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Lecture W2
Week1
Jan. 10, 2001
For the first meeting of next year (Jan. 8), review Chapter 14 in Harris and read chapter 15
on Potentiometry.
Read Statistics Chapters 1 & 2 for Wed. Jan 8th.
Harris homework for Monday of second week winter quarter:
Appendix D: 1a,b,c,d,2a,b,c,3b,g,4b,g; Ch. 14:5,6.
An important part of Chapter 14 is the connection between redox chemistry and
thermodynamics.
Let’s look at a sample redox system.
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Gold dissolves in hydrochloric acid solution in a reaction with MnO2 as follows:
3 MnO2(s) + 2Au(s) + 12H+(aq) + 8C1-(aq) == > 3Mn2+(aq) + 2AuCl4-(aq) + 6 H2O(l)
Which element is reduced and which oxidized? manganese is reduced from +4 to +2 and
gold is oxidized from 0 to +3.
Conditions of equilibrium for redox reactions can be described using the same concepts of
thermodynamics and equilibrium we discussed at the beginning of the quarter.
What is the equilibrium constant expression?
Keq = [Mn2+]3 [AuCl4-]2
[H+]12 [Cl-]8
Remember, G, the change in Gibbs Free Energy, is a measure of the “tendency to
react” of a system. When the value of G for a given change is negative, the change
occurs spontaneously.
G = Go + RT ln Q = Go + RT ln [ products ]
[reactants]
Go is the free energy change (free energy of products minus free energy of reactants)
when all the substances are in their standard states – when temperature is 298K and
pressure is 1 bar. Conditions in most analytical labs are pretty close to this standard.
The standard state is just a reference so that by comparing free energy of each substance to
this state, we can find differences between free energies of reactants and products.
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For ions in solution the standard state is 1 M concentration.
Remember that Q is the reaction quotient, the value of the reaction expression under any
set of conditions. Only at equilibrium does Q = Keq.
Also at equilibrium, G = 0 so
0 = Go + RT ln Q =>
eliminating the log term gives
Go = - RT ln Keq = - RT ln [ products ]eq
[ reactants ]eq
-G*/RT
Keq = e
.
This is the equation from chapter 6 page 102 that we derived early in the quarter,
So given Go values for each substance involved in a reaction we can calculate Go by
Go = Σ Gfo (products) – Σ Gfo (reactants)
and thermodynamic tables list values for the free energy of formation of nearly all
substances.
We can then use Go of formation values to calculate equilibrium constants. Or we can
use Keq to calculate Go for a reaction.
In addition to free energy, another property is available for describing the tendency of one
substance to react with another when the reaction involves changes in oxidation state.
This property, is called the oxidation potential or oxidation-reduction potential.
Any oxidation-reduction reaction, theoretically at least, can be set up so that the transfer of
electrons from one element to another will take place through a wire.
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For a reaction between zinc and copper, the arrangement is very simple: pieces of the two
metals are connected by a wire and submerged in copper sulfate solution.
The overall reaction is
Zn(s) + Cu2+(aq) == > Zn2+(aq) + Cu(s)
The piece of zinc slowly dissolves, fresh copper from the solution plates out on the copper
metal, and a current flows through the wire.
The process occurring at the zinc electrode is
Zn(s) == > Zn2+(aq) + 2e-
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Is oxidation or reduction occurring at this electrode?
Oxidation. The electrode at which oxidation occurs is the anode.
The liberated electrons move along the wire to the copper electrode, where they are used in
the reaction
Cu2+(aq) + 2e- == > Cu(s)
The electrode at which reduction occurs is called the cathode.
Anode and oxidation both start with vowels. Cathode and reduction both start with
consonants.
Reactions of this kind, showing the processes that occur as electrons are produced or
consumed at an electrode, are the half-reactions we talked about last week, and are
sometimes called electrode reactions.
As we have seen, addition of two half-reactions gives the complete oxidation-reduction
reaction.
