Electrolysis is the process by which a compound can be broken down or split into its elements by passing an electrical current through the sample. ‘Electro’ refers to the use of electricity and ‘lysis’ refers to breaking down or splitting. Electrolysis can only be used if the compound can conduct electricity, and for a compound to be able to conduct electricity, it must contain charged particles.
Covalent compounds do not contain any charged particles. Their electrons are all used up in bonding and they do not contain any ions. Covalent compounds cannot conduct electricity and therefore cannot be broken down using electrolysis.
Ionic compounds contain charged particles in the form of positively charged ions known as cations, and negatively charged ions known as anions. Therefore, ionic compounds can be broken down into their elements using electrolysis.
For an ionic substance to undergo electrolysis, the ions must be free to move. When an ionic compound is in a solid phase, the ions are regularly arranged and fixed into position in a giant ionic lattice. As the ions cannot move unless the structure is broken down, solid ionic compounds cannot conduct electricity and therefore cannot undergo the process of electrolysis.
When the solid ionic compound is heated to its melting point, the substance becomes molten. The lattice is broken down and the ions are released from their fixed positions. As the ions are now able to move, the molten ionic compound will conduct electricity and the process of electrolysis can be used.
Another way in which the ionic lattice can be broken down is by adding water to dissolve the solid. When the solid dissolves, the water particles are able to overcome the ionic bonds and the ionic lattice breaks down. The ions become free to move and the ionic solution formed can conduct electricity.
Ionic compounds will only conduct electricity when they are molten or in aqueous solution. Therefore, electrolysis can be used to separate ionic compounds into their elements, only when they are molten or in a solution.
The process of electrolysis involves passing an electrical current through the molten compound or aqueous solution. The substance which contains the ions and through which the electrical current is passed, is known as the electrolyte.
The electricity is passed into the electrolyte using two electrodes which are connected to an external direct current (DC) power supply.
One of the electrodes is positively charged and the other is negatively charged. The positively charged electrode is known as the anode and the negatively charged electrode is known as the cathode. These electrodes are usually made out of an inert (non-reactive) substance such as carbon. Over time the carbon electrodes will wear away and will need to be replaced.
Alternatively, platinum electrodes can also be used. Platinum is inert and is a metal so will conduct electricity. Platinum will also last longer than carbon electrodes. However, platinum is much more expensive so most people opt for the cheaper carbon option.
The ions in the molten compound or aqueous solution are free to move, so when the electrical current is passed through the electrolyte, the ions move towards the oppositely charged electrodes and lose or gain electrons to be converted back into the atoms again.
The metal ions are always positively charged and move towards the negative electrode. The non-metal ions are always negatively charged and move towards the positive electrode.
Oxidation and reduction reactions taking place in electrolysis
Oxidation is the name for the process which occurs when electrons are lost by a chemical species. A species which has been oxidised has lost electrons. Reduction is the name for the process which occurs when electrons are gained by a chemical species. A species which has been reduced has gained electrons.
A redox reaction is a chemical reaction in which both oxidation and reduction occurs. As electrolysis involves the simultaneous losing and gaining of electrons, we can say that the reaction is a redox reaction. This loss and gain of electrons by the ions provides the movement of electrons needed for the electrical current to flow. Negatively charged anions move towards the positively charged anode where they lose electrons to become atoms again. The anions are oxidised. The electrons flow through the wires to the negatively charged cathode. At the cathode, positively charged cations are reduced by gaining electrons to become atoms again.
We represent the oxidation and reduction reactions that occur at the electrodes during electrolysis using ionic half-equations which show whether a species has gained or lost electrons. In ionic half-equations, electrons are shown as e–.
Example – molten aluminium oxide
When molten aluminium oxide (Al2O3) is electrolysed, the oxide anions move towards the positive anode where they are oxidised by losing two electrons to become oxygen atoms. The aluminium cations move towards the negative cathode where they are reduced by gaining three electrons to become aluminium atoms. Write the ionic half-equation to represent the reaction at each electrode.
- Write the formulae of the reactant and products. You must make sure that the number of ions and atoms is equal.
| Anode | Cathode |
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- Add enough electrons to the correct side to balance the charge on the ion. The side of the equation on which the electrons are included, and the number of electrons gained or lost, depends upon the charge on the ion.
If the ion is positively charged, reduction will occur. As the electrons are gained, they should appear on the left side of the equation along with the ion. If the ion is negatively charged, oxidation will occur. As the electrons are lost, they should be included on the right side of the equation opposite to the ion.
| Anode | Cathode |
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The two ionic half-equations can then be combined to determine the overall chemical equation for the reaction. When combining the two half-equations, the total number of electrons being lost and gained must be equal. This may mean that you have to multiply one or both of the half-equations to make the number of electrons balanced. For example, the two half-equations for aluminium reduction and oxide ion oxidation can be combined to determine the overall equation for the redox reaction that occurs in the electrolysis of aluminium oxide.
