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Oxidation number

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Intros
Lessons
  1. What is the oxidation number?
  2. Identifying reduction and oxidation
  3. Rules for oxidation numbers.
  4. Assigning oxidation numbers in reactions.
  5. What has been reduced and oxidized in the reaction?
  6. Redox and disproportionation reactions
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Examples
Lessons
  1. Find the oxidation numbers of both elements in the following compounds.
    Find the oxidation numbers of each atom in the following chemical compounds.
    1. H2O2
    2. MgH2
    3. Na2O
    4. NCl3
  2. Recognise and distinguish regular redox reactions from disproportionation reactions.
    1. H2 + O2 \enspace → \enspace 2H2O
      2H2O2 \enspace → \enspace 2H2O + O2
      H2 + Cl2 \enspace → \enspace 2HCl
Topic Notes
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Introduction to Oxidation Numbers

Welcome to our exploration of oxidation numbers, a crucial concept in understanding redox reactions! Oxidation numbers are essential tools that help us track electron movement in chemical reactions. They provide a way to assign charges to atoms in molecules or ions, even when actual charges aren't present. In redox reactions, these numbers allow us to identify which species are being oxidized (losing electrons) and which are being reduced (gaining electrons). Our introduction video is designed to make this concept crystal clear, walking you through the basics and providing real-world examples. By watching this video, you'll gain a solid foundation in assigning oxidation numbers and recognizing their patterns in various compounds. This knowledge is invaluable for tackling more complex redox problems and understanding electron transfer in chemical reactions. So, let's dive in and demystify oxidation numbers together!

Understanding Oxidation Numbers and Basic Rules

Oxidation numbers, also known as oxidation states, are a fundamental concept in chemistry that helps us understand the behavior of atoms in chemical reactions and compounds. These numbers represent the degree of oxidation of an atom in a chemical compound, indicating the number of electrons an atom has gained or lost relative to its neutral state. The significance of oxidation numbers lies in their ability to track electron transfer in redox reactions, predict the products of chemical reactions, and balance complex redox equations.

To assign oxidation numbers correctly, chemists follow a set of basic rules. Let's explore these rules and their applications:

1. Elemental Form Rule: Atoms in their elemental form always have an oxidation number of zero. This applies to both monoatomic and diatomic elements. For example, Na, Fe, O2, and Cl2 all have an oxidation number of 0 in their elemental form.

2. Oxygen Oxidation Number Rule: In most compounds, oxygen has an oxidation number of -2. This rule is based on oxygen's high electronegativity, which allows it to attract electrons from other elements. However, there are exceptions to this rule:
- In peroxides (e.g., H2O2), oxygen has an oxidation number of -1.
- In compounds with fluorine (e.g., OF2), oxygen can have a positive oxidation number due to fluorine's higher electronegativity.

3. Halogen Oxidation Number Rule: Halogens (F, Cl, Br, I) typically have an oxidation number of -1 in compounds. This is due to their tendency to gain one electron to achieve a stable electron configuration. However, when halogens bond with oxygen or other more electronegative elements, they can have positive oxidation numbers. For instance, in ClO- (hypochlorite ion), chlorine has an oxidation number of +1.

4. Hydrogen Oxidation Number Rule: In most compounds, hydrogen has an oxidation number of +1. This is because hydrogen tends to lose its single electron when bonding with other elements. An important exception is when hydrogen bonds with metals in metal hydrides (e.g., NaH), where it has an oxidation number of -1.

5. Sum of Oxidation Numbers Rule: In a neutral compound, the sum of all oxidation numbers must equal zero. For ions, the sum must equal the charge of the ion. This rule is crucial for verifying the correctness of assigned oxidation numbers and balancing redox equations.

