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Lecture Comments (3)

1 answer

Last reply by: Professor Hovasapian
Fri Jan 5, 2018 3:29 AM

Post by Matthew Stringer on January 3 at 09:41:53 PM

Great lecture!

0 answers

Post by Rafael Mojica on April 24, 2014

You used a "+" which really confused. I did the calculations and you ended up multiplying.


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
  • Electrolysis 3:16
    • Electrolysis: Part 1
    • Electrolysis: Part 2
    • Galvanic Cell Example
    • Nickel Cadmium Battery
    • Ampere
    • Example 1
    • Example 2

Transcription: Electrolysis

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

Today, we are going to close off our discussion of electrochemistry by discussing the process of electrolysis.0004

In short, electrolysis is the reverse of the standard spontaneous galvanic cell.0011

If we have two sets of components from which to make a galvanic cell, we take a look at the reduction potential table; we find out which one is going to be oxidized and which one is going to be reduced.0018

The one that is reduced--we keep it that way; the one that is oxidized, the one with the lower reduction potential--we flip that.0030

We balance each half-reaction based on that process that we did before; we add the two reactions; we add the two cell potentials; and what we get is this particular reaction with a positive cell potential.0036

That reaction is spontaneous: in other words, any time we bring those components together (either directly or by connecting them with a wire), electrons will spontaneously flow from the wire, from one compartment to the other.0048

That is a galvanic cell; that is a voltaic cell; that is a battery.0063

Well, sometimes, what we want to do is actually run that reaction in reverse.0067

Because it's spontaneous in one direction (the galvanic cell), the reverse of that we call an electrolytic cell (electrolysis).0073

Electrolysis literally means "splitting by electricity"; it refers to the process (my presumption is) of using an electric current to actually split up water into hydrogen and oxygen gas.0082

But again, it's just a term; think of it as just the reverse of the galvanic cell--electrolysis is the reverse of the galvanic cell.0097

It is making a non-spontaneous process run.0108

Well, if a spontaneous process happens without our doing anything (just bringing the components together), and, let's say, the cell potential for a galvanic cell is .5 volts, the electrolysis reaction (the electrolytic reaction) is the one that runs in reverse, and it has a negative cell potential (-.5 volts).0113

What we have to do is: we actually have to do work: because, in a galvanic cell, electrons want to flow naturally this way, we actually have to put energy in the system, and use a power source to keep those electrons from moving this way, and actually push them back farther.0133

So, not only do we have to add .5 volts to keep them from moving in the direction they want to move; we have to put more than .5 volts to actually push them back to where they are coming from.0150

That is it: the primary purpose of electrolysis...well, any time you recharge a battery, that is what you are doing: you are converting a galvanic cell (which is a standard battery) into an electrolytic cell.0160

You plug that into your outlet, and a current actually forces electrons that move from the anode to the cathode back the other way, and then the battery is ready to go again.0173

Another major process that electrolysis is used for is for plating metal.0187

That is what we are actually going to talk about in some of our examples.0192

So, let's go ahead and get started: I'm going to write out some of these things, so that you have them written down.0196

Let's see: OK, so let's say, for example...now we said that the maximum work that we can actually gain from a galvanic cell is...0204

Well, you know what, before I get into that, let me actually write down what it is that I just talked about.0222

So, electrolysis is running the reaction (I'm sort of changing some of these things as I go)...running the spontaneous reaction of a galvanic cell in reverse; that is it.0228

We have to do work in order to push electrons in the non-spontaneous direction.0273

There we go...non-spontaneous direction.0302

OK, let's actually draw out a couple of circuits, just so we have a physical idea of what is happening, so we can see electrons flowing one direction or the other direction.0305

It is one thing to sort of hear it, but we really should see it.0313

Another way of describing this: I'll give another; it's about the same thing, but I'll go ahead and write it down anyway.0321

Electrolysis is forcing a current through a cell to produce a chemical reaction for which the cell potential is negative.0332

The fact of the matter is: when I am dealing with the reduction potentials that I see in a table, I can actually write them in any order that I want to--I mean, they are just reduction potentials.0366

