Oxidation-Reduction
Titration
Dr. A.K.M. Shafiqul Islam
August 24, 2007
Oxidation Reduction Reaction



Reactions of metals or any other organic
compounds with oxygen to give oxides are
labeled as oxidation.
The removal of oxygen from metal oxides to
give the metals in their elemental forms is
labeled as reduction.
In other words, oxidation is addition of oxygen
and removal of hydrogen whereas reduction is
addition of hydrogen and removal of oxygen.
Oxidation Reduction Reaction
Reactions of metals or any other organic compounds
with oxygen to give oxides are labeled as oxidation.
The removal of oxygen from metal oxides to give the
metals in their elemental forms is labeled as
reduction.
In other words, oxidation is addition of oxygen and
removal of hydrogen whereas reduction is addition of
hydrogen and removal of oxygen.
Chemical reaction



Reduction-Oxidation reaction commonly known
as RedOx reaction
Ox1 + Red2 ⇌ Red1 + Ox2
Chemical reaction based oxidation and reduction
reaction is known as RedOx reaction
Fe2+ + Ce4+ → Fe3+ + Ce3+
(1)
Electrochemical Cells
There are two kinds of electro chemical cells, galvanic or
electrolytic.
In galvanic cells, the chemical reaction
spontaneously to produce electrical energy.
occurs
In a electrolytic cell, electrical energy is used to force
the non spontaneous chemical reaction.
Galvanic cells are of importance in our further
discussion as we will be discussing the spontaneous
chemical reaction to produce electrical energy.
If a solution containing Fe2+ is mixed with another solution
containing Ce4+, there will be a redox reaction situation due
to their tendency of transfer electrons. If we consider that
these two solution are kept in separate beaker and
connected by salt bridge and a platinum wire that will
become a galvanic cell. If we connect a voltmeter between
two electrode, the potential difference of two electrode can
be directly measured.
The Fe2+ is being oxidised at the platinum wire (the anode):
Fe2+ → Fe3+ + e-
The electron thus produced will flow through the wire to the
other beaker where the Ce4+ is reduced (at the cathode).
Ce4+ + e- → Ce3+
The reaction involve electron transfer
Fe2+ → Fe3+ + eCe4+ + e- → Ce3+
(2)
(3)

