alkanes chemical properties

Higher and foundation tier- 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 the 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, compounds containing metals and non-metals will conduct electricity if they are molten (melted to form a liquid) or when they dissolve in water to form a solution.

Firstly let's look at a rather unusual property of water to help us explain what happens during the electrolysis of solutions. Water has some quite amazing properties. One of these properties of water is called auto-ionisation. Water is a very poor conductor of electricity, but it does conduct. This means that it must contain ions, but water is a covalent compound with the formula H2O, so where do these ions that allow water to conduct electricity come from?
Well it's to do with this property called auto-ionisation, which basically means that a few water molecules at any one time will break apart to form hydrogen ions (H+) and hydroxide ions (OH-) e.g. In the water molecule shown below, one of the bonds holding the water molecule together breaks and forms the H+ and OH- ions, this reaction is reversible and the water molecule can reform. However at any one time there will be a stable concentration of H+ and OH- ions. It is the presence of these ions that allows water to conduct electricity.

water auto-ionisation equation

or we can simply show this as:

H2O(l)OH-(aq) + H+(aq)

The electrolysis of sodium chloride and copper chloride solutions

When these two ionic compound are dissolved in water to be able 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, the substance that is dissolved and most importantly the ions present due to the dissociation of the water molecules, the hydrogen ions, H+ and the hydroxide ions, OH-.

ions present sodium chloride solution copper chloride solution
ions present from the solid which dissolves (the solute) sodium ions Na+ chloride ions Cl- copper ions Cu2+ chloride ions Cl-
ions from from dissociation of water molecules hydrogen ions H+ hydroxide ions OH- hydrogen ions H+ 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 a greenish-yellow gas, chlorine, produced at the anode.

electrolysis of copper chloride solution

However the results for the sodium chloride solution 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 chlorine gas is produced, as expected. So how do we explain this? Well remember that you need to consider not only the ions present from the compounds which dissolve in the water but also the hydrogen ions (H+) and the hydroxide ions (OH-) from the auto-ionisation of the water molecules, as shown below:

 ions present
  in a solution of copper and sodium chloride

At the cathode we have the sodium ions (Na+) and the hydrogen ions (H+). We already know that reduction 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)

Since hydrogen is lower down the reactivity series than sodium, 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.


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 an 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 the properties 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. This means 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 chlorine, bromine or iodine gas. As shown in the image below:

products of solutions of halides

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 ions:

anions (negatively charged ions formula
sulfate SO42-
nitrate NO3-
carbonate CO32-

The ions in the table, often called group ions, are stable and are not normally discharged at the anode. So what will be discharged? Well don't forget the hydroxide ion (OH-) from the water. It will be discharged instead of any of the group ions in the table above. The hydroxide ion will discharge to give off oxygen gas according to the equation below:

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

Consider the electrolysis of the following three solutions, calcium nitrate solution, potassium carbonate solution and lithium sulfate solution: Despite the fact that three different solutions are used the products of electrolysis of all three solutions are the same, hydrogen gas at the cathode and oxygen gas at the anode, as shown below:

products of electrolysis of group ions

The reason for this is simply based on the rules we have covered so far: At the cathode: the metal lowest in the reactivity series is discharged first. In all cases the metals present, calcium, potassium and lithium are all above hydrogen (from the water) in the reactivity series. Therefore hydrogen ions are discharged (or reduced in this case) to release hydrogen gas:

2H+(aq) + 2e → H2(g)
At the anode: nitrate, carbonate and sulfate are all stable group ions and are not discharged, hydroxide ions (OH-) from the water are discharged (or oxidised in this case) instead to release oxygen gas:
4OH-(aq) 2H2O(l) + O2(g) + 4e

Key Points

Practice questions

Check your understanding - Questions on electrolysis of solutions

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