At the equivalence point:

Both oxidation and reduction occur during a titration.

The equivalence point is based on the

concentration of the oxidized and reduced form of all species involved.

A_ox + B_red = A_red + B_ox Ce⁴⁺ + Fe²⁺ = Ce³⁺ + Fe³⁺

For a REDOX titration, the equivalence point is the point where E_forward = E_reverse.

E_cell = 0

Since the E values are concentration dependent, we must rely on the Nernst equation

E = E_o - log ^0.0592 n

[A_red] [A_ox]

At the equivalence point:

E_A^o - 0.0592

n_A log [A_red]

[A_ox] E_B^o - 0.0592

n_B log [B_red] [B_ox]

=

Since A_red=B_ox and B_red=A_ox at the equivalence point, we can reduce this to:

E_eq = n_AE_A^o + n_BE_B^o n_A + n_B

E_eq = n_AE_A^o + n_BE_B^o n_A + n_B

Determine the E_eq for the following reaction:

Fe²⁺ + Ce⁴⁺ = Fe³⁺ + Ce³⁺

E

E

E_eq = 1.70V + 0.771V = 1.24 V 2

Determining E_eq becomes much more complicated for more complex systems.

The inclusion of one additional species, like H⁺ is relatively common.

Lets look at another example.

6Fe²⁺ + Cr₂O₇^2- + 14H⁺ = 6Fe³⁺ + 2Cr³⁺ + 7H₂O

First, obtain both half reactions:

E_eq = E^o_Fe -

E_eq = E^o_Cr - 0.0592

1 log [ Fe²⁺] [ Fe³⁺] E^o = 0.771V

0.0592

6 log [ Cr³⁺]² [ Cr₂O₇^2-][ H⁺]¹⁴ E^o = 1.33 V

7 E_eq = E^o_Fe + 6E^o_Cr - log _{[ Fe}3+^{[ Fe}] [ Cr^{2+] [ Cr}₂O₇^2-3+] [ H^{]2 +}]¹⁴

At the equivalence point:

[ Fe

] = 6 [Cr

O

] [ Fe

] = 3 [Cr

]

E_eq = E^o_Fe + 6E^o_Cr 7

0.0592

- 7 2 [ Cr³⁺] [ H⁺]¹⁴ log

log

E_eq = 2 [ Cr³⁺]

[ H⁺]¹⁴ 1.25V - 8.46 x10^-3

So for this reaction, the equivalence point is dependent on both [ Cr

] and [ H

].

This explains why we commonly work in 1M acid and with dilute solutions.

If the reaction was conducted under these conditions then:

log

E_eq = 2 [ Cr³⁺]

[ 1 ]¹⁴ 1.25V - 8.46 x10^-3

E_eq shows only a small dependence on Cr³⁺.

As with acid/base titrations, we can get either to the following types of curves.

Based on the type of reaction.

buffer region

over titration

equivalence point

Ecell

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.

Between 0% and 100% titration, we can use the Nernst equation for Fe²⁺/Fe³⁺.

E_Fe = 0.771 - 0.0592 log

There is no significant level of Ce⁴⁺ to work with anyway.

We can simplify our calculations by using the % titration.

E_Fe = 0.771 - 0.0592 log %Fe²⁺

%Fe³⁺

[ Fe²⁺] [ Fe³⁺]

Fe³⁺ Fe²⁺ Fe²⁺/Fe³⁺ E

10 90 9 0.715

20 80 4 0.735

30 70 2.33 0.749

40 60 1.5 0.761

50 50 1 0.771

60 40 0.67 0.781

70 30 0.43 0.793

80 20 0.25 0.807

90 10 0.11 0.828 ^0.7

0.72 0.74 0.76 0.78 0.8 0.82 0.84

0 20 40 60 80 100

There was only a change of 0.113 V from 10 to 90% titration

1.70V + 0.771V 2

0.6 0.7 0.8 0.9 1 1.1 1.2 1.3

0 20 40 60 80 100 120

Note the large jump in E at the equivalence point.

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.

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.

At the equivalence point:

Solution volume = 200 ml [Ce³⁺] = 0.05M

At 10% overtitration, we’ve added an additional 10ml of our Ce⁴⁺ solution so:

[Ce³⁺] = 0.0476 M [Ce⁴⁺] = 0.00476 M

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⁺.

E = 1.40 - 0.0592 log = 1.34

At 20% overtitration, E would be 1.36.

0.0476 0.00476

0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4

0 20 40 60 80 100 120

[ Ind_red] [ Ind_ox] 0.0592

n log

or > 10

<

N N

3 Fe

(Phen)₃Fe³⁺ + e = (Phen)₃Fe²⁺

pale blue red E^o = 1.06V

NH HSO₃

E^o = 0.80V irreversible reversible

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.

HCl

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.

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 complex.

C₅H₅N^.SO₃ + CH₃OH C₅H₅N(H)SO₄CH₃ This is done because the SO₃ complex can

react with water as well as many other species.