Lesson video

Hello, my name's Dr.

Warren, and I'm so pleased that you can join me today for this lesson on half-equations used in electrolysis.

It's part of the electrolysis unit.

I'm here to work with you through this lesson and guide you all the way, especially through the tricky parts.

The learning outcome for this lesson is, I can write half-equations and balanced ionic equations.

And here are the key words.

Reduction, a type of reaction in which a substance gains electrons or loses oxygen.

Oxidation, a type of reaction in which a substance loses electrons or gains oxygen.

Half-equation, a chemical equation used to show the electrons lost in oxidation or the electrons gained in reduction.

Ironic equation, a balanced equation showing all the ions involved in the reaction.

Spectator ions, the ions that are not involved in a chemical reaction.

You may wish to pause the video now and copy down these keywords and their meanings so that you can refer to them later on during the lesson.

In today's lesson, we have two learning cycles, half-equations and iconic equations.

So let's get started with our first learning cycle on half-equations.

Reduction is a reaction where a substance loses oxygen.

Reduction can also be defined as a reaction where a substance gains electrons.

And when we're talking about half-equations, we are talking about electrons.

That's something to remember.

So here we have a sodium ion.

We add an electron, and we get a sodium atom.

So during the electrolysis of molten sodium ions, reduction occurs at the cathode, the negative electrode, because we have gained an electron.

In chemical equations, an electron is always shown as e-, where e stands for the electron with a negative charge.

So if we were going to write this as a chemical equation, we would have Na+ plus e- gives Na.

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

Let's have a quick check of understanding.

The reduction of potassium ions at the cathode can be shown as K+ plus an electron gives K.

True or false? Well done if you picked true.

Let's have a look at the reason why.

Well, each potassium ion gains one negatively charged electron to form a potassium atom.

Or B, each potassium ion loses one negatively charged electron to form a potassium atom.

Well done if you chose A.

That is the correct answer.

Oxidation is a reaction where a substance gains oxygen.

Oxidation can also be used to describe a reaction where a substance loses electrons.

In this example we've got here, we've got a chloride iron, dot and cross diagram, and it loses an electron to form a chlorine atom and electron on its own.

This happens during the electrolysis of chloride ions.

Oxidation occurs at the anode, which is the positive electrode.

The equation for the reaction is shown here.

The chloride ion goes to a chlorine atom plus an electron.

And we can also write that Cl- goes to Cl plus e-.

But it's important to realise that chlorine exists as a diatomic molecule.

So in practise at the anode, two atoms combine to form chlorine gas, which is then collected.

Half-equations show the electrons lost during oxidation, but they must always show a substance in its standard chemical form.

That means we need to show the substance chlorine as chlorine molecule.

So what we end up with is two chloride ions make a chlorine molecule plus two electrons.

The half-equation must always be balanced to reflect the element produced at the anode.

So here, we've got a diatomic molecule, so there must be two electrons.

Other diatomic molecules produced at the anode include bromine, Br2, iodine, I2, and oxygen, O2.

So quick check for understanding, true or false.

The oxidation of bromide ions at the anode can be written as 2Br goes to Br2 plus 2e-.

Is that true or false? Well done if you chose false.

Now let's have a look at the reason why.

Is it A, the equation must show 2BR- signs, not two BR atoms, or B, each bromine atom shares an electron to become a bromine molecule? Well done if you chose A.

It's a half equation.

We need to show all the electrons.

So the equation must show 2Br- ions, not the atoms. Excellent work.

A useful way to remember what's happening to electrons during oxidation and reduction is to use the anacronym OIL RIG.

Oxidation is a loss of electrons, reduction is gain of electrons.

And during electrolysis, oxidation occurs at the positive electrode, the anode, reduction occurs at the negative electrode, which is the cathode.

So OIL RIG will help you to remember what is happening at the electron level.

Let's have a look at an example that we've covered in a previous lesson.

During the electrolysis of molten aluminium oxide, the reactions at the electrodes can be written as half equations.

