contents

Chemistry

4.11 - Electrolysis - Aqueous Solutions

11 min

Recap & ions in solution

The discharge rules

Copper sulfate solution

Sodium chloride solution

note 1 of 6

What forms in an aqueous electrolysis

A dissolved ionic compound has ions that are free to move, so its solution can be electrolysed without melting it.
Water breaks up slightly into hydrogen and hydroxide ions:
- H₂O(l) ⇌ H⁺(aq) + OH^-(aq)
This means four kinds of ion are present: the positive and negative ions of the dissolved compound, plus H⁺ and OH^- from the water.
At each electrode, only one of the two competing ions is discharged to form an element, while the other stays in solution.
Inert electrodes (such as platinum or graphite) are used so they don't react.

Three example solutions

Dilute sulfuric acid contains H⁺, SO₄^2- and OH^-. Hydrogen forms at the cathode and oxygen at the anode. OH^- is discharged in preference to SO₄^2-, so water is decomposed.
Copper(II) chloride contains Cu²⁺, Cl^-, H⁺ and OH^-. Copper is deposited at the cathode and chlorine forms at the anode.
Concentrated sodium chloride contains Na⁺, Cl^-, H⁺ and OH^-. Hydrogen forms at the cathode and chlorine at the anode. Na⁺ and OH^- stay in solution, leaving sodium hydroxide.

Solution	Ions present	At the cathode	At the anode	Left in the solution
Dilute sulfuric acid	H⁺, SO₄^2-, OH^-	Hydrogen	Oxygen	Sulfate ions
Copper(II) chloride	Cu²⁺, Cl^-, H⁺, OH^-	Copper	Chlorine	Water, as the compound is used up
Concentrated sodium chloride	Na⁺, Cl^-, H⁺, OH^-	Hydrogen	Chlorine	Sodium hydroxide

A beaker labelled copper(II) chloride solution holding OH-, H+, Cl- and Cu2+ ions, with Cl- going to the positive electrode (anode) making chlorine (Cl2) and Cu2+ to the negative electrode (cathode) making copper (Cu).

Industrial use of sodium chloride electrolysis

Electrolysing concentrated sodium chloride produces three useful products:

Sodium hydroxide - Used to make soap.
Chlorine - Used for bleach and plastics.
Hydrogen - Used as a fuel.

note 2 of 6

The discharge rules at each electrode

At the cathode

The positive ions are the metal ion and the hydrogen ion (H⁺).
The ion of the less reactive element is discharged:
- If the metal is more reactive than hydrogen, hydrogen gas is produced.
- If the metal is less reactive than hydrogen, the metal is deposited.

At the anode

The negative ions are the non-metal ion and the hydroxide ion (OH^-):

If halide ions (Cl^-, Br^-, I^-) are present in a concentrated solution, the halogen (chlorine, bromine, or iodine) is formed.
If no halide is present (or if the solution is very dilute), oxygen is formed from the hydroxide ions. Sulfate and nitrate ions are not discharged.

Two charts: negative electrode (cathode) asks is the metal more reactive than hydrogen, yes hydrogen formed, no metal formed; positive electrode (anode) asks are halide ions present, yes halogen formed, no oxygen formed.

note 3 of 6

Worked example - Products of potassium bromide

Predict the products of electrolysing concentrated potassium bromide solution, and what is left in the beaker.

Step 1 - List every ion present

From potassium bromide: K⁺ and Br^-
From the water: H⁺ and OH^-

Step 2 - Work out the cathode product

The positive ions are K⁺ and H⁺.
Potassium is more reactive than hydrogen, so hydrogen ions are discharged.
Hydrogen gas is formed at the cathode.

Step 3 - Work out the anode product

The negative ions are Br^- and OH^-.
Bromide is a halide and the solution is concentrated, so bromide ions are discharged.
Bromine is formed at the anode, turning the solution around it orange-brown.

Step 4 - Check what is left behind

K⁺ and OH^- are not discharged and remain in solution, forming potassium hydroxide (making the solution alkaline).

note 4 of 6

Half equations at aqueous electrodes

A half equation shows the reaction at a single electrode, including the electrons transferred.
Cathode - Positive ions gain electrons (reduction).
Anode - Negative ions lose electrons (oxidation).

