Reduction - Oxidation Titrations Theory and Application

2. Reduction - Oxidation Titrations
Theory and Application
Ashraf M. Mahmoud, Associate P
1

3. Oxidation-Reduction (Redox)
Definitions & Terms
2 Fe2+ + Cl2 2Fe3+ + 2Cl
2 Fe2+ - 2e 2 Fe3+ (Loss of electrons: Oxidation)
Cl2 + 2e 2Cl- (Gain of electrons: Reduction)
 In every redox reaction, both reduction and oxidation must occur.
 Substance that gives electrons is the reducing agent or reductant.
 Substance that accepts electrons is the oxidizing agent or oxidant.
Reaction of ferrous ion with chlorine gas
Overall, the number of electrons lost in the oxidation half
reaction must equal the number gained in the reduction
half equation.
4

4. Oxidation Number (O.N)
 The O.N of a monatomic ion = its electrical charge.
 The O.N of free uncombined elements = zero
 The O.N of metals in combination is usually positive
 The O.N of an element in a compound may be calculated by
assigning O.N to the remaining elements of the compound
using the above rules and the following additional rules:
 The O.N. oxygen = -2 (except