And there we have the special electrolysis cell for the electrolysis of molten aluminium oxide.

We have our graphite cathode.

So what happens at the cathode? Well, the Al3+ ions receive three electrons.

The Al3+ ions are reduced to aluminium atoms. So aluminium 3+, Al3+, plus three electrons, 3e-, gives Al, aluminium.

The charges in the half-equation must always balance.

And there you get your molten aluminium at the bottom of the electrolysis cell ready to be tapped off.

Oxidation occurs at the anode.

So this time at the anode, the O2- ions lose to electrons and are oxidised to oxygen atoms. And we can write this O2- goes to O plus 2e-.

So those electrons are lost and go around the circuit.

The half-equation is 2O2- gives O2 plus four electrons because oxygen exists in a standard state as the O2 molecule.

And you can see the oxygen gas there being produced at the graphite anode.

The atoms and charges in the half-equation must always balance.

That's really important.

So on each side of the equation, your charges and electrons balance.

A quick check for understanding.

During the electrolysis of sodium chloride solution, hydrogen is formed at the cathode.

Which of the following half-equations is correct? A, B, or C? Well done if you chose C.

Hydrogen gas is formed, and the electrons must, the electron charges must balance with the hydrogen plus ions.

So C is correct.

Well done if you got that right.

So we come to our first task.

Please, can you complete the following half-equations and state whether oxidation or reduction is taking place? So just work through A to F saying if oxidation or reduction is occurring.

Question 2, during electrolysis, oxidation occurs at the anode.

Can you explain why? Pause the video and have a go at these questions, and then when you're ready, we'll look at the answers together.

Right now, question 1, part A, Na+ plus e- gives Na, and that is reduction.

The sodium ion has gained an electron.

Part B, 2O2- gives O2 plus four electrons.

Oxidation has happened because the O2- has lost an electron, and we need to make the balances charge.

C, Mg2+ plus two electrons gives magnesium.

That is reduction.

D, 2Br- is bromine plus two electrons.

That is oxidation.

E, Al3+ plus three electrons gives aluminium, which is reduction.

Again, we need to make those charges balance.

3+ and 3- balanced out.

F, 2I- gives I2 plus two electrons.

And again, that's oxidation.

So if you've got all those right, well done.

You've really got that sorted.

So question 2, during electrolysis, oxidation occurs at the anode.

Can you explain why? Well, first of all, we need to state that the anode is positively charged.

During electrolysis, it attracts negative ions.

At the anode, the negative ions lose electrons.

Loss of electrons is oxidation.

So when you're answering a question like this, we have to explain something, just remember to write things in a logical order.

So if you've got that right in all those points, very well done.

Excellent work.

Right, question 3, the diagram shows what happens during the electrolysis of copper sulphate.

And you've got your diagram there, and you're being told you've got some fitting at the anode, and you've got a solid forming on the cathode.

So can you go through each part of the question? A, name the ions present in the electrolyte.

B, what is formed at the cathode? C, what type of reaction occurs at the cathode? D, write a half-equation for the cathode reaction.

Then for E, what is formed at the anode? What type of reaction takes place at the anode? And then balance the half-equation for the anode reaction.

Pause the video, have a go at this question, and then we'll look at the answers together.

Right, so the names of the ions present in the electrolyte are Cu2+ and SO42- coming from the copper sulphate.

And because it's an aqueous solution, we've got H+ and OH-.

You must have all four ions to get the correct answer.

B, what is formed at the cathode? Well, that's copper.

We saw that orange-brown solid.

What type of reaction occurs at the cathode? Reduction, the copper ions have gained electrons.

So write a half-equation for the cathode reaction.

Cu2+ plus 2e- gives Cu.

What is formed at the anode? That fizzing with oxygen.

What type of reaction occurs at the anode? That is oxidation.

Now balancing the half-equation for the anode reaction, this equation is a little bit trickier, but we should be able to manage it.

