AP Chemistry Oxidation Reduction Reactions

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

In the last couple of lessons, we have been talking about aqueous solution reactions and stoichiometry.0005

We discussed precipitation reactions; we discussed acid-base reactions; today, we're going to discover the third major class, which is oxidation-reduction reactions.0011

Let's jump on in: a couple of basic definitions, and we'll just start right in with the chemistry.0021

OK, so oxidation is the loss of electrons; that's it--nothing more than that; that is all it means.0027

That means some atom has lost a couple of electrons: 1, 2, 3, 4, 5, 6--there are a bunch of different oxidation states.0039

And, of course, you see it--for example, magnesium 2+; that just means that magnesium atom has lost two of its electrons, and now it's carrying a 2+ charge.0048

That charge up there, at the top, is the oxidation state, or oxidation number; either one of those is fine.0059

So, oxidation is a loss of electrons, and reduction is the gaining of electrons; that's it.0065

They are complementary processes; so, when something is oxidized, something is reduced.0077

The thing that is gaining is gaining it from the thing that lost it; so they always come in pairs, just like acid-base.0083

Every time you have an acid giving up a proton, generally--generally, there is a base that is taking the proton.0090

It's that kind of a relationship.0095

In fact, oxidation-reduction and acid-base chemistry are actually quite parallel.0097

In acid-base chemistry, the particle that is bouncing back and forth like a tennis ball is a proton--just a hydrogen atom with its electron missing.0102

In the case of this oxidation-reduction reaction, the particle that is bouncing back and forth is the electron.0111

So, the chemistry, the underlying chemistry, is actually essentially the same; it's just the species that are different.0117

That is one way of looking at it, if you want a more unified view.0123

An example of this might be--well, since we're talking about oxidation and reduction, let's deal with oxygen.0126

Oxygen has stolen two electrons from something; now it's carrying a 2- charge; its oxidation state is -2; oxygen has been reduced; magnesium has been oxidized; that is what we say.0133

Let's jump right into an example here and give you an idea of what it is that is really going on, chemically, at the atomic level, so that you have an intuitive sense of what it is that is going on.0146

Where are these electrons going? Where are they coming from?0156

We'll take, as an example (actually, let me write the equation first, and then we'll balance it)--let's take aluminum metal, and we will drop it in a solution of hydrochloric acid.0159

It's a strong acid, and we want to see what happens.0172

Well, something actually does happen; you end up forming aluminum chloride, and you end up bubbling off hydrogen gas.0178

Hydrogen gas: when something is a gas, you will see a little (g) written at the bottom right; you will also see it, in older notation, as a little arrow going up.0187

Just like, for a precipitate, we had an arrow going down, a gas is an arrow going up.0196

I'll often use this notation; I hope it doesn't confuse you.0201

Now, let's go ahead and balance this: let's see...2: let's go ahead and put a 2 here; let's put a 6 here, a 3 here, and a 2 here; so 2 aluminum atoms plus 6 hydrogen chlorides form 2 molecules of aluminum chloride and bubbles off hydrogen gas.0206

Now, let's actually discuss what it is that is going on.0231

Well, here is what happens: aluminum has three electrons in its valence shell; and another aluminum--I'll put three electrons in its valence shell.0235

In other words, it has three electrons that it can give up if it were to do so, if the right thing came along to actually want to oxidize it.0246

Now, I'm going to put in--as you know, HCl dissociates completely, so--I'm going to draw out 6 hydrogen ions here (that's 4, that's 5, that's 6), and the Cl^-s I'm going to ignore, because they don't really do anything; they're not involved in the chemistry, so I'll leave them off.0254

I just want to show you what it is that is going on.0276

Here is what is happening: H, in a battle for electrons--in a tug-of-war--the hydrogen ion will win, between hydrogen and aluminum.0278

I'll discuss why a little bit later, when I talk about something called reduction potential and activity series, but for the time being, just so we understand what is happening with oxidation-reduction reaction--any time you have species floating around, with certain species, there is going to be a battle of electrons going on.0287

