So the nerds equation reveals the quantitative connection between the concentrations of compounds and sell potential. So the nerds equation is utilized When are concentrations of our compounds differ from one molar. We're going to say that the nerds equation equals our self potential here. And this self potential represents self potential under nonstandard conditions, meaning that our concentration isn't equal to one moller. Our temperature wouldn't be 25 degrees Celsius are ph we're not equal seven. Um Our pressure wouldn't be one atmosphere. All of those values represent standard conditions, so one atmosphere ph of seven, a temperature of 25 degrees Celsius as well as a concentration of one molar. When we have all of these conditions met, that means we're dealing with self self potential under standard conditions. So that's E zero cell. So this represents our self potential under standard conditions minus R. T. Divided by N times F times Ln A divided by a. So here again we said that this represented our standard cell potential are here is our gas constant. It is equal to 8.314 jewels. Over moles times K. Also remember here that when we talk about jules, we're talking about energy, A jewel is equal to volts times columns. Remember a vault is equal to jules over columns. So substituting jewels over columns, times columns. And that's how it equals jewels. So we can say this or 8.314 volts times cool. Ums over moles times K and equals the number of electrons transferred within our redox reaction F equals Faraday's constant, which is 9.649 times 10 to the four columns over moles of electrons A represents our activities. So activity here would just be our activity coefficient times the concentration of the ion if necessary. Many times we may not be given some type of coefficient. So you can just say activity, you could substitute in concentration for that value. Now we're going to say that this expression here represents products. Overreact ints in terms of our redox reaction and it's equal to cube our reaction quotient. Now, if we were to take a look at the portion that's r times T divided by F here at 25 degrees Celsius. Remember we have our our constant, we have our temperature 25 degrees Celsius. We add to 73.15 to this so that gives me to 98.15 kelvin F is our Faraday's constant here. We would see that from this mold would cancel out kelvin's would cancel out what we have left at the end is jules over columns, which is equal to volts. So our T over F reduces all the way down 2.257 volts. So that means that our nerds equation becomes now self potential under nonstandard conditions equals self potential under standard conditions minus 0.257 volts divided by n. The number of electrons transferred times. Ellen F. Q. Remember Q is just your equilibrium expression now here If we multiply Ln by 2.303 we can attain the log function. Now when we multiply uh this portion here by 2.303 we get a new value of 0.5916 volts divided by N. Now log of Q. This is true because we say that log of X equals Ln of X, divided by Ln of 10. Ellen of 10 Equals 2.303. So when I multiply both sides by 2.303, We can see that multiplying by 2.303 helped us to establish this new relationship here of this value. Now this is important. We're gonna say the self potential calculated from the nurse equation is the maximum potential at the instant the circuit the cell circuit is connected. So that's the moment that the current or the flow of electrons moves from the an ode to the cathode. So we're talking about basically the transferring of electrons from one electrode to another electrode. We're gonna say as the cell discharges and current flows, the electrolyte concentrations will change. What's going to happen here is that Q will begin to increase and as a result our self potential over time will decrease until it gets to a point where the self potential of my electrochemical cell equals zero. That's when we have a dead battery because at that moment the battery has reached equilibrium. So over time the reaction will reach equilibrium and then Q which is our reaction quotient equal K. Which is our equilibrium constant. The cell potential like I said, would equal zero, we have a dead battery at that moment. Now, as a result of this, once we've reached equilibrium, we can substitute in K instead of Q. So now our new equation can become S. L. Equals sl under standard conditions minus R TNF times lnk. So here if we were to work this out we could have zero equals e cell -. Remember this value here would be .0257 volts divided by N. Now times allen of K. So what we can do here is we subtract cell potential from both sides. So that would be negative. Standard cell potential equals negative 0.257 volts divided by N times L N F K multiply both sides by N, Then divide both sides. Now by negative .0257V. So we get at this point alan f k Equals and Times yourself potential under standard conditions divided by .0257V. To get rid of this island, we take the inverse of the natural log. So that means that K equals E. To the end. Times your self potential. Under standard conditions divided by .0257V. And that's when we're dealing with L N. If we were to substitute and log instead. So if we're dealing with log function, then It would be .05916V divided by N. Times Log of K. And this is the instance if we Did the same exact um mathematical conversions in this case because we're dealing with log we'd find out K at the end equals 10 to the end times yourself potential. Understanding conditions divided by .05916V. So this is how we connect our equilibrium constant to our standard cell potential with these two formulas. One when we're dealing with Ellen and one more we're dealing with the log. Now we could also say that when you're at equilibrium we can talk about the connection to Gibbs free energy and your equilibrium constant K. So here we say that delta G under nonstandard conditions equals delta G. Under standard conditions minus R T L n K. When we've reached equilibrium, this delta G equals zero. So that means zero equals delta G zero minus R T L N K. Subtract Delta G0 from both sides. So negative delta G zero equals negative R. T. Lnk, divide both sides. Now by negative R T. So Ellen K equals delta G zero divided by R T. And so K equals E. To the delta G zero divided by R T. So we get this value at the end in terms of our connection between gibbs free energy and your your equilibrium constant K. So just keep in mind some of the connections that we've seen here in terms of how things are connected to one another. These two equations are just a way of us connecting cell potential two K. And then from K. Two delta G. And we know that from earlier. We could also connect to self potential from delta G as well. When we have delta G equals negative end times. Faraday's constant times self potential. Later on we'll talk about this connection between the three variables. But for now just realize we deal with the nurse the equation when we have concentrations that are not equal to one Moeller. And remember nerds equation can be written in two different ways. We can write it in this version when we're dealing with L. N. Or we can have the nurse equation in this version. When we're dealing with the log To go from Ellen to log, you just use two, you just multiply by 2.303 to go from your allen function to your log function. So keep in mind the intricate dynamics involved in the equation and how it connects your self potential in standard conditions to nonstandard conditions

