Activity

Summary

In this activity, students will gain a deeper understanding of how quantitative chemical concepts relate to each other.

Grade Level

High school

Objectives

By the end of this lesson, students will

• Better understand how thermodynamics relate to reduction potentials.

Chemistry Topics

This lesson supports students’ understanding of

• Reduction potentials
• Thermodynamics
• Electrochemistry
• Hess’s Law
• Gibb’s free energy

Time

Teacher Preparation: none
Lesson: 30 minutes

Materials

• Calculator

Safety

This is a reading and calculations exercise, no safety precautions are noteworthy.

Teacher Notes

• A teacher could modify this lesson by having students determine the steps necessary on their own and to look up thermodynamic values themselves.

Downloads

• Teacher Guide.pdf
  
• Teacher Guide.doc
  
• Student Guide.pdf
  
• Student Guide.doc
  

Submitted by
Brinn Belyea
Torrey Pines High School
San Diego, California

Thanks to
Ward’s Science

For the Student

Problem

Which has a more favorable reduction potential, chlorine or fluorine?

Analysis

Don’t look at a table of reduction potentials to answer this question, answer it with chemistry knowledge.

Student 1 says: Fluorine has a higher reduction potential because it is more electronegative than chlorine.

Student 2 says: Chlorine has a lower electron affinity than fluorine, so it should have the higher reduction potential.

Which student is right?

Electronegativity doesn’t have units, so it’s not a good concept to base an answer on—it’s not a measurable quantity. So student 1’s logic has a flaw.

Student 2 is pretty impressive because they remember that chlorine is an exception to the electron affinity trend, which predicts that fluorine should have a lower electron affinity. This doesn’t mean they are automatically right.

What do you know?

The reduction half reactions for fluorine and chlorine are:

F₂ + 2 e^- ⇾ 2 F^-

Cl₂ + 2 e^- ⇾ 2 Cl^-

And what happens in each process?

A bond breaks. You might predict that it’s easier (less energy) to break a Cl–Cl bond because it’s longer than an F–F bond. Combining that with the lower electron affinity of chlorine should give chlorine a higher (more favorable) reduction potential.

However, the F–F bond is an exception to the general rule that shorter bonds are stronger and thus require more energy to break. Fluorine’s bond energy is 154 kJ/mol and chlorine’s is 239 kJ/mol. The electron affinity of F is -327.8 kJ/mol and Cl is -348.7 kJ/mol. Overall, the reduction half reaction for F is -174 kJ/mol and Cl is -110 kJ/mol.

So fluorine has a lower energy (more favorable) reduction half reaction. 

What are the values?

F₂ + 2e^- ⇾ 2F^-         E^o = 2.87 V

Cl₂ + 2e^- ⇾ 2Cl^-    E^o = 1.36 V

Do the data predict this? Quick math, comparing fluorine to chlorine values:

2.87 V/1.36 V = 2.11

174 kJ/110 kJ = 1.58

Why the difference?

Three issues. One is rounding. Two is the ions aren’t in the gaseous state, they are aqueous, so a state change is also unaccounted for in these calculations. Three is that you calculated ΔH^o values for the half reactions, finding ΔG^o results in a more accurate calculation (remember: ΔG^o = -nFE^o).

Your Turn

Find the energy associated with each ion’s state change from gas to aqueous to calculate a more accurate ΔH^o value.

Then, using ΔG^o = ΔH^o - T ΔS^o, calculate the ΔG^o values.

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