- [Instructor] So what we have here is a redox reaction.
Things are getting oxidized and reduced,
thus the name redox,
but we wanna balance this redox reaction.
And when we talk about balancing a redox reaction,
we want to make sure we conserve mass
and charge on both sides of this reaction.
So how do we do that?
Well, the first step is to assign oxidation numbers
or oxidation states to each of the constituent elements
on either side of the reaction.
Then we'll know who's getting oxidized
and who's getting reduced.
And then we can set up the half reactions
which we can then balance.
All right, so let's first look at this aluminum
right over here.
It has an oxidation state or an oxidation number of zero.
It's just aluminum by itself.
And then we can go to this hydrogen.
This is really just a proton right over here.
It has a plus one charge.
And so it's hypothetical charge, you could say.
Well, that would be plus one as well,
which would be its oxidation number or its oxidation state.
And then when we get onto the right side of this reaction,
we see that the aluminum now has a positive three charge.
So its oxidation number would also be positive three.
And we can see here that the hydrogen,
and actually now we have two,
so we're gonna have to deal with that later on,
we only have one here.
But now each of these hydrogens
have an oxidation number of zero.
They're not taking up electrons or giving away electrons.
You just have two hydrogens bonded to each other.
All right, so now that we've assigned oxidation numbers,
we can figure out who's getting oxidized
and who is getting reduced.
So if you look at the aluminum,
aluminum goes from an oxidation number of zero
to plus three.
So if your oxidation number is increasing,
that means you're getting oxidized.
Oxidized.
You might also remember OILRIG,
oxidation is losing electrons.
Because you're losing electrons,
you're having a more positive charge.
But your oxidation number is going up,
so you're getting oxidized.
And if you look at the hydrogen,
we're going from a plus one to a zero.
If your oxidation number is reducing,
you are being reduced.
So reduced.
So now let's set up both of the half reactions,
the oxidized half reaction for aluminum
and then the reduction for hydrogen.
So first, for the aluminum,
we have an aluminum solid and its half reaction.
It is going to now a form of aluminum
with a plus three charge in an aqueous state.
So let's first balance this
for just the number of aluminums we have.
We have one aluminum on the left, one aluminum on the right.
So that seems balanced.
Now let's try to balance it for charge.
So we have no charge here.
So we should have a total of no charge on the right
but we have plus three here.
And so what we need to do is add some electrons.
So let me add three electrons right over here
and there we have it.
We have balanced it for charge.
And now let's think about the hydrogen.
So if we take,
we have a proton on the left.
Let me just do it like this.
And it's an aqueous solution.
And then on the right,
we just have hydrogen molecules, neutral hydrogen molecules.
And so first, let us balance it for the number of hydrogens.
We have two on the right, one on the left.
So we're gonna have to put a two right over here.
So we've balanced for the number of hydrogens
and now let's balance for charge.
So let's see.
On the right-hand side, we have no net charge
while on the left-hand side, right over here,
we have a positive two net charge.
So in order to balance this,
I have to put two electrons on the left-hand side.
So let me do that here.
So two electrons, I'm gonna add over here.
And now it looks like it is balanced for charge.
Now, the next thing we wanna do
is we want to balance the number of electrons
that we have on the right-hand side
and on the left-hand side in these half reactions.
And so how can you do that?
Let's see, you have three here and you have two here.
The least common multiple, involving a little bit
of your elementary school mathematics here
of three and two is six.
So we can get these both to six.
And how could we do that?
Well, we could multiply this top half reaction by two.
So let me do that.
So if I multiply this top half reaction by two,
and if I multiply this bottom half reaction by three.
Why does that work?
Well, two times three electrons
is gonna give us six electrons.
Three times two electrons is going to get us six electrons.
And so let me now rewrite this.
So if I multiply by two, we're gonna have two aluminums.
So we're gonna have two aluminums, the solid state.
And then on the right hand side of this half reaction,
I'm multiplying everything by two.
So now I have two aluminums plus three in aqueous solution.
And then I'm multiplying these electrons by two as well.
So plus six electrons.
And now let me do this down here.
Three times two electrons,
that is going to give us six electrons.
And then I'm going to multiply this three times this two.
So that's gonna give us six hydrogen protons,
I guess I could say.
So plus six hydrogen protons
that are in an aqueous solution.
And then on the right hand side, if I multiplied by three,
I have three hydrogen molecules,
each of them with two hydrogen atoms.
And so now I have balanced
the number of electrons on both sides.
And now I'm gonna add these two half reactions.
And if I add the two half reactions, what do I get?
And let me do it down here.
So on the left-hand side, I'm just going to add up
all of this stuff right over here.
So I'm gonna get six electrons plus two aluminums,
plus six hydrogen protons, plus six hydrogen protons.
And then on the right-hand side,
I am going to add up all of this stuff.
So I have those two aluminums,
now with a plus three charge in an aqueous solution.
I'm going to have the three hydrogens,
three hydrogen molecules I should say.
There's actually six hydrogens here.
And then I have the six electrons,
plus six electrons.
And I have six electrons on the left,
six electrons on the right.
I could cancel those out.
And then what I have here is our balanced redox reaction.
Balanced for both mass and charge.
And we are done.