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Introduction to redox reactions
Video udskrift
What we have depicted right over
here is a combustion reaction. We have the hydrocarbon
methane right over here. You take that. You take some molecular oxygen. You give them enough heat. And then, they are
going to combust. And they're going to produce
carbon dioxide, water, and then more energy than you put in. This is an exothermic reaction. More energy comes
out than you put in. This is why fires
keep spreading. This is why combustion
is used to power things. But that's not what we're going
to focus on in this video. In this video, we want
to think about-- which of these components, of
these molecules, what's being oxidized and
what is being reduced? And to do that,
let's first think about the oxidation
states of the input atoms and the oxidation
states of the outputs-- of the different constituents
of these molecules. So I encourage you to
now pause this video and try to figure
that out on your own. So I'm assuming you've
given a go at it. Now, let's work
through this together. Let's first think
about the methane. And I have a bunch of
electronegativities here based on the Pauling scale
listed out here, but let's just
visualize methane. Methane is a carbon
bonded to 4 hydrogens. In our oxidation state world--
even though, this, in reality, is a covalent bond--
we pretend like they're hypothetically ionic bonds. So we have to give the electron
pair to one of the parties to the bond. If we look between
carbon and hydrogen, carbon is more
electronegative than hydrogen. So we will assume that carbon
will be hogging the electrons and that hydrogen will be
giving away the electrons. So carbon's going to take--
in this hypothetical world-- hypothetically take an electron
from each of these hydrogens. It is going to have an
oxidation state of negative 4, an electron from
each of 4 hydrogens. Negative 4. And once again, we write
the sign after the number probably so that we don't get
these confused with exponents. Now, let's think
about the hydrogens. Each of those
hydrogens is having-- in this hypothetical world--
an electron taken away from it. We could say it has an
oxidation state of plus 1, which we could write as 1 plus. Or we could just
write a positive right over here-- a plus sign. Now, we have molecular oxygen--
oxygen bonded to oxygen. Well, all oxygens
are created equal. Or we'll assume that these
oxygens are created equal, that they're not different
isotopes or anything like that. In this reality,
there's no reason why one oxygen would
hog any electrons from the other oxygen. In this world, the oxygen
has an oxidation state-- when it's in this
molecular oxygen form-- it has an
oxidation state of 0 or an oxidation number of 0. Now, let's think about
this side-- the products. Now, what's happening
here with carbon dioxide? Carbon dioxide is
a carbon double bonded to 2 different oxygens. We see oxygen is one of the
most electronegative elements out there, definitely more
electronegative than carbon. So in our hypothetical
ionic bond world, we would say that oxygen
would take 4 electrons. These bonds would all go
to the different electrons. Each oxygen will take 2
electrons from the carbon. The carbon will
lose 4 electrons. We could say the carbon
loses 4 electrons. You lose 4 electrons. That gives you a hypothetical
positive charge of positive 4. Each of these oxygens
is gaining 2 electrons. So it gives them each
a hypothetical charge of negative 2. And we see it nets out. Positive 4. 2 times negative
2 is negative 4. It all adds up. This is a neutral molecule. And we saw that over here--
for negative 4 plus 4 times positive 1. It all nets out to be neutral. And that makes sense because
these are neutral molecules. And then, finally,
you have water. And we've seen that
multiple times already. Oxygen-- what, in reality,
are covalent bonds with the hydrogens-- in
our hypothetical ionic bond world, oxygen is a good
bit more electronegative. So we're assuming
it's going to take the electrons from
the hydrogens. So each of the hydrogens
loses an electron, giving it an
oxidation number of 1. I could even write it
like this, if you like. An oxidation number
of positive 1. The oxygen has
gained 2 electrons. So that gives it an oxidation
number of negative 2. So now that we've
done that, let's think about who is getting oxidized
and who is being reduced. So let's first
focus on the carbon. The carbon starts off at an
oxidation number of negative 4. The reaction takes place. And then, carbon now has an
