Electrolysis

Welcome to one of the most exciting topics in GCSE Chemistry! Electrolysis literally means "splitting with electricity". It is a powerful chemical technique used to break down ionic compounds into their constituent elements. In these notes, we will break down the fundamental rules of electrolysis step-by-step so you can confidently predict products, write half equations, and ace your exams.

What you'll learn

How ionic compounds conduct electricity when molten or dissolved, but not when solid.
How to predict the products formed at the anode and cathode during the electrolysis of molten binary compounds.
[Higher Tier Only] How to predict the products of aqueous solutions where multiple ions compete at the electrodes.
[Higher Tier Only] How to write balanced half equations representing the loss or gain of electrons at each electrode.

What is Electrolysis?

To understand electrolysis, we must first look at the experimental setup. Electrolysis is the decomposition of a liquid or solution using a direct current (DC) of electricity. For this process to work, we need three key components:

The electrolyte: The liquid or solution containing ions that conducts electricity.
The electrodes: Solid rods (usually made of unreactive materials like graphite or platinum) that dip into the electrolyte to complete the electrical circuit.
The power supply: A DC source that drives the movement of electrons in the wires and ions in the electrolyte.

Definition

Electrolysis

Electrolysis is the chemical decomposition of an electrolyte caused by the passage of a direct electric current through it.

Definition

Electrolyte

An electrolyte is an ionic compound in a molten state (liquid) or dissolved in water (aqueous solution) that conducts electricity and is decomposed by it.

The two electrodes have opposite charges:

The anode is the positive electrode.
The cathode is the negative electrode.

Tip

Remembering electrode names

To remember which electrode is which, use the phrase PANIC:

Positive Anode
Negative Is Cathode

How Ions Move During Electrolysis

When the power supply is turned on, the electrodes become charged. Because opposite charges attract:

Cations (positive ions, like metal ions $Mn+\text{M}^{n+}$ or $H+\text{H}^+$ ) are attracted to the negative cathode.
Anions (negative ions, like non-metal ions $Xn−\text{X}^{n-}$ or $OH−\text{OH}^-$ ) are attracted to the positive anode.

As these ions reach the electrodes, chemical reactions occur, and the compound is decomposed.

Diagram showing a standard electrolysis cell with a positive anode, negative cathode, and the movement of cations and anions in an electrolyte

Common Mistake

Confusing ion flow with electron flow

A very common exam mistake is stating that ionic solutions conduct electricity because "electrons flow through the solution". This is incorrect!

In the external wires, electricity is conducted by moving electrons.
In the electrolyte, electricity is conducted by moving ions.

Electrons cannot travel freely through water or liquid ionic compounds.

Prerequisites: Why Must the Electrolyte Be Liquid or Aqueous?

You might wonder why we cannot perform electrolysis on solid ionic compounds like solid table salt ( $NaCl(s)\text{NaCl(s)}$ ).

In a solid ionic compound, the oppositely charged ions are held tightly together in a rigid, giant three-dimensional ionic lattice by strong electrostatic forces of attraction. Because the ions are locked in fixed positions, they cannot move. Therefore, solid ionic compounds do not conduct electricity.

To undergo electrolysis, we must break this lattice apart. We can do this in two ways:

Melting it: Heating the solid until it turns into a liquid ( $molten\text{molten}$ ). The thermal energy overcomes the strong electrostatic forces, allowing the ions to move freely.
Dissolving it: Mixing the solid with water ( $solution\text{aqueous solution}$ ). Water molecules surround the ions and pull them out of the lattice, allowing them to move freely.

Key Idea

The Golden Rule of Conductivity

Ionic compounds only conduct electricity when molten or in aqueous solution because only then are the ions free to move to carry the charge.

Electrolysis of Molten Binary Ionic Compounds

A binary ionic compound contains only two elements: a metal cation and a non-metal anion (for example, molten lead(II) bromide, $PbBr2(l)\text{PbBr}_2\text{(l)}$ ).

Because there are only two types of ions present in a molten binary compound, predicting the products is straightforward:

The metal ions (cations) move to the cathode and form metal atoms/liquids.
The non-metal ions (anions) move to the anode and form non-metal molecules/atoms.