Let's apply these rules to some examples:

Example 1: H2O (water)
- Oxygen follows rule 2: oxidation number = -2
- Hydrogen follows rule 4: oxidation number = +1
- Applying rule 5: 2(+1) + (-2) = 0 (neutral compound)

Example 2: KMnO4 (potassium permanganate)
- K is in group 1, so its oxidation number is +1
- Oxygen follows rule 2: 4 × (-2) = -8
- Applying rule 5 for a neutral compound: +1 + x + (-8) = 0, where x is Mn's oxidation number
- Solving for x: x = +7 (Mn's oxidation number)

Example 3: PCl5 (phosphorus pentachloride)
- Chlorine follows rule 3: 5 × (-1) = -5
- Applying rule 5: x + (-5) = 0, where x is P's oxidation number
- Solving for x: x = +5 (P's oxidation number)

Understanding and applying these rules for assigning oxidation numbers is crucial in various aspects of chemistry. They help in identifying oxidizing and reducing agents, predicting the products of redox reactions, and balancing complex chemical equations. By mastering the concept of oxidation numbers, chemists

Calculating Oxidation Numbers in Compounds and Ions

Understanding how to find the oxidation number of elements in various compounds and ions is a crucial skill in chemistry. Let's dive into this topic with some step-by-step examples and explanations to help you master this concept.

First, let's review the basic rules for assigning oxidation numbers:

  1. The oxidation number of a free element is always 0.
  2. For monatomic ions, the oxidation number equals the charge on the ion.
  3. The sum of oxidation numbers in a neutral compound is 0.
  4. In polyatomic ions, the sum of oxidation numbers equals the charge on the ion.
  5. Some elements have fixed oxidation numbers in most compounds (e.g., Group 1 metals are +1, Group 2 are +2).

Now, let's apply these rules to some common molecules:

Example 1: H2O (Water)

To find the oxidation number of O in H2O:

  1. We know H typically has an oxidation number of +1 in compounds.
  2. The compound is neutral, so the sum of oxidation numbers must be 0.
  3. Let x be the oxidation number of O.
  4. Equation: (+1) + (+1) + x = 0
  5. Solving: 2 + x = 0, so x = -2

Therefore, the oxidation number of O in H2O is -2.

Example 2: CO2 (Carbon Dioxide)

To determine the oxidation number of C in CO2:

  1. We know the oxidation number of O2 is typically -2 in compounds.
  2. The compound is neutral, so the sum must be 0.
  3. Let x be the oxidation number of C.
  4. Equation: x + (-2) + (-2) = 0
  5. Solving: x - 4 = 0, so x = +4

Thus, the oxidation number of C in CO2 is +4.

Example 3: NO3- (Nitrate Ion)

To find the oxidation number of N in NO3-:

  1. We know O typically has an oxidation number of -2.
  2. The ion has a -1 charge, so the sum of oxidation numbers must equal -1.
  3. Let x be the oxidation number of N.
  4. Equation: x + 3(-2) = -1
  5. Solving: x - 6 = -1, so x = +5

Therefore, the oxidation number of N in NO3- is +5.

When dealing with more complex compounds or ions, it's helpful to break them down into smaller parts. For instance, in the compound NaHCO3 (sodium bicarbonate):

  1. Na has an oxidation number of +1 (Group 1 metal)
  2. H is typically +1
  3. O is usually -2
  4. Let x be the oxidation number of C
  5. Equation: (+1) + (+1) + x + 3(-2) = 0
  6. Solving: 2 + x - 6 = 0, so x = +4

It's important to note that some elements, like carbon, can have variable oxidation numbers.

Oxidation Numbers in Redox Reactions

Understanding redox reactions is crucial in chemistry, and oxidation numbers play a pivotal role in identifying oxidation and reduction processes. Let's dive into how these numbers help us decipher the complex world of electron transfer in chemical reactions.

Oxidation numbers, also known as oxidation states, represent the degree of oxidation of an atom in a chemical compound. They are essential tools in tracking electron movement during redox reactions. By assigning these numbers to atoms in a reaction, we can easily identify which species are being oxidized (losing electrons) and which are being reduced (gaining electrons).

To demonstrate how oxidation numbers work in practice, let's consider the oxidation of carbon monoxide (CO) by oxygen (O) to form carbon dioxide (CO). This reaction is not only important in understanding air pollution but also serves as an excellent example of how oxidation numbers change during a redox reaction.

The reaction can be written as: 2CO + O 2CO

Now, let's assign oxidation numbers to each atom:

In CO: Carbon has an oxidation number of +2, and oxygen has -2.
In O: Each oxygen atom has an oxidation number of 0.
In CO: Carbon has an oxidation number of +4, and each oxygen has -2.

Tracking the change in oxidation number for carbon, we see it increases from +2 to +4. This increase indicates that carbon is being oxidized. Conversely, oxygen in O starts with an oxidation number of 0 and ends up at -2 in CO, showing that it's being reduced.

This example clearly illustrates how the carbon monoxide oxidation number changes during the reaction, helping us identify the oxidation and reduction processes. The change in oxidation number is a key indicator of electron transfer in redox reactions.

In this reaction, oxygen acts as the oxidizing agent. An oxidizing agent is a species that causes oxidation in other substances by accepting electrons. It gets reduced in the process. Here, oxygen accepts electrons from carbon monoxide, causing the carbon to be oxidized while itself being reduced.

On the flip side, carbon monoxide serves as the reducing agent in this reaction. A reducing agent is a substance that causes reduction in other species by donating electrons. It gets oxidized in the process. In our example, CO donates electrons to oxygen, reducing it while being oxidized itself.

Understanding these concepts is crucial in various fields, from environmental science to materials engineering. For instance, in catalytic converters used in automobiles, similar redox reactions are employed to convert harmful carbon monoxide into less harmful carbon dioxide.

It's important to note that while the oxidation number system is incredibly useful, it's a theoretical concept. In reality, electrons are not always transferred completely between atoms. However, this system provides a practical way to keep track of electron movement in chemical reactions.

As you delve deeper into chemistry, you'll encounter more complex redox reactions. The principles we've discussed here will serve as a solid foundation for understanding these reactions. Remember, practice is key in mastering the art of assigning and tracking oxidation numbers.

In conclusion, oxidation numbers are invaluable tools in identifying oxidation and reduction in chemical reactions. By tracking changes in these numbers, we can easily determine which species are oxidizing agents and which are reducing agents. This knowledge is fundamental in understanding a wide range of chemical processes, from simple lab experiments to complex industrial applications. As you continue your chemistry journey, keep these concepts in mind, and you'll find that the world of redox reactions becomes much clearer and more fascinating.

Special Cases: Variable Oxidation States and Disproportionation

When it comes to oxidation numbers, there are some fascinating special cases that we need to explore. Let's dive into the world of variable oxidation states and disproportionation reactions, shall we? These concepts are crucial for understanding more complex chemical processes and the behavior of certain elements.

First, let's talk about variable oxidation states. You might have noticed that some elements, particularly transition metals, can have multiple oxidation states. This means they can lose or gain different numbers of electrons depending on the compound they're in. For example, iron can have oxidation states of +2 or +3, while manganese can range from +2 to +7! This variability makes transition metals incredibly versatile in chemical reactions and biological processes.

Why do transition metals exhibit this behavior? It's due to their electronic configuration. They have partially filled d-orbitals, which allow them to lose or gain electrons more easily than other elements. This property is what makes them so important in catalysis and many industrial applications.

Now, let's move on to an intriguing phenomenon called disproportionation. In a disproportionation reaction, a single element in one oxidation state is simultaneously oxidized and reduced, forming two different oxidation states. It's like the element is playing both roles in a redox reaction!

A classic example of disproportionation is the decomposition of hydrogen peroxide. Hydrogen peroxide (H2O2) is an interesting compound where the oxidation state of oxygen is -1. This is unusual because oxygen typically has an oxidation state of -2 in most compounds. When hydrogen peroxide decomposes, something remarkable happens:

2 H2O2 2 H2O + O2

In this reaction, the oxygen in hydrogen peroxide (with an oxidation state of -1) splits into two different forms. Some of it is reduced to water, where oxygen has its usual -2 oxidation state. At the same time, some of it is oxidized to molecular oxygen (O2), where the oxidation state is 0. This simultaneous oxidation and reduction of the same element is the hallmark of a disproportionation reaction.

Disproportionation reactions are not limited to hydrogen peroxide. They occur in various systems and play important roles in chemistry and biochemistry. For instance, the disproportionation of chlorine in water produces hypochlorite and chloride ions, a reaction crucial for water treatment.

Understanding these special cases is essential for grasping the full complexity of oxidation-reduction chemistry. Variable oxidation states, particularly in transition metals, allow for a wide range of chemical behaviors and reactions. This variability is what makes transition metals so important in fields like catalysis, where they can facilitate reactions by easily changing their oxidation states.

As for disproportionation, it's a reminder that chemical reactions aren't always straightforward. Sometimes, an element can act as both an oxidizing and reducing agent in the same reaction. This concept challenges our understanding of redox reactions and showcases the dynamic nature of electron transfer in chemistry.

In conclusion, special cases like variable oxidation states and disproportionation reactions add depth and complexity to our understanding of oxidation numbers. They demonstrate that chemistry is full of surprises and that elements can behave in unexpected ways. By mastering these concepts, you'll be better equipped to understand and predict chemical behavior in various systems. Remember, whether you're dealing with the variable oxidation states of transition metals or the disproportionation of hydrogen peroxide, these special cases are windows into the fascinating world of electron transfer and chemical reactivity.

Practice Problems and Common Mistakes

Ready to put your oxidation number skills to the test? Let's dive into some practice problems and explore common mistakes students often make when assigning oxidation numbers. Don't worry if you find some challenging we'll guide you through the process and help you avoid pitfalls along the way!

Practice Problem 1: Determine the oxidation number of sulfur in H2SO4.

Practice Problem 2: What is the oxidation number of chromium in K2Cr2O7?

Practice Problem 3: Assign oxidation numbers to all elements in the reaction: Cu + 2AgNO3 Cu(NO3)2 + 2Ag

Practice Problem 4: Identify the oxidation number of nitrogen in NH4NO3.

Practice Problem 5: Determine the oxidation number of manganese in KMnO4.

Now, let's discuss some common mistakes students make when assigning oxidation numbers and how to avoid them:

1. Forgetting the sum rule: Remember, the sum of all oxidation numbers in a neutral compound must equal zero. For ions, the sum should equal the charge of the ion. Always double-check your calculations using this rule.

2. Misidentifying polyatomic ions: When dealing with compounds containing polyatomic ions like sulfate (SO4^2-) or nitrate (NO3^-), treat the entire ion as a unit. The oxidation number of the entire polyatomic ion is its charge.

3. Overlooking the oxidation number of oxygen: While oxygen typically has an oxidation number of -2, there are exceptions. In peroxides (O2^2-), oxygen has an oxidation number of -1. In OF2, it's +2. Always consider the compound's context.

4. Confusion with transition metals: Transition metals can have multiple oxidation states. To determine the correct oxidation number, consider the overall charge of the compound and the oxidation numbers of other elements present.

5. Mishandling fractional oxidation numbers: In some cases, you may encounter fractional oxidation numbers. Don't round these off they're perfectly valid and necessary for maintaining the sum rule.

To improve your skills in assigning oxidation numbers, try these tips:

1. Start with the elements that have fixed oxidation numbers, such as Group 1 (+1), Group 2 (+2), and fluorine (-1).

2. Use the sum rule to your advantage. If you know the oxidation numbers of all elements except one, you can easily calculate the unknown.

3. Practice with a variety of compounds and reactions. The more diverse examples you encounter, the better you'll become at recognizing patterns and exceptions.

4. When in doubt, write out the full electron configuration of the element in question. This can help you visualize how many electrons are gained or lost.

5. Remember that oxidation numbers are a theoretical concept used to track electron movement. They don't always represent actual charges on atoms.

As you work through oxidation number examples, you'll develop a better intuition for how to tell oxidation numbers quickly and accurately. Don't get discouraged if you make mistakes they're an essential part of the learning process!

Here are some additional practice problems to further hone your skills:

Practice Problem 6: Determine the oxidation number of phosphorus in H3PO4.

Practice Problem 7: What is the oxidation number of iron in Fe2O3?

Practice Problem 8: Assign oxidation numbers to all elements in the reaction: Zn + 2HCl ZnCl2 + H2

Practice Problem 9: Identify the oxidation number of sulfur in Na2S2O3.

Practice Problem 10: Determine the oxidation number of chlorine in NaClO.

Remember, the key

Conclusion and Further Study

In this article, we've explored the crucial concept of oxidation numbers in redox reactions. We've covered the rules for assigning oxidation numbers, their importance in understanding electron transfer, and how they relate to oxidation and reduction processes. The introductory video provided a solid foundation for grasping these concepts, making it an essential starting point for your learning journey. Remember, mastering oxidation numbers is key to excelling in redox reactions. To reinforce your understanding, we encourage you to practice more problems and explore related topics. Why not try calculating oxidation numbers for complex molecules or delve deeper into redox reactions? The more you engage with these concepts, the more comfortable you'll become. Don't hesitate to revisit the video or seek additional resources if you need a refresher. Keep up the great work, and soon you'll be a pro at handling oxidation numbers in various chemical contexts!

