For more information, please see full course syllabus of AP Chemistry

For more information, please see full course syllabus of AP Chemistry

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### Cell Potential & Concentration

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Cell Potential & Concentration 0:29
- Example 1: Question
- Example 1: Nernst Equation
- Example 1: Solution
- Cell Potential & Concentration
- Example 2
- Manipulating the Nernst Equation
- Example 3

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### Transcription: Cell Potential & Concentration

*Hello, and welcome back to Educator.com; welcome back to AP Chemistry.*0000

*Today, we are going to continue our discussion of the relationship between cell potential and thermodynamics.*0003

*We are going to talk a little bit, though, about cell potential and concentration.*0010

*Let's start off with a little bit of a qualitative discussion--Le Chatelier's Principle--and then we will quantify it.*0015

*We will introduce a new equation--a very, very important equation in chemistry and physical chemistry and physics, called the Nernst Equation.*0021

*Let's just go ahead and get started here.*0027

*Let's just start with an example.*0031

*Example 1: For the following cell reaction--for the cell reaction which is: 2 aluminums, plus 3 manganese 2+ ions, going to 2 aluminum 3+ ions, plus 3 manganese ions--the standard cell potential for (let me write it a little bit--get a little bit more room here) the cell is equal to 0.48 volts.*0036

*OK, so under standard conditions: that is what this little circle on top means--so when you see that circle on top, you are talking about standard conditions.*0079

*What that means is that all of the ions are in 1 Molar concentration; any gases that are present--1 atmosphere pressure; temperature--25 degrees Celsius; so, standard conditions.*0086

*Well, what happens to the cell potential if, maybe, this is not 1 Molar, and this is not 1 Molar--if this is 10 Molar and this is 3 Molar?*0100

*What happens--does it reduce the cell potential (in other words, does it make it want to go that way), or does it increase the cell potential (does it make it more spontaneous as written)?*0109

*OK, so let's say: If...or actually, let's say: What happens to a cell when the conditions are as follows (in other words, nonstandard)?*0119

*What happens if the aluminum ion concentration is now set at 2.0 Molar (moles per liter), and the manganese ion concentration stays at 1.0 Molar?*0154

*OK, so again, this cell potential that we calculated is based on standard conditions (1 Molar concentration for each); but what happens if we actually arrange this in such a way, and in the anode compartment--the oxidation--we actually put 2 Molar aluminum ion, and leave the cathode compartment 1 Molar in the manganese ion concentration?*0170

*Is this going to drop the cell potential?--is it going to raise the cell potential?--what is it going to be?*0199

*Well, raising the cell potential means pushing the equation to the right; dropping the cell potential means moving the equation to the left; that is what it means in terms of cell potential.*0204

*Think of this sort of as an equilibrium constant kind of thing; it rises as you move to the right.*0216

*OK, well, Le Chatelier's Principle tells us immediately what is going to happen.*0221

*If I increase (this is sort of normal) the concentration of the aluminum ion concentration, the system is going to want to adjust in such a way to bring it back down.*0226

*In order to bring this back down, it has to shift the equilibrium to the left.*0237

*When it shifts the equilibrium to the left, it's going to drop the cell potential.*0241

*So, under these conditions, now the cell potential (and notice: we no longer have a little 0 on top of it--that little degree sign--because now it's no longer standard) will decrease.*0245

*This is our qualitative: we know from Le Chatelier's Principle that this is going to happen--it will decrease.*0259

*Now, we want to attach some numbers to it: how much is it going to decrease--is it going to go to .28? is it going to go to .36? is it going to go to negative? what is going to happen?*0264

*Or is it going to go to like .46, .45--is it going to be just a little bit of a drop?*0274

*So, now that we know qualitatively what is going to happen, let's see what is going to happen quantitatively.*0278

*OK, recall: from thermodynamics, we had this following relation: we said that the free energy change of a particular reaction is equal to the standard free energy change, plus RT times the ln(Q), and Q is the reaction quotient.*0283

*Remember, it's the same as the equilibrium expression: products over reactants--except at any given moment.*0301

*OK, well, we also know from our last lesson that ΔG is equal to -nF times the cell potential, right?*0306

*We will also give the standard version of that: ΔG standard equals -nF, for the standard cell potential.*0320

*Well, let's go ahead and put this into here and see what we get.*0329

*We get -nFE _{cell} is equal to -nFE_{standard cell}, plus RT ln(Q).*0338

*Let's divide both sides by negative nF, by that thing right there; and we get that the cell potential is going to be the standard cell potential, minus RT over nF, times ln(Q)--a profoundly important equation.*0354

*This particular equation is called the Nernst Equation, and it establishes a relationship between the cell potential of a cell, the cell potential of the cell under standard conditions, and the reaction quotient, which is the concentrations at which you are running that particular cell (which are nonstandard--not 1 Molar/not 1 atmosphere pressure/not 25 degrees Celsius...things like that).*0379

*This equation is very, very important; it tells us when things are not normal (not standard)--can I still calculate the cell potential of that galvanic cell?*0411

*Of course I can; there it is; so let's see what we can do.*0419

*Let me write down what it actually is: it expresses a relation between cell potential and the concentrations of the cell components.*0427

*That is it--that is what the Nernst Equation does.*0463

*So, for our first example, we had 2 aluminum + 3 manganese 2+ (manganese is not 3+ here...2+) going to 2 Al ^{3+} + 3 manganese (so aluminum becoming aluminum ion; manganese ion becoming manganese; aluminum oxidizing; manganese reducing).*0466

*That cell potential standard, we said, was 0.48 volts.*0491

*We said that the aluminum ion concentration was 2.0 Molar (right?), and how about if we...let's go ahead and change the manganese 2+ concentration; instead of leaving it as 1, let's do 0.5 Molar.*0496

*So, let's change both concentrations.*0512

*Well, Q (the reaction quotient) is going to be products over reactants; these are solids, so they don't show up in the equilibrium expression--the reaction quotient expression.*0515

*So, it just becomes: Al ^{3+} squared (right?--we have to include the stoichiometric coefficients), over Mn^{2+} cubed.*0529

*Therefore, we have the Nernst equation, which is E _{cell}=E_{standard cell} - RT/nF ln(Q), which equals 0.48 (that is our standard cell potential right there), minus R (which is 8.3145--remember, we don't use the .08206; we use the 8.3145 when dealing with thermodynamic considerations)...*0543

*Temperature is the absolute temperature, so 25 degrees Celsius--I'm sorry; I should have specified--this was at 25 degrees Celsius--the temperature can be anything; it's not a problem; so 25 is 298 Kelvin)...*0575

*The number of moles that were transferred was 6 (right?--the number--that is for the balanced equation: 6 moles of electrons were transferred, because 6 moles were what had to be canceled)...*0590

*Moles, times the farad, which is 96,485 coulombs per mole, times the logarithm of 2.0 (the concentration of aluminum) squared, over the concentration of manganese cubed...*0602

*Then, when we do the math, we end up with 0.48; and this thing--this whole thing--becomes 0.01 to one decimal place.*0621

*We end up being equal to 0.47 volts.*0632

*Under standard conditions, the cell potential is 0.48 volts; if I construct this galvanic cell with a concentration of 2 molarity aluminum ion and .5 molarity manganese ion, and if I measure the cell potential, it has actually gone down--not a lot, but it has gone down.*0639

*.47: so, I can use concentration to control voltage; that's very, very important.*0658

*This, of course, confirms that...remember, we said that the E _{cell} is going to be less than the E_{standard cell} qualitatively; now we see it quantitatively.*0669

*We have a number that demonstrates that .47 is below .48.*0678

*The reaction has shifted to the left.*0682

*OK, now let me make a statement which...I think I have maybe mentioned it a couple of times, but let's formalize it.*0686

*E _{cell} is the maximum potential before any current actually flows--before any electrons flow--before any current flows.*0696

*OK, once we open the circuit and allow electron flow, the electrons flow spontaneously (that is what a positive cell potential means--that means, once you open the circuit, it will spontaneously flow in the direction as written), and the cell discharges until it reaches equilibrium.*0721

*At equilibrium, the E _{cell} is equal to 0.*0783

*Here is how I would like you to think about it: Let's just go ahead and take the cell that we have been dealing with--so I have 2 aluminum, plus 3 manganese, going to 2 aluminum ion, plus 3 manganese metal (that's not an ion), and we said that the standard cell potential for that is equal to 0.48, under standard conditions.*0793

*It's different if the concentrations are different, which we just did in the problem.*0825

*Here is what this means: this measurement is a measurement of, again, potential.*0829

*It is what could happen; once we open the circuit and allow the electrons to actually flow through the wire from aluminum to manganese, turning aluminum to aluminum ion and manganese ion to manganese metal--once that happens, as the electrons start flowing, think about it this way.*0835

*I'll go...it is as if there are a whole bunch of electrons that are just piled up over here, that really, really want to get over here; that is the whole idea--that is what potential measures.*0852

*It measures the desire for these electrons to actually go across the wire to the other compartment.*0865

*Well, here is what happens: like anything else, as electrons flow over here--well, now these electrons--the number of electrons diminishes on this side; here, the number of electrons rises.*0873

*Well, at some point, there is going to be an equality.*0886

*It is going to...this is what a battery is: a battery is a collection of electrons on one side of that battery; and once you close the circuit, it allows electrons to flow.*0893

*That electron flow does work for you; but at some point, the battery is going to die.*0905

*A dead battery just means that this reaction has reached equilibrium; this potential is a measure of its capacity--of how badly it wants to reach equilibrium.*0909

*Once we open the faucet, once we open the circuit, the electrons start to flow.*0921

*Well, as they start to flow, this cell potential drops: .47, 46, 45, 43, 42, 41; that means--it is because there are fewer electrons, because all the electrons are moving over here.*0926

*But, at some point, it's going to reach a point where the electrons and everything else have reached an equilibrium point, and now nothing else moves.*0939

*No electron wants to go this way; no electron wants to go that way.*0946

*When that happens, the cell potential is 0, because that is what cell potential measures--it measures the potential of electrons to move from one compartment to the next.*0950

*So, at equilibrium, the cell potential is 0--just like, at equilibrium, free energy equals 0.*0960

*Do you remember when we did thermodynamics?--same thing.*0972

*Cell potential; free energy; same thing.*0975

*All right, so now, in other words, when the cell does that...that means a dead battery, basically.*0981

*So, a dead battery means that the reaction has reached equilibrium; electrons are not flowing anymore in any direction.*0987

*OK, let's see: let's see if we should do...yes, let's go ahead and do this one.*0998

*OK, Example #2: Describe (give us a little bit more practice in these equations...) the cell based on the following half-reactions.*1005

*We have (and again, don't let the complicated nature of these fool you--it's very, very simple what is going on, OK? Just follow along, and everything will be good) Cr _{2}O_{7}^{2-} + 14 H^{+} + 6 electrons goes to Cr^{3+} (2 of them) + 7 H_{2}O; and the standard reduction potential is 1.33 volts.*1031

*We also have zinc 2+, plus 2 electrons, going to zinc metal, and that is 0.76, negative (I'm going to make this a little clearer).*1064

*The standard reduction potential is equal to -0.76 volts.*1080

*Notice how I have written these: I have given the half-reactions as reductions--reduction potentials: this is what you see in a table of reduction potentials.*1085

*I have to decide which one gets flipped and gets oxidized.*1093

*So, in this particular case...well, this one is 1.33; this one is -.76; clearly, this one is bigger than this; this stays the same--this gets flipped.*1098

*But we are not done yet, because we are going to do this under nonstandard conditions.*1111

*So, the Cr _{2}O_{7}^{2-} concentration is equal to 1.80 molarity; the hydrogen ion concentration is equal to 0.60 molarity; the chromium 3+ concentration is equal to 0.2 molarity; and the zinc ion concentration is equal to 0.1 molarity.*1114

*Clearly, nonstandard conditions: so, we said Cr _{2}O_{7}^{2-} + 14 H^{+} + 6 electrons goes to 2 chromium 3+, plus 7 H_{2}O.*1142

*The standard reduction potential is 1.33 volts.*1161

*OK, now we said that the zinc...since it was less, we have to flip it; so when we flip it, it becomes zinc metal, going to zinc ion plus two electrons; and because we flipped it, it becomes a positive 0.76 volts.*1167

*Well, now we need to equalize the electrons: 2 and 6, so I'm just going to multiply this equation by 3.*1184

*When I multiply by 3, the equation I get is: 3 zinc metal goes to 3 zinc ion, plus 6 electrons.*1191

*And now (let me do this in red), I have (oops, let's try...there we go; that is better) this equation, and I have this equation, and I'm going to cancel 6 electrons on the right with 6 electrons on the left.*1201

*I'm going to be left with: Cr _{2}O_{7}^{2-} + 14 hydrogen ion + zinc metal goes to 2 chromium 3+, plus 7 H_{2}O, plus 3 zinc ion.*1219

*Chromium, here, has gone from chromium 6+ to chromium 3+; it has been reduced.*1241

*Zinc has gone from zinc 0 to zinc 2+; it has been oxidized.*1249

*The E _{cell} for this reaction...well, I just add these two together--this one and this one--because I added these reactions together; it's 2.09.*1253

*2.09 volts--that is a lot for an oxidation-reduction reaction.*1263

*That is my reaction; this is my cell potential; now I'm going to write the actual line notation for this.*1269

*Zinc is the anode; it is the thing being oxidized.*1276

*Zinc is going to zinc 2+.*1280

*In the other compartment, I have Cr _{2}O_{7}^{2-} turning into Cr^{3+}; and here, because neither one of these--the dichromate ion or the chromium ion--are a conducting metal (they are in ionic form, so they are still aqueous), I'm going to go ahead and use an inert platinum electrode.*1284

*This is my line notation, describing this reaction with this cell potential.*1305

*My anode is zinc; my cathode is platinum dichromate chromium; everything is good--that is a nice, complete description.*1311

*Now, let's go ahead and actually calculate the real cell potential, based on these concentrations--because you notice: this is the standard cell potential that we got from our table of reduction potential values by adding.*1319

*It is not the actual potential of the cell with these concentrations.*1333

*We are going to use the Nernst Equation.*1337

*E _{cell} is equal to E_{cell standard} - RT over nF ln(Q).*1340

*OK, here is what is interesting: Q...if you look back on that equation, zinc is an ion; chromium is an ion on the product side; dichromate is an ion; and hydrogen is an ion.*1353

*All four of those things actually show up in the reaction quotient.*1368

*Water doesn't show up--it's a liquid; and the zinc metal doesn't show up, because it is a metal.*1374

*The reaction quotient is the following: zinc ion raised to the 3 power, chromium ion raised to the 2 power, divided by dichromate ion raised to the 1 power and hydrogen ion (you are going to love this one) raised to the 14 power, because its coefficient is 14; it's kind of crazy.*1381

*OK, so that equals 2.09 (that is our standard reduction potential), minus R (which is 8.3145), times temperature (again, this is happening at standard temperature...right, if we didn't specify the temperature, then you can just assume that it's standard--25 degrees Celsius; 298 Kelvin; the concentrations are nonstandard, but the temperature happened to stay at 25)...*1406

*OK, the number of electrons transferred--go back to the equation that you just looked at: when you balanced the equations, how many moles of electrons did you have to transfer--did you have to get to make them equal? 6.*1435

*So, 6 moles of electrons, times 96,485 coulombs per mole of electrons, times the logarithm of (OK, here is where it gets interesting) 0.1 cubed, times 0.2 squared (these are the concentrations of the ions--remember, we listed them at the beginning of the problem), over 1.8, times (this is the best) 0.6 to the power of 14 (that always kills me).*1447

*We end up with 2.09 + 0.02.*1477

*So, we end up with a cell potential of 2.11 volts.*1484

*This particular cell has a greater cell potential than the standard cell potential, which is 2.09.*1491

*It is actually going to be even more wanting-to-go-forward; you are going to get more work out of it.*1498

*2.11 Joules per coulomb: for every coulomb of charge transferred, the maximum work that you are going to get out of this is 2.11 Joules.*1505

*OK, now let's go ahead and do a little bit more manipulation on that equation.*1515

*Let me rewrite the equation one more time.*1520

*Let's go back to blue.*1522

*We said that E _{cell} is equal to E_{standard cell}, minus RT over nF times ln(Q).*1525

*Oh, by the way, I should let you know, because I don't want this to confuse you if you happen to look in your books and see this Nernst equation looking slightly different than what it is that I have written here.*1534

*R is a constant, 8.3145; temperature--if we keep the temperature at 25 degrees Celsius, that is 298 Kelvin; F is a constant--it is 96,485 coulombs per mole; n is the only thing that actually changes.*1545

*A lot of books--what they do is: they take this number, this number, and this number, and they go ahead and combine them into one number, and then they change this natural logarithm to a common logarithm, base 10.*1563

*So, you will see this equation written as...I don't even know...it looks something like this, if I'm not mistaken.*1577

*I don't like it like this, because I think it's too much manipulation, but I think it is something of the nature of -0.0592, divided by n, times the log of Q (this is a base 10 log, not natural log).*1585

*I like to stick to the equations as-is, and I don't like to combine my constants; I like to see what constants I'm working with.*1598

*It doesn't matter; it's just a simple little multiplication and division problem.*1605

*This confuses me, because I'm not sure where this came from; here, I see everything that is part of the equation; work with this.*1609

*ln, log: at this point in your studies, ln and log--you should have no fear of this.*1616

*It is just a natural logarithm (base e) versus a common logarithm (base 10).*1621

*I would say don't learn that equation: this is the one you want to work with--the one as written--the one as derived.*1626

*OK, back to what we were doing: now, we said: at equilibrium, the E _{cell} equals 0.*1635

*So, if we set 0 over here, we get 0 is equal to E _{standard cell} minus RT over nF ln(Q).*1644

*Here you go: E _{cell} equals -RT over nF, and at equilibrium, Q becomes K, right?--it is equal to K at equilibrium, so this is ln(K).*1658

*This is our final equation that is really, really important.*1673

*This equation is the analog of this equation from thermodynamics.*1678

*Well, because--remember, at equilibrium, free energy is related to the equilibrium constant, well, free energy, under electrochemical cell situations, is just the cell potential.*1691

*There is a relationship between the equilibrium constant and the actual standard cell potential; this gives us a way, again, of finding experimentally an equilibrium constant for a reaction.*1707

*Let's do our final example.*1723

*Example: Calculate K, the equilibrium constant, for the above reaction.*1728

*OK, so let's write the reaction again: we have: Cr _{2}O_{7}^{2-} + 14 H^{+} + 3 Zn going to 2 Cr^{3+} + 3 Zn^{2+}, and it's 7 H_{2}O, right?*1740

*And the standard cell potential for this is equal to 2.09 volts: 2.09, not 2.11; 2.11 was the cell potential under nonstandard conditions--whatever we happened to...the different concentrations.*1762

*We are using standard; so we said that the standard cell potential is equal to RT over nF ln(K).*1777

*We get ln(K) (I'm sorry; I flipped them around) nF E _{standard cell}, divided by RT.*1792

*OK, that is equal to...6 moles were transferred; 96,485 coulombs per mole; 2.09 Joules per coulomb; over 8.3145, times 298 Kelvin.*1806

*We get: K is equal to e raised to the power of this thing, which is equal to 488.*1836

*I'm not even going to write this number down; I'm just going to leave it at e to the 488.*1852

*Huge--this equilibrium constant is huge.*1857

*You know what that means--a huge equilibrium constant: that means the reaction is so far to the right--that means, once this reaction is done, there is virtually none of this left; everything has turned to chromium 3+; everything has turned to 7 zinc +; and there is a whole bunch of water floating around in that cell.*1861

*Now, at equilibrium, yes, there is always a little bit of reactant left; but in this case, you know that this is so huge that it is not even noticeable--that equilibrium is actually a theoretical state.*1880

*This is definitely completion at its best.*1892

*This is very typical: large K values, large equilibrium constants, are very, very common for oxidation-reduction reactions...very common for redox reactions.*1895

*When you take a piece of wood, and you burn it, is there any wood left at the end?*1916

*No; that is an oxidation-reduction reaction; you are taking organic matter, and you are converting it; you are oxidizing the carbon.*1924

*You are turning it into carbon dioxide and water.*1931

*When you ingest the food that you ingest, and your body converts it to glucose so that it can oxidize that glucose aerobically (under oxygen conditions), and it converts it to carbon dioxide and water, it is an oxidation-reduction reaction.*1935

*That carbon in the glucose is being oxidized to carbon dioxide.*1950

*So, the fact that these things release so much energy is confirmed by the fact that the equilibrium constant is going to be very, very huge.*1959

*You are not going to have a lot of reactant left over; it is going to be all product; that is the whole idea.*1972

*So again, we have all of these wonderful, wonderful equations that we have been working with; we have the thermodynamic versions of those equations; we have the electrochemical version of these equations; and they are all just expressing a relationship between these things.*1977

*Free energy; cell potential; equilibrium; all of these things are tying together.*1993

*Remember when we started back discussing equilibrium: we moved from equilibrium; we moved to thermodynamics; and now, we closed off with electrochemistry.*1998

*So, we only have one more thing to discuss regarding electrochemistry; the next time, we are going to talk about something called an electrolytic cell.*2007

*We have been talking about a galvanic cell, which is a spontaneous process; well, what if I don't want to do the spontaneous?*2015

*What if I want to push energy into the system and make the electrons go in the direction opposite to where they want to go normally?*2020

*If they want to flow this way, I want to push them back the other way and run the other reaction.*2028

*Well, that is what we are going to...that is called an electrolytic cell--a very, very important process, industrially and for all other things.*2033

*It is how you recharge your batteries: when a battery discharges, you push the electrons back through the wire and back into the original compartment they came from, like a NiCad battery.*2040

*We recharge the battery, and now the battery is ready to go again.*2050

*Until then, thank you for joining us, and we will see you next time at Educator.com; goodbye.*2054

1 answer

Last reply by: Professor Hovasapian

Tue Mar 1, 2016 1:44 AM

Post by RHS STUDENT on February 29, 2016

Sir, from the two examples you have given to us, the voltage of the cell seems to be inversely proportional to the concentration of ion in either compartment. Is it because with less ion presence in the solution, less ion "move"to the other compartment to balance the charge in the other compartment?

1 answer

Last reply by: Professor Hovasapian

Fri Feb 26, 2016 3:28 AM

Post by Tammy T on February 21, 2016

Hello professor Hovasapian,

-At27:55, would E*cell= +(RT/nF)lnK instead of = negative that whole expression?

-Also, I don't understand how at equilibrium, Ecell=0V but E standard cell E*cell does not equal 0V. Wouldn't E*cell the same galvanic cell as in Ecell but under standard condition?

Thank you for your lecture!

Bests,

Tammy T