Let's look at the classic example of molten lead(II) bromide, $PbBr2(l)\text{PbBr}_2\text{(l)}$ , using inert electrodes (electrodes made of graphite or platinum that do not react with the electrolyte or products).

At the Cathode (-): Positive lead ions ( $Pb2+\text{Pb}^{2+}$ ) are attracted here. They gain electrons to become grey lead metal:

\text{Pb}^{2+}\text{(l)} + 2\text{e}^- \rightarrow \text{Pb(l)}

At the Anode (+): Negative bromide ions ( $Br−\text{Br}^-$ ) are attracted here. They lose electrons to form brown bromine gas:

2\text{Br}^-\text{(l)} \rightarrow \text{Br}_2\text{(g)} + 2\text{e}^-

Example

Predicting the products of molten sodium chloride electrolysis

Predict the products formed at the cathode and anode during the electrolysis of molten sodium chloride, $NaCl(l)\text{NaCl(l)}$ , using inert electrodes.

Identify all the ions present in the molten compound. Since this is a molten binary compound, the only ions present are sodium ions ( $Na+\text{Na}^+$ ) and chloride ions ( $Cl−\text{Cl}^-$ ). There is no water present.
Determine which ion goes to the cathode and predict the product. The positive sodium ions ( $Na+\text{Na}^+$ ) are attracted to the negative cathode. They gain electrons to form neutral sodium metal ( $Na(l)\text{Na(l)}$ ).
Determine which ion goes to the anode and predict the product. The negative chloride ions ( $Cl−\text{Cl}^-$ ) are attracted to the positive anode. They lose electrons to form neutral, diatomic chlorine gas ( $Cl2(g)\text{Cl}_2\text{(g)}$ ).

Electrolysis of Aqueous Solutions: Competing Reactions

When we dissolve an ionic compound in water to make an aqueous solution, things get a little more complicated.

Water molecules ( $H2O\text{H}_2\text{O}$ ) dissociate (split up) slightly into hydrogen ions ( $H+\text{H}^+$ ) and hydroxide ions ( $OH−\text{OH}^-$ ):

\text{H}_2\text{O(l)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{OH}^-\text{(aq)}

This means an aqueous solution always contains four different ions:

The metal cation from the solute.
The non-metal anion from the solute.
Hydrogen ions ( $H+\text{H}^+$ ) from water.
Hydroxide ions ( $OH−\text{OH}^-$ ) from water.

Because only one substance can be released (discharged) at each electrode, these ions must compete! We use a set of rules to determine which ion "wins" and gets discharged.

Rules at the Cathode (Negative Electrode)

Both the metal cation ( $Mn+\text{M}^{n+}$ ) and the hydrogen ion ( $H+\text{H}^+$ ) are attracted to the cathode. Only one will be discharged, depending on the reactivity of the metal relative to hydrogen:

If the metal is more reactive than hydrogen (e.g., sodium, potassium, calcium, magnesium, aluminium):
- Hydrogen gas ( $H2(g)\text{H}_2\text{(g)}$ ) is produced. The metal ions remain in solution.
If the metal is less reactive than hydrogen (e.g., copper, silver, gold, platinum):
- The metal is produced as a solid coating on the electrode.

Tip

Reactivity Series Hint

Remember that hydrogen sits low in the reactivity series. Most metals you encounter in school (like iron, zinc, and sodium) are more reactive than hydrogen, so you will often get hydrogen gas at the cathode! Copper is the most common exception you will test.

Rules at the Anode (Positive Electrode)

Both the non-metal anion ( $Xn−\text{X}^{n-}$ ) and the hydroxide ion ( $OH−\text{OH}^-$ ) are attracted to the anode. Which one is discharged depends on whether a halide (Group 7 ion) is present:

If halide ions are present (chlorine $Cl−\text{Cl}^-$ $Cl^{-}$ , bromine $Br−\text{Br}^-$ $Br^{-}$ , or iodine $I−\text{I}^-$ $I^{-}$ ) in reasonable concentration:
- The corresponding halogen ( $Cl2\text{Cl}_2$ , $Br2\text{Br}_2$ , or $I2\text{I}_2$ ) is produced.
If no halide ions are present (e.g., sulfate $SO42−\text{SO}_4^{2-}$ $SO_{4}^{2 -}$ , nitrate $NO3−\text{NO}_3^-$ $NO_{3}^{-}$ , carbonate $CO32−\text{CO}_3^{2-}$ $CO_{3}^{2 -}$ ):
- Oxygen gas ( $O2(g)\text{O}_2\text{(g)}$ ) is produced from the discharge of hydroxide ions.