Understanding Oxidation Numbers

What is the oxidation number? Identifying reduction and oxidation

Step 1: Introduction to Oxidation Numbers

Oxidation numbers, also known as oxidation states, are a concept used in chemistry to indicate the degree of oxidation of an atom in a chemical compound. They are essential in understanding redox (reduction-oxidation) reactions, which are reactions where the transfer of electrons occurs between chemical species. In a redox reaction, one species is oxidized (loses electrons) and another is reduced (gains electrons).

Step 2: Objectives of Learning Oxidation Numbers

The primary objectives of learning about oxidation numbers include:

  • Recalling the oxidation states of common and predictable elements.
  • Using oxidation numbers to determine the oxidation state of atoms in chemical reactions.
  • Identifying which atoms are oxidized and which are reduced in a redox reaction.
  • Determining the reducing and oxidizing agents in a reaction.

Step 3: Understanding Redox Reactions

Redox reactions involve two simultaneous processes: reduction and oxidation. Reduction is the gain of electrons, which decreases the oxidation number of an atom. Oxidation is the loss of electrons, which increases the oxidation number of an atom. These processes are complementary and must occur together in a redox reaction. Identifying the changes in oxidation numbers helps in understanding which species are oxidized and which are reduced.

Step 4: Finding Oxidation Numbers

To identify which species are oxidized and which are reduced, it is crucial to determine the oxidation numbers of the atoms involved in the reaction. The oxidation number is a theoretical charge that an atom would have if all bonds to atoms of different elements were 100% ionic. By calculating the oxidation numbers, we can track the transfer of electrons in the reaction.

Step 5: Using Oxidation Numbers to Identify Reducing and Oxidizing Agents

Once the oxidation numbers are determined, we can identify the reducing and oxidizing agents in the reaction. The reducing agent is the species that donates electrons (is oxidized), and the oxidizing agent is the species that accepts electrons (is reduced). Understanding the role of these agents is crucial in analyzing redox reactions.

Step 6: General Rules for Assigning Oxidation Numbers

There are several general rules for assigning oxidation numbers:

  • The oxidation number of an element in its elemental form is zero.
  • The oxidation number of a monatomic ion is equal to its charge.
  • Oxygen usually has an oxidation number of -2, except in peroxides where it is -1.
  • Hydrogen has an oxidation number of +1 when bonded to non-metals and -1 when bonded to metals.
  • Halogens usually have an oxidation number of -1, except when bonded to oxygen or other halogens.

Step 7: Applying the Rules to Predict Oxidation States

By applying these general rules, we can predict the oxidation states of atoms in more complex compounds. This helps in understanding the electron transfer process in redox reactions. For example, in water (H2O), oxygen has an oxidation number of -2, and each hydrogen has an oxidation number of +1.

Step 8: Practical Examples and Exercises

To solidify the understanding of oxidation numbers, it is beneficial to work through practical examples and exercises. This involves identifying the oxidation numbers in various compounds and reactions, determining which species are oxidized and reduced, and identifying the reducing and oxidizing agents.

Conclusion

Understanding oxidation numbers is fundamental in the study of redox reactions and electrochemistry. By mastering the rules and concepts associated with oxidation numbers, one can accurately analyze and predict the behavior of chemical species in redox reactions.

FAQs

Here are some frequently asked questions about oxidation numbers:

1. Why is the oxidation number of oxygen usually -2?

Oxygen typically has an oxidation number of -2 because of its high electronegativity. In most compounds, oxygen attracts electrons strongly, gaining two electrons to achieve a stable electron configuration. However, there are exceptions, such as in peroxides (where it's -1) and compounds with fluorine (where it can be positive).

2. What is the oxidation number of oxygen in O2?

In O2 (molecular oxygen), the oxidation number of each oxygen atom is 0. This is because O2 is an element in its free state, and all elements in their elemental form have an oxidation number of 0.

