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- [Voiceover] When you are working with standard reduction potentials, it's important to realize that voltage is an intensive property and by the end of the video, you'll understand what I mean by the word intensive. We're going to use this first half reaction here as an example. So this is the reduction of silver ion to silver metal, so we have Ag plus plus an electron gives us solid silver. The standard reduction potential is positive .80 volts for this half reaction. Now, we're going to start by calculating the standard change in free energy, so from an earlier video, we know if we have the voltage, we can find the standard change in free energy, delta G zero. So for this half reaction, it would be equal to negative. Remember that n is the number of moles of electrons, and here we have one mole of electrons, so we have one mole, so we plug that into our equation. Next, we have F, which is Faraday's constant, and from an earlier video, we know that's 96,500 coulombs per mole, so it's the charge of one mole of electrons. We need to multiply that by the voltage which was .80. So this is .80. Instead of writing volts, I'm going to write joules per coulomb here, so we can see our units cancel, so moles of electrons cancels out, charge, coulombs cancels out, and so we should get our answer in terms of joules, so this is equal to negative 77 kilojoules is the change in free energy that accompanies the formation of one mole of silver. What about if we want to form two moles of silver? So we need to multiply everything in our half reaction by two, so we get two Ag plus, so two moles of silver ions plus two moles of electrons should give us two moles of solid silver. All right, what would be the standard change in free energy that accompanies the formation of two moles of solid silver? Well, if, let me change colors here, if this is the standard change in free energy for the formation of one mole of solid silver, then we should just be able to multiply that number by two, right? So, negative 77 kilojoules times two gives us negative 154 kilojoules, so that's the standard change in free energy that accompanies the formation of two moles of silver, so free energy is what's called an extensive property, so it depends on how much you're dealing with here, so we're dealing with two moles of electrons in the formation of two moles of silver, so we have to change the standard change in free energy accordingly. All right, what about the voltage? So we have our standard change in free energy. Let's use this equation again to solve for the voltage, so we're going to plug in negative 154 kilojoules in for our standard change in free energy, and we need to make that into joules, so that's negative 154,000 joules, and that's equal to, let me change colors again here, this would be equal to negative, right, so we have a negative sign, and then n is the number of moles of electrons, well now we're saying two moles of electrons, so this would be negative two here, and then, multiply that by Faraday's constant, so we have Faraday's constant, which is 96,500, and then we can solve for the voltage. So we can solve for E zero. We get a voltage, a voltage of .80 volts. So it's the same voltage to form two moles of silver to form one mole of silver, right? Let me highlight that. So, let me change colors here. So we had a voltage of .80 volts to form two moles of silver. That's the same voltage to form one mole of silver, so voltage is an intensive property. It's the same. It doesn't matter how much silver you're forming, how many moles of electrons you're using. The voltage is the same, and it's important to remember that when you're doing your standard reduction potentials, because if you need to, if you need to do something like what we did up here, multiply a half reaction by two, you don't multiply the voltage by two because voltage is an intensive property.