Let’s see how to
identify the oxidizing and reducing agents
in a redox reaction. So here, we’re forming
sodium chloride from sodium metal
and chlorine gas. And so before you assign
oxidizing and reducing agents, you need to assign
oxidation states. And so let’s start with sodium. And so the sodium atoms are
atoms in their elemental form and therefore have an
oxidation state equal to 0. For chlorine, each
chlorine atom is also an atom in its elemental form,
and therefore, each chlorine atom has an oxidation
state equal to 0. We go over here to the
right, and the sodium cation. A plus 1 charge on sodium,
and for monatomic ions, the oxidation state is equal
to the charge on the ion. And since the charge
on the ion is plus 1, that’s also the oxidation state. So plus 1. We’re going to circle the
oxidation state to distinguish it from everything else
we have on the board here. And for chloride anion,
a negative 1 charge. Therefore, the oxidation
state is equal to negative 1. And so let’s think about
what happened in this redox reaction. Sodium went from an
oxidation state of 0 to an oxidation state of plus 1. That’s an increase in
the oxidation state. 0 to plus 1 is an increase
in oxidation state, so therefore, sodium, by
definition, is being oxidized. So sodium is being
oxidized in this reaction. We look at chlorine. Chlorine is going from
an oxidation state of 0 to an oxidation
state of negative 1. That’s a decrease in
the oxidation state, and therefore, chlorine
is being reduced. So each chlorine atom
is being reduced here. Now, before we assign
oxidizing and reducing agents, let’s just go ahead
and talk about this one more time, except showing
all of the valence electrons. So let’s also assign some
oxidation states using this way because there are two ways
to assign oxidation states. So let’s assign an oxidation
state to sodium over here. So if you have your electrons
represented as dots, you can assign an
oxidation state by thinking about how
many valence electrons the atom normally
has and subtracting from that how many electrons
you have in your picture here. So for sodium,
being in group one, one valence electron
normally, and that’s exactly what we
have in our picture. Each sodium has a valence
electron right here. So 1 minus 1 gives
us an oxidation state equal to 0, which is what
we saw up here, as well. So sodium has an oxidation
state equal to 0. Notice that I have two
sodium atoms drawn here, and that’s just what
the two reflects in the balanced
equation up here. Let’s assign an oxidation
state to each chlorine atom in the chlorine molecule. And so we have a bond between
the two chlorine atoms, and we know that bond
consists of two electrons. Now, when you’re assigning
oxidation states and dot structures, you want
to give those electrons to the more
electronegative elements. In this case, it’s
the same element, so there’s no difference. And so we give one electron to
one atom and the other electron to the other atom, like that. And so assigning
an oxidation state, you would say chlorine normally
has seven valence electrons, and in our picture
here, this chlorine atom has seven electrons around it. So 7 minus 7 gives us an
oxidation state equal to 0. And of course, that’s what
we saw up here as well, when we were just using
the memorized rules. And so it’s the same for
this chlorine atom over here, an oxidation state equal to 0. So sometimes it just helps
to see the electrons. We’ll go over here
for our products. We had two sodium chlorides, so
here are two sodium chlorides. And let’s see what happened
with our electrons. So the electron in
magenta, this electron over here in magenta
on this sodium, added onto one of
these chlorines here. And then this electron
on this sodium added onto the other
chlorine, like that, and so sodium lost
its valence electron. Each sodium atom lost
its valence electron, forming a cation. And when we calculate
the oxidation state, we do the same thing. Sodium normally has
one valence electron, but it lost that
valence electron. So 1 minus 0 is equal to plus 1
for the oxidation state, which is also what we saw up here. And then when we
do it for chlorine, chlorine normally has
seven valence electrons, but it gained the
one in magenta. So now it has eight around it. So 7 minus 8 gives
us an oxidation state equal to negative 1. And so maybe now
it makes more sense as to why these
oxidation states are equal to the charge on the
polyatomic monatomic ion here. And so now that
we’ve figured out what exactly is happening
to the electrons in magenta, let’s write some half
reactions and then finally talk about what’s
the oxidizing agent and what’s the reducing agent. So let’s break down the
reaction a little bit more in a different way. So you can see we have two
sodium atoms over here. So we’re going to
write two sodiums. And when we think
about what’s happening, those two sodium
atoms are turning into two sodium ions
over here on the right. And so we have two
sodium ions on the right. Now, those sodium atoms
turned into the ions by losing electrons, so each
sodium atom lost one electron. So we have a total of two
electrons that are lost. I’m going to put
it in magenta here. So those 2 electrons are lost,
and this is the oxidation half reaction. You know it’s the
oxidation half reaction, because you’re losing
electrons here. So remember, LEO the lion. So Loss of Electrons
is Oxidation. So this is the
oxidation half reaction. We’re going to write the
reduction half reaction. The chlorine molecule gained
those two electrons in magenta. So those two
electrons in magenta we’re going to put
over here this time. The chlorine
molecule gained them, and that turned the chlorine
atoms into chloride anions. And so we have two
chloride anions over here. And so those are,
of course, over here on the right, our
two chloride anions. And so here we have
those two electrons being added to
the reactant side. That’s a gain of electrons,
so this is our reduction half reaction, because LEO
the lion goes GER. Gain of Electrons is Reduction. And so if we add those two
half reactions together, we should get back the
original redox reaction, because those two electrons
are going to cancel out. It’s actually the
same electrons. These two electrons in magenta
that are lost by sodium are the same electrons that
are gained by chlorine, and so when we add all of
our reactants that are left, we get 2 sodiums
and Cl2, so we get 2 sodiums plus chlorine gas. And then for our products,
we would make 2 NaCl, so we get 2 NaCl for
our products, which is, of course, our original
balanced redox reaction. So finally, we’re able
to identify our oxidizing and reducing agents. I think it was necessary
to go through all of that, because thinking
about those electrons and the definitions are really
the key to not being confused by these terms here. And so sodium is
undergoing oxidation, and by sodium
undergoing oxidation, it’s supplying the two electrons
for the reduction of chlorine. Therefore, you could
say that sodium is the agent for the
reduction of chlorine, or the reducing agent. So let’s go ahead
and write that here. So sodium, even though
it is being oxidized, is the reducing agent. It is allowing
chlorine to be reduced by supplying these
two electrons. And chlorine, by
undergoing reduction, is taking the electrons
from the 2 sodium atoms. That allows sodium
to be oxidized, so chlorine is the agent
for the oxidation of sodium, or the oxidizing agent. Let me go ahead and
write that in red here. Chlorine is the oxidizing agent. And so this is what students
find confusing sometimes, because sodium is
itself being oxidized, but it is actually
the reducing agent. And chlorine itself
is being reduced, but it is actually
the oxidizing agent. But when you think
about it by thinking about what happened
with those electrons, those are the exact
same electrons. The electrons that
are lost by sodium are the same electrons
gained by chlorine, and that allows sodium to be
the reducing agent for chlorine, and that is allowing
chlorine at the same time to oxidize sodium. And so assign your
oxidation states, and then think about
these definitions, and then you can assign
oxidizing and reducing agents.