Lecture 19. Redox Reactions
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Transcript: English(auto-generated)
00:09
so-so, chemists, time-wise, it's like 2019, it seems like yesterday is the first time.
00:23
Now we're already very different, so we're in progress here. And we are continuing our journey into chemistry with getting on our final topic of the entire course, which is redox reactions. So that's the last part, and once again, our three main topics.
00:45
These are the main types of reactions that you may know about, because they are so central to chemistry that you need to have the knowledge for them. And these are the presentation reactions, the acid-based reactions, and finally the redox reactions.
01:02
And we will be continuing talking about the redox reactions until the end of the lecture. We will have a lecture on Wednesday, and Wednesday is a funny day, because some folks will be traveling to their homes elsewhere,
01:30
and that's fine, but for those of you who are here, I strongly recommend you try to make that comment, because I'm going to discuss some very important points, basically about the term of the topic. So it's kind of like a little bit of a review section.
01:45
So there's other points that are very specific, specifically target points that are typically hard for people to kind of like digest. So it's very useful in your own bias. Next week thereafter, after Thanksgiving, we have midterm.
02:01
It's midterm number two. Midterm number two is a little awkward, because of the time, because of its kind of late in the season, so technically we have our midterm this week because of Thanksgiving the whole day is we shifted it up and that leads to a very strange situation that the midterm
02:21
and the final are separated by only two weeks, so that's very close. So I apologize to that, we couldn't figure out another way to do it, but we were appreciating this problem. So midterm, then we have only one week of lectures left, and then it's the week of the finals. So we're seeing the end of the tunnel
02:44
coming very close. All right. Before we do that, though, we have to talk about redox reactions, so let's get it started. Here is a beautiful molecule. You may recognize it, this is methane. And methane has, of
03:02
course, a carbon atom right here in the middle. Now, carbon is, of course, a very ubiquitous atom when it comes to organic compounds, in fact, organic compounds are carbon-containing compounds. Here is another one. This is carbon right here, and at first I said, well this is the
03:21
same kind of carbon atom, it's just a different box. But we would like to have a way to kind of maybe classify differences between these atoms. One thing I should say, for instance, is that this carbon has four bonds, four single bonds, and this one has two double bonds. Clearly this carbon sees the sliding of the environment.
03:40
and because it seems slightly different, I could insinuate that the electron distribution around the carbon atoms, the central carbon atoms in this case, is a little bit different compared to the situation here. So these two carbon atoms, in terms of the intrinsic atoms, are exactly the
04:02
same, but in terms of their surrounding, they're different. They're different bonds. So there's many ways in chemistry in which we can describe those differences. And one way in particular is relative to us with our bond redox reactions. One way to kind of classify such differences, the immediate environment of the
04:24
atom in terms of the electron distribution is a quantifier, which is called oxidation state. So oxidation state is a funny descriptor. It is a descriptor like electro negativity. It is a number that is
04:40
as chemists came up with, just to simplify our thinking about particular processes. So, as you will see, it has to do with losing and gaining electrons, but it's not the same as charge. So we'll get back to that point. So for now, consider it as just a trick for chemists to keep track of certain
05:03
processes. So it's really an artificial descriptor. So for instance, chemists would call the oxidation state of the atom this carbon atom, minus 4. So the value of the oxidation state is minus 4, and just to indicate the differences between the two situations, this is plus 4. And you'll see that these numbers will get some meaning in
05:27
a few seconds. But you clearly see, even though the carbon atoms are exactly the same, their immediate surrounding, in terms of where electrons are close to it, defines that the oxidation state here is minus 4 and here is plus 4.
05:43
Now, if the atom is just by itself, or a part of the elemental material, for instance in the face of carbon, the elemental material of carbon is a diamond. Only carbon atoms. So in its elemental form, the oxidation state is always zero. However, when the element is part of a compound where there are different atoms associated with it, then
06:10
it is generally non-zero. So if the atom is part of the elemental material, the oxidation state is zero, but if it's not part of the elemental material, in the case of carbon, that
06:24
would be diamond. This is including a diamond, so these are compounds that are not elemental, then the carbon has generally an oxidation state that is not zero. And that holds for every element. Not just carbon. So to make a long story short, the oxidation state is not zero.
