electrolysis of solutions header image.

Higher and foundation tiers- Foundation tier students are NOT required to write ion-electron half equations for reactions happening at the electrodes. You only need to be able to predict what will be produced at the anode and cathode.

You should have a good idea of how electrolysis works before you read this page; click the link on electrolysis if you need to revise this first. Ionic compounds; that is compounds containing metals and non-metals will conduct electricity if they are molten (melted to form a liquid) or when they are dissolved in water to form a solution. When an ionic compound dissolves in water the giant ionic lattice of ions that make up any solid ionic compound is broken down by the water molecules and what we end up with is ions that are free to move within the solution. This process is called hydration and it is outlined in the image below:

sodium chloride lattice structure being solvated by water molecules

The ionisation of water molecules-autoionisation

Firstly let's look at a rather unusual property of water to help us predict the products of the electrolysis of solutions. Water is a very weak electrolyte; what this means is that at any one time a very small number of water molecules will break up or dissociate to form hydrogen ions (H+) and hydroxide ions (OH-); in fact around 2 molecules in every billion water molecules present at room temperature will break up to form hydrogen ions and hydroxide ions. This self-ionisation of water molecules is often called the autoionisation of water. During this autoionisation process one water molecule will transfer a hydrogen ion (H+) onto another water molecule. This results in the formation of a hydroxide ion (OH-) and an ion called a hydronium ion (H3O+); we can show this as:

Equations for the autoionisation of water molecules.

This autoionisation reaction is extremely rapid in both directions and no molecules stay ionised for very long, but there is always a constant concentration of ions present in water due to this ionisation reaction.

The electrolysis of copper chloride and sodium chloride solutions

When trying to predict the products of electrolysis of solutions not only do you need to consider the ions present from the dissolved solute but you must also consider the hydrogen ion (H+) and the hydroxide ion (OH-) from the autoionisation of the solvent water molecules.

As an example consider the electrolysis of concentrated solutions of sodium chloride solution and copper chloride using inert electrodes; that is electrodes that take no part on the reactions at the anode and cathode. What will form at the anode and cathode when these two solutions are electrolysed? When these two ionic compounds are dissolved in water; to work out what will happen at the anode and cathode when the solutions are electrolysed you need to consider all the ions present. This includes the ions from the solute; that is the substance that is dissolved and most importantly the ions present due to the dissociation of the water molecules, that is the hydrogen ions (H+) and the hydroxide ions (OH-). The table below list the ions present in these two solutions:

ions present sodium chloride solution copper chloride solution
ions present from the ionic solid which dissolves (the solute) sodium ions Na+ chloride ions Cl- copper ions Cu2+ chloride ions Cl-
ions from the dissociation of water molecules hydrogen ions H+ and hydroxide ions OH- hydrogen ions H+ and hydroxide ions OH-

The diagrams below show the results from the electrolysis of these two solutions. From the work you did on the electrolysis of molten ionic compounds you would expect a metal to form at the cathode and a non-metal to form at the anode. For the electrolysis of the copper chloride solution this is indeed what happens. The cathode is coated with a brown layer of copper metal and bubbles of greenish-yellow chlorine gas are produced at the anode.

Products for the electroysis of copper chloride solution.  The diagram shows the products for the electrolysis of copper chloride solution.

However the results for electrolysis of the sodium chloride solution (shown below) are probably not what you were expecting. No sodium metal is produced at the cathode; instead bubbles of hydrogen gas are produced. At the anode as expected chlorine gas is produced. So how do we explain this?

Diagram shows the products forming at the anode and cathode for the electroylsis of sodium chloride solution

Well remember that you need to consider not only the ions present from the ionic compound that dissolve in the water but also the hydrogen ions (H+) and the hydroxide ions (OH-) from the autoionisation of the water molecules. At the cathode there will be the sodium ions (Na+) from the sodium chloride solute and the hydrogen ions (H+) from the autodissociation of the water molecules. You should already know from the work you have done on the electrolysis of molten ionic compounds that reduction always takes place at the cathode; that is the ions pick up electrons and turn back into atoms. So the two possible options at the cathode are the reduction of hydrogen ions and the reduction of sodium ions:

2H+(aq) + 2e → H2(g)
Or
Na+(aq) + e → Na(s)

Only one ion can be discharged at any one time; so which ion will be reduced? Will it be the sodium ion or the hydrogen ion? The answer is fairly simple if you think about it. The ion which requires the least amount of energy to be reduced will be the one which is discharged. Since hydrogen is lower down the reactivity series than sodium it will take a lot less energy to reduce a hydrogen ion to a hydrogen atom than to reduce a sodium ion to a sodium atom. Or another way of looking at it is to say that it will take a lot less energy to force an electron back onto a hydrogen ion than a sodium ion or we could say the sodium ions are more stable than the hydrogen ions. This means that hydrogen will be produced or discharged before the sodium ions. In general we can say that:

The rule for predicting which "metal" will be produced at the cathode is simple: The metal LOWEST in the reactivity series is produced first.

