Balancing redox reactions in acid | Redox reactions and electrochemistry | Chemistry | Khan Academy
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Balancing redox reactions in acid | Redox reactions and electrochemistry | Chemistry | Khan Academy


Our goal is to balance this
redox reaction in acid. And before we get
into the steps, let’s talk about
the fact that this is a redox reaction by
assigning some oxidation states. And so we start over here
with the dichromate anion. And we know that oxygen has an
oxidation state of negative 2. We have seven oxygens. So negative 2 times 7 gives me
negative 14 as our total here. We know that the total
for the entire anion has to equal
negative 2, which is the charge on the
dichromate anion. Therefore, we must have plus 12
for all of our chromiums here. So plus 12 and minus
14 give us negative 2. Since we have two chromiums,
each one must be plus 6. And so that’s the oxidation
state for chromium here. We go over here to
the chloride anion. So the charge is negative
1, so our oxidation state is negative 1. Chromium ion over here. So plus 3. And then, finally, chlorine over
here, so oxidation state of 0. So if we look at
chlorine, chlorine went from an oxidation
state of negative 1 to an oxidation state of 0. That’s an increase in
the oxidation state. Therefore, chlorine
was oxidized here. Look at chromium. Chromium went from
plus 6 to plus 3. That’s a decrease in
the oxidation state, or a reduction in
the oxidation state. Therefore, chromium was reduced. And so this is a redox reaction
because something is oxidized and something is reduced. In terms of balancing
it, our first step is to write the
different half reactions. And so we’re going to break
those into an oxidation half reaction and a
reduction half reaction. So let’s go ahead and
get some space down here. And let’s go ahead and
write our half reactions. And so we had a chloride anion
going to chlorine like that. And we said that this was our
oxidation half reaction, so put that way over
here on the right. So that’s out oxidation. Our reduction half
reaction involved chromium. So we had the chromate anion
here, so Cr2O7 2 minus, going to chromium
3 plus like that. So this is our
reduction half reaction. So that’s step one. Write the different
half reactions. Step two. Balance the atoms other
than oxygen and hydrogen. And so, if you look at
our first half reaction, we have one chlorine on the left
and two chlorines on the right. So we need to balance it
by putting a 2 over here on the left like that. We go down here to the
reduction half reaction, and we have two chromiums on the
left and only one on the right. And so we have to put a 2
right here to balance it. So step two is done. Step three. Balance the oxygens
by adding water. So if I look at my
oxidation half reaction, there are no oxygens
so I don’t need to worry about doing anything
to this half reaction at the moment. I go down to the
reduction half reaction, and I do have to
balance my oxygens. So if I go over
here and I can see that I have seven
oxygens on the left side and none on the right. And so I need to do that
by adding the water. And since I have seven
oxygens on the left, I need seven oxygens
on the right. So I’m going to go ahead and
add seven water molecules. And that now gives
me seven oxygens on the right of
my half reaction. So that’s this step,
step three, right here. Step four. Balance the hydrogens
by adding some protons. So let me go ahead and
I’ll use red for this. So step four. Balance the hydrogens
by adding protons here. Once again, the
oxidation half reaction, we don’t have to do
anything, because we don’t have to balance
oxygen or hydrogen here. But, again, we go down to
our reduction half reaction and we have the oxygens
balanced by adding water, but by adding water, now
we have some hydrogens on the right side. So we can see we have a total of
14 hydrogens on the right side. So 7 times 2. And so we’re going to balance
that by adding protons. And so we need to add protons
to the left side of our half reaction. So we need to add 14. So 7 times 2 is 14. So we go ahead and add 14 H plus
to the left side of our half reaction. So step four is done. Step five. Balance the charges
by adding electrons. So let’s get some more
space here, first of all. OK. So we’re going to balance the
charges by adding electrons. So first let’s
analyze what kinds of charges that we have here. So we’ll start with the top
oxidation half reaction. So we have the chloride anion,
which is a negative 1 charge. And we have two of them. So we have two negative charges. Notice, these are
not oxidation states. That’s what gets people
confused sometimes. These are charges. Over here on the right,
we have no charges. So that’s a neutral
chlorine molecule here. So we have two negative
charges on the left and 0 for a charge on the right. So we need to figure out how to
balance those charges by adding electrons, and so it makes
sense that we would have to add two electrons
to the right over here, because that now gives us a
total charge of negative 2 on the right. So that’s one way to
think about it– just getting these
numbers equal here. This negative 2 right
here and this negative 2. Another way to do it
would be, of course, you know the electrons have
to go on the right side because this is the
oxidation half reaction. And one of the ways to
remember that– LEO the lion. So Loss of Electrons
is Oxidation. And so if you’re
losing electrons they must go on the product
side of your half reaction. And so our top half reaction,
