AP Chemistry Potential, Work, & Free Energy

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

Today, we are going to continue our discussion of electrochemistry, and we are going to talk about cell potential, work, and free energy--the relation between the potential, the work, and the free energy.0004

Before I actually get into that, though, I would like to spend a little bit of time talking about something called line notation.0014

It is a shorthand notation that actually describes a galvanic cell--a voltaic cell.0021

Let's go ahead and take a look at that first, and make sure we have a nice, complete description of this cell, so we are solid with that; and then we will continue on and discuss the relationship between thermodynamics and the electrochemistry.0027

OK, here we go.0041

A complete description of a galvanic cell includes these things...so a complete description of a galvanic cell (also called voltaic cell) contains three things.0044

You will see this list...sometimes it's listed as 4; sometimes it's listed as 5; it depends on how many things you actually want to put together and separate and be specific.0071

I tend to think of it as three things, and you will see why in a minute.0078

OK, so the first thing that it needs is the cell potential; so we have to know what the cell potential is, and that cell potential comes from, remember, the oxidation and reduction reactions, arranged and balanced and added--that is what gives you the cell potential.0082

The individual potentials are called the standard reduction potentials for the individual components.0099

OK, so the cell potential, and balanced equation: so we need a balanced equation, also.0105

And again, that comes from adding the two half-reactions--the reduction half-reaction and the oxidation reaction.0111

OK, #2: We need to know the direction of electron flow (in other words, the designation of what is the anode and what is the cathode).0118

Direction of electron flow...when the circuit is open: OK, so we measure a potential--that doesn't mean we have opened the circuit; we have measured the pressure (that is the pull or the push on the electrons) once we actually open that circuit and allow the electrons to flow.0126

Potential is a measure of that pressure; current means the actual flow of electrons--we will talk more about current when we actually open the circuit in a little bit.0147

The direction of electron flow, which is a designation of anode and cathode...0157

And, as a quick reminder, electrons always flow from the anode to the cathode, because "anode" is where oxidation takes place...so anode to cathode...electron flow.0172

Oxidation takes place at the anode; reduction takes place at the cathode.0190

They both start with vowels; they both start with consonants--that is how I remember it.0197

OK, and 3 (which is probably the most important thing, from a practical point of view): We need the physical nature of each electrode, of each compartment.0202

So, the physical nature--in other words, what makes it up--what components are we using? Are they ions? Are they solids? What is an ion--what is a solid?--things like that--so, the physical nature of each electrode and the ions present in each compartment.0212

Now, when we say "ions," we are talking about the ions that are involved in the oxidation-reduction; we are not talking about the counter-ions.0246

OK--like, for example, if we are dealing with zinc and copper, we put a solution; we put the zinc metal in; one of the compartments is going to contain zinc sulfate; let's say the other contains copper sulfate--we are only concerned with the zinc and copper, because those are the ones that are going to transform.0252

The sulfate is just a spectator ion--it's not going to do anything; those don't matter.0266

OK, and if none of the substances participating in the reaction is a conducting solid, then an inert material (more often than not, it's going to be platinum) must act as the electrode.0272

So, if one of the things involved is not a conducting metal (like zinc--a conducting metal--you can use that as the electrode)--if not--then you will have to use platinum.0325

OK, so let me give the definition of line notation.0333

Line notation is a shorthand for describing an electrochemical cell (in other words, a galvanic or voltaic cell).0340

There is another type of cell called electrolytic, which we will actually talk about next time...actually, not next time--not next lesson, but the lesson after that.0373

That is also considered electrochemical: "electrochemical" is just your general term for it.0381

OK, so let's do an example.0386

Let's go ahead and...you know what, let me--that's fine; I'll do it over here.0389

Example #1: For the reaction that we saw before (the reaction which was 2 aluminum ion, plus 3 magnesium ion, becomes 2 pieces of aluminum metal--by 2 pieces, I mean 2 moles or 2 atoms--plus 3 magnesium ion)--in this case, the aluminum is going to reduce; the magnesium is going to oxidize; and here is what the shorthand notation looks like.0397

It's going to be Mg, single line, Mg²⁺ (oops, let's make this a little more clear), double line, Al³⁺, single line, Al.0430

