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Lecture 2: Electrochemical Thermodynamics
1. Meaning of Standard Potential table values and reference potentials
2. Relation between reaction potential and free energy
3. Nernst equation with mass balance
4. Detailed thermodynamics
Review of last lecture
Last lecture, we talked about the definition of a battery and how it stores electricity via chemical
reactions, and how two key properties for energy storage are voltage and capacity. Both of these
are determined by the reactions- the capacity is a related directly to how much reaction has
occurred by Faraday’s Law, and the voltage comes from the exact nature of the reactions.
Voltage is a measure of the energy of the electrons that are produced or consumed in each of the
chemical reactions- it’s electrical potential energy, as you learned about in physics. Higher
voltage means more energy per electron. Today, we’ll talk in more detail about how the voltage
is determined, and to do that we’ll get into some basic thermodynamics.
Let’s go over what we did last time to find the standard cell potential of the Daniell cell.
Zn (s)
ZnSO4
in H2O
CuSO4
in H2O
Cu (s)
Barrier
Zn ↔ Zn 2+ + 2e −
Cu 2+ + 2e − ↔ Cu
Last time, what we did was simply to look up these reactions in the Table of Standard Electrode
Potentials and to add them together.
Zn + CuSO 4 ↔ ZnSO 4 + Cu
E0cell = 0.34 – (-0.76) = 1.10 V
If you made this cell with all components at standard state and connected it up to a voltmeter,
you would measure 1.1 V. How much current is flowing? None, because the current that goes
through a voltmeter is negligible, and without an external circuit to let electrons flow this
reaction cannot proceed. Zero current means the system is at rest, and a system at rest is a system
at equilibrium. We call the potential of the cell in this condition the equilibrium or open-circuit
potential. As a general rule, when you see the word “equilibrium” you should be thinking
“thermodynamics”. Thermodynamics is what nature wants to do, kinetics and transport are how
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it can get there. This applies in other fields like heat transfer, as well as in chemistry. The
equilibrium potential is the thermodynamic voltage limit. We’ll go into this further soon.
What happens when you start bringing current through the external circuit? The electronic
current that you want for useful work needs to be balanced by ionic current through the cell. If
there is resistance to current flow within the battery, V = IR, which means that some of the
voltage will be lost to Joule heating. As the current increases, more and more power is converted
to heat instead of useful work. Thus, the equilibrium potential gives us a theoretical limit to the
cell voltage. Practice question: is the equilibrium potential always a maximum voltage? When
would it be a minimum?
1. Meaning of Standard Potential table values and reference potentials
We just added together two standard potentials to get the cell voltage. Where did the standard
potentials come from? How was this table generated? The short answer is that somebody made
cells with all of these reactions at standard state and measured the voltage. This sounds simple,
but imagine that this was your job. If you made the Daniell cell without knowing the table in
advance, how would you figure out the potential contribution from each electrode? Is it possible
to measure the potential of just one of the two reactions? The answer is that you really can’t
measure this reaction alone, because circuits have to be completed and that means there has to be
a second reaction to balance the electrons. The way we deal with this problem is with the use of a
reference electrode. Look at the Table of Standard Electrode Potentials- what does it say at the
top? Potential vs NHE. The NHE is the normal hydrogen electrode, and practically, it means that
the potential of each of the reactions in this table has been measured with respect to a NHE. The
NHE has hydrogen gas at 1 atm and protons at unit activity.
[draw image of NHE on board]
The NHE is accepted as zero potential only by convention. It is not a terribly practical reference
electrode because it’s hard to work with, so you’ll often see literature use a different reference
electrode. There are practical experimental reasons for choosing one reference electrode over
another, but this doesn’t change the fact that you always need a reference. If you’ve taken some
thermodynamics, you probably talked about how there always needs to be a reference state.
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None of the thermodynamic functions of state- energy, enthalpy, Gibbs, Helmholtz (U, H, G, A)have meaning absolutely, only as differences. As we’ll next talk about, voltage is a
thermodynamic function and thus it has no meaning as an absolute value, only as a difference
from some other voltage.
2. Relation between reaction potential and free energy
We’ve said that the voltage produced in a cell is related to the nature of the chemical reactions
that are occurring. To understand why, let’s go back to chemistry class and the definition of a
reaction equilibrium constant. When you took chemistry, you probably wrote out generic
equations like this.
aA + bB cC + dD
 − ∆G O
s
K = ∏ a i i = exp
i
 RT
or



