This page summarises the key ideas of oxidation, reduction and redox reactions from GCSE chemistry course. It also shows you how to write and combine ion-electron half-equations to form overall equations for the reactions taking place.

Oxidation and reduction

Metal oxides

Reactive metals will react with oxygen to produce metal oxides.

metal_(s) + oxygen_(g) → metal oxide_(s)

No doubt at some time in your science lessons you have held a piece of magnesium ribbon in a Bunsen flame and cautiously observed the bright flash from the burning magnesium.

Oxidation and reduction

The product of this combustion reaction is magnesium oxide. No doubt from your GCSE chemistry you will recall that we defined oxidation as the addition of oxygen to a substance and reduction as the removal of oxygen from a compound. While these definitions of oxidation and reduction are useful, they are by no means the only way in which we can talk about substances being oxidised and reduced, for example:

In the example above the magnesium metal has had oxygen added to it to form the compound magnesium oxide, so we would say that the magnesium has been oxidised. We can write a word equation to show this simple reaction:

magnesium_(s) + oxygen_(g) → magnesium oxide_(s)

A symbolic equation for this reaction is:

2Mg + O_2(g) → 2MgO_(s)

Now magnesium oxide is an ionic compound which contains Mg²⁺ and O^2- ions. Now if we consider how these ions are formed; the magnesium ions are formed when the magnesium atoms lose two electrons to form the magnesium cation; as shown below:

Mg -2e → Mg²⁺

Or the preferred way to show oxidation is with the electrons on the right hand side of the equation:

Mg → Mg²⁺ + 2e

The magnesium atoms are losing 2 electrons to form magnesium ions (Mg²⁺). The addition of oxygen or the loss of electrons both amount to the same thing in this example. This loss of electrons (or addition of oxygen) is called oxidation. So if losing electrons is another way to define oxidation, then if a substance gains electrons it will be reduced.

The oxygen atoms in this reaction have gained electrons from the magnesium atoms to form oxide ions (O^2-). We can show this as:

O + 2e → O^2-

Or since oxygen is a diatomic gas we can write:

O₂ + 4e → 2O^2-

The gain of electrons is called reduction, and when a substance gains electrons we say it has been reduced. Reactions where one substance is reduced and another is oxidised are called redox reactions.

Burning copper metal

If a square of copper metal is held with a pair of tongs in a hot Bunsen flame for about 30 seconds the shiny bronze coloured copper turns black. Unlike when magnesium metal was oxidised no flames or bright flashes are produced. However the copper metal, just like the magnesium metal is reacting with the oxygen in the air; it is being oxidised. An equation for this reaction is:

2Cu_(s) + O_2(g) → 2CuO_(s)

Copper(II) oxide is an ionic compound which contains Cu²⁺ ions and O^2- ions. The copper ions are formed when the copper atoms are oxidised, that is they lose two electrons to form copper ions:

Cu → Cu²⁺ + 2e

The oxide ions are formed when the oxygen atoms in the oxygen molecule each gain electrons (reduction):

O₂ + 4e → 2O^2-

A substance which donates electrons is called a reducing agent. In this example the copper atoms are the reducing agent. An oxidising agent is an electron acceptor; it accepts electrons and is itself reduced.

The equations above are called ion-electron half equations (half-equations). To get the overall equation, add the half-equations together, ensuring the number of electrons lost and gained is the same.

2Cu → 2Cu²⁺ + 4e   oxidation

O₂ + 4e → 2O^2-   reduction

2Cu + O₂ → 2CuO   overall equation

Example 2 - The displacement reaction between copper(II) sulfate solution and iron

When an iron nail is placed in a beaker containing a solution of copper(II) sulfate a displacement reaction occurs. The more reactive iron displaces the copper ions (Cu²⁺) from the solution, equations for this displacement reaction are shown below:

iron_(s) + copper(II) sulfate_(aq) → iron(II) sulfate_(aq) + copper_(s)

Fe_(s) + CuSO_4(aq) → FeSO_4(aq) + Cu_(s)

If we now write an ionic equation it may allow us to better understand what has been oxidised and what has been reduced:

Fe_(s) + Cu²⁺_(aq) + SO₄^2-_(aq) → Fe²⁺_(aq) + SO₄^2-_(aq) + Cu_(s)

The sulfate ion (SO₄^2-) is unchanged in the reaction, so it is a spectator ion, that is it is an ion that takes no part in the reaction and remains unchanged during the reaction. Removing spectator ions gives the net ionic equation:

Fe + Cu²⁺ → Fe²⁺ + Cu

In this reaction the iron atoms lose 2 electrons (oxidation):

Fe → Fe²⁺ + 2e

The copper ions (Cu²⁺) gain 2 electrons (reduction) to form copper metal:

Cu²⁺ + 2e → Cu

In this reaction the iron atoms are the reducing agent and the copper ions (Cu²⁺) are the oxidising agent.

Example 3 - The reaction of sodium and chlorine

Sodium is an alkali metal with an electronic configuration of 1s²2s²2p⁶3s¹, so when it reacts it will lose its outer 3s¹ electron and it will be oxidised. Chlorine is a reactive halogen in group 7 with an electronic configuration of 1s²2s²2p⁶3s²3p⁵. To gain a noble gas electronic configuration, chlorine only has to gain 1 electron, so it is a strong oxidising agent. By contrast, sodium will be looking to lose an electron, so metals are typically good reducing agents.

If a small piece of sodium metal is placed on a pile of sand in a flask filled with dry chlorine gas you might expect a violent reaction to occur; however the reaction is slow to start with. To start the reaction a few drops of water are added. A violent reaction between the sodium and chlorine gas then starts and the flask quickly fills with dense white or colourless fumes of sodium chloride.

A half equation for the oxidation of sodium is:

Na → Na⁺ + e

A half equation for the reduction of chlorine gas (Cl₂) is:

Cl₂ + 2e → 2Cl^-

Since chlorine is a diatomic gas it requires 2 moles of electrons to form 2 moles of chloride ions. This means the sodium half-equation must be multiplied by 2:

oxidation half equation: 2Na → 2Na⁺ + 2e

reduction half equation: Cl₂ + 2e → 2Cl^-

Cancelling the electrons gives the overall equation:

overall equation: 2Na + Cl₂ → 2NaCl

Self-check: Complete the activity below to test your understanding of oxidation, reduction and redox.

Complete the activity below by deciding if each statement shows an oxidation, reduction or redox process.

Key Points

The image opposite may help you in understanding the terms: reduction, oxidation, reducing agent and oxidising agent.

• Oxidation is the loss of electrons, or the addition of oxygen, or the removal of hydrogen from a substance. The addition of an electronegative element such as chlorine can also be considered an oxidation reaction.
• Reduction is the gain of electrons, or the removal of oxygen, or the addition of hydrogen to a substance.
• An oxidising agent is an electron acceptor. It oxidises another substance by removing electrons from it, and is itself reduced.
• A reducing agent is an electron donor. It reduces another substance by supplying electrons to it, and is itself oxidised.
• Remember the word "OILRIG" - Oxidation is a loss of electrons; reduction is a gain of electrons.
• To get the overall equation from two half-equations, ensure the number of electrons lost equals the number gained.

Practice questions

Check your understanding - Redox reactions.

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