# Oxidation-reduction (redox) reactions

Examples of oxidation reduction (redox) reactions, oxidizing and reducing agents, and common types of redox reactions.

## What is a redox reaction?

A redox (or oxidation-reduction) reaction is a type of chemical reaction that involves a transfer of electrons between two species.
Great question! A chemical species is a term that refers to a set of atoms, molecules, or ions with the same chemical formula.
For example, in the thermite reaction:
$\text{Fe}_2 \text{O}_3+2 \text{Al} \rightarrow 2 \text{Fe}+\text{Al}_2 \text O_3$
the four species involved in this reaction are the reactants $\text{Fe}_2 \text{O}_3$ and $\text{Al}$, as well as the products $\text{Fe}$ and $\text{Al}_2 \text O_3$.
We can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products.
Chemists use oxidation numbers (or oxidation states) to keep track of how electrons are being shared within an atom or molecule. It is important to understand how to find the oxidation number before you can understand redox reactions.
If you would like a refresher on what oxidation numbers are and how to find them, see this video.
A very exothermic thermite reaction produces heat, light, and molten iron globules.
Lots of light and heat released = chemical reaction! This is a thermite reaction, which is also a redox reaction. The reaction is: $\text{Fe}_2 \text{O}_3+2 \text{Al} \rightarrow 2 \text{Fe}+\text{Al}_2 \text O_3.$ Thermite Reaction image by Nikthestunned at Wikimedia Commons, CC-BY-SA 3.0.
Redox reactions are everywhere! Your body uses redox reactions to convert food and oxygen to energy plus water and $\text {CO}_2$, which we then exhale. The batteries in your electronics also rely on redox reactions, which you will hear more about when we learn electrochemistry. Can you find other examples of redox reactions happening around you?

## An example and important terms

Redox reactions have some associated terms you should be comfortable using. We will go over these terms using the following example reaction:
$2 \text{Fe}_2 \text O_3(s)+ 3 \text C(s) \rightarrow 4 \text {Fe}(s) + 3 \text{CO}_2(g)$
Here are some questions we want to be able to answer:
$1.~$ Is this reaction a redox reaction, and how do we know?
$2.~$ If this is a redox reaction, what is being reduced and oxidized?
Here are two ways to remember oxidation and reduction:
$1.~$OIL RIG: "Oxidation Is Loss" and "Reduction Is Gain"
$2.~$LEO the lion says GER: "Loss of Electron is Oxidation" and "Gain of Electron is Reduction"
$3.~$ What is the reducing agent in this reaction?
$4.~$ What is the oxidizing agent in this reaction?
Question $1$:
Based on the title of this article, we can make an educated guess about the first part of this question. Yes, this is probably a redox reaction, but how do we know that for sure? We need to show there is an electron transfer occurring, and we can do that by checking if any oxidation numbers change from the reactants to the products.
If we find the oxidation numbers for each atom in the reactants and products, we get the following:
$\text C(s)$ and $\text {Fe}(s)$: The oxidation number for pure elements is $0$, so we know the oxidation number for $\text C(s)$ and $\text {Fe}(s)$ is $0$.
$\text{Fe}_2 \text O_3(s)$: The oxidation number for $\text O$ is $-2$ and the charge on the compound is neutral, so the oxidation state of the iron atoms is $+3$.
$\text{CO}_2(g)$: The oxidation of $\text O$ is also $-2$ and the charge on the compound is neutral, so the oxidation number of the $\text C$ is $+4$.
$2 \text{Fe}_2 \text O_3(s)+ 3 \text C(s) \rightarrow 4 \text {Fe}(s) + 3 \text{CO}_2(g)$
$\purpleC{~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow}~~~~~~~~~$ $\purpleC{+3, -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4, -2}~~~~~~~~~\text{(Oxidation numbers)}$
We can use the oxidation numbers to answer the second part of question $1$, because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons.
Question $2$:
Carbon being oxidized because it is losing electrons as the oxidation number increases from $0$ to $+4$.
Iron is being reduced because it is gaining electrons when the oxidation number decreases from $+3$ to $0$.
Question $3$:
The reducing agent is the reactant that is being oxidized (and thus causing something else to be reduced), so $\text C(s)$ is the reducing agent.
Reducing agents and oxidizing agents can be a tricky concept! For a more thorough explanation, you can watch this video on oxidizing and reducing agents.
Question $4$:
The oxidizing agent is reactant that is being reduced (and thus causing something else to be oxidized), so $\text{Fe}_2 \text O_3(s)$ is the oxidizing agent.