3. How do you find the oxidation number of an element in a compound?

To find the oxidation number of an element in a compound: 1. Use known oxidation numbers for other elements. 2. Remember that the sum of oxidation numbers in a neutral compound is 0. 3. For ions, the sum equals the charge of the ion. 4. Solve for the unknown oxidation number using these rules.

4. Can oxygen have a positive oxidation number?

Yes, oxygen can have a positive oxidation number in rare cases. For example, in oxygen difluoride (OF2), oxygen has an oxidation number of +2 because fluorine is more electronegative than oxygen.

5. What is the oxidation number of carbon in carbon monoxide (CO)?

In carbon monoxide (CO), the oxidation number of carbon is +2. This can be calculated by knowing that oxygen typically has an oxidation number of -2 in compounds, and the sum of oxidation numbers in a neutral molecule must be 0. Therefore, carbon must have an oxidation number of +2 to balance out the -2 of oxygen.

Prerequisite Topics for Understanding Oxidation Number

To fully grasp the concept of oxidation number in chemistry, it's crucial to have a solid foundation in several prerequisite topics. Understanding these fundamental concepts will greatly enhance your ability to work with oxidation numbers and apply them in various chemical contexts.

One of the key prerequisites is an introduction to chemical reactions. This foundational knowledge is essential because oxidation numbers play a vital role in describing electron transfer in chemical reactions. By understanding how atoms interact and form bonds, you'll be better equipped to interpret and assign oxidation numbers to elements in different compounds.

Another critical prerequisite topic is balancing redox equations. Oxidation numbers are integral to redox reactions, where electrons are transferred between species. Mastering the art of balancing redox equations will significantly improve your ability to work with oxidation numbers and understand their practical applications in chemistry.

Additionally, familiarity with calculating cell potential in voltaic cells is highly beneficial. This knowledge helps in predicting redox reaction products and understanding the relationship between oxidation numbers and electrochemical processes. By grasping the concepts of cell potential, you'll gain insights into how oxidation numbers influence the direction and spontaneity of redox reactions.

The importance of these prerequisite topics cannot be overstated when it comes to mastering oxidation numbers. They provide the necessary context and theoretical framework for understanding how oxidation numbers are assigned, how they change during reactions, and their significance in various chemical processes.

For instance, when you're introduced to chemical reactions, you learn about the transfer of electrons between atoms. This fundamental concept directly relates to oxidation numbers, as they represent the degree of oxidation of an atom in a compound. Similarly, balancing redox equations requires a deep understanding of how oxidation numbers change during a reaction, making it an essential skill for working with more complex oxidation-reduction processes.

Furthermore, the ability to calculate cell potential and predict redox reaction products demonstrates the practical application of oxidation numbers in electrochemistry. This knowledge allows you to understand how oxidation numbers relate to the flow of electrons in electrochemical cells and how they can be used to determine the feasibility of redox reactions.

By building a strong foundation in these prerequisite topics, you'll be well-prepared to tackle the complexities of oxidation numbers. You'll find it easier to assign oxidation states, understand their changes in chemical reactions, and apply this knowledge to solve real-world chemistry problems. Remember, a solid grasp of these fundamentals will not only help you in understanding oxidation numbers but will also enhance your overall comprehension of chemistry and its various interconnected concepts.

In this lesson, we will learn:

  • To recall the oxidation states of common elements.
  • How to find the oxidation number of atoms in full equations.
  • How to use oxidation numbers to show what has been reduced or oxidized in reactions.
  • To identify disproportionation reactions, using oxidation numbers to do so.

Notes:

  • Redox reactions will always have a chemical that is being oxidized and a chemical that is being reduced. You need to find the oxidation number of the atoms in the reaction to see what is being oxidized or reduced.
    The oxidation number is an assigned number that shows the number of electrons gained or lost by an atom compared to its original outer shell configuration.
    There are some key rules of oxidation numbers that help in most situations:
    • ANY ATOM in elemental form always has an oxidation state of zero.
    • Oxygen atoms in a compound have an oxidation state of -2.
    • The oxidation state of ions or atoms in a molecule will add up to the overall charge.
    • Halogen atoms in a compound have an oxidation state of -1.
    • Hydrogen atoms have an oxidation state of +1 unless they are bonded to a metal atom, where they will be -1. Compounds with the formula MHn, where M is a metal, are called hydrides.
    • These oxidation numbers are often written using Roman numerals, especially when writing outside of a chemical equation. Make sure you know Roman numerals for the numbers 1 through 8!