Equation (2) & (3) are called half reactions

No half reaction can occur by itself


There must be an electron donner (reducing agent) and
an electron accepter (oxidizing agent)
Fe2+ is the reducing agent and Ce3+ is the oxidizing agent
Before we start the discussion of the oxidation reduction
titration curve construction, we should understand the
Nernst equation which was introduced by German scientist,
Wlater Nernst in 1889.
This equation express the relation between the potential of
metal and metal ion and the concentration of the ion in the
solution.
Lets consider the following chemical reaction :aA + bB
⇋ cC + dD
The change in free energy is given by the equation
ΔG = ΔGo + 2.3RT log [C]c x [D]d / [A]a x [B]b
From the relation ship of the free energy and cell
potential, we can get
ΔG = -nFE
In a standard states, free energy will be
ΔGo = -nFEo
Hence, the above equation can be written as
-nFE = -nFEo + 2.3RT log [C]c [D]d / [A]a [B]b
After dividing both side with –nF, we can get the expression as
E = Eo – 2.3 RT/ nF log [C]c [D]d / [A]a [B]b
At 25ºC, the equation can be written as :-
E = Eo – 0.059/n log [C]c [D]d / [A]a [B]b
This equation is known as the “Nernst equation” that
correlate the electrode potential with the concentration
of the ionic species.
Redox Titration Curve
Redox titration is monitored by observing the change of electrode
potential. The titration curve is drawn by taking the value of this
potential against the volume of the titrant added. Unlike other
titration curve the p values are substituted by electrode potential
values in the curve.
The redox reaction is rapid and the system is always in equilibrium
throughout the titration. The electrode potential of the two half
reaction are always identical.
If we consider the oxidation of Fe (II) with standard Ce (IV), then we
can write the equation as follows :Fe2+ + Ce4+ ⇋ Fe3+ + Ce3+
The electrode potential of the two half reaction will be always
identical.
For iron, the electrode potential will be
E = EºFe – 0.059 log [Fe2+] / [Fe3+]
For cerium the electrode potential will be
E = EºCe – 0.059 log [Ce3+] / [Ce4+]
Therefore, either of the electrode potential could be utilised to
calculate the potential of the solution.
The utilisation of the either equation is based on the stage of
titration. Prior to the equivalence point, the concentration of
Fe(II) and Fe(III) are appreciable compare to Ce(IV) ion which is
negligible because of the presence of large excess of Fe(II).
Beyond the equivalence point, the concentration of Ce(IV) and
Ce(III) is readily computed from the addition and the electrode
potential for the Ce(IV) could be used.
At equivalence point, the concentration of the oxidised and
reduced forms of the two species are such that their attraction
for electron are identical. At this point, the reactant species
concentration and product species concentrations ratios are
known and they are utilised to calculate the potential of the
solution.
# 50.0 ml of 0.05M Fe2+ is titrated with 0.1M Ce4+ in a sulphuric
acid media at all times. Calculate the potential of the inert
electrode in the solution at various intervals in the titration and
plot the titration curve. Use 0.68V as the formal potential of the
Fe2+- Fe3+ system in sulphuric acid and 1.44V for the Ce3+- Ce4+
system.
Initial step : After addition of 5.0 ml of Ce4+
As because the Ce4+ is too small, we are considering the iron (Fe)
electrode potential to calculate the solution potential.
[Fe3+] = 5.0ml X 0.10M / (50.0 + 5.0) ml
= 0.5 mmol / 55.0ml
Similarly, [Fe2+] = (50.0ml X 0.05M – 5.0 ml X 0.1M) / 55.0 ml
= 2.0 mmol / 55.0 ml
Substituting the values to the standard electrode potential
equation, we can get
E = EºFe – 0.059 log [Fe2+] / [Fe3+]
E = Eº – 0.059 log 2.0 / 0.5
E = 0.68 -0.036 = 0.64 V
Step-2 : Equivalence point
At equivalence point in the titration of Fe(II) and Ce(IV), the
potential of the solution is controlled by both the half reaction.
Eeq=E0Ce -0.059 log [Ce3+] / [Ce4+]
Eeq = E0Fe - 0.059 log [Fe2+] / [Fe3+]
Adding the two expression, we can get
2Eeq = E0Ce + E0
Fe
- 0.059 log [Ce3+] [Fe2+]/ [Ce4+][Fe3+]
At equivalence the concentration of Fe3+ = Ce3+ and Fe2+ = Ce4+
We can get ,
2Eeq = E0Ce + E0
Fe
- 0.059 log [Ce3+] [Ce4+]/ [Ce4+] [Ce3+]
Eeq = (E0Ce + E0
Fe)
/2
Eeq = (1.44 + 0.68) / 2 = 1.06 V
Step-3 : After addition of 25.1 ml Ce4+
At this stage, the concentration of Fe(II) is very small and we can
neglect the value and for convenience, we will utilise the Ce(IV)
electrode potential to calculate the solution potential.
[Ce3+] = (25.0ml X 0.1M ) / (50.0 + 25.1) ml
= 2.5 mmol / 75.1 ml
[Ce4+] = (0.1 ml X 0.1M) / 75.1 ml
E = EºCe – 0.059 log [Ce3+] / [Ce4+]
 E = 1.44 – 0.059 log 2.5 / 0.01
= 1.30 V
Redox Titrations