So, Mg on the left; Mg²⁺ next to it; double line; Al³⁺ and Al; so now, let's talk about what these things mean.0448

Let me actually write it over again: Mg, Mg²⁺, a line, Al³⁺, Al; and it's OK if the lines are slanted; it's not a problem--they don't have to be vertical.0457

OK, here is what is going on: the anode is on the left--is at the left, with the electrode (the actual electrode--the physical metal piece) at the far left.0475

So, this is the anode: it's on the left-hand side; we are demonstrating electron flow from left to right--that is the whole idea: electrons move that way.0500

Anode...and the actual electrode itself, the magnesium metal, is going to be on the far left.0512

The cathode is on the right, with the electrode itself on the far right.0519

Aluminum is the electrode; this aluminum-aluminum ion combination is considered the cathode; this metal--the electrode itself (the physical electrode)--that is on the far right.0537

OK, now, phase boundary: phase boundary is a single line.0550

A phase boundary means--well, we have three phases: solid, liquid, and gas; in this case, the phase boundary is going to be between solid and aqueous, aqueous and solid.0557

Magnesium metal is sitting in a solution that contains magnesium ion; so there is magnesium ion that is in one phase (it's aqueous), and here it's magnesium solid.0566

If you want, you can go ahead and put "aqueous," if you want to be absolutely--make sure that everything is there.0578

You put Mg; you can put a little s subscript for solid; you can put aq; but that is what this means any time you see an ion, unless you are specifically talking about a gas phase ion (which, for most chemistry, you don't).0583

I think the only people that actually deal with gas-phase ions are going to be physical chemists, and since we are not doing high-end physical chemistry here, it's not a problem.0596

We know we are talking about an aqueous solution--we know we are talking about a galvanic cell, a wet cell.0604

OK, so the phase boundary is a single line (sorry about that--it might be nice if I actually finish what I'm writing before I get into a discussion of it)--so this single line here--it's a phase boundary between Mg and Mg⁺, aluminum solution and solid aluminum metal.0610

Now, the compartments are separated by a double line; so you see the double line right there; OK.0628

This double line represents the porous disk, the one that allows counter-ion flow; it represents the porous disk or the salt bridge, depending on what the physical arrangement is.0652

And again, nowadays, a porous disk is what we use most of the time.0670

OK, so let's take a look at another example, Example #2.0675

Now, for the reaction (slightly more complicated reaction--this is the permanganate and chlorate--so let's do): 2 MnO₄^- + 6 H⁺ + 5 ClO₃^- goes to 2 Mn²⁺ + 3 H₂O + 5 ClO₄^-...0680

So, this says that, when you mix permanganate and chlorate in a solution, the permanganate (or the manganese in the permanganate) is going to reduce to manganese 2+; the chlorine is going to oxidize to chlorine 7+ (it's going to go from 5+ to 7+).0711

So, this looks like...so we know that the anode is going to be the chlorate-perchlorate solution, and the anode is going to have the permanganate and manganese solution.0730

Well, neither one of those is actually a conducting metal; so we are going to have to use an inert electrode; so it looks like this.0744

So we write: Pt, single line (that is a solid), and then we write ClO₃^-, ClO₄^-.0751

Now notice: I went from left to right; I didn't write ClO₄, ClO₃; again, we are talking about the direction of...as we go from left to right, what turns into what.0761

So, in this case, ClO₃ is going to turn into ClO₄; if you are not exactly sure what, just look up here: ClO₃ comes first; ClO₄ is next; so this is on the left; this is on the right.0773

That is the anode; the cathode compartment...the MnO₄ becomes Mn²⁺, so this is written MnO₄^-, Mn²⁺.0783

Notice: I didn't include...0794

Oh, I need to finish it off with a platinum electrode; again, none of these is a...these are all aqueous solutions: aqueous ion, aqueous ion...none of these is a conducting metal, which is why I have to use the platinum electrode.0796

This is the line notation for the cell that is based on this reaction.0809

That is what we are doing: this gives us a notation for the actual physical arrangement of the cell that is based on (a galvanic cell that is founded on) this reaction.0815

Now, notice: I didn't do anything about the H₂O, and I didn't do anything about the H⁺.0825

They are just there; they are not actually...I don't want to say they are not involved in the process--they actually make the process happen; yes, they are involved, but they are not directly involved.0831

