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Chapter 10 Notes
Up until now we have dealt with pretty well behaved things: ideal gases, ideal solutions,
dilute ideal solutions… As long as we have kept within certain limits, such as low
concentrations of solutes, we have been able to get by without have to worry too
explicitly about activity coefficient and nonideality. However, there is one type of
molecule which is so common in solutions (mostly water) that we cannot ignore it,
however, it is very ill-behaved indeed. These are ions. Think about what happens when
you dissolve salt (NaCl) in water. The sodium and the chloride ions separate from one
another giving Na+ and Cl-. The reason this reaction goes forward is because it results in
an increase in entropy as the salt crystals dissolve. However, salt does not dissolve in
many other liquids like oil. This is because in addition to an increase in entropy, there
needs to be interactions between the ions and the solution that help to stabilize the ions in
the soluble form. Water is very polar and can always align itself in such a way that their
more positive hydrogens are pointing at the Cl- and their more negative oxygens are
pointing at the Na+. This lowers the Gibbs free energy of the system making it possible
for the anions (negative ions like Cl-) and cations (positive ions like Na+) to separate in
solution.
In addition to discussing ions themselves, we will also discuss reactions the either form
or use up ions by adding or removing electrons from molecules in solution.
Chapter 10.1
With many kinds of solvation reactions, reduction/oxidation reactions or ionization
reactions we end up taking a neutral species (such as NaCl) and converting to two charge
species (such as Na+ and Cl-).
NaCl(s)  Na+(aq) + Cl-(aq)
As always, we can talk about the Gibbs free energy, enthalpy and entropy of such
reactions. However, if we try to break these reactions down into the Gibbs energy,
enthalpy and entropy of formation of each species in the reaction, we end up defining
things like the Gibbs energy of a sodium cation in solution. Yet there is no way to
generate just a sodium cation in solution (that would violate charge balance). Thus, we
can only determine the Gibbs energy of formation of pairs of ions. This does not cause
any real physical chemistry problems, but it causes a book keeping problem: how do we
write down the Gibbs energy of formation of a Sodium cation in a book so that we can
use it in any reaction where sodium cations are formed?
The answer is that we come up with a sort of standard ion, H+ (really something more
like H3O+) and we define the free energy of formation of all other ions relative to this
one. For example, I can measure the Gibbs energy change when I dissolve HCl in water
to form H+ and Cl-.
HCl(g)  H+(aq) + Cl-(aq)
G01
I can now just define the Gibbs energy of formation of H+ as zero and then determine the
relative Gibbs energy of formation of Cl-.
G01 = G0F(H+) + G0F(Cl-) - G0F(HCl)
G0F(H+) = 0 (definition)
G0F(Cl-) = G01 + G0F(HCl)
Knowing this, I can measure the Gibbs energy of dissolving NaCl in solution forming
Na+ and Cl-.
NaCl(s)  Na+(aq) + Cl-(aq)
G02
G02 = G0F(Na+) + G0F(Cl-) - G0F(NaCl)
G0F(Na+) = G02 + G0F(NaCl) - G0F(Cl-)
I now have a value for the Gibbs energy of formation for Cl-, so I can determine the value
for the Gibbs energy of formation of Na+. I can just keep this up, relating one ion pair to
another until I have Gibbs energies of formation for all individual ions relative to H+.
These values I can put in a book. This works because whenever we create one ion, we
always create another. Thus, we only need the relative Gibbs energy of formation for
ions -- the combination of the Gibbs energies for the two ions together is all that must be
absolutely correct. We will use the same trick later in the chapter to talk about the
standard Gibbs energy (or enthalpy or entropy) for reactions in which an electron is
transferred from one species to another, since this always involves the formation or
consumption of two ions. Once you have generated this table of Gibbs energies,
enthalpies and entropies of formation for ions, you use them just as you do for any
chemicals.
In other words, for any reaction involving ions, you can take all the Gibbs energies (or
enthalpies or entropies) of formation of the products and add them up and then just
subtract the the Gibbs energies of formation of the reactants, just as you have done in the
past.
You do not need to worry about the standard Gibbs energies of solvation and the Born
equation.
Chapter 10.2
Ok, as I mentioned above, one of the problems with ions is that they have very strong
interactions with each other over long distances. Thus, even very dilute solutions behave
in a nonideal way. Thus with ions, activity coefficients tend to be rather different from
one. For this reason, we much more frequently need to calculate activity of ions then of
most other chemical species (frankly, we rarely bother with neutral species). It is not
unusual for the activities of ions to differ from their concentrations by nearly a factor of
two. Immediately we run into the ion pair problem again. The problem is that we cannot
measure Gibbs free energies or chemical potentials of individual ions, only ion pairs:
G so ln     
We can expand the chemical potentials in the usual way:
G so ln     RT ln   b      RT ln   b 