H is going to take; a second H is going to take the second electron; a third is going to take the other electron; and then the same thing is going to happen here.0304

This H is going to take that one; this H is going to take that one; this H is going to take that one.0312

What you end up with is the following: you end up with an aluminum that has lost three electrons, another aluminum that has lost three electrons, and now, before any reaction takes place, you have H, H, H...you have six hydrogen atoms, because now the ions have taken an electron, and they become hydrogen atoms.0317

Let's see if I have...1, 2, 3, 4...oops, I have one extra.0337

Now, as we know, hydrogen is a diatomic gas; hydrogen atoms don't exist like that--they bind with other hydrogen atoms.0343

So, let me connect those two; this one binds with that one; this one binds with that one.0351

I end up with my net reaction: well, 3 (let me write them individually)--I end up with a hydrogen₂ molecule, a hydrogen₂ molecule, a hydrogen₂ molecule, an aluminum ion, and an aluminum ion.0357

Of course, I still have my 6 chlorides floating around, and because aluminum chloride is, generally, soluble, we leave it like this; this, when we write it as AlCl₃, is the molecular form; it's not really the ionic form.0376

That is all that is happening here; when you have species where one can actually take an electron from the other, that species will oxidize that other.0393

Itself, it will become reduced.0402

Now, aluminum has been oxidized from aluminum metal to aluminum ion; it has lost electrons.0405

Hydrogen ion has been reduced to hydrogen atoms, and then ultimately to hydrogen gas, which is an elemental form.0412

That is all that is going on with oxidation-reduction.0423

One thing takes; one thing loses.0426

OK, let's see; I suppose the next question we want to ask ourselves is, when we see an equation like this, how can we tell?0429

How can one tell whether oxidation-reduction has taken place?0440

Well, we tell by looking at the oxidation states, the numbers of the individual atoms in the compound.0452

I want to tell whether oxidation-reduction has taken place; well, the answer is: you consider oxidation states of individual atoms in an equation.0460

In an equation--that means, in a chemical equation, you have reactants and products; so you check the oxidation states of the reactants--individual elements--and the products, and you see if there has been a change.0489

Now, it's called reduction because...reduction is the gain of electrons, but what ends up happening is: the oxidation state goes to a lower number.0502

It might either be from a positive down to something less positive, or 0, or negative, or it might be from 0 down; that is why it's called reduction.0512

You might think that oxidation should actually be called an increase, as opposed to a reduction, because the oxidation state is actually going up.0521

For example, aluminum had a 0 oxidation state (all elements in their natural form are 0), and here, over on the right side, it has a 3+.0528

So, numerically, the number has actually gone up; we call it oxidation because, historically, when we were first learning about stuff like this, and discovering it, the compounds that were doing the oxidizing all had oxygen in them.0537

That was what was noticeable, so they called it oxidation; but oxygen doesn't have to be the only thing oxidizing--anything that takes electrons is an oxidizer.0550

Let's just sort of do some examples in this particular...well, actually, let's go back to this aluminum; and so, let's rewrite the equation.0560

2Al + 6HCl goes to 2AlCl₃ + 3H₂.0571

Aluminum is the metal; usually there is an S right there to let us know that it is the metal; in elemental form, it has a 0 oxidation state; it has all of its electrons--it is neutral.0581

HCl; well, we know that HCl is -, and we know that H is +; that is just how HCl comes together.0591

Over here, again, we know that Cl is always -, so Cl is -; there are three of them, for a total of 3-; therefore, there is one Al, so it's balanced by a 3+.0600

Aluminum has gone from a 0 oxidation state to a 3+; that tells us that this is an oxidation-reduction reaction; it has lost electrons.0617

In this case, it has lost 3 electrons; let's take a look at hydrogen; hydrogen has gone from a +1...and now hydrogen gas in its elemental form; everything in its elemental form, as it is in nature, has a 0 oxidation state.0625