What is being oxidized is the chlorine, going from chlorine +5 to chlorine +7; here, the manganese is going from manganese +7 to manganese 2.0840

It is only the species that are involved in the oxidation-reduction that we are concerned with.0849

But, we do need the actual reaction itself.0853

OK, and again, if somebody didn't give us the reaction--if they just gave us this line notation--it is not a problem; we can recover the equation--it's very, very easy.0857

You just take a look at a table of standard reduction potentials (the ones that have the list from positive to negative or negative to positive--different books actually put them in different orders).0865

We look up the ClO₃ to ClO₄; we write that half-reaction with its reduction potential.0876

We look up the MnO₄^-, Mn²⁺ reaction--that is in there, and again, it is all written as a reduction--with its electric potential.0882

We decide which one is higher; the one that is higher stays as written; the one that is lower--we flip it, and that becomes the oxidation reaction.0890

We balance it (remember, H's, H₂O's, electrons, everything; cancel; add everything; add this)...well, that is where this comes from.0898

We can go from equation to line notation; we can go from line notation to equation--not a problem.0908

These things are what is important, what is involved in there.0913

OK, so let's go ahead and do one more example here.0917

Example 3: Give a complete description of a cell based on the following reaction...oh, I'm sorry; not based on the following reaction--that is what we are doing; based on the following substances.0925

So here, they are not telling us anything; they are just giving us a couple of substances to work with; we have to do everything--which is nice; we get a nice chance to see the whole process.0957

Cl₂, Cl^-, and Br₂, Br^-: so we want to construct a cell that uses these components.0967

Well, OK: let's see: let's see what is going on.0978

The first thing we want to do is check the table of standard reduction potentials.0982

I'll say "consult" (how is that?): we want to consult a table of standard reduction potentials; and remember, they are all written as reductions--that is why is called a standard reduction potential (a table of standard reduction potentials).0991

Here is what we find: when we look at the Br₂, Br^- combination, we get the following: Br₂ + 2 electrons goes to 2 Br^-.1011

It tells me that the standard reduction potential for that equals 1.09 volts.1027

If I check the chlorine-chloride combination in the table (I just run down the table until I find what I am looking for), it tells me that chlorine, plus 2 electrons, becomes 2 chloride ions, and the standard reduction potential for that is 1.36 volts.1033

Well, which one is higher? 1.36 volts is higher.1052

So, that stays as written: that is the reduction reaction.1055

That means this one has to be flipped; when we flip it, that becomes the oxidation reaction: 2 Br^- goes to Br₂ + 2 electrons.1060

Because we flipped it, the standard reduction potential changes and becomes negative.1079

OK, well, now that we have a reduction reaction and we have an oxidation reaction, we know that chlorine is going to be the one that is going to be reduced.1086

Bromide is going to end up oxidizing to bromine.1095

Well, everything balances; there are no oxygens, no hydrogens; now, we just need to make sure that the electrons balance to cancel.1101

We have 2 electrons on the left, 2 electrons on the right; everything is good--we don't have to do anything; we can just add straight.1109

If not, then we would have to go through that process of balancing the reduction half-reaction, balancing the oxidation half-reaction, adding the reactions (remember that process?)...1116

That cancels that, and my overall reaction is going to be the following--I have: chlorine, plus 2 bromides, is going to be converted into 2 chlorides, plus bromine.1126

The cell potential is going to be the sum of those two, 1.36-1.09; and I end up with...it looks like it's going to be 0.27.1138

And again, I hope you check my arithmetic; I'm notorious, notorious, for not being able to do arithmetic.1160

OK, so here we go: that is the first part--remember, we said "a complete description."1165

We have our balanced reaction, and we have our cell potential; so we are good--that is the first part.1170

Now, we want to designate the anode: the anode is where oxidation takes place.1176

What is going to be oxidized is: the bromide goes to bromine (right?--oxidation: bromide goes to bromine--so that is going to be the Br^-).1183

The cathode (yes, we can write this; that is not a problem: Br to Br₂) is where reduction takes place, and the reduction reaction is: Cl₂ to Cl^-.1193

So, electron flow is that way.1212

Again, pictures are perfectly fine--pictures, I think, are actually better.1218

OK, now--the last thing we need is, of course, the physical description: we need the line notation: what does this look like?1222

