If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains ***.kastatic.org** and ***.kasandbox.org** are unblocked.

Main content

Current time:0:00Total duration:14:02

In the previous
video, we saw how to balance redox reactions
in acidic solution. In this video, we're going
to balance a redox reaction in basic solution. And these are a
little bit harder. But we're going to approach it
the same way that we balanced the reactions in
the acidic solution. So we're going to, once
again, in step four, add some protons here. And we're going to go
ahead and add the half reactions together. But after we do
that, we're actually going to add some hydroxide
anions in here as well. Since this takes place
in basic solution, the hydroxide ions are necessary
to neutralize the protons that we added in step four. So let's start off by starting
with oxidation states, just like we did in
the previous video. So if I start with the
hypochlorite anion here, I know that oxygen has an
oxidation state of negative 2. I know the total has to
equal negative 1 here. So that means that
chlorine must have an oxidation state of plus 1. Plus 1 and minus 2
give you negative 1. If I go to this chromium
compound over here, the easiest way
to approach it is to remember that the
hydroxide anion has a charge of negative 1. And you have four
of them, giving you a total of negative 4. In terms of oxidation
states, a total has to add up to equal minus
1 here, the charge on the ion. Therefore, chromium must have
an oxidation state of plus 3. Plus 3 and minus 4
give you negative 1. So chromium's oxidation
state is plus 3 here. Move over here to the right
to the chromate anion. So oxygen is negative 2. I have four of them
for a total of minus 8. I need to get a total of minus
2, the charge on the ion. And so for an oxidation
state for chromium, it would be plus 6 because plus
6 and minus 8 give you minus 2. So chromium's oxidation
state is plus 6 here. And then finally, for the
chloride anion negative 1 charge, therefore an
oxidation state of negative 1. In terms of what
is being oxidized and what is being
reduced, you need to look for an increase in the
oxidation state for oxidation. So going from an
oxidation state of plus 3 to an oxidation state
of plus 6 for chromium means chromium was oxidized. And if we look at
the chlorine, it's going from an oxidation state
of plus 1 all the way over here to an oxidation
state of negative 1. That's a decrease in
the oxidation state or a reduction in
the oxidation state. So chlorine is being reduced. And so we know that
something is being oxidized and something is being reduced. And so we can go ahead
and do step one now. We're going to write
our half reactions. So one oxidation half reaction
and one reduction half reaction. This time, let's go ahead
and write the reduction half reaction first. It doesn't really
matter which one you do. So we'll go ahead and write
the one using chlorine since chlorine was reduced here. So the hypochlorite anion going
to the chloride anion here. So this is the
reduction half reaction. The oxidation half reaction is
the one that involved chromium. So we have this
chromium compound, Cr(OH)4 with a negative charge,
going to chromate, CrO42 minus. So this is our
oxidation half reaction. So let's go to step
two, balance the atoms other than oxygen and hydrogen. So the atoms other than
oxygen and hydrogen. So for the reduction
half reaction that applies to chlorine. We have one chlorine on the left
and we have one on the right. So chlorine's balanced. In the oxidation half reaction,
that applies to chromium. So one chromium on the left
and one chromium on the right. So in this case, step two
is already done for us. But always be careful because
sometimes you actually have to balance
these atoms and it will affect your final answer. So let's go to step
three, balance the oxygens by adding water. So if we look at our
reduction half reaction, we can see there's one
oxygen on the left side and we have no oxygens
on the right side. So we need to add one water
molecule to the product side of our reduction
half reaction. Now we have one
oxygen on the right and one oxygen on the
left and so oxygen is balanced for our
reduction half reaction. Let's look at our
oxidation half reaction. Remember, this four here
applies to everything in the parentheses. So we have four
oxygens on the left and we have four
oxygens on the right. And so oxygen is
already balanced. Next, we go to step four,
balance the hydrogens by adding protons. So if I look at my
reduction half reaction, we added a water molecule, and
that introduced two hydrogens on the right side
of my half reaction. So therefore, I need two
hydrogens on the left side because right now
I don't have any. And I'm going to balance
them by adding them in the form of protons. I have to add two protons to the
left side of my reduction half reaction. And now I have two
hydrogens on the left and two hydrogens on the right. For my oxidation half
reaction, remember, this four applies to the hydrogen as well. So I have four hydrogens
on the left and none on the right side of my
oxidation half reaction. So I need four on the right. I'm going to add those
in the form of protons. So 4H plus on the right side. Next, we're going to add
electrons to balance charge. So step five, we need to
analyze the total charges on both sides of
my half reactions. So let's look at the
reduction half reaction first. I have two protons so
that's two positive charges on the left side here. And I also have
a negative charge on the hypochlorite anion here. So I have two positive charges
and one negative charge. So, of course, that's
a total of plus 1. And I have a habit
of writing my charges like how I write
oxidation states. So just remember that these are
charges not oxidation states. And a total of plus
1 on the left side. And on the right side, I have
about one negative charge on the chloride anion and
that's it in terms of charges. And so I have a negative one
charge on the right side here. So I need to get those
charges to be equal. And I can only get that by
adding electrons, which are, of course, negatively charged. So if I want to get both sides
of my reduction half reaction equal, the only thing I
could do is add two electrons to the left side right
here because that gives me two more negative charges. So that takes us positive
one to a negative one. So now I have a total
of negative 1 charge on the left side of