06:42
The oxidation state is a descriptor that chemists use to identify differences in the surrounding of the atom in terms of where the electrons are. So let's look at this table. This is a table that contains elements, but this is the truncated form
07:03
of the periodic table, and you see here numbers in red. And those numbers are oxidation states. So let's highlight a couple. Let's start with fluorine. Fluorine, if it is part of a compound, has oxidation state minus one.
07:23
If fluorine is in its elemental form, which is fluorine gas, F2, the oxidation state is zero. But if it's part of a compound, or if it's bonded to something else, the oxidation state is minus one. Sodium, if it's bonded to something, typically in form of an ionic compound,
07:43
its oxidation state is plus one. So you see here, this is very close to what you would expect based on their charges, if they are ions. So we know that the charge of a sodium ion is one plus. The oxidation state is plus one.
08:01
Fluorine, we just saw, has oxidation state minus one. We know that fluorine, when it is an ion, has the charge one minus. Let's look at another one. Here is beryllium beryllium has oxidation state plus two when it is part of a compound. The beryllium ion has charge two plus. So, again, this is
08:26
the same as the charge of the ion. And in terms of iron, we know iron, we have seen, actually can have two different types of charges. It can have two different kinds of oxidation states. Those oxidation states are
08:43
plus three and plus two. Those are the most common. So we actually got more, so these are the most common oxidation states of the iron element. So unlike fluorine and sodium, which only have one oxidation state, this element here, iron, has at least two oxidation states. And the same holds for carbon. Carbon does
09:06
not have one oxidation state, it has multiple. And the most common ones are plus four and minus four. And even hydrogen, the simplest element we have, can have two oxidation states.
09:20
Plus one and minus one. So we see that some elements have one oxidation state and others have two. So how do we go about this? We need a set of rules. Yes. These guys are the noble gases, okay,
09:43
and so they typically don't have an oxidation state, but you will see, actually, an example where we will highlight oxidation state from xeons. So not all noble gases are completely without oxidation states. But basically, this table here, we will simply not use those elements in the discussion of problems.
10:06
Okay. And actually the reason for that is that these elements simply don't form compounds in general. They form compounds in very extreme situations. So a vehicle in noble gases, because they don't form compounds in something else, they don't react.
10:21
So that means they're always in their elemental form, and that means the oxidation state is always zero. Okay. So when an atom is in its elemental form, the oxidation state is zero. This is basically our definition. In other words, carbon in its elemental form, diamond, the carbon atom here has an oxidation state of zero.
10:42
Oxygen in its elemental form is oxygen gas, O2, it has an oxidation state of zero. And iron, solid iron, in its elemental form, is also an oxidation state of zero. And also every element in its elemental form has an oxidation state of zero.
11:01
If you have a monoatomic ion, so when you look at an ionic form of an element, for instance, sodium chloride, sodium atom is not in its elemental form, it's an ionic form, then the oxidation state is the same as the charge of the ion. And we saw, all right, fluorine,
11:21
minus one, sodium, plus one, right? So these are the charges of the elemental ions. That means that barium 2+, has an oxidation state of plus two, and fluorine minus one has an oxidation state of R.
11:46
Do you want some H2O? Okay. Fluorine,
12:00
like I said, has an oxidation state of minus one, and so that means that fluorine is part of a compound, for instance, HF, and this F has an oxidation state of minus one. The same holds for this F, which is part of a compound, a covalent compound, but this is before covalent. It doesn't matter if the F is associated with an ionic compound, or if the covalent compound, SF6, definitely is a covalent compound.
12:26
Yet, the oxidation state of fluorine is classified as minus one. It doesn't mean that the fluorine is charged as one, this is why I'm saying that this minus one is just a descriptor for chemists to keep track of chemical reactions. And you'll see how
12:43
it comes into play. So don't confuse these oxidation states with it being charges. They're not. They're related, but they're not the same. Okay. Oxygen, if it's part of a compound, its oxidation state is minus two. Except when oxygen is in a peroxide. We will not consider those cases.