our oxidation half reaction, is now balanced. Let’s go down to our
reduction half reaction. So LEO the lion goes GER. So Gain of Electrons
is Reduction. So we already know we’re
going to have to add electrons to the reactant side
of this half reaction. Let’s see if we can figure
out how many electrons we’re going to need to use. So we have 14 positive
charges from the protons. And then we have
two negative charges from the dichromate anion here. So we have 14 positive charges
and two negative charges which gives us a total
of 12 positive charges on the left side. On the right side, we
have a chromium ion. So this is a charge of 3
plus, and I have two of them. So 2 times positive 3
gives me positive 6. So I have positive 6 on the
right side of my half reaction. So I have positive 12 on the
left, positive 6 on the right. I need to add some electrons
to balance out that charge. I already know I’m going to
add them to the reactant side. I know that from LEO
the lion goes GER. Or I could just
think about the fact that, if I have 12
positive charges, I would need to add six
negative charges to get me to a total charge of plus 6. So I need to add six electrons
to the left side over here. So let me go plus six
electrons like that. And now, we have the
charges balanced. So this step is done. So step five is done. We move on to step six. So make the number
of electrons equal. So what does that mean? Well, let’s focus in
here on the electrons that we just added to
our half reactions. So if I go back up to here,
we added two electrons to the oxidation
half reaction, and we had six for the
reduction half reaction. But we know that that number
has to be the exact same number because the electrons that are
lost from our oxidation half reaction are the exact
same electrons that are gained in our
reduction half reaction. So that is why we have to make
these electrons equal in terms of the number. And so the way to do
that would, of course, be to multiply my first
half reaction by 3. Because if I multiply my
first half reaction by 3, that would give me a total
of six electrons, which is what we’re looking for,
making the number of electrons equal. So let’s go ahead and do that. I’m going to rewrite
our first half reaction. I’m going to multiply
everything in our half reaction through by 3, so
everything in parentheses. So if I take 3 and multiply
that by 2 chloride anions, I would, of course, get 6. So we have six chloride
anions like that. And then the three would go
in front of the chlorine. So I have 3Cl2, and then
3 times two electrons gives me six
electrons like that. OK. Let’s go ahead and rewrite
our reduction half reaction because we have a lot
of stuff going on here. Some I’m just going to
rewrite exactly what we have. Six electrons plus 14 protons
plus the dichromate anion like that. And then I have two chromium
ions and seven the waters. OK. So now I have my
two half reactions. I’ve made the number
of electrons equal. And we’re ready
for the last step. We just take our
two half reactions and we’re going to add
them back together, and that’s going to give us
our overall balanced redox reaction. So let’s go ahead
say we did this. And let’s go ahead and
get some more room here so we can add those
half reactions. So all I’m going to do
is just take everything on the reactant side
and add that together. So let’s go ahead
and just rewrite everything on our reactant side. So we have 6Cl minus and then
six electrons, 14 protons. We have the
dichromate anion here. And then I’m going to take
everything on my product side. So I’m going to take
all of this stuff right here and put that
on my product side. So I have 3Cl2 plus six
electrons plus 2 chromium ions plus 7 water
molecules like that. And so now it’s just a
little bit easier, I find, for students to see that your
electrons are going to cancel, right? You have these six electrons
on the reactant side. You have these six electrons
on the product side. So you can go ahead
and take those out. And we’re left with
our final answer. So you can go ahead
and rewrite it if you want to make
it look better. We have 6Cl minus plus
14 H plus plus Cr2O7 2 minus yields 3Cl2 plus
2Cr 3 plus plus 7 H2O. And this should be
our final answer. So I always like to
box my final answer, so it just makes it easier
for your instructor to grade. And the nice thing
about redox reactions is, when you’re finished, you
can always check yourself. Because you know that
you need to balance both the atoms and the charge. So let’s go ahead and
check that real fast. Let’s first start with chlorine. So we have six
chlorines on the left, then over here on the right
we have 3 times 2, which is 6. So the chlorines balance. Hydrogen. We have 14 on the left,
and then 7 times 2 gives us 14 on the right. The chromiums. We have two on the left
and then over here we have two on the right. Oxygens. Seven oxygens on the left
and then over here this seven applies to this oxygen. And so the atoms are
balanced properly. Let’s next check
charge because you have to have the correct charge. Let’s see, we have
6 negatives charges. I’ll just put minus 6 here. And then I have 14 positives. And then I have a
negative 2 right here. So when I add all those
up, I get a total of plus 6 on the left side. So 14 minus 8. Over here on the right
side, the only charge is this chromium ion. It’s 3 plus, and I
have two of them. So 2 times 3 gives me plus 6. And so the charge
balance as well. And so this is the
correct answer. It balances both in terms of
atoms and in terms of charge.