Zn + CuSO 4 ↔ ZnSO 4 + Cu
where
∆G O = ∑ s i GiO
i
We can get GiO , the standard-state Gibbs energy for species i, from the CRC tables or NIST
website or something like that. ∆G O is the standard Gibbs energy of reaction, and a is activity,
which for an ideal component is equal to concentration. [remind non-chemists of ideal as in
“ideal gas law”] So K is the ratio of concentrations at equilibrium, called the reaction equilibrium
constant. So if we look at definition of K, we see that if ∆G O is negative, K is greater than one.
That means this reaction is favorable, that thermodynamics cause it to move in the forward
reaction. The general principle is that energy always wants to be minimized
Now, how does this relate to our standard cell potential? We can see how, explicitly, by
the following equation, which is really just a definition:
∆G o = − nFU o or
∆G o = −nFE o
Both U and E are used, depending on whose notation we follow. Here we see that the standard
potential of a battery is related to the Gibbs free energy of the chemicals that are reacting.
Notice- negative sign means opposite convention for non/spontaneous reaction, and factor of F, n
relates the number of electrons transferred.
Zn ↔ Zn 2+ + 2e −
Cu 2+ + 2e − ↔ Cu
Let’s tie this information about the free energy back with what we just learned about reference
potentials. Look at the half reactions for the Daniell cell. Volt is J/C. So if one coulomb of
electrons leaves zinc for something else, leaving behind ½ coulomb of zinc ions, they will be
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0.76 J lower in energy than they would have been if they had started on hydrogen for something
else, leaving behind protons. Conversely, if one coulomb of electrons leaves copper for
something else, leaving behind ½ coulomb of copper ions, they will be 0.34 J higher in energy
than they would have been if they had started on hydrogen for something else, leaving behind
protons. This assumes that all the ion concentrations are at standard state. Remember, energy
wants to be minimized, so the dissolution of zinc and the deposition of copper are favored
(versus hydrogen). Practice question: H 2 ↔ 2 H + + 2e − has a standard electrode potential of zero.
Does this mean that H+ and H2 are energetically equal? Why not?
For many practical battery applications, it’s enough to accept that “this is the voltage of this
reaction” and to move on and work with it. For other purposes, however, you might want to take
a more physical interpretation based on the chemistry. For example, understanding how certain
atoms influence voltage to be higher or lower has important applications in developing new
battery materials. Look at the standard potential for the following three reactions:
F2 + 2e ↔ 2 F −
E = 2.87
Cl2 + 2e ↔ 2Cl −
E = 1.36
Br2 + 2e ↔ 2 Br −
E = 1.08
The more electronegative we go in the periodic table, the higher the voltage gets. This means that
these elements have a strong attraction for electrons. If a coulomb of electrons left somewhere
else and formed fluorine ions, they would be 2.87 J lower in energy than if they’d run to protons
and formed hydrogen. Practice question: ignoring all other factors, would the halogens be good
positive or negative electrodes? How could we take advantage of their electronegativity?
3. Nernst equation, with mass balance
So far we’ve only discussed standard conditions: 1 atm, 25C, and molality of 1mole/kg. If you
are working at non-standard concentrations, temperatures, or pressures you will need to use
something called the Nernst equation. First I want to show you the Nernst equation, how it is
used, and what it means conceptually. In the next section we’ll get involved in the
thermodynamics of where it comes from.
U =Uo −
RT 
ln
nF 
∏c
i
i
si




Go through the units on RT/F.
This uses the same convention for writing reactions that we use for Faraday’s Law.
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∑ si M izi → ne −
i
Let’s continue our example of the Daniell cell, changing our notation.
Zn ↔ Zn 2 + + 2e −
Zn − Zn 2 + ↔ 2e−
Cu 2 + + 2e − ↔ Cu
Cu − Cu 2 + ↔ 2e −
If we write out the Nernst equation for each of these reactions we get
U = 0.34 −
8.3143x298
8.3143x298  cCu 2+ 
−1
ln cCu cCu
ln 

2 + = 0.34 +
2 x96487
2 x96487
 1 
U = −0.76 +
8.3143x298 Zn 2+
ln
2 x96487
1
[
]
Concentration of a solid metal never changes, so we can set it equal to one and assume it
absorbed into the reference potential- remember, we only ever care about differences. Here we
see that the potential of each of these reactions depends on the concentration of the copper and
zinc in the liquid phase. Because the components in this cell are separated, as the cell is
discharged the copper and zinc in the solution will be depleted or increased. That is, the cell
potential is given by
U = 1.1 +
8.3143x298 Cu 2 +
ln
2 x96487
Zn 2 +
The change of the concentration of the copper and zinc in the solutions can be related to the
amount of charge that has been passed using Faraday’s law. For example, the number of moles
of Cu2+ that has reacted is given by
N
Cu 2 +
=
It
2 x96487
If the concentration of Cu2+ starts at 1mole/L, then the concentration of Cu2+ in the solution can
be related to the amount of charged passed by
It
[Cu ] = 1 − 2 × 96487
× vol
2+
Combining, we arrive at
U = 1.1 +
8.3143x 298
It
ln 1 −
2 × 96487
2 × 96487 × vol
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How much does the concentration change for a given current, and what effect does this have on
the equilibrium cell potential? This is just a combination of material balance and
thermodynamics.
U = 1.10V +
RT 
It 
ln1 −

2 F  nF 
To review: we talked about how the values in the Table of Standard Electrode Potentials are
generated, and what a voltage means in terms of reaction energy. For non-standard conditions,
the Nernst equation can be used to find an electrode potential. Why do we care about this for
batteries? Thermodynamics limit our maximum potential on discharge, and our minimum on
charge. In the last section, we’ll look at the equlibrium potential in more detail. This section is a
little more advanced, so if you feel lost don’t worry about it. If you like it, take more of it next
semester.
4. Detailed thermodynamics
Remember the definition of Gibbs free energy. We’ll skip the derivation here.
dG = VdP − SdT + ∑ µ i dni
i
The last term is the chemical potential. Roughly speaking, the chemical potential is a measure of
the interaction energy of the different particles in a system including the van der Waals and
electrostatic interactions between species. Mathematically, it is defined as
 ∂G 

 ∂n  T , p ,
µ =
Just as equality of temperature defines thermal equilibrium, and equality of pressure defines
mechanical equilibrium, so equality of chemical potential defines chemical equilibrium. For a
battery in which the temperature and pressure are fixed, we can look at the chemical potential to
examine the chemical equilibrium of the reactions that are occurring in the system. We saw last
time that it is the electrochemical reactions that define the thermodynamics of the system. In
particular, the definition of the cell potential is given by
(
)
FU = F φ α − φ α ' = µ eα− − µ eα−'
Here is how we use this method to calculate the potential of an electrochemical cell.
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1. Draw a picture of the cell and label all of the phases. Write down the reactions that occur
at each electrode surface, and make sure they are balanced.
α
Pt (s)
δ
β
CuSO4
in H2O
ZnSO4
in H2O
Zn (s)
γ
ε
α'
Cu (s) Pt (s)
Barrier
Zn ↔ Zn 2+ + 2e −
Cu 2+ + 2e − ↔ Cu
Here, the dashed line indicates there may be a junction region, which is just a region over
which there is a difference in composition or type of solution. We sometimes use a set of
shorthand in electrochemistry, shown below.
Zn/Zn2+, SO42-/ Cu2+, SO42- /Cu
The slash means there is a phase boundary, and the commas mean that two components
are in the same phase. The middle phase is an aqueous solution, but the water is implied.
If you were using a solvent other than water you would write it explicitly. The platinum
represents the wires that connect the cell. You need to have the same metal to compare,
and platinum is considered inert- there isn’t actually any in a real battery.
2. Express the equilibrium reactions in terms of the electrochemical potentials of each of the
species.
β
δ
α
µ Zn
= µ Zn
2+ + 2 µ −
e
γ
ε
α'
µ Cu
= µ Cu
2+ + 2µ −
e
3. Write the equilibrium cell potential, U, as (by definition)
(
)
FU = F φ α − φ α ' = µ eα− − µ eα−'
FU =
β
µ Zn
2
−
δ
µ Zn
2
2+
−
γ
µ Cu
2
+
ε
µ Cu
2
2+
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4. And our last step is to express each of the terms in terms of the electrochemical potential
at the standard state plus a concentration dependent term.
We write the chemical potential in terms of the activity:
µ A = µ Ao + RT ln c A f A
We sometimes use different superscripts depending on how we define what is called a secondary
reference state, which we won’t get into here.
FU =
0
µ Zn
2
θ
µ Zn
−
2+
2
−
RT ln cδ 2+ f Zn2+
Zn
2
−
0
µCu
2
+
θ
µCu
2
2+
+
RT ln cε 2+ f Cu2+
Cu
2
We typically group the standard values into a term we call FU θ . To obtain the Nernst equation
from this point simply set the activity coefficients to one, and ignore any effects in the liquid
junction.
θ
FU = FU −
θ
FU =
0
µ Zn
2
RT ln cδ 2+ f Zn2+
Zn
2
−
θ
µ Zn
2
2+
−
0
µ Cu
2
+
+
RT ln c ε 2+ f Cu 2+
Cu
2
θ
µ Cu
2+
2
From this we see that the values
U = 1.10V +
(
RT
− ln c δ 2+ + ln c ε 2+
Zn
Cu
2F
)
ε
RT c Cu 2 +
U = 1.10V +
ln
2 F c δ 2+
Zn
When we don’t see Nernstian behavior, it’s because of the things that we neglected. In
particular, it’s because we set the activity coefficients equal to 1. Another way to put this is that
the solution is ideal if the activity coefficients are equal to 1; as with most things in life, solutions
are not ideal. The departure of the measured potentials from the Nernstian potentials provide
insight on the activity coefficients.