It has gone from +1 to 0; these numbers here are just stoichiometric coefficients, and also used to balance the charge, the number of electrons that were transferred.0638

It's 2 aluminums that lost a total of 6 electrons.0648

6 hydrogen each gained 1 electron, for a total of 6 electrons.0652

That is why it is--so charges have to balance, too; but that is all we're doing--just look at the oxidation state of each individual atom, and then that is it!0657

If something is lost and gained, it's an oxidation-reduction reaction.0665

Now, let's go ahead and take a look at some rules for assigning oxidation states, because it's not just random; how does one decide what an oxidation state is?0670

For example, how did I know that Cl was always a -1? OK, well that is what we're going to do here now.0679

The rules for assigning oxidation state: when you are faced with a compound, what is the oxidation state of each element in that compound?0684

Oxidation state, also known as oxidation number...0700

OK, #1: An atom or diatomic molecule (well, I'll say diatomic element, and what I mean by that is--like oxygen, nitrogen, hydrogen, bromine, chlorine, iodine--these things are diatomic in their natural form; they're not individual atoms; there are two of them stuck together, so I'll just say diatomic element)...0706

An atom or a diatomic element in elemental form (in other words, in its natural form) has an oxidation state of 0.0738

So, o/s, oxidation state--always; so aluminum metal is 0; hydrogen gas--0; oxygen gas--0; chromium metal--0; mercury, with liquid mercury--0.0752

For a monoatomic ion, the oxidation state is the charge on the ion; that's it.0769

So, a chloride ion--oxidation state is -1; magnesium 2+ ion--oxidation state is +2; chromium 6+ ion--oxidation state is +6; no problem.0788

3 (probably the most important): Oxygen is always -2 in both ionic and covalent compounds.0806

So, any time you are faced with an oxygen (we'll talk about one exception in a second here), you're always just going to automatically assign it a -2 charge.0826

Oxygen is one of the most highly electronegative elements on there; in fact, it's the second most electronegative, after fluorine.0835

It will always take electrons from everything--oxidation; that is where it comes from.0841

An exception is the peroxide ion, which is O₂^2-.0849

Well, 2 oxygens carrying a 2- charge; that means each oxygen has a -1 charge; so, in this case, each oxygen has a -1 oxidation state.0863

So, what we're looking at when we're assigning oxidation state is the charge on one atom, not the combined thing.0874

We use the charge--we use the total number to decide what the total charge is--but we're looking for oxidation state for each individual atom.0881

#4: Hydrogen is a +1 when bonded to a nonmetal, and it is -1 when bonded to a metal.0890

You know, most of the time, it's just going to be H⁺, the hydrogen ion.0911

But, occasionally, you will see it bonded to a metal; -1 charge; sodium hydride--notice, we call it sodium hydride; sodium is the positive; hydride--it's always the anion that has the -ide ending; that is an H^- ion--very, very different from H⁺, and different from the hydrogen atom; three different types of chemistry.0917

OK, 5: F is always -1; that is because it's the most electronegative; it will never give up its electrons.0942

The other halogens (halogens being bromine, iodine, astatine, chlorine--I forgot chlorine) are usually -1, but when bound to oxygen (because oxygen is more electronegative than the other halogens, with the exception of fluorine), they take on variable oxidation states, depending on the number of oxygens.0954

We'll do some examples in a minute, so no worries.1015

6: The sum of the oxidation states of a neutral compound equals 0; so in a neutral compound, the oxidation states have to add up to 0.1021

And, they equal the charge on the polyatomic ion when it's not neutral; so, the only time you are going to see a combination of things like sulfate, of atoms, SO₄, SO₃, CO₃ for carbonate--more than one atom that actually carries a charge--it's going to be a polyatomic ion.1049

Well, the individual oxidation states of the atoms have to add up to the total charge on the ion; and again, we'll do an example in a minute.1078

