>> Welcome to Atoms and Elements, Part 3.
In this video lecture, we are going to look at ionic compounds
as oxidation and reduction reactions to form the ions.
And so this will just be a very preliminary introduction
to oxidation and reduction that we'll continue to build
on throughout the course.
Let's get started.
All right.
So we, in our last video,
we learned that ionic compounds form
from the transfer of electrons.
So we learned that the metals lose electrons, right?
To form the cations.
And the non-metals gain electrons to form the anions.
All right?
Now, the overall charge on each side
of the reaction must be equal.
So in chemical reaction, we have reactants on the left
and products on the right, and then there's our reaction arrow.
So you can think of that reaction arrow
like an equal sign.
All right?
And we can think of, chemical reactions are like a blending
of chemistry, English and math.
So when we look at a chemical reaction, we can treat it
like an algebraic expression.
[ Pen scratching ]
All righty.
And then there's a little mnemonic
that can help us understand oxidation and reduction.
There's two of them.
There's "LEO says GER" or the other one is "OIL RIG".
So as I explain oxidation and reduction,
I'll show you how these sayings can help you.
Right? So first we'll focus on oxidation, right?
So here it is, LEO, loss of electrons, whoops.
Loss of electrons is oxidation.
Right? LEO.
All right?
Or the OIL would be oxidation is loss.
Right? So there's our LEO.
So here we have sodium, as a pure element and we know
that pure elements have no charge,
and we know that it likes, it has one valence electron,
so here is the lost electron.
Alrighty. So, one thing
that sometimes students think is a little strange is
that we lost the electron, but we put the plus.
That's because in chemical reactions,
we only have addition, we never have subtraction.
The subtraction would be in the charge, all right?
So here's our lost electron, so here's LEO,
the lost electron means this is an oxidation reaction.
And here we go about balancing the charge, right?
The sodium was a zero, was neutral on the left,
treat the reaction arrow as an equal sign;
there's our sodium ion, that was plus 1,
and then there's the plus from our reaction.
And so you can think of it just as "and", right?
So everything is like "and, and, and".
And then the electron, of course, has a minus 1 charge.
So plus 1 and minus 1 equals zero, right?
So the overall charge on the left is zero,
and the overall charge on the right is zero.
So there's our oxidation reaction.
Now we've been working
to memorize the variable charge metals,
and this will help us understand why they have more
than one charge, right?
So here we have the pure element iron,
and pure elements are always zero.
But then the iron can either lose 2 electrons or 3 electrons.
Right? So here's those lost electrons, right?
Right is oxidation.
So there's our LEO, and then we look, so that's why
like for sodium, we just say
that this is a sodium ion, and everybody knows.
But if we say iron ions, people will be like, "Well which one?"
So that's why we have to have those Roman numerals, right?
This is iron(II), and then this is iron(III).
And if we were to go deeper into the chemistry, we would learn
that iron(II) and iron(III) have very different chemistries.
So it's important to distinguish between them
with the Roman numerals when we discuss them.
And then just to double check, right?
Overall charge, we started with a zero, with the pure element,
and then either way, right?
So if we, we'll just do this one here, right?
Iron was plus 3, and then we had 3 electrons, so once again,
the charge on the left equals the charge on the right.
All right?
So that's oxidation at its simplest level,
and that's a great place to start.
As we can see here with the iron, right?
This iron plus 2, iron plus 3 is involved in a lot
of biochemical reactions, so we'll be coming back
to this, again in the future.
Now let's look a little more closely at reduction.
So now this is right, gaining, so this would be our non-metals,
like Sulphur is a non-metal, right?
Here's our GER, gaining electrons is reduction.
So there's our GER, right?
Or RIG, for the RIG, reduction is loss of electrons.
So there's the RIG version.
All right?
So GER, OIL RIG or LEO says GER.
Okay. So now we're gaining electrons,
so here they are, right?
Gained electrons, right there.
And that's pretty much it.
So let's just double check,
see how we're doing on our charge, right?
So Sulphur is pure, it's zero,
and now it's gaining 2 electrons,
so that would be minus 2.
Our reaction arrow is our equal, and then we're forming sulfide,
the anion sulfide, which has a minus 2,
so I think at this point, we can just cover this up,
and minus 2 equals minus 2, right?