       RT ln       RT ln b b 


As you can see, it is not really possible to separate the activity coefficients of the cation
and anion. All we get is their product. Thus we define a new ionic activity coefficient
which is a geometric average of the anion and cation coefficients. For a simple
monovalent ion pair like NaCl this is just:
     
For a salt with one or more multivalent ions like Mn+ Xm- it becomes
    n m n m
1
This is what we typically use. There are problems and ambiguities with this though. It is
pretty straightforward for something like NaCl where you end up with two monovalent
ions. But what about CaCl2? Here you get the average of the activity coefficient for a
monovalent and a divalent ion. It gets even more confusing when there are other ions in
solution as well. The thing to remember is that this is just a way of keeping track of how
a particular ion pair behaves.
Next we will talk about a way of estimating what these average activity coefficients are.
These are just estimates and designed only to give a rough indicator of the actual
situation. The calculations are useful in that they tell you roughly how large an non-
ideality you have to deal with. However, they usually do not give very precise
corrections. (They are more accurate for very low concentration ions.) The method we
will use is called the Debye-Huckel limiting law. The law relates the total ionic strength,
which is just a measure of how ionic the solution is, and the concentration of a particular
ion pair of interest, to the average activity coefficient for the ion pair. Before we can use
this law, we need to learn about ionic strength.
Calculating ionic strength.
As stated above, the ionic strength of a solution is a measure of the amount of ions
present. As you might guess a divalent ion (a 2+ or 2- ion, like Ca2+) has does more to
make the solution ionic than a monovalent ion (e.g., Na+). This must be taken into
account. The other very critical thing to remember is that the ionic strength of a solution
depends on the concentrations of all the ions in the solution, not just the ion pair that you
are calculating the activity coefficient for. Thus, if you are calculating the average
activity coefficient of dissolved CaCl2, but there is also dissolved NaCl present, the ionic
strength you use has contributions from all the ions.
1
The formula for ionic strength is I   mi zi2 . Ionic strength is technically unitless,
2 i
though the values behave as though they were in molal units (or, for aqueous solutions,
one often uses molarity as a good approximation). As usual, they are really
concentrations relative to the standard state (1 molal), though I have left the standard
molalities out of the equation for simplicity. The easiest way to see how to apply this
formula is to consider a few examples. First consider 100 mM NaCl. Upon dissolving,
one obtains 100 mM Na+ and 100 mM Cl-. Thus
1
1
1
1
2
2
2
2
I  Na  1  Cl   1  0.1M 1  0.1M  1  0.1
2
2
2
2
Notice that for a simple salt of two monovalent ions, the ionic strength is just the
concentration of the salt. This is not true for a salt with one or more multivalent ions like
MgCl2. For a 100 mM solution of this salt:
1
1
1
1
2
2
2
2
I  Mg 2 2  Cl   1  0.1M 2  0.2M  1  0.3
2
2
2
2
Note that the Mg cation is divalent and thus it has a big effect since the charge is squared.
Also note that the chloride anion is present at twice the concentration since there are two
chloride ions per molecule of salt. What is the ionic strength of a solution of 100 mM
NaCl plus 100 mM of acetic acid which has been titrated with NaOH until the pH of the
solution is 4.75 (the pKA of acetic acid)? When the pH equals the pKA, that means that
half of the acetic acid has been converted to the conjugate base, sodium acetate. Acetic
acid is uncharged and does not contribute to the ionic strength. However sodium acetate
ionizes completely to form acetate anions and sodium cations. Since half was converted,
there are 50 mM of each. Then we must add in the 100 mM of NaCl. So there is 50 mM
acetate anion, 150 mM sodium anion, and 100 mM chloride anion:
1
1
1
1
2
2
2
2
I  Na  1  Cl   1  acetate 1  0.15M 1
2
2
2
2
1
1
2
2
 0.1M  1 0.05M  1  0.15
2
2




 


 
 
