Chapter 16 Redox Titrations
Problems: 3, 4, 6, 8
Overhead Fig 16-1 adjusted for example
Just as calculating titration curves for acids and bases helped to tie lots of ideas
together and help to give a better understanding of acid/base reaction, let us now look
at redox titration to try to tie up some of the concepts we talked about in the Redox
chapters
16-1 Theory of Redox Titrations
Rather than following the book, I am going to use an example from a lab we
could do. We will use the titration of ascorbic acid (vitamins C) with iodine.
Since this reaction was done in neutral water, pH 7, the most appropriate E to
use is Eo' for pH 7. Thus we will use:
Dehydroascorbate + 2e- +2H+ ºAscorbate Eo'=0.08 and
I3- +2e- º 3I- Eo'= .54
Net rxn: Ascorbate + I3- W 3I- + Dehydroascorbate + 2H+
The physical set up looks something like Figure 16-1 in your text. We will start
with ascorbate in the beaker and have I-3 in the buret.
Now let’s look a bit at the chemistry. Vitamin C or Ascorbic acid and dehydroascorbic
Acid have the structures:
2
What does a titration curve for a redox reaction look like? What are we going to
use for the Y axis? Well, since all redox titrations involve the flow of electrons, all
redox titrations can be monitored by looking at the electrical potential of the solution.
As you found in the last chapter, all you need to monitor the potential of a solution is a
reference electrode, and in inert electrode. So we set up an apparatus like figure 16-1
and monitor E as a function of titrant volume.
So a titration curve will look some thing like this:
As the titration occurs, what do you have in
the solution? A mixture of Ascorbate,
dehydroascorbate, I3- and I-. The K’s of
redox equations are usually very large (do
you remember how to calculate the K for
this reaction? Give it a try)
As a result you can usually assume that the
reaction goes to 100% completion so you
can calculate the [] of each of the ions
using a simple reaction table, rather than
worrying about Keq. So where does the
Figure 2 Potentiometric titration curve
potential you measure come from?
Well there are two ½ reactions going on:
Dehydroascorbate + 2e- +2H+ ºAscorbate Eo’=0.08 and
I3- +2e- º 3I- Eo'= .54
And they are going on simultaneously, so either one can be used to calculate the
potential of the cell vs the reference electrode. The trick is that only 1 will give you a
reasonable value, the other will make your calculator give you an error because you
are either multiplying or dividing by 0.
There are only three equations you need to know when working with Redox
titration curves. One equation is applied to get the titration curve up to the equivalence
point, a second to get the E at the equivalence point, and a third (that looks a whole lot
like the first) is used after the equivalence point.
Now notice that the book uses a calomel electrode as a reference electrode, and
will occasionally throw in a factor of E=E+-Ecalomel to describe the actual measured
potential. In this example I will assume we are using a hydrogen electrode as our
reference. This means that the reference potential is 0.00 so we don’t have to put it
into any of our equation so it won’t mess us up.
3
The next point is important we have a platinum wire as an inert electrode to
detect the potential of the solution in the beaker. What potential will it be detecting the
potential of the Iodine ½ reaction or the potential of the ascorbic acid half reaction??
The answer is both. Since these reagents are mixed together, they are in
equilibrium with each other, thus Eacorbate= Eiodine and any point in the titration.
Since the two redox systems we are titrating together are in equilibrium, the
potential of the two pairs are equal. As a result the potential of the of the reactant
mixture can be found by calculating the E for either of the redox pairs. Because of this
we always choose the redox pair that is the easiest to calculate.
Let's define concentrations and volumes for this example. I am starting with a 40
mls of a .045 M (.09N) solution of ascorbate and I am titrating it with 0.05 M (.1N) I3-.
Location of E.P.
As usual lets first determine how much titrant is need to get to the equivalence
point.
If I have 40 mL. of .045 M solution, I have 1.8 mMoles of ascorbic acid. The
reaction stoichiometry is 1:1 so I will need 1.8 mMoles of I3.
I therefore need 1.8/.05, or 36 mL. of the iodine solution to reach the
equivalence point.