Half equations with inert electrodes

At the cathode:
- 2H⁺ + 2e^- → H₂
- Cu²⁺ + 2e^- → Cu
At the anode:
- 4OH^- → O₂ + 2H₂O + 4e^-
- 2Cl^- → Cl₂ + 2e^-

Non-inert copper anode

When the anode is made of copper, copper atoms lose electrons and dissolve into solution as ions:

Cu → Cu²⁺ + 2e^-

note 5 of 6

Worked example - Half equations for dilute sulfuric acid

Dilute sulfuric acid is electrolysed. Write the half equation at each electrode and state which is oxidation and which is reduction.

Step 1 - Name the ion discharged at each electrode

The solution contains H⁺, SO₄^2- and OH^-.
H⁺ is discharged at the cathode and OH^- at the anode.

Step 2 - Build the cathode equation

Hydrogen gas is H₂, requiring two H⁺ ions and two electrons:

2H⁺ + 2e^- → H₂

Step 3 - Build the anode equation

Four OH^- ions form one O₂ molecule and two H₂O molecules, releasing four electrons:

4OH^- → O₂ + 2H₂O + 4e^-

Step 4 - Identify oxidation and reduction

The hydrogen ions gain electrons, so the cathode reaction is reduction.
The hydroxide ions lose electrons, so the anode reaction is oxidation.

note 6 of 6

Worked example - Electrolysing copper(II) sulfate

Copper(II) sulfate solution is electrolysed using inert carbon electrodes. What forms at each electrode, and what happens to the solution?

Step 1 - List every ion present

From copper(II) sulfate: Cu²⁺ and SO₄^2-
From water: H⁺ and OH^-

Step 2 - Apply the cathode rule

Copper is less reactive than hydrogen, so copper ions are discharged.
A pinkish-brown layer of copper builds up on the cathode.

Step 3 - Apply the anode rule

There are no halide ions, so hydroxide ions are discharged.
Oxygen gas bubbles off at the anode.

Step 4 - Identify what happens to the solution

As Cu²⁺ ions are removed from the solution and deposited on the cathode, the blue colour of the solution gradually fades.

key terms

TermDefinition

Inert electrodeAn unreactive electrode, such as platinum or graphite, that does not take part in the electrolysis reaction.

Half equationAn equation that shows the electrons lost or gained by one substance in a reaction, such as the reaction at one electrode. It is also called an ion-electron equation.

end of notes

Key points to remember

contents

Chemistry

4.11 - Electrolysis - Aqueous Solutions

11 min

Recap & ions in solution

The discharge rules

Copper sulfate solution

Sodium chloride solution

note 1 of 6

What forms in an aqueous electrolysis

A dissolved ionic compound has ions that are free to move, so its solution can be electrolysed without melting it.
Water breaks up slightly into hydrogen and hydroxide ions:
- H₂O(l) ⇌ H⁺(aq) + OH^-(aq)
This means four kinds of ion are present: the positive and negative ions of the dissolved compound, plus H⁺ and OH^- from the water.
At each electrode, only one of the two competing ions is discharged to form an element, while the other stays in solution.
Inert electrodes (such as platinum or graphite) are used so they don't react.

Three example solutions

Dilute sulfuric acid contains H⁺, SO₄^2- and OH^-. Hydrogen forms at the cathode and oxygen at the anode. OH^- is discharged in preference to SO₄^2-, so water is decomposed.
Copper(II) chloride contains Cu²⁺, Cl^-, H⁺ and OH^-. Copper is deposited at the cathode and chlorine forms at the anode.
Concentrated sodium chloride contains Na⁺, Cl^-, H⁺ and OH^-. Hydrogen forms at the cathode and chlorine at the anode. Na⁺ and OH^- stay in solution, leaving sodium hydroxide.

Solution	Ions present	At the cathode	At the anode	Left in the solution
Dilute sulfuric acid	H⁺, SO₄^2-, OH^-	Hydrogen	Oxygen	Sulfate ions
Copper(II) chloride	Cu²⁺, Cl^-, H⁺, OH^-	Copper	Chlorine	Water, as the compound is used up
Concentrated sodium chloride	Na⁺, Cl^-, H⁺, OH^-	Hydrogen	Chlorine	Sodium hydroxide

Industrial use of sodium chloride electrolysis

Electrolysing concentrated sodium chloride produces three useful products:

Sodium hydroxide - Used to make soap.
Chlorine - Used for bleach and plastics.
Hydrogen - Used as a fuel.

note 2 of 6

The discharge rules at each electrode

At the cathode

The positive ions are the metal ion and the hydrogen ion (H⁺).
The ion of the less reactive element is discharged:
- If the metal is more reactive than hydrogen, hydrogen gas is produced.
- If the metal is less reactive than hydrogen, the metal is deposited.

