Lesson 1: Redox Reactions

Part b: Oxidation States

Part a: Oxidation and Reduction
Part b: Oxidation States
Part c: Balancing Redox Reactions
Part d: Corrosion

 

The Big Idea

Oxidation states are the bookkeeping system of electrochemistry. They reveal how electrons transfer during chemical reactions and help us to identify which substances are oxidized and reduced.

 
 

Why Do We Need Oxidation States?

Oxidation and reduction was introduced in Lesson 1a using two simple examples.

Mg(s)  +  2 H⁺(aq)    →    Mg²⁺(aq)  +  H₂(g)
 
2 Na(s)  +  Cl₂(g)    →    2 NaCl(s)

In both cases, the metal (Mg and Na) was oxidized. The metal lost electrons to become an ion (Mg²⁺ and Na⁺). It was a relatively easy task determining that these metals lost electrons since they were present as neutral atoms on the reactant side of the equation and as an ion on the product side of the equation. The change from a neutral atom to a positively charged ion requires the loss of one or more electrons.

Mg(s)    →    Mg²⁺(aq)  +  2 e^-
 
Na    →    Na⁺  +  e^-

Similar reasoning was applied to determine that the other element (H and Cl) was reduced.
 
When there are only two elements present in the reaction, the task of determining the substances that are oxidized and reduced is relatively simple. But the waters quickly get muddy when the reaction looks like this:

16 H⁺(aq)  +  2 Cr₂O₇^2-(aq)  +  C₂H₅OH(l)    →    4 Cr³⁺(aq)  +  2 CO₂(g)  +  11 H₂O(l)

When there are covalently bonded atoms in polyatomic ions (e.g., Cr₂O₇^2-) and in molecular compounds (e.g., C₂H₅OH and CO₂), we need a strategy that allows us to keep track of electrons in order to detect their transfer from one element to another element. That strategy involves what we refer to as oxidation states (sometimes referred to as oxidation numbers).
 
 

What are Oxidation States?

An oxidation state (or oxidation number) is a number assigned to every element in a reaction that provides a measure of the degree of oxidation of that element.

 
The assignment is based on the assumption that the electrons in a covalent bond belong to the more electronegative element. Once oxidation states are assigned to every element in both reactants and products, we can inspect the numbers and tell which element is being oxidized and which is being reduced. An element that increases its oxidation number during the reaction is an element that has been oxidized. And an element that decreases its oxidation number is one that has been reduced.

While we observe in the above reaction that carbon is increasing its oxidation number, it is more common to refer to the reactant that contains carbon as the reactant that is being oxidized. So we would claim that the substance C₂H₅OH is being oxidized. Similarly, which Cr is the element that is decreasing its oxidation number, it is more common to refer to the Cr₂O₇^2- as the substance that is reduced.

 

Every redox reaction will have one element that is increasing its oxidation number (it is the element that is oxidized) and one element that is decreasing its oxidation number (it is the element that is reduced). This means that oxidation numbers become a bookkeeping tool used in redox reactions to identify reducing agents and oxidizing agents. To use this tool, you need to know the rules for assigning oxidation states to elements.

 
 
 

Rules for Assigning Oxidation Numbers

The following four rules can be used to assign oxidation states to the elements in the reactants and the products of a reaction.

1. Elements in the uncombined state have an oxidation number of 0.
   (Examples:  Mg in Mg    or    Br    in    Br₂    or    S in S₈.)
2. The oxidation number of a monatomic ion is equal to its charge; this rule applies to monatomic ions present in binary ionic compounds.
   (Examples: Ca of Ca²⁺ has an ox. # of +2.    Ca in CaCl₂ has an ox. # of +2.    P of P^3- and the P of Mg₃P₂ has an ox. # of -3.)
3. There are certain elements that always (or almost always) have the same oxidation number when present in compounds or ions. This includes ...
   • Group 1 metals have an oxidation number of +1.
   • Group 2 metals have an oxidation number of +2.
   • Fluorine has an oxidation number of -1.
   • Oxygen has an oxidation number of -2 (exception: it’s -1 in peroxides).
   • Hydrogen has an oxidation number of +1 (exception: it’s -1 in hydrides).
4. The sum of the oxidation numbers of all atoms in a compound is 0. And the sum of the oxidation numbers of all atoms in a polyatomic ion equals the charge of the ion.

 
 
 

Examples of Assigning Oxidation States

The examples below demonstrate how to use the above rules to determine the oxidation state of an element in a compound or an ion.
 
Example 1:  CO₂



 
Example 2:  C₂H₆O



 
Example 3:  NO₃^-



 
Example 4:  Cr₂O₇^2-


   
 

Identifying the Substance Being Oxidized and Reduced

Once the skill of assigning oxidations states to elements has been acquired, you can put it to use to analyze a redox reaction and determine which substance is oxidized and which substance is reduced. As an example, consider the following reaction:
   
We can determine the oxidation states of all elements in the reactants and products as follows:

• N of N₂ is 0 (Rule 1)
• H of H₂ is 0 (Rule 2)
• H of NH₃ is +1 (Rule 3)
• N of NH₃ is -3 (Rule 4)

 
This analysis can be represented as follows:
   
The element hydrogen is increasing its oxidation number. The reactant H₂ is the substance that is oxidized. The element nitrogen is decreasing its oxidation number. The reactant N₂ is the substance that is reduced.
   
 

Identifying the Oxidizing Agent and Reducing Agent

We introduced the terms oxidizing agent and reducing agent in Lesson 1a. We can now associate these terms with oxidation number. The oxidizing agent is the reactant that is reduced; it contains the element whose oxidation number decreases. The reducing agent is the reactant that is oxidized; it contains the element whose oxidation number increases. In the ammonia synthesis reaction (above), the N₂ is the oxidizing agent and the H₂ is the reducing agent.
 
