Lesson video

Hello, and welcome to today's lesson.

I'm Mrs. Adcock.

And today, we are going to be looking at redox reactions.

We are going to be thinking about how we can construct half equations and balanced ionic equations.

Today's lesson outcome is, I can write half equations and balanced ionic equations.

Some of the keywords we will be using in today's lesson include oxidation, reduction, half equations, and redox reactions.

Here, you can see each of those keywords written in a sentence.

It would be a good idea to pause the video now and read through those sentences.

You might also like to make some notes so that you can refer back to them later in the lesson if needed.

Today's lesson on redox reactions is split into two main parts.

First of all, we are going to be looking at half equations, and then we are going to move on to look at ionic equations.

Let's get started on the first part of our lesson on half equations.

First of all, we are going to look at oxidation.

Oxidation is a reaction where substance gains oxygen, but oxidation can also be used to describe a reaction where a substance loses electrons.

Oxidation is when a substance gains oxygen, but oxidation is also when a substance loses electrons.

Let's have a look at an example.

This reaction shows the oxidation of calcium.

We have a reaction here where calcium is reacting with zinc oxide, and our products are calcium oxide + zinc.

This is an example of a displacement reaction.

Calcium is more reactive than zinc, so the calcium displaces the zinc, and we can see in our products we produce calcium oxide and zinc.

Underneath that word equation, you can see a symbol equation.

We've got calcium reacting with the zinc oxide, and that's producing calcium oxide that has the formula, CaO and zinc, which has the symbol Zn.

Calcium has been oxidised in this reaction because it has gained oxygen.

We can see that the calcium's gone from Ca, and in the products, we've got CaO.

So the calcium has gained oxygen and become calcium oxide.

So the calcium has been oxidised.

Calcium has also been because it has lost electrons.

So just remember, oxidation is to do with gaining oxygen or losing electrons.

It has gone from a calcium atom in our reactants to a Ca2+ ion in the calcium oxide which is one of our products.

So the calcium has lost two electrons and formed a positive two ion.

Because it's lost electrons, we can say the calcium has been oxidised.

Time for a check for understanding.

Which statements about oxidation are true? A, oxidation is the loss of oxygen.

B, oxidation is the gain of oxygen.

C, oxidation is the loss of electrons, or D, oxidation is the gain of electrons.

See if you can remember what oxidation is, both in terms of oxygen and electrons.

Oxidation is the gain of oxygen.

So well done if you chose B.

And in terms of electrons, oxidation is the loss of electrons.

Well done if you got both of those correct, and chose B and C as your answers.

Now, we're going to have a look at reduction.

Reduction is a reaction where a substance loses oxygen, but reduction can also be defined as a reaction where a substance gains electrons.

So reduction is when a substance loses oxygen or gains electrons.

Let's look at an example.

Here is an example showing reduction.

We've got our calcium and that's reacting with the zinc oxide again.

And then we get our products, calcium oxide and zinc.

Here's the balanced symbol equation for that reaction.

The zinc oxide has been reduced, and that's because it has lost oxygen.

So remember, reduction is the loss of oxygen.

So zinc oxide, which has the molecular formula, ZnO in our reactants has lost oxygen in this reaction, and in the products, we can see we formed zinc, which has the symbol Zn.

We can also say that zinc oxide has been reduced because it has gained electrons with the Zn2+ ions in the zinc oxide becoming zinc atoms in our products.

The zinc ions, which have a charge of positive two, have gained two electrons and become zinc atoms. Just a quick recap, oxidation is the gain of oxygen or loss of electrons, and reduction is the loss of oxygen or the gain of electrons.

Which statements about reduction are true? A, reduction is the loss of oxygen, B, reduction is the gain of oxygen, C, reduction is the loss of electrons, or D, reduction is the gain of electrons.

When you are answering this question, try to think what reduction is both in terms of oxygen and electrons.

Let's see how you got them.

Reduction is the loss of oxygen.

So well done if you chose A and got that part correct.

And reduction is also the gain of electrons.

Time for another check for understanding.

In a reaction sodium ions, sodium has the symbol Na and sodium ions have a plus 1 charge.