So when we have chemical reactions, the charge,
the overall charge for each side balances.
Now, you know, this is a little bit of continuity.
I don't know what to do about it,
so we're just going to go for it.
Amongst the non-metals, we have what are called
diatomic elements.
And you'll learn more about these in a future video.
But the main thing is that these elements are all
in their pure form.
In their pure form, they're like Siamese twins.
They are going to stay together.
So when we look at the- and they're non-metals,
so they want to gain electrons.
So here we have, we'll do chlorine.
Here's chlorine in its pure form, so it's neutral,
and then we've learned, right, that chlorine
when it reduces, it forms chloride.
Right? But because of the 2 here, we actually have to allow,
we have to count the atoms, and so,
to balance the charge, right?
Of our neutral chlorine molecule,
forms 2 chlorine anions.
We are going to have to gain 2 electrons.
All right?
And once again, right?
The charge is going to balance, because we have zero
and then we would have minus 2 equals minus 2, right?
Because we have 2 chlorides altogether.
Right? The 2 in front says we have 2.
Alrighty. Okay.
So there is the basics of oxidation and reduction.
So now let's put this altogether and practice it a bit
to make sure we understand what's going on.
All right.
So, as I mentioned, whether the electrons are gained or lost,
they are always added to the reaction.
So, gained electrons, right, appear on the reactant side.
So if we had copper, 2 plus, and it gained 1 electron,
it would go to copper plus, right?
So there's our reduction.
And just making a plug for those variable charge metals here,
right?
Has to have, copper 2 has to have the Roman numeral,
because it can exist in more than one oxidation state.
So even though this is reduction, oxidation
and reduction go together, so we call them oxidation states.
And then the loss of electrons, right?
Appears on the product side.
So let's look at an oxidation reaction.
So if we had iron 2 plus, oxidizing to iron 3 plus, right?
There's our lost electron,
and we've already named those on the previous page.
All right.
So this would be a good place to pause the video
and answer these next set of questions, on your own,
and then restart the video and see how you're doing.
What we want to do is we want to add the correct number
of electrons to either the reactant or the product side.
I left a big gap, so try not to give it away.
Classify the reactions as oxidation or reduction,
and then for our ions,
let's just practice our ion nomenclature.
All right, so I'll give you a minute to pause.
All right, let's see how you did.
Okay. So step 1, add the correct number of electrons, right?
So, remembering that electrons, right?
Electrons are negative, so if we have a zero,
and a positive, right?
We're going to have to add 2 electrons to the right,
so these are the lost electrons.
So since they're lost electrons, this is an oxidation reaction,
and then this would be a magnesium ion.
Okay. Now let's look at the next one.
Here is nitrogen, and now we have nitride.
So we have zero on the left, and we have minus 3 on the right,
I have two of them, minus 6.
So overall, there's minus 6 on the right hand side,
and there's zero on the left hand side.
So we are going to have to add 6 electrons.
So here they were lost, and here they're gained.
All right?
So gaining electrons, GER, right?
That's reduction.
Okay. So the nitrogen would gain 6 electrons
to make two nitride ions.
And when nitride, as soon as we see the
"-ide" we know it's an ion.
[pen scratching] Okay, so it's not so essential, right?
Here we have the magnesium element and the magnesium ion.
So then it's important to write "ion".
But as soon as we have the suffix,
-ide, everyone will know.
Okay, let's look at the last one.
Here we have chromium 6 and chromium 3.
So to balance the charge, right?
This has more positive, so we would need
to add 3 electrons here.
So once again, we're gaining electrons,
so gained electrons is reduction, and last
but not least, just to kind
of really emphasize why we need those Roman numerals, right?
So this would be chromium(VI) and as soon as we write the
"VI", everybody knows it's an ion,
and then here chromium 1, 2, 3; chromium(III).
All right?
So if there's a Roman numeral, or a suffix, we don't have
to worry about saying "ion" but for those fixed charge,
it's good for clarity.
Okay, so that finishes up our introduction
to oxidation and reduction.
Now would be a great time
to work a couple post-video practice problems
to reinforce your understanding.