As the oxide half-equation shows that four electrons are lost and the aluminium half-equation shows that three electrons are gained, it is necessary to multiply both half-equations to balance the total number of electrons being lost and gained.
We multiply the oxide half-equation by three and the aluminium half-equation by four to make the total number of electrons lost and gained twelve. When we multiply the half-equations, everything in the equation must be multiplied by the same number.
For example, multiplying the oxide half-equation by three, gives the equation:
and multiplying the aluminium half-equation by four, gives the equation:
As the number of electrons being gained and lost is equal, they can be cancelled out from the overall equation for the reaction. The overall equation for this redox reaction is therefore:
Electrolysis of aluminium oxide produces aluminium metal at the cathode and oxygen gas at the anode. We will now look at the process in more detail using some common examples of experiments used to investigate electrolysis.
Electrolysing molten ionic compounds
Lead(II) bromide
As lead(II) bromide is an ionic compound, it has a giant lattice structure where the lead(II) and bromide ions are tightly packed together. Lead(II) bromide cannot conduct electricity in solid form. To separate the lead from lead(II) bromide using electrolysis, the compound must be in molten form. As lead(II) bromide has such a high melting point, it takes a lot of energy to melt the solid and break down the ionic lattice. The lead(II) bromide is melted to become a molten liquid electrolyte and this is placed into a suitable vessel such as an evaporating dish.
The positive and negative electrodes are placed into the molten electrolyte and connected to an external power supply. The power supply is switched on and the electrical current begins to flow. The negative bromide ions move towards the positive electrode and the positive lead ions move towards the negative electrode as shown in the diagram below:
Lead(II) bromide contains lead(II) ions (Pb2+) and bromide ions (Br–). The Br– ions move towards the positive electrode where they lose one electron each to become bromine atoms. This means that the bromide ions are oxidised to become bromine atoms.
The bromine atoms join up in covalently bonded pairs to become diatomic molecules of bromine gas (Br2). This Br2 gas is discharged from the positive electrode. Br2 gas is toxic so it is important that this experiment is always done in a fume cupboard in a well-ventilated room.
The ionic half-equation representing the oxidation of bromide to bromine is:
The Pb2+ ions move towards the negative electrode where they gain two electrons each to become Pb atoms. Pure Pb atoms are formed which sink to the bottom of the vessel and gather as Pb liquid. The ionic half-equation representing the reduction of lead(II) to lead metal is:
The overall ionic equation for the electrolysis of lead(II) bromide is:
The electrolysis of lead(II) bromide produces two products – pure lead metal as a liquid and bromine gas.
Electrolysing ionic solutions
Molten ionic compounds contain only the metal ions and non-metal ions. The products of electrolysis therefore are easy to predict. The metal ions will produce a liquid metal and the non-metal ions will produce a gas. Predicting the products of electrolysis of ionic solutions, however, can be a little more difficult.
The ions contained in the solution are not just the metal and no-metal ions of the ionic compound. When the ionic compound is dissolved in water to form the aqueous solution, we also have to consider the hydrogen ions (H+) and the hydroxide ions (OH–) present in the water.
The identity of the products formed at each electrode depend upon the ions present and their reactivities compared to the H+ or OH– ions. If the metal is more reactive than hydrogen, they will remain as ions and the H+ instead will move to the negative electrode where it will be reduced to hydrogen atoms. The hydrogen atoms will then covalently bond as pairs forming diatomic molecules of hydrogen gas (H2).
If the non-metal ion present is a halide ion, molecules of halogen gas will be produced at the positive electrode. If no halide ions are present, oxygen gas will be produced at the positive electrode instead.
A common way of investigating electrolysis reactions is to examine how the amount of product formed is affected by the duration of electrolysis or the size of the current used by the power supply. You could also investigate how the concentrations of the solutions used affects the extent of electrolysis.
The amount of product formed can be measured as either the volume of gas or mass of solid produced. There is a directly proportional relationship in these experiments. When the duration of electrolysis or the current used is doubled, the amount of product formed also doubles.
If the duration and current are both doubled, the amount of product formed quadruples. The amount can be the volume of gas produced, for example oxygen, or it could be the mass of the solid produced. If the time taken is doubled or the current which is flowing is doubled, this conclusively results in the amount of product being doubled. If time and current are both doubled, the amount produced will increase four times.
You must be able to describe the experiments and predict the products of the electrolysis of sodium chloride, dilute sulphuric acid and copper(II) sulphate.
