In oxidation-reduction reactions, electrons are transferred from one reactant to another. Another name for oxidation-reduction reactions is redox reactions, where the “red” comes from reduction and the “ox” comes from oxidation. Redox reactions and the concepts of oxidation-reduction come up in many areas of chemistry. Getting a good foundation on redox chemistry will help you in this course, when you study transition metals and coordination chemistry (later in Unit 3) and electrochemistry (in Unit 4), as well as in organic chemistry.

Example of redox reaction

$C{{e}^{4+}}+F{{e}^{2+}}\to C{{e}^{3+}}+F{{e}^{3+}}$

In this reaction Ce⁴⁺ gained 1 electron. You can tell that because its charge got less positive (a charge getting less positive or more negative indicates a gaining of electrons). Fe ²⁺ lost 1 electron – you can tell that because its charge got more positive. Since Ce⁴⁺ gained 1 electron and Fe ²⁺ lost 1 electron, we can conclude an electron was transferred from Fe²⁺ to Ce⁴⁺.

Another way we can identify redox reactions is to realize that oxidation states change in redox reactions. To make use of that, though, you have to have an understanding of what is meant by oxidation state and how to identify it, which is the topic of the next objective.

Objective 4: Assign oxidation states to the elements in a compound or ion.

In any chemical formula (whether element, compound, or ion) you might encounter, an oxidation state or oxidation number can be assigned to each element in the formula. In a neutral element or compound, the oxidation state of each atom must add to zero. In an ion, the oxidation state of each atom must add to the overall charge of the ion.

In these notes, I’ll describe how to determine the oxidation state of any element in a chemical formula. The link in the Unit 3 – additional study resources section of this site called Assigning Oxidation Numbers that provides a two page summary of what is explained below and organizes it in a slightly different way that may also be helpful to you.

Oxidation States for Elements

The oxidation state of any pure element is zero.

Examples:

In elemental sodium (Na), the oxidation state of Na is zero.
In elemental iron (Fe), the oxidation state of Fe is zero.
In elemental oxygen (O₂), the oxidation state of O is zero.

Oxidation States for monoatomic ions

The oxidation state of any monoatomic ion is the ion’s charge.

Examples:

In the iron(II) ion (Fe²⁺ or Fe⁺²), the oxidation state of Fe is +2 or 2+. Note: either “2+” or “+2” is acceptable.
In the iron(III) ion (Fe³⁺ ), the oxidation state of Fe is +3 or 3+.
In the oxide ion (O^2- or O^-2), the oxidation state of O is -2 or 2-.

Oxidation States for Ionic Compounds

The oxidation state of each element in an ionic compound is its ionic charge. Common ionic charges for elements in binary ionic compounds (made up of a metal cation and a nonmetal anion) are shown here.

Examples:

In CaO:

Ca exists as a Ca²⁺cation. The oxidation state of Ca is +2.
O exists as a O^2-cation. The oxidation state of O is -2.

In Al₂O₃:

Al exists as a Al³⁺cation. The oxidation state of Al is +3.
O exists as a O^2-cation. The oxidation state of O is -2.

Al and O combine in a 2:3 ratio in Al₂O₃because 2 Al ions (with a +3 charge) and 3 oxide ions (with a -2 charge) add overall to zero. In a neutral element or compound, the oxidation state of each atom must add to zero.

Also, oxidation state is per atom or per ion. That means the oxidation state of Al in the Al₂O₃compound is +3, not +6. And oxygen has an oxidation state of -2, not -6.

In FeCl₃:

Fe exists as a Fe³⁺cation. The oxidation state of Fe is +3.
Cl exists as a Cl^–cation. The oxidation state of Cl is -1.

In FeCl₂:

Fe exists as a Fe²⁺cation. The oxidation state of Fe is +2.
Cl exists as a Cl^–cation. The oxidation state of Cl is -1.

Oxidation States for Covalently Bonded Compounds

Covalent compounds are generally those made up of all nonmetals. In these compounds, oxidation numbers are assigned with the following priorities:

The top priority is that fluorine is assigned an oxidation state of -1
The second priority is that oxygen is assigned an oxidation state of -2.
The second priority is that hydrogen is assigned an oxidation state of +1.
Other elements: the more electronegative element takes priority, and in a neutral compound the oxidation states must add to zero.