For the polyatomic ion, if ionic...I know that is a little redundant, but OK.1089

Let's see--let's do an example.1099

We'll call this Example 1: we want to determine the oxidation states of sulfur in the following compounds (so we have several different compounds, each of them containing sulfur).1105

This is going to be an example of using the rules that we just came up with.1128

It's also going to be an example of the fact that one compound can have many different oxidation states--and others, maybe only one oxidation state (like oxygen; -2).1132

Sulfur, in the following compounds: we have one which is H₂S (hydrogen sulfide or hydrosulfuric acid, depending if you bubble it into water); H₂SO₄ (this is sulfuric acid); we'll do H₂SO₃ (sulfurous acid); and we'll have a covalent compound, SO₂ (sulfur dioxide).1144

Let's start with H₂S.1170

H is bonded to a nonmetal, so it's carrying a +1 charge; there are 2 of them, so the total charge is +2; well, that means it has to be balanced by a -2, because this is a neutral compound.1173

There is only on S, so that S carries that -2 charge.1186

The oxidation state is -2.1190

Now, I'm going to write this as 2-; now, there are some teachers that actually prefer the number to come before the negative sign; it doesn't matter, as long as you know what is going on--it really doesn't matter.1192

Now, H₂SO₄: well, H₂SO₄ consists of 2 H⁺s and an SO₄^2-.1210

So, when we do the oxidation state of sulfur here, we're going to have to look at the SO₄^2-, not worry about the H⁺; we know this is a +1, but we're concerned about the sulfur.1220

Well, we said that the sum of the oxidation states has to equal the charge on the ion itself.1230

What did we say about oxygen?--oxygen is always a -2.1236

Well, there are 4 oxygens, so the total charge is going to be -8; that means the charge on sulfur, whatever it is, plus a -8, has to equal the -2.1240

Or, in other words, -8 plus the charge on sulfur is equal to -2.1256

Well, let me just do some basic algebra here; -2 plus 8 is going to be 6; +6.1262

So, in the case of sulfuric acid, sulfur is highly oxidized; it has actually lost 6 of its electrons.1271

OK, let's do H₂SO₃: same thing--we have an SO₃^2-; well, now, again, oxygen is always a -2; there are three oxygens; so 3 times -2 is -6 for the oxygens.1281

Now, that means -6 plus the charge, the oxidation state, on sulfur, has to equal -2.1297

Now, the oxidation state of sulfur is just -2 plus 6, so it's equal to 4.1307

So, the sulfite ion is not as oxidized; it has lost four electrons, but it hasn't lost 6 electrons--more oxidized--the sulfate is.1314

So here, S is 4+.1326

OK, now let's look at SO₂, a covalent compound.1329

Oxygen is always a -2; there are 2 of them; well, since there are 2 of them, the total charge is -4; this is a neutral compound, so this has to be a +4, and since there is only one sulfur, that means sulfur is carrying the +4 charge.1333

So, as we see, sulfur +4, sulfur +4, sulfur +6, sulfur 2-; that is a huge range.1351

Here, sulfur has been reduced; it has stolen electrons from something--from hydrogen, specifically.1357

Here, it has been oxidized; it has lost its electrons to oxygen.1363

That is all we're doing here; that's all we're doing.1368

Let's actually just throw in one more example for the fun of it; let's do a permanganate ion, MnO₄^-.1371

We know that that ion is MnO₄^-; we know that oxygen is always a -2; there are 4 of them, so oxygen is carrying a total of -8.1378

Well, I know that -8 plus something should give me the charge on the total ion, -1; therefore, -8 plus what equals -1?1388

Well, that something is equal to +7; therefore, we say we have manganese 7+.1400

That means manganese has lost 7 electrons to oxygen; totally oxidized--completely oxidized, in fact.1410

This is one of the reasons that makes this a very powerful oxidizing agent, in and of itself.1418

When manganese comes in contact with other things, it has lost its electrons; but when it comes in contact with something that actually has electrons to give up, it will definitely go out of its way to rip those electrons away.1422

