0:15

So when we look at acid-base reactions,

Â we're frequently looking at proton transfer, or

Â H+ transfer between two species, either a proton donor or a proton acceptor.

Â When we look at redox reactions, or oxidation-reduction reactions, we're

Â looking at the transfer of electrons, and when we change the number of electrons

Â on a particular substance, we see a change in what we call the oxidation number.

Â And so, we want to talk about how we assign oxidation numbers to species, and

Â what that tells us about those substances in the reaction.

Â One important thing to note is that oxidation number is merely a bookkeeping

Â tool for us to use to be able to determine what is oxidized and what is reduced.

Â So this is an example of a redox reaction.

Â When we have a redux reaction we have to have two halves.

Â We have to have one oxidation, and we have to have at least one reduction.

Â And so, we can't have two oxidations or two reductions, or just an oxidation or

Â just a reduction.

Â They have to happen in that pair.

Â When we have an oxidation, what we see is that we have something like magnesium,

Â going to Magnesium 2+, it has lost electrons.

Â And if we have a reduction we have something that is gaining electrons.

Â So we have something going from O2 to say O2 minus,

Â where it has gained those electrons.

Â And what we can do is then take these two half reactions, and

Â figure out what the total reaction is, which in

Â this case is magnesium plus oxygen, yielding magnesium ions and oxide ions.

Â And so this is one of the reactions that was used in old flashbulbs,

Â to provide a source of light when taking photos.

Â 1:52

Now that we see our electrons ar ethe same on both sides,

Â we can actually cancel them out.

Â So, our overall reaction won't actually show those electrons, and

Â we can rewrite it to look like magnesium plus oxygen yielding magnesium oxide, but

Â we still need to be able to divide it up to determine what those number of

Â electrons are to make sure those are balanced as well as all the elements.

Â We won't get into balancing redox reactions in this course, but

Â we will look at how we identify what's being oxidized and what's being reduced.

Â So when we're looking at redox reactions,

Â we have to be able to show the two half reactions.

Â Each half reaction explicitly shows the electrons involved in

Â the overall reaction.

Â When we get to the overall reaction,

Â the number of electrons on the left should be equal to the number on the right, so

Â the electrons don't appear in the overall reaction.

Â And every time we look at a redox reaction,

Â we have to have one oxidation and one reduction.

Â 2:50

When we're looking at redox reactions,

Â sometimes it's a challenge to keep track of which one is which.

Â And so here are a couple of pneumonics that can help you remember remember what's

Â going on in this processes.

Â The first one is called OILRIG, so

Â oxidation is loss of electrons, reduction is gain of electrons.

Â Another option is LEO the lion says GER, so

Â lose electrons oxidation, gain electrons reduction.

Â Mean the same thing,

Â it's just a little mnemonic to help us remember what we're looking at.

Â 3:19

Before we can determine that what species is being oxidized and reduced, and

Â write our oxidation and

Â reduction half reactions, we first have to assign oxidation numbers.

Â And remember I said that oxidation numbers are merely a means of

Â bookkeeping of keeping track of our electrons.

Â And what we see is that the oxidation number is

Â the number of charges the atom would have in a molecule or

Â ionic compound if the electrons were transferred completely.

Â And when we're looking at an ionic compound,

Â the electrons are transferred completely.

Â The charge on the ion is the same as the oxidation number.

Â However, for molecular compounds that are covalent bonds.

Â What we see is the electrons are being shared.

Â And so we basically have to pretend for

Â a moment that instead of being shared, the electronics are transferred completely.

Â They will be able to assign oxidation numbers to every substance, and

Â as a result determine our reduction and oxidation half reaction.

Â 4:11

So let's go through the rules that we use to determine the oxidation numbers of

Â a species.

Â The first one is this,

Â probably the simplest, free elements with no charge have an oxidation number of 0.

Â This includes both monatomic and diatomic elements, but

Â any element that doesn't explicitly show a charge will have an oxidation number of 0.

Â When I see something that already has a charge, some monatomic ion,

Â what I see is that the ionic charge is equal to the oxidation number.

Â So for lithium plus 1, we have an oxidation number of plus 1.

Â For fluoride, we have oxidation number of minus 1.

Â For titanium our oxidation number is 4.

Â So, those are also fairly easy to identify.

Â 5:05

When we get to oxygen, we see that it usually has an oxidation number of

Â minus 2, which is the same as the charge it would have in an ionic compound.

Â Even if we have an oxygen in a molecular compound,

Â it's oxidation number is generally minus 2.

Â There are occasions where we have the oxidation of minus one for oxygen.

Â So something like H2O2 is a peroxide, or

Â a compound involving our peroxide ion, O2, 2 minus.

Â In that case, the oxidation number will be minus 1.

Â 5:35

For hydrogen, the oxidation number is typically plus 1.

Â Things like H2O, HCL, CH4, all of these has an oxidation number of plus 1.

Â If it is paired with a metal compound, lithium, calcium, sodium,

Â magnesium, then the hydrogen will actually have an oxidation number of minus 1.

Â We don't see these binary metal compounds or metal hydrides very often.

Â For fluorine, it will always have an oxidation number of minus 1.

Â And for other halides, they can have a value of minus 1 or

Â they can also have positive values.

Â So if I'm looking at something like HCl,

Â than chlorine has an oxidation number of minus one.

Â However, if I'm looking at something like KClO3 or HIO4, basically are acids formed

Â from our polyatomic ions or the derivative salts from those polyatomic ions,

Â then my halogen can have a different oxidation number.

Â And what I'm going to have to do is determine the oxidation number of

Â the halogen, based on the oxidation number of the other elements.

Â 6:39

So a few of the rules that apply when we're assigning oxidation numbers to

Â elements was the free element or an ion or in a compound.

Â For a neutral molecule the sum of the oxidation states must be equal to 0.

Â So if we're looking at something like sodium chloride,

Â sodium has an oxidation number of plus 1,

Â chloride has an oxidation number of minus 1, plus 1, minus 1 equals 0.

Â If we're looking at an ion, such as a polyatomic ion, then what we know is

Â the sum of the oxidation states must be equal to the net charge of the ion.

Â So in this case, I'm looking at permanganate ion.

Â I see I have a minus 1 charge, so I know that tells me that I'm

Â the sum of my oxidation numbers will be equal to minus 1.

Â I know that each oxygen has an oxidation state of minus 2.

Â So, here I have 4 times minus 2 to represent those four oxygens.

Â We still report the oxidation state as belonging to a single atom of

Â that element, so we still say the oxidation number is minus 2.

Â But it contributes a minus 8 to the overall charge of that polyatomic ion.

Â And then we look at what's left, and we see that manganese must be a plus 7, so

Â that the rest of the statement is equal to minus 1.

Â Now what the, most of the time,

Â our oxidation numbers are going to be equal to integers or whole numbers.

Â And this is actually an example we did look at in our chart of rules

Â because typically oxygen is a minus 2 oxidation number, or a minus 1.

Â On very rare occasions, we can have a fractional oxidation number,

Â such as this one for O2 minus, which have an oxidation number of minus one half.

Â 8:22

The first thing I'm going to notice about the substance is,

Â what is the overall charge, is it neutral or does it have a charge?

Â Because that's going to tell me some information about the sum of

Â the oxidation numbers.

Â For SO2, I see that the su,

Â the charge on the molecule is neutral, so I know that it's going to be equal to 0.

Â When I look at my oxygen know that oxygen is going to have an oxidation number of

Â minus 2, and that means I'm going to have x for my oxidation number of sulfur.

Â So I have x plus 2,

Â because there are two oxygen, times the oxidation number of minus 2.

Â So I have x minus 4 equals 0, and x equals 4.

Â So I was able to solve for

Â the value of x, and solve for our oxidation of sulfur as plus 4.

Â 9:12

When I look at my next example,

Â the dichromate ion, I notice that I have a charge of minus 2, so

Â I know that the sum of my oxidation numbers is going to be equal to minus 2.

Â Here, we have chromium,

Â which we didn't have any rules that apply to the oxidation of chromium.

Â So I'm going to call it x.

Â When I look at oxygen, it's a minus 2 oxidation number for each of them.

Â So I can say I have 2, from our subscript dichromium, times x

Â because that's the oxidation number of chromium, plus 7 times minus 2.

Â Because, I have seven oxygens, each with an oxidation number minus 2,

Â and that's going to be equal to a total of -2.

Â So I have 2x minus 14 equals minus2.

Â I add 14 to both sides, I get 2x equals 12, and x equals 6.

Â So now, I add the oxidation number of chromium is equal to plus 6.

Â I can go back and check my work.

Â I have 2 times plus 6, gives me 12.

Â Minus 2 times 7 gives me minus 14.

Â 12 minus 14 equals minus 2, so I know I've set my oxidation numbers up correctly.

Â 10:23

One more example where for our oxidation is positive.

Â We have out polyatomic cation here.

Â We know that we have a plus charge there, so

Â I know that the sum of my oxidation numbers is going to be equal to plus 1.

Â When I look at my nitrogen,

Â 10:38

I may not know what the oxidation number's going to be there, but

Â I do know that the oxidation number on hydrogen is going to be plus one.

