Cracking the AP Chemistry Exam
Part IV
Content Review for the AP Chemistry Exam
Chapter 7
Big Idea #5: Laws of Thermodynamics and Changes in Matter
VOLTAGE AND FAVORABILITY
A redox reaction will be favored if its potential has a positive value. We also know from thermodynamics that a reaction that is favored has a negative value for free-energy change. The relationship between reaction potential and free energy for a redox reaction is given by the equation below, which serves as a bridge between thermodynamics and electrochemistry.
∆G° = −nFE°
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∆G° = |
Standard Gibbs free energy change (J/mol) |
|
n = |
the number of moles of electrons exchanged in the reaction (mol) |
|
F = |
Faraday’s constant, 96,500 coulombs/mole (that is, 1 mole of electrons has a charge of 96,500 coulombs). |
|
E° = |
Standard reaction potential (V, which is equivalent to J/C) |
From this equation we can see a few important things. If E° is positive, ∆G° is negative and the reaction is thermodynamically favored, and if E° is negative, ∆G° is positive and the reaction is thermodynamically unfavored.
Let’s take a look at an example.
Calculate the ΔG value for the below reaction under standard conditions:
Zn(s) + 2Ag+ → Zn2+ + 2Ag(s) E° = + 1.56 V
For this reaction, two moles of electrons are being transferred as the silver is reduced and the zinc is oxidized. So, n = 2. With that in mind:
ΔG° = − (2)(96500)(1.56)
ΔG° = − 301,000 J/mol
As the ΔG° value is negative, we would expect this reaction to be favored under standard conditions, which is in line with the positive value for the cell.
Note that your units on the free energy are in J/mol and not kJ/mol; this will always be the case with this equation as one volt is equivalent to one joule/coulomb, so the answer comes out in joules instead of kilojoules.