Calculating Activity Coefficients.
Now we actually will use the Debye-Huckel limiting law itself. There are three very
important things about applying the Debye-Huckel theory. First, it only applies to ions.
Molecules that are not charged have an activity coefficient of 1.0 according to this theory
(in reality, that is not true, but their activity coefficients will be much closer to 1 than will
that of an ion). Second, the charges that appear in the equation are only those of the salt
you are calculating the activity coefficient for. Finally, all you can only ever calculate
are average activity coefficient of the two ions which make up the salt you are
considering. For MgCl2, you cannot use this theory to calculate the activity coefficient
of Mg2+ separately from Cl-, you can only calculate the geometric average of the two
activities,     2  3 .
1
The Debye-Huckel limiting law is log     z  z  A I where A=0.509 for water at 25
C. (A is an empirical constant.) In the acetate/acetic acid example given above, the
sodium acetate ions would have an average activity coefficient given by
log     (1)( 1) (0.509) 015
.  0197
.
   0.635
Notice that the ionic strength is that of the whole solution, while the charges are those of
the sodium acetate ions we are calculating the activity coefficient for.
Don't worry about the extended Debye-Huckle law. It is the same thing with yet another
fudge factor in it to make it more accurate at higher concentrations.
Chapter 10.3 through 10.6
Now let's consider reactions in which ions are formed or consumed by the transfer of an
electron (a negative charged particle) from one chemical species to another. The first
thing to do is to consider the concept of a half reaction. A half reaction is not a real
reaction in that it can never happen by itself. It has the form
X + e-  Xfor example. Of course, the electron must come from somewhere. That somewhere is
another half reaction:
Y  Y+ + eIn these reactions, X is getting reduced (receiving electrons) and Y is getting oxidized
(giving up electrons). If we add these equations together, we get a real reaction:
X + Y  X- + Y+
An aside -- note the similarity between this formalism and the one we use for acids and
bases:
X + H+  XH+
YH  Y- + H+
Which when added gives:
X + YH  XH+ + YA realistic example is:
H2O + AH  H3O+ + AReduction/oxidation reactions are really very similar to acid/base reactions, one uses
electrons and the other uses protons. Neither protons nor electrons can exist in solution
by themselves -- an acid needs a base to receive the proton, a reductant needs an oxidant
to receive the electrons. For acids and bases we have the pKA which is the pH at which
half of the acid has been converted to the conjugate base. For reduction/oxidation
reactions we have the midpoint potential which is the voltage at which half of the reduced
compound has bee turned into the oxidized version of the same compound. I will try to
point out the parallels as we go along.
Electrochemical reactions are often run in electrochemical cells (typical flashlight
batteries are good examples of such cells). There is some language and nomenclature we
will need to get down before we can talk about them.
Electrochemical Cell -- a system with two electrodes connected electrically which each
come in contact with an electrolyte (the solution that the chemistry is occurring in).
These electrodes can either be in the same solution or in two different solutions that are
connected by some means (such as a salt bridge).
Galvanic Cell -- an electrochemical cell in which a spontaneous reaction generates
electricity.
Galvanic
Cell
Electrolytic Cell -- an electrochemical cell in which electricity (an external voltage) is
used to drive a nonspontaneous reaction.
Electrolytic
Cell
Anode -- This is the electrode that takes up electrons from the solution (do not think of it
in terms of positive or negative because that changes depending on whether it is a
galvanic or electrolytic cell).
Cathode -- This is the electrode that gives off electrons into the solution.
Redox couple -- The oxidized and reduced forms of a molecule, usually symbolized, for
example, as Cu2+/Cu.
Electrochemical Cell Notation -- There are many ways to physically hook up cells. I am
not going to worry much about this, but to do you homework you will need to know a
few of the conventions about writing out the components of electrochemical cells. In the
figure for the electrochemical cell below, we would write:
Zn(s)|ZnSO4(aq)||CuSO4(aq)|Cu(s)
The single lines represent phase boundaries, and the double lines represent the salt bridge
between electrode compartments.
As always with generation or consumption of ions, we have the problem that we can only
measure processes for formation or consumption of pairs of ions. For this reason, we
again define a standard half reaction and call its Gibbs free energy zero. For consistency
we pick the formation of H2(gas) from H+:
2H+(aq) + 2e-  H2(g)
rG0 = 0
Just as before with the free energy of formation, we can use pairs of equations (which we
can measure) to determine the relative standard reaction Gibbs energies for any other half
reaction and we can write them in tables, such as the one in your book.
Finally, we need to somehow relate the electrical world (the voltage on your 1.5 volt
battery in your flashlight) to the Gibbs free energy change associated with a particular
electrochemical reaction. What is voltage? The units of voltage are (you guessed it)
Volts. A volt is a Joule per Coulomb. What's a Coulomb? A Coulomb is 6.24x1018
charges. A Coulomb is the world of charged particles what a mole is the world of
molecules. It is just a number, like a dozen, that we use to measure charges in. So
voltage is a measure of the amount of energy involved in moving a certain number of
charges from one place to another. This is not very unlike a molar free energy. A molar
Gibbs free energy is the amount of free energy it takes to convert one mole of a substance
from one form or state to another.
It should come as no great surprise that the reaction free energy for an electrochemical
reaction is just proportional to the Voltage it can generate.
rG = -FE
or form standard conditions:
rG0 = -FE0
Here E is the voltage (actually, it is the zero current potential -- the maximum voltage
that could be produced by the system with not current flowing). The proportionality
constant has two parts. Part (F, the Faraday constant) takes into account the fact that
Gibbs free energy is on a per mole basis, while voltage is on a per coulomb basis and a
mole is a different number than a coulomb (6.02 x 10 23 vs. 6.24 x 1018). The value of F
is just 6.02 x 10 23 / 6.24 x 1018 = 9.65 x 104 C/mole. The other part of the
proportionality comes from the fact that the Gibbs free energy is based on the amount of
the chemical being oxidized or reduced while the voltage is based on the number of
electrons that flow during the oxidation/reduction reaction. Thus  is the number of
electrons involved in the oxidation or reduction reaction. For example, it is different to
oxidize Fe+ to Fe2+ than it is to oxidize Cu to Cu2+. The former involves the transfer of
one electron for every iron, the later requires the transfer of 2 electrons for every Cu
atom. Notice also that there is a minus sign in the equation. This is because the
conventions for charge movement (current) were developed before anyone knew about
electrons. People just assumed that the things carrying the charges were positive. The
assumption was not correct, but the convention stuck. For this reason we have to throw a
negative sign into the equation.
Ok, now how do we use this equation to do something useful?
Oxidation/Reduction and half reactions.
We talked about half reactions before. In reactions that involve oxidation and reduction,
it is usually the case that you can think of the reaction as occurring in two steps: one
molecule gives up electrons (becomes oxidized) and another molecule picks up electrons
(becomes reduced). The two parts are the half reactions. They can be dealt with
mechanically (adding their free energies or multiplying their equilibrium constants to
give the total reaction) in the same way as any other set of reactions can. One difference
is that in writing the quotient in the free energy expression (RTlnQ), you do not include
the electrons as part of Q.
Using the relationship between reaction Gibbs free energy and zero current potential, E.
 G
E   r where the Greek letter  in the denominator is the number of electrons in the
F
reaction. The important point is that in most respects E behaves just like rG multiplied
by a constant. The E’s for two coupled reactions can be added together. If you reverse
the direction of a reaction, you just reverse the sign of E. To convert between the
reaction free energy and the zero current potential, you need to know the number of
electrons involved. You can get this from looking at the half reactions for the
oxidation/reduction reaction.
There are two points that differ between E and rG:

First, if you multiply a chemical equation by a constant, the reaction free energy
multiplies with it, but the zero point potential does not. This is because the zero point
potential is proportional to the reaction free energy divided by the number of
electrons. When you multiply a chemical reaction by a constant value both the
reaction free energy and the number of electrons change by that constant value, the
ratio does not.

Second, because of negative sign in the relationship between reaction free energy and
zero current potential, a spontaneous reaction has a negative reaction free energy but
a positive zero point potential.
These issues are summarized below using the example of an oxidation/reduction reaction
involving Zn and Cu.
Reaction (or half reaction)
Zn(s)
--> Zn2+ + 2e-
Cu2+ + 2e-
--> Cu(s)
Zn(s) + Cu2+ --> Zn2+ + Cu(s)
2Zn(s) + 2Cu2+ --> 2Zn2+ + 2Cu(s)
rG (see note1)
rGZn
rGCu
rG1=
rGZn+rGCu
rG2=
rGZn+rGCu
)
=2rG1
E
Q (see note2)
E Zn  
 r GZn
2F
QZn 
E Cu  
 r GCu
2F
QCu 
 r G1
2F
 E Zn  E Cu

2F
a Zn s
a Cu s
4F
2
1

2
Zn
Cu
2 2
2
2 2
1
The Nernst Equation:
Using the relationship between the reaction free energy and the zero current potential,
one can derive a general expression for the zero current potential as a function of the
standard zero current potential and the concentrations of the components in the solution:
 r G   r G   RT ln Q
 r G  FE
FE  FE   RT ln Q
RT
E  E 
ln Q
F
This final equation is called the Nernst Equation. You can use this equation to calculate
the zero current potential given the standard potential (midpoint potential) and the actual
concentrations of the reactants and products involved.
Here again, we can see a parallel between oxidation/reduction and acid/base problems.
For a half reaction such as
We have

Cu  Cu 
2
Note1: this reaction is spontaneous. Therefore rG1 is negative and E1 is positive.
Note2: the activity of a pure solid is one (this is the standard state of a solid). Therefore
a{Zn(s)} = a{Cu(s)} = 1.0 and these terms do not appear in the final equilibrium
constants.
X + e-  X-

 Zn 2
Zn   Q Q
Q 
Cu 
Zn   Q
Q 
Cu 
1
2  r G1 
  r G1   E
2
2
E1 
E2 
Zn 
2
1
E  E 
RT X 
ln
vF
X
This looks rather like:
pH  pK A  log
base
acid
E0 in this case is the midpoint potential, the potential where half of the compound is
oxidized, just as the pKA is the pH where half of the acid has been converted to the
conjugate base.
Chapter 10.7
One application of this is the calculation of the solubility constants for sparingly soluable
salts. Generally
MX  M+(aq) + X-(aq)
KS = aM+ aX-
Note that MX does not appear in the equilibrium equation because the pure solid is the
standard state (so the activity is one). Since the activity of M+ and X- should be about the
same (due to the reaction stochiometry), we can see that
S  KS
where S is the concentration of the salt that will dissolve.
One can calculate the solubility now from electrochemical data by simply finding
reactions that we can add together in order to get the overall reaction, converting the zero
point potential to a reaction Gibbs free energy and that to an equilibrium constant and
then relating the equilibrium constant to the amount of salt dissolved as described above.
The example in the book is:
AgCl(s)  Ag+(aq) + Cl-(aq)
We can get this by adding together the following two half reactions:
AgCl(s) + e-  Ag(s) + Cl-(aq)
Ag(s)  Ag+(aq) + e-
E0 = +0.22 V
E0 = -0.80 V
So the total reaction has an E0 of -0.58 V
We can convert this to the equilibrium constant using the Nernst equation (see above) and
the conversion between equilibrium constants and standard reaction Gibbs free energy
(see chapter 9). The result is
ln K S 
E 
RT / F
K S  1x10 10
S  1x10 5
 23
moles
kg
Chapter 10.8
The measurement of pH is certainly a common thing, so it is worth having some idea
how it works. In principle one can see that:
H  aq   e   12 H 2  g 
PH 2
RT
ln
F
aH 
E0 is zero by definition for this reaction. If we were to set up an electrode with the
pressure of hydrogen gas at the standard 1 bar:
E  E 
E
RT
RT
RT
ln a H  
log a H   
pH
F
2.303F
2.303F
For T=25C, we find that:
E = (59.16 mV) pH
Thus, if we had a hydrogen electrode at 25 C and measured the potential generated in a
solution of some unknown pH using a solution of pH 0 (1 M H+) as a reference at the
other electrode, the potential we measured would be proportional to the pH as given
above:
Pt|H2(g)|1M H+||saturated KCl||X M H+|H2(g)|Pt
This represents two hydrogen electrodes (using platinum as the metal to transfer the
electrons since it will not become oxidized or reduced at these potentials), one in a
solution of pH 0 and the other in a solution of unknown pH. The potential generated is
proportional to the pH. There is something very important to notice here. The
proportionality constant is directly dependent on temperature. Thus, the pH meter will
have to be recalibrated or adjusted for any temperature used (modern pH meters measure
the temperature of the solution and do this calibration automatically).
Hold on, you say, you have never seen a pH meter with a hydrogen electrode. True. Gas
electrodes are a pain and no one actually uses this reaction to measure pH. Instead other
reactions involving H+ ions are used. Which ones depend on the electrode you buy and it
is not worth our time to go through this in great detail. Most of the modern electrodes in
general chemistry laboratories are actually combination electrodes where more than one
reaction is going on. However, the principle is always the same and the result is the
same. One gets a potential reading out which is proportional to the pH of the solution via
the Nernst equation. That proportionality constant is always dependent on temperature in
a linear fashion.
Chapter 10.9
All this chapter says is that since we can convert between the electrochemical cell
potential (zero current potential, E) and the Gibbs free energy, we can use the same
relationships as always to get enthalpies and entropies. My suggestion is not to
memorize a bunch of new formulas. Just convert E to rG and then treat it as you always
have in terms of enthalpy and entropy. There is really no more to it than that.