Â Now, typically when hydrogen has a positive oxidation number,

Â it is written first.

Â Amino compounds,

Â these are NHx compounds, are an example of where they are not written first.

Â And so, remember, the only time hydrogen is going to be a minus 1 is when

Â it's paired with one of these metals, so

Â it's written second, but it's still a positive 1 value.

Â And I'm going to assign x as my oxidation number for nitrogen.

Â So we have x plus 4 times plus 1.

Â So we have x plus 4 equals 1 or x equals minus 3.

Â 11:22

So now we have an oxidation number of minus three for nitrogen plus four times

Â plus one of our hydrogen, which gives us a total of plus one, which is the overall

Â oxidation number of the molecule, or the overall charge on the molecule.

Â Now, let's set an example to let you assign the oxidation number for

Â chlorine in perchlroric acid, HCLO4, our correct answer is 7.

Â Remember, when we're looking at our compound we want to go through and

Â assign the oxidation numbers that we know.

Â Hydrogen would be plus 1, chlorine we don't know, so we're going to call it x.

Â And Oxygen's going to have an oxidation number of minus 2.

Â Because this is a neutral compound,

Â the sum of our oxidation number must be equal to 0.

Â So we have 1 from our hydrogen, plus x and

Â we only have one chlorine so we don't need a coefficient in front of the x.

Â Plus 4 times minus 2 equal to 0.

Â So we have 1 plus x minus 8 equals 0.

Â X minus 7 equals 0, or x equals 7.

Â So we have a plus 7 for the oxidation state of chlorine.

Â When I go back and look at my problem, I see I have plus 1

Â plus 7 minus 4 times 2, and that in fact does equal 0.

Â So, I know I've done the calculation correctly.

Â 12:48

Now let's look at an example and define which species, in being oxidized and

Â which species is being reduced.

Â So here we have iron plus copper ion, yielding iron ion plus copper solid.

Â Now we can go through and assign our oxidation states.

Â We have a free element, so our oxidation state will be 0 for iron.

Â For our ions, it's going to be the same as the charge.

Â So we have plus 2 for our oxidation state.

Â For iron, we also have a plus 2.

Â And for copper over here, we have a 0.

Â Now what we want to look at is how these substances change.

Â How does iron change as it goes from left to right, and how does the copper change?

Â So let's look first at our iron, and what we see is that

Â as we go from iron to iron 2 plus, that we've actually lost electrons, and

Â so that tells us we're dealing with an oxidation process.

Â When I go from copper to plus to copper, we're gaining electrons because we're

Â gaining negative charge, and the oxidation number goes from plus 2 to 0.

Â So our charge is going down, our oxidation number is going down,

Â we're gaining electrons, and that's considered to be a reduction.

Â 14:00

Now we know the process, it's undergoing oxidation.

Â We know that iron is being oxidized, and copper is being reduced.

Â There's another type of terminology that we also use,

Â when we're looking at redox reaction.

Â 14:35

So when we look at redox reactions,

Â there are actually kind of some subtypes that are all redox reactions.

Â And it doesn't mean that every time we see these reactions they're always redox, but

Â just to cover a few of them so you're familiar with them.

Â One is a combination, two or

Â more substances combine to form a single product.

Â So if we have something such as Na plus Cl2 going to NaCl,

Â that would be a combination reaction.

Â We can have decomposition reactions,

Â where a substance is breaking down into its components.

Â We can have a displacement reaction where an ion and

Â a compound is replaced by an ion or an atom of another element.

Â So we frequently see this when we look at the reaction of a metal and an acid for

Â example, and what we get on the other side is zinc chloride plus hydrogen gas.

Â All right, so that would be a displacement reaction, and

Â then we have a disproportionation reaction, it's an element in

Â one oxidation state, which is simultaneously oxidized and reduced.

Â 15:38

In this reaction, we're going to look at what happens when we combine silver

Â nitrate and solid copper.

Â This is an oxidation reduction reaction,

Â which is also known as a redox reaction, because we have the transfer of electrons.

Â So in our test tube we have a piece of copper wire, and

Â we're going to pour in some silver nitrate to cover that wire.

Â 16:06

And what we're going to see,

Â is that the copper that's in the wire is actually going to go into solution.

Â And the silver nitrate that's in solution is actually going to form solid silver and

Â plate onto the copper wire.

Â 16:25

As we can see, the solution is turning blue, and that is from the formation of

Â the copper two ions that have been oxidized from the copper 0 state.

Â And we also see the formation of solid silver which happens from the reduction

Â from the silver plus 1 ion to solid silver which has a 0 oxidation state.

Â In the next module, we're going to look at Combustion Reactions,

Â which is one type of a redox reaction.

Â