## Common types of redox reactions

Since redox reactions are an important class of reactions, we want to be able to recognize them. There are a few special types of redox reactions you might want to be familiar with. For each of these examples, take a minute to figure out what is getting reduced and oxidized!
$1.~$Combustion reactions
A combustion reaction is a redox reaction between a compound and molecular oxygen ($\text O_2$) to form oxygen-containing products. When one of the reactants is a hydrocarbon, the products include carbon dioxide and water.
A hydrocarbon is a compound that is made up of only carbon and hydrogen atoms. Some examples of hydrocarbons are the compounds methane, $\text {CH}_4$, and benzene, $\text C_6 \text H_6$.
The following reaction is the combustion of octane, a hydrocarbon. Octane is a component of gasoline, and this combustion reaction occurs in the engine of most cars:
$2\text C_8 \text H_{18} + 25 \text O_2 \rightarrow 16 \text{CO}_2(g) + 18\text H_2 \text O$
$2.~$Disproportionation reactions
A disproportionation reaction (or auto-oxidation reaction) is a reaction in which a single reactant is both oxidized and reduced. The following reaction is for the disproportionation of hypochlorite, $\text{ClO}^-$:
$3\text{ClO}^-(aq)\rightarrow \text{ClO}_3^-(aq) + 2\text{Cl}^-(aq)$
If we analyze the oxidation numbers for chlorine, we see that the reactant $\text{ClO}^-$ is being oxidized to $\text{ClO}_3^-$ (where the oxidation number increases from $+1$ to $+5$). At the same time, the chlorine in some other molecules of $\text{ClO}^-$ are being reduced to $\text{Cl}^-$(where the oxidation number decreases from $+1$ to $-1$). Oxygen has an oxidation number of $-2$ in both $\text{ClO}^-$ and $\text{ClO}_3^-$, so it does not get oxidized or reduced in the reaction.
$3.~$Single replacement reactions
A single replacement reaction (or single displacement reaction) involves two elements trading places within a compound. For example, many metals react with dilute acids to form salts and hydrogen gas. The following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid:
$\text{Zn}(s)+2\text{HCl}(aq) \rightarrow \text{ZnCl}_2(aq)+\text H_2(g)$
For another example of a single replacement redox reaction, see this video on a reaction with iron.