That is what makes permanganate ion a powerful oxidizing agent.1433

Oxygen has taken the electrons already; it's not going to give anything up; so it doesn't matter--it's this manganese that has a +7 charge that wants to get back down to a +5, +2, +3.1439

It wants to take electrons away.1449

OK, so now that we have talked a little bit about oxidation state--what oxidation-reduction is--I want to discuss something called an activity series, and I also want to discuss something called the reduction potential.1451

Now, those of you in your classes: at this particular stage of the game, activity series will be introduced to you.1463

It has to do with: if I put some metal in contact with some other solution, will the metal replace that other metal?1469

Will magnesium replace zinc, or will zinc replace magnesium in a single-replacement reaction?1477

Well, I'm going to discuss activity series and reduction potential simultaneously; the reason is this: reduction potential--towards the end of the year, when we study electrochemistry, it's going to be very, very important.1483

I would like you to actually start to think of this oxidation-reduction stuff in terms of reduction potential, not necessarily activity series.1498

But, I will show you what both of them mean, because I want you to have both of them at your disposal now.1508

I don't want you to wait until later, having gotten used to activity series, and then all of a sudden be hit with reduction potential when everything is flipped backwards.1512

I want to develop the idea--I want you to start thinking in terms of reduction potential now, so that by the time you get to electrochemistry, it will be just second nature.1521

So, over here, we have (that's OK--I'll just keep it in blue) activity series and reduction potential.1532

This, over here on the left, is called an activity series; and over here is called a table of reduction potentials.1540

Notice, here you have no numbers; over here, you have these numbers--these reduction potentials.1548

Let me tell you about the activity series first.1557

This basically says the following: let me mark off hydrogen, just as a reference; and the reason I do that is because, over here, if we can find hydrogen--notice, it is going to be 0.1560

Hydrogen is sort of a...it is what the chemists have chosen as a reference point, above and below.1575

These things--if I take, for example, chromium and aluminum--when you are looking at an activity series, anything that is lower on the activity series means that it has a stronger affinity for electrons, which means, if I put a chromium ion near aluminum metal, the chromium is going to take the electrons from the aluminum.1582

The way that you are probably going to be seeing this in class is the following: If I take aluminum metal, and I add it to, let's say, chromium...let's just use chromium chloride, the question is: is there going to be a single-replacement reaction?1613

Will aluminum replace this?--well, yes, it will.1632

You will actually end up getting Al₂Cl₃, aluminum chloride, plus chromium metal, solid.1635

Here, chromium is in ionic form, right?--well, the activity series--the way it is explained is that anything that is higher up on the activity series will replace the metal that is lower on the activity series.1643

But what is really happening here is that the thing that is lower on the activity series is actually oxidizing the thing that is higher on the activity series.1656

That is what activity series is; it's a measure of the extent to which something wants to be oxidized.1667

Things on top oxidize easily; things as you go lower--they oxidize less easily; or, because oxidation and reduction are complementary processes--as you go lower, they have a greater tendency to reduce.1673

In other words, they have a greater tendency to take electrons.1687

So again, the thing that is lower on the activity series will actually take electrons from what is higher.1690

Chemically, what that looks like is this: you have aluminum metal and chromium ion; chromium is lower than aluminum; chromium doesn't want to stay in ionic form--it will take electrons from aluminum, and it will convert itself to chromium metal.1697

Aluminum ion will now go from aluminum metal to aluminum ion. That is what is happening.1711

Now, let me express this in terms of reduction potential.1717

These numbers up here are actual measures of the strength--how badly something wants to be reduced.1723

Activity series is the reverse of reduction potential, but reduction potential is written numerically from big number to small number.1731

What that means is that--let me see if I can find aluminum on here somewhere; there it is, so let me put that there; and let me see if I can find chromium.1740

