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## Chemistry library

### Course: Chemistry library > Unit 16

Lesson 1: Oxidation-reduction reactions- Oxidation and reduction
- Oxidation state trends in periodic table
- Practice determining oxidation states
- Unusual oxygen oxidation states
- Balancing redox equations
- Oxidizing and reducing agents
- Disproportionation
- Worked example: Balancing a simple redox equation
- Worked example: Balancing a redox equation in acidic solution
- Worked example: Balancing a redox equation in basic solution
- Redox titrations
- Oxidation–reduction (redox) reactions

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# Worked example: Balancing a simple redox equation

AP.Chem:

TRA‑2 (EU)

, TRA‑2.C (LO)

, TRA‑2.C.1 (EK)

A redox equation can be balanced using the following stepwise procedure: (1) Divide the equation into two half-reactions. (2) Balance each half-reaction for mass and charge. (3) Equalize the number of electrons transferred in each half-reaction. (4) Add the half-reactions together. In this video, we'll use these steps to balance the redox equation for the reaction between aluminum metal and hydrogen ions. Created by Sal Khan.

## Want to join the conversation?

- How would I balance Br- +ClO3- >>> Br2 + Cl-(2 votes)
- We first need to identify which atoms are being reduced and which are being oxidized. The bromine begins with an oxidation state of -1 and proceeds to 0. This is an increase which means it is being oxidized. Automatically this means the chlorate is being reduced, but we can prove this too with oxidation states. The oxygens will have a constant oxidation state of -2 which means only the chlorine is changing. It starts with +5 and proceeds to -1. A decrease in the oxidation number means a reduction.

Next we separate the reaction into two half-reactions; an oxidation and reduction reaction.

Oxidation: Br^(-) → Br2

Reduction: ClO3^(-) → Cl^(-)

We then balance the half-reactions similarly to a normal chemical equation except here in a redox reaction we have to balance the mass AND the charge. Starting with the oxidation reaction, to balance the mass we simply add a '2' coefficient in front of the bromide.

2Br^(-) → Br2

Afterwards we balance the charge by adding electrons to one side of the equation. We can do this since in a redox reaction we have transfers of electrons. Initially the left side has a net charge of -2 and the right is 0. So to balance this we need to add two electrons to the right so both sides are -2. That balances the oxidation half-reaction.

2Br^(-) → Br2 + 2e^(-)

For the reduction half-reaction the chlorines are already balanced, but not the oxygens. We solve this by adding water molecules to the right side of the equation. We can do this because most redox reactions are assumed to be in a water solution.

ClO3^(-) → Cl^(-) + 3H2O

So while this balances the oxygens, it creates a new issue where hydrogen is introduced and is not balanced now. We solve this by adding hydrogen cations to the left side of the equation.

ClO3^(-) + 6H^(+) → Cl^(-) + 3H2O

Now that the masses are balanced, we balance the charge. The left side has a net charge of +5 while the right side is -1. Therefore we add six electron to the left side. This balances the reduction half-reaction.

ClO3^(-) + 6H^(+) + 6e^(-) → Cl^(-) + 3H2O

The last step is to add the two half-reactions into a single reaction again to get the overall redox reaction. In the overall reaction we want the electrons to cancel each other out from the two reactions since we can't have electrons in the final equation since they're not actually chemicals. The issue is that the oxidation reaction has two electrons, but the reduction reaction has six. We need the number of electrons to be the same so they cancel completely. We need the least common multiple of 2 and 6, which is 6. This means we need to multiple the entire oxidation by a certain number to have six elections; which is three here.

3x(2Br^(-) → Br2 + 2e^(-)) = 6Br^(-) → 3Br2 + 6e^(-)

Now we can finally add the two equations together and see that the six electrons from both equations cancel out.

* 6Br^(-) + ClO3^(-) + 6H^(+) + 6e^(-) → 3Br2 + Cl^(-) + 3H2O + 6e^(-)

* 6Br^(-) + ClO3^(-) + 6H^(+) → 3Br2 + Cl^(-) + 3H2O

And there's the final balanced redox reaction. Specifically thought it's balanced in an acidic solution. It is possible to balance it in a basic solution is needed.

Hope that helps.(7 votes)

- Do you always balance the number of atoms before the charges?(1 vote)
- You don't absolutely have to balance mass before change, but it is generally easier to do so in redox reactions.(5 votes)

- How I can balance Mg+Fe(2)O(3)>>>Fe+MgO(1 vote)
- I won't just give you the answer, but it's pretty straightforward if you balance them in a certain order.

Balance the elements in this order: Iron, oxygen, then magnesium.

Hope that helps.(4 votes)

- Near the end of the video the 6e- is added to 2Al(s) and 6H+(aq) why? same goes for the stuff added together after that.(1 vote)
- Balancing redox reactions are slightly different from regular reactions in that not only do we have to balance mass, but we also have to balance charge. We do this primarily by adding electrons as an intermediate step which we represent with the symbol e^(-).

Once we've added electrons to each half reaction to balance the charge, we have to multiply each reaction to get the electrons to their LCM. The reason we need to is that when we add the half reactions back into a full reaction the electrons will cancel out. And we need to do this because electrons can't show up in the overall reaction's equation because they're not truly reactants/products. Instead what's really happening is just that electrons (six in this case) are moving from the aluminum atoms to the hydrogen atoms.

