Representation of reactions at electrodes as half equations (HT only)

What you'll learn

Why reactions at the cathode are reductions and reactions at the anode are oxidations.
How to read half equations by looking at where the electrons are.
How to write balanced half equations for metals, hydrogen, halogens and oxygen.
How to avoid the common GCSE traps with charges, diatomic gases and electron placement.

Before half equations: what electrolysis is doing

Electrolysis is the use of electricity to break down an ionic substance. The ionic substance must be molten or dissolved in water so that its ions can move.

An ion is an atom or group of atoms with an electrical charge. Positive ions are called cations. Negative ions are called anions.

Definition

Key electrolysis words

An electrolyte is the liquid or solution containing mobile ions.
An electrode is a conductor that allows charge to enter or leave the electrolyte.
The cathode is the negative electrode in electrolysis.
The anode is the positive electrode in electrolysis.

In electrolysis, cations move to the cathode and anions move to the anode. Ions move through the electrolyte; electrons move through the external circuit and electrodes.

Electrolysis cell showing cations moving to the cathode and anions moving to the anode

Reduction and oxidation at electrodes

This bit is Higher Tier: you need to describe the electrode reactions using half equations.

Definition

Reduction and oxidation

Reduction means gain of electrons. Oxidation means loss of electrons. A reaction involving both is called a redox reaction.

At the cathode, positive ions gain electrons. So cathode reactions are reductions.

At the anode, negative ions lose electrons. So anode reactions are oxidations.

Key Idea

OIL RIG at electrodes

OIL RIG means Oxidation Is Loss of electrons and Reduction Is Gain of electrons. In electrolysis: cathode = gain = reduction, and anode = loss = oxidation.

Example

Classifying an electrode reaction

Consider this half equation:

2Cl−(aq)→Cl2(g)+2e−2\text{Cl}^{-}\text{(aq)} \to \text{Cl}_{2}\text{(g)} + 2\text{e}^{-}

The electrons are on the right-hand side, so electrons are being produced.
Producing electrons means the chloride ions have lost electrons.
Loss of electrons is oxidation, so this reaction happens at the anode.

Common Mistake

Mixing up where electrons go

Do not decide reduction or oxidation from the charge of the electrode alone. Look at the electrons: electrons on the left means gain, and electrons on the right means loss.

What a half equation shows

A half equation shows what happens to one species at one electrode. It includes the electrons gained or lost.

For example, hydrogen ions can gain electrons at the cathode to form hydrogen gas:

2H+(aq)+2e−→H2(g)2\text{H}^{+}\text{(aq)} + 2\text{e}^{-} \to \text{H}_{2}\text{(g)}

This equation is “half” of the overall redox process because it only shows the reduction at one electrode.

Example

Reading the hydrogen half equation

2H+(aq)+2e−→H2(g)2\text{H}^{+}\text{(aq)} + 2\text{e}^{-} \to \text{H}_{2}\text{(g)}

The electrons are on the left, so the hydrogen ions gain electrons.
Gain of electrons is reduction, so this is a cathode reaction.
The charge is balanced: the left side has $2 (+ 1) + 2 (- 1) = 0$ , and hydrogen gas is neutral.

Tip

Electron side shortcut

For GCSE electrolysis half equations: cathode equations usually have electrons on the left, and anode equations usually have electrons on the right.

How to write cathode half equations

At the cathode, positive ions become neutral atoms or molecules by gaining electrons.

For a metal ion, the charge tells you how many electrons are needed. A 2+ ion needs two electrons. A 3+ ion needs three electrons.

Metal ions at the cathode

Example

Writing the copper cathode half equation

Write the half equation for copper ions forming copper metal at the cathode.

Start with the reacting ion and product: copper ions become copper atoms.
$Cu2+(aq)→Cu(s)\text{Cu}^{2+}\text{(aq)} \to \text{Cu}\text{(s)}$
Balance the atoms. There is one copper atom on each side, so the atoms are already balanced.
Balance the charge. The left side is 2+ and the right side is neutral, so add two electrons to the left.
$Cu2+(aq)+2e−→Cu(s)\text{Cu}^{2+}\text{(aq)} + 2\text{e}^{-} \to \text{Cu}\text{(s)}$
Check the meaning: electrons are on the left, so copper ions are reduced at the cathode.

Hydrogen at the cathode

Hydrogen is a diatomic molecule, meaning each molecule contains two atoms. So hydrogen gas is written as H₂, not just H.

Example

Writing the hydrogen cathode half equation

Write the half equation for hydrogen ions forming hydrogen gas.

The product is hydrogen gas, so write H₂ on the product side.
$H+(aq)→H2(g)\text{H}^{+}\text{(aq)} \to \text{H}_{2}\text{(g)}$
Balance hydrogen atoms by putting two hydrogen ions on the left.
$2H+(aq)→H2(g)2\text{H}^{+}\text{(aq)} \to \text{H}_{2}\text{(g)}$
The left side now has a total charge of 2+, so add two electrons to the left to make the total charge zero.
$2H+(aq)+2e−→H2(g)2\text{H}^{+}\text{(aq)} + 2\text{e}^{-} \to \text{H}_{2}\text{(g)}$

How to write anode half equations

At the anode, negative ions lose electrons. The electrons usually appear on the product side.

Halide ions at the anode

Halide ions are ions from Group 7 elements, such as chloride ions, bromide ions and iodide ions. The halogen products are diatomic molecules: Cl₂, Br₂ and I₂.

Example

Writing the chlorine anode half equation

Write the half equation for chloride ions forming chlorine gas.

The product is chlorine gas, which is Cl₂, so two chloride ions are needed.
$2Cl−(aq)→Cl2(g)2\text{Cl}^{-}\text{(aq)} \to \text{Cl}_{2}\text{(g)}$
The left side has a total charge of 2−, while chlorine gas is neutral.
Add two electrons to the right-hand side so both sides have total charge 2−.
$2Cl−(aq)→Cl2(g)+2e−2\text{Cl}^{-}\text{(aq)} \to \text{Cl}_{2}\text{(g)} + 2\text{e}^{-}$
Electrons are on the right, so chloride ions have been oxidised at the anode.

Common Mistake

Forgetting diatomic gases

Hydrogen, oxygen and the halogens are diatomic in these equations: H₂, O₂, Cl₂, Br₂ and I₂. Forgetting the small 2 often makes the half equation impossible to balance correctly.

Hydroxide ions forming oxygen

In aqueous electrolysis, oxygen can be produced at the anode from hydroxide ions, OH⁻.

The key half equation is:

4OH−(aq)→O2(g)+2H2O(l)+4e−4\text{OH}^{-}\text{(aq)} \to \text{O}_{2}\text{(g)} + 2\text{H}_{2}\text{O}\text{(l)} + 4\text{e}^{-}

Example

Writing oxygen from hydroxide ions

Build the half equation for hydroxide ions forming oxygen gas at the anode.

Oxygen gas is O₂, and water may also be formed, so start with hydroxide ions producing oxygen and water.
Use four hydroxide ions. This gives four oxygen atoms and four hydrogen atoms on the left.
On the right, O₂ uses two oxygen atoms, and 2H₂O uses the other two oxygen atoms and all four hydrogen atoms.
The left side has a total charge of 4−. The substances on the right are neutral, so add four electrons to the right.
$4OH−(aq)→O2(g)+2H2O(l)+4e−4\text{OH}^{-}\text{(aq)} \to \text{O}_{2}\text{(g)} + 2\text{H}_{2}\text{O}\text{(l)} + 4\text{e}^{-}$

You may also see the same hydroxide equation written like this:

4OH−(aq)−4e−→O2(g)+2H2O(l)4\text{OH}^{-}\text{(aq)} - 4\text{e}^{-} \to \text{O}_{2}\text{(g)} + 2\text{H}_{2}\text{O}\text{(l)}

This means the same thing: four electrons are removed from the hydroxide ions. The version with electrons on the right is usually clearer because it directly shows oxidation.

