Oxidation-Reduction Reactions

Many of the most important reactions in chemistry, from the batteries that power our phones to the process of metabolism, involve the transfer of electrons from one substance to another. These are known as oxidation-reduction or redox reactions.

The process is defined by two complementary actions:

Oxidation is the loss of electrons.
Reduction is the gain of electrons.

A classic mnemonic to remember this is OIL RIG: Oxidation Is Losing, Reduction Is Gaining.

Crucially, oxidation and reduction are a matched pair; one cannot happen without the other. The substance that loses electrons is called the reducing agent (as it causes another substance to be reduced), and the substance that gains electrons is called the oxidizing agent.

The Problem: Tracking Electrons in Covalent Bonds

For simple ionic reactions, it’s easy to see the electron transfer. When sodium metal reacts with chlorine gas to form NaCl, the sodium atom clearly loses an electron (it is oxidized) and the chlorine atom clearly gains one (it is reduced).

But what about reactions between nonmetals, like the formation of carbon dioxide from carbon and oxygen? \[ \mathrm{C(s) + O_2(g) \longrightarrow CO_2(g)} \] The bonds in CO₂ are covalent, meaning the electrons are shared, not fully transferred. How can we talk about “losing” and “gaining” in this context?

The Solution: Oxidation Numbers

To handle this, chemists developed a powerful bookkeeping tool called the oxidation state, or oxidation number. An oxidation number is the hypothetical charge an atom would have if all its bonds to other atoms were treated as if they were 100% ionic.

This concept is based on the idea of unequal sharing. In a covalent bond between two different atoms, the more electronegative atom “pulls” the shared electrons closer to itself. We can formalize this by imagining the bond breaks and the more electronegative atom gets all the electrons from that bond. This process is called heterolytic cleavage.

In heterolytic cleavage, the more electronegative atom (Cl) takes both shared electrons, resulting in ions.

This is in contrast to a bond between two identical atoms, like in H₂, where the electrons are shared perfectly equally. If this bond were to break, each atom would take one electron back, a process called homolytic cleavage. In this case, no charge develops.

In homolytic cleavage, each identical atom retains one electron.

The oxidation number, therefore, is a formal charge assigned to an atom based on this hypothetical “winner-takes-all” scenario for electrons. It allows us to track the flow of electrons in any reaction, whether the bonds are ionic or covalent.

If an atom’s oxidation number increases during a reaction, it has been oxidized (it has “lost” electron density).
If an atom’s oxidation number decreases, it has been reduced (it has “gained” electron density).

To determine which species is oxidized and which is reduced, we must first learn the set of rules for assigning oxidation numbers.

A Note on Charge

It’s easy to confuse the different types of charge we use for bookkeeping. Here’s a quick summary:

Ionic Charge: A real, measurable charge on an ion (e.g., the charge on Na⁺ is +1).
Oxidation State: A hypothetical charge assuming all bonds are 100% ionic. It is excellent for tracking electrons in redox.
Formal Charge: A hypothetical charge assuming all bonds are 100% covalent. It is best for evaluating Lewis structures.

Rules for Assigning Oxidation Numbers

Assigning oxidation numbers follows a specific set of rules. We can group these into two categories. The first set contains four foundational rules that are always true and provide the basis for all our calculations.

Foundational Rules

The oxidation number of an atom in its pure, elemental form is always 0.
- Examples: An atom in a piece of solid zinc, Zn(s), has an oxidation number of 0. Each hydrogen atom in a molecule of hydrogen gas, H₂(g), also has an oxidation number of 0.
The oxidation number of a monatomic ion is equal to the charge of the ion.
- Examples: For Na⁺, the oxidation number of the sodium atom is +1. For Cl⁻, the oxidation number of the chlorine atom is −1.
The sum of the oxidation numbers of all atoms in a polyatomic ion is equal to the overall charge of the ion.
- Example: In the hydroxide ion (OH⁻), oxygen has an oxidation number of −2 and hydrogen has +1. The sum, (−2) + (+1) = −1, equals the ion’s charge.
The sum of the oxidation numbers of all atoms in a neutral compound is 0.
- Examples: In H₂O, the two hydrogen atoms (+1 each) and one oxygen atom (−2) sum to 0. In NaCl, the sodium atom (+1) and chlorine atom (−1) sum to 0.