The potential difference between the electrodes of our zinc-copper cell can be measured by
adding a galvanometer to the circuit. The amount of the potential difference depends on a
great many variables, but we arrange to keep most of these constant.
Thus we fix the activity of both Cu2+ and Zn2+ at 1;
we make sure that metal of the electrodes is pure and has a clean surface;
we hold the temperature at 25 C and the pressure at 1 bar;
and we arrange to have as small a flow of current as possible.
You can see that these are the same standard state conditions that we used for free energy.
Under such conditions the measurement of potential difference is reproducible and may be
compared with potential differences measured similarly for other oxidation-reduction
reactions.
It is this spontaneous potential difference that is a measure of “tendency to react” for redox
reactions.
If we set up a number of cells similar to the zinc-copper cell, using various metals in
contact with solutions of metal ions, we find that the metals can be arranged in a series
according to their ability to displace one another from solution and according to the size of
the potential difference produced by different pairs.
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Thus zinc displaces copper, copper displaces silver, and silver displaces gold; and the
potential difference of a zinc-silver or a zinc-gold cell is greater than the potential
difference of a zinc-copper cell.
On the other hand, the reverse reactions do not take place appreciably: silver placed in
copper sulfate solution or in zinc sulfate solution causes no detectable reaction.
Experiments of this sort give us what is called the electromotive series of metals, in which
we express the chemical reactivities of various metals with respect to one another.
The electromotive series can be made quantitative by assigning a potential difference to
each half-reaction. This requires that one half-reaction be chosen as a standard and given
an arbitrary potential of zero, so that other half-reactions can be measured against it.
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A convenient choice is the hydrogen couple,
1/2 H2(g) == > H+(aq) + e-
Eo = 0.00 volt
If we arrange a cell with zinc as one electrode and hydrogen as the other (by letting
hydrogen at 1 bar and 25oC bubble over a platinum rod), and use a solution containing H+
and Zn2+ both at an activity of 1, we obtain the potential difference for the overall reaction,
Zn(s) + 2H+(aq) == > Zn2+(aq) + H2(g)
Eo = + 0.762 volt
and we use this number as the potential for the zinc electrode reaction:
Zn(s) == > Zn2+(aq) + 2e-
Eo = + 0.762 volt
There is no way of measuring potentials for half-reactions independently. We get them
only as differences between pairs of half-reactions, so that the actual numbers are no more
than relative voltages compared with the hydrogen electrode.
Our book presents these values as standard reduction potentials in Appendix H.
Zn2+(aq) + 2e- == > Zn(s) Eo = - 0.762 volt
When we reverse the direction of a half reaction, we reverse its sign.
Notice that just like standard free energies, the values are based on an arbitrary standard
and are thus really differences or delta values.
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Also the standard potentials refer to a set of standard conditions, usually with all substances
at unit activities and temperature at 298K and pressure at 1 bar, so we need to have an
equation to calculate situations other than unit activity.
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There is something you should note. The + and -signs given to the voltages are purely
arbitrary, and unfortunately are not uniform from one reference to another.
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Harris says the reduction of iron(III) to iron(II) is +0.771 v
Fe3+ + e- == > Fe2+
Eo = +0.771 v
This makes the oxidation
Fe2+ == > Fe3+ + e-
Eo = -0.771 v
Fortunately Hem also uses the same convention. However, many geochemistry text books
use an alternative approach in which
Fe2+ == > Fe3+ + e-
Eo = +0.771 v
So you must be careful in reading books and articles in which redox chemistry is discussed
and be aware of which convention is being used.
It is just something you have to learn to live with by learning to think about what is going
on rather than just blindly plugging numbers into equations.
This introduction to reduction potential is summarized in Hem starting on page 20. He also
cautions readers about the sign differences between different references on page 21.
Since the voltage values are the same, it is just the sign that is different.
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Harris uses Ecell = Ereduction - Eoxidation and writes reactions as reductions.