We use a similar argument to explain what happens at the cathode during the electrolysis of the copper chloride solution. The two options at the cathode are the reduction of the copper ions (Cu2+) and the hydrogen ions (H+)

Cu2+(aq) + 2e → Cu(s)
Or
2H+(aq) + 2e → H2 (g)

The rule for predicting which "metal" is produced at the cathode is: The metal lowest in the reactivity series is discharged first. Copper is below hydrogen in the reactivity series so it is discharged at the cathode before hydrogen.

At the anode - predicting products of electrolysis

The ions present at the anode; the positive terminal; will obviously be negatively charged non-metal ions. Most of the solutions you are likely to meet will be solutions of metal halides e.g. metal chloride, metal bromide and metal iodide solutions. In this case the anode product will simply be halogen molecules: chlorine, bromine or iodine gas. As shown in the image below:

Diagram shows the products for electrolysis of solutions of sodium halides, that is sodium chloride, sodium bromide and sodium iodide.

However you may also come across solutions which contain negative ions other than the halides (chlorides, bromides and iodides). These solutions could contain the following group ions:

Group ion formula
sulfate SO42-
nitrate NO3-
carbonate CO32-

The ions in the table; often called group ions are stable ions and are not normally discharged at the anode. So what will be discharged in their place? Well there are two possible equations which are used to describe what happens at the anode; both of which you are likely to meet in the exam and ultimately it is up to you to decide which one you are going to use. Before we look at these equations consider the electrolysis of the following three solutions:

The diagram below shows the results of electrolysing these three solutions; in every case the products at the cathode and anode are the same; Hydrogen gas at the cathode and oxygen gas at the anode. You may notice from the diagram below that the hydrogen gas is produced much faster than the oxygen gas is produced at the anode. predicting the products of electrolysis of solutions containing group ions So any equations used to describe the formation of these products would have to explain these observations. The first equation below explains the production of oxygen gas at the anode. It uses the hydroxide ion produced by the autoionisation of water as the source of the oxygen gas produced and it also assumes that it is discharged instead of any group ion present.

4OH-(aq)2H2O(l) + O2(g) + 4e

This equation would be accepted as a suitable answer in your exam as it explains the formation of oxygen gas. However it can be improved upon. Another observation which so far has not been mentioned is the fact that as well as oxygen gas being produced at the anode if a few drops of universal indicator are added close to the anode it will turn red showing the presence of an acid; that is hydrogen ions (H+). The equation above does not explain this observation. Another possible equation which you may see which offers a better explanation of what is actually happening at the anode is shown below:

2H20(l) → O2(g) + 4H+(aq) + 4e

This equation offers a better description of what is actually happening at the anode since it shows the formation of oxygen gas and also the acidic hydrogen ions (H+). Here water molecules are oxidised and are basically being "ripped to pieces" at the anode. It really is a matter of personal choice as to which equation you decide to use as both are given credit at gcse level chemistry.

The electrolysis of the three solutions above; calcium nitrate solution, potassium carbonate solution and lithium sulfate all produced oxygen gas at the anode. So what about the cathode product? Well this was covered earlier; the metal ion present in each solution is above hydrogen in the reactivity series so it will not be discharged. Hydrogen gas will be produced instead. Recall the rule:

The equation we used above to describe the formation of hydrogen was:

2H+(aq) + 2e → H2(g)
 

This equation is perfectly acceptable and explains the observation that hydrogen gas is produced at the cathode. However an observation that is not mentioned above is that if universal indicator is added to the solutions being electrolysed then at the cathode it turns purple indicating the presence of an alkaline solution; that is the formation of hydroxide ions (OH-). So where are these hydroxide ions coming from?
An equation which perhaps better describes the reaction taking place at the cathode is shown below:

2H2O(l) + 2e → H2(g) + 2OH-(aq)

Both of these equations offer an acceptable description of the reduction reaction taking place at the cathode, as with the anode reaction it is up to you to decide which of the equations you intend to use.

The table below may help you work out the products produced at the cathode and anode when various solutions are electrolysed.

At the cathode At the anode
If the metal is below hydrogen in the reactivity series then the metal is produced at the cathode. If the solution contains halide ions (Cl-, Br- or I-) then the halogen (chlorine, bromine or iodine) is produced at the anode.
If the metal is more reactive than hydrogen in the reactivity then hydrogen is produced instead of the metal at the cathode. If the solution contains group ions such as sulfate, nitrate or carbonate then these ions are not discharged at the anode. Oxygen gas is produced by the discharge of water or hydroxide ions.
The equations which describe the reactions taking place at the cathode are:
2H+(aq) + 2e → H2(g)
or
2H20(l) + 2e → H2(g) + 2OH-(aq)
The equations which describe the reactions taking place at the anode are:
4OH-(aq)2H2O(l) + O2(g) + 4e
or
2H20(l) → O2(g) + 4H+(aq) + 4e

Key Points

Practice questions

Check your understanding - Questions on electrolysis of solutions

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