Hierarchical Rules

For atoms within a compound or polyatomic ion, we assign oxidation numbers using a set of rules that are applied in a priority order. You should start at the top of the following list and work your way down. The first rule that applies takes precedence over any rules below it.

Table 1: Hierarchical rules for assigning oxidation numbers

Notation: Oxidation State vs. Ionic Charge

It is crucial to distinguish between the notation for an oxidation state and a real ionic charge.

Ionic Charge

Ionic charge is a measurable physical property. The convention is to write the numeral first, followed by the sign as a superscript.

Charge of an iron(II) ion: Fe²⁺
Charge of a chloride ion: Cl⁻

Oxidation State

An oxidation state is a theoretical “bookkeeping” number. The notation depends on the context.

1. In a Name (Stock System): The most common method, where the oxidation state is a Roman numeral in parentheses.

For FeCl₂: iron(II) chloride

2. On a Chemical Formula (Formal IUPAC Notation): The most formal way to show an oxidation state on a formula is with a superscript Roman numeral.

For FeCl₂: Fe^IICl^−I₂

While this is the formal rule, it is rarely used in general chemistry textbooks due to its complexity.

3. For Calculation and Discussion (the Common Convention): When assigning and working with oxidation states, the universal convention is to write the sign first, followed by the numeral.

Oxidation state of iron in FeCl₂: +2
Oxidation state of chlorine in FeCl₂: −1

For clarity, this book will use the common sign-then-numeral convention when discussing the value of an oxidation state.

Example: Identifying the Oxidizing and Reducing Agents

Let’s walk through the process of assigning oxidation numbers and identifying the oxidizing and reducing agents in a classic single-displacement reaction between solid zinc and an aqueous solution of copper(II) chloride.

Step 1: Assign Oxidation Numbers to Each Atom

We start with the balanced molecular equation and apply the rules to assign an oxidation number (O.N) to every atom. The numbers are written directly below the corresponding element.

\[ \begin{array}{lcccccl} \text{} & \mathrm{Zn(s)} & + & \mathrm{CuCl_2(aq)} & \longrightarrow & \mathrm{ZnCl_2(aq)} & + & \mathrm{Cu(s)} \\ \text{} & 0 & & +2 ~-\!\!1~~~~~~~ & & +2~-\!\!1~~~~~~~ & & ~~0 \end{array} \]

Zn(s) and Cu(s) are pure elements, so their oxidation numbers are 0.
In CuCl₂ and ZnCl₂, chlorine is a halide. Since it’s not bonded to oxygen or fluorine, its oxidation number is −1.
To ensure the compounds are neutral, the copper in CuCl₂ and the zinc in ZnCl₂ must each have an oxidation number of +2.

Step 2: Identify Oxidation and Reduction

Next, we track how the oxidation numbers change from the reactant side to the product side to see which element lost electrons (was oxidized) and which gained them (was reduced).

Zinc’s oxidation state increased from 0 to +2. It lost electrons and was oxidized. Therefore, Zn(s) is the reducing agent.
Copper’s oxidation state decreased from +2 to 0. It gained electrons and was reduced. Therefore, CuCl₂(aq) is the oxidizing agent.

Analyze the Net Ionic Equation

This process is often even clearer when we look at the net ionic equation, which removes the spectator ions (in this case, Cl⁻).

Complete Ionic Equation

\[ \begin{array}{lcccccccccc} \text{} & \mathrm{Zn(s)} & + & \mathrm{Cu^{2+}(aq)} & + & \mathrm{2~Cl^-(aq)} & \longrightarrow & \mathrm{Zn^{2+}(aq)} & + & \mathrm{2~Cl^-(aq)} & + & \mathrm{Cu(s)} \\ \text{} & 0 & & +2~ & & -1~ & & +2~ & & -1~ & & 0 \end{array} \]

Net Ionic Equation

\[ \begin{array}{lccccc} \text{} & \mathrm{Zn(s)} & + & \mathrm{Cu^{2+}(aq)} & \longrightarrow & \mathrm{Zn^{2+}(aq)} & + & \mathrm{Cu(s)} \\ \text{} & 0 & & +2 & & +2 & & 0 \end{array} \]

The net ionic equation shows the core of the redox reaction: solid zinc transfers two electrons to a copper(II) ion, becoming a zinc ion itself and producing solid copper.

