Lesson video

Hello, my name's Mrs. Clegg, and today we're gonna be looking at ionic equations and we'll use the reactions of group one and group seven elements to do that.

Let's get started.

So here's our lesson outcome.

By the end of today's lesson, we should be much more comfortable and be able to write balanced half equations and ionic equations for group one and group seven elements.

Here are our keywords to watch out for today.

Oxidation, half equation, reduction, redox and ionic equation.

There's quite a few words today.

And they are written into a sentence and you'd might like to pause the video and make some notes here to be able to refer back to during the lesson.

I've divided today's lesson up into two parts, oxidation and reduction and then redox and ionic equations.

So let's get started with oxidation and reduction first of all.

So oxidation occurs when a metal atom loses outer shell electrons to become a stable positive ion, and here we've got potassium, potassium atom in group one, one outer electron, you can see there, and when it reacts, it will form the potassium ion and it's lost an electron, so it will be positively charged and we show that with these square brackets and the positive sign in super script, plus the electron.

And the loss of electrons to form positive ions can be shown using an equation, and we'll do this now.

So in chemical equations, an electron is represented by the symbol e with a minus, that sort of super script there.

Here's our potassium atom again reacting to form an ion and an electron.

And to save us writing it out long hand, we can use symbols, like this.

So we've got the potassium atom, K, and then it forms the potassium ion, K plus, plus a negative electron.

And this equation is known as a half equation, because it shows half of the reaction with regards to the transfer of electrons.

So let's have a quick check.

So oxidation is the loss of electrons to form.

Let's look at the answer.

So C, positive ions, well done.

Another one.

The oxidation of sodium, in group one, can be shown as this.

Is that true or is that false? It's false, and why is false? Why do think it was false? So here's some statements to help you.

Which one of those would help? And so A should help there.

Each sodium atom loses one negatively charged electron to form a positively charged sodium ion.

Now, similar to balancing chemical equations, multiple electrons can be lost from an atom and they can also be shown in half equations using coefficients.

So we've got magnesium here, and remember, magnesium is in group two, so it has two outer electrons and when it reacts, it forms magnesium ion and because it's lost two negatively charged electrons, it's got two plus in super script.

Magnesium loses two electrons, remember, magnesium is in group two so it had two outer electrons.

So it loses two electrons.

Here's aluminium, how many does that lose? And we could represent this in mole, so one mole of aluminium atoms would lose three moles of electrons.

So let's have a quick check.

How many electrons would a calcium atom lose? Always helpful to have a periodic table beside you.

Which group is calcium in? That might help you work out how many electrons are on the outer shell.

Let's look at the answer.

So it is B, because it has two, it's in group two, so two electrons to lose.

Reduction occurs when a non-metal gains outer shell electrons to become a stable negative ion.

So here we've got fluorine in group seven.

If you count around, there's seven electrons in the outer shell.

So fluorine gains an electron, a negatively charged electron and therefor the fluoride ion that's formed is negatively charged.

So this gain of electrons to form negative ions can also be shown using an equation.

So let's have a look at this.

So the reduction reaction going on here is shown like this.

So we've got the fluorine atom, an electron, which is being donated from another atom, another element, and fluoride ion has been formed, negatively charged.

However, it's important to remember that group seven elements exist as diatomic molecules.

So this is not the correct half equation.

Dia, remember means two, so fluorine always goes round as f2 in pairs.

They must always show the substance in its standard chemical form.

So for group seven elements, that means the half equation must be balanced to reflect that the element is diatomic.

So if we look at fluorine from before, we now need to balance it.

And so we've got two fluorine atoms making a fluorine molecule.

We've gained two negatively charged electrons and made two fluoride ions.

Other diatomic elements are nitrogen, oxygen and hydrogen.

So the same thing would apply to them.

Let's have a quick check.

Which of these atoms would be reduced and why? So hopefully you've noticed oxygen, and it's a non-metal so it gains electrons to form an oxide ion.

Let's do another question.

So true or false? Chloride ions gain electrons to become chlorine molecules.

Is that true or is that false? It is false, and why? So can you justify your answer? And the answer is B, so each chlorine atom gains an electron to become a negatively charged chloride ion.