It's 4OH-, gives 2H2O, plus O2 plus 4e-.

So if you've got that last one right, that was brilliant 'cause that wasn't easy.

But what you need to do is make sure that all the atoms balance like in a normal balance equation, but also the charges do as well.

So very, very well done if you got that correct.

So that brings us to the end of our first learning cycle on half-equations, and we're now going to progress to look at ionic equations.

So chemical reactions can be represented using ionic equations.

And an ionic equation is a balance equation showing the substances in a reaction which undergo oxidation and reduction.

And you may have come across this before in some of your earlier topics.

So they are very useful for showing what is happening during electrolysis.

An example is electrolysis of brine, which is concentrated sodium chloride solution.

And if you look at that animation as we have seen in previous lessons, you can see which ions are attracted towards which electrodes and how the electrons are lost or gained during the reaction.

So thinking about the electrolysis of brine, it involves the overall equation of two sodium chlorides, aqueous, plus two waters, which is our liquid giving sodium hydroxide, which is aqueous, which is what is left in the electrolysis cell at the end, plus chlorine gas, plus hydrogen gas.

So that's what's actually happening in that overall reaction in the electrolysis of brine.

So the ions present are as follows, including the ionised water.

So when we're trying to develop analytic equations, it's always good to write out all the ions present.

So we have 2Na+.

We have 2Cl-.

We have 2H+, 2OH-.

And then on the other side of the equation, we have 2Na+, we have 2OH-, we have Cl2 gas, and H2 gas.

Really important to include the state symbols as well so we know which are ions and which are dissolved ions in aqueous solution and which are gases or liquids or anything else that's involved.

So the ions present, we can see here the chloride iron goes to chlorine gas.

And we have that oxidation at the anode.

So our half-equation, which we've already looked at in the first learning cycle, shows the two chloride minus ions going to chlorine gas and two electrons.

If we now look and see what else is happening, we have our 2H+ going to H2 gas, and this is reduction at the cathode.

So if we take our half-equation that we've already looked at, we have 2H+ plus two electrons, gives hydrogen.

So the sodium and hydroxide ions do not change.

They're not oxidised, and they're not reduced, but they do remain in solution the whole time.

So our Na+, we have it on both sides of the equation as products and reactants.

We also have a OH-.

We have two of them as a product and two of them as a reactants.

They are there in solution the whole time.

And these are known as spectator ions.

Think of whether it's spectator sport or if you go and watch a football match, you are a spectator, you are watching.

The reason we call them a spectator ions is they appear to be watching what is going on.

And because they're there on both sides of the equation, like we do in math, we can just cancel them out.

They take no part in the reaction.

So we cancel through all of those sodiums and hydroxide ions from the equation.

And what is left is the remaining substances that form the ionic equation for the electrolysis of brine.

So what is actually going on in the electrolysis of brine is 2Cl- plus 2H- gives Cl2 plus H2.

They are the products that are formed at the electrodes.

Okay, so that's taken us through what is actually happening in terms of ions during the electrolysis of brine.

So let's just recap.

Looking at our little animation again, we have an H+ ion and an Na+ ion.

We have a Cl- ion and an OH- ion.

The Na+ and the OH- ions are present in the electrolyte, but they are not involved in the reactions at the electrodes.

And if you actually look at the animation, you can see that they are in the electrolytes at the end as well as at the start.

So they're called spectator ions, and they are not included in the ionic equation.

At the end of the reaction, sodium plus and OH- ions are there, and that's why it's now an alkaline solution of sodium hydroxide.

Quick check of understanding.

The ionic equation for the electrolysis of copper bromide is Cu2+ plus 2Br- goes to Cu plus Br2.

Which of the following are spectator ions? Cu+, H+, Br-, or OH-.

Hint, look at the equation.

Well done if you chose B, H+, and D, OH-.

The clue is they do not appear in the ionic equation.

So they must be spectator ions.

So they'll be there in the reactants and the products, and they can cancel each other out.

So well done.

