Cell Potential, Gibbs Free Energy & the Equilibrium Constant

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Cell Potential & the Equilibrium Constant

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So here we say that galvanic and voltaic cells are able to produce electricity because they are not yet at equilibrium. Now recall that the chemical reaction will eventually reach equilibrium and at that point Q will equal K. In addition to this, once we've reached equilibrium, we would say that our self potential under non stand conditions would also equal zero. So here we have our initial nerds equation here we have your cell potential are nonstandard conditions equals self potential under standard conditions minus 0.5916 volts divided by N. The number of electrons transferred times log of Q. Q. Being your reaction quotient. And remember this represents your voltage and an exact moment once we've reached equilibrium, Like we've said yourself potential under nonstandard conditions with equal zero and your Q would equal your case. So now we use your equilibrium constant as your new variable. Because of this, we can try to solve for what K would equal in relation to your cell potential under standard conditions. So here we would subtract e cel from both sides which would give us a negative here and a negative here. And then you would just divide both sides by negative one. To make it positive. At this point. We could then multiply both sides by N. To give me n times yourself potential under standard conditions equals 0.5916 volts times log of K. Here we're reformatting the equation to show the relationship that the equilibrium constant has towards yourself potential Under standard conditions, we would then divide both sides by 0.5916 volts. So we'd have log of K Equals and Times yourself potential under standard conditions divided by .05916V. You want to get rid of the log of K. So you just take the inverse log. So that would become K equals 10 to n. Times E. Cell divided by 100.5916 volts. So this is the equation would use when we are given your self potential understanding conditions and asked to find your equilibrium constant K. Now we're gonna say the relationship between your self potential under standard conditions and your equilibrium constant K and gives free energy can be seen in this format. So when we have K and delta G, we connect them by this equation here, Delta G equals negative. Our tee times L n f K. Remember that R equals 8.314 jewels over moles times K. Remember here that the units could also be changed to volts times columns over moles times K. Then if we have our self potential under standard conditions in delta G, they're connected by this formula here. Delta G equals negative end number of electrons transferred times Faraday's constant times self potential under standard conditions. And then finally K. and Esa can be connected by this equation that we saw up above. Remember at 25°C. This portion here We can get as a standard value. And when we multiply by 2.303 that changes Ln into log. Then finally realized that we have cell potential here and here. And those cell potentials connect to this equation here. So remember if we're dealing with standard conditions, we're dealing with this circle here, that would mean that our concentrations are equal to one Moeller. Therefore we wouldn't rely on the Nerdist equation to help us determine what this cell potential is at all. We would just use the cell potential of my cathode minus the self potential of Maya node. So this is dealing with standard cell potential soul. No nurse equation needed. So keep in mind the relationship that Q and K have with each other in terms of reaching equilibrium, how there's a transition from your Q. Value to your K value. Remember the nurse equation is really utilized when we have concentrations that are different from when Mueller here in this triangle, we're assuming that we are at equilibrium now. So Q. Has been transitioned into K. We're dealing with one molar concentrations And therefore we use the simplified version to help us determine the overall standard cell potential and then look at the connection that it has to give free energy. So keep in mind these connections as we delve deeper and deeper into looking at calculations that show the interconnectedness of these different variables

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Calculations 1

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So here we need to determine the equilibrium constant K. For the following reaction, we have one mole of gold plus one ion reacting with one mole of Siri um solid to produce one mole of serum three plus. Aquarius plus one mole of gold solid. Were in addition to this, we're told that the half reactions are determined as each one of the following with their given cell potentials. We'll hear from these cell potentials. We can determine what our overall standard cell potential will be. We know that we're gonna figure out the standard cell potential because we're not given any concentrations. So we assume that they've approached unity where they are equal to one Moeller. Therefore the nerds equation isn't allowed to be used in addition to this, we're going to have to figure out which one of them is being oxidized versus which one is being reduced. If we take a look here at the overall reaction, we see that we have gold plus one. So its oxidation numbers plus one and here it's neutral. It's oxidation number has decreased in value. Therefore it has been reduced and represents the cathode. So it's been reduced and it represents the cathode. Then if we look at syria and we see that it's at zero initially and then it goes up to plus three. So its oxidation number has increased. Therefore it's been oxidized and represents your anodes. Next, we're gonna say at this point, we can just say that our standard cell potential equals your cathode minus your an ode. So that's 1.690V -A 2.336V minus of a minus really means you're adding them. So that's 4.0 to 6 volts. This is a very large voltage for our cell potential, which means that we should expect a very large K or equilibrium constant. Now we know that from the previous page, that the relationship between the equilibrium constant and your self, potential your standard cell potential is K equals 10 To the end times some potential. And standard conditions divided by .05916V. So we can plug in the values that we know. So we put 10, we have to determine the number of electrons transferred here. They're not the same number. You'd have to multiply this equation here by three to get the correct number of electrons transferred. And remember multiplying or dividing does nothing at all in terms of the cell potential. So it's still stay that same value. And because we multiply this by three, that means the balance equation will be three here and three there. So we have three here times yourself potential which was 4.0-6V Divided by .05916V equals 10 to 2 oh 4.158, which gives us a very large equilibrium constant of 1.44 times 10 to 204. This tells me that this reaction is incredibly spontaneous because it gives us a value for your standard cell potential that's much greater than zero. Um and it also gives us a. K. That is much greater than one if, depending on the calculator that you're using, you may get an error if you're not using us high enough or advanced enough calculator. This is an incredibly large value for K. And that's the connection that we can have between your equilibrium constant K and your standard cell potential.

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Calculations 1

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From the two half reactions, provided the equilibrium constant is calculated as 6.79 times 10 to the 30. Now from these two half reactions, we have one where the electrons are reacting. So this is our reduction reaction. So this would represent our cathode. And then here we have the electrons as products which represents an oxidation. So this is our an ode here. It has determined the standard cell potential for basically the cathode compartment. Alright. So they gave us the equilibrium constant. K. Remember that that shares a connection with our standard cell potential here. The overall standard cell potential equals 0.5916 volts divided by the number of electrons transferred times log of K. The number of electrons transferred has to be the same here we have three. And here we only have one. So we have to multiply this by three to make it three electrons. Remember multiplying our re are half reaction or dividing our half reaction does nothing to the self potential value, it stays the same. The only thing that can change is that if we reverse the reaction itself. All right. So here we're going to say that we have .05916V divided by three. And this is times log of 6.79 times 10 to the 30 when we plug that in. That gives me .608V. So we have the overall cell potential standard cell potential. That's equal to cathode minus an ode. This value here is the original self potential of this reaction when it was written as a reduction. So we have here .608V equals e positive minus 0.308 volts here. So we do here is we add this to both sides. So that comes out to .916V as the missing self potential for the half reaction that occurs at the cathode. So remember, the key to this question is realizing what is the connection between our overall standard cell potential and our equilibrium constant. K. Knowing that allows us to then isolate this equation and solve for the one missing variable, which gives us the self potential of our cathode compartment.