Well, OK: bromine and chlorine are not conducting metals, and certainly bromide and chloride are just aqueous ions; so we are going to need an inert material.1228

We are going to use platinum, so: platinum; I'll put a single line here; the thing that is going to be oxidized--remember, the anode is on the left, where oxidation takes place, so that is going to be here.1241

It is going to be Br^- becoming Br₂; remember, we are going from the thing that is becoming (I'm doing it from your perspective); the thing that is becoming, that is turning into, goes to the right; so we are moving to the right.1254

Bromide becomes bromine, and then we have a double line (let me make that double line a little bit better), and then here, chlorine is becoming chloride; so here, Cl₂, Cl^-, single line, and then another platinum electrode.1272

That is the line notation.1291

This is a complete description, based on just substances that they give us; they just threw it out: they said, "Make me a galvanic cell based on chlorine and bromine."1293

Look through the standard reduction potentials; find the equations; pick which one is higher; that stays reduction; the other one--we have to flip it, because that is going to be oxidized.1303

We balance each one (in this case, they are already balanced); we add the potentials for a cell potential; we designate the anode; we designate the cathode; and then, we write the physical description as the line notation.1312

Standard, standard, standard; OK.1324

Now, we can start our discussion of cell potential, free energy, work...things like that.1328

Now, we are going to start discussing the connection between potential and thermodynamics.1334

All right, now let's see: let us reintroduce this volt--we have been talking about cell potential being volts, and we said that it is Joules per coulomb.1340

Well, so J/C; that is equivalent to V; well, let's think about what this means.1359

We know what a Joule is: a Joule is a unit of work--a Joule is a unit of energy.1365

A coulomb is a unit of charge; so, a coulomb doesn't mean 1 positive or 1 negative charge; those are just specific charges that we use to designate.1370

A coulomb is actually a unit of charge, but it isn't 1; it isn't + and -.1385

I'll explain what that means in just a minute.1392

Let me just talk about what this actually means--what is a volt?--what is a Joule per coulomb?1395

Well, here is what it means: for a 1-volt cell (so let's say we have some galvanic cell, and we measure the potential, and it turns out to be 1 volt), that is equivalent to 1 Joule per coulomb.1401

OK, this means: as 1 coulomb of charge flows through the wire, 1 Joule of work is done by the cell.1420

So, remember what we said: we said that, if we put two substances together that differ in a reduction potential, when we add them together, the cell potential that we get is a measure of a spontaneous reaction.1456

In other words, if we opened up that circuit and allowed the electrons to flow, they will flow naturally, spontaneously, without us doing anything.1470

Well, the flow of electron through a wire allows us to do work--basically, everything that you enjoy in your life is a flow of electrons (electricity).1476

A potential of 1 volt means that, for every coulomb of charge that passes through the wire (and I'll tell you what a coulomb is in just a minute, in terms of actual number of electrons)--as that much charge passes through the wire, 1 Joule of work is done.1486

So, if I have a 2-volt potential, that means, for every coulomb of charge (it's 2 Joules per coulomb) that passes through the wire, that cell can give you 2 Joules of work.1508

If I have a 15-volt cell...a 12-volt battery: a 12-volt battery means that, for every coulomb of charge that passes from the battery through whatever it is that you are running (like, for example, your starter engine on your car)--12 volts--that means it does 12 Joules of work on that whatever-it-is (on the starter).1520

That is the whole idea: so that is what a volt means--it is the amount of energy that the cell does every time 1 coulomb of charge is transferred.1543

I'll define what a coulomb means in just a minute; don't worry.1554

OK, and I should also say: this is the maximum work--the maximum amount of energy you can extract from a 1-volt cell.1557

"Maximum" means--well, in thermodynamics, you remember what "maximum" means; "maximum" means that is sort of an ideal; we never reach that ideal, and the reason we don't reach that ideal is because some of that work is actually dissipated as heat.1590

The reason it is dissipated as heat is because the second law of thermodynamics tells us that the entropy of the universe always has to increase in any spontaneous process.1606

Let me say that again--it's profoundly important: In any spontaneous process (which this is), the entropy of the universe has to increase.1615

Well, in order for the entropy of the universe to have to increase, we can't extract the maximum work available from a process.1627

Back in thermodynamics, we said that, let's say, we have a free energy difference of -50 Joules; well, that means -50 Joules is free energy that I can use to do work.1635

Unfortunately, if I am lucky, I get maybe 60% of that; all of the other 40% of those 50 Joules--they end up being dissipated as heat.1645

That is necessary, because it is a fundamental law of nature that the entropy of the universe has to increase for any spontaneous process.1656

This is why there is no such thing as a perpetual motion machine; a perpetual motion machine means that you get back what you put in; it will never happen that way.1665

You are always going to lose something as heat; so every cycle that you pass in a perpetual motion machine, you are going to be increasing the entropy of the universe; you are going to be losing heat; you are going to get less energy back, less energy back; at some point, the machine will just stop.1672

It is OK, theoretically; we can talk about the potential, and we can talk about the maximum work available, as long as you remember: the maximum work that you get won't be that.1686

So, for a 1-volt cell--yes, theoretically, you should be able to get 1 Joule for every coulomb; you are probably not going to get 1 Joule.1697

You will be lucky if you get about .5 or .6 Joules, and I do mean lucky.1704

That is all this means; OK.1708

Now, let us actually talk about cell potential and what this means in terms of cell potential.1712

Cell potential is equal to...well, as we said, it's equal to a Joule per coulomb.1721

Well, a Joule is a unit of work, and a coulomb is a unit of charge; so a Joule is a unit of work, and a coulomb is a unit of charge.1727

Well, work is W; charge is q; this is not the same q as heat; remember, this is different: when we are talking about electricity, q means charge.1742

And because the cell is actually doing work on the surroundings, that means the cell is doing work, so we are always looking at work from the system's point of view.1753

So, when the cell does work--when the electrons start to flow and we open up the circuit--the cell is doing work on something else.1767

That means that energy is leaving the system; so this is actually negative.1775

This is sort of our basic equation that we start with; this particular equation is not altogether that important; we are going to be fiddling with this a little bit.1781

Let's rearrange this and write: -W=q, times the cell potential, or the potential; and now, let's go ahead and divide by a -1; so work equals -qE.1789

So, the total work that we get is equal to...the magnitude of the work is equal to...the charge transferred, times the cell potential.1806

OK, so let me write what this is: this is the total charge transferred--the total charge that passes through the wire--all of the charge that goes through as the batter (the galvanic cell) discharges.1818

Once all of the electrons are done--no more flow--that is the total charge.1840

This is, of course, the cell potential.1845

If we multiply these two numbers, we get the actual work that the cell can do.1849

OK, now, let's talk about charge and electrons--specifically, how many electrons equal a charge--how many coulombs and electrons...what is the relationship?1855

Here is what it is: the charge on 1 mole of electrons is called a farad (after Michael Faraday), and carries a charge of 96,485 coulombs.1867

I know it's kind of a strange number; don't worry about where the number comes from.1898

So, we have 96,485 coulombs per 1 mole of electrons.1904

So now, we have a better idea of what a coulomb means, because we are very familiar with what a mole is: a mole of electrons is 6.02x10²³.1915

That means, when 6.02x10²³ electrons pass through that wire, that means that 96,485 coulombs of charge have passed through that wire.1925

Well, for a 1-volt cell, 1 coulomb means it has done 1 Joule of work.1938

That is the relationship: moles of electrons are related to coulombs, because coulomb is a unit of charge.1945

1 mole of electrons equals this many coulombs.1951

"Per coulomb of charge"--that is how many Joules: that is the cell potential.1955

So, for a 2-volt cell, that is 2 Joules per every coulomb of charge; well, that means every coulomb of charge has a certain number of electrons associated with it.1961

That is the relationship; so this is just another conversion factor.1975

OK, so volt equals Joules per coulomb, and a farad equals coulombs per mole of electrons: 96,485 coulombs per every mole of electron.1978

OK, now let's go ahead and put this equation together with this equation.1995

q...now, we said that work is equal to -q, times the cell potential; and we said that q was the total charge transferred.2003

That is the total amount of charge; OK.2020

That (total charge transferred) is equal to the moles of electron that are actually transferred, times the number of coulombs per mole, because moles cancel to leave us coulombs.2023

Well, q is equal to n, which is the number of moles, times F, which is 96,485 coulombs per mole.2046