oxidation number of positive 4. So how does something go from an
oxidation number of negative 4 to positive 4? Well, the way to
increase your charge or your hypothetical charge
is to lose electrons. Every time you lose an electron,
this becomes less negative. And eventually, it'll
become positive. So you have to lose 8 electrons. So I'll write plus 8
electrons right over here. You take these electrons,
give it to this carbon. You're going to get to
this side of this reaction. And the way I'm
writing right now, these are called
"half reactions" where I'm independently
focusing on each of the elemental components
of these reactions. So here, you have carbon. In this reaction, carbon--
in our hypothetical oxidation number world-- has
lost 8 electrons. What do we call it
when you are losing these hypothetical electrons? Well, we can remind ourselves
OIL RIG-- oxidation is losing electrons, reduction is gaining. Or LEO-- losing
electrons is oxidation-- the lion says GER-- gaining
electrons is reduction. So it's clear right over here
that carbon is being oxidized. It is losing electrons. Oxidation is losing electrons. Carbon is oxidized. Let's think about the hydrogen. On the left-hand side,
you have 4 hydrogens that each have an
oxidation number of plus 1. On the right-hand side,
you have 4 hydrogens. We're writing it a
slightly different way. We could write it
like this-- 2H2's that each have an
oxidation number of plus 1. The oxidation numbers for the
hydrogens has not changed. The hydrogens has neither
been oxidized nor reduced. Let's think about the oxygens. On the left-hand side, you have
2O2's neutral oxidation number. And on the right-hand
side, what do you have? You have 4 total oxygens. I'll combine these together. I'll just write this
as 4 total oxygens. And what's each of
their oxidation numbers? Well, we see it's a negative 2. So what happened to
each of these 4 oxygens? I could have written
4O here instead of 2O2. Either way, I'm
just really trying to account for the oxygens. Here I have 4 oxygens with
a neutral oxidation number-- with an oxidation number of 0. And here I have 4 oxygens with
a negative oxidation number. How do you go from
0 to negative? Each of them must have
gained 2 electrons. If you have 4 oxygens, each
of them gained 2 electrons, we could actually write
the reaction like this. Actually, let me
write it like this. Let me move this part. So cut and paste. Let me move it over to
the right a little bit because what I want to show is
the gaining of the electrons. So plus 8 electrons. So what happened to oxygen? Well, oxygen gained electrons. What is gaining electrons? Reduction is gaining-- RIG. GER-- gaining
electrons is reduction. Oxygen has been reduced. Now, what oxidized carbon? Well, carbon lost
electrons to the oxygen. So carbon oxidized
by the oxygen, which is part of the motivation
for calling it "oxidation." And what reduced the oxygen? Well, oxygen took those
electrons from the carbon. So oxygen reduced by the carbon. And this type of
reaction-- where you have both oxidation
and reduction taking place, and really they're two
sides of the same coin. One thing is going
to be oxidized if another thing is being
reduced, and vice versa. We call these oxidation
reduction reactions. Or sometimes "redox" for short. Take the "red" from "reduction"
and the "ox" from "oxidation," and you've got "redox." This is a redox reaction. Something is being oxidized. Something else is being reduced. Not everything is being
oxidized or reduced, and we can see that very
clearly when we depict it in these half reactions. And one way to check that
your half reactions actually makes sense is you can
actually sum up the two sides. If you take this left side right
over here and you sum them up, you should get all the
constituents right over here minus the electrons. And the same thing over here. You should get all
the constituents that you have on the right-hand
side minus the electrons. So one way to think about it,
if you added all this stuff on the left-hand side, you would
get this plus this 8 electrons on the left-hand side. If you added all of this up, you
would have the plus 8 electrons on that side as well. And so the 8 electrons
you could kind of say, oh, let's take them away
from both sides. You would get your
original reaction. And so I'm going
to leave you there. This is a very, very
powerful tool in chemistry because it really helps us
think about what's actually going on inside of some
type of a reaction.