About James Carlton

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89 thoughts on “Balancing redox reactions in acid | Redox reactions and electrochemistry | Chemistry | Khan Academy

  1. OMG thank you! Btw I thought you have Pakistani ethnicity because of you're last name than I went on you're Wikipedia and it says that you're actually a Bengali, and I was like OMG, I am a bengali too and I am so proud because you're so smart and you're helping so many students and you are one of us. Haha.

  2. Don't we have in the end on the left side 2*7 O=14O –> on the left side we have only 7*O=7O so don't we have to multiply the H2O and H+ with 2 that we get it completly balanced?

  3. I was out with my bio class on a field trip so I missed this in school, and the trip made me sick therefor i lost even more days of school. But thanks to you i know i can catch up again, this is my first time on your channel hope you have more vids like this one! thanx

  4. Sir, how did the chemical equation Cr2O7 ^ 2  —>   Cr^3 become the reduction reaction? It also lost electrons, right?

  5. I dont understand, why didn't he come up with the charge of 7H2O in the reduction half reaction…. He just did got the charge of 2Cr wich was 6+, why not the one of h2o

  6. guys you don't know how thankful I am right now . thank you so very much our book added two extra steps which made it complicated

  7. Because Cl2 is covalently bonded, there is no exchange of electrons and the charge of each Cl is zero.

  8. The Best Explanation!!♥♥♥
    Thanks a lot.
    My teachers confused me in those last two steps!!! But now I got it!!!

  9. This method is not working for me for all problems. Try

    ClO3- + I2 = Cl- + IO3-

    This method is useless to me for that problem

  10. oh wow the method my teacher taught us is much much simpler, basically just shortcuts through most of the legwork, i'll try and write it out if you're looking for something more brief

    1. write out oxidation states, as you do, as well as adding 1/2 in front of things like Cl2 etc
    2. count how much each thing got reduced or oxidized; switch the numbers, and multiply them into the equation
    3. count the total charge on each side, by counting, and add H+ or OH- to balance it
    4. count the oxygens, and add water where appropriate, and then count the H's to check

  11. Khan academy is best way for learning from youtube i got first in class from khan academy thanks ☺️☺️☺️☺️

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