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# Electrolytic reactions

## Electrolysis of molten ionic compounds

Electrolysis separates molten ionic compounds into their constituent elements.

Molten ionic compounds contain free-flowing metal cations and non-metal anions.

The metal cations are reduced at the cathode to form neutral metal atoms. These atoms are usually deposited as solids on the cathode.

The non-metal anions are oxidised to form neutral non-metal atoms or molecules, which are often liberated as gases.

In the electrolysis of molten aluminium oxide ($\ce{Al2O3}$), aluminium cations are reduced to form aluminium atoms at the cathode.

Oxide anions are oxidised to form oxygen gas at the anode.

Aluminium oxide is therefore separated into its constituent elements (aluminium and oxygen).

Aluminium (left) is produced through electrolysis of aluminium oxide (right). Oxygen gas is released at the anode.

## Half equations

The transfer of electrons during electrolysis can be described using half equations.

Reduction of cations occurs at the cathode. This reaction can be described by adding electrons to the charged cations.

$$\ce{Na^+ + e^- -> Na {(s)}}$$

Oxidation of anions occurs at the anode. This reaction can be described by the breakdown of the anion into a neutral atom or molecule and electrons.

$$\ce{2Cl^- -> Cl2 + 2e^-}$$

In the case of molten sodium chloride (table salt), sodium metal is formed on the cathode while chlorine gas is released at the anode.

When combining two half equations to form a chemical equation, the numbers of electrons on both sides have to agree. Combining the above two half equations above gives:

$$\ce{2Na^+ + 2Cl^- -> 2Na + Cl2}$$

## Electrolysis of aqueous ionic compounds

In the electrolysis of aqueous ionic compounds, water competes with the dissolved ions for oxidation and reduction at the electrodes.

In solution, water partially dissociates into hydrogen ($\ce{H^+}$) and hydroxide ($\ce{OH^-}$) ions.

$$\ce{H2O -> H^+ + OH^-}$$

Water exists in all three forms ($\ce{H2O}$, $\ce{H^+}$, $\ce{OH^-}$) in the electrolyte.

At the cathode, the metal cations of the aqueous ionic compound have to compete with the hydrogen ions from water for reduction. The cathode product depends on the metal ion present.

If hydrogen ions are reduced, hydrogen gas is discharged. If the metal cations are reduced instead, the metal is deposited. The element leaves the electrolyte as a gas or solid.

The reduction of hydrogen is shown.

$$\ce{2H^+ + 2e^- -> H2}$$

At the anode, the non-metal anions of the aqueous ionic compound have to compete with hydroxide ions for oxidation.

Either oxygen or the non-metal will be discharged. The oxidation of hydroxide to oxygen is shown.

$$\ce{4OH^- -> 2H2O + O2 + 4e^-}$$

## Discharge of cations in electrolysis

The reactivity series lists metals according to reactivity. This serves as a guide to determine what will be discharged at the cathode.

Metals higher on the reactivity series (more reactive) have a greater tendency to exist in cation form.

If a metal is more reactive than hydrogen, it will remain in the solution as an ion while hydrogen is discharged.

If the metal is less reactive than hydrogen, it will be discharged while hydrogen ions remain in solution.

In the case of aqueous sodium chloride, the following two reactions compete.

$$\ce{2H^+ + 2e^- -> H2}$$

$$\ce{Na^+ + e^- -> Na}$$

In this case, the reduction of hydrogen prevails, and sodium cations are left in solution while hydrogen gas is discharged. This occurs because sodium is more reactive than hydrogen.

If copper cations were present instead, copper would be discharged instead of hydrogen.

## Discharge of anions in electrolysis

At the anode, the non-metal anion must compete with hydroxide ions for oxidation and discharge. Either the non-metal or oxygen gas is released.

Halides (the anions from halogens) are usually discharged at the anode instead of oxygen when they are highly concentrated in the electrolyte.

In the case of concentrated aqueous sodium bromide, the following two reactions compete.

$$\ce{4OH^- -> O2 + 2H2O + 4e^-}$$

$$\ce{2Br^- -> Br2 + 2e^-}$$

In this case, the oxidation of bromide ions prevails because a bromide ion is a halide. Bromine gas is discharged instead of oxygen.

Sulfates ($\ce{SO4^2-}$) and nitrates ($\ce{NO3^-}$) tend to remain in solution during electrolysis. The oxidation of hydroxide to oxygen gas generally prevails.

The electrolysis of substances such as sodium sulfate ($\ce{Na2SO4}$) and potassium nitrate ($\ce{KNO3}$) almost always results in the discharge of oxygen gas instead of sulfate or nitrate.

## Effect of concentration on electrolysis

The concentration of ions influences the discharge of matter at the electrodes.

As the concentration of particular ions increases, their discharge becomes more likely. As the concentration of particular ions decreases, their discharge becomes less likely.

A dilute solution of a halide salt (e.g. sodium chloride, potassium bromide) will result in the discharge of oxygen and hydrogen at the electrodes, leaving the salt ions in the solution.

If the concentration of the salt in water is increased, then the halogen will discharge at the anode instead of oxygen.

Concentrated solutions of aqueous halides tend to discharge the halide as halogen molecules. Dilute solutions of aqueous halides tend to discharge oxygen gas.

Electrolysis of concentrated sodium chloride solution (left) gives chlorine (right).

## Electroplating

Electroplating uses electrolysis to coat an object in pure metal.

The electrodes in electroplating are not inert. The electrodes participate in the reactions. The metal used for plating is the anode.

The oxidised metal atoms leave the anode, enter the electrolyte solution, and are plated onto the cathode to produce a layer of pure metal.

The electrolyte used has to contain cations of the metal to be plated. The electroplating of copper requires copper (II) sulfate solution.

Copper is purified through electroplating, using impure copper as the anode and another metal as the cathode.

In the electroplating of copper, the copper atoms in the anode are oxidised to form cations ($\ce{Cu^2+}$), which enter the electrolyte solution.

The copper ions are then reduced to atoms and deposited onto the metal cathode.

The cathode is plated at the rate at which the anode dissolves.