The final balancing check

A correct half equation must balance in two ways:

Atoms: the same number of each type of atom on both sides.
Charge: the total charge must be the same on both sides.

Electrons do not get state symbols. The reacting ions are usually aqueous, shown as (aq), in solutions. Products might be solid (s), liquid (l) or gas (g).

Example

Correcting the electron number

A student writes:

Al3+(aq)+e−→Al(s)\text{Al}^{3+}\text{(aq)} + \text{e}^{-} \to \text{Al}\text{(s)}

The atoms are balanced because there is one aluminium atom on each side.
Check the charge: the left side is $+ 3 + (- 1) = + 2$ , but aluminium metal is neutral. The charge is not balanced.
Aluminium ions have a 3+ charge, so they need three electrons to become neutral.
$Al3+(aq)+3e−→Al(s)\text{Al}^{3+}\text{(aq)} + 3\text{e}^{-} \to \text{Al}\text{(s)}$

Exam technique

In the exam

Decide which electrode is involved: cathode = reduction = electrons gained, anode = oxidation = electrons lost.
Balance atoms first, then balance total charge using electrons.
Check for diatomic gases: H₂, O₂, Cl₂, Br₂ and I₂.

Self review

Check yourself

Why are reactions at the cathode reductions?
Write the half equation for magnesium ions forming magnesium at the cathode.
In the hydroxide-to-oxygen half equation, why are four electrons produced?

Representation of reactions at electrodes as half equations (HT only)

What you'll learn

Why reactions at the cathode are reductions and reactions at the anode are oxidations.
How to read half equations by looking at where the electrons are.
How to write balanced half equations for metals, hydrogen, halogens and oxygen.
How to avoid the common GCSE traps with charges, diatomic gases and electron placement.

Before half equations: what electrolysis is doing

Electrolysis is the use of electricity to break down an ionic substance. The ionic substance must be molten or dissolved in water so that its ions can move.

An ion is an atom or group of atoms with an electrical charge. Positive ions are called cations. Negative ions are called anions.

Definition

Key electrolysis words

An electrolyte is the liquid or solution containing mobile ions.
An electrode is a conductor that allows charge to enter or leave the electrolyte.
The cathode is the negative electrode in electrolysis.
The anode is the positive electrode in electrolysis.

In electrolysis, cations move to the cathode and anions move to the anode. Ions move through the electrolyte; electrons move through the external circuit and electrodes.

Electrolysis cell showing cations moving to the cathode and anions moving to the anode

Reduction and oxidation at electrodes

This bit is Higher Tier: you need to describe the electrode reactions using half equations.

Definition

Reduction and oxidation

Reduction means gain of electrons. Oxidation means loss of electrons. A reaction involving both is called a redox reaction.

At the cathode, positive ions gain electrons. So cathode reactions are reductions.

At the anode, negative ions lose electrons. So anode reactions are oxidations.

Key Idea

OIL RIG at electrodes

OIL RIG means Oxidation Is Loss of electrons and Reduction Is Gain of electrons. In electrolysis: cathode = gain = reduction, and anode = loss = oxidation.

Example

Classifying an electrode reaction

Consider this half equation:

2Cl−(aq)→Cl2(g)+2e−2\text{Cl}^{-}\text{(aq)} \to \text{Cl}_{2}\text{(g)} + 2\text{e}^{-}

The electrons are on the right-hand side, so electrons are being produced.
Producing electrons means the chloride ions have lost electrons.
Loss of electrons is oxidation, so this reaction happens at the anode.

Common Mistake

Mixing up where electrons go

Do not decide reduction or oxidation from the charge of the electrode alone. Look at the electrons: electrons on the left means gain, and electrons on the right means loss.

What a half equation shows

A half equation shows what happens to one species at one electrode. It includes the electrons gained or lost.

