrink and the Cd electrode will grow.
A major similarity between the two cells is that always, always, ALWAYS, oxidation occurs at the anode and reduction occurs at the cathode. No matter what the cell is, that is how you label your electrodes.
Here’s a table showing big picture differences between the two cells:
Faraday’s Law and Electrolysis Problems
An important problem that you may be given regarding electrolytic cells is calculating the mass accumulated of a certain metal. To do this we’ll want to do a chain of conversions using the definition of an amp (1A = 1C/s) and Faraday’s constant (1 mol e- = 96485 coulombs). Let’s take a look at an example problem:
Determine the mass of chromium that can be produced when a solution of Cr(NO3)2 is electrolyzed for 60 minutes with a current of 15 amps.
Let’s start by converting from amps to coulombs:
60 minutes * 60s/min * 15 C/s = 54000 coulombs
Next, we can use this value of charge to convert to moles of electrons:
54000 coulombs * 1 mol e-/96485C = 0.559 mol e-
And then we can use stoichiometry to convert from mol e- to moles of chromium and then use the molar mass of chromium (52.0 g/mol) to convert to grams:
0.559 mol e- * (1 mol Cr/2 mol e-) * (52g/mol Cr) = 14.55g Cr produced.
This sort of dimensional analysis can also be done in one loooong chain as follows:
You should be prepared to solve for any of these unknowns (except for Faraday’s constant and stoichiometry stuff). This means a problem could ask about finding the time necessary to produce a certain mass or the amperage required for a certain time to produce a mass. To do these problems, you can simply manipulate the chain we just did.
For example, if a problem asked for a time but gave a mass, you could start with the mass, convert to moles, then convert to moles of electrons, then charge, and then to amperage. Let’s look at the units: