AP Chemistry 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 Al³⁺ + 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: Al³⁺ squared (right?--we have to include the stoichiometric coefficients), over Mn²⁺ 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) Cr₂O₇^2- + 14 H⁺ + 6 electrons goes to Cr³⁺ (2 of them) + 7 H₂O; 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 Cr₂O₇^2- 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 Cr₂O₇^2- + 14 H⁺ + 6 electrons goes to 2 chromium 3+, plus 7 H₂O.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: Cr₂O₇^2- + 14 hydrogen ion + zinc metal goes to 2 chromium 3+, plus 7 H₂O, 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 Cr₂O₇^2- turning into Cr³⁺; 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: Cr₂O₇^2- + 14 H⁺ + 3 Zn going to 2 Cr³⁺ + 3 Zn²⁺, and it's 7 H₂O, 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