Hope that helps.(1 vote)

- Why does H2 have no charge?

I would have thought the charge is +2 because each H has a charge of +1.(1 vote)- Hydrogen has the possibility of taking on several different charges as an ion including +1 or -1. But two neutral atoms of hydrogen will form a neutral molecule of diatomic hydrogen. Each atom has 1 proton and 1 electron, which is neutral, so two of those together would still be neutral.

Hope that helps.(1 vote)

## Video transcript

- [Instructor] So what we
have here is a redox reaction. Things are getting oxidized and reduced, thus the name redox, but we wanna balance this redox reaction. And when we talk about
balancing a redox reaction, we want to make sure we conserve mass and charge on both sides of this reaction. So how do we do that? Well, the first step is to
assign oxidation numbers or oxidation states to each
of the constituent elements on either side of the reaction. Then we'll know who's getting oxidized and who's getting reduced. And then we can set up the half reactions which we can then balance. All right, so let's first
look at this aluminum right over here. It has an oxidation state or
an oxidation number of zero. It's just aluminum by itself. And then we can go to this hydrogen. This is really just a
proton right over here. It has a plus one charge. And so it's hypothetical
charge, you could say. Well, that would be plus one as well, which would be its oxidation
number or its oxidation state. And then when we get onto the
right side of this reaction, we see that the aluminum now
has a positive three charge. So its oxidation number
would also be positive three. And we can see here that the hydrogen, and actually now we have two, so we're gonna have to
deal with that later on, we only have one here. But now each of these hydrogens have an oxidation number of zero. They're not taking up electrons
or giving away electrons. You just have two hydrogens
bonded to each other. All right, so now that we've
assigned oxidation numbers, we can figure out who's getting oxidized and who is getting reduced. So if you look at the aluminum, aluminum goes from an
oxidation number of zero to plus three. So if your oxidation number is increasing, that means you're getting oxidized. Oxidized. You might also remember OILRIG, oxidation is losing electrons. Because you're losing electrons, you're having a more positive charge. But your oxidation number is going up, so you're getting oxidized. And if you look at the hydrogen, we're going from a plus one to a zero. If your oxidation number is reducing, you are being reduced. So reduced. So now let's set up both
of the half reactions, the oxidized half reaction for aluminum and then the reduction for hydrogen. So first, for the aluminum, we have an aluminum solid
and its half reaction. It is going to now a form of aluminum with a plus three charge
in an aqueous state. So let's first balance this for just the number of aluminums we have. We have one aluminum on the
left, one aluminum on the right. So that seems balanced. Now let's try to balance it for charge. So we have no charge here. So we should have a total
of no charge on the right but we have plus three here. And so what we need to
do is add some electrons. So let me add three
electrons right over here and there we have it. We have balanced it for charge. And now let's think about the hydrogen. So if we take, we have a proton on the left. Let me just do it like this. And it's an aqueous solution. And then on the right, we just have hydrogen molecules,
neutral hydrogen molecules. And so first, let us balance
it for the number of hydrogens. We have two on the right, one on the left. So we're gonna have to
put a two right over here. So we've balanced for
the number of hydrogens and now let's balance for charge. So let's see. On the right-hand side,
we have no net charge while on the left-hand
side, right over here, we have a positive two net charge. So in order to balance this, I have to put two electrons
on the left-hand side. So let me do that here. So two electrons, I'm gonna add over here. And now it looks like it
is balanced for charge. Now, the next thing we wanna do is we want to balance
the number of electrons that we have on the right-hand side and on the left-hand side
in these half reactions. And so how can you do that? Let's see, you have three
here and you have two here. The least common multiple,
involving a little bit of your elementary school mathematics here of three and two is six. So we can get these both to six. And how could we do that? Well, we could multiply this
top half reaction by two. So let me do that. So if I multiply this
top half reaction by two, and if I multiply this bottom
half reaction by three. Why does that work? Well, two times three electrons is gonna give us six electrons. Three times two electrons is
going to get us six electrons. And so let me now rewrite this. So if I multiply by two, we're
gonna have two aluminums. So we're gonna have two
aluminums, the solid state. And then on the right hand
side of this half reaction, I'm multiplying everything by two. So now I have two aluminums
plus three in aqueous solution. And then I'm multiplying these
electrons by two as well. So plus six electrons. And now let me do this down here. Three times two electrons, that is going to give us six electrons. And then I'm going to multiply
this three times this two. So that's gonna give us
six hydrogen protons, I guess I could say. So plus six hydrogen protons that are in an aqueous solution. And then on the right hand
side, if I multiplied by three, I have three hydrogen molecules, each of them with two hydrogen atoms. And so now I have balanced the number of electrons on both sides. And now I'm gonna add
these two half reactions. And if I add the two half
reactions, what do I get? And let me do it down here. So on the left-hand side,
I'm just going to add up all of this stuff right over here. So I'm gonna get six
electrons plus two aluminums, plus six hydrogen protons,
plus six hydrogen protons. And then on the right-hand side, I am going to add up all of this stuff. So I have those two aluminums, now with a plus three charge
in an aqueous solution. I'm going to have the three hydrogens, three hydrogen molecules I should say. There's actually six hydrogens here. And then I have the six electrons, plus six electrons. And I have six electrons on the left, six electrons on the right. I could cancel those out. And then what I have here is
our balanced redox reaction. Balanced for both mass and charge. And we are done.