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Unformatted text preview: Both oxidation and reduction occur during a
The equivalence point is based on the
concentration of the oxidized and reduced form
of all species involved.
Aox + Bred = Ared + Box
Ce4+ + Fe2+ = Ce3+ + Fe3+ For a REDOX titration, the equivalence point is
the point where Eforward = Ereverse.
Ecell = 0
Since the E values are concentration dependent,
we must rely on the Nernst equation
E = Eo - 0.0592
n [A ]
ox At the equivalence point:
EAo log red = EBo log red
Since Ared=Box and Bred=Aox at the
equivalence point, we can reduce this to:
Eeq = n E o + nBEBo
Eeq = A A
nA + nB nAEAo + nBEBo
nA + nB Determine the Eeq for the following reaction:
Fe2+ + Ce4+ = Fe3+ + Ce3+ Eo Fe3+/Fe2+
Eo Ce4+/Ce3+ = +0.771V
= +1.70 V nFe = 1, nCe = 1
Eeq = 1.70V + 0.771V
= 1.24 V
2 Determining Eeq becomes much more
complicated for more complex systems.
The inclusion of one additional species, like H+ is
Lets look at another example.
6Fe2+ + Cr2O72- + 14H+ = 6Fe3+ + 2Cr3+ + 7H2O [ Fe2+] [ Cr3+]2 7 Eeq = EoFe + 6EoCr - log [ Fe3+] [ Cr O 2-] [ H+]14
27 [ Fe2+]
Eeq = EoFe - .0592 log
o = 0.771V
Eeq = EoCr -0.0592 log
6 [ Cr3+]2
[ Cr2O72-][ H+]14 Eo = 1.33 V Eeq = EoFe + 6EoCr
7 - Eeq = 1.25V - 8.46 x10-3 At the equivalence point:
[ Fe3+] First, obtain both half reactions: 2 [ Cr3+]
[ H+]14 log So for this reaction, the equivalence point
is dependent on both [ Cr3+] and [ H+]. = 6 [Cr2O72-]
= 3 [Cr3+] This explains why we commonly work in 1M acid
and with dilute solutions. As with acid/base titrations, we can get either to
the following types of curves. If the reaction was conducted under these
conditions then: Eeq = 1.25V - 8.46 x10-3 log 2 [ Cr3+]
[ H+]14 2 [ Cr3+]
[ 1 ]14 Eeq shows only a small dependence on Cr3+. Based on the type of reaction. over titration Ecell equivalence
buffer region percent titration Unlike acid/base titrations, we really can’t do
much with this region.
While some Fe3+ must be present, we can only
guess what the concentration is.
No Ce4+ or Ce3+ are present, so we don’t have
a complete reaction. Fe3+ Fe2+
10 Fe2+/Fe3+ 0.715
0.828 We can simplify our calculations by using the %
EFe = 0.771 - 0.0592 log %Fe3+
%Fe E 9
0.11 Between 0% and 100% titration, we can use the
Nernst equation for Fe2+/Fe3+.
EFe = 0.771 - 0.0592 log
There is no significant level of Ce4+ to work
with anyway. 0.84
0.8 There was only
a change of
0.113 V from 10
to 90% titration 0.78
0 20 40 60 80 100 1.3
1.1 1.70V + 0.771V
0.9 Note the large
jump in E at the
0 At greater than 100% titration, the predominate
change is that Ce4+ is being added and diluted
into a solution of Ce3+.
All Fe2+ has been converted to Fe3+ and no
longer figures into the calculations.
We just need to keep track of the amounts of
Ce3+ and Ce4+ as well as the total volume of
the system. At the equivalence point:
[Ce3+] = 200 ml
= 0.05M At 10% overtitration, we’ve added an additional
10ml of our Ce4+ solution so:
[Ce4+] = 0.0476 M
= 0.00476 M 20 40 60 80 100 120 So far, we’ve not needed to worry about the
actual concentrations. However, to know the
volumes involved, we now need them.
Lets assume that we started with 100ml of a
Fe2+ solution and our titrant was 0.1M Ce4+.
Now we can determine the E for overtitration. We can again use the Nernst equation to
determine the E for our system.
In this case, however, we’ll use the values for
Ce3+ and Ce4+.
0.0476 E = 1.40 - 0.0592 log 0.00476
At 20% overtitration, E would be 1.36. 1.4
0 20 40 60 80 100 120 <
n log or > 10 [ Indred]
[ Indox] (Phen)3Fe3+ + e = (Phen)3Fe2+
N pale blue
3 Eo = 1.06V red HSO3 Eo = 0.80V irreversible reversible NH Starch-I If the goal is to reduce your analyte to a
single oxidation state, you can use
either a Jones or Walden Reductor.
Both are columns containing a metal.
You slowly wash your sample
through the column with water. If the goal is to oxidize your analyte to a single
form, no material is available that can be used
as a column.
You must have a method to remove any
excess oxidizing reagents prior to titration.
HCl Often, the results in the preparation step being
more complicated than the rest of the method. Ag+ Methanol is added to remove the reactive
C5H5N.SO3 + CH3OH C5H5N(H)SO4CH3 This is done because the SO3 complex can
react with water as well as many other species. ...
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