13:04
So, in H2O, this oxidation state of this atom here is minus two. And it doesn't matter where the oxygen is, so in nitrogen, NO2, the oxygen is also having an oxidation state of minus two. It doesn't have a charge.
13:21
There's an oxidation state of minus two. When hydrogen is part of a covalent compound, for instance, H2O, that's a covalent molecule, we know that, then its oxidation state is plus one. And the same for NH3. So when the hydrogen is covalently bonded to another element, then its oxidation state is plus one.
13:45
However, when oxygen- sorry, when hydrogen is part of a compound that involves a metal, okay, like magnesium hydride, a sodium hydride, then the oxidation state is minus one. Yes? Hydrogen peroxide is plus one for hydrogen.
14:07
It's different for oxygen. But hydrogen has been hydrogen peroxide and is still plus one. Because it's covalently bonded. Okay. So these are a set of rules, and with these set of rules you can do almost everything, yes.
14:22
The relationship on the level of ions, so as long as you have ionic compounds, then it is one-to-one. Whereas if you have covalent compounds, then that relation stops working properly. Because with covalent compounds you don't have charge separation.
14:44
Now we keep this method anyway because it really helps us find out how electrons flow in a chemical reaction as you will see. So in addition to these basic rules, we have two more rules that will help us to find oxidation state of all elements.
15:03
Rule number one. The oxidation state of all atoms in a neutral molecule add up to zero. So if you have a molecule that is neutral, that doesn't have a charge, so it's not an ion, but a neutral molecule, then if you add up the oxidation state, all these guys should add up to zero. This rule is going to be very helpful for us.
15:26
And if you have not a neutral molecule, but you have an ion, a polyatomic ion, like this is sulfur, SO4 2-. Then, if you add up the oxidation states of the atoms in the polyatomic ion, they should add up to the charge of that ion. So if
15:44
you add up the sulfur, plus the four oxygens, they are oxidation states, you should retreat minus two. Because that's the charge of the ion. All right? So this list of rules actually will help us determine the oxidation states of a whole bunch of elements.
16:02
Okay. So let's put it into practice and do a couple of quick exercises to see if we get this. Here are three. HBr that's the first. Okay. So H has an oxidation state of plus one
16:22
when it is bonded to a non-metal. Okay? So this is a non-metal so in this case, this H has an oxidation state of plus one. Now this whole thing must be neutral, because it is. That means that this guy must have an oxidation state of minus one. Okay? Because they have to add up to zero.
16:44
So this Br, actually, I don't know at the onset what the oxidation state is, but I use what I do know about hydrogen, and then from using the rule that they must add up to zero, I can determine that bromine has an oxidation state in this compound of minus one.
17:00
It doesn't mean that every bromine always has an oxidation state of minus one. Because in this compound, as you will see in a second, it does not have an oxidation state of minus one. So let's do that. What do I use here? Well, I use the same trick as before this H must have an oxidation state of plus one because it's not bonded to a metal. Okay?
17:22
So this one has plus one once again. I also know that oxygen, when it's bonded to another element, it has an oxidation state of minus two. Except when it's peroxide, but I promise you, we will not talk about peroxide. For all intents and purposes, the oxidation state of oxygen is minus two in the compound.
17:45
Okay? So this is minus two. Now what do I do now to determine the oxidation state of bromine? I take this one, four times minus two should add up to zero and that allows me to calculate the oxidation state of bromine. So what I do is I have one times plus one,
18:02
four times minus two when I add up the unknown oxidation state, X, of bromine, I should add up to zero. So I can quickly solve for X and if you do that, you find X is seven. See, X is seven here, it's minus one there. Very different.
18:21
So the oxidation state of bromine in this compound is plus seven. So I use what I know for hydrogen, and what I know for oxygen, and that is how I determine the oxidation state of the element that I do not know how to determine the oxidation state. In terms of notation, this is how it works. You always indicate the oxidation state of a single atom.
18:45
And then, to calculate what the missing oxidation state of the other atom is, you consider the subscripts. Okay? But you don't put here minus eight. It always means the oxidation state of a single atom. Irrespective of the subscripts.
