4.9 Oxidation-Reduction (Redox) Reactions

This is the last reaction to learn for this unit! Before moving on to the rest of the study guide, be sure to review both precipitation and acid-base reactions.

What is a Redox Reaction?

The last type of reaction in this unit is oxidation-reduction reactions, commonly called redox. These reactions deal with the transferring➡️ of electrons, which causes molecules to change oxidation states. 

An oxidation state is a measure of the degree of oxidation of an atom in a chemical compound. It is represented by a positive or negative number that expresses the number of electrons that an atom has gained or lost in a compound relative to its elemental state on the periodic table.

When a molecule loses an electron, it’s oxidized, and its oxidation number increases. When a molecule gains an electron, it’s reduced, and its oxidation number decreases. Electrons travel from the oxidized species to the reduced species. Okay, wait...that was a lot of information in three sentences. Here is a quick recap of what you should understand about redox reactions so far:

• Redox reactions involve the transfer of electrons from one atom to another.
• Oxidation is the process of losing electrons or increasing the oxidation state of an atom.
• Reduction is the process of gaining electrons or decreasing the oxidation state of an atom.
• Acids donate hydrogen ions to bases in acid-base reactions, but here, the substance that is oxidized "donates" electrons to the substance that is reduced. Writing out the chemical equation of redox reactions reveals which species are oxidized and which are reduced by illustrating the transfer of electrons between molecules.

💡 Mnemonic Device Time! There are two different ones, use the one that seems the most catchy to you:

• OIL RIG = "oxidation is loss" and "reduction is gain"
• LEO says GER "loss of electrons = oxidation" and "gain of electrons = reduced"

Assigning Oxidation Numbers

How do we know the oxidation state of an atom? Here are a few rules that you should remember and familiarize yourself with:

1. Free elements (ex. Br₂, Na, P₄) have an oxidation number of 0.
2. Neutral molecules also have oxidation numbers of 0, so their elements’ oxidation numbers must sum to 01. In the compound IF₆, let x = the oxidation number for iodine and y = the oxidation number for fluorine. The following must be true since IF₆ is a neutral molecule: x + 6y = 0
3. Monatomic ions have an oxidation number equal to their charge (ex. Na⁺ has an oxidation number of +1, Ba²⁺ has an oxidation number of +2, Cl⁻ has an oxidation number of -1, and Al⁺³ has an oxidation number of +3).
4. Oxygen is -2, except in hydrogen peroxide (H₂O₂) and peroxide ion (O₂⁻²), where it’s -1
5. Hydrogen is +1, except in metal hydrides (ex. LiH, BaH₂), where it’s -1
6. Fluorine is -1. Other halogens are usually -1, but they vary (ex. Br₂O₃)
7. Oxidation numbers can be fractions, but it’s very rare (ex. superoxide, O₂⁻ = -½)

Redox Practice Problem

Redox reactions are important in a variety of chemical processes, including the production of electricity, the corrosion of metals, and the metabolism of cells. The scope of redox reactions in this unit is understanding how to balance their chemical equations. We come back to redox in the very final unit of this course: applications of thermodynamics.

On the AP examination, you may be given a redox reaction and be asked to either balance it in an acidic or basic solution. Let's go over both scenarios.

Balancing Redox in an Acidic Solution

2Mg (s) + O₂ (g) → 2MgO (s)

Step 1

First, assign oxidation numbers to all atoms that take part in the reaction. Mg and O₂ are both neutral molecules, so they must have oxidation states of 0. MgO is made up of an ionic bond between Mg²⁺ and O²⁻, and we know their oxidation numbers have to sum to 0 since MgO is a neutral compound. Since oxygen typically has an oxidation number of 2-, we know that Mg will have an oxidation state of 2+ in this compound.

With this information, we can write half-reactions for each reactant. A half-reaction just shows the transfer of electrons for each molecule specifically, showing either the oxidation or reduction process.

Step 2

Start writing the half-reaction equations by illustrating the change in oxidation numbers. Neutral 2 moles of Mg becomes 2 moles of Mg²⁺, and 1 mole of neutral O₂ becomes 2 moles of O²⁻:

2Mg → 2Mg²⁺

O₂ → 2O²⁻

However, just like any other equation, we need to balance the products and reactants. We’re dealing with the change in charge, so not only do we need to ensure the conservation of mass, but we need to conserve charge❗❗

Step 3

We can conserve charge by adding electrons, negative subatomic particles, to the appropriate side. In the first half-reaction equation, the reactants have a charge of 0, but the products have a charge of +4 (2 moles of 2+ ions). We need to add a charge of -4 (or 4 electrons) to the products to make it so that both sides have 0:

2Mg → 2Mg²⁺ + 4e⁻

In the second equation, the reactants also have a charge of 0, but the products have a charge of -4. Thus, we need a way to add a charge of +4 to balance the products out. However, we can only add electrons. The way around this is to add a charge of -4 (or 4 electrons) to the neutral side. Now, both sides have an equal charge of -4.

O₂+ 4e⁻ → 2O²⁻

Step 4

Since both of our half-reactions conserve both mass and charge, and we can see the transfer of electrons, we can add the two. Adding the two half-reactions allows us to see the full balanced redox reaction: 2Mg + O₂ + 4e⁻ → 2Mg²⁺ + 2O²⁻ + 4e⁻

Wait...but there are four electrons on both sides of the equation! When we have similar terms, we can cancel them out.

2Mg + O₂ → 2Mg²⁺ + 2O²⁻

Congratulations! You made a redox reaction!🥳

However, we’re not done yet. In ionic bonds, electrons are transferred completely from one element to another. MgO is an example of an ionic bond. Alternatively, in non-ionic bonds, electrons are shared between molecules. HCl and BrNO₃ are some examples. 

Scientists👨‍🔬👩‍🔬 decided to deal with this caveat by giving molecules with non-ionic bonds oxidation numbers instead of ionic charges. These oxidation numbers reflect the maximum number of electrons a molecule would give or accept if they were in an ionic bond.

Balancing Redox in a Basic Solution 

Above, we balanced a really easy equation in an acidic solution. You could also be asked to balance redox in a basic solution. Balancing in a basic solution follows the same steps, BUT there is an additional step at the end since OH⁻ ions are present. That extra step is to form H₂O with the present H⁺ ions and oxygen atoms, and then add that mass onto the other side with OH⁻.

Steps for Balancing Redox

⬇️Here are the general rules for making redox equations: 

1. Assign oxidation numbers and determine which element is being oxidized and which element is being reduced.
2. Write the half-reactions for the oxidation and reduction processes.
3. Balance elements in the half-reactions other than O and H.
4. Balance the oxygen atoms by adding the appropriate atoms of water molecules.
5. Balance the hydrogen atoms by adding hydrogen ions.
6. Balance the half-reactions for charge by adding electrons as necessary. Most of the time, the hydrogen ion is on the same side as the electrons.
7. Combine the half-reactions by adding them together and canceling out any species that appear on both sides of the equation.
8. If the reaction is being balanced in a basic solution, add the appropriate number of hydroxide ions to neutralize all hydrogen ions and convert them to water molecules. Remember, whatever you add to one side, you must add to the other as well.
9. Check your work by making sure the number of atoms of each element is balanced on both sides of the equation, and that the total charge on the reactant side is equal to the total charge on the product side.

🎥 Watch AP Chemistry teacher Wes Winter review the different types of reactions, including oxidation-reduction reactions.