In a reaction, these sodium ions become sodium atoms. You can see there, the symbol is Na, and we've got no charge now because we have formed an atom.

Why is this an example of reduction? When sodium ions becomes sodium atoms, is it reduction because? A, the sodium ions have lost oxygen, B, the sodium ions have lost electrons, or C, the sodium ions have gained electrons.

Now, just think carefully about whether those statements are to do with reduction or oxidation before you choose your answer.

The correct answer is C.

In this reaction, we know it's an example of a reduction reaction because the sodium ions have gained electrons to go from Na+ to forming sodium atoms with no charge.

The sodium ions have lost oxygen.

We don't know from the information we were given whether there was a loss of oxygen, and B, the sodium ions have lost electrons.

That answer is false because to go from a positive charge to no charge, the ions must have gained electrons.

Just remember that electrons are negatively charged.

We can write half equations showing the electrons that have been lost during oxidation.

Half equations are chemical equations that show the electrons have been lost during oxidation or the electrons that have been gained during reduction.

Here's an example of a half equation.

This is showing us the oxidation of calcium.

We have a calcium atom and that is going to lose 'cause remember, we lose electrons in oxidation.

The calcium atom loses two electrons to form a calcium ion with a positive two charge.

Calcium atoms lose two electrons each to become calcium ions.

This oxidation half equation can be rearranged, and it is more commonly represented as we can see there.

So normally, when we are drawing oxidation reactions, we have the atom on the left hand side.

So in this case, we've got calcium, and then the calcium loses two electrons to form that calcium ion, and we've got those two electrons that have been lost on the right hand side.

Let's have a go then at writing half equations for oxidation.

I'll do one first and then it'll be your turn to have a go.

Write a half equation for the oxidation of lithium.

First of all, we need to think about what oxidation is.

So oxidation is the loss of electrons.

So the lithium is going to lose electrons.

And when atoms lose electrons, they form ions with a positive charge.

Next, we're going to identify the atom and the ions involved.

We have a lithium atom, and in this reaction because it's oxidation, our lithium atom is going to become a lithium ion.

You can use a periodic table to help you identify that lithium is in group one.

Lithium therefore has one electron in its outer shell.

When lithium reacts, it will lose this one outer electron and form an ion with a positive one charge.

Once we've identified our atom and the ion that's going to be formed, we then need to add the number of electrons that have been lost.

In this case, the lithium atom has lost one electron.

So on the right hand side there, we are going to have our lithium ion and the one electron that has been lost.

We've now drawn our half equation for the oxidation of lithium.

Your turn to have a go.

What you need to do is write a half equation for the oxidation of calcium.

Pause the video now, have a go at answering this question.

And then when you come back, we'll go over the answer.

First of all, we are going to identify that this is an oxidation reaction, and oxidation is the loss of electrons.

So the calcium atom is going to lose electrons to form a positively charged ion.

Secondly, you need to identify that atom and the ion involved.

Hopefully, you've got this part correct.

We've got calcium.

Calcium is in group two of the periodic table.

Therefore, it has two electrons in its outer shell.

When it reacts, it will lose these two electrons and form an ion with a positive two charge.

So we have a calcium atom.

And because this is an oxidation reaction in our products, we have a Ca2+ ion.

Finally, we are going to add the number of electrons that have been lost.

In this case, the calcium lost two electrons, so we've completed our half equation for the oxidation of calcium.

Hopefully, you got that one correct.

Well done if you correctly worked out the charge on the calcium ion and added the correct number of electrons.

We can also write half equations showing the electrons that have been gained during reduction.

Let's have a look.

In our example, we've got the reduction of zinc ions.

So we have our zinc ions there with a positive two charge.

They are going to gain electrons because this is reduction, and reduction is gain of electrons.

So our zinc ions gain two electrons because they had a positive two charge, so they're going to gain two electrons and form our zinc atom.

The zinc ions gain two electrons each to become zinc atoms. Let's have a go at writing half equations for reduction.

Once again, I will do one first, and then it'll be your turn to have a go.

Write a half equation for the reduction of barium ions.

First of all, we are going to think about this is reduction, and reduction is the gain of electrons.