At the anode

The negative ions are the non-metal ion and the hydroxide ion (OH^-):

If halide ions (Cl^-, Br^-, I^-) are present in a concentrated solution, the halogen (chlorine, bromine, or iodine) is formed.
If no halide is present (or if the solution is very dilute), oxygen is formed from the hydroxide ions. Sulfate and nitrate ions are not discharged.

note 3 of 6

Worked example - Products of potassium bromide

Predict the products of electrolysing concentrated potassium bromide solution, and what is left in the beaker.

Step 1 - List every ion present

From potassium bromide: K⁺ and Br^-
From the water: H⁺ and OH^-

Step 2 - Work out the cathode product

The positive ions are K⁺ and H⁺.
Potassium is more reactive than hydrogen, so hydrogen ions are discharged.
Hydrogen gas is formed at the cathode.

Step 3 - Work out the anode product

The negative ions are Br^- and OH^-.
Bromide is a halide and the solution is concentrated, so bromide ions are discharged.
Bromine is formed at the anode, turning the solution around it orange-brown.

Step 4 - Check what is left behind

K⁺ and OH^- are not discharged and remain in solution, forming potassium hydroxide (making the solution alkaline).

note 4 of 6

Half equations at aqueous electrodes

A half equation shows the reaction at a single electrode, including the electrons transferred.
Cathode - Positive ions gain electrons (reduction).
Anode - Negative ions lose electrons (oxidation).

Half equations with inert electrodes

At the cathode:
- 2H⁺ + 2e^- → H₂
- Cu²⁺ + 2e^- → Cu
At the anode:
- 4OH^- → O₂ + 2H₂O + 4e^-
- 2Cl^- → Cl₂ + 2e^-

Non-inert copper anode

When the anode is made of copper, copper atoms lose electrons and dissolve into solution as ions:

Cu → Cu²⁺ + 2e^-

note 5 of 6

Worked example - Half equations for dilute sulfuric acid

Dilute sulfuric acid is electrolysed. Write the half equation at each electrode and state which is oxidation and which is reduction.

Step 1 - Name the ion discharged at each electrode

The solution contains H⁺, SO₄^2- and OH^-.
H⁺ is discharged at the cathode and OH^- at the anode.

Step 2 - Build the cathode equation

Hydrogen gas is H₂, requiring two H⁺ ions and two electrons:

2H⁺ + 2e^- → H₂

Step 3 - Build the anode equation

Four OH^- ions form one O₂ molecule and two H₂O molecules, releasing four electrons:

4OH^- → O₂ + 2H₂O + 4e^-

Step 4 - Identify oxidation and reduction

The hydrogen ions gain electrons, so the cathode reaction is reduction.
The hydroxide ions lose electrons, so the anode reaction is oxidation.

note 6 of 6

Worked example - Electrolysing copper(II) sulfate

Copper(II) sulfate solution is electrolysed using inert carbon electrodes. What forms at each electrode, and what happens to the solution?

Step 1 - List every ion present

From copper(II) sulfate: Cu²⁺ and SO₄^2-
From water: H⁺ and OH^-

Step 2 - Apply the cathode rule

Copper is less reactive than hydrogen, so copper ions are discharged.
A pinkish-brown layer of copper builds up on the cathode.

Step 3 - Apply the anode rule

There are no halide ions, so hydroxide ions are discharged.
Oxygen gas bubbles off at the anode.

Step 4 - Identify what happens to the solution

As Cu²⁺ ions are removed from the solution and deposited on the cathode, the blue colour of the solution gradually fades.

key terms

TermDefinition

Inert electrodeAn unreactive electrode, such as platinum or graphite, that does not take part in the electrolysis reaction.

Half equationAn equation that shows the electrons lost or gained by one substance in a reaction, such as the reaction at one electrode. It is also called an ion-electron equation.

end of notes