Excellent work.

An ionic equation includes both half equations for the reduction and oxidation reactions occurring at the electrodes.

Hydrogen ions are gaining electrons so they're being reduced.

So remember a half-equation, 2H+ plus 2e- goes to H2 gas.

At the same time, chloride ions are losing electrons and so are being oxidised.

And again, we've seen this half-equation before, 2Cl- goes to Cl2 plus 2e-.

Here, we can cancel out the electrons and combine the two equations.

So we've cancelled out both electrons at each side of the equation, and we've ended up with 2Cl- plus 2H+ gives H2 plus Cl2.

So we can get to that ionic equation by combining the two half-equations as well.

So that's two different routes.

There we have our oxidation taking place for the chloride ions of lost electrons, and we have our reduction taking place.

The hydrogen ions have gained electrons to give hydrogen gas.

Right, another quick check for understanding.

Ionic, an ionic equation summarises the half-equations occurring at the electrodes.

True or false? False.

An ionic equation is just another way to describe a half-equation.

Or B, an ionic equation combines the half-equations at the electrodes to show which substances are oxidised and reduced in a reaction.

Well done if you chose B.

It's really important to understand the differences between the ionic and half-equations 'cause they tell us different things.

Okay, so we've got another example here.

So during the electrolysis of sodium sulphate in aqueous solution, the overall equation for the reaction is sodium sulphate plus four waters gives two hydrogens plus oxygen, plus water, plus sodium sulphate.

So that's the overall equation for the reaction.

We can now write out the ions including those for the ionised water.

So we've got sodium plus ions, sulphate ions, hydrogen ions, hydroxide ions on one side, and then on the other side, we've got hydrogen gas plus oxygen gas plus water.

But then we've got sodium plus ions, sulphate ions.

It's really important that those equations are balanced.

So if we look carefully at that equation, we can find that the sodium plus ions and the sulphate minus ions do not react at the electrodes.

They are not oxidised, they're not reduced, they're on both sides of the equations.

Therefore, they must be spectator ions.

And what we can find is we can cross them out.

They cancel each other out.

They don't take part at their reaction.

So during the electrolysis of sodium sulphate, the half-equations at the electrodes are, cathode, 2H+ plus 2e- gives H2 gas.

The hydrogen ions are reduced.

At the anode, 4OH- goes to O2 plus 2H2O plus four electrons.

The OH minus ions are oxidised.

So that's your two half-equations.

We can combine the half equations to get our ionic equation, which is 4H+, 4OH- gives 2H2 plus O2 plus 2H2O.

Again, remember to put your state symbols in as this will really help.

Another quick check for understanding.

In the electrolysis of molten lead bromide, the ionic equation is Pb2+ plus 2Br- gives Pb plus Br2.

Which of the following is correct? H+ and OH- are spectator ions.

Pb2+ is a spectator ion.

Or C, there are no spectator ions.

Well done if you chose C.

Remember, it's the electrolysis of molten lead bromide.

So that means there's only Pb2+ and Br- present.

The Pb2+ is reduced, and the Br- is oxidised.

So all the ions are involved in the reaction at the electrodes.

So there are no spectator ions present.

Now this brings us to our second task.

For question 1, we'd like you to complete the table.

So you've got your electrolyte, and you have got to write down, first of all, all the ions present and then the spectate, the spectator ions.

So to work out the spectator ions, you're really gonna have to think hard.

Remember the reactivity series and the electrolysis rules and work out what's happening at the electrodes and then write down the ions that are not taking part in the reaction.

So to work out the spectator ions is quite tricky.

You'll have to really think about that.

In question 2, we'd like you to complete the half equations and state at which electrode they occur.

So pause the video and have a go at these questions, and when you're ready, we will look at the answers together.

Right, so for the first question, we have zinc chloride solution.

It's got the aq, which means we've got a Zn2+, we've got Cl-, we've got H+ and OH-.

So at the electrodes, we will have chlorine made because it's a halogen.