So we have our barium ions, and they are going to gain electrons to form barium atoms. Then we are going to identify the atom and the ion involved, and start constructing our half equation.

We have our barium ion on the left hand side, and this is going to form a barium atom.

Finally, we're going to add the number of electrons that have been gained.

Barium had a positive two charge, therefore it will gain two electrons.

We've added these to our half equation, and our half equation is now complete for the reduction of barium ions.

Your turn to have a go.

Pause the video now and have a go at writing a half equation for the reduction of potassium ions.

Potassium has the symbol K, and we can see there that potassium ions have a charge of positive one.

First of all, hopefully, you thought about whether this was reduction or oxidation.

The question was about reduction, and reduction is the gain of electrons.

Our potassium ions will gain electrons to form potassium atoms. We then want to start constructing our half equation.

When we start constructing our half equation, we are going to place the ions on the left hand side because this is reduction where we start with our potassium ions and these are going to gain electrons to form potassium atoms which are on the right hand side.

Finally, we are going to add the number of electrons that have been gained.

In this case, the potassium ions had a positive one charge, so they will gain one electron each to form potassium atoms. Well done if you correctly constructed that half equation for the reduction of potassium ions.

You're doing really well today.

Here's another question to have a go at.

What does this half equation show? Is it A, oxidation, B, reduction, or C, neutralisation? Look carefully at that half equation and decide whether it's showing electrons that have been gained or electrons that have been lost, and use this to help you work out whether it's oxidation, reduction, or perhaps it's neutralisation.

The correct answer is A, oxidation.

We have potassium atoms here that have lost electrons to form potassium ions.

Remember, we draw half equations for oxidation, showing the loss of electrons, and we also draw half equations for reduction, showing the gain of electrons.

We don't draw half equations for neutralisation.

Well done if you chose A, oxidation.

Time for our first practise task of today's lesson.

First of all, for each half equation, decide whether it shows oxidation or reduction of the species.

So you've got A, B, C, and D, so four half equations there and you need to decide whether they show oxidation or reduction.

Once you've done that, then for question two, you need to, for each reaction, decide which species has been oxidised and which has been reduced.

Look carefully at those three reactions in question two and decide which species, that means which atom or ion has gained electron and which one has lost electrons, and use that to help you work out what's been oxidised and what's been reduced.

Pause the video now.

Have a go at answering this question, and then come back when you're ready to go over the answers.

Let's see how you got on.

Question one A, we've got silver ions there, and they are gaining one electron to form silver atoms. So because we have the gain of electrons, A is a half equation showing reduction.

In B, we've got copper atoms, and these copper atoms have lost two electrons to form copper ions.

So B is oxidation.

The copper atoms have lost electrons.

C is reduction, and that's because the calcium ions have gained electrons.

Remember, reduction is gain of electrons.

And in D, we've got oxidation.

Oxidation is the loss of electrons, and the aluminium atoms have lost electrons to form aluminium ions.

If you are struggling to distinguish between oxidation and reduction, then some people use the acronym OIL RIG, and this stands for oxidation is loss of electrons and reduction is gain of electrons.

For question two, you needed to decide which species had been oxidised and which had been reduced.

So let's see how well you did in this question.

The calcium atoms have been oxidised.

They have lost electrons and formed calcium ions and the iron ions have been reduced.

So they have gained electrons to become ion atoms. Well done if you got that question correct.

B, the magnesium has been oxidised because the magnesium atoms have lost electrons to form magnesium ions, and the copper ions have been reduced because the copper ions have gained electrons to form copper atoms. Finally, question C, it looks more difficult because we have numbers in front of our atoms and ions, but you answer this question in exactly the same way.

The aluminium has been oxidised because the aluminium atoms have lost electrons to form aluminium ions, and the zinc ions have been reduced because they have gained electrons to form zinc atoms. Hopefully, you got all of those questions correct.

Well done if you did.

Don't worry if you just made a couple of errors.

Hopefully, you're happy with oxidation is the loss of electrons and reduction is the gain of electrons.

For the final part of this practise task, you're going to try and construct half equations yourself.

There are four questions here and you need to construct a half equation for each one.

A is the reduction of magnesium ions to magnesium atoms. B is the reduction of sodium ions to sodium atoms. C is the oxidation of aluminium atoms to aluminium ions, and D is the oxidation of beryllium atoms to beryllium ions.

Pause the video now.

Think carefully when you're doing these half equations whether they are reduction or oxidation half equations.

Have a go at answering these questions, and then come back when you're ready to go over the answers.

How did we get on writing those half equations? For A, the reduction of magnesium ions, you should have magnesium ions with a positive two charge.

They are going to gain two electrons, and then we'll have an arrow towards our product, which is magnesium atoms. The reduction of sodium ions, you should have your sodium ion that's going to gain one electron, and then we have our sodium atoms that are formed.

C, the oxidation of aluminium atoms. We have the aluminium atoms and then these form aluminium ions with a positive three charge, and we've got the three electrons that have been lost.

In C, you might have put those aluminium atoms minus those three electrons that have been lost, and that will produce are aluminium ions.

Either of those is correct, but the standard way of writing them would be the first one that we looked at.

D, we've got the beryllium atoms, and they lose to electrons to form our beryllium ions with a positive two charge, And although that's the preferred way of writing our oxidation half equations, you may have written it like this, which is also correct, where you've got the beryllium atoms and they've lost two electrons each to form beryllium ions with a positive two charge.

Well done if you correctly constructed those half equations.

You've worked really hard on the first part of our lesson on half equations.

Now, we're going to move on to have a look at ionic equations.

In a displacement reaction, a more reactive metal will replace a less reactive metal in a compound.

And we've got an example here where we've got magnesium, and that's more reactive than copper.

So the magnesium replaces the copper in the compound, and you could say the magnesium displaces the copper in the compound.

We've got magnesium, and that's going to react with copper nitrate.

Because the magnesium is more reactive than copper, we will then form magnesium nitrate + copper.

Underneath, you can see the balanced symbol equation for this reaction.

We've got magnesium which has the symbol Mg.

That reacts with copper nitrate, and we can see that nitrate part has a formula of NO3.

So we've got the copper nitrate, and these react together to form magnesium nitrate and copper.

Both oxidation and reduction in terms of electrons occur in displacement reactions.

Let's have a look at that a bit more closely.

In this displacement reaction, the one we've just been focused on, magnesium atoms lose electrons to become magnesium ions, and the magnesium ions have a positive two charge.

Just have a think, if the magnesium loses electrons, is this oxidation or reduction? Because the magnesium atoms lost electrons, this is oxidation, and the half equation for the oxidation of magnesium is shown there at the bottom.

We've got magnesium atoms, and they form Mg2+ ions, and we've got the two electrons that have been lost.

In the same displacement reaction, copper ions are going to gain electrons to become copper atoms. And how many electrons do you think each copper ion is going to gain to form copper atoms? We've got here at the bottom the half equation for the reduction of copper, and this is copper ions which have a positive two charge, are going to gain two electrons.

So well done if you've got that correct.

And they are going to gain two electrons to form copper atoms. So in this displacement reaction, we had oxidation of the magnesium and we've had reduction of the copper ions.

A reaction where both reduction and oxidation take place is known as a redox reaction.

The magnesium atoms have been oxidised and the copper ions have been reduced.

This is an example of a redox reaction.

What is a redox reaction? Let's see if you can remember.

Is it A, a reaction where reduction takes place, B, a reaction where oxidation takes place, or C, a reaction where both reduction and oxidation take place? Have a look at the word redox and see if you can work out which is the correct answer.

The correct answer is C, a redox reaction is a reaction where we have both reduction and oxidation taking place.

Ionic equations are used to show the species that have been oxidised and reduced in a reaction.

Ionic equations do not show ions that are unaltered in the reaction, and these ions are known as spectator ions.

We've got an example here.

And in our example, we have zinc and copper sulphate reacting together, and they form zinc sulphate and copper.

This is another example of a displacement reaction.

The zinc is more reactive than the copper, so the zinc will displace the copper in the compound.

We have a balanced symbol equation written underneath.

Now, the sulphate ions are unaltered in this reaction, so the sulphate ions, they are ions which are SO4 with a negative 2 charge.

