Chemguide: Support for CIE A level Chemistry


Learning outcomes 6.3(a), 6.3(b), 6.3(c), 6.3(d), 6.3(e)

These statements deal with standard electrode potentials and their uses. I am dealing with these in one block for reasons I will explain below.

Before you go on, you should find and read the statements in your copy of the syllabus. Just skim quickly through these to start with, and then come back to individual statements as you need to.


Essential background work

This topic can be a total nightmare. There are almost as many different ways of approaching this as there are teachers and writers, and most of these ways manage to make the whole thing completely confusing. Even the A level Coursebook makes this feel more difficult and confusing than it is.

CIE's syllabus and the teacher support material offer a sensible approach, but some of their earlier exam questions and mark schemes up to 2010 seemed to take a different approach, and what they wanted sometimes seemed inconsistent with the teacher support material. More recent questions (including the one in the specimen paper for the new syllabus) have been more reasonable.

So what I would like you to do is to get comfortable with this topic without any reference to what the syllabus is asking point-by-point. Once you are comfortable, we will come back and look at the individual statements.

On Chemguide, there is a set of five pages about this. They are intended to be read in order, and by the end of them you should have a clear idea of what is happening, and a simple visual way of using electrode potentials with confidence.

The section starts with an introduction to redox equilibria, and you will find a link to the next page at the bottom of each page you finish. There are a few important points before you start:

  • You must take your time over this. I would suggest that you don't try to do more than one page at a time, and that you don't move on to the next page until you are completely confident about the current one.

  • On the first page, about 3/4 of the way down, you will find a section about cell conventions, including a quick way of drawing a cell. Up to now, CIE have never used this method, but you should read about it anyway, because it isn't difficult. You will need to know the convention about the sign of the right-hand electrode.

  • You will also find similar cell diagrams on the third page of the sequence. These are getting a bit complicated, and you don't need them. Just skip over them, but take care not to miss any important bits of surrounding text.


When you have finished this, you will have covered all of what the syllabus and the teacher support material asks, apart from some very simple calculations.

It is now time to go through point-by-point to see how it all fits the syllabus, and to try to tie it into the sort of questions that get asked.

Don't go on unless you are sure that you understand what you have read so far.


Statement 6.3(a)(i)

The definition of standard electrode potential is just over 3/4 of the way down the first page in the sequence you have just read. Don't forget to learn the conditions relating to the word "standard" from further up that page.


Statement 6.3(b)

This is about the standard hydrogen electrode. You will find that on the same page.


Statement 6.3(c)

You also will find the measurement of the standard electrode potentials of metals in contact with their ions on the same page.

Measuring the standard electrode potentials of non-metals and their ions (such as chlorine and chloride ions) is discussed on the third page in the sequence, along with ions of the same element in different oxidation states.


Statements 6.3(a)(ii), 6.3(d) and 6.3(e)(i)

These statements are about overall cell potentials. The standard cell potential is the emf measured when you connect two half-cells together under standard conditions.

The standard electrode potential is actually a special case of a standard cell potential where one of the half-cells is a hydrogen electrode.

You can also imagine a cell made up of two half-cells like this:

The E° values for the two half-cells are:

The negative sign of the zinc E° value shows that it releases electrons more readily than hydrogen does. The positive sign of the copper E° shows that it releases electrons less readily than hydrogen.

So you can work out that in the combined cell in the diagram, the zinc electrode will be relatively more negative, and the copper one relatively more positive. There will be more build up of electrons on the zinc than there are on the copper.

Calculating the cell potential

There is a very simple formula for doing this. To find the standard cell potential, you subtract the E° value of the left-hand half-cell from the E° value of the right-hand half-cell.

cell = E°right - E°left

So in this case:

cell = +0.34 - (-0.76) volts = +1.10 volts

Now, remember a convention that you came across early in the series of pages that you have read. The sign of E°cell when calculated in this way represents the sign of the right-hand electrode as you have drawn it.

That means that the copper electrode (the right-hand electrode) is the positive one. We have already worked this out of course, but the calculation gives you another check on it.

Suppose you had taken the same cell, but drawn it the other way around - with the zinc on the right and the copper on the left. If you had done the same sum, you would get an answer of -1.10 volts. Now that the zinc is the right-hand electrode, the answer shows that this is the negative electrode.

The direction of electron flow in the external circuit

This is from statement 6.3(e)(i), but just follows logically on from what we have been talking about. The two half-cells we have been using as an example are:

The position of the zinc equilibrium must be further to the left because it is the more negative. That means that there is a greater build-up of electrons on the zinc at equilibrium than on the copper.

If electrons are allowed to flow through the external circuit, then they will obviously flow from where there are more of them to where there are fewer. They will flow from the zinc to the copper.

Calculating the cell potential from a given equation

If you are given a diagram of a cell, it is really easy to calculate the cell potential. If you are given an equation for an overall reaction, it suddenly gets much more confusing.

A CIE question asked you to work out the E° for this reaction (although they made you work out the equation to start with!):

The first thing you have to do is to split the equation into two bits so that you can find the electron-half-equations, and the E° values which go with them.

In this case, the Fe3+ ions are being converted into Fe2+ ions, and the Cu is being converted into Cu2+ ions.

In an exam, you will be given a Data Booklet which will give you the relevant half-equations and their E° values. You can find that Data Booklet towards the end of the syllabus.

E° values are listed twice - once in alphabetical order, and once in reverse electrochemical series order. Find these, and look at the list in alphabetical order.

Now find the two equations you will need, and write them down exactly as they are listed in the table.

  • For the iron equation, you will need Fe3+ ions on one side, and Fe2+ ions on the other. For now, don't worry about which side of the equation they are on.

  • For the copper equation, you will need copper atoms on one side, and Cu2+ ions on the other. Again, don't worry about which side of the equation they are on.

The two equations you should have found are:

Before we look at the calculation, just check that you can see why these values lead to the reaction we are interested in.

The iron equilibrium is the more positive, and so will move to the right. The less positive (more negative) copper equilibrium will move to the left. That is exactly what the original equation is saying.

Now all you have to do is to take one from the other in order to find the E° of the reaction.

But there is obviously a problem here. Which do you take away from which? There isn't a right-hand and left-hand half-cell any more.

For reasons which are beyond A level, the E° value of a reaction has to be positive. So when you do sums of this sort, you do the subtraction which leads to a positive answer. Make sure that you do this. The mark scheme for this question wouldn't allow an answer of -0.43 v. The answer had to be +0.43 v (or just 0.43 v).

0.43, of course, is what you get by subtracting 0.34 from 0.77.


Note:  Being able to have either a positive value or a negative value in some circumstances, but only a positive one in others is quite confusing!

You need to remember that getting a negative or a positive value by using the equation E°cell = E°right - E°left is simply a convention that quickly enables you to discover the sign of the right-hand electrode as you have drawn it. If you are interested in the voltage produced by a cell reaction, that will always be recorded as positive.




Statement 6.3(e)(ii)

This statement asks you to make predictions about the feasibility of a reaction. You will already have read about this in the set of pages I asked you to look at at the start of this page. It would pay you to read about it again now. You will find it on this page.


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© Jim Clark 2011 (last modified April 2014)