Nernst Equation Calculations

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Nernst Equation Calculations 1

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So here we need to determine the cell potentials of the following concentration cells here, it's written liners on notation for us. So we're gonna do here is we're going to deconstruct it and give us both of our half reactions. So here we have our anodes are physical break and our cathode for the an ode we have our silver solid producing silver ion here. The charge is zero and the overall charge here is plus one. So I add one electron to this side. For my other half reaction, I have copper to ion producing copper solid. The overall charge here is plus two. The overall charge here is zero. I add electrons with the more positive side. So I had two electrons to this side so that it becomes zero overall, just like this side is zero. Overall notice that my electrons do not match. They need to match because we have the transferring of an equal number from one electrode to another. So I'd have to multiply this half reaction times two. Which gives me two silver solids produces. Here, we should do reversible arrows, gives me two silver Plus two electrons. So these are reversible reactions. So we put double arrows. Alright, so now we have our two half reactions. So the first one deals with the annual where oxidation occurs. So here that would represent this potential for the silver. Now we're going to say that e minus which is the potential of our A note equals the potential under standard conditions minus 0.5916 over N. Times log of Q. So notice here we're using the equation because we have a concentration at least one concentration that is not equal to one molar. So these are nonstandard conditions. So we're gonna plug in 0.799 volts minus 0.5916 volts divided by number of electrons transferred Which is two electrons. And then remember Q equals products. Overreact ints. So our products are two moles of silver ion. Remember we ignore solids and liquids. So this reacting, we would ignore it because it's a solid. So here it would just be log of a G L A G squared G positive squared. Which would be 20.10 squared because again Q. Here equals products Overreact ints. It ignores solids and liquids. So cue here will just be a G positive squared When we punch all that in. That gives us a number of .976V. Now we do the same thing for the compartment that is our cathode. So that's gonna be E positive equals E. Positive Under standard conditions minus 0.5916 volts divided by n. Times log of Q. Here, my positive understanding conditions is 0.339 volts minus 0.5916 volts divided by two electrons transferred. And then Q. Is products overreact ints. So based on this half reaction, our product is a solid so we're going to ignore it. So that's one over and then we have copper two ion. There's just a coefficient of one here. So we don't have to raise that concentration to any power. So it would be log of one over the concentration of c. U. two plus which is one molar. realise here that this just becomes log of one divided by one. So this is one inside of here, log of one is equal to zero. So all of this drops out. So this is just .339V. So we've found the self potential of the anodes of the cathode. Now we can find out the overall cell potential if we want. So overall cell potential equals cathode minus an out or positive minus negative here. So that's 0.339 volts minus 0.976 volts. Which gives me negative 0.637 volts. As our final answer. What this value is telling me is because my self potential overall is less than zero. This is a non spontaneous process which would mean that the only way that silver would be oxidized by copper is if a battery were used because that battery would force the electrons to leave the silver and go towards the copper electrode. So this would represent an electrolytic cell because it's non spontaneous. Now that you've seen the basic setup for a question like this attempted to example, two try to approach to determine which one is the katha. Which one is the anodes try to set up the nurse equation for each compartment of the electoral of the electrochemical cell, and from there determine what the overall cell potential is and describe what it means in terms of spontaneity, based on the overall cell potential that you determine. Once you do that, come back and see how your answer matches up with mine.

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Nernst Equation Calculations 1

