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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### Electrolysis

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

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### 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: M ^{3+} + 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

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.