In most redox reactions
something is oxidized and something else
is being reduced. But in some redox
reactions, one substance can be both oxidized and
reduced in the same reaction. And those are called
disproportionation reactions. And here we have one the
most famous examples, the decomposition of
hydrogen peroxide. And so over here on the left
is hydrogen peroxide, H202. Which when you add a catalyst
like potassium iodide will turn into water and oxygen. So let's go ahead and
assign some oxidation states or oxidation numbers
in this reaction. And we'll start with the
oxygen and the water molecule. And there are a couple ways
to assign oxidation states. The simplest is just, of
course, to memorize the rules that you'll see in any
general chemistry textbook. And when you have
oxygen in water, the oxidation state is
negative two for oxygen. When you have O2, so
over here on the right, the oxidation state
of oxygen is zero. And over here on the
left, hydrogen peroxide is one of those
weird examples where oxygen has an oxidation
state of negative one. So you could, of
course, memorize these. Or there's, of course, a
different way of figuring out the oxidation state,
and that involves drawing out your dot
structures and thinking about electronegativity. And so let's start with, once
again, the water molecule here. And we know that these bonds
consists of two electrons, right? So I'm going to go ahead and
draw in these electrons here. We know that oxygen is more
electronegative than hydrogen. So when you're thinking
about it that way, you could think
about those electrons going towards the oxygen. So you treat it
like an ionic bond even though it
isn't an ionic bond. And now you can see
the oxygen's going to get all of those electrons. Oxygen normally has
six valence electrons. So oxygen normally has
6 valence electrons. Here it's being
surrounded by eight. So 6 minus 8 gives
us negative 2. And so that is, of
course, the same number that we figured out
we memorized it. But thinking about
the electronegativity helps you understand oxidation
states a little bit more. Let's next do oxygen in
the O2 molecule here. And so once again, we
think about these bonds. Each bond consists
of two electrons, and so we have a double
bond in this case. All right. So in this case, we have
oxygen bonded directly to another oxygen
atom, and of course they both have this exact
same electronegativity. And so we have to share
all those electrons. And so since we have
four total electrons, each oxygen is going to get two. And so we can go and divide
up those electrons that way. So once again, oxygen normally
has six electrons around it. It is now surrounded by six. So 6 minus 6 gives you 0. So an oxidation
state of 0, which is, of course, exactly
what we got up here. All right, let's go ahead and
do the hydrogen peroxide example over here. So once again, we draw in
our electrons in these bonds. So we have it look like that. And we consider the oxygen. Let's go ahead and do the
oxygen on the left here. So oxygen versus hydrogen,
oxygen is more electronegative. Oxygen gets those electrons. And then over here, the two
electrons between those oxygen atoms, once again, an
equal electronegativity, and so therefore
each oxygen gets one. And so you can see
that is our situation. So this time we have 6
minus-- and if you count up those electrons in there that
I've circled, you'll get seven. So 6 minus 7 gives you an
oxidation state of negative 1 for oxygen in hydrogen peroxide. And so either way,
the memorization way is, of course,
faster, but sometimes the dot structure
way is very useful. So now we have our
oxidation states and we can analyze this
a little bit better. And you can see that we
have a case where oxygen is on the left, has an
oxidation state of negative 1. And let's say this
oxygen atom turned into one of the O2 oxygen atoms. So it's going from an
oxidation state of negative 1 to an oxidation state of 0. So you have oxygen going
from negative 1 to 0. And that is an increase
in the oxidation state. That's an increase in
the oxidation state, therefore by definition,
we know that oxygen being oxidized here. So this is an example
of an oxidation. Once again, just
look at the numbers. Negative 1 is increasing
to 0, so that's oxidation. All right, let's think about
what else is happening, right? So we could start out with an
oxygen state of negative one. And let's say that
oxygen this time is going to an oxidation
state of negative two, like the oxygen and water. And so that, of
course, is a reduction or a decrease in
the oxidation state. You're going from
negative 1 to negative 2. So a decrease in the oxidation
state is, of course, reduction. So in this case, oxygen
is being reduced. And so you have a
substance that is being both oxidized and
reduced in the same reaction. And so once again, that's called
a disproportionation reaction.