A Point of Precision: Identifying the Agent

A common point of confusion is identifying the exact chemical species that is the oxidizing or reducing agent. The rule is simple but critical:

While we track the oxidation number of a single element to see the change, the agent is the entire molecule or ion from the reactants that contains that element.

Consider the following redox reaction: \[ \mathrm{2~\color{red}{Fe^{2+}}(aq)} + \mathrm{\color{green}{H_2O_2}(aq)} + \mathrm{2~H^+(aq)} \longrightarrow \mathrm{2~\color{red}{Fe^{3+}}(aq)} + \mathrm{2~H_2O(l)} \]

Let’s analyze the two key elements:

Iron (Fe): The oxidation number of the iron atom increases from +2 to +3. The iron atom is oxidized. The species on the reactant side that contains this iron is the Fe²⁺ ion. Therefore, Fe²⁺ is the reducing agent.
Oxygen (O): The oxidation number of each oxygen atom in H₂O₂ is −1. In the product H₂O, its oxidation number is −2. The oxygen atom is reduced. The species on the reactant side that contains this oxygen is the hydrogen peroxide molecule. Therefore, H₂O₂ is the oxidizing agent.

The Takeaway: When asked to identify an agent, always write the complete formula for the reactant species that “delivers” the element undergoing the change.

In the last example, we identified Zn(s) as the reducing agent and Cu²⁺(aq) as the oxidizing agent by carefully tracking oxidation numbers. This method is always the definitive way to determine the role of each reactant.

As you practice, however, you will start to notice patterns. Certain substances, like permanganate and dichromate ions, appear frequently as powerful oxidizing agents, while others, like pure metals, are common reducing agents. The following table lists some of the most common agents you will encounter in general chemistry and the products they typically form.

Table 2: Common oxidizing and reducing agents and their typical reaction products.

A Special Case: Disproportionation Reactions

In the examples we have seen so far, the oxidizing agent and the reducing agent have been two separate substances. However, in a special class of reactions called disproportionation reactions, a single substance acts as both the oxidizing agent and the reducing agent.

This occurs when an element in an intermediate oxidation state is simultaneously oxidized to a higher state and reduced to a lower state, forming two different products.

Example: Decomposition of Hydrogen Peroxide

Hydrogen peroxide (H₂O₂) is a classic example. In this molecule, oxygen has an intermediate oxidation state of −1. When it decomposes, it forms water and oxygen gas:

\[ \begin{array}{lcccccl} \text{} & \mathrm{2~H_2O_2(aq)} & \longrightarrow & \mathrm{2~H_2O(l)} & + & \mathrm{O_2(g)} \\ \text{O.N. (O):} & -1 & & -2 & & 0 \end{array} \]

Let’s track the change in the oxygen atom’s oxidation state:

−1 → −2 (in H₂O): The oxidation number decreased. This is a reduction.
−1 → 0 (in O₂): The oxidation number increased. This is an oxidation.

Because the oxygen in hydrogen peroxide is both oxidized and reduced, H₂O₂ is both the reducing agent and the oxidizing agent in this reaction.

Example: Chlorine Gas in a Basic Solution

Another common example is the reaction of chlorine gas with sodium hydroxide:

\[ \begin{array}{lcccccl} \text{} & \mathrm{Cl_2(g)} & \text{...} & \longrightarrow & \mathrm{NaCl(aq)} & + & \mathrm{NaClO(aq)} & \text{...}\\ \text{O.N. (Cl):} & 0 & & & -1 & & +1 \end{array} \]

Here, the elemental chlorine (oxidation state of 0) is converted into chloride (oxidation state of −1) and hypochlorite (oxidation state +1). Since the chlorine atom is both oxidized and reduced, Cl₂ is the substance that undergoes disproportionation.

A Deeper Look: Why Is It Still Called “Oxidation”?

Ever wondered why we’re stuck with the word “oxidation” when a reaction might have nothing to do with oxygen? The story behind that name starts with simple studies of rust and fire and leads to the complex chemistry happening in our own bodies.

To understand the full story and see how the definition expanded to include reactions like the conversion of alcohol to vinegar, read our deep dive: A Brief History of Oxidation and Reduction.