Thus for our zinc\copper system
Zn(s) + Cu2+(aq) == > Zn2+(aq) + Cu(s)
Cu2+(aq) + 2e- == > Cu(s)
Eo = +0.339 volt
Zn2+(aq) + 2e- == > Zn(s)
Eo = -0.762 volt
Ecell = +0.339 - (-0.762 ) = + 1.101 volt
If the net cell voltage is positive, then the net cell reaction is spontaneous in the forward
direction. If the net cell voltage is negative, then the reaction is spontaneous in the reverse
direction.
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The geochemical convention writes reactions as oxidations and reverses the signs
Cu(s) == > Cu2+(aq) + 2e-
Eo = -0.339 volt
Zn(s) == > Zn2+(aq) + 2e-
Eo = +0.762 volt
Ecell = Eoxidation - Ereduction = 0.762 - (-0.339) = 1.101 v
So, again, you will not have to worry about this with our textbooks, but be aware and
careful using outside references.
The standard reduction potential is independent of the number of moles of reactant and
product shown in the balanced half reaction. Thus
5 Fe3+ + 5 e- == > 5 Fe2+
Eo = +0.771 v
At the first of the year we said the free energy of a system was related to the amount of
work (energy) we could get out of that system. When G = -69.8 kJ we said that we could
get up to 69.8 kJ of energy from this system.
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Thus work = -G. But in chapter 14 it is shown that work = q*E (charge * potential).
And q = n (mole) * F (Faraday constant 96,485 C/mol)
1 joule of energy is gained or lost when one coulomb of charge is moved through a
potential difference of 1 volt.
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G = -nFE shows us that electric potential difference is directly proportional to free
energy difference.
Since
G = Go + RT ln Q = Go + RT ln [ products ]
[reactants]
we can substitute to get
-n FE = -n FEo + RT ln Q = -n FEo + RT ln
[ products ]
[reactants]
dividing by –nF, gives the Nernst Equation, (I have substituted Eh for E – the system
potential as per Hem).
The h is included to remind you that the potential difference is measured relative to the
standard hydrogen electrode.
Eh = Eo - RT ln
[ products ]
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nF
[reactants]
use the base 10 log and inserting activities we get
Eh = Eo - 2.303 RT log
nF
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Which at 25oC (298 K) becomes
Aproducts
Areactants
Eh = Eo - 0.05916v log
n
Aproducts = Eo - 0.05916v log Q
Areactants
n
As a redox system approaches equilibrium, Eh goes to 0.
At equilibrium Eh = 0 and Q = Keq
Eo = - Go/nF = 0.05916v/n log Keq
nEo/0.05916 = log Keq
Keq = 10nE°/0.05916
Since Eh is proportional to -G, a positive Eh means a spontanteous process while a
negative Eh means the process is spontaneous in the opposite direction.
Again this whole development of redox potential is paralleled in Hem on pages 21 & 22.
There are two differences. First he uses kilocalories as his energy unit instead of the
modern use of kilojoules.
Second, in Hem the Nernst equation is given as
Eh = Eo + 0.05916v log
n
[oxidized] = Eo + 0.05916v log [reactants]
[reduced]
n
[products]
The important difference is the plus sign because Hem has reactants over products, so this
is the same convention we are using in Harris, but be careful which concentrations or
activities go where.
The Nernst equation corrects electrode potentials for concentrations other than unit
activity. The equation can be applied to half reactions or to complete redox equations.
We will try an example of each type.
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Here is a zinc half-reaction
Zn2+ + 2e- < == > Zn(s)
Eo = -0.762 v
What is the half cell potential in a 3.0 x 10-2 M solution of zinc chloride?
What is n? 2
What is the product?
What is the reactant?
The electrons do not appear in the Nernst Equation, except as n.
E = Eo - 0.05916v * log
n
Aproducts = -0.762 - 0.05916 v * log
1
.
Areactants
2
AZn2+
What do we need to do next? Calculate the ionic strength.