So now, we can put this q in here, and we get that work_max is equal to -n, times F, times E.2057

Here we go: this was our first primary, important equation.2073

The maximum work that is possible is equal to the total number of moles of electrons that are transferred, times the farad, times the potential.2078

This gives us Joules, right?--so n is moles; F is coulombs per mole of electrons; and the cell potential is Joules per coulomb.2087

Mole cancels mole; coulomb cancels coulomb; and we are left with Joules--that is the total amount of work that is possible for us, upon that much transfer of electrons.2101

OK, now the next part: you remember when we discussed free energy--thermodynamics: the work_max is also equal to the ΔG of the particular reaction (the free energy change is equal to work_max).2110

Well, work_max is equal to free energy; work_max is equal to -nFE; what we get is: ΔG is equal to -nFE of the cell, and for standard conditions (which is--you remember--25 degrees Celsius, 1 Molar concentration, 1 atmosphere pressure for gases...)...that version...A minus nF...there we go; and this is actually cell: this is the cell potential.2127

So, what this says: the maximum cell potential is directly related to the free energy of the cell reaction.2167

There is a relationship between the balanced cell reaction, the oxidation-reduction reaction that is going to take place (it's just a reaction)...there is a relationship between the free energy of that reaction and the cell potential that we calculated based on standard reduction potentials.2201

The constant of proportionality equals the farad times the number of moles that are actually transferred in that particular reaction.2218

For example, in the aluminum-magnesium, there are 6 moles of electrons that are transferred (right?--because of the 6 electrons, we have to cancel 6); that is what this n is--it's the number of moles of electrons that are transferred per reaction.2226

This is the important equation that we want to talk about.2241

This is actually pretty great, because this actually gives us an experimental means of calculating free energies for the reactions.2245

All we have to do is basically run the reaction or calculate the cell potential; multiply by F; multiply by n; and we actually get the ΔG.2252

This is really, really great.2261

Let's finish off with an example here--Example 3: Calculate ΔG standard for the cell Mg, Mg², Al³⁺, Al.2262

OK, let's calculate the free energy for this: well, when we calculated this, we got a standard: the cell potential was 0.71 volts.2290

Therefore, the ΔG standard is equal to -nFE; OK, now let me write the equations down, so we remember what we are talking about, as far as what n is going to be.2302

We had: aluminum 3+, plus 3 electrons, went to aluminum, and magnesium went to magnesium 2+, plus 2 electrons.2315

The total number of moles of electrons that were transferred (remember, we had to multiply this equation by 2 and this equation by 3): 6.2329

6 moles of electrons transferred for one cycle of this.2338

All right, it's the balanced reaction: when we balance this reaction, we get: 2 Al goes to 2 Al; 3 Mg goes to 3 Mg; 2x3 is 6 electrons--3x2 is 6 electrons.2342

Equalize the number of electrons: that is the number of moles--that is what n is.2357

So, this is going to be -6 moles...well, actually, you know what, I'm going to go ahead and write out the units.2361

-6 moles, times 96,485 coulombs per mole, times 0.71 Joules per coulomb: coulomb goes with coulomb; mole goes with mole; and we end up with ΔG=-4.11x10⁵ Joules, or 411 kilojoules.2371

That is actually quite a lot; so, if I have a cell that is made up of aluminum and magnesium, I am going to get that much energy, maximum; that is how much work I can do: 411 kilojoules.2406

Now again, I am not going to get all of that; some of that is going to be lost as heat.2427

That is still...I can still get a fair amount of that; even if I end up with 30% of that, that is still a lot of energy that I can use up.2430

Now, some things to think about: ΔG is negative: that confirms that this is a spontaneous process.2436

A positive cell potential is a spontaneous process as written.2444

If the cell potential is positive as written, the reaction will happen without you doing anything to it.2449

We arranged it that way; that is how we want it; that is what is happening here.2455

That is a galvanic cell; it is a spontaneous discharge once you close the circuit.2459

ΔG is negative--spontaneous: so it confirms that this is the case.2464

There we go: OK, this sort of closes off our basic discussion of thermodynamics.2469

We will actually continue on in the next lesson and get a little bit deeper.2476

Until then, thank you for joining us here at Educator.com.2479

We'll see you next time; goodbye.2482