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So here we need to determine the spontaneity in self potential based on the given cell notation. Alright, so here we have our platinum solid which represents our inert electrode and we're using this because here in our compartment we have the presence of KBR So we have bromide becoming B R two. And here they're not separated by a phase diagram because this is bromide dissolved in a solvent and the solvent happens to be bro me then on this side where we have the cathode, we have iron two ion becoming iron solid. So that is the particular ion that's undergoing reduction. So from this information, we're going to write our two half reactions. So we have B r minus acquis produces br two liquid here. Again, they're not separated by phase boundary because they're in direct contact with one another. Um They're mixed together because bromine is just dissolved in bromine, bromine liquid. Here we have F E two plus gives me f E solid. Alright, now we have to determine them of electrons involved. So here we have a plus two charge and here we have zero. So we have electrons with more positive side. So that's two electrons. Then on this side here it's -1 and here at zero here, that would mean one electron on this side. But the electrons don't match. So they have to multiply by two. So this here is actually two electrons and this is to what this is saying is that we have to bromide ions and they connect together to give me br to liquid and they do this by releasing an electron each This is my actual balanced half reaction. All right now we have this as our an 02. That e negative equals the negative. Under standard conditions minus 20.5916 volts divided by N times log of Q. And then this is a positive equals E. Positive under standard conditions minus 0.5916 volts divided by n times log of Q. Alright, so now we plug in the values that we know. So the negative would be 1.078V -15916V divided by number of electrons transferred which is to and then log of Q. Q equals products over reactant. It ignores solids and liquids. The product is a liquid so it's ignored. So that's just one over BR -2. So that would be one over 10.1, which is going to be squared Here when we plug that in. That gives me .960V over here. E positive Under standard conditions is negative 0.440 volts minus 0.5916 volts divided by number of electrons transfer which is two and then Q here with equal products over reactant signore the product because it's a solid Divided by FE two plus. So that'd be one divided by one. All of this cancels out because log of one is equal to zero. Now that we have those values, we can find the overall cell potential, so overall cell potential equals the positive minus the negative, So that's negative .440V -160V, which gives me negative 1.40V. So as the just like the example up above, we got a value that's less than zero. So we can say that this process is a non spontaneous process and this represents an electrolytic cell. Remember for it to be spontaneous, you would want your overall cell potential to be greater than zero by being greater than zero, would be spontaneous and represent a galvanic or voltaic cell.

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Nernst Equation Calculations 2

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consider a standard voltaic cell based on the reaction of two moles of H plus ions reacting with one mole of 10 solid to produce one mole of 10 to ion plus hydrogen gas here, it says which of the following actions would change the E. M. F. Of the cell. So E M F. Is referring to our electro motive force which is basically our self potential here. Because we're talking about increasing or decreasing the ph that means that we're adjusting the concentration of H Plus. So we're assuming here that we're not dealing with the standard concentration of one molar. Therefore we're talking about cell potential under nonstandard conditions. So here we're gonna say using the nerds equation that are self potential under nonstandard conditions equals our self potential under standard conditions minus 0.5916 volts divided by N times log of Q. Remember Q is our reaction quotient? It equals products. Overreact ints we ignore solids within our expression, so it would be the concentration of S and two plus hydrogen here is a gas, we normally put p there to represent pressures involved for that gas divided by H. Plus and because there's a two here and B squared, we don't include the tin because it's a solid. Now we're talking about adjusting the ph we're talking about changing concentrations as well as pressures. So for the first one it says increasing the ph of the um at the cathode. Alright, realize here that whether we're increasing or decreasing the ph that means that we're having a direct impact here on the concentration of H plus, Remember that there is an inverse relationship between our reaction quotient and the overall cell potential. If our Q. Increases, that means that our overall cell potential will decrease and vice versa. So they're inversely related to each other. So doing either one of these things will affect the concentration of H. Plus. Therefore change the value of Q. And therefore change the value of our self potential or our electro motive force. If we were increasing the ph, that would mean that we're dropping the concentration of H. Plus. Because remember if you increase the ph you're becoming more basic. So this H plus would decrease which would cause an increase in Q. Which would therefore cause a decrease in your self potential. This one here lowering the ph would increase the concentration of H. Plus which would then four decrease the constant, will decrease our reaction quotient and increase our self potential. Next increasing the concentration of sn two plus at the anodes. So you're increasing the numerator here in the expression which will increase Q. And therefore decrease your self potential, increasing the hydrogen gas pressure at the cathode. So again here you're increasing H. Two now so you're increasing Q. Which will lead to a decrease in your self potential. So here the answer would be E. Because all of these actions would result in a changing of your E. M. F. Value. Just realize here because we're changing concentrations were dealing with non ideal conditions or nonstandard conditions. Therefore, we introduce the nurse equation and try to relate our reaction quotient Q to the overall cell potential.

5

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Nernst Equation Calculations 2

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Mhm. So here it says, consider the following half cell reaction at 25 degrees Celsius will be the value for e the half cell potential. For standard conditions for the reaction. Here, we need to notice that we have reversed the reaction. Our product is now a reactant and our reactant are now products. So we've reversed the reaction in addition to that, realized that we've multiplied by two. So we've multiplied this reaction by two. Now, when it comes to these changes, we need to understand what will the effect be on my standard cell potential here, realize that you can multiply or divide your half reaction by any number that will not cause a change in your self potential. So yourself potential would still stay the same. But if you reverse your half reaction, that reverses the sign of your cell potential. So by reversing it now it's going to become positive .260V giving us option B as an answer. Multiplying or dividing by any number does nothing to the cell potential. So we would say that this could not be an answer. Nor this. So, again, reversing the reaction reverses the sign for yourself potential. Multiplying and dividing by any number causes no effect