So that means that there will be an OH- ion as spectator.

And we have hydrogen plus, hydrogen formed because zinc is above hydrogen reactivity series, so the spectator ion will be Zn2+.

On the next row, we have zinc chloride liquid.

This means it's molten zinc chloride, which means we've only got Zn2+ and Cl- ions present.

So there are no spectator ions.

Both ions take place in reactions at the electrodes.

Copper nitrates.

Okay, so we've got Cu2+ and NO3- from the copper nitrate.

H+ and OH- from the water are spectator ions.

Well, copper is formed because it is below hydrogen in the reactivity series.

That means that our positive spectator ion is going to be H+.

There's a nitrate present, so that means we're gonna get oxygen from the OH-.

So our spectator ion is the NO3- ion.

So very well done if you've got that bit right.

And our final one in this table is silver sulphate.

So the ions present are Ag+ and SO42-, both coming from the salt, H+ and OH- from the water.

So this is similar to the line above.

Silver is not very reactive.

So that is gonna be formed, which means that our positive ion, H+, is gonna be the spectator ion, and the sulphate is also gonna be the spectator ion because it's the OH- that gets discharged to form oxygen.

So that was quite a tricky question.

Well done if you got it right.

For question 2, we'd like you to complete the half-equations and state which electrode they occur at.

So the first one, 2H+, we need to balance those charges.

We've got two of them, so there must be two electrons.

And that occurs at the cathode.

4OH-, the clue here is we have to balance our 4e-.

So we've got the charges balance to both sides.

That occurs at the anode because we are losing electrons.

Then we've got Zn2+ plus two electrons.

Again, we have to balance that charge, and we then form zinc at the cathode.

So very well done if you've got those equations balanced correctly.

Right, in our next question, we would like you to complete the tables and the ionic equations below.

So you're told in the first table, you've got the electrolyte of sodium chloride or liquids.

That's molten sodium chloride.

So look at the anode and the cathode, work out what the product is, if it's oxidation, reduction, half-equation, and then the ionic equation.

And then for the second one, the electrolyte is copper iodide aqueous.

So pause the video, have a go at the question, and then when you are ready, we'll look at the answer together.

Let's have a look at the answer for the sodium chloride molten.

So at the anode, our product is chlorine.

We have oxidation taking place.

The chloride ions have lost electrons.

So the half-equation is 2Cl- goes to Cl2 plus 2e-.

At the cathode, we're going to have sodium produced because there's only sodium ions present.

It's positive ions.

It's reduction.

And the half-equation is Na+ plus an electron gives sodium metal.

Now to form the ionic equation, we need to combine these and make sure the electrons cancel out.

So we have 2Na+, which is in the liquid state, plus 2Cl-, in the aqueous state, gives 2Na in the solid state, plus Cl2 in the gas state.

So very well done if you got that correct.

Quite a tricky question.

Right, we'll now move to the second example, which was copper iodide solution.

So at the anode, our product is iodine because it was a halide iodine present.

We have oxidation.

The half-equation is 2I- goes to I2 plus two electrons.

At the cathode, we have copper formed because copper is not very reactive.

It's below hydrogen in the reactivity series.

So we have reduction, and we have Cu2+ plus 2e- gives Cu.

We then combine our ionic equations, and we get Cu2+ in the aqueous state, plus 2I- in the aqueous state, gives Cu in the solid state, plus I2 in the gas state.

And that completes the ionic equation.

So very well done if you've got that correct including all of the state symbols.

That's great work.

So that brings us to the end of today's lesson.

Let's have a look at our key learning points.

A half-equation describes what happens to one reactant in a reaction, such as the process at the individual electrode.

Half-equation show balanced charges as well as balanced atoms. A balanced ionic equation shows all reacting ions involved in a reaction.

Ions not involved in a reaction do not appear in a balanced ionic equation.

They are called spectator ions.

I hope you have enjoyed today's lesson, and I look forward to learning with you again very soon.