 
 

Next Up

Oxidation and reduction always occur together. They are the two halves of the whole reaction. In Lesson 1c, we will learn how to write balanced chemical equations for redox reactions by using the half-reaction method. But before you tap the link and navigate forward, use one or more of the reinforcement suggestions in the Before You Leave section.
 
 
 

Before You Leave - Practice and Reinforcement

Now that you've done the reading, take some time to strengthen your understanding and to put the ideas into practice. Here's some suggestions.
 

• Try our Concept Builder titled Oxidation States. All three activities provide a great follow-up to this lesson.
• Try our Concept Builder titled Oxidation-Reduction. All three activities provide a great follow-up to this lesson.
• The Check Your Understanding section below includes questions with answers and explanations. It provides a great chance to self-assess your understanding.
• Download our Study Card on Oxidation States. Save it to a safe location and use it as a review tool. (Coming Soon.)

 
 
 

Check Your Understanding of Oxidation State

Use the following questions to assess your skill at assigning oxidation states and identifying the element that is oxidized and reduced. Tap the Check Answer buttons when ready.
 
1. The substance that is oxidized contains the element whose oxidation state is _____.

1. negative
2. positive
3. increasing
4. decreasing

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2. The substance that is reduced contains the element whose oxidation state is _____.

1. negative
2. positive
3. increasing
4. decreasing

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3. Use the rules for assigning oxidation states to determine the oxidation state for each element in the following formulae:

1. O₂ 
   
   Check Answer
   Answers: 0
   
   This is Rule 1.
   
    
   
    
2. H₂O 
   
   Check Answer
   Answers: H is +1; O is -2.
   
   This is Rule 3.
   
    
   
    
3. BF₃ 
   
   Check Answer
   Answers: F is -1; B is +3.
   
   Use Rule 3 for F. Then use Rule 4 for B.
   
    
   
    
4. OH^- ion 
   
   Check Answer
   Answers: O is -2; H is +1.
   
   This is Rule 3. You can use Rule 4 as a check; the sum of -2 + 1 should equal the ion's charge.
   
    
   
    
5. H₂SO₄ 
   
   Check Answer
   Answers: H is + 1; O is -2; S is +6
   
   Use Rule 3 for H and O
   Then use Rule 4 for S. The 2*(+1) + 4*(-2) + X = 0 where X is the ox. # for S. Solve for X.
   
    
   
    
6. PO₄^3- ion 
   
   Check Answer
   Answers: O is -2; P is +5
   
   Use Rule 3 for O
   Then use Rule 4 for P. The 4*(-2) + X = -3 where X is the ox. # for P. Solve for X.
   
    
   
    
7. KMnO₄ 
   
   Check Answer
   Answers: K is + 1; O is -2; Mn is +7
   
   Use Rule 3 for K and O
   Then use Rule 4 for Mn. The +1 + 4*(-2) + X = 0 where X is the ox. # for Mn. Solve for X.
   
    
   
    
8. Mg(ClO₃) ₂ 
   
   Check Answer
   Answers: Mg is + 2; O is -2; Cl is +5
   
   Recognize that it's an ionic compound with ions Mg²⁺ and ClO₃^-.
   Use Rule 3 for Mg and O. (You can also use Rule 2 for Mg.)
   Then use Rule 4 for Cl. The polyatomic ClO₃^- has a -1 charge. So the sum of Cl and O atoms must equal -1. The 3*(-2) + X = -1 where X is the ox. # for Cl. Solve for X.
   
    
   
    
9. Al₂(SO₄)₃ 
   
   Check Answer
   Answers: Al is +3; O is -2; S is +6
   
   Recognize that it's an ionic compound with ions Al³⁺ and SO₄^2-.
   Use Rule 2 for Al.
   Use Rule 3 for O.
   Then use Rule 4 for S. The polyatomic SO₄^2- has a -2 charge. So the sum of S and O atoms must equal -2. The 4*(-2) + X = -2 where X is the ox. # for S. Solve for X.
   
    
   
    
10. Na₂CO₃ 
    
    Check Answer
    Answers: Na is + 1; O is -2; C is +4
    
    Use Rule 3 for Na and O.
    Then use Rule 4 for C. The 2*(+1) + 3*(-2) + X = 0 where X is the ox. # for C. Solve for X.
    
     
    
     
11. NH₄F 
    
    Check Answer
    Answers: H is +1; F is -1; N is -3
    
    Use Rule 3 for H and F.
    Then use Rule 4 for N. The 4*(+1) + (-1) + X = 0 where X is the ox. # for N. Solve for X.
    
     
    
     

 

 
4. Consider the following reaction. Determine oxidations states for all elements in each reactant and product. Then answer the given questions.
   

1. The substance being oxidized is _____.
2. The substance being reduced is _____.
3. The oxidizing agent is _____.
4. The reducing agent is _____.

 


 
5. Consider the following reaction. Determine oxidations states for all elements in each reactant and product. Then answer the given questions.
   

1. The substance being oxidized is _____.
2. The substance being reduced is _____.
3. The oxidizing agent is _____.
4. The reducing agent is _____.

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6. Consider the following reaction. Determine oxidations states for all elements in each reactant and product. Then answer the given questions.
   

1. The substance being oxidized is _____.
2. The substance being reduced is _____.
3. The oxidizing agent is _____.
4. The reducing agent is _____.

​
 


 
7. Consider the following reaction. Determine oxidations states for all elements in each reactant and product. Then answer the given questions.
   

1. The substance being oxidized is _____.
2. The substance being reduced is _____.
3. The oxidizing agent is _____.
4. The reducing agent is _____.

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