my half reaction. And that, of course, is
now equal to the negative 1 on the right side
of my half reaction. So I have balanced the
charges by adding electrons. Let's do the oxidation
half reaction now. So I have only
one of these ions. So I have a negative 1
charge on the left side. On the right side,
I have 1 chromate, but each chromate has two
negative charges like that. And then I also have
four positive charges from my protons. So negative 2 plus 4 gives
me a total of plus 2. So I have plus 2 on the
right side and negative 1 on the left side. So once again, my goal is
to balance those charges by adding electrons only. And if I add three electrons
to the right side of this half reaction, I would change this
total charge from a plus 2 to a negative 1. And so now I have negative 1 on
the right and I have negative 1 on the left, and so my
charges are balanced. Another way to think about
where those electrons go is to remember that Leo
the lion goes "ger." So oxidation, Leo, loss
of electrons is oxidation. So that's why you have
to show your electrons and your oxidation half
reaction being lost or given off on the product
side, if you will. And then Leo the lion goes
ger, so gain of electrons is reduction. So you have to show the
electrons for your reduction half reaction being on your
reactant side of your half reaction because they're
being gained here. So that's another way just to
double check yourself or think about it. So we're on to step six now. Make the number of
electrons equal. So for step six, make the
number of electrons equal. So the reason you
have to do that is because the electrons that are
being lost in the oxidation half reaction are the exact same
electrons that are being gained in the reduction half reaction. So we had three
electrons being lost, but only two electrons
being gained. And that doesn't make any sense. Right? They're the same electrons,
they should be the same number. So we need to make those
number of electrons equal. And the way to do
that, of course, is to think about lowest
common denominator. So we have a two and the three. So we could make
those six electrons. So we need to figure
out a way to make both of our half reactions
have six electrons. And we can do that by
multiplying our first half reaction by a factor of 3. And multiplying our second
half reaction by a factor of 2. And that will give us
equal numbers of electrons. So let's go ahead and
do that math here. So let's go ahead and
take 3 times 2 electrons, that gives me 6 electrons. And then 3 times 2 protons
gives me 6 protons. 3 times the 1, the coefficient
in front of hypochlorite anion, gives me 3 CLO minus. The 3 times the 1 in front
of the chloride anion gives me 3Cl minus. And then finally,
there's a 1 in front of the water as a
coefficient as well. So 3 times 1 gives me 3 waters. So let's go ahead and do the
oxidation half reaction now. Let's get some more room here. So I take the 2, I
multiply it through. There's a coefficient
of 1 in front of here, in front of this ion. So now I would have
to Cr(OH)4 minus. The 2 applies to
the 1 coefficient in front of chromate as well. So I would have 2CrO42
Minus The 2 applies. 2 times 4 gives me 8 protons. So I have 8 protons here. And then, finally, 2
times the 3 electrons gives me those 6 electrons. And so now we have six electrons
from a reduction half reaction and six electrons from an
oxidation half reaction. Next, we're going to add
our half reactions together. So once again, we're
still pretending like we are in acid here. So let's go ahead and add
our half reactions together. We're going to take
all of our reactants and put them on one side. So let's go ahead
and do that first. So we have 6 electrons plus 6
protons plus 3ClO minus plus 2Cr(OH)4 minus. And so on the right side,
so we take all of this. And we add it all together
for our product side. So let's see what we have. We have 3Cl minus plus 3H2O
plus 2CRO42 minus plus 8H plus plus 6 electrons. So finally. All right. Now that we've
written that out, I think it's easier
to see that you have six electrons on the left
and six electrons in the right. So you can cancel those out. You also have some
protons on both sides. You have six protons here
and eight protons here. So you could cancel
out these six protons and you could make this two
protons on the right side. So let's go ahead and rewrite
what we have once again. So let's see what
we have now, now that we've simplified
it a little bit. We have 3ClO minus plus 2CR(OH)4
minus yields 3Cl minus plus 2H2O, sorry, 3H2O plus
2CrO42 minus plus 2H plus. All right. So now we have to remember
that this reaction was actually done in base. So we're going to add
in some hydroxide anions to neutralize the protons. So if we look at the
number of protons we have, we have two protons
over here on the right. So we need to add
two hydroxide anions. So we're going to add
two hydroxide anions to neutralize those protons. And since this is a
balanced equation, what we do to the right side we
have to the left side as well. So since I added two
hydroxides to the right, I need to add in two
hydroxides to the left. Now let's think about what
adding those hydroxides to those protons will do. So they will
neutralize each other. If you take H plus and OH
minus, you will get H2O. And so we're going to
get water molecules. And we're going
to get two of them since we're neutralizing
two protons with two hydroxide anions. And so actually
on the right side, we're going to
get two H2Os here, so two water molecules
in the right side. So let's go ahead and
write the final answer here including everything
that we've done. So on the left side, we have
3ClO minus plus 2Cr(OH)4 minus. And now we added in those two
hydroxides, so plus 2OH minus. And then on the right side,
we had 3Cl minus plus-- well, over here, we have three waters. And we just made two more. And so we really
have a total of five. So 5H2O. And then, once again, we have
the 2 chromates over here. So 2CrO42 minus. And that should take
care of everything. So we can go ahead and
box our final answer. So we got there. It took us a long time, but
we have our final answer. And remember, you can always
check your final answer in terms of number
of atoms and charge. So both atoms and
charge should balance. And so you can go ahead and
check this one on your own just like I showed you how to
do it in the previous video.

AP® is a registered trademark of the College Board, which has not reviewed this resource.