Example

Predicting the products of aqueous sodium chloride electrolysis

Predict the products formed at each electrode during the electrolysis of concentrated aqueous sodium chloride ( $NaCl(aq)\text{NaCl(aq)}$ ), and identify the species left behind in solution.

List all species and ions present in the solution. The ions present are sodium ions ( $Na+\text{Na}^+$ ) and chloride ions ( $Cl−\text{Cl}^-$ ) from the salt, plus hydrogen ions ( $H+\text{H}^+$ ) and hydroxide ions ( $OH−\text{OH}^-$ ) from the water.
Apply the cathode rules. The positive ions ( $Na+\text{Na}^+$ and $H+\text{H}^+$ ) migrate to the negative cathode. Since sodium is highly reactive (much more reactive than hydrogen), the $H+\text{H}^+$ ions are discharged instead. Therefore, hydrogen gas ( $H2(g)\text{H}_2\text{(g)}$ ) is produced at the cathode.
Apply the anode rules. The negative ions ( $Cl−\text{Cl}^-$ and $OH−\text{OH}^-$ ) migrate to the positive anode. Since chloride ( $Cl−\text{Cl}^-$ ) is a halide ion, it is discharged. Therefore, chlorine gas ( $Cl2(g)\text{Cl}_2\text{(g)}$ ) is produced at the anode.
Identify the remaining solution. The ions left behind in the beaker are sodium ( $Na+\text{Na}^+$ ) and hydroxide ( $OH−\text{OH}^-$ ). Together, they form sodium hydroxide ( $NaOH(aq)\text{NaOH(aq)}$ ), which is an alkaline solution.

Reactions at the Electrodes and Half Equations (Higher Tier Only)

During electrolysis, chemical reactions occur at the surface of the electrodes. These reactions involve the transfer of electrons and are classified as redox reactions (reduction and oxidation occurring simultaneously).

Definition

OIL RIG

Oxidation is the loss of electrons.
Reduction is the gain of electrons.

At the electrodes:

At the cathode (-), positive ions gain electrons. This is reduction.
At the anode (+), negative ions lose electrons. This is oxidation.

We can represent these processes using half equations. Half equations show the gain or loss of electrons ( $e−\text{e}^-$ ) for a single species.

1. Reduction at the Cathode

Positive ions arrive at the negative cathode and gain electrons to form neutral atoms.

If hydrogen is discharged:

2\text{H}^+\text{(aq)} + 2\text{e}^- \rightarrow \text{H}_2\text{(g)}

If a metal (like copper) is discharged:

\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)}

2. Oxidation at the Anode

Negative ions arrive at the positive anode and lose electrons to form neutral elements.

If a halide (like chlorine) is discharged:

2\text{Cl}^-\text{(aq)} \rightarrow \text{Cl}_2\text{(g)} + 2\text{e}^-

If hydroxide ( $OH−\text{OH}^-$ ) is discharged to produce oxygen gas:

4\text{OH}^-\text{(aq)} \rightarrow 2\text{H}_2\text{O(l)} + \text{O}_2\text{(g)} + 4\text{e}^-

Common Mistake

Hydroxide Half Equation

The half equation for the discharge of hydroxide ions ( $OH−\text{OH}^-$ ) is complex and a very common source of lost marks. Make sure you memorise it exactly, noting that $4e−4\text{e}^-$ are lost to balance the charges!

Example

Writing half equations for the electrolysis of aqueous copper(II) sulfate

Write balanced half equations for the reactions occurring at each electrode during the electrolysis of aqueous copper(II) sulfate, $CuSO4(aq)\text{CuSO}_4\text{(aq)}$ , using inert electrodes.

