Identifying Oxidation and Reduction

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Objective

Students will be able to identify which atoms are losing electrons and which atoms are gaining them in a chemical reaction.

Big Idea

Oxidation involves losing electrons, while reduction involves gaining them. Together, these processes are known as a redox reaction.

Introduction

In the last lesson students began to learn how to assign oxidation numbers to atoms. In this lesson students will use the skill from that previous lesson to identify which atoms are being oxidized and which are being reduced.

Like the previous lesson, this lesson aligns to the NGSS Disciplinary Core Idea of HS-PS1-2: Construct and revise an explanation for the outcome of a simple chemical reaction based on the outermost electron states of atoms, trends in the periodic table, and knowledge of the patterns of chemical properties because redox reactions focuses on a chemical reaction in which the outermost electrons are transferred from one element to another.

This lesson aligns to the NGSS Practices of the Scientist of Using mathematics and computational thinking because students will need to use math in many cases in order to assign to the correct oxidation number to an element.

It aligns to the NGSS Crosscutting Concept of Stability and Change because understanding oxidation numbers is the first step in constructing an explanation for how redox reactions occur—oxidation numbers are the method for tracking change to elements when they enter into compounds, and for how electron changes occur in redox reactions.

In terms of prior knowledge or skills, students should have already had a lesson about how to assing oxidation numbers.

There are no special materials needed for this lesson.

 

Do Now/Activator

10 minutes

Do Now: Students enter the classroom and this prompt is waiting for them: "Find a classmate who has completed the oxidation number assignments from the last class, and compare answers. Highlight any disagreements or confusion. If you did not do it, do NOT copy a partner. Instead, get help or work on it now."

I reason that this is a good way to start class because the lesson today builds on the skill of assigning oxidation numbers. I want students to self-identify if they are struggling with this skill.

Activator: I briefly check in with the class. I explain that today’s lesson builds on yesterday’s lesson. Some students would like me to review how we used the rules for assigning oxidation numbers in the previous class, and so I review the oxidation number notes from yesterday’s class. I note that students who were absent or really confused from yesterday’s class can get additional help during the independent work time in today’s class.

Mini-lesson and Guided Practice

15 minutes

Mini-lesson: I begin the mini-lesson by instructing students to record these notes from a PowerPoint slide:

  • Oxidation is the loss of electrons
  • Reduction is the gain of electrons
  • Redox reaction is when one atom gains, and one atom loses, electrons
  • Reducing agent is the element that is oxidized; it reduces another element
  • Oxidizing agent is the element that is reduced; it oxidizes the other element

I then review each of the terms. I note that loss and gain of electrons is the subject of today’s class. I point out that you cannot have loss without gain, which is why we have the combined term of redox. I point out that the terms reducing agent and oxidizing agent are confusing because they are opposites; the element that is oxidized is called a reducing agent because it allows reduction to occur. I note that this unit deals a lot with opposites—the gain of electrons is called reduction, and the oxidation number actually decreases even though there is a gain of electrons because electrons are negatively charged. It all makes sense once you think about it, I note, but it is not immediately intuitive.

I then light a piece of magnesium on fire, after warning students not to stare at the flame. The shiny silver metal burns brightly and I am left with a piece of brittle white substance that crumbles into a white powder. On the board I record this reaction:

2Mg + O2 --> 2 MgO

I then ask students to tell me what the oxidation numbers are for Mg and O before and after the reaction. On my document camera I make this table, and students help me fill it in:

 

Mg

O

Before

0

+2

After

0

-2

 

I then note that we can tell which element was reduced and which element was oxidized based on the oxidation numbers and the definitions from our notes. Mg lost electrons and O gained electrons, so Mg as oxidized and O was reduced. I note that this is our task for today—to show how electrons go from one element to another using oxidation numbers.

Guided Practice: I then ask the class to wrestle with problems 81c and 81d on page 509 in their text book in order to see what they understand. After several minutes we discuss these problems.

In the first equation H goes from a +1 to 0, and O goes from -2 to 0.

2 H2O(l) → 2H2(g) + O2(g)

This shows that H is reduced and O is oxidized.

In the second equation Na goes from 0 to +1 and H goes from +1 to 0 in the H2 molecule and stays +1 in the NaOH. The O stays the same at -2.

2Na(s) +2H2O(l) → 2NaOH(aq) + H2(g)

This shows that Na was oxidized, H was reduced in some cases, and O was neither oxidized nor reduced.

My instructional method is known as gradual release—I gradually release them to work independently. Based on the mood of the room, I can tell that students should get more guided practice before I release the class to do independent work. I ask them to do the first two problems in the Redox Reaction Practice. I then ask a student to explain his work as shown in this guided practice video.

 

Application

25 minutes

Student Activity: Now that I have gradually released students, I instruct them to complete the Redox Reaction Problems and the Analyzing a Redox Reaction practice problems. While that is happening, I divide up this time for myself into discrete 10 minute intervals. For the first, I give an overview of what students missed yesterday, and I have them work some practice problems from yesterday. For the second interval, I walk around and answer questions and offer encouragement. I am especially looking for students who are forgetting to divide the total positive charge by the subscript as this is a mistake students frequently make. It does not change which element is reduced and which is oxidized, but it does affect oxidation number. In the final segment I stop class for Catch and Release.

Catch and Release Opportunities: I invite a student to go to the front of the class and discuss how she worked a problem. Stopping class for a student to model this work is important because it helps us remember the big picture. Like in the Guided Practice video above, if she does not articulate every point that I want her to make, I ask some guiding questions. This is essentially re-teaching, and it gives students a break and gives them a chance to ask questions and hear a new voice teaching today’s skill.

Debrief

10 minutes

To wrap this lesson up I show the answers for the first question in the Analyzing a Redox Reaction with Answers . I want students to either recognize that they are on the right path, or I would like them to make sure they get help from a peer or me during study hall.

In looking at the student work I see that this student is able to name the element that is oxidized and the element that is reduced. However, on the student work part 2, I do notice that I will need to go back and remind students about some of the rules. I need to remind them that an ion has an oxidation number equal to its charge (notice the errors in problem 1 and 4, for example).

I also wonder in looking at student work if I rushed introducing the oxidizing agent and reducing agent. This unit comes at the end of the school year, and in retrospect, I wish I had left this part of the lesson out because I noticed that students started to forget everything they learned when this material was thrown in. For this unit’s purposes, I simply want them to understand how a voltaic cell works, which entails electrons flowing from one metal to another. Towards that end, I feel good about where most students are with their understanding.