  • Because oxidation number is related to electrons being lost or gained, you can make general predictions on oxidation numbers based on if an element is a metal or non-metal.
    • Non-metals are more electronegative than metals, so in general they will form negative ions in reactions through gaining electron(s) which decreases their oxidation number.
    • Metals are less electronegative than non-metals, so they generally form positive ions in reactions by losing electron(s) which will increase their oxidation number.

  • Just like charge of molecules/ions, the overall change in oxidation state during a reaction will be zero. Because the changes are caused by electrons being lost or gained, any positive change in oxidation state will be mirrored by a negative change in another reactant.

  • WORKED EXAMPLE: Using the rules of oxidation numbers, it is possible to predict the oxidation state of other less predictable elements in a redox reaction. For example:

    Fe2O3 (s) + 3 CO (g)  \, →  \, 2 Fe (s) + 3 CO2 (g)

    Using the rules above, the oxidation numbers of iron and carbon in the reactants and products can be found.
    • Oxygen atoms have an oxidation number of -2 so three of them in Fe2O3 is a total of -6, and the compound is neutral, so the two iron atoms must, combined, have an oxidation number of +6. There are two of them, so Fe in Fe2O3 has an oxidation number of +3.
    • Oxygen has an oxidation number of -2 and there is only one of them in CO, which is neutral, so the C atom in CO must have an oxidation number of +2 to counteract this to bring it back to zero.
    • Fe in the products is elemental iron so it has oxidation number 0. Atoms in any elemental sample will have oxidation number 0.
    • Oxygen has an oxidation number of -2 and there are two O atoms in CO2, which is -4 in total due to the oxygen. The molecule is neutral, so the one C atom in CO2 must have a +4 oxidation number to balance the molecule out.

  • Working out the oxidation numbers in the products and reactants lets you work out the change in oxidation state of the atoms in the reaction – remember, the overall change will be zero.
    • Iron has changed from +3 oxidation state in Fe2O3 to 0 in its elemental form. Therefore iron has been reduced (△\triangleOS = -3) in this reaction. The reaction with carbon monoxide is what caused this change, so carbon monoxide is the reducing agent in this process.
    • Carbon has changed from +2 oxidation state in the CO reactant to +4 in the CO2 products. Therefore carbon has been oxidized (△\triangleOS = +2) in this reaction. The reaction with iron oxide caused this, so iron is the oxidizing agent, which was itself reduced in the reaction.

    There were two iron atoms in the reaction which is a total change in OS of -6. There were three carbon atoms in the reaction which is a total change in OS of +6.
    These changes in oxidation state balance to zero.

  • Disproportionation reactions are a rare but important type of redox reaction where identical atoms in a chemical compound undergo both reduction and oxidation. One of the most common disproportionation reactions is the decomposition of hydrogen peroxide:

    2H2O2\enspace → \enspace 2H2O2 + O2
    Oxidation states: \quad \qquad -1 \qquad \quad \enspace -2 \quad 0 \qquad \qquad \qquad \qquad \enspace \enspace

    In the reactants, hydrogen peroxide has its oxygen atoms in an oxidation state of -1. This is because the oxygen-oxygen bond does not ‘assign’ electrons to either atom. Only the oxygen-hydrogen bonds do that, where one on each oxygen atom gives them an oxidation state of -1 each.
    In the products:
    • One oxygen atom went from a peroxide molecule to elemental oxygen. This takes it from an oxidation state of -1 to 0. This oxygen atom was oxidized.
    • The other oxygen atom went from peroxide to a water molecule. This oxygen atom went from an oxidation state of -1 to -2 and was therefore reduced.
    Disproportionation reactions require an atom to be able to take three different oxidation states: 
    • The reactants will have an atom in a middle oxidation state, where both the reduced and oxidized atoms start. For oxygen this is -1 when it is bonded to another oxygen atom.
    • One higher oxidation state for the oxidized atom. For oxygen, this is 0 in elemental oxygen, O2 (g).
    • One lower oxidation state for the reduced atom. For oxygen, this is -2 such as in H2O.