Chapter 5 Notes
Ionic strength and the Debye-Huckle law
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 and all biochemical solutions) 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 that either form or use
up ions by adding or removing electrons from molecules in solution.
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 H O ) +
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) ΔG - 01
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.
ΔG = ΔG (H ) + ΔG (Cl) - ΔG (HCl)- 0
1 F F F
ΔG (HF) = 0 (definition)
ΔG (CF) = ΔG + ΔG (H1l) 0F
Knowing this, I can measure the Gibbs energy of dissolving NaCl in solution forming Na and +
NaCl(s) Na (aq) + Cl(aq) ΔG - 0
ΔG = 2G (Na ) F ΔG (Cl) - ΔG (FaCl)- 0F
ΔG (NF ) = ΔG + ΔG (N2Cl) - ΔGF(Cl) 0F -
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 Gibbs
energies of formation of the reactants, just as you have done in the past.
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 the 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:
ΔG soln= μ ++ μ −
We can expand the chemical potentials in the usual way: 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 M X it becomes
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 CaCl ? 2ere 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 a 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 measu2+ of the amount of ions present.As
you might guess a divalent ion (a 2+ or 2- ion, like Ca ) 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 CaCl , but2there is also dissolved
NaCl present, the ionic strength you use has contributions from all the ions.
The formula for ionic strength is . Ionic strength is sometimes stated as having
units of molal (or molar) and other times stated as being unitless (this is the case on your book),
depending on the book you read. In most cases, ionic strength is considered unitless, and is
calculated from concentrations relative to the standard state (1 molal). I have left the standard
molalities out of the equation below for simplicity, but in principle, all of the concentrations are
divided by 1 M. 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 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 MgCl For 2.100 mM
solution of this salt:
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 pK of A
acetic acid)? When the pH equals the pK , thaA 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
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 ever calculate is average activity coefficient of the two ions
which make up the salt you are c2+sidering. For MgCl , -ou c2nnot use this theory to calculate
the activity coefficient of Mg separately from Cl, you can only calculate the geometric average
of the two activities, .
The Debye-Huckel limiting law is whereA=0.509 for water at 25 C. (Ais
an empirical constant.) In the acetate/acetic acid example given above, the sodium acetate ions
would have an average activity coefficient given by
Log γ €= −|(1)(−1)|(0.509)(0.15) €= − 0.197
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. Remember this. It trips up
many students on exams.
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.Ahalf reaction is not a real reaction in that it can never
happen by itself. It has the form
X + e X
for example. Of course, the electron must come from somewhere. That somewhere is another
Y Y + e -
In 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 + Y + -
Arealistic example is:
H 2 + AH H O +A 3 + -
Reduction/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. The big difference is that water can act as both a proton acceptor and donor. It cannot
act as either an electron acceptor or donor under normal circumstances (you can take electrons
from water, forming hydrogen ions and oxygen gas, but this is not something that happens until
you get above a fairly high ambient potential). For acids and bases we have the pK which is Ahe
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 been 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. Electrolytic Cell -- an electrochemical cell in which electricity (an external voltage) is used to
drive a nonspontaneous reaction. 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
Cathode -- This is the electrode that gives off electrons into the solution.
Redox couple -- Th2+oxidized and reduced forms of a molecule, usually symbolized, for
example, as Cu /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 your 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)|ZnSO (aq4||CuSO (aq)|4u(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 H (gas) from H : +
2H (aq) + 2e H (g) Δ2 = 0 r 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 6.24x10 charges.A Coulomb is to the world of
charged particles what a mole is to the world of molecules. It is just a number, like a dozen, that
we use to measure charges in. So voltage (energy per charge) tells us 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. Amolar 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.
or for standard conditions:
ΔGr= -νFE 0
Here E is the voltage (actually, it is the zero current potential -- the maximum voltage that could
be produced by the system with no 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
23 18 23 18 4
(6.02 x 10 vs. 6.24 x 10 ). The value of F is just 6.02 x 10 / 6.24 x 10 = 9.65 x 10 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 ν is the number of electrons
involved in the oxidation or reduction reaction. For example, it is different to oxidize Fe to
Fe than it is to oxidize Cu to Cu . 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.
where the Greek letter ν in the denominator is the number of electrons in the
reaction. The important point is that in most respects E behaves just like ΔG multirlied 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 ΔG: r
• 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) Δ r (see note1) E Q (see note2)
Zn(s) --> Zn + 2e Δr Zn
Cu + 2e --> Cu(s)
Zn(s) + Cu --> Zn + Cu(s) ΔGr= 1
Δr +ZnG r Cu
2Zn(s) + 2Cu --> 2Zn + 2Cu(s) ΔGr= 2
2(ΔGr+ΔZn) r Cu
=2€ΔG r 1
Note1: this reaction is spontaneous. Therefore ΔG is nrga1ive and E is positiv1.
Note2: the activity of a pure solid is one (this is the standard state of a solid). Therefore a[Zn(s)]
= 1.0 and[Cu(s)] = 1.0 and these terms do not appear in the final ratios.
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: 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
X + e X
This looks rather like:
E in this case is the midpoint potential, the potential where half of the compound is oxidized,
just as the pK As the pH where half of the acid has been converted to the conjugate base.
Speaking of pH, it turns out that you actually measure pH by measuring electrochemically
generated voltages for reactions involving H O . T3e measurement of pH is certainly a common
thing, so it is worth having some idea how it works. In principle one can see that:
E 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:
For T=25C, we find that:
E = (59.16 mV) pH
Thus, if we had a hydrogen electrode at 25 C and measured th+ 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|H 2g)|1M H ||saturated KCl||X M H |H (g)|Pt2 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
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 dep