When we do a galvanic cell, the one with the higher potential is left alone as reduction, and it's the one with the lower potential that is flipped.0377

That always gives us a positive cell potential; a positive cell potential always means "spontaneous as written."0385

Well, it's just a reaction--it's just something that we write down on paper; if I wanted the reaction to actually run the other way, all I would have to do is flip it.0391

And if I flip it, I flip, and the cell potential becomes negative; so now, as written, there is a negative cell potential.0399

If I want that reaction to proceed as written with that negative cell potential, I have to do work on it; that is all electrolysis is.0406

These are just fancy terms for a process that you sort of understand in terms of the process itself.0415

I do work on it, as opposed to the spontaneous reaction doing work for me--the current running through the wire.0420

I have to push current through a wire to make electrons go in the opposite direction.0428

OK, so let's take a look at our basic cell.0433

Here is the water line; here is our porous disk; and let's go ahead and use zinc and copper.0437

These are our electrodes; we have some zinc ion--and, of course, zinc is a conducting metal, so this is zinc; and this is going to be copper; there is some copper ion here.0448

When we measure the cell potential, we have electrons that are going to flow in this direction.0460

Zinc is going to be oxidized; copper ion is going to be reduced to copper; and you are going to have just this natural buildup of copper metal.0468

1.10 volts: OK.0477

This is the anode; oxidation takes place; this is the cathode where reduction takes place; this is a galvanic cell.0480

OK, now, the spontaneous reaction is: zinc plus copper 2+ going to zinc 2+ plus copper; that is what is happening--that happens spontaneously if you bring zinc metal and copper ion in close proximity to each other, or connect them via a wire.0491

Now, let's take a look at the electrolytic cell version of this.0513

We have the same setup.0517

Actually, let me draw it a little bit lower, so we have more room for what I'm going to draw here--you know what, that is just fine; I'll just put it over here.0520

Something like this: here is our water line; this is our...now, this thing is (my drawing is a little elementary, but again, it's not the artwork that is important here)...0531

So again, we have a zinc ion solution; we have a copper ion solution; we have our copper metal electrode; we have our zinc metal electrode; and now, what we have--this thing is a power source.0548

It is no longer something that actually measures cell potential; it is a power source that is greater than 1.10 volts.0564

Basically, it is just a power source here; I'm pushing...so, the electrons want to move this way; and the potential for that, the push on the electrons (or the pull, depending on your perspective--you can think about it either way) is 1.10 volts (Joules per coulomb).0577

Well, I have to supply a power source of more than 1.10 volts, pushing the electrons in that way, so that I could actually...when I open the circuit, the electrons don't come this way spontaneously; they are pushed back that way.0595

I want to push them back: that is the whole idea.0612

They are pushing this way; I have to push back harder than they push to drive them back into this particular compartment.0614

Here, the electron flow is like that; now, what is going to happen is that this is going to be oxidized, and this is going to be reduced.0623

Zinc is going to be reduced; now, zinc is going to actually plate out: that is what plating means--taking the ion in solution, reducing it, and putting, basically, a coating of metal on whatever it is that you happen to have dipped in the solution.0633

This is a very, very important industrial process.0651

So now, what was the anode has now become the cathode, and what was the cathode has now become the anode.0654

That is what has happened: this is the electrolytic cell.0664

Its spontaneous reaction is...sorry--its reaction--not spontaneous--is zinc ion, plus copper, going to zinc metal, plus copper ion.0668

The cell potential for this is negative 1.10 volts.0682

As written, this is negative--because again, I can write it this way if I want to; this is negative.0687

If I want to run this reaction that happens to have a negative cell potential, I have to do work on the system.0694

I have to have a power source--an external power source--that will push electrons opposite the direction they want to go in.0702

Forgive me for repeating myself, but I find that that constant repetition of these basic things really, really helps to cement the ideas in your mind, because this is how you want to think about it.0709

OK, now what we are going to do is: we are going to use this process to actually force a current into a solution to plate out some metal.0719

OK, let's see: let's talk about a NiCad battery--nickel-cadmium battery--the kind of batteries that are in your computers and calculators, that you can recharge.0731