Consider the following redox reaction between copper and nitric acid.

\[\mathrm{Cu(s)} + \mathrm{2~NO_3^-(aq)} + \mathrm{4~H_3O^+(aq)} \longrightarrow \mathrm{Cu^{2+}(aq)} + \mathrm{2~NO_2(g)} + \mathrm{6~H_2O(l)}\]

Nitric acid is a strong acid and completely ionizes into hydronium and nitrate. Nitrate becomes nitrogen dioxide and nitrogen is reduced (oxidation state: 5 → 4) whereas copper is oxidized to copper(II). Hydrogen and oxygen experience no change in oxidation state.

Recognizing Redox in Familiar Reaction Types

In an earlier chapter, you learned to classify reactions by their patterns. Now that you have the tool of oxidation numbers, we can re-examine these patterns to see which ones are fundamentally driven by the transfer of electrons.

Single Displacement reactions are always redox reactions. An element (oxidation state 0) always displaces an ion, changing its own oxidation state and that of the ion it displaces.
Synthesis and Decomposition reactions are often redox reactions, especially when a pure element is involved as a reactant or a product. For example, in 2 Na + Cl₂ → 2 NaCl, the elements (state 0) form ions (+1, −1). However, a reaction like CaO + CO₂ → CaCO₃ is a synthesis reaction, but it is not a redox reaction as no oxidation states change.
Double Displacement reactions are never redox reactions. In these reactions, such as precipitation or acid-base neutralizations, ions simply swap partners. The oxidation state of every element remains unchanged throughout the process.

Balancing Redox Reactions by the Half-Reaction Method

When balancing chemical equations, we are taught to conserve mass. In redox reactions, however, there is another crucial quantity that must be conserved: charge. The half-reaction method is a systematic way to ensure that both mass and charge are balanced. In this chapter, we will apply this method to reactions in neutral aqueous solutions. In General Chemistry II, you will learn how to extend this method for more complex reactions that occur in acidic or basic conditions.

Example: Balancing via the Half-Reaction Method

Consider the following net ionic equation:

\[ \mathrm{Cu^+(aq)} + \mathrm{Fe(s)} \longrightarrow \mathrm{Fe^{3+}(aq)} + \mathrm{Cu(s)} \]

At first glance, this might look balanced. There is one copper atom and one iron atom on each side. A closer look at the charges, however, reveals a problem. The reactant side has a total charge of +1, while the product side has a total charge of +3. To fix this, we follow a series of steps.

Step 1: Identify What is Oxidized and Reduced

First, we assign oxidation states to every species to determine where the electrons are moving.

On the reactant side: Cu in Cu⁺ is +1. Fe as a pure element is 0.
On the product side: Fe in Fe³⁺ is +3. Cu as a pure element is 0.

Now we can see the change.

Iron’s oxidation state increased from 0 to +3. A loss of electrons leads to an increase in oxidation state (Oxidation Is Loss).
Copper’s oxidation state decreased from +1 to 0. A gain of electrons leads to a decrease in oxidation state (Reduction Is Gain).

Step 2: Write the Unbalanced Half-Reactions

Now we can split the overall reaction into two separate half-reactions.

\[ \begin{alignat*}{2} &\text{Ox. half-reaction:} \quad &\mathrm{Fe(s)} &\longrightarrow \mathrm{Fe^{3+}(aq)} \\ &\text{Red. half-reaction:} \quad &\mathrm{Cu^+(aq)} &\longrightarrow \mathrm{Cu(s)} \end{alignat*} \]

Step 3: Balance Atoms and Charge in Each Half-Reaction

Next, we balance each half-reaction individually. The atoms are already balanced, so we only need to balance the charge by adding electrons (e⁻).

For the oxidation half-reaction, iron loses 3 electrons. Since oxidation is the loss of electrons, we show them as a product. \[ \text{Oxidation:} \quad \mathrm{Fe(s)} \longrightarrow \mathrm{Fe^{3+}(aq)} + 3~\mathrm{e^- } \] (Check: The charge is now 0 on both sides.)

For the reduction half-reaction, copper gains 1 electron. Since reduction is the gain of electrons, we show them as a reactant. \[ \text{Reduction:} \quad \mathrm{Cu^+(aq)} + \mathrm{e^-} \longrightarrow \mathrm{Cu(s)} \] (Check: The charge is now 0 on both sides.)