Chromium, chromium, where is chromium...aha; all right, there is chromium.1751

Notice, here, that chromium--the reduction potential is -.74; the aluminum is -1.66; even though this is negative, relatively speaking, .74 is higher up, is a bigger number, than -1.66.1757

This means that, in a battle of electrons, chromium will beat aluminum; chromium will take electrons from aluminum.1777

Anything with a higher reduction potential will take electrons from something with a lower reduction potential.1786

The activity series is reversed, because what they noticed, historically, is that for example, when I mix aluminum and chromium, the aluminum will actually replace the chromium in this example.1793

Let's say I had aluminum chloride (Al₂Cl₃), and let's say I drop in some chromium metal--let's say I reverse the position of these.1808

So now, chromium metal goes into a solution of aluminum chloride; is anything going to happen?1817

No; the reason something is not going to happen is because chromium is lower; in other words, chromium already has all of the electrons it wants.1823

Aluminum is already ionic; it has no electrons to give up; nothing is going to happen.1832

There is no reaction.1837

So, activity series--the thing that is lower on the activity series will take electrons from what is higher on the activity series.1839

Or, the thing that is higher will replace the thing that is lower in a single-replacement reaction.1847

The important thing is not so much the single-replacement reaction; what is important is that you recognize that they switch precisely because chromium takes electrons from aluminum.1857

This reduction potential--this is why I want you to get used to this, as opposed to the activity series.1867

This gives you a numerical measure of just how strongly it will...anything on top will take electrons from anything on the bottom.1872

Please, know these very, very well.1883

Activity series--you will see it as activity series, but it is reduction potential; this is what is going on--oxidation-reduction.1885

It's not just about a single-replacement reaction.1892

That is just incidental.1894

OK, so let's do an example here: let's just finish off with the example that we did.1897

Example (yes, that's fine) #2: we want to write balanced molecular, total ionic, and net ionic equations for aluminum plus hydrochloric acid.1908

OK, so aluminum plus hydrochloric acid: well, is a reaction taking place?--yes, a reaction will take place, because if you look on the activity series or the reduction potential (let's just deal with the activity series), H is lower than Al.1941

That means it will oxidize Al; it will take electrons; so this--they will switch places.1960

It turns into AlCl₃--again, I'm putting together an ionic compound, so I have to make sure that it goes together properly--plus hydrogen gas.1967

Now, we just go ahead and finish it off; this is going to be 2; this is going to be 6; this is going to be 2; and this is going to be 3.1981

This is the molecular formula; it gives me the stoichiometry.1990

Total ionic will let me break it up into its ions.1993

Well, I'm going to have 2Al, plus 6 hydrogen ion, plus 6 chloride ion, going to 2Al³⁺ + 6Cl^- + 3H₂ gas.1997

Now, I cancel what is on the left and what is on the right, which is chloride and chloride, and I am left with the actual reaction that takes place.2016

2 aluminum atoms--metal--react with 6 hydrogen ions to create 2 aluminum ions plus 3 molecules of H₂ gas.2024

That is your net ionic.2038

Aluminum is oxidized--2 aluminum ion; hydrogen ion is reduced to hydrogen gas.2041

This only happens precisely because the reduction potential of hydrogen is higher than the reduction potential of aluminum.2048

If you look back on your reduction potentials chart, you will see that H is 0; the aluminum is -1.2, 1.3, something like that.2055

OK, let's do one more here: let's do Example 3 (let me do this in blue).2064

Again, we're going to write balanced molecular, ionic, and net ionic equations for the following reaction: what happens when I put together magnesium metal in a solution of chromium sulfate--chromium (3) sulfate?2078

6, 6...that is correct; all right, so, let's see!2100

Mg + Cr₂(SO₄)₃; so, the question is, will magnesium replace chromium?2106

Well, yes; if I look on the reduction potentials chart, chromium has a higher reduction potential than magnesium does.2117

Therefore, it will switch; it will take these electrons, and the magnesium ion will end up binding with the sulfate for magnesium sulfate.2123

