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Rationalizing Reduction Potential (2 Favorites)

ACTIVITY in Reduction, Reduction Potentials, Hess's Law, Electron Affinity, Gibb's Free Energy , Bond Energy. Last updated April 26, 2019.


Summary

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

Grade Level

High school

AP Chemistry Curriculum Framework

  • Big Idea 3: Changes in matter involve the rearrangement and/or reorganization of atoms and/or the transfer of electrons.
    • 3.8 The student is able to identify redox reactions and justify the identification in terms of electron transfer.
    • 3.12 The student can make qualitative or quantitative predictions about galvanic or electrolytic reactions based on half-cell reactions and potentials and/ or Faraday’s laws.
  • Big Idea 5: The laws of thermodynamics describe the essential role of energy and explain and predict the direction of changes in matter.
    • 5.15 The student is able to explain how the application of external energy sources or the coupling of favorable with unfavorable reactions can be used to cause processes that are not thermodynamically favorable to become favorable.

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.

For the Student

Lesson

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- Eo = 2.87 V

Cl2 + 2e- ⇾ 2Cl-Eo = 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 ΔHo values for the half reactions, finding ΔGo results in a more accurate calculation (remember: ΔGo = -nFEo).

Your Turn

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

Then, using ΔGo = ΔHo - T ΔSo, calculate the ΔGo values.