Electrolysis of sodium chloride solution
Sodium chloride is an ionic compound. At room temperature it is a solid with its ions packed closely together in a regular giant ionic lattice structure. Sodium chloride in solution can be electrolysed to produce chlorine gas, hydrogen gas and sodium hydroxide solution, as shown by the diagram below.
Sodium chloride solution contains sodium ions (Na+), chloride ions (Cl–), hydrogen ions (H+) and hydroxide ions (OH–). During electrolysis, the Cl– move towards the positive electrode where they are oxidised to form chlorine atoms (Cl). The chlorine atoms covalently bond to form diatomic molecules of chlorine gas (Cl2) which is released from the positive electrode.
Chlorine gas is toxic so this experiment must always be completed in a fume cupboard in a well-ventilated laboratory. The electrolysis should also be run in a school laboratory for no more than 30 seconds to minimise the volume of chlorine gas produced.
The ionic half-equation for this reaction is:
As sodium is more reactive than hydrogen, the hydrogen ions move towards the negative electrode instead of the sodium ions. The hydrogen ions are reduced and gain electrons to become hydrogen atoms. These hydrogen atoms covalently bond to form diatomic molecules of hydrogen gas (H2) which is then released at the negative electrode.
The presence of hydrogen gas can be tested using a lit splint. When hydrogen is present, the lit splint will produce a squeaky pop sound as the hydrogen reacts with the heat and oxygen in the air. The half-equation for the reaction that occurs at the negative electrode is:
The overall equation for the reaction is:
Two products of sodium chloride electrolysis are chlorine gas from the positive electrode and hydrogen gas from the negative electrode. The sodium chloride solution also contains Na+ and OH– ions which combine to produce the third product sodium hydroxide (NaOH):
All three products are useful:
- Chlorine is a toxic gas and can be added to water and other chemicals to produce chlorine water and chlorine compounds. Very low concentrations of chlorine compounds are used to eliminate bacteria in drinking water and swimming pools. Chlorine can also be used to produce solvents and to make plastics like polyvinyl chloride (PVC)
- Sodium hydroxide can be mixed with fats and oils to make soaps in a process known as saponification
- Sodium hydroxide can also be reacted with chlorine to produce the active ingredient in household bleaches – sodium chlorate (NaClO). The reaction also produces sodium chloride and water as shown by the equation:
Electrolysis of sulphuric acid
When a dilute solution of sulphuric acid is electrolysed, hydrogen gas and oxygen gas are produced. The diagram below shows the set-up for the electrolysis of sulphuric acid.
Hydrogen ions (H+) move towards the negative electrode where they are reduced to form hydrogen atoms. These hydrogen atoms then pair up to form covalently bonded diatomic molecules of hydrogen gas (H2), which is released from the cathode.
The ionic half-equation for the reaction occurring at the negative electrode is:
The hydroxide ions (OH-) from the solution move towards the positive electrode where they are oxidised to form oxygen atoms. The oxygen atoms pair up to form diatomic molecules of oxygen gas (O2).
The ionic half-equation for the reaction occurring at the positive electrode is:
The two gases can be collected by placing inverted test tubes over each electrode.
As the concentration of H+ ions in the solution is roughly twice that of the OH- ions (due to the presence of the acid), the amount of hydrogen gas produced should be roughly twice that of the oxygen gas. This is also shown by the ratio of gases produced in the overall ionic equation for the reaction:
Electrolysis of copper(II) sulphate solution
The electrolysis of copper(II) sulphate solution, produces copper metal and oxygen gas. As copper is less reactive than hydrogen, it is the copper cations in the copper(II) sulphate solution that are attracted towards the negative electrode. The hydrogen ions remain in solution. At the negative electrode the copper(II) ions are reduced to form copper atoms, as shown in the ionic half-equation:
The copper atoms produced are deposited onto the surface of the negative electrode which is usually made of carbon. This can be easily seen as a brown deposit. The amount of copper produced can be easily determined, as the mass of the carbon electrode can be measured before and after the experiment. The difference between the two masses can be taken as the mass of copper produced and deposited onto the electrode.
The hydroxide ions present in the solution move towards the positive electrode where they are oxidised to produce oxygen gas. The sulphate ions (SO42-) remain in solution. The ionic half-equation for the reaction that occurs at the positive electrode is:
Before the copper(II) sulphate solution undergoes electrolysis it has a blue colour due to the presence of the copper (II) ions. As the copper(II) ions are reduced, their concentration in the solution decreases and the blue colour of the solution fades until it eventually becomes colourless when no more copper(II) ions remain. The H+ and SO42- ions remaining in the solution combine to produce a solution of sulphuric acid (H2SO4).