That is all that is going on here.2134

Or, in the activity series, chromium is lower than magnesium; therefore, it will take it; magnesium will replace the chromium.2137

Let's put it together: magnesium is 2+; SO₄ is 2-; we end up with magnesium sulfate, and chromium ends up just becoming chromium metal.2143

Let me put a 2 here to balance the chromium; I have 3 SO₄, 3Mg, 3Mg; I think that takes care of it.2155

Yes, OK; so we have our molecular formula.2167

This is solid; this is solid; these are metals; these are aqueous; aqueous; and again, they won't always be aqueous--these are not always soluble.2171

We always still have to check solubility.2182

So, the total ionic would be: 3 magnesium metals plus 2 chromium 3+ ions, plus 3 sulfate ions, goes to 3 magnesium ions plus 3 sulfate ions plus 2 pieces of chromium metal.2185

Chromium metal--metals and ions are not the same; chromium is 3+.2212

Now, let's cancel what is on both sides (sorry about the stray lines--hopefully you can see everything).2219

Our net reaction is: 3 pieces of magnesium metal react with 2 chromium 3+ ions; they produce 3 magnesium ions, plus 2 pieces of (sorry; this is chromium metal--of course it is; you saw that up there) chromium metal.2227

Chromium oxidized magnesium to turn it into magnesium ion.2248

Chromium was reduced in the process--gained electrons to become chromium metal.2253

It went from solution to falling out of solution; this went from a solid, sinking in solution, to all of a sudden dissolving and disappearing into solution.2258

That is what is going on here.2266

This is the chemistry--this equation right here--oxidation-reduction.2267

OK, now let's do an example of a stoichiometry problem, given oxidation-reduction conditions.2275

We have Example #4: Will an aqueous solution of cobalt (2) sulfate oxidize zinc metal?2288

That is the question; if so, find the mass of cobalt formed when 2.6 grams of zinc is dropped into a solution--dropped into--I'll tell you specifically, it's going to be 160 milliliters of a 0.95 Molar cobalt (2) sulfate--so it's going to be CoSO₄.2321

Will an aqueous solution of cobalt sulfate oxidize zinc metal? If so, find the mass of cobalt formed when 2.6 grams of zinc is dropped into 160 milliliters of a .95 (let's make my 9 a little clearer here) Molar cobalt sulfate.2368

Well, if I look on the activity series or the reduction potential--so, cobalt is below zinc on the activity series; it's higher than zinc on the reduction potential, which means that it has a higher reduction potential than zinc--yes, cobalt will oxidize zinc.2385

So, let's go ahead and write it out.2401

I have zinc metal, solid, plus cobalt sulfate; it will turn into zinc sulfate (zinc carries a +2 charge, so there is no balancing to do here), and we have that it will form cobalt metal.2405

Good--nice and easy: when we do the ionic, the sulfates cancel, and what we end up with is a net ionic of the following: zinc metal, solid, plus cobalt 2+, which is aqueous, turns into zinc ion plus cobalt metal.2426

That is the reaction; this is what is happening--and again, the stoichiometry is 1:1--not a problem.2447

Now, let's go ahead and see what is going on.2452

We have 2.6 grams of the zinc metal, so let's find out how many moles that is.2455

2.6 grams (oops, let me write a little shorthand; I want to specify zinc here so I don't have to keep writing it), times...well, one mole of zinc is 65.39 grams, giving me 0.0398 mol of zinc.2460

OK, now, as far as the cobalt is concerned: I have cobalt sulfate; one mole of cobalt sulfate releases one mole of cobalt ion, so I have (let me put cobalt there) 0.160 liters, which is the 160 milliliters, times 0.95 (that is the molarity--that is moles per liter), so that I can find the number of moles of cobalt and compare the two to see which one runs out first.2487

That equals 0.152 mol of cobalt ion; and again, it's moles of cobalt sulfate, but cobalt sulfate, as you can see from this--one cobalt sulfate releases one cobalt ion; that is what this is--1--that is what these subscripts tell me.2516