Equipment for
obtaining a titration
curve for a redox
titration.
RedOx Titration Curve
1.8
1.6
E, volts
1.4
1.2
1.0
0.8
0.6
0
20
40
60
80
100
120
mL Ce4+
140
160
180
200
Redox Titrations
The net balanced redox equation is the
sum of the two half-cell reactions.
It may be necessary to multiply one or
both half-cells by some coefficient so that
the same number of electrons are lost by
the substance that is oxidized as are
gained by the substance reduced.
Balancing simple redox reactions
Cu0(S)
2Ag +(aq) + 2e-
Cu2+(aq) + 2e2Ag (S)
Cu0(S) + 2Ag +(aq) + 2eCu2+(aq) + 2Ag (S) + 2e-
Cu0(S) + 2Ag +(aq)
Cu2+(aq) + 2Ag (S)
Number of e-s involved in the overall reaction is 2
Balancing complex redox reactions
Fe+2(aq) + MnO4-(aq)
Oxidizing half:
Fe+2(aq)
Reducing half:
MnO4-(aq)
Balancing atoms:
Balancing
-(aq)+
MnO
4
oxygens:
Mn+2(aq) +
Fe+3(aq)
Fe+3(aq) + 1eMn+2(aq)
Mn+2(aq) + 4H2O
Balancing complex redox reactions
Balancing hydrogens:
MnO4-(aq)+8H+
Oxidation
numbers: Mn = +7,
O = -2
Reaction happening in an acidic medium
Mn+2(aq) + 4H2O
Mn = +2
Balancing electrons:
The left side of the equation has 5 less electrons than the right side
MnO4-(aq)+8H++ 5eMn+2(aq) + 4H2O
Reducing Half
Balancing complex redox reactions
Final Balancing act:
Making the number of electrons equal in both half reactions
[Fe+2(aq)
[MnO4-(aq)+8H++ 5e5Fe+2(aq)
MnO4-(aq)+8H++ 5e-
Fe+3(aq) + 1e- ]× 5
Mn+2(aq) + 4H2O]×1
5Fe+3(aq) + 5eMn+2(aq) + 4H2O
5Fe2++MnO4-(aq)+8H++ 5e5Fe3+ +Mn+2(aq) + 4H2O + 5e-
Redox Titrations


Many of the end point indicators used in
redox titrations are substances that are
either oxidized or reduced at given
potentials.
Much like acid-base indicators, the
electrical potential at the end point
oxidizes or reduces the indicator to one of
the colored forms to signal the end point.
RedOx Indicators
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
If the titrant is highly colored, this color
may be used to detect end point.
0.02 M potassium permanganate is deep
purple. A dilute solution is pink. The
product of its reduced form (Mn2+)
colorless.
Titration of unknown sample of Iron
Vs KMnO4
The unknown sample of iron contains, iron in Fe2+
oxidation state. So we are basically doing a redox
titration of Fe2+ Vs KMnO4
5Fe2++MnO4-(aq)+8H+
5Fe3+ +Mn+2(aq) + 4H2O
Vinitial
Vfinal- Vinital= Vused (in mL)
Important requirement:
The concentration of
KMnO4 should be
known precisely.
KMnO4
Vfinal
End point:
Pale Permanent
Pink color
250mL
250mL
250mL
Redox Indicators
Titrations Involving Iodine
One of the most common redox titrations
involve either using iodine (I2) as a mild
oxidizing agent or iodide (I-) as a mild
reducing agent.
Iodine as oxidizing agent:
I2 + 2 e - = 2 I When iodine is used as the titrant the
method is known as iodimetry.
Titrations Involving Iodine
I2 is not very soluble in water (only about
1.3 x 10 -3 mol/L). It solubility is increased
in the presence of excess iodide by the
formation
of the triiodide (I3-) species,
I2 + I - = I 3 –
So it is really the triiodide species, though it
will commonly be referred to as iodine
that is involved in the chemical reactions.
Titrations Involving Iodine

Due to the difficulty is maintaining the
concentration of I2, (limited solubility in water,
appreciable vapor pressure of the I2.) iodometric
methods are more commonly used. The amount
of I2 produced by the action of the oxidizing
analyte on excess iodide is usually titrated with
standardized sodium thiosulfate, Na2S2O3.
Titrations Involving Iodine
Iodide as reducing agent:
2 I- = I2 + 2 e When iodine is produced by the addition of
an oxidizing analyte to an excess of
iodide, the method is as iodometry.
Titrations Involving Iodine
Thiosulfate (S2O3-2) is commonly used in
titration reactions involving iodine, both
for iodimetric and iodometric methods.
The iodimetric methods generally involve
an excess of standard I2 (as I3 -) followed
by back titration with standard thiosulfate.
Titrations Involving Iodine
In the iodometric methods, an excess of
iodide is added to the sample of an
oxidizing analyte and a stoichiometric
amount of iodine (I2 or I3- ) is produced.
This iodine is titrated with a standard
solution of thiosulfate. This reaction is
shown on the following slide.
Titrations Involving Iodine
Titrations Involving Iodine
The reaction with iodine needs to occur in
a solution whose pH < 9 to prevent side
reactions which produce iodates (IO3-).
Generally acetic acid is added to the
analyte mixture before titration to assure
the proper pH. In some cases,
appropriate pH buffers may also be added.
Titrations Involving Iodine
The titrant solution of thiosulfate cannot
be prepared directly. It is made to an
approximate concentration and then
standardized with a primary standard
oxidizing agent by iodometry. The
thiosulfate solution is unstable if the pH <
5, undergoing the following
disproportionation reaction.
S2O3-2 + 2 H+ < == > H2SO3 + S
Titrations Involving Iodine