Examples:

In CO:

The oxidation state of O is -2. (there is no F so O is the next priority)
The oxidation state of C is +2 (to give an overall of zero)

In CO₂

The oxidation state of O is -2. (there is no F so O is the next priority)
The oxidation state of C is +4 (to give an overall of zero)

Oxidation States for covalently bonded polyatomic ions

The same rules apply as for covalently bonded neutral compounds, except , the oxidation state of each atom must add to the overall charge of the polyatomic ion.

In NO₃^–

The oxidation state of O is -2. (there is no F so O is the next priority)
The oxidation state of N is +5 (to give an overall of -1)

In the thiosulfate ion S₂O₃^2-

The oxidation state of O is -2. (there is no F so O is the next priority)
The oxidation state of S is +2 (to give an overall of -2). The 2 S atoms must provide a total of +4 to give the polyatomic ion a total of -2 when combined with the 3 O atoms. Therefore, S is assigned +2.
Another way to determine S is by an algebraic equation:

2S + 3(-2) = -2. Solving for S gives on oxidation state of -2.

Peroxides and superoxides

There are 2 exceptions for oxygen as well, where O does not have an oxidation state of -2:

The peroxide ion (O₂^2-) is a polyatomic ion in which the oxidation state of O is -1. The most familiar peroxide compound is hydrogen peroxide (H₂O₂).
The superoxide ion (O₂^–) is a polyatomic ion in which the oxidation state of O is -1/2. An example of a superoxide salt is potassium superoxide (KO₂).

Ionic compounds with polyatomic ions

For these compounds, combine the rules above. Here’s an example:

Sodium carbonate: Na₂CO₃

Sodium carbonate is made up of Na⁺ ions and CO₃^2-ions. The oxidation state of Na is +1. We will have to figure out the oxidation states of C and O in a covalent polyatomic ion CO₃^2-

In CO₃^2-

The oxidation state of O is -2. (there is no F so O is the next priority)
The oxidation state of C is +4 (to give an overall of -2)

Therefore, in Na₂CO₃_,the oxidation states are:

Na: +1
C: +4
O: -2

Objective 4 Practice:

Identify the oxidation states of all elements in:

Potassium dichromate (K₂Cr₂O₇)

Oxygen difluoride (OF₂)

HNO₃

Mg(C₂H₃O₂)₂

Objective 1: Identify oxidation-reduction (redox) reactions

Objective 2: Define the terms oxidation and reduction in terms of loss or gain of electrons and impact on oxidation state

Objective 3: Identify which element is oxidized and which element is reduced in a redox reaction

Objective 6: Identify oxidizing and reducing agents in redox reactions.

Let’s look at an example reaction:

$Zn(s)+2{{H}^{+}}(aq)\to Z{{n}^{2+}}(aq)+{{H}_{2}}(g)$

First, we need to determine whether or not it is a redox reaction (not all reactions are). The first step to doing that is to determine all oxidation states in the reactants and products. If you understand the previous section, you should be able to do that.

Using these oxidation states, we can determine that $Zn(s)+2{{H}^{+}}(aq)\to Z{{n}^{2+}}(aq)+{{H}_{2}}(g)$ is a redox reaction because the oxidation state of at least one element changes when going from reactant to product. If the oxidation states of all elements are unchanged going from reactant to product, it is NOT a redox reaction.

Next, let’s look at some facts about this redox reaction

$Zn(s)+2{{H}^{+}}(aq)\to Z{{n}^{2+}}(aq)+{{H}_{2}}(g)$

Zinc is oxidized

A species is oxidized when its oxidation number increases (more + or less -). In this reaction, zinc loses two electrons and its oxidation number is changed from 0 to +2.

Hydrogen ion is reduced

A species is reduced when its oxidation number decreases (less + or more -). Here, each of the H⁺ions gains an electron and the oxidation number of H is changed from +1 to 0.

Oxidizing and Reducing Agents

H⁺ is the oxidizing agent

The reactant that is reduced is the oxidizing agent. In this example, H⁺is the oxidizing agent because it oxidizes Zn by taking electrons from it. The oxidizing agent causes another species to be oxidized.