Identify the ions present and which are discharged. The ions are $Cu2+\text{Cu}^{2+}$ , $SO42−\text{SO}_4^{2-}$ , $H+\text{H}^+$ , and $OH−\text{OH}^-$ . At the cathode, copper ( $Cu2+\text{Cu}^{2+}$ ) is less reactive than hydrogen, so it is discharged. At the anode, no halide is present (only sulfate, $SO42−\text{SO}_4^{2-}$ ), so hydroxide ( $OH−\text{OH}^-$ ) is discharged to produce oxygen gas.
Write the cathode half equation (reduction). Copper ions ( $Cu2+\text{Cu}^{2+}$ ) gain two electrons to form solid copper atoms:

\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)}

Write the anode half equation (oxidation). Hydroxide ions ( $OH−\text{OH}^-$ ) lose electrons to produce oxygen gas and water:

4\text{OH}^-\text{(aq)} \rightarrow 2\text{H}_2\text{O(l)} + \text{O}_2\text{(g)} + 4\text{e}^-

Inert vs. Non-Inert Electrodes

The electrodes used in electrolysis can be classified as either inert or non-inert.

Inert electrodes (e.g., graphite or platinum) are chosen because they conduct electricity but are chemically unreactive. They simply provide a surface for the transfer of electrons without participating in the reaction themselves.
Non-inert electrodes (e.g., copper electrodes) are active participants in the electrolysis process. They can dissolve into the electrolyte or have metal from the electrolyte deposit on them.

Purification of Copper (Non-Inert Electrodes)

This technique is particularly important for separate Chemistry (J248) students.

A brilliant industrial application of non-inert electrodes is the purification of copper. Copper wire used in electrical wiring must be extremely pure to conduct electricity efficiently.

In this process:

The anode is made of impure copper (which we want to purify).
The cathode is made of a thin sheet of pure copper.
The electrolyte is an aqueous solution of copper(II) sulfate ( $CuSO4(aq)\text{CuSO}_4\text{(aq)}$ ).

As current passes through the cell:

At the Anode: Copper atoms in the impure anode lose electrons and dissolve into the solution as copper ions ( $Cu2+\text{Cu}^{2+}$ ):

\text{Cu(s, impure)} \rightarrow \text{Cu}^{2+}\text{(aq)} + 2\text{e}^-

(The anode slowly dissolves and decreases in mass, leaving behind "anode sludge" containing impurities). 2. At the Cathode: Copper ions ( $Cu2+\text{Cu}^{2+}$ ) in the solution are attracted to the pure copper cathode, gain electrons, and deposit as pure copper metal:

\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s, pure)}

(The cathode increases in mass as pure copper builds up on it).

The concentration of copper ions in the electrolyte remains constant because the rate of copper dissolving at the anode equals the rate of copper depositing at the cathode.

Diagram showing the purification of copper using an impure copper anode, a pure copper cathode, and copper(II) sulfate solution as the electrolyte

Exam technique

In the exam

Identify the state first. Before predicting products, always check whether the question specifies molten or aqueous. If it is molten, there are only two ions. If it is aqueous, there are four!
Use PANIC and OIL RIG. Write these mnemonics down on your scrap paper immediately. They will prevent you from accidentally swapping anode/cathode reactions under exam pressure.
State symbols matter. In half equations, always write correct state symbols. For example, oxygen gas is $O2(g)\text{O}_2\text{(g)}$ and copper deposited on a cathode is $Cu(s)\text{Cu(s)}$ .
Remember the diatomic gases. Hydrogen ( $H2\text{H}_2$ ), chlorine ( $Cl2\text{Cl}_2$ ), bromine ( $Br2\text{Br}_2$ ), and oxygen ( $O2\text{O}_2$ ) always form diatomic molecules. Don't forget to balance your half equations to reflect this.

Self review

Check yourself

Explain why solid sodium chloride does not conduct electricity, whereas sodium chloride solution does.
Predict the products formed at each electrode during the electrolysis of aqueous magnesium sulfate, $MgSO4(aq)\text{MgSO}_4\text{(aq)}$ , using inert electrodes.
[Higher Tier Only] Write a balanced half equation for the reaction occurring at the anode during the electrolysis of aqueous sodium bromide, $NaBr(aq)\text{NaBr(aq)}$ .

Electrolysis

What you'll learn

How ionic compounds conduct electricity when molten or dissolved, but not when solid.
How to predict the products formed at the anode and cathode during the electrolysis of molten binary compounds.
[Higher Tier Only] How to predict the products of aqueous solutions where multiple ions compete at the electrodes.
[Higher Tier Only] How to write balanced half equations representing the loss or gain of electrons at each electrode.

What is Electrolysis?

The electrolyte: The liquid or solution containing ions that conducts electricity.
The electrodes: Solid rods (usually made of unreactive materials like graphite or platinum) that dip into the electrolyte to complete the electrical circuit.
The power supply: A DC source that drives the movement of electrons in the wires and ions in the electrolyte.

Definition

Electrolysis

Electrolysis is the chemical decomposition of an electrolyte caused by the passage of a direct electric current through it.

Definition

Electrolyte

An electrolyte is an ionic compound in a molten state (liquid) or dissolved in water (aqueous solution) that conducts electricity and is decomposed by it.

The two electrodes have opposite charges:

The anode is the positive electrode.
The cathode is the negative electrode.

Tip

Remembering electrode names

To remember which electrode is which, use the phrase PANIC:

Positive Anode
Negative Is Cathode

How Ions Move During Electrolysis

When the power supply is turned on, the electrodes become charged. Because opposite charges attract:

Cations (positive ions, like metal ions $Mn+\text{M}^{n+}$ or $H+\text{H}^+$ ) are attracted to the negative cathode.
Anions (negative ions, like non-metal ions $Xn−\text{X}^{n-}$ or $OH−\text{OH}^-$ ) are attracted to the positive anode.

As these ions reach the electrodes, chemical reactions occur, and the compound is decomposed.

Diagram showing a standard electrolysis cell with a positive anode, negative cathode, and the movement of cations and anions in an electrolyte

Common Mistake

Confusing ion flow with electron flow

A very common exam mistake is stating that ionic solutions conduct electricity because "electrons flow through the solution". This is incorrect!

In the external wires, electricity is conducted by moving electrons.
In the electrolyte, electricity is conducted by moving ions.

Electrons cannot travel freely through water or liquid ionic compounds.

Prerequisites: Why Must the Electrolyte Be Liquid or Aqueous?

You might wonder why we cannot perform electrolysis on solid ionic compounds like solid table salt ( $NaCl(s)\text{NaCl(s)}$ ).

To undergo electrolysis, we must break this lattice apart. We can do this in two ways:

Melting it: Heating the solid until it turns into a liquid ( $molten\text{molten}$ ). The thermal energy overcomes the strong electrostatic forces, allowing the ions to move freely.
Dissolving it: Mixing the solid with water ( $solution\text{aqueous solution}$ ). Water molecules surround the ions and pull them out of the lattice, allowing them to move freely.

Key Idea

The Golden Rule of Conductivity

Ionic compounds only conduct electricity when molten or in aqueous solution because only then are the ions free to move to carry the charge.

Electrolysis of Molten Binary Ionic Compounds

A binary ionic compound contains only two elements: a metal cation and a non-metal anion (for example, molten lead(II) bromide, $PbBr2(l)\text{PbBr}_2\text{(l)}$ ).

Because there are only two types of ions present in a molten binary compound, predicting the products is straightforward:

The metal ions (cations) move to the cathode and form metal atoms/liquids.
The non-metal ions (anions) move to the anode and form non-metal molecules/atoms.

At the Cathode (-): Positive lead ions ( $Pb2+\text{Pb}^{2+}$ ) are attracted here. They gain electrons to become grey lead metal:

\text{Pb}^{2+}\text{(l)} + 2\text{e}^- \rightarrow \text{Pb(l)}

At the Anode (+): Negative bromide ions ( $Br−\text{Br}^-$ ) are attracted here. They lose electrons to form brown bromine gas:

2\text{Br}^-\text{(l)} \rightarrow \text{Br}_2\text{(g)} + 2\text{e}^-

Example

Predicting the products of molten sodium chloride electrolysis

Predict the products formed at the cathode and anode during the electrolysis of molten sodium chloride, $NaCl(l)\text{NaCl(l)}$ , using inert electrodes.

