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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 metals we have AG plus plus an electron gives us solid silver the standard reduction potential is positive 0.8 zero volts for this half reaction 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 right 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 point which was 0.8 zero so this is 0.8 zero 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 and let's figure out let's figure out what it is this would be one times ninety six thousand five hundred times point H zero and it should be the negative of that so negative seventy seven thousand two hundred you which I'll just say is negative seventy seven kilojoules so we get negative seventy-seven kilojoules is the change in free energy that accompanies the formation of one mole of silver all right what about if we want to form two moles of silver so we need to multiply everything on our half reaction by two so we get two AG plus so two 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 seventy-seven 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 and 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 all right 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 one hundred and fifty four thousand 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 a 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 right so we have Faraday's constant which is 96 thousand five hundred and then we can solve for the voltage so we can solve for e zero so let's plug that in on our calculator we have negative negative 150 four thousand right we need to divide that by negative two divide that by negative 2 and then divide that by Faraday's number so Faraday's constant I should say so ninety six thousand five hundred which gives us a voltage of 0.8 zero all right so if we round that get a voltage voltage of point eight zero volts so it's it's the same voltage to form two moles of silver as it was to form one mole of silver all right let me highlight that so let me let me change colors here so we've we had a voltage of 0.8 zero 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 we did up here multiply a half reaction by two you don't multiply the voltage by two because voltage is an intensive property