## Balancing a simple redox reaction using the half-reaction method

Redox reactions can be split into reduction and oxidation half-reactions. Chemists use half-reactions to make it easier to see the electron transfer, and it also helps when balancing redox reactions. Let's write the half-reactions for another example reaction:
If you want an explanation of half-reactions, oxidation, and reduction in a different format, check out this video for an example of a redox reaction with iron.
$\text {Al}(s)+ \text {Cu}^{2+}(aq) \rightarrow \text {Al}^{3+}(aq) + \text{Cu}(s)$
Is the above reaction balanced? Our atoms appear to be balanced: we have $1\,\text {Al}$ atom and $1\,\text {Cu}$ atom on each side of the arrow. However, when we add up the charges on the reactant side we get a $2+$ charge, which is not the same as the $3+$ charge on the product side. We need to make sure both the atoms and the charges are balanced! We will use the half-reaction method to balance the reaction.
Reduction half-reaction: The reduction half-reaction shows the reactants and products participating in the reduction step. Since $\text {Cu}^{2+}$ is being reduced to $\text{Cu}(s)$, we might start by writing out that step:
$\text {Cu}^{2+}(aq) \rightarrow \text{Cu}(s)$
However, this is not the correct half-reaction, because it is not charge-balanced. There is a net charge of $2+$ on the reactant side and $0$ on the product side. We can balance the charges by including the electrons being transferred, and then we will get our reduction half-reaction:
$\text {Cu}^{2+}(aq) +2e^-\rightarrow \text{Cu}(s)~~~~~~~~~~~~~~~~~~\blueD{\text{Reduction half-reaction}}$
The balanced half-reaction tells us that $\text{Cu}^{2+}$ is gaining $2 e^-$ per copper atom to form $\text{Cu}^0$. So where are those electrons coming from? We can follow the trail of electrons to the oxidation half-reaction.
Oxidation half-reaction: The oxidation half-reaction shows the reactants and products participating in the oxidation step. This reaction will include the oxidation of $\text {Al}(s)$ to $\text {Al}^{3+}$, and we will also want to make sure the half-reaction is charge-balanced:
$\text {Al}(s) \rightarrow \text {Al}^{3+}(aq) +3e^-~~~~~~~~~~~~~~~~~~~\blueD{\text{Oxidation half-reaction}}$
The oxidation half reaction tells us that each atom of $\text {Al}(s)$ is losing $3e^-$ to form $\text {Al}^{3+}$.
We will combine the balanced half-reactions to get the balanced overall equation, but there is one more thing to check. The electrons must cancel out in the overall equation. Another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction. Otherwise we would have stray electrons floating around! That means we need the number of electrons being transferred in each half-reaction to be equal.
We can multiply the reduction half-reaction by $3$ and multiply the oxidation half-reaction by $2$ so both reactions involve the transfer of $6$ electrons:
Now that we have the same number of electrons in each half-reaction, we can add them together to get our overall balanced equation:
Lastly, we can check to see if any reactants and products appear on both sides. Since that is not the case here, we are done! Our reaction is balanced for both mass and charge.
This reaction is occurring under neutral conditions $([\text H^+]=[\text{OH}^-])$. Sometimes you need to balance redox reactions under acidic conditions (where $([\text H^+]>[\text{OH}^-])$, or under basic conditions ($([\text H^+]<[\text{OH}^-])$. The procedures for balancing redox reactions under acidic and basic conditions have a few extra steps compared to the example we just went through.
For more on balancing under acidic conditions, see this video on balancing redox reactions in acid.
For more on balancing redox reactions under basic conditions, see this video on balancing redox reactions in base.

## Summary

We can identify redox reactions by checking for changes in oxidation number. Redox reactions can be split into oxidation and reduction half-reactions. We can use the half-reaction method to balance redox reactions, which requires that both mass and charge are balanced. Three common types of redox reactions are combustion, disproportionation, and single replacement reactions.

Kotz, J. C., Treichel, P. M., Townsend, J. R., and Treichel, D. A. (2015). Oxidation-Reduction Reactions. In Chemistry and Chemical Reactivity, Instructor's Edition (9th ed., pp. 125-131). Stamford, CT: Cengage Learning.
Kotz, J. C., Treichel, P. M., Townsend, J. R., and Treichel, D. A. (2015). Chlorine Compounds. In Chemistry and Chemical Reactivity, Instructor's Edition (9th ed., pp. 850-851). Stamford, CT: Cengage Learning.

## Try it!

#### Problem 1

What is the balanced form of the following redox reaction?
$\text {Al}(s)+\text {H}^+(aq) \rightarrow \text{Al}^{3+} (aq)+ \text H_2(g)$
$\text {Al}(s)\rightarrow \text{Al}^{3+} (aq)+3e^-$
$2\text {H}^+(aq) + 2e^- \rightarrow \text H_2(g)$
We can multiply the reduction half-reaction by $3$ and multiply the oxidation half-reaction by $2$ so both reactions involve the transfer of $6$ electrons: