# 19.5 Nonstandard Cell Potentials the Nernst Equation

- 19.1 Oxidation Reduction Reactions and Oxidation States
- 19.2 Balancing Oxidation Reduction Reactions
- 19.3 Galvanic Cells
- 19.4 Standard Cell Potentials aka emf or Voltage
- 19.5 Nonstandard Cell Potentials the Nernst Equation
- 19.6 Reduction Potentials and the Relationship between Cell Potential, Delta G, and the Equilibrium Constant
- 19.7 Electrolytic Cells
- 19.8 Electrolysis Calculations

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Okay. The next question. Just like when we calculate Delta G we have a way to calculate Delta G standard. And where do you usually get the Delta G standard values? Out of a book. It's where you also get the E cells. You look at these tables of reduction potentials and calculate it from the values out of a book. That's where I got them. So, but those are the standard values. But what if you're not under standard conditions? Well then you have an equation. You look up the value in the book and then use an equation to get you a fudge factor. We have the same thing here. We're going to have a fudge factor. So, and the equation is called the Nernst equation, as E equals E standard minus RT over NF, natural log of Q. It's called the Nernst equation. But we're usually a little nicer to you than this. What you're usually given is not this rough form. We usually assume that the temperature is 298 Kelvin, F here is called Faraday's constant, we'll go talk about it later, but it's 96,500 and R here is the gas constant. And if we're always doing this at 25 degrees Celsius, 298 Kelvin, then that's a constant temperature. And what's a constant times a constant divided by a constant? A constant. And so sometimes we'll give this to you in this form. We'll say E equals E standard minus 0.0257 over n, natural log of Q. And it makes it a little bit easier in the plug-in and chug-in. However, natural logs kind of suck. And they're not so bad now because we have calculators. But back when they were doing this math, they didn't have calculators. And using a natural log really sucks. But for log base 10, regular logs, they had tables and they could calculate any log just by matching it up on the table. And so originally, what they actually did, is so that they could use a log base 10- turns out that a log base 10 is the same thing as natural log with it with this extra factor of 2.303- and when you factor that in you get negative 0.05916 over n, log base 10 of Q. Doesn't really matter which version you use. They're both on your sheets. It could be given to you in either way on your on your exam on the front page. Which is why I've given you both. I'm personally going to use this one because I like log base 10. I can do a lot of those in my head. I can't do natural log for anything in my head. Okay. So the next question on your handout. So it says, first thing on the back page there, "Calculate the cell potential, from Part A- right here- from the last question." So calculate the cell potential when the conditions are changed a little bit. And it's saying the conditions are now the zinc ion concentration is 0.001 molar, and the cobalt ion concentration is 0.1 molar. So how would I know that this negative 0.48 volts doesn't apply? Well, because the negative 0.48 volts is for what conditions? Standard. Are these standard? What would these have to be if I want it to be under standard conditions? They'd have to be one molar both. And they're not. So this doesn't apply. The standard value under these conditions is- our value is not going to be negative 0.48. That's the standard value. We're not under standard conditions. Okay. We need to use this to get there. And so this equation, the way it works here, we got our E, we still use that standard value and that standard value is still negative 0.48, that we use from the data that came out of a book. But now we got this extra fudge factor to take into account 0.05916 over n. Now n. What does n usually represent? Moles. In this context, n means moles of electrons transferred in the balanced reaction. So if you look here, let's use our lovely green marker, for zinc two plus turning into zinc, how many electrons would that be? That's two electrons gained. For cobalt two, or cobalt turning the cobalt 2 plus, how many electrons is that? Two. That's two electrons lost. In this case two electrons gained, two electrons lost. this was two electrons transferred. And so we're going to plug in a big fat two. If you recall, the reaction we balanced way back at the beginning of the day- beginning the night rather- and one of the half reactions, was five electrons the other half reaction was two electrons, how many electrons overall did we balance that out to? So one half reaction was five electrons that were being gained, and the other one was two electrons lost. And we balanced it out to the least common multiple of 10 electrons. 10 electrons in that case would've been n. So n is total # of electrons transferred to the balanced reaction, whatever it's balanced to. Well in this case it's two. Great. Makes it easy when it's something like this. So it's two. And then we've got the log term. In this case log. And if we look at the balanced reaction, again you should know that ions are aqueous but elemental metals, as long as it's not mercury, are solids. And so what's Q for this reaction? What's the expression for Q? Cobalt 2 plus over zinc two plus. Which in our case solids again don't show up. In our case cobalt 2 plus is 0.1 molar. Zinc 2 plus is 0.001 molar. Make sure, notice I didn't have any coefficients to worry about- the coefficients were 1, but notice if I had coefficients I might have to square some numbers in here or something like that. But we don't have to in this case. Cool. And from here it's plug and chug. So can anybody work this all out and get me a nice, lovely answer? Awesome. Cool. I had nice round numbers for powers of ten, and that's when logs are a little bit easier. So, but- awesome. It's just plug and chug. Let's the calculator do the work for you. So that's the Nernst equation. This is how we calculate the potentials when we're not under standard conditions. You still need that standard value. We already had it calculated, but if we hadn't already had it calculated, we'd have to calculate that first and then add the fudge factor in there as well. Question. Equations totally gonna be on the front page of your exam. For sure. You do not have to memorize this equation at all. Totally provided. Moles of electrons transferred in the balance reaction. So here, two electrons gained, two electrons lost, that's two moles of electrons transferred. So yeah. If you notice like one of the ones we did earlier had- In fact let's write it out. We had two Cr3+ plus three Cu go into three Cu2+ plus two Cr. What do we call this one? And if you look for chromium three plus to chromium, how many electrons is that? Yeah. With two of them each, it's six total. For copper going to copper two plus, there's three even, that's at six electrons as well. Six electrons gained there. Six electrons lost there. Six electrons, six moles of electrons transferred over all. N would be six for this reaction. Not a problem. Okay. So we just did the calculations, but we can do more than just the calculations. So if you look, let's say this reaction right here reaches equilibrium. Let's say it reached equilibrium. When it reached equilibrium, what would delta G be? Zero. Guess what the voltage would be? Zero. And again it's not the value you get from a book that goes to zero, it's the non standard value that equals zero at equilibrium. Cool. It goes to zero. So let's say I let this thing reach equilibrium, if I add a bunch of zinc ions in the solution will it still be at equilibrium? No it's not going to be at equilibrium anymore. By adding a bunch of zinc ions, which way is it going to shift? To the right. And when we say "shift right," that actually means the reaction is now spontaneous to the right. And so is the E cell still zero at that point? No. The E cell would now be? Positive. And delta G would now be? Negative. Because it's spontaneous going left to right. Cool. So you should know that anything that causes the equilibrium to shift more right, it makes the voltage get more positive and delta G get more negative. Anything that would cause a shift to the left, exactly the opposite. A shift to the left would make the voltage get more negative and delta G get more positive. So that's qualitatively, conceptually what you need to understand about this principle as well? We can do the calculation and do the math but qualitatively it comes in handy as well. If you guys recall this reaction right here- So. Question you might get conceptually now, might give you the reaction, the balanced reaction, here and they might ask you some things? What would happen to the equilibrium if I added more solid? Which way would the equilibrium shift? It doesn't. Solids and liquids can't shift it. Only aqueous or gas. Notice, solids and liquids don't show up in the equilibrium constant expression. So if I add more solid, it's still at equilibrium. I haven't changed anything. However, though. Which way would the equilibrium shift if I added more I-? To the right. And so the voltage would go up or down? It would go up. It gets more positive. And notice more positive, means the same thing ad less negative by the way. So- but yes. And that's an increase in potential. Anything that shifts right, increases potential decreases delta G. But this chapter is about potential. So what would happen if I increase the amount of Mn2+ in the solution? Shift left. So what would happen to the potential? It goes down. Now the tricky of all tricky questions guys. Think it about before you answer. What would happen to the potential, the voltage, if I increase the pH? Right. If I increase the pH, am I making the solution more acidic or more basic? More basic. So what's happening to him? If I'm making the solution more basic? I'm removing him. And the equilibrium is going to shift to the left and the voltage is going to go down. So be careful. If pH is increasing, H+ is decreasing and make sure you get your shift right. They love doing that to you on something that as H+ on it. They can technically do it with OH as well, but you'll see it much more commonly with H+. Yeah.