Step 4: Equalize the Number of Electrons

A core principle of redox is that the number of electrons lost in the oxidation must equal the number of electrons gained in the reduction. Right now, we have 3 electrons lost but only 1 electron gained.

To fix this, we must find the least common multiple (which is 3) and multiply the half-reactions as needed. The oxidation reaction is fine, but we need to multiply the entire reduction half-reaction by 3.

\[ \begin{alignat*}{2} &\text{Ox. half-reaction:} \quad &\mathrm{Fe(s)} &\longrightarrow \mathrm{Fe^{3+}(aq)} + 3~\mathrm{e^-} \\ &\text{Red. half-reaction:} \quad &3~\mathrm{Cu^+(aq)} + 3~\mathrm{e^-} &\longrightarrow 3~\mathrm{Cu(s)} \end{alignat*} \]

Step 5: Combine the Half-Reactions

With the electrons balanced, we can add the two half-reactions together. The electrons on opposite sides of the arrow will cancel out, just like spectators in a net ionic equation.

\[ \mathrm{Fe(s)} + 3~\mathrm{Cu^+(aq)} + \cancel{3~\mathrm{e^-}} \longrightarrow \mathrm{Fe^{3+}(aq)} + 3~\mathrm{Cu(s)} + \cancel{3~\mathrm{e^-}} \]

This gives us our final, balanced net ionic equation:

\[ 3~\mathrm{Cu^+(aq)} + \mathrm{Fe(s)} \longrightarrow 3~\mathrm{Cu(s)} + \mathrm{Fe^{3+}(aq)} \]

Step 6: Final Check

Always finish by verifying your work.

Mass Balance: Reactants have 3 Cu and 1 Fe. Products have 3 Cu and 1 Fe.
Charge Balance:
- Reactants have a total charge of 3
- Products have a total charge of 3

The equation is now correctly balanced for both mass and charge. Note that a properly balanced reaction will not contain any electrons. They should all cancel out.

Combustion Reactions

Combustion is a familiar process we see in everything from a burning candle to a gas stove. At its core, combustion is a type of redox reaction that involves the rapid combination of a fuel with an oxidant, typically producing heat and light.

While the most common oxidant is oxygen gas (O₂), the defining feature of combustion is a high-temperature, exothermic redox process.

A photograph of a fireplace with logs on fire illustrating a combustion reaction of a fuel (wood) with an oxidant (oxygen in the air).

Combustion of a fuel with an oxidant can produce flames. (Source: Wikipedia)

Typical Combustion: The Role of Oxygen

In general chemistry, most combustion examples involve an organic fuel (like a hydrocarbon) reacting with O₂ gas. When a compound containing carbon and hydrogen burns completely, the products are always carbon dioxide and water.

A familiar example is the combustion of methane, the primary component of natural gas: \[ \mathrm{CH_4(g)} + 2~\mathrm{O_2(g)} \longrightarrow \mathrm{CO_2(g)} + 2~\mathrm{H_2O(g)} \] Let’s prove this is a redox reaction by tracking the oxidation states:

Reactants: In CH₄, H is +1, so C must be −4. In O₂, oxygen is in its elemental form, so its state is 0.
Products: In CO₂, O is −2, so C must be +4. In H₂O, O is −2.

The carbon atom’s oxidation state increased dramatically from −4 to +4 (oxidation). The oxygen atoms’ oxidation state decreased from 0 to −2 (reduction). This confirms that combustion is indeed a redox process.

This principle applies to other fuels as well. The cellular respiration that powers our bodies is a slow, controlled combustion of glucose: \[ \mathrm{C_6H_{12}O_6(s)} + 6~\mathrm{O_2(g)} \longrightarrow 6~\mathrm{CO_2(g)} + 6~\mathrm{H_2O(l)} \] Combustion doesn’t always have to involve carbon. The combustion of ammonia or hydrogen gas are also important industrial redox reactions: \[ 4~\mathrm{NH_3(g)} + 5~\mathrm{O_2(g)} \longrightarrow 4~\mathrm{NO(g)} + 6~\mathrm{H_2O(g)} \] \[ 2~\mathrm{H_2(g)} + \mathrm{O_2(g)} \longrightarrow 2~\mathrm{H_2O(g)} \]

Expanding the Definition: Combustion Without Oxygen

While we associate fire with oxygen, any powerful oxidant can take its place. The defining characteristic of combustion is the rapid, exothermic redox reaction, not the specific oxidant used.