OK, let's see; now that we have .0398 moles of zinc, .152 moles of cobalt, well, let's do our limiting reactant.2537

We can see, just from here, because this is 1:1, that .152 moles of cobalt requires .152 moles of zinc.2547

But we only have .0398 moles of zinc; so, it's the zinc that is limiting.2558

Don't worry if we did that a little fast; we'll actually do it numerically in the next lesson, when we do a series of examples in stoichiometry.2568

Again, 1:1; I have .152 moles of cobalt; I need .152 moles of zinc; I only have .0398 moles of zinc--not enough--therefore, that is what is limiting.2576

The zinc is limiting, so I have to use the .0398 in my further calculations.2586

All right, now we take 0.0398 moles of zinc metal, and the mole ratio of zinc to cobalt (this is ion--no, I'm sorry, this is zinc metal); moles of cobalt is 1:1, right?--one mole of zinc produces one mole of cobalt.2593

That is 1:1, so I end up producing 0.0398 moles of cobalt metal; and 0.0398 moles of cobalt times its molar mass, which is 58.93 grams per one mole--I end up with 2.35 grams of cobalt produced.2631

There we go: we started off with some zinc; we started off with some cobalt sulfate; we drop the zinc in there.2667

Cobalt, because it is higher, has a higher reduction potential than zinc (or it's lower on the activity series); it will take electrons from zinc, turning it into zinc ion.2675

In the process, the cobalt ion will turn into cobalt metal and fall to the bottom--hopefully, fall to the bottom.2685

It will actually stick to the zinc; you can scrape it off later--not a problem.2691

Theoretically, we can form 2.35 grams of cobalt.2695

OK, now I just want to give you some words of caution when dealing with activity series and reduction potential.2700

Here is what happens: let's use our example of zinc and magnesium.2709

Zinc and magnesium: all right, if I take zinc metal, solid, and if I mix it with magnesium metal, there is going to be no reaction.2714

The reason there is no reaction is, well, zinc has the electrons it wants; magnesium has the electrons it wants; they're neutral--there is no real desire for electrons at this point.2731

Metals generally tend to want to give up electrons, instead of taking them; there is nothing going to happen here--there is no oxidation-reduction.2740

When you see oxidation-reduction potentials, it depends on what species you are mixing.2747

In this previous example, we had cobalt ion: it's missing electrons; therefore, since zinc metal has electrons, it will take them.2753

But here, nothing is missing; so there is going to be no reaction.2764

However, if I do zinc ion plus magnesium metal, now zinc is missing electrons; magnesium has electrons to give.2767

Here, because zinc has a higher reduction potential--it's lower on the activity series--it will actually take the electrons that magnesium has to give.2779

It becomes zinc metal plus magnesium ion.2788

Well, another possibility is: how about if we do zinc metal plus magnesium ion?2794

Well, again, zinc has a higher reduction potential than magnesium--but zinc already has its electrons, and magnesium doesn't even have electrons to give up any more.2800

Even if something were in the neighborhood that would pull it, there is no reaction here.2810

One more possibility: what if I had both ions floating in solution--zinc ion and magnesium ion?2816

Well, yes, zinc wants electrons--it is missing 2 electrons; it is higher than that in reduction potential; but magnesium doesn't have any electrons to actually give up.2823

Therefore, there is no reaction.2836

When you look at these oxidation-reduction reactions, make sure that the thing that wants to be reduced is in ionic form, and the thing that is being oxidized actually has electrons to give up.2838

The charts that I gave you are specific, in terms of the species that are involved.2851

Zinc ion, plus magnesium, will produce an oxidation-reduction reaction; no other combination will do that.2857

So, it isn't just zinc and magnesium metal; no, it's very, very specific about how this chemistry works.2864

OK, thank you for joining us today at Educator.com for oxidation-reduction chemistry.2871

We'll see you next time; take care.2877