Lesson Explainer: Oxidization and Reduction Science

In this explainer, we will learn how to identify and analyze oxidization and reduction reactions.

Oxidization and reduction are two chemical processes that can be thought of as opposite to each other. Both can be explained in a variety of ways.

One of the simplest examples is the burning of carbon. When charcoal, a form of carbon, is burned, carbon is oxidized to produce carbon dioxide. The following equation shows the oxidization of carbon:

Based on this, we can describe oxidization as simply the addition of oxygen.

The opposite reaction would be the loss or removal of oxygen. We can describe this process as reduction.

When a stream of hydrogen gas is passed over a heated sample of copper oxide, the hydrogen gas removes oxygen from the copper oxide to produce copper metal.

The chemical equation for this reaction is

As copper oxide has lost oxygen, we could describe this reaction as a reduction.

However, the hydrogen has gained oxygen, and so we could also describe this reaction as an oxidization.

In this reaction, copper oxide has been reduced and hydrogen gas has been oxidized.

We can also describe each substance in the reaction above using the terms “oxidizing agent” and “reducing agent.”

For the reaction between copper oxide and hydrogen, we can consider the copper oxide as an oxidizing agent because it oxidizes the hydrogen gas.

We can also consider the hydrogen as a reducing agent because it reduces the copper oxide to copper metal.

From the diagram above, we can see that an oxidizing agent gets reduced and a reducing agent gets oxidized.

We can also explain oxidization and reduction in terms of the gain or loss of hydrogen.

In the following reaction diagram, we can see that the ammonia gas loses hydrogen and becomes oxidized, while the bromine gains hydrogen and becomes reduced.

We can also define oxidization as the loss of hydrogen and reduction as the gain of hydrogen.

Let’s clarify our understanding with some definitions.

As we have seen, oxidization and reduction reactions occur in pairs. In reactions, we can identify the oxidizing and reducing agents.

Up until this point, we have considered oxidization in terms of the addition of oxygen or loss of hydrogen. However, it is also possible to define oxidization and reduction in terms of the gain or loss of electrons.

Describing oxidization and reduction in terms of the gain or loss of electrons allows us to describe reactions that do not involve oxygen or hydrogen.

Consider a simple combination reaction in which sodium metal reacts with chlorine:

The sodium metal has an electronic configuration of 2, 8, 1, whereas the chlorine gas has an electronic configuration of 2, 8, 7.

During a chemical reaction, sodium and chloride ions are formed.

The atom of sodium gives an electron to the chlorine atom to form a positive sodium ion with an electronic configuration of 2, 8.

The chlorine atom receives the electron and becomes a chloride ion with an electronic configuration of 2, 8, 8.

The sodium atom has lost an electron, and the chlorine atom has gained an electron. We can define oxidization and reduction in similar terms.

Oxidization is the loss of electrons.

In the chemical reaction above, the sodium atoms lose electrons and become oxidized, with the chlorine atoms acting as an oxidizing agent.

Conversely, the chlorine atoms gain electrons and become reduced, with the sodium atoms acting as a reducing agent.

In another example, we can see the displacement of copper from copper sulfate by the more active magnesium metal:

To better understand the movement of electrons, we can use half-equations to focus on the important parts of the chemical reaction.

In the example involving magnesium and copper sulfate, the magnesium atoms lose electrons and become oxidized. We can show this using the following half-equation:

This half-equation explicitly shows the number of electrons involved. As magnesium can form ions, then it loses two electrons.

Now, let’s consider only the copper ions, which form part of the ionic compound copper sulfate. We can see that they initially exist as ions with a charge.

During the chemical reaction, the copper ions gain two electrons and so become reduced. We can represent this by the following half-equation:

Notice how the number of electrons in both of these half-equations is the same.

We can combine these two half-equations to give us an overall ionic equation for this chemical reaction:

Finally, as we can consider oxidization and reduction in terms of electrons, we can also define oxidizing and reducing agents in terms of electrons.

Example 3: Describing an Oxidizing Agent in Multiple Ways

Which of the following descriptions can be used for an oxidizing agent?

1. 1, 2, and 5
2. 1, 3, and 4
3. 2 and 5
4. 2, 3, and 4
5. 1 and 5

Answer

It is possible to define oxidization and reduction in multiple ways. For example, we can define them in terms of the gain or loss of oxygen, hydrogen, or electrons.

Similarly, oxidizing and reducing agents can be defined in the same ways.

An oxidizing agent oxidizes another chemical and during the process lose electrons itself. These electrons are gained by the oxidizing agent, and so option 2 is part of our correct answer.

Oxidization could also be an increase in the percentage of oxygen in a substance, which would require an oxidizing agent to donate oxygen. Therefore, we can consider option 3, the donation of oxygen, as also correct.

The final way of defining oxidization is through the loss of hydrogen. For an oxidizing agent, this would mean removing hydrogen from another chemical species. Therefore, option 4 is also correct.

We have now established that options 2, 3, and 4 can describe an oxidizing agent. Therefore, the correct answer is D.

Oxidization and reduction are concurrent processes (i.e., they both occur at the same time). The oxidization of a substance cannot occur without the reduction of another.

Let’s consider one final reaction.

Iron(III) chloride reacts with hydrogen gas to form iron(II) chloride and hydrochloric acid:

Let’s just focus on what happens to the iron during this reaction.

In the reactant, , iron exists as a ion, whereas in the product, , iron exists as a ion. The iron ion has gained an electron to become an iron ion, and so it has been reduced.

This electron is gained from a hydrogen atom, making hydrogen a reducing agent. As a reducing agent, the hydrogen loses an electron and is oxidized to a hydrogen ion. This hydrogen ion then forms part of the product, the hydrochloric acid.

The chloride ions remain unchanged during the course of the reaction and are neither oxidized nor reduced.

We can write half-equations for the oxidization and reduction processes for this reaction.

For the reduction of , we can write

For the oxidization, we can write

We can represent this reaction as an ionic equation, without the chloride ions: