Lesson 17.3 Cell Potential, Free Energy, & Equilibrium
Suggested Reading

Zumdahl Chapter 17 Sections 17.3 & 17.4
Essential Questions

What is the relationship between cell potential, free energy, and
equilibrium?
Learning Objectives:.


Relate cell potential to the equilibrium constant and free energy.
Use the Nernst equation to calculate voltage at nonstandard conditions.
Introduction
Some of the most important results of electrochemistry are the relationships
among cell emf, free-energy change, and the equilibrium constant. Let jump
right in.
Cell Potential and Free-Energy
The free-energy change ∆G for a reaction equals the maximum useful work of
the reaction, where
∆G = wmax
For a voltaic cell, this work is the electrical work where,
wmax = -nFE∘cell
Recall that n is the number of electrons transferred in the redox reaction
and F is Faraday's constant.Thus, when reactants and products are in their
standard states you have
∆G∘ = -nFE∘cell
With this equation, emf (cell potential) measurements can be used to obtain
thermodynamic information. Lets look at some examples.
Example: Calculating the Free-Energy Change from Electrode Potentials
Using standard electrode potentials, calculate the standard free-energy
change at 25∘C for the reaction
Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s)
Solution:
Use a table of standard potentials to obtain E∘cell. The cell reaction equals
the sum of the half-reactions after they have been multiplied by factors so
that the electrons cancel in the summation. The half-reactions and their
corresponding half-cell potentials are
Note that each half reaction involves two electrons, so there is no need to
balance the number of electrons before summing, & n = 2. Now we can
plug into the equation ∆G∘ = -nFE∘cell and solve for ∆G∘.
∆G∘ = -2(96,500C)(1.56V) = -3.01x105 J
Recall that coulombs x volts = joules. The standard free energy change is
therefore -301 kJ.
Is the reaction spontaneous as written? Yes, ∆G∘ because is negative. In
fact, ∆G∘ is must always be negative for a voltaic cell if it is to be useful.
These cells rely on spontaneous reactions to do work. Thus, for a
spontaneous reaction, Ecell is positive and ΔG is negative.
Example: Calculating Cell Potential from Free-Energy
Suppose a reaction of zinc metal and chlorine gas is utilized in a voltaic cell
in which zinc ions and chloride ions are formed in aqueous solution
Zn(s) + Cl2(g) → Zn2+(aq) + 2Cl-(aq)
Calculate the standard emf for the cell at 25∘C from standard free energies
of formation.
Solution:
Calculate ∆G∘ and substitute into ∆G∘ = -nFE∘cell and solve for emf. Begin by
writing the equation with the ∆G∘f beneath.
Zn(s) + Cl2(g) → Zn2+(aq) + 2Cl-(aq)
∆G∘f (kJ)
0
0
-147
2x-131
Now,
= [-147 + 2 x (-131)]kJ = -409 kJ = -4.09 x 105 J
n can be found by splitting up the reactions.
Zn(s) + → Zn2+(aq) + 2eCl2(g) + 2e- → 2Cl-(aq)
Each half-reaction involves two electron, so n = 2. Now you can substitute
into ∆G∘ = -nFE∘cell and solve.
-4.09 x 105 J = -2 x 96,500 C x E∘cell = 2.12 V
Cell Potential and the Equilibrium
Constant
The measurement of cell potential gives you yet another way to obtain
equilibrium constants. Combining ∆G∘ = -nFE∘cell, with the equation from
lesson 16.2 ∆G∘ = -RTlnK, gives
-nFE∘cell = -RTlnK
or
nFE∘cell = RTlnK
rearranging gives
This equation can be further simplified by substituting values for the
constants R & F giving
Lets look at another example.
Example: Calculating the Equilibrium Constant from Cell Potential
The standard cell potential for the following voltaic cell is 1.10 V:
Zn(s) ❘ Zn2+(aq) ❘❘ Cu2+(aq) ❘ Cu(s)
Calculate the equilibrium constant for the cell reaction.
Solution:
First, write the equation for the cell reaction (line notation was covered in
17.1).
Zn(s) + Cu2+ ⇌ Zn2++ Cu(s)
Next, substitute the cell potential into the equation relating cell potential and
the equilibrium constant above and solve for K. We can see very easily
from the equation that n = 2 (However, if this is unclear to you, write the
half-reactions).
The number of significant figures equals the number of decimal places in
37.2 (one).
Nernst Equation
The cell potential depends on the concentrations of ions and on gas
pressures. You can relate cell potentials for various concentrations of ions
and various gas pressures to standard electrode potentials by means of the
Nernst equation. Nernst also formulated the 3rd law of thermodynamics, for
which he received the Nobel Prize in 1920. Recall that the free-energy
change, ∆G, is related to the standard-free energy change by the following
equation from lesson 16.2,
∆G = ∆G∘ + RTlnQ
where Q is the reaction quotient. You can apply this equation to a voltaic
cell. If you substitute ∆G∘ = -nFE∘cell and ∆G = -nFEcell into this equation you
get
-nFEcell = -nFE∘cell + RTlnQ
This rearranges to give the Nernst equation, which gives us a direct
relationship between cell potential and the reaction quotient. This equation
takes the following forms:
If you substitute 298 K for T and put in the values for R and F, you get the
simpler form
You can show from the Nernst equation that the cell potential decreases as
the reaction proceeds as the reactants are converted to products.
Eventually the cell potential goes to zero when the cell reaction comes to
equilibrium.
Read the following web page about calculating cell potential at standard and
non standard conditions, BUT ONLY READ THE SECOND TOPIC ENTITLED
DETERMINING NON-STANDARD STATE CELL POTENTIALS.
http://www.chem.purdue.edu/gchelp/howtosolveit/Electrochem/Electrochemic
al_Cell_Potentials.htm#Standard
HOMEWORK: Practice exercises 20.10 – 20.12, 20.14