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🔋Unit 9

4 min read•july 9, 2021

Sander Owens

Dylan Black

In the last section, we explored the idea of **electromotive force** from redox reactions. This is the “pull” on electrons from a reducing agent to an oxidizing agent. The higher this electromotive force, the more spontaneous the reaction will be. We can think of this generally as thinking about how strongly each species either “wants” to get rid of electrons or “wants” to gain electrons. This thinking about atoms/ions having “wants” isn’t always the best realistically, but it can help you visualize how electromotive force may work. We measure electromotive force in volts, as is the potential for cells as we’ll see.

When we are dealing with a galvanic cell, the reduction and the oxidation occur on separate sides. We have an **anode** where oxidation occurs and a **cathode** where reduction occurs. Electrons will travel through a wire from the anode to the cathode. The force these electrons feel as they are pulled through the wire is known as the **cell potential** and is the same as the electromotive force caused by the redox reaction that is occurring. We can calculate standard cell potential (standard meaning 1M concentrations, 298K, and 1 atm) by using the equation E°cell = Ecathode - Eanode.We can also think of this as meaning that the standard cell potential for a cell is the reduction potential for the species in the cathode minus the reduction potential of the species in the anode. Let’s take a look at an example:

For this reaction we see that AgBr is reduced into Ag and Br- so the cathode will be AgBr and at our anode Hg is oxidized. Plugging into E°cell = Ecathode - Eanode we find: E°cell = 0.071 - 0.140 = **-0.069V**.

Another strategy for finding cell potential is to negate the oxidation reaction and then add your potentials. For example, looking at the last problem we could have also acknowledged that the actual half reaction for Hg was 2Hg + 2Br- → Hg2Br2 + 2e- which has a potential of -0.140 which when added to +0.071 yields the same result.

In the last problem we were given specific electrochemical data, but for some problems you’ll need to use a table of **Standard Reduction Potentials**. This is essentially a very long list of reduction reactions with their corresponding potentials that you have to use to find the correct reactions, understand which one is for the cathode and which is for the anode, and then subtract. Luckily, on the AP Exam you will not be given this table and will instead be given specific necessary data, but it’s still a useful thing to be familiar with. Below is a sample of a table of standard reduction potentials:

Image From __Grade12UChem__

The values in this table are all reductions, but note that for many substances, the reduction has a negative potential. This indicates to us that substances like Li and Na are very good reducing agents. This is because they are suuuuper hard to reduce (as shown by their negative potentials) but if we flip the reductions to turn into oxidations, the negative potentials become positive.

**Standard Cell Potentials and Spontaneity**

The major conclusion we want to draw from finding cell potential is the spontaneity of the redox reaction taking place. If E°cell is positive, the reaction is spontaneous. This is because the electromotive force is enough to pull the electrons off of the reducing agent and onto the oxidizing agent. Conversely, if the E°cell is negative, the reaction is nonspontaneous. Therefore, we can predict the sign on ΔG° for a reaction given E°cell. To reiterate:

If E°cell > 0, the reaction is spontaneous

If E°cell < 0, the reaction is nonspontaneous

**Calculating ΔG° Using E°cell**

While we can look at the sign on E°cell to predict the sign of ΔG°, we also have a way of directly calculating ΔG° from cell potential. The formula for ΔG° using E°cell is as follows:

In this equation, ΔG° is the standard Gibbs Free Energy change, E°cell is the standard cell potential, n is the number of moles of electrons transferred in the reaction, and F is Faraday’s constant which is 96,485 coulombs/mol e-. A **coulomb **is a measure of electric charge. Let’s look at an example:

Suppose we have a galvanic cell for which E°cell = 1.02V. In the reaction that occurs, 1 mol of electrons are transferred. At 298K, what is ΔG° for this cell? What is the equilibrium constant, K?

First we can plug into ΔG° = -nFE°cell to find ΔG°:

ΔG° = -1(96,485)(1.02) = -98414.7 J = -98.414 kJ

Therefore ΔG° = -98414.7 J or -98.414 kJ.

Next we can use our relationship between ΔG° and K to find the equilibrium constant:

K = e^(98414.7 /(8.314)(298)) = 1.78 * 10^17

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