19:06
Okay. So let's do this one. This is hydrogen bonded to calcium is a metal, so now in this case, hydrogen has an oxidation state of minus one and there's two of them and that means that calcium is having an oxidation state of plus two. This is
19:22
to be expected. Calcium has typically a charge of 2+. This is a very plausible outcome. Calcium has an oxidation state of plus two in this compound. Okay. Let's do another three.
19:41
Interesting compound here. Where do I start? Let's first see that this looks to me like an ionic compound, this here is a polyatomic anion this is magnesium magnesium has always only one oxidation state, which is corresponding to its charge. It must be 2+. Magnesium, in its ionic form, its oxidation state is 2+.
20:05
I also know oxygen, when it's bonded to something else, has an oxidation state of minus two. So I use that as well and I have only one element missing, which is a phosphorus. So when I do the same trick 7 times minus two
20:21
plus two times plus two equals two times the missing oxidation state of this element. So I have the following equation 2 plus 2 which is form of magnesium plus 7 times minus two, which is this one here, plus 2 times the missing oxidation state of the element, phosphorus, should add up to
20:46
zero. Only one unknown in this equation, so I can quickly solve for x, and x is 5. In this case. And 5, indeed, is a common oxidation state for phosphorus. So even if you don't know that, it should be fine. All right. So now here is an interesting situation.
21:13
This is not a neutral molecule. Okay? This is a polyatomic anion. This is nitrate. It has a minus charge.
21:21
So what I do is the same trick as before, but I don't put in zero, I put minus one, because it adds up to the charge of the polyatomic anion. So let's do that. I'm going to start again with oxygen because I know oxygen always has oxidation state of minus two, when it's bonded to something else.
21:41
So minus two, this one, nitrogen, I don't know the oxidation state of, but I do know that one times nitrogen plus three times oxygen, which has minus two, must add up to minus one. Okay? Because the charge is minus one. Or one minus.
22:01
That means again I can solve for x, and in this particular case, x equals 5. The oxidation state is 5. Up to the nitrogen. This does not add up to zero because this is a polyatomic anion, and the rule is, if it is so, then the oxidation states add up to the charge of the polyatomic anion.
22:26
The charge of this molecule is zero. So it adds up to zero, the charge of this guy is minus one, adds up to minus one. Okay. Final one. Exotic compound. One that's xenon. Xenon is a normal gas. It doesn't really form a lot of stable compounds. This is actually
22:40
one of them that is stable, with a little bit of an exotic case, but it doesn't matter for our purposes. We can just try to calculate the oxidation state anyway using the same set of tricks. Oxygen, once again, minus two. Fluorine is always minus one in compounds. So I have fluorine here as well, I indicate the oxidation states of the single atom here. Okay? Then
23:05
I'm trying to calculate the oxidation state of xenon by considering four times minus one plus one times minus two plus x equals zero because this is a neutral compound. Minus two, that's for the oxygen, plus four times minus one,
23:22
that is four times minus one of fluorine, plus the unknown, x equals zero, that means x equals six, and xenon has an oxidation state of plus six. So you can see by using the rules we just went through, looking at hydrogen, oxygen, and fluorine,
23:42
we're using the fact that it should add up to four times minus one. Either zero or the charge of the form is zero, you should be able to derive the oxidation state. It only happens in a particular form. Now this is the first step to understanding what happens in an oxidation reduction reaction. So let's have a look at that right now.
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This is what happens. I have a compound A, okay, to be specified, and a compound B. This is before the reaction, after the reaction, this happens. Okay? So A apparently has lost two electrons
24:24
in this particular example and B took them out. So what this really is, is basically the transfer of electrons from one compound to another compound. And you can compare this with an acid-base reaction
24:42
where we have a proton that goes from one compound to the next. It goes from an acid to a base. Here, it's not a proton that travels, but it is a electron, or two electrons, or three electrons. So electrons are being channeled from one compound to the next. Now, if you do that, we can look at this a little bit more closely and realize
25:09
that this guy is losing electrons. This one is losing electrons. Before the reaction happened, it had more, after the reaction has been completed.
25:20
It has less. That step, the losing of electrons, is called oxidation. So oxidation means the compound loses electrons. That's what it is. If you lose electrons, you're being oxidized. So A is being oxidized. And because A is being oxidized, it has a name
25:42
and it's called the reducing agent. So the reducing agent becomes oxidized. That means the reducing agent loses electrons. Compound B actually, the situation is reversed because compound B gains electrons
26:00
and the process of gaining electrons is called reduction. And that means that B is being reduced. B is called the oxidizing agent. It is the agent that makes this step possible A makes this step possible. That's where the names come from.
26:23
So the oxidizing agent is being reduced which means it gains electrons the reducing agent is being oxidized, meaning it loses electrons. That's it. Okay? So in an oxidation and reduction reaction, these two steps happen at the same time.
26:45
Okay? So one is gaining, the other one is losing. It happens at the same time. That's why it's called redox. Oxidation and reduction. What happens is, one or more electrons are transferred from one compound to the next.
27:02
This diagram can be summarized as follows the reducing agent is oxidized, that means it loses electrons the oxidizing agent is reduced that means it gains electrons. Oxidizing means if you are oxidized, you lose electrons, if you are reduced, you gain electrons. Okay. Now let's look at this.
27:28
We'll look at this closer, and some pictures will be helpful. So here we have sodium sodium here is solid, that means it's in its elemental form
27:40
this is chlorine, CO2 it's chlorine gas, it's also in its elemental form. I can make these things react in a particular way, and then after the reaction, I have, lo and behold, NACI. sodium chloride. sodium chloride looks very different, and salt.
28:00
Actually, you can eat this, no problem, but you cannot really eat this, and certainly you shouldn't eat salt, certainly. And it will explode from the inside out. So it's remarkable that these elements here in their elemental form once they exchange electrons, they become something completely different.
28:22
So let's look at this, this is an oxidation reduction reaction, and I'll show you why that is so. So let's look at sodium first. Sodium is in its elemental form that means it has oxidation state zero after the reaction, it's right here now it is an ion
28:42
that means it has oxidation state 1, because the charge is goblets. If you look at this, you see it has become more positive. The oxidation state has grown. That means it has lost electrons. If something becomes more positive, between the charge and oxidation state, that means you
29:05
will lose electrons. That means it is being oxidized, because being oxidized means you will lose electrons. So in this process, sodium is being oxidized, which loses electrons. Chlorine also has oxidation state zero at the very beginning,
29:25
but after the reaction we see it is in an ionic form. meaning it has charge minus one, and that means oxidation state minus one. Sorry. Charge, one minus, oxidation state minus one. It's your first.
29:42
So before zero, after the reaction, oxidation state minus one, it has become more negative, it means it gained electrons. That means it is being reduced. So this is an oxidation reaction. Oxidation reduction. A reduction reaction. So in a reduction reaction, you can say that
30:02
the oxidation state of reducing oxidizing agents are changing. If the oxidation states do not change, it's not going to be a reduction reaction. But if they do change, this is a reduction reaction. That is a definite reduction. The oxidation states of the reducing oxidizing agents has to change.
30:26
Okay. Another example. This is a combustion reaction. We've seen it many times. The combustion of methane. CH4 is reacting with oxygen to form CO2 and water. So let's look at the oxidation states of all these elements before and after the reaction.
30:48
Let's see what happens. So let's start with hydrogen on the metal. When hydrogen is bonded to a nonmetal, oxidation state is plus one. So what is the oxidation state of this carbon?
31:05
minus four, right? Because I have four times plus one, the whole thing needs to be zero, that means the oxidation state equals minus four. The oxidation state of oxygen is zero because it is in its elemental form.
31:20
Okay. So this is for the reagents, let's look at the products. Let's first look at carbon dioxide. The oxygen is part of a compound, that means oxidation state minus two, two times minus two, this guy is? Plus four. That's right. Because again, if you add four times minus two,
31:40
so two times minus two is minus four, you have to add four to make a neutral compound. Add up to zero. So the carbon here must be having an oxidation state of plus four. Let's look at this oxygen here, minus two, and this H here is plus one because it's part of a compound, not bonded to the metal, so plus one.
32:02
Okay. So let's make a table of carbon before and after. The oxidation state before equals minus four, after plus four, it becomes less negative, more positive, that means it loses electrons. And that means it is being oxidized.