For example, hydrogen gas burns explosively not only with oxygen, but also with fluorine gas, which is an even stronger oxidant: \[ \mathrm{H_2(g)} + \mathrm{F_2(g)} \longrightarrow 2~\mathrm{HF(g)} \] Other powerful oxidants can sustain combustion with various fuels. Chlorine trifluoride and nitrous oxide are two examples used in rocketry: \[ \mathrm{Fe(s)} + 2~\mathrm{ClF_3(g)} \longrightarrow \mathrm{FeF_3(s)} + \mathrm{Cl_2(g)} \] \[ \mathrm{C(s)} + 2~\mathrm{N_2O(g)} \longrightarrow \mathrm{CO_2(g)} + 2~\mathrm{N_2(g)} \] These examples show that the principles of oxidation and reduction are the true foundation of combustion, regardless of the specific fuel or oxidant involved.

Practice

Practice

Write out the oxidation numbers for each atom in the following reaction

\[\mathrm{Zn(s)} + \mathrm{Cl_2(g)} \longrightarrow \mathrm{ZnCl_2(s)}\]

and determine the following

Species oxidized
Species reduced
Oxidizing agent
Reducing agent

Solution

\[\begin{alignat*}{3} & \mathrm{Zn(s)} ~+~ && \mathrm{Cl_2(g)} ~ \longrightarrow ~ && \mathrm{ZnCl_2(s)} \\ & ~(0) && ~(0) && (2)(-1) \end{alignat*}\]

Species oxidized: Zn(s)
Species reduced: Cl₂(g)
Oxidizing agent: Cl₂(g)
Reducing agent: Zn(s)

Practice

For the single displacement reaction shown below

\[\mathrm{Cu(s)} + 2~\mathrm{AgNO_3(aq)} \longrightarrow \mathrm{Cu(NO_3)_2(aq)} + 2~\mathrm{Ag(s)}\]

determine the following:

Species oxidized
Species reduced
Oxidizing agent
Reducing agent

Solution

\[\begin{alignat*}{5} & \mathrm{Cu(s)} &&~+~ 2~\mathrm{AgNO_3(aq)} &&~ \longrightarrow ~ && \mathrm{Cu(NO_3)_2(aq)} &&~+~ 2~\mathrm{Ag(s)} \\ & ~(0) &&~~ (+1)(+5)(-2) && &&~ (+2)(+5)(-2) &&~~ (0) \end{alignat*}\]

Species oxidized: Cu(s)
Species reduced: AgNO₃(aq)
Oxidizing agent: AgNO₃(aq)
Reducing agent: Cu(s)

Practice

Consider the following net ionic equation for a reaction in an acidic solution:

\[\mathrm{MnO_4^-(aq)} + 5~\mathrm{Fe^{2+}(aq)} + 8~\mathrm{H^+(aq)} \longrightarrow \mathrm{Mn^{2+}(aq)} + 5~\mathrm{Fe^{3+}(aq)} + 4~\mathrm{H_2O(l)}\]

Identify the following:

Species oxidized
Species reduced
Oxidizing agent
Reducing agent

Solution

\[\begin{alignat*}{7} & \mathrm{MnO_4^-(aq)} &&~+~ 5~\mathrm{Fe^{2+}(aq)} &&~+~ 8~\mathrm{H^+(aq)} &&~ \longrightarrow ~ && \mathrm{Mn^{2+}(aq)} &&~+~ 5~\mathrm{Fe^{3+}(aq)} &&~+~ 4~\mathrm{H_2O(l)} \\ & ~(+7)(-2) &&~~ (+2) &&~~ (+1) && &&~ (+2) &&~~ (+3) &&~~ (+1)(-2) \end{alignat*}\]

Species oxidized: Fe²⁺(aq)
Species reduced: MnO₄⁻(aq)
Oxidizing agent: MnO₄⁻(aq)
Reducing agent: Fe²⁺(aq)