(It isn't always this easy, will examine again in a minute)
E at the initial point hooray, you can't calculate this one. To calculate an E
using the Nernst equation, you need to have good, well defined concentrations for both
the oxidized and the reduced species in a redox pair. For the initial point, no titrant as
in solution, so it can't help you. You only have the analyte, and since none of it has
been converted to its redox partner, the concentration for one of the two redox species
is not well defined. This is can't be calculated.
E between the initial point and the equivalence point I said earlier the E of
the system can be determined using the E of either Redox Pair, and you choose the
most convenient pair. Between the initial point and the equivalence point you have
measurable quantities of ascorbate and dehydroascorbate. However all the Iodine
(actually triiodide) you add is converted into iodide, so the concentration of Iodine is not
measurable. Thus the ascorbate-dehydroascorbate is the system you use to get the E
of the cell. Bottom line between start and EP use analyte
4
The analyte is the Vitamin C/Dehydroascorbic acid pair. So we will focus on this
½ reaction and ignore the other. Let’s write down this ½ reaction:
Dehydroascorbate + 2e- +2H+ ºAscorbate Eo'=0.08
Note: that I am writing down this reaction as the reduction reaction, even
though in my net reaction it is actually an oxidation! Get used to this. Because our
reduction potentials are for the reduction reactions, all of our math is always set up for
reduction reactions, even if the reaction is reversed and turns into an oxidation in the
net reaction!
E=Eo -(.059/n) log ([reduced]/[oxidized])
The potential for the ascorbate couple is:
Second Note: Although H+ occurs in the half reaction, I am ignoring it in this equation.
In this problem we are using Formal potentials for a solution at pH 7. I am assuming
that the solution is buffered, and the pH does not change. If this is so, then it may be
removed from the cell potential equation.
Let’s figure out the moles of reactants, and fill in a reaction table.
Moles Ascorbate = 40 ml x .045 M = 1.8 mmoles
Moles I3- = 10 ml x .05M = .5 mmoles
Reaction table:
Initial
Reaction
Net
Ascorbate +
1.8
-.5
1.3
I3 - W
.5
-.5
0.0
3I- + Dehydroascorbate +
0
0
+.5(3/1)
+.5
1.5
.5
2H+
1x10-7
1x10-7
1x10-7
Since our solution was buffered at pH 7, the H+ concentration does not change, so this
column of our reaction table can be ignored
[Ascorbate]=1.3 mMoles/(40+10) ml = .026 M
[Dehydroascorbate]=.5 mMoles/(40+10) ml = .01M
5
Note: I have converted both numbers into moles by dividing by the volume. A
sharp student might have figured out that I could skip this step because the volume
term will drop out of the Nernst equation. However as you will see in when we
calculate the point after the equivalence point, this doesn’t always happen, so I don’t
want you getting into a bad habit here, so you should always be sure to convert to
molarity so don’t skip this step and have it bite you in the ...... on a tricky test question.
Plugging these numbers into the Nernst Equation
At the equivalence point
The equation for the equivalence point may be fairly easily derived, if the
stoichiometry of the reaction is not too difficult. For those who are interested look on
pages 424 of the book. I will not ask you to derive the equation, but you do need to
memorize it. The equation is:
Thus the E at the equivalence point is the mean of the to Eo's weighted by the
number of electrons in each half reaction.
For this example:
6
Beyond the equivalence point
Now all the Ascorbate has been turned into dehydroascorbate, so only a very
tiny amount remains. On the other hand measurable amount of both Iodine and Iodide
exist in solution, so it is most convenient to use the E of this half reaction to determine
the E of the Cell. Let's try the point after 40 mL. of titrant have been added.
Let’s figure out the moles of reactants, and fill in a reaction table.
Moles Ascorbate = 40 ml x .045 M = 1.8 mmoles
Moles I3- = 40 ml x .05M = 2.00 mmoles
Reaction table:
Initial
Reaction
Net
Ascorbate +
1.8
-1.8
0
I3 - W
2
-1.8
0.2
3I- + Dehydroascorbate +
0
0
+1.8 (3/1)
+1.8
5.4
1.8
2H+
1x10-7
1x10-7
1x10-7
Again our solution was buffered at pH 7, the H+ concentration does not change, so this
column of our reaction table can be ignored
[I3-] = .2 mmoles/(40+40) ml = .0025M
[I-] = 5.4 mM /(40+40 ml) ml = .0675 M
And E is determined using the Nernst of the I3-/I- pair.