Nickel-cadmium battery: OK, so again, a nickel-cadmium battery has an anode; it has a cathode; here are the two reactions that take place.0753

The oxidation: cadmium is being oxidized under basic conditions to cadmium hydroxide, plus 2 electrons.0769

The reduction process is: nickel (4) oxide, plus 2 molecules of water, plus 2 electrons (because it is reduction)--it becomes nickel (2) hydroxide, plus 2 OH-.0785

So, these are the two reactions that take place in a NiCad battery.0809

Spontaneously, what will happen is: cadmium will oxidize to cadmium 2+; nickel will reduce from nickel (4) to nickel (2).0815

Now, when I take this battery and I plug it into the wall outlet (plug my computer into the wall outlet)--now, electrons are going to be flowing from the wall outlet, and they are going to be pushing electrons back that way.0823

They are actually going to take cadmium 2+, and they are going to be converting them back to cadmium.0842

They are going to take...let me see...this goes to this, so this is going to go back to that, and you are basically going to have a brand-new battery all over again.0851

Let's go ahead and do a problem here.0867

Oh, before we do, here is the question: How much current?0874

This is the practical question that we are going to be ultimately concerned with in electrolysis, and these are going to be the types of problems that you are going to see on the AP exam.0881

So, now that we have a little bit of background on what electrolysis is and what is going on, now let's ask some actual quantitative, numerical questions about it.0890

How much current do we need (in other words, how many electrons), and for how long do we need to run that current to recover the cadmium that became cadmium 2+?0898

So, there is a certain amount of cadmium metal that is there, that is turning into cadmium ion--it is turning into cadmium ion in the form of cadmium hydroxide.0941

The question is: How much of a current do I actually need to run, and for how long, in order to recover a given amount of cadmium?0951

Well, OK: so let's go ahead and introduce something new, a new unit.0960

It is called the ampere, and its symbol is A, and it is the number of coulombs that are transferred per second.0967

This is equivalent to C/s.0980

So, if we talk about something that is 5 amps, well, that means there are 5 coulombs of charge travelling through that wire per second.0983

If there is something which is 20 amps, that is 20 coulombs per second.0991

OK, what I am going to draw out next is the general path that you are going to follow when you solve these problems.0997

Grams to moles (this is basic stoichiometry problem) to moles of electrons to coulombs...because, remember, a coulomb is not the same as an electron; you have 96,485 coulombs per one mole of electrons, so a coulomb is actually a lot of charge.1011

So, when we are talking about, let's say, a 20-amp line (20 coulombs per second), that is a lot of electrons; it is actually a very dangerous line.1045

Coulombs, and then current, plus time: these are the things that are going to be involved in any problem regarding an electrolytic cell.1054

Basically, there is going to be a certain amount of current that we are going to run through the wire, via our external power source, for a certain amount of time.1071

Well, we said that current is coulombs per second; so coulombs per second, times the time that you run that current, gives you the number of coulombs.1081

Well, the relationship between coulombs and moles is the faraday, the 96,485 coulombs per mole of electrons.1095

Once we have the mole of electrons, there is a relationship that exists between the mole of electrons and the moles of...I'll just say...species--whatever species we happen to be interested in (in this case, let's say cadmium--the moles of cadmium).1104

That relationship is expressed via the half-reaction (right?--you have a half-reaction; there is a certain number of electrons that are transferred in that half-reaction to take it from the ion to the metal).1118

Or, let's say iron 2+ to iron 3+: there is one mole of electrons transferred; from copper metal to copper (2) ion is 2 moles of electrons transferred; from aluminum ion to aluminum metal, there are 3 moles of electrons.1129

That is the relationship there.1143

And then, of course, from moles to grams, that is just the molar mass.1144

Notice these double arrows: in these problems, I can be given any one of these, and I can go to any other one of these.1148

OK, so they might ask for how many grams of something I need to recover, based on a certain current and how long I move in this direction, or I move in this direction, depending on what it is that I am looking for.1155

Let's just go ahead and do an example, and I think it will make a lot more sense; but this is the path that you are following.1167