 = ½ { .030*22 + .060*12} = 0.090 M
from table 8-1 Zn2+ has an ion radius of 600 pm and the values for the activity coefficient
in the table are 0.485 at 0.05 and 0.405 at 0.10.
The difference is 0.080. We are 80% of the way from 0.05 to 0.10 at 0.090. 80% of 0.080
= 0.064. 0.485 – 0.064 = 0.421
Half way gives  = 0.421 makes a substantial difference
E = -0.762 - 0.02958 v * log
1
. = - 0.762 - 0.056 = - 0.818 v
0.030*0.421
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Calculate the equilibrium constant for the reaction
2 Fe3+ + 3I- < == > 2 Fe2+ + I3Which substance is being reduced? Iron(III) +3 to +2
Which is being oxidized? Iodide -1 to 0
Half reactions at found in Appendix H
2 Fe3+ + 2 e- == > 2 Fe2+
I3- + 2 e- == > 3 I-
Eo = 0.771 v
Eo = 0.535 v
Eocell = Eoreduction – Eooxidation = 0.771 - (+0.535) = +0.236 v
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Keq = 10nE°/0.05916 = 102*0.236/0.05916 = 9.5 x 107
Let’s suppose I have a solution that has the following concentrations
[Fe3+] = 1.0 x 10-2 M
[Fe2+] = 1.0 x 10-2 M
[I3-] = 1.0 x 10-2 M
[I-] = 1.0 x 10-3 M
Ignoring activity, will the reaction go towards products or reactants?
I’m going to solve this two different ways.
First what is n? 2
Using the Nernst equation we get
Eh = Eocell - 0.05916v * log Aproducts
n
Areactants
Eh = 0.236 v - 0.05916v * log [Fe2+]2[I3-]
2
[Fe3+]2 [I-]3
Eh = 0.236 v - 0.05916v * log [10-2]2[10-2]
2
[10-2]2 [10-3]3
Eh = 0.236 v - 0.02958v * log 107
Eh = 0.236 v - 0.02958 *7 = 0.236 – 0.207 = 0.029 v
Eh is positive, therefore ΔG is negative and the reaction proceeds in the direction written –
towards products.
The second method takes advantage of the fact that we know the equilibrium constant.
Keq = 9.5 x 107
We calculated Q to be 107. Q < Keq so, in order to get to equilibrium, the numerator must
increase and the denominator must decrease, i.e. the reaction must proceed towards
products.
One last example.
Calculate the equilibrium constant for the reaction
10
-
2 MnO4 + 3 Mn
2+
+ 2 H2O < == > 5 MnO2(s) + 4 H
+
First what substance is being reduced? Permanganate +7 to +4
Which is being oxidized? Manganese ion +2 to +4
Now what are the two half reactions?
From Appendix H we find
MnO4- + 4 H+ + 3 e- < == > MnO2(s) + 2 H2O
Eo = 1.692 v
Multiplying by two gives
2 MnO4- + 8 H+ + 6 e- < == > 2 MnO2(s) + 4 H2O
Eo = 1.692 v
notice that it does not change the Eo value.
The second half reaction is the oxidation but it will be in the table as a reduction.
MnO2(s) + 4 H+ + 2 e- < == > Mn2+ + 2 H2O
Eo = 1.230 v
Multiplying by three gives
3 MnO2(s) + 12 H+ + 6 e- < == > 3 Mn2+ + 6 H2O
Eo = 1.230 v
if we reverse the oxidation half-reaction and add them together we get.
2 MnO4- + 8 H+ + 6 e- < == > 2 MnO2(s) + 4 H2O
3 Mn2+ + 6 H2O < == > 3 MnO2(s) + 12 H+ + 6 e2 MnO4- + 3 Mn2+ + 2 H2O < == > 5 MnO2(s) + 4 H+
Eocell = Eoreduction – Eooxidation = 1.692 - (+1.230) = +0.462 v
Keq = 10nE°/0.05916 = 106*0.462/0.05916 = 7 x 1046
Will the reaction proceed? How fast will it go?