Zinc is the reducing agent

The reactant that is oxidized is the reducing agent. In this example, Zn is the reducing agent because it reduces H⁺ by giving it electrons. The reducing agent causes another species to be reduced.

Key things to remember when identifying oxidizing and reducing agents

The oxidizing and reducing agents are always reactants in the reaction. In our example, H⁺ is the oxidizing agent. The oxidizing agent is not H. The oxidizing agent is not H₂. If it were either of these, then H (or H₂) would have to be reduced to a -1 state or lower.

When identifying agents, be specific about the species and its oxidation state. For example, iron exists in three oxidation states:

0 (as an element)
+2 (as an ion)
+3 (as an ion)

Of these, only Fe²⁺ and Fe³⁺ can be oxidizing agents. Fe cannot be an oxidizing agent, since being one would require being reduced to Fe^–.

As neutral elements, metals generally tend to lose electrons and be oxidized (good reducing agents), while nonmetals generally tend to gain electrons and be reduced (good oxidizing agents).

Mnemonics

Here are a couple common mnemonics (memory aids) people use to keep oxidation and reduction straight (the most common mistake is reversing some aspect):

OIL RIG: Oxidation Is Loss, Reduction Is Gain (you have to remember it is loss and gain of electrons)

LEO (the lion) says GER (pronounced Grrrr) : Loss of Electrons is Oxidation, Gain of Electrons is Reduction

If these help you, great! Use them. If they don’t, no worries, they aren’t necessary.

Objective 1,2,3,6 Practice

Answer the following for the reaction

$C{{e}^{4+}}+F{{e}^{2+}}\to C{{e}^{3+}}+F{{e}^{3+}}$

Half Reactions

Any redox reaction can be split into two half reactions — an oxidation half reaction (in which the oxidized species loses electrons (is oxidized) and a reduction half reaction (in which the species that is reduced gains electrons). The electrons are shown in the half reactions. The half reactions add to make the overall redox reaction.

Half reactions example

For the redox reaction

$Zn(s)+2{{H}^{+}}(aq)\to Z{{n}^{2+}}(aq)+{{H}_{2}}(g)$

The oxidation half reaction is

$Zn(s)\to Z{{n}^{2+}}(aq)+2{{e}^{-}}$

It shows the oxidation of Zn to Zn 2+. Note there is a loss of electrons, corresponding to oxidation. The electrons are a product in the oxidation half reaction.

The reduction half reaction is

$2{{H}^{+}}(aq)+2{{e}^{^{-}}}\to {{H}_{2}}(g)$

It shows the oxidation of H⁺ to H₂ . Note there is a gain of electrons, corresponding to reduction. The electrons are a reactant in the reduction half reaction.

In a balanced redox reaction, the half reactions will have the same number of electrons, so they cancel when added together to make the overall reaction. The overall redox reaction will not show electrons.

Objective 1,2,3,6 Practice

Answer the following questions about this reaction:

$2Na(s)+C{{l}_{2}}(g)\to 2NaCl(s)$

Answer the following questions about this reaction:

${{H}_{2}}S{{O}_{4}}(aq)+2NaOH(aq)\to N{{a}_{2}}S{{O}_{4}}(aq)+2{{H}_{2}}O$

Objective 7: Balance oxidation-reduction reactions using the half-reaction method in both acidic and basic solutions.

In redox reactions, it is customary to focus on the species being oxidized and reduced. For example, a redox reaction might be described as permanganate ion and iron (II) react to yield iron (III) and manganese (II) in solution.

$Mn{{O}_{4}}^{-}(aq)+F{{e}^{2+}}(aq)\to F{{e}^{3+}}(aq)+M{{n}^{2+}}(aq)$

To balance this equation, you will not be able to follow the procedure you have followed in the past to balance equations. Note there is oxygen in the reactant side but not on the product side. This does not mean that oxygen atoms are destroyed in the reaction! Rather, these reactions occur in aqueous solution and water is involved. Typically, they occur in either acidic or basic solution as well so either hydrogen or hydroxide ions are involved as well. The hydroxide or water can both be reactants or products involved in balancing the overall equation. A procedure for balancing these redox equations (the half-reaction method) is described below. There are also detailed, step-by-step examples in Openstax (see example 4.7 in Openstax section 4.2) and in the document “balancing oxidation-reduction reactions ” in the course D2L site.