32:20
So the carbon atom here is oxidized in this process. Let's look at hydrogen. Hydrogen has plus one here, plus one there. That's funny. It doesn't change. So is it an oxidizing agent or a reducing agent? Neither.
32:40
This is a spectator element. Like a spectator ion in a precipitation reaction, we can have elements here in the redox reaction that don't do anything. They don't change the oxidation state. They are spectators. Oxygen is zero before the reaction takes place, after minus two. So it changes, becomes more negative,
33:04
means it must have gained electrons. And if you gain electrons, you are being reduced. So combustion reactions are redox reactions.
33:20
Now let's look at this a little bit more closely, and this will be important for the next lecture, actually the next lecture, the next lecture, when we look at path reactions. I have here carbon has minus four here, plus four here. The difference is eight. Eight electrons. Okay? So eight electrons
33:40
must have been transferred from carbon and they go somewhere else. Where do they go? Well, they go to the oxygens. Because the oxygens actually have more negative oxidation states. So carbon basically transfers eight electrons to oxygens nearby. And we'll see again, when we talk about path reactions, how we can understand the transfer of electrons. You can keep track of them.
34:09
Okay. Here's another example I have here a copper wire and I dunk it into this solution which contains two things a solution of silver nitrate
34:23
okay, so the silver nitrate is there, and a little bit of ammonia in there. A little bit of ammonia. So what happens here is the following. The copper wire actually as you can see, the copper is going to lose electrons. So it's being oxidized.
34:41
and this guy here, the silver ion in the solution takes those electrons. So they are gaining electrons. They are being reduced. So silver is reduced, the copper is being oxidized. And in this particular case, you can see the beautiful blue color, and that is because they added a little bit of ammonia to the solution
35:00
okay, and this ammonia then will make a complex with copper ions. So I can detect whether or not copper ions are formed because of complexes with ammonia. You can hear this beautiful blue color. At the same time, if you look at this vinyl here, it looks very cruddy, and that's because silver deposits sit on this
35:21
wire. Part of the copper, which was a solid, has gone in the solution to become the copper ions, and in return, silver deposits the silver ions became salt-silver deposits. So it looks like a crony, silver coated wire coming out. This is a typical reduction
35:41
reaction. Okay. Now let's do a couple of examples just to get our fingers back. Okay. New situation. I have zinc metal, so that's
36:05
the products are zinc chloride and hydrogen gas. The question is, is this a redox reaction? And if so, could you tell me what the oxidizing agent, the reducing agent, and the substances being oxidized and reduced are? The first
36:25
thing you would like to do is to write out what you can read in words into a chemical equation. So let's do that. I see zinc here on the reagent side, and hydrochloric acid. So let's put that down. Zinc, hydrogen gas. and hydrochloric acid. It forms zinc chloride, right there, and hydrogen gas, H2.
36:47
So already balanced for you, there's a two here to balance the number of hydrogens and chlorine. So this is the balanced equation. What I have to do is to determine whether or not this
37:03
is a redox reaction. So I have to look at the oxidation states before and after. Let's do that. I see here, this is an elemental form, that means zinc has oxidation state zero. Zinc is in its elemental form.
37:21
This H here is bonded to Cl, this is not a metal, that means this has oxidation state plus one, and that means that chlorine has oxidation state minus one. I could already tell you that, because this is basically in solution to the ionic form, where Cl is minus, 1, and Cl is minus. And then the hydrogen is positive. So I can tell you that way.
37:44
Okay. On the other side, I see zinc now is in ionic form, that means this is 2+, because Cl is in ionic form as well, that means it has oxidation state minus 1, zinc must have 2+. So 2 times minus 1, zinc must be 2+. And then hydrogen here, what is the oxidation state? Zero, because it is in its elemental form.
38:10
H2 is in the elemental form of hydrogen. Okay. So the next question then is, is this a redox reaction? Yes it is. There it is.
38:22
It's because zinc is changing its oxidation state and also hydrogen is changing its oxidation state. So what we get is the following. Zinc goes from zero to plus two. Is it gaining or losing? Yes.
38:42
It is being oxidized, that means it's losing electrons, it is a reducing agent. So zinc is a reducing agent, it is being oxidized because it loses electrons. The hydrogen went from plus one to zero that's more negative.
39:02
So it needs a more negative charge added to it to become zero, that means it is being reduced, the definition of something that is being reduced, is the oxidizing agent. And then finally, chlorine does not carry the oxidation state. And that, of course, is a spectral element. Yes. Yes, it's possible. It's possible. But
39:30
usually, usually we zoom in on the elements. Usually we zoom in on the elements. So for these examples, zooming in on the elements gives a ratio. Okay. Another example. Liquid silicon tetrachloride is reacting with water and it forms hydrochloric acid and solid
39:52
silicon dioxide. The same question. Is it a redox? And if so, can you identify and oxidize it and lose the agents? So the first thing to do, again,
40:03
is to write down in the form of a chemical equation what is going on. This is silicon tetrachloride. This is water, this is HCl, which is a product, and this is silicon dioxide. Okay? I have balanced that for you, there's a 4 there, and there's a 2 there to make the number of oxygens in the
40:22
hydrogen. Okay. So the next step is to assign oxidation states. Let's do that. We'll see if we can do it quick this time. Chlorine and silicon. Chlorine, typically, like chlorine, when it's part of a co-vated compound, it has oxidation state minus 1. Okay? It's very similar to fluorine. So I can safely assume that, and
40:47
this is why I showed you the example. Chlorine, like fluorine, typically has oxidation state minus 1 in this part of the compound. That means that the silicon here must have oxidation state plus 4. 4 times minus 1 needs a plus 4 to make it 0.
41:03
Oxygen always matters, too. That's part of the compound. This H must be plus 1. Okay? On the other side, the Cl must be minus 1, and the H must be plus 1. Because they're ionic. And then oxidation states are similar to the charge of the ions. And then here on this side, the H must be minus 1.
41:23
I see oxygen, that's minus 2, okay, times 2, I need a plus 4 there to make it 0. Is this a redox? The answer is yes. Okay? That means no. And that is because there is no change of oxidation state. So there's no oxidizing or oxidation state. Okay. Well,
41:48
let's look at one more example of silicon tetrachloride. Okay? Another reaction of silicon tetrachloride, it's reacting with magnesium metal. and it will form adhesin chloride with solid silicon. Same question. Is this a redox? And if so, can you identify the oxidation state?
42:10
Okay. Then I'm going to start by writing down the equation. Silicon tetrachloride, it just gave you a previous slide, it's now reacting with magnesium, which is solid,
42:21
which is elemental magnesium, which is ionic compound, magnesium chloride, and solid silicon. If you look at this reaction, do you think it's a redox or not? Yeah, you can see it already, right? Because here, you see magnesium is elemental, and here it's not. So there must be a change of oxidation state. Well, let's have a look. That's a sinus oxidation state. In the previous slide, I already
42:45
did the one for silicon tetrachloride, so plus 4 minus 1, this is elemental. That means there's no change of oxidation state. And this is 0 for magnesium. On this side, magnesium, in its ionic form, it has oxidation state, plus 2. The chlorine
43:01
has oxidation state. Minus 1. Plus 2 minus 1. And this is elemental silicon, must be 0. So this is definitely a redox because there's changes in oxidation state. And let's see if we can identify what is being reduced, what is being oxidized.
43:22
So let's look at it, let's look at silicon. It goes from plus 4 to 0. Oxidized or reduced? Who says oxidized? Who says reduced? Okay. So this is a 50-50 split here. Let's look at this closely. Plus 4 to 0. It
43:41
was more positive, becomes more negative. You gain electrons, that means you are being reduced. You gain electrons if you are being reduced. And that means if you are being reduced, you are the oxidizing agent. Another example, of course, magnesium
44:00
from 0 to plus 2, it lost electrons. It becomes more positive. It loses electrons, and if you are losing electrons, you are being oxidized. And that means that you act as the reducing agent. Chlorine, again, once again, is a spectacular element. It doesn't do anything. It doesn't say it's oxidation state.
44:26
Okay. So make sure you can identify these two processes. Being reduced or being oxidized. That is the recipe for next time. See you on Wednesday.
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