For example, hydrogen ions can gain electrons at the cathode to form hydrogen gas:

2H+(aq)+2e−→H2(g)2\text{H}^{+}\text{(aq)} + 2\text{e}^{-} \to \text{H}_{2}\text{(g)}

This equation is “half” of the overall redox process because it only shows the reduction at one electrode.

Example

Reading the hydrogen half equation

2H+(aq)+2e−→H2(g)2\text{H}^{+}\text{(aq)} + 2\text{e}^{-} \to \text{H}_{2}\text{(g)}

The electrons are on the left, so the hydrogen ions gain electrons.
Gain of electrons is reduction, so this is a cathode reaction.
The charge is balanced: the left side has $2 (+ 1) + 2 (- 1) = 0$ , and hydrogen gas is neutral.

Tip

Electron side shortcut

For GCSE electrolysis half equations: cathode equations usually have electrons on the left, and anode equations usually have electrons on the right.

How to write cathode half equations

At the cathode, positive ions become neutral atoms or molecules by gaining electrons.

For a metal ion, the charge tells you how many electrons are needed. A 2+ ion needs two electrons. A 3+ ion needs three electrons.

Metal ions at the cathode

Example

Writing the copper cathode half equation

Write the half equation for copper ions forming copper metal at the cathode.

Start with the reacting ion and product: copper ions become copper atoms.
$Cu2+(aq)→Cu(s)\text{Cu}^{2+}\text{(aq)} \to \text{Cu}\text{(s)}$
Balance the atoms. There is one copper atom on each side, so the atoms are already balanced.
Balance the charge. The left side is 2+ and the right side is neutral, so add two electrons to the left.
$Cu2+(aq)+2e−→Cu(s)\text{Cu}^{2+}\text{(aq)} + 2\text{e}^{-} \to \text{Cu}\text{(s)}$
Check the meaning: electrons are on the left, so copper ions are reduced at the cathode.

Hydrogen at the cathode

Hydrogen is a diatomic molecule, meaning each molecule contains two atoms. So hydrogen gas is written as H₂, not just H.

Example

Writing the hydrogen cathode half equation

Write the half equation for hydrogen ions forming hydrogen gas.

The product is hydrogen gas, so write H₂ on the product side.
$H+(aq)→H2(g)\text{H}^{+}\text{(aq)} \to \text{H}_{2}\text{(g)}$
Balance hydrogen atoms by putting two hydrogen ions on the left.
$2H+(aq)→H2(g)2\text{H}^{+}\text{(aq)} \to \text{H}_{2}\text{(g)}$
The left side now has a total charge of 2+, so add two electrons to the left to make the total charge zero.
$2H+(aq)+2e−→H2(g)2\text{H}^{+}\text{(aq)} + 2\text{e}^{-} \to \text{H}_{2}\text{(g)}$

How to write anode half equations

At the anode, negative ions lose electrons. The electrons usually appear on the product side.

Halide ions at the anode

Halide ions are ions from Group 7 elements, such as chloride ions, bromide ions and iodide ions. The halogen products are diatomic molecules: Cl₂, Br₂ and I₂.

Example

Writing the chlorine anode half equation

Write the half equation for chloride ions forming chlorine gas.

The product is chlorine gas, which is Cl₂, so two chloride ions are needed.
$2Cl−(aq)→Cl2(g)2\text{Cl}^{-}\text{(aq)} \to \text{Cl}_{2}\text{(g)}$
The left side has a total charge of 2−, while chlorine gas is neutral.
Add two electrons to the right-hand side so both sides have total charge 2−.
$2Cl−(aq)→Cl2(g)+2e−2\text{Cl}^{-}\text{(aq)} \to \text{Cl}_{2}\text{(g)} + 2\text{e}^{-}$
Electrons are on the right, so chloride ions have been oxidised at the anode.

Common Mistake

Forgetting diatomic gases

Hydrogen, oxygen and the halogens are diatomic in these equations: H₂, O₂, Cl₂, Br₂ and I₂. Forgetting the small 2 often makes the half equation impossible to balance correctly.