So a useful way sometimes to help us remember what happens to the electrons in oxidation and reduction is to think about an oil rig.

Sounds a bit strange, doesn't it? But OILRIG stands for oxidation is loss of electrons, reduction is gain of electrons.

Oxidation is loss, reduction is gain.

So remember that.

So let's do a task.

So try and work out whether we've got oxidation or reduction happening in each of these equations.

So remember OILRIG.

So pause the video and come back when you're ready.

So let's have a look at the answers.

There we go.

So lithium, remember, is in group one, so it loses one electron, so that's oxidation.

Oxidation is loss.

If we go down, iodine is in group seven.

Now, iodine remember, forms diatomic molecules, so we have to balance the equation there so to reflect that and iodine atoms have seven outer electrons, so each of those atoms wants to gain an electron.

So this is the balanced equation, so there have gained electrons, so it's reduction.

This is also a useful exercise in remembering the symbols as well.

So if you're stuck with the symbols, again, it's useful to have a periodic table to just check that you've got them right.

Let's have a look at this part of the task.

So complete the half equation now and state whether oxidation or reduction has happened.

And number three there, write the half equations for the oxidation of magnesium and the reduction of oxygen and explain what is happening as we go along.

So pause the video and come back when you're ready.

So let's have a look at the answers.

So the sodium ion is gaining a negatively charged electron there to form a sodium atom.

So the sodium has gained an electron, so that is reduction and reduction is gain.

So pause the video and check your answers and join us when you're ready.

And for part three.

So we look at the oxidation of magnesium first of all.

There's the half equation, so the magnesium atom is losing its two outer shell electrons to form a magnesium ion.

And here's the reduction of oxygen.

So oxygen, remember, is a diatomic molecule.

So we have to remember that and balance the equation.

Each oxygen atom in the oxygen molecule gains two outer shell electrons to form two oxide ions.

So fantastic if you got all of that correct.

If not, perhaps, rewind the video and go through it again.

It's quite tricky.

So let's look at the second part of our lesson today, redox and ionic equations.

So a redox reaction is one in which one substance in the reaction is reduced and the other one is simultaneously oxidised.

In other words, it happens at the same time.

So there's reduction, oxidation and there's no overall gain or loss of charge in a reaction.

So the number of electrons lost by the substance undergoing oxidation must equal the number of electrons that are gained by the substance undergoing reduction.

So changes of state may also occur during a redox reaction.

These are shown using state symbols, so remember those, solid, liquid, gas and aqueous.

Aqueous means that the solute is dissolved in water, so for example, potassium that you can see whizzing around in the water there, forming potassium hydroxide, which is aqueous and hydrogen gas.

A little check.

What state symbol would you use for the element chlorine? It is of course a gas.

Well done.

You might recall that in displacement reactions, a more reactive element displaces a less reactive one.

If you remember, the group seven elements, we looked at displacement reactions there.

And these reactions are also redox reactions.

Remember, one element is oxidation while the other is simultaneously reduced.

So chlorine for halogen and potassium iodide, a metal halide salt, react together and iodine is formed and potassium chloride.

So iodine is the displaced halogen and potassium chloride is the resulting metal halide salt.

Group one metals are often found in metal halide salts, such as potassium iodide, potassium chloride, potassium bromide et cetera.

The reactions of group seven elements decreases as we go down the group.

So we've got fluorine at the top there, most reactive and astatine towards the bottom there, much less reactive.

So we got chlorine reacting with potassium bromide to produce bromine and potassium chlorine.

Now, what is really helpful is that the halogens are coloured in solution and the halide salts are colourless.

So we can always see what is happening during a displacement reaction with group seven elements.

So as bromine is less reactive than chlorine, it's displaced and we see this distinct colour change occurring there.

So let's have a quick check.

Bromine can displace iodine from a compound.

Is that true or is that false? It is true and why? Bromine is much more reactive than iodine and again, if you've got your periodic table in front of you, you'd be able to see that bromine is above iodine in the periodic table under group seven.

So redox reactions can be represented using ionic equations and these actually only show the substances that have been oxidised, reduced or changed state in a reaction.

So for example, the displacement reaction we've just been looking at between chlorine and potassium bromide, that's the equation for the reaction, but an ionic equation would look like this.

So we've actually removed the ions which are not involved.

Those are called spectator ions.

So we'll just have a quick look at those.

The spectator ions are ions which are not involved, they haven't been oxidised, reduced or changed state.

So it's potassium and potassium has been removed from that equation.

So redox displacement reactions can be shown as ionic equations.

So the full equation, including the state symbols, looks like this and then to write the ionic equation, we write out all the ions present.

So you can see we've written, we've changed potassium iodide to the potassium ions and the iodide ions and we've changed potassium chloride to the potassium ions and the chloride ions.

Then work out which ions are spectator ions and remove them.

So which are the same on either side of the equation? And if you have a look, it's the potassium.

That has not changed at all.

And then you write out the remaining substances to give the ionic equation.

So let's have a quick check.

Which ions or ion is or are spectator ions in the following reaction between sodium hydroxide and hydrochloric acid? Pause the video and come back when you're ready.

The sodium ions are spectator ions and the chloride ions are spectator ions.

They have not changed at all during the reaction.

An ionic equation includes both half equations for the reduction and half equation for the oxidation reaction.

So in this example, chlorine atoms are gaining electrons and are being reduced.

Remember, OILRIG.

And the iodide ions are losing electrons and are being oxidised.

And the electrons cancel each other out on either side of the equation to give us this.

And so we can see that chlorine is being reduced to form chloride ions and we can see the iodide ions are being oxidised to produce iodine.

Let's have a quick check.

So bromine is being oxidised in the following reaction.

Is this true or is this false? This is false and why is it false? So here's two statements to help you.

Why do you think it's false? Well done if you said B, because bromine is being reduced in the reaction, because each of the atoms, each of the bromine atoms is gaining an electron.

Well done if you got that correct.

So halide salts form a precipitate with silver ions.

When you put the two reactants together, you will see an insoluble substance, a solid forming in the solution, and this is really helpful, because it helps us to identify if we've got chloride, bromide or iodide ions present.

And we can show those using ionic equations.

So if we've got silver nitrate and we add it to a solution to see if there's halide ions present, we should get a precipitate form if there are any present.

And they also have slightly different coloured precipitates.

So chloride ions will form a white precipitate of silver chloride.

And there we've go the equation at the bottom.

Bromide ions form a creamy precipitate and iodide ions form a yellow precipitate of silver iodide, and there's the equation.

So for task B, what I'd like you to do, is complete the displacement and ionic equations here for all of these reactions.

Pause the video and come back when you're ready.

Good luck.

Let's have a quick check of the answers.

So you might like to pause the video again and just check that yours are correct.

Fantastic if you got all those correct, well done.

So question two.

Jun has three unknown halide solutions.

Jun knows that one's a chloride, one's a bromide and one's an iodide.

So suggest a method that he could use to identify the halide in each solution.

That's the first part of the question.

And then complete the results table and give the ionic equations for the reactions that have actually happened underneath.

So pause the video and I'll see you in a moment.

Let's look at the answers.

So for the first part, which was the method, you might have written something like this.

Probably won't be exactly the same, but something similar to that.

So here's the results table.

You might've put yours in different orders, but that doesn't matter, as long as you got the information correct.

So where you've written a yellow precipitate, that's an iodide, white precipitate, chloride and a cream precipitate, bromide, and that you've also got the correct equations below.

Don't forget the state symbols as well.

Well, we've now come to the end of our lesson.

We've done an awful lot today and quite a lot of tricky things, so brilliant, fantastic.

Well done for sticking with us.

So let's have a look at the summary.

These are the things we've been focusing on.

So reduction is the gain of electrons and oxidation is the loss of electrons.

Remember, OILRIG, that might help you in the future.

And then displacement reactions are often examples of redox reactions.

And remember, a redox reaction is where you simultaneously got one substance being reduced and the other oxidised at the same time.

And then half equations show which elements are being oxidised and which are reduced.

And we also looked at ionic equations where you remove the spectator ions, those that are not actually taking part in the reaction, they're not reduced or oxidised or have changed of state.

Well done.

As I said before, lots of tricky things there to learn.

I look forward to seeing you next time.