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Unit 4

4.9 Redox Reactions

4 min read•august 22, 2020

Dalia Savy

Hope Arnett

AP Chemistry 🧪

269 resources

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What is a Redox Reaction?

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

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.

Writing redox reactions reveals which species is which 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"

Example Problem

Balancing in an Acidic Solution

2Mg (s) + O2 (g) --> 2MgO (s)

Step 1

Identify the charges of each species. Mg and O2 are both neutral molecules. MgO is made up of an ionic bond between Mg2+ and O2-, and we know their charges have to sum to 0 since MgO is a neutral compound.

With this information, we can write half reactions for each reactant. A half reaction just shows the transfer electrons for each molecule.

Step 2

Start writing the half reaction equations by illustrating the change in charge. Neutral 2 moles of Mg becomes 2 moles of Mg2+, and 1 mole of neutral O2 becomes 1 mole 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 a conservation of mass, but we need to keep a conservation of charge❗❗

Step 3

We can conserve charge by adding electrons to the appropriate side. In the first equation, the reactants have a charge of 0, but the products have a charge of +4 (2 moles of 2+ ion). 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

Finally, we’ll add the two reactions and simplify by cancelling out similar terms.

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.

Key Facts

💭Remember these rules for assigning oxidation numbers:

Free elements (ex. Br2, Na, P4) have an oxidation number of 0
Neutral molecules also have oxidation numbers of 0, so their elements’ numbers must sum to 0

Ex. In the compound IF6, let x = the oxidation number for iodine and y = the oxidation number for fluorine. This must be true: x + 6y = 0

Monatomic ions have an oxidation number equal to their charge (ex. Na+ = oxidation number of +1, Ba2+ = +2, Cl- = -1, Al3+ = +3)
Oxygen is -2, except in hydrogen peroxide (H2O2) and peroxide ion (O2-2), where it’s -1
Hydrogen is +1, except in metal hydrides (ex. LiH, BaH2), where it’s -1
Fluorine is -1. Other halogens are usually -1, but they vary (ex. Br2O3)
Oxidation numbers can be fractions, but it’s very rare (ex. superoxide, O2- = -½)

⬇️Here are the general rules for making redox equations:

If the molecule has an ionic bond, each element’s oxidation number is their ionic charge. If not, determine their oxidation numbers.
Start writing the half reaction for each molecule by writing the change in charge.
Make sure the elements are balanced
Finalize the half reaction. Balance them by adding electron(s) to the appropriate side.
Add the half reactions together and simplify the equation.

Balancing 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-.

🎥 Watch: AP Chemistry - Redox Reactions

Browse Study Guides By Unit

⚛️Unit 1 – Atomic Structure & Properties

🤓Unit 2 – Molecular & Ionic Bonding

🌀Unit 3 – Intermolecular Forces & Properties

🧪Unit 4 – Chemical Reactions

4.0Unit 4 Overview: Chemical Reactions

4.1Introduction for Reactions

4.2Net Ionic Equations

4.3Representations of Reactions

4.4Physical and Chemical Changes