So voltage, which is represented by our variable e, represents the amount of work done in an electrochemical cell as electrons travel from one electrode to another. Remember that one volt is equal to joules over coulombs. Now, here we have half reactions written as reductions. They're written as reductions because the electrons are all written as reactants. We have the oxidation number of our reactants decreasing as it becomes a product, and here we've highlighted this particular equation because this one here represents the SHE electrode. So this is our reference electrode. When we talk about the potential or voltage of a particular half-reaction, it's in reference when compared to the SHE electrode. So here this is 1.507V, which means it has that much of a likelihood of undergoing a reduction than the SHE electrode. Realize here that the larger the potential of a half reaction, the more it wants to be reduced. So the more likely reduction will occur, and the smaller your e value then the more likely oxidation will occur.

So, what we can say here from this chart is that hydrogen will be SHE electrode when it's being compared to these bottom three because it has a higher voltage or higher cell potential, it wants to be reduced more so than the bottom three reactions, and then when comparing it to these top reactions because they have cell potentials that are larger than 0, they have a greater tendency of being reduced instead of the SHE electrode. So, by comparing the cell potentials between different half-reactions, you can determine which one wants to be oxidized and which one wants to be reduced. This is important because this will help us determine the overall cell potential and overall voltage that can be generated from a galvanic or voltaic cell.

So, we're going to say here when combining 2 half-reactions together, the cell potential for the total net reaction is given when the concentrations approach unity. So it means that they're going to approach 1 molar. So as the concentrations approach 1 molar, we're going to say that our cell potential, or e_{cell}, equals e^{+} minus e^{-}. We're going to say e^{+} here represents the cathode electrode. e^{-} here represents the anode electrode. So, this is basically saying cathode minus anode will give us our overall cell potential. Realize that this equation is this simple, again, when the concentrations are equal to 1 molar. If the concentrations are not equal to 1 molar, then this equation cannot be used to figure out our overall cell potential. In those cases, we'd have to rely on the Nernst equation in order to find our correct cell potential.

So, for now, just realize that if you are not given the concentrations of the ions reacting within a given redox reaction, we assume that the concentrations have approached unity, therefore equal to 1, and therefore our equation is just simply this here. Now that we've gotten down the basics in terms of cell potential and how the value of e can help us determine which one is being reduced versus oxidized, attempt to do the practice question here on the bottom. Here, we're given 2 half-reactions. We're given a half-reaction for lithium and a half-reaction for silver. Based on their given cell potential values, determine what my overall cell potential would be for this particular galvanic cell. Once you do that, come back and see if your answer matches up with ours.