In this case, each arrow represents a conversion factor: 1, 2, 3, 4: there are four conversions: 1, 2, 3, 4, depending on where you are.1173

Remember, back in stoichiometry, going from moles to grams or grams to number of molecules (you have to go from grams to moles, moles to number of molecules, molecules to number of atoms).1184

It is a path that you are following, and each step is a conversion factor.1195

OK...oh, yes, while I'm at it, I think I want to discuss a very, very small possible error that I made in the last discussion, in the last lesson.1200

When I discussed the notion of 96,485 coulombs per mole of electrons, I can't remember, but I think I may have called that a farad instead of a faraday.1216

This is a faraday; a farad is actually a unit used in electricity--a unit of capacitance.1233

So, if in fact I made that mistake, forgive me; this is actually called a faraday.1240

Both are actually named after Michael Faraday.1245

OK, so Example 1: How long must a current of 6 amps be applied to cadmium hydroxide to recover 1.2 grams of cadmium?1248

So, let's say that I happen to have 1.2 grams of cadmium in my NiCad battery; those 1.2 grams have been completely converted to cadmium hydroxide.1293

I want to recharge that battery: I have a power source that runs 6 amps.1301

How long do I have to run that current for to get back all of my 1.2 grams, in order to completely recharge my battery?1310

OK, well, here is what we have: grams, moles; moles, coulombs; current, time.1318

OK, what is it that they actually give us (let me do this in red)?--they give us 1.2 grams.1327

What else do they give us?--they give us the number of amps.1333

With that, we could actually do this, because the rest are just standard conversion factors.1336

Well, the first thing we need to do, though, is: we need to find out...well, let's do that when we actually do the problem; I think it will make more sense if we do it step by step.1341

OK, so I have 1.2 grams of cadmium; I need to go from grams to moles of cadmium; well, cadmium happens to be 112.14 grams per mole of cadmium.1351

Now, I need to go from moles of cadmium to the moles of electrons that need to be transferred.1369

Well, this is where the equation comes in: our cadmium goes to cadmium 2+ (right?--cadmium hydroxide is cadmium 2+).1376

So, there are two moles of electrons transferred per every one mole of cadmium that is either oxidized or reduced.1391

We write: 1 mole of cadmium, 2 moles of electrons: this is just a standard stoichiometric conversion--I am just using this half-reaction.1399

Remember, this coefficient and this coefficient is a mole ratio.1413

Instead of talking about a mole of some other atom or molecule, it's a mole of electrons.1417

That is it; that is all that is going on here.1423

So now, we need to go from moles of electrons to coulombs; so here, I have mol of electrons, and I have 96,485 coulombs; and I have to go from coulombs to amps and time.1425

In this case, I'm trying to actually go to time via current.1446

So, my amperage is: for every 1 second, I have 6 coulombs; and there you go.1452

Grams of cadmium, grams of cadmium; mol of cadmium, mol of cadmium; mol of electron, mol of electron; coulomb, coulomb; and then, when you do the multiplication--well, what you end up getting is 343 seconds, which we can convert to minutes.1464

It ends up being 5.7 minutes.1483

So, if I have a battery that has 1.2 grams of cadmium, I discharge that battery, and I want to recharge that battery--if I have a 6-amp power source, I have to run that power source for 5.7 minutes, and I will recharge my battery.1487

That is it; that is all that is happening here.1504

OK, let's do another example.1507

Current and time...amperage...you know what, I am actually going to change one little thing here on my little...what I call a solution map.1512

"Solution map" is a pictorial representation of where we are going and where we are coming from.1522

I'm going to just leave time here, and the relationship between time and coulombs is the ampere.1528

Let me go back to blue.1537

There we go: the ampere is the relationship between coulombs per second--coulomb and time--so I won't put that here.1540

OK, so let's go to Example #2.1547

That is it: as long as you are familiar with that pictorial--with that solution map--everything should be fine.1556

OK, a solution containing a 3+ ion is electrolyzed by a current of 5.0 amps for 10 minutes.1563

So now, we are giving you the amperage, and we are giving you the time.1601

OK, the question is: What is the identity of the metal if 1.18 grams of metal is plated out during this time interval?1604