Identify all the ions present in the molten compound. Since this is a molten binary compound, the only ions present are sodium ions ( $Na+\text{Na}^+$ ) and chloride ions ( $Cl−\text{Cl}^-$ ). There is no water present.
Determine which ion goes to the cathode and predict the product. The positive sodium ions ( $Na+\text{Na}^+$ ) are attracted to the negative cathode. They gain electrons to form neutral sodium metal ( $Na(l)\text{Na(l)}$ ).
Determine which ion goes to the anode and predict the product. The negative chloride ions ( $Cl−\text{Cl}^-$ ) are attracted to the positive anode. They lose electrons to form neutral, diatomic chlorine gas ( $Cl2(g)\text{Cl}_2\text{(g)}$ ).

Electrolysis of Aqueous Solutions: Competing Reactions

When we dissolve an ionic compound in water to make an aqueous solution, things get a little more complicated.

Water molecules ( $H2O\text{H}_2\text{O}$ ) dissociate (split up) slightly into hydrogen ions ( $H+\text{H}^+$ ) and hydroxide ions ( $OH−\text{OH}^-$ ):

\text{H}_2\text{O(l)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{OH}^-\text{(aq)}

This means an aqueous solution always contains four different ions:

The metal cation from the solute.
The non-metal anion from the solute.
Hydrogen ions ( $H+\text{H}^+$ ) from water.
Hydroxide ions ( $OH−\text{OH}^-$ ) from water.

Because only one substance can be released (discharged) at each electrode, these ions must compete! We use a set of rules to determine which ion "wins" and gets discharged.

Rules at the Cathode (Negative Electrode)

If the metal is more reactive than hydrogen (e.g., sodium, potassium, calcium, magnesium, aluminium):
- Hydrogen gas ( $H2(g)\text{H}_2\text{(g)}$ ) is produced. The metal ions remain in solution.
If the metal is less reactive than hydrogen (e.g., copper, silver, gold, platinum):
- The metal is produced as a solid coating on the electrode.

Tip

Reactivity Series Hint

Rules at the Anode (Positive Electrode)

If halide ions are present (chlorine $Cl−\text{Cl}^-$ $Cl^{-}$ , bromine $Br−\text{Br}^-$ $Br^{-}$ , or iodine $I−\text{I}^-$ $I^{-}$ ) in reasonable concentration:
- The corresponding halogen ( $Cl2\text{Cl}_2$ , $Br2\text{Br}_2$ , or $I2\text{I}_2$ ) is produced.
If no halide ions are present (e.g., sulfate $SO42−\text{SO}_4^{2-}$ $SO_{4}^{2 -}$ , nitrate $NO3−\text{NO}_3^-$ $NO_{3}^{-}$ , carbonate $CO32−\text{CO}_3^{2-}$ $CO_{3}^{2 -}$ ):
- Oxygen gas ( $O2(g)\text{O}_2\text{(g)}$ ) is produced from the discharge of hydroxide ions.

Example

Predicting the products of aqueous sodium chloride electrolysis

Predict the products formed at each electrode during the electrolysis of concentrated aqueous sodium chloride ( $NaCl(aq)\text{NaCl(aq)}$ ), and identify the species left behind in solution.

List all species and ions present in the solution. The ions present are sodium ions ( $Na+\text{Na}^+$ ) and chloride ions ( $Cl−\text{Cl}^-$ ) from the salt, plus hydrogen ions ( $H+\text{H}^+$ ) and hydroxide ions ( $OH−\text{OH}^-$ ) from the water.
Apply the cathode rules. The positive ions ( $Na+\text{Na}^+$ and $H+\text{H}^+$ ) migrate to the negative cathode. Since sodium is highly reactive (much more reactive than hydrogen), the $H+\text{H}^+$ ions are discharged instead. Therefore, hydrogen gas ( $H2(g)\text{H}_2\text{(g)}$ ) is produced at the cathode.
Apply the anode rules. The negative ions ( $Cl−\text{Cl}^-$ and $OH−\text{OH}^-$ ) migrate to the positive anode. Since chloride ( $Cl−\text{Cl}^-$ ) is a halide ion, it is discharged. Therefore, chlorine gas ( $Cl2(g)\text{Cl}_2\text{(g)}$ ) is produced at the anode.
Identify the remaining solution. The ions left behind in the beaker are sodium ( $Na+\text{Na}^+$ ) and hydroxide ( $OH−\text{OH}^-$ ). Together, they form sodium hydroxide ( $NaOH(aq)\text{NaOH(aq)}$ ), which is an alkaline solution.

Reactions at the Electrodes and Half Equations (Higher Tier Only)

Definition

OIL RIG

Oxidation is the loss of electrons.
Reduction is the gain of electrons.

At the electrodes:

At the cathode (-), positive ions gain electrons. This is reduction.
At the anode (+), negative ions lose electrons. This is oxidation.

We can represent these processes using half equations. Half equations show the gain or loss of electrons ( $e−\text{e}^-$ ) for a single species.

1. Reduction at the Cathode

Positive ions arrive at the negative cathode and gain electrons to form neutral atoms.

If hydrogen is discharged:

2\text{H}^+\text{(aq)} + 2\text{e}^- \rightarrow \text{H}_2\text{(g)}

If a metal (like copper) is discharged:

\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)}

2. Oxidation at the Anode

Negative ions arrive at the positive anode and lose electrons to form neutral elements.

If a halide (like chlorine) is discharged:

2\text{Cl}^-\text{(aq)} \rightarrow \text{Cl}_2\text{(g)} + 2\text{e}^-

If hydroxide ( $OH−\text{OH}^-$ ) is discharged to produce oxygen gas:

4\text{OH}^-\text{(aq)} \rightarrow 2\text{H}_2\text{O(l)} + \text{O}_2\text{(g)} + 4\text{e}^-

Common Mistake

Hydroxide Half Equation

Example

Writing half equations for the electrolysis of aqueous copper(II) sulfate

Write balanced half equations for the reactions occurring at each electrode during the electrolysis of aqueous copper(II) sulfate, $CuSO4(aq)\text{CuSO}_4\text{(aq)}$ , using inert electrodes.

Identify the ions present and which are discharged. The ions are $Cu2+\text{Cu}^{2+}$ , $SO42−\text{SO}_4^{2-}$ , $H+\text{H}^+$ , and $OH−\text{OH}^-$ . At the cathode, copper ( $Cu2+\text{Cu}^{2+}$ ) is less reactive than hydrogen, so it is discharged. At the anode, no halide is present (only sulfate, $SO42−\text{SO}_4^{2-}$ ), so hydroxide ( $OH−\text{OH}^-$ ) is discharged to produce oxygen gas.
Write the cathode half equation (reduction). Copper ions ( $Cu2+\text{Cu}^{2+}$ ) gain two electrons to form solid copper atoms:

\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)}

Write the anode half equation (oxidation). Hydroxide ions ( $OH−\text{OH}^-$ ) lose electrons to produce oxygen gas and water:

4\text{OH}^-\text{(aq)} \rightarrow 2\text{H}_2\text{O(l)} + \text{O}_2\text{(g)} + 4\text{e}^-

Inert vs. Non-Inert Electrodes

The electrodes used in electrolysis can be classified as either inert or non-inert.

Inert electrodes (e.g., graphite or platinum) are chosen because they conduct electricity but are chemically unreactive. They simply provide a surface for the transfer of electrons without participating in the reaction themselves.
Non-inert electrodes (e.g., copper electrodes) are active participants in the electrolysis process. They can dissolve into the electrolyte or have metal from the electrolyte deposit on them.

Purification of Copper (Non-Inert Electrodes)

This technique is particularly important for separate Chemistry (J248) students.

A brilliant industrial application of non-inert electrodes is the purification of copper. Copper wire used in electrical wiring must be extremely pure to conduct electricity efficiently.

In this process:

The anode is made of impure copper (which we want to purify).
The cathode is made of a thin sheet of pure copper.
The electrolyte is an aqueous solution of copper(II) sulfate ( $CuSO4(aq)\text{CuSO}_4\text{(aq)}$ ).

As current passes through the cell:

At the Anode: Copper atoms in the impure anode lose electrons and dissolve into the solution as copper ions ( $Cu2+\text{Cu}^{2+}$ ):

\text{Cu(s, impure)} \rightarrow \text{Cu}^{2+}\text{(aq)} + 2\text{e}^-

\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s, pure)}

(The cathode increases in mass as pure copper builds up on it).

The concentration of copper ions in the electrolyte remains constant because the rate of copper dissolving at the anode equals the rate of copper depositing at the cathode.

