so voltage which was represented by a variable E represents the amount of work done in electrochemical cell as electrons travel from one electrode to another. Remember that one vault is equal to jewels over coolants. Now here we have half reactions written as reductions, the written as reductions because the electrons are all written as reactant. We have the oxidation number of our reactions 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. Then the she electrode realize here that the larger the potential of a half reaction, the more wants to be reduced. So the more likely reduction will occur and the smaller your value then the more likely oxidation will occur. So what we can say here from this chart is that hydrogen will the she electrode when it's being compared to these bottom three because it has a higher voltage or higher self 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 zero. 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 gonna say here when combining two half reactions together the cell potential for the for the total net reaction is given when the reaction, when the concentrations approach unity. So that means that they're gonna approach one. So here one Molar. So as the concentrations approach one Moeller, we're gonna say that we can just say that our self potential or S. L. Equals E plus minus E minus. We're gonna say E plus here represents the cathode electrode E minus. He represents the anodes electrode. So this is basically saying cathode minus an ode will give us our overall cell potential. Realize that this equation is this simple again, when the concentrations are equal to one molar, if the concentrations are not equal to one molar then this equation cannot be used to figure out our overall cell potential. In those cases we have to rely on the nurse 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 one 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 were given two half reactions were given 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 mine.

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Standard Potential

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So in this example it says determine the electrical or electric potential that results from a given galvanic cell. In the illustration, we show that this half reaction is transferring electrons. So this half reaction here is transferring electrons. They're moving away from this half reaction to the silver half reaction. The movement of electron tells us which one is the an old and which one is the cathode. Remember at the an ode we always have oxidation when we have a loss of electrons. So this here represents my an ode here. This represents my cathode. Now we don't necessarily need this illustration for the electrons moving from the an ode to the cathode to tell them apart. There's a few things we can look at here. We have lithium in its neutral natural state, so its oxidation number zero and then here we have lithium plus one. So it's oxidation now is plus one. We could have also said that the oxidation number goes from zero to plus one. It is increased because it has undergone oxidation. Which means that this has to be the anodes on the other side. We have plus one and zero for the oxidation numbers, we can see that the oxidation number decreases. Therefore it has to represent reduction and therefore the cathode. In addition to this, we could have looked at the potential is given for each half reaction. Remember up above we said that the larger your value is, the more likely reduction will occur and the small your potential is the more likely oxidation will occur. So this one having the larger E value tells me that this has to be reduction because we don't we're not giving any concentrations for the ions. We assume that the concentrations have approached unity therefore equal one. And because of that we can simply say that our overall cell potential equals cathode which is positive minus E minus. So that would be 0.799 volts minus A -3.040V and a minus of a minus really means plus. So overall this is 3.839V of electricity that have been generated from the movement of electrons from the annual compartment to the cathode compartment. Remember earlier on we talked about the value of your cell potential could determine if this is a spontaneous process or not because our overall cell potential is greater than zero. We know that this is a spontaneous process which makes sense because a galvanic or voltaic cell represents a spontaneous electrochemical cell. So remember this is just the fundamentals to help us determine what the overall cell potential is. Later on, we talked about concentrations different from one Moeller. We'll have to utilize the use of the nerds equation with the nurse equation. We have to be very careful in terms of concentrations of ions as well as the moles of ions to determine the correct overall cell potential for Now just assume that the concentrations equal unity because they're not given to us and therefore we simply use this version to help us determine the overall cell potential. Mhm.

Standard Potential Calculations

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example

Standard Potential Calculations 1

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So here in states use the standard half cell potentials listed below to calculate the standard cell potential for the following reaction occurring in an electrochemical cell at 25 degrees Celsius here, it says, assume the concentrations have approached unity. So here we're given concentrations of our ions. So we have to assume that they have approached one Moeller. Therefore we don't have to use the energy equation. And can simply say that our self potential equals r cathodes minus an out. Now. Typically we would say that the smaller value would represent oxidation and therefore my anodes and that the larger cell potential would represent reduction and therefore my cathode. But we cannot say that when we're given the overall equation, when we're given the overall equation, we actually have to look at the overall equation and determine what's being oxidized and what's being reduced. And in that regard, then we say, who's the cap and who's the anodes. Now, if we look, we say that chlorine goes from its natural state, which is an oxidation number of zero to its charged form -1. So here chlorine goes from being zero to minus one. So its oxidation number reduced itself, Therefore, reduction has occurred. And therefore that represents the cathode, which happens to be the larger E value. But again, we can only say that the E value given the larger one is the cathode only when they don't give us an overall equation because they give us an overall equation, we actually have to use the overall equation to determine who is truly the cathode and who is truly the anodes. Now here, iron goes from being zero for its oxidation number two plus three. So it's oxidation number increased. Therefore underwent oxidation and represents the anodes. Now that we know who's the catholic, whose piano? Based on the overall equation, we can say that our standard cell potential. So the little zero here equals cathode minus a node. So that's 1.396 volts minus a minus 0.40 volts. So minus of a minus really means that we're adding them together. So we get at the end is 1.436 volts for this reaction. And what that tells me is that this is a spontaneous process in which chlorine would naturally become reduced when in the presence of iron solid. So just remember we can say that the larger E value equals cathode, smaller E value equals the anodes. When we are not given the overall equation. When you're given the overall equation, you actually have to look at that instead to determine who's been reduced and therefore represents the cathode and who's been oxidized and therefore represents the adult. Now that we've seen this basic example, move on to example to attempt it on your own. Once you do come back and see if your answer matches up with mine

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example

Standard Potentials Calculations 1

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So for the voltaic cell with the overall reaction, we have our standard cell potential as being equal to 1.10 volts. So our overall cell potential is this value here, it says, given that the standard reduction potential of zinc to ion to zinc solid is negative 0.6762 volts, calculate the standard reduction potential four copper two plus plus two electrons giving us copper solid. Alright, so in this question we have our redox reaction, zinc is going from its neutral natural state where oxidation number is equal to zero to its charge state. Remember for an eye on its charge equals its oxidation number. So here it's plus two, it goes from an oxidation number of zero to an oxidation number of plus two. Since its oxidation number has increased its undergone oxidation and because it represents oxidation, it's going to be the anodes. Then we're going to say within this equation we have copper two plus. So oxidation numbers plus two and it goes to zero. Right? So it's being reduced. Therefore it represents the cathode. Where reduction occurs here, we assume our concentrations have gone to unity, so they're equal to one Moeller. Therefore my standard cell potential equals cathode minus anodes. So my standard cell potential is 1.10 volts. We don't know what the potential of our cathode is. That's what we're looking for, minus a minus 0.762 volts. So remember minus of a minus really means positive. So now subtract .762V from both sides. So the potential here for my cathode would be equal to .338V. So that would be our final answer. So just remember for questions like this, it's important to be able to look at the redox reaction and determined from that what's been oxidized and therefore represents the anodes and what's been reduced and therefore represents the cathode. Once you've figured that out, just remember to fall back on the equation fill in the given values and sulfur, the missing variable, which in this case was just the potential of my cathode.