OK, so let me just draw a pictorial of this and see what is happening.1644

I'm only going to draw one compartment: we have this thing here; we have an external power source (I'll just call it p.s.); and we have some metal ion that happens to be 3 charge.1646

I'm going to run 5 amps of current through this thing--push 5 amps (5 coulombs of charge for every second) into here; and I'm going to do that for 10 minutes.1661

As I do that, the metal ion is going to be reduced; so it is going to start plating out.1674

It is going to start actually depositing right on the metal itself.1679

Well, let's say I have already weighed out this metal electrode before anything plated out, and then after the process was done--after the 10 minutes--I take it out, I dry it, and I weigh it again.1684

The difference in the weight is 1.18 grams.1696

That is what it is saying: this deposit is 1.18 grams of the metal--it's actually plated out during this time interval.1699

We want to know what that metal is: well, the only thing we know about it is that it's a 3+ ion.1707

OK, well, first of all, let's just write the equation, so we understand the mole--the number of moles--of electrons that are transferred.1712

Metal goes to metal 3+ ion, plus 3 electrons: I could have written it the other way--it doesn't really matter.1719

I could have written: M3+ + 3 electrons goes to M; it really doesn't matter.1728

Well now, when they say "identity" of something, and when they give you the grams, here is what you need to identify something.1733

Molar mass: you need to know its molar mass: "this is a metal"--"this is an element"--something; you need to know its grams per mole.1739

We need to know grams per mole.1747

Well, we have the grams already; that is 1.18; what we need to know is how many moles of this metal are actually deposited.1758

1.18; so, this is what we need.1767

OK, let's see what we can do.1773

If you want to go back and refer to that pictorial...let's see--what do they give us?--they give us 10 minutes, and they give us 5 amps.1777

So, time: we have 10 minutes, times 60 seconds per minute (and the reason is because amperage is coulombs per second; the second is the time unit for the amp), times 5 coulombs per second, plus 96,485 coulombs per mole of electrons, times 1 mole of metal; 1 mole of metal comes from the transfer of 3 moles of electrons.1788

That is it: min, min, s, s, C, C, mol of electron, mol of electron; I am left with moles of metal, and my final answer is going to be 0.0104 moles of metal.1834

Now, let's go ahead and go here; so we have 1.18 grams, which consists of 0.0104 moles, and I end up with something like 113.8 grams per mole.1851

When I look at a periodic table, I am going to be somewhere in the cadmium or indium range.1871

Well, cadmium doesn't have a 3+ ion; as it turns out, indium does have a 3+ ion; it just happens to be underneath the aluminum column.1878

So, in this particular case, our answer is going to be indium.1891

Now, granted, this is a real-world process; it is not going to be exactly what the molar mass is...you are not going to get 111 or 112 point something--it is going to be in that range.1895

This is the thing: there are tolerances in science.1908

Yes, science is exact (or we try to make it as exact as possible); but in things like this, again, it's not going to be...1911

If you end up with 113.8, and you have maybe 3 or 4 different things, and you are saying, "Well, wait a minute; which one is it? Is it cadmium or indium? Or maybe it's the one before cadmium; maybe it's the one after indium," well, you have other information at your disposal.1919

The other information is that cadmium doesn't form a 3+ ion; indium does form a 3+ ion.1934

The thing right after indium doesn't form a 3+ ion; so, indium is our guess.1941

OK, so that is it as far as electrochemistry is concerned.1948

We have discussed galvanic cells; we have discussed cell potential (work); we have made the connection between cell potential and thermodynamics, and cell potential and the equilibrium constant.1951

And we have, finally, discussed electrolysis; so, hopefully, everything that is important has been discussed.1964

With that, I will see you next time, and we will begin our discussion of light and quantum mechanics.1973

We are actually going to be going back to talk about some things...what I consider to be in the middle part of chemistry: the quantum mechanics, the bonding, the solutions...some of the things that I actually skipped, because I wanted to spend a fair amount of time on the kinetics, the equilibrium, the electrochemistry, and things like that.1978

So, until then, take care; thank you for joining us here at Educator.com; goodbye.1998