Diagram showing the purification of copper using an impure copper anode, a pure copper cathode, and copper(II) sulfate solution as the electrolyte

Exam technique

In the exam

Identify the state first. Before predicting products, always check whether the question specifies molten or aqueous. If it is molten, there are only two ions. If it is aqueous, there are four!
Use PANIC and OIL RIG. Write these mnemonics down on your scrap paper immediately. They will prevent you from accidentally swapping anode/cathode reactions under exam pressure.
State symbols matter. In half equations, always write correct state symbols. For example, oxygen gas is $O2(g)\text{O}_2\text{(g)}$ and copper deposited on a cathode is $Cu(s)\text{Cu(s)}$ .
Remember the diatomic gases. Hydrogen ( $H2\text{H}_2$ ), chlorine ( $Cl2\text{Cl}_2$ ), bromine ( $Br2\text{Br}_2$ ), and oxygen ( $O2\text{O}_2$ ) always form diatomic molecules. Don't forget to balance your half equations to reflect this.

Self review

Check yourself

Explain why solid sodium chloride does not conduct electricity, whereas sodium chloride solution does.
Predict the products formed at each electrode during the electrolysis of aqueous magnesium sulfate, $MgSO4(aq)\text{MgSO}_4\text{(aq)}$ , using inert electrodes.
[Higher Tier Only] Write a balanced half equation for the reaction occurring at the anode during the electrolysis of aqueous sodium bromide, $NaBr(aq)\text{NaBr(aq)}$ .

Electrolysis

What you'll learn

What is Electrolysis?

How Ions Move During Electrolysis

Prerequisites: Why Must the Electrolyte Be Liquid or Aqueous?

Electrolysis of Molten Binary Ionic Compounds

Electrolysis of Aqueous Solutions: Competing Reactions

Rules at the Cathode (Negative Electrode)

Rules at the Anode (Positive Electrode)

Reactions at the Electrodes and Half Equations (Higher Tier Only)

1. Reduction at the Cathode

2. Oxidation at the Anode

Inert vs. Non-Inert Electrodes

Purification of Copper (Non-Inert Electrodes)

Recap questions

You've reached the end

Electrolysis

What you'll learn

What is Electrolysis?

How Ions Move During Electrolysis

Prerequisites: Why Must the Electrolyte Be Liquid or Aqueous?

Electrolysis of Molten Binary Ionic Compounds

Electrolysis of Aqueous Solutions: Competing Reactions

Rules at the Cathode (Negative Electrode)

Rules at the Anode (Positive Electrode)

Reactions at the Electrodes and Half Equations (Higher Tier Only)

1. Reduction at the Cathode

2. Oxidation at the Anode

Inert vs. Non-Inert Electrodes

Purification of Copper (Non-Inert Electrodes)

Recap questions

You've reached the end

Electrolysis

What you'll learn

What is Electrolysis?

How Ions Move During Electrolysis

Prerequisites: Why Must the Electrolyte Be Liquid or Aqueous?

Electrolysis of Molten Binary Ionic Compounds

Electrolysis of Aqueous Solutions: Competing Reactions

Rules at the Cathode (Negative Electrode)

Rules at the Anode (Positive Electrode)

Reactions at the Electrodes and Half Equations (Higher Tier Only)

1. Reduction at the Cathode

2. Oxidation at the Anode

Inert vs. Non-Inert Electrodes

Purification of Copper (Non-Inert Electrodes)

Recap questions

You've reached the end

Electrolysis Revision Guide

Electrolysis

What you'll learn

What is Electrolysis?

How Ions Move During Electrolysis

Prerequisites: Why Must the Electrolyte Be Liquid or Aqueous?

Electrolysis of Molten Binary Ionic Compounds

Electrolysis of Aqueous Solutions: Competing Reactions

Rules at the Cathode (Negative Electrode)

Rules at the Anode (Positive Electrode)

Reactions at the Electrodes and Half Equations (Higher Tier Only)

1. Reduction at the Cathode

2. Oxidation at the Anode

Inert vs. Non-Inert Electrodes

Purification of Copper (Non-Inert Electrodes)

Recap questions

You've reached the end

Electrolysis Revision Guide