This disproportionation reaction is
prevented by using freshly boiled
deionized water as the solvent and adding
a small amount of NaOH. Although the
thiosulfate needs to be stored in a basic
solution, as mentioned earlier, its reaction
as a reductant titrant needs to occur in an
acid solution.
Chapter 16 – Iodine Methods

The indicator for both iodimetric and
iodometric titrations is a starch solution; in
the presence of iodine, it shows a blue or
purple color. The starch indicator should
not be added when I2 is in large excess
because the desorption of I2 from the
starch molecule is not reversible.
Starch Indicator

The repeating
amylose unit in
the starch
molecule. Starch
is a polymer of
amylose.
Starch Indicator

The starch-iodine
complex where the
sugar chain forms a
helix about I6 units.
Iodimetric titrations: Titrations with
standard iodine (actually I3)
Species analyzed
SO2
Oxidation reaction
SO2 + H2O < == > H2SO3
H2SO3 + H2O < == > SO24- + 4H+ + 2e-
H2S
H2S < == >
Zn2+, Cd2+,
Hg2+, Pb2+
M2+ + H2S  MS(s) + 2H+
MS(s) < == > M2+ + S + 2e-
S(s) + 2H+ + 2e-
Iodimetric titrations: Titrations with standard iodine
(actually I3)
Species analyzed
Cysteine, glutathione,
2emercaptoethanol
Aldehydes
2e-
Oxidation reaction
2RSH < == > RSSR + 2H+ +
H2CO + 3OH- < == >
HCO2- + 2H2O +
Glucose (and other reducing sugar)
O
RCH + 3OH- < == > HCO2- + 2H2O + 2eAscorbic acid
(or vitamin C)
See next slide
Ascorbic acid (or vitamin C)
Iodometric titrations: Titrations of iodine (actually
I3) produced by the analyte
Species analyzed
Reaction
HOCl
HOCl + H+ + 3I- < == > Cl
Br2
Br2 + 3 I- < == > 2 Br
IO3-
2 IO3- + 16 I
-
+ 12 H+ < == > 6 I3- + 6 H2O
IO4-
2 IO4- + 22 I
-
+ 16 H+ < == >
O2
H2 O
H2O2
-
-
+ I3- + H2O
+ I3-
8 I3- + 8 H2O
O2 + 4 Mn(OH)2 + 2 H2O < == > 4 Mn(OH)3
2 Mn(OH)3 + 6 H+ + 6 I - < == > 2 Mn2+ + 2 I3- + 6
H2O2 + 3 I
-
+ 2H+ < == > I3- + 2 H2O
Iodometric titrations: Titrations of iodine (actually
I3) produced by the analyte
Species analyzed
Reaction
O3
O3 + 3 I
NO2-
2 HNO2 + 2 H
-
+ 2 H+ < == >
+
+ 3I
O2 + I3- + H2O
< == >
-
2 NO + I3- + 2 H2O
S2O82-
S2O82- + 3 I
-
< == >
2 SO42- + I3-
Cu2+
2 Cu2+ + 5 I
-
< == >
2 CuI(s) + I3-
MnO4-
MnO2
2 MnO4- + 16 H+ + 15 I
MnO2(s) + 4 H
+
-
+ 3I
< == > 2 Mn2+ + 5 I3- + 8 H2O
-
< == > Mn2+ + I3- + 2 H2O