Hydroxide ions forming oxygen

In aqueous electrolysis, oxygen can be produced at the anode from hydroxide ions, OH⁻.

The key half equation is:

4OH−(aq)→O2(g)+2H2O(l)+4e−4\text{OH}^{-}\text{(aq)} \to \text{O}_{2}\text{(g)} + 2\text{H}_{2}\text{O}\text{(l)} + 4\text{e}^{-}

Example

Writing oxygen from hydroxide ions

Build the half equation for hydroxide ions forming oxygen gas at the anode.

Oxygen gas is O₂, and water may also be formed, so start with hydroxide ions producing oxygen and water.
Use four hydroxide ions. This gives four oxygen atoms and four hydrogen atoms on the left.
On the right, O₂ uses two oxygen atoms, and 2H₂O uses the other two oxygen atoms and all four hydrogen atoms.
The left side has a total charge of 4−. The substances on the right are neutral, so add four electrons to the right.
$4OH−(aq)→O2(g)+2H2O(l)+4e−4\text{OH}^{-}\text{(aq)} \to \text{O}_{2}\text{(g)} + 2\text{H}_{2}\text{O}\text{(l)} + 4\text{e}^{-}$

You may also see the same hydroxide equation written like this:

4OH−(aq)−4e−→O2(g)+2H2O(l)4\text{OH}^{-}\text{(aq)} - 4\text{e}^{-} \to \text{O}_{2}\text{(g)} + 2\text{H}_{2}\text{O}\text{(l)}

This means the same thing: four electrons are removed from the hydroxide ions. The version with electrons on the right is usually clearer because it directly shows oxidation.

The final balancing check

A correct half equation must balance in two ways:

Atoms: the same number of each type of atom on both sides.
Charge: the total charge must be the same on both sides.

Electrons do not get state symbols. The reacting ions are usually aqueous, shown as (aq), in solutions. Products might be solid (s), liquid (l) or gas (g).

Example

Correcting the electron number

A student writes:

Al3+(aq)+e−→Al(s)\text{Al}^{3+}\text{(aq)} + \text{e}^{-} \to \text{Al}\text{(s)}

The atoms are balanced because there is one aluminium atom on each side.
Check the charge: the left side is $+ 3 + (- 1) = + 2$ , but aluminium metal is neutral. The charge is not balanced.
Aluminium ions have a 3+ charge, so they need three electrons to become neutral.
$Al3+(aq)+3e−→Al(s)\text{Al}^{3+}\text{(aq)} + 3\text{e}^{-} \to \text{Al}\text{(s)}$

Exam technique

In the exam

Decide which electrode is involved: cathode = reduction = electrons gained, anode = oxidation = electrons lost.
Balance atoms first, then balance total charge using electrons.
Check for diatomic gases: H₂, O₂, Cl₂, Br₂ and I₂.

Self review

Check yourself

Why are reactions at the cathode reductions?
Write the half equation for magnesium ions forming magnesium at the cathode.
In the hydroxide-to-oxygen half equation, why are four electrons produced?

Representation of reactions at electrodes as half equations (HT only)

What you'll learn

Before half equations: what electrolysis is doing

Reduction and oxidation at electrodes

What a half equation shows

How to write cathode half equations

Metal ions at the cathode

Hydrogen at the cathode

How to write anode half equations

Halide ions at the anode

Hydroxide ions forming oxygen

The final balancing check

Representation of reactions at electrodes as half equations (HT only) Revision Guide

Representation of reactions at electrodes as half equations (HT only)

What you'll learn

Before half equations: what electrolysis is doing

Reduction and oxidation at electrodes

What a half equation shows

How to write cathode half equations

Metal ions at the cathode

Hydrogen at the cathode

How to write anode half equations

Halide ions at the anode

Hydroxide ions forming oxygen

The final balancing check

Representation of reactions at electrodes as half equations (HT only) Revision Guide