These are spectator ions.

So we don't include them in the overall ionic equation.

The overall ionic equation will be the zinc atoms and the copper ions, and then we will show in the products the zinc ions and the copper atoms and we have missed out those spectator ions, which in this case are the sulphate ions.

We're going to have a go now at writing overall ionic equations for displacement reactions.

I will show you an example and then it'll be your turn to have a go at writing an ionic equation yourself.

We've got magnesium, and that's going to react with silver nitrate to form magnesium nitrate and silver.

We know this is a displacement reaction because magnesium is more reactive than silver and the magnesium has displaced silver in the compound.

Here, we have a balanced symbol equation for the reaction.

And first of all, what we are going to do is write out all of the atoms or ions that are involved in this reaction.

We have magnesium atoms, and these react with that ionic compound silver nitrate, and that contains silver ions and nitrate ions.

And then in our products, we have the ionic compound magnesium nitrate.

So this contains magnesium ions and nitrate ions, and also our other product is those silver atoms. Then we are going to discard any spectator ions.

So remember, these are ions that are unaltered in the reaction, so they are the same in the reactants and products.

In this case, if you look, you should be able to notice that the nitrate ions are unaltered in this reaction.

So we are going to discard of those.

Finally, we form our overall ionic equation.

We've disposed of those spectator ions, and then we have the rest of our atoms and ions forming our overall ionic equation.

Your turn to have a go.

You need to write an ionic equation for the following displacement reaction.

Pause the video now and have a go at answering this question.

Hopefully, you haven't found this too hard.

Let's see how you got them.

First of all, write out all the atoms or ions in the reaction.

We've got sodium atoms, and these react with the ionic compound lithium bromide, which contains lithium ions and bromide ions.

We then form sodium bromide, which contains sodium ions and bromide ions, and then we also form lithium atoms. Discard of any spectator ions.

Hopefully, you identified the bromide ions as spectator ions.

They are the same in the reactants and products.

Therefore, your overall ionic equation should look like this.

What do ionic equations show? Do they show A, the species that is oxidised, B, the species that is reduced, or C, the spectator ions? Choose any answers that you think are correct.

The correct answers are A and B.

So ionic equations show the species that is oxidised and the species that is reduced, but they do not show the spectator ions.

Time for our final practise task of today's lesson.

For question one and two, you've been given equations for two different reactions.

And then below that, you've got ionic equations for those reactions, and you need to complete the ionic equations for the reactions shown in one and two.

If you pause the video now and have a go at answering those two questions.

Let's go over the answers.

In question one, we've got calcium atoms, and they will react with the zinc ions, and then we will produce the calcium ions + zinc atoms. So well done if you got that correct.

We haven't included the oxygen because the oxygen is a spectator ion in this reaction.

Question two.

To complete the ion equation, we should have shown the ion atoms. This is going to react with the copper ions, and then we form ions which have a positive two charge and also copper atoms. Those were two tricky questions, so well done if you've got question one and question two correct.

We are going to write ionic equations for these reactions that we have been given.

So remember, first of all, write out all the atoms and ions that are involved, and then discard of any spectator ions, and then you will form the overall ionic equations.

Take your time having a go at these difficult questions, and then come back when you're ready to go over the answers.

Let's see how you got them.

For A, we've got the ion atoms and they react with the copper ions, which have a positive two charge, and then we form copper atoms and iron ions with a positive two charge.

In B, we've got sodium atoms. These react with silver ions with a positive one charge, and this forms sodium ions and silver atoms. C, we've got magnesium atoms react with lead ions with a positive two charge, and these react to form magnesium ions with a positive two charge and lead atoms. And in D, we've got potassium, and that reacts with lithium ions with a positive one charge to form potassium ions with a positive one charge and lithium atoms. You've worked really hard in today's lesson, so well done.

Let's just summarise some of the key points that we have covered in today's lesson.

Oxidation is the loss of electrons and reduction is the gain of electrons.

Half equations can be written to show which species are oxidised and which are reduced.

And ionic equations can be written for displacement reactions.

I hope you've enjoyed today's lesson and I hope that you're able to join me for another lesson soon.