Half reaction method (Acidic Solution)

Write half reactions for the species being oxidized and reduced.
Balance each half reaction
1. Balance all but H and O
2. Balance O by adding H₂O
3. Balance H by adding H⁺
4. Balance charge by adding electrons
Multiply the half reactions so the number of electrons in each half reaction is equal.
Add the half reactions together.
Check the balanced equation to confirm that the atoms and charges are balanced.

Half reaction method (Basic Solution)

Do steps 1 through 4 for acidic solution
Add OH^– ions to each side. You should add the number of hydroxide ions necessary to neutralize all the H⁺ to water.
Write out the equation after the neutralization takes place.
Check the balanced equation to confirm that the atoms and charges are balanced.

Example — Half reaction method

$Mn{{O}_{4}}^{-}(aq)+F{{e}^{2+}}(aq)\to F{{e}^{3+}}(aq)+M{{n}^{2+}}(aq)$

In the video below, this redox reaction is balanced first in acidic solution then in basic solution.

Objective 5: Predict the products for the single replacement reaction of a metal with an acid, water, or salt and use the activity series to predict whether a reaction will occur.

This document on single replacement reactions (which is also available in your course D2L site) provides detailed step-by-step instructions, examples, and practice questions with answers on balancing single replacement reactions. These reactions were covered in CHEM 151, so you may also find materials from your CHEM 151 class useful for review,

How the single replacement reactions relate to redox chemistry

Single replacement reactions are redox reactions — an example is

$Mg+2AgCl\to 2Ag+MgC{{l}_{2}}$

In this reaction, magnesium replaces silver in the ionic compound with chloride ion. Magnesium is oxidized (its oxidation state goes from zero to +2) and silver is reduced (its oxidation state goes from +1 to zero). If you examine the Activity Series of Metals (in the document on single replacement reactions), you can see that the higher up metals on the table want to be oxidized more – therefore, magnesium (higher on the series than silver) will be oxidized, going to a state of +2 and forming the magnesium chloride product.

What’s next?

This concludes our look at oxidation – reduction reactions for now. The concept of oxidation states will be importand in out next topic (transition metals and coordination chemistry). In the topic of electrochemistry (in Unit 4) , you will see how redox reactions are used in a wide variety of applications to generate electricity and to plate metals.

Example of redox reaction

Objective 4: Assign oxidation states to the elements in a compound or ion.

Oxidation States for Elements

Examples:

Oxidation States for monoatomic ions

Examples:

Oxidation States for Ionic Compounds

Examples:

Oxidation States for Covalently Bonded Compounds

Examples:

Oxidation States for covalently bonded polyatomic ions

Peroxides and superoxides

Ionic compounds with polyatomic ions

Objective 4 Practice:

Objective 1: Identify oxidation-reduction (redox) reactions

Objective 2: Define the terms oxidation and reduction in terms of loss or gain of electrons and impact on oxidation state

Objective 3: Identify which element is oxidized and which element is reduced in a redox reaction

Objective 6: Identify oxidizing and reducing agents in redox reactions.

Zinc is oxidized

Hydrogen ion is reduced

Oxidizing and Reducing Agents

H+ is the oxidizing agent

Zinc is the reducing agent

Key things to remember when identifying oxidizing and reducing agents

Mnemonics

Objective 1,2,3,6 Practice

Half Reactions

Half reactions example

Objective 1,2,3,6 Practice

Objective 7: Balance oxidation-reduction reactions using the half-reaction method in both acidic and basic solutions.

Half reaction method (Acidic Solution)

Half reaction method (Basic Solution)

Example — Half reaction method

Objective 5: Predict the products for the single replacement reaction of a metal with an acid, water, or salt and use the activity series to predict whether a reaction will occur.

How the single replacement reactions relate to redox chemistry

What’s next?

H⁺ is the oxidizing agent