Notice here that the numerator of the Nernst quotient has a cubed term, while
the denominator was to the 1 power. This is a case where you had to change to
molarity. If you didn’t and used just moles instead of moles/volume, the volumes terms
don’t cancel, and you would have been up the creek without a paddle.
Example 2
With the exception of the [I-]3 term the above titration was fairly simple because
the number of electrons in both half reactions was equal. Things get a little messier if
they aren't. Let's just briefly look at the example of:
2Fe3+ +Sn2+ º 2Fe2+ + Sn4+
Fe3+ + e- W Fe2+ E=0.77
Sn4+ + 2e- W Sn2+ E=0.15
Say we are starting with 40 mL. of .196 M Fe3+. First where is the equivalence
7
point when you titrate with .1 M Sn2+?
The tricky part of this problem is keeping track of the stoichiometry. Each mole
2+
of Sn reacts with 2 moles of Fe3+, therefore if you start with 40(.196) or 7.85 mMoles
of Fe3+, you only need 3.92 mMoles of Sn2+, or 39.2 mL. of .1M Sn2+. You have to
remember to keep this factor of two in all the appropriate places in the calculations.
For points on the curve, let’s try 10.00, 39.2, and 44.2 mls of Sn2+
10 ml
Starting with moles and a reaction table
40 ml of .196M Fe3+ = 7.84 mmole Fe3+
10 ml of .100M Sn2+ = 1.00 mmole Sn2+
2Fe3+ +
Initial 7.84
Rxn -1(2/1)
Net 5.84
Sn2+ º
1.00
-1
0.00
2Fe2+ +
0.00
+1(2/1)
2.00
Sn4+
0.00
+1
1.00
[Fe3+] = 5.84 mmoles/(40+10) ml = .1168M
[Fe2+] = 2.00mmole/(40+10) ml = .040M
Using the Analyte ½ reaction
Fe3+ + e- 6Fe2+
(And ignoring the actual stoichiometry in the net reaction)
The Nerst Equation is:
E = .77 - .059/1 log (.04/.1168)
= .77 -.059log(.3425
=.77+.027
=.797 V
39.2 ml (equivalence point)
Let’s just prove this is the equivalence point using moles and the reaction table:
40 ml of .196M Fe3+ = 7.84 mmole Fe3+
39.2 ml of .100M Sn2+ = 3.92 mmole Sn2+
2Fe3+ +
Sn2+ º
2Fe2+ +
Initial 7.84
3.92
0.00
Rxn -3.92(2/1)
-3.92
+3.92(2/1)
Net 0.00
0.00
7.84
The E at the equivalence point is:
(2(.15)+.77)/3 =.357
Sn4+
0.00
+3.92
3.00
8
44.2 ml
Starting with moles and a reaction table
40 ml of .196M Fe3+ = 7.84 mmole Fe3+
44.2 ml of .100M Sn2+ = 4.42 mmole Sn2+
2Fe3+ +
Initial 7.84
Rxn -7.84
Net 0
Sn2+ º
4.42
-7.84(½)
0.5
2Fe2+ +
0.00
+7.84
7.84
Sn4+
0.00
+7.84(1/2)
3.92
[Sn2+] = .5 mmoles/(40+44.2) ml = .005938M
[Sn4+] = 3.92mmole/(40+44.2) ml = .04656M
Using the Titrant ½ reaction
Sn4+ + 2e- 6Sn2+
(And ignoring that this reaction is actually an oxidation in the net reaction)
E = .15 - .059/2 log (.005938/.04656)
= .15 + .026
= .176 V
16-2 Redox Indicators
A redox indicator is a compound that changes color when if goes from its
oxidized state to its reduced state.
Just as in acid/base indicators we see distinct color when we have 10:1 or 1:10
ration of indicator forms. Since we want our indicator to change color as close to our
equivalence point as possible, you choose an indicator the has an Eo that closely
matches the E of your equivalence point ( to +/- .06V) . A list of indicators is given in
Table 16-1
Many times. However indicators are not needed. Frequently the redox reagent
themselves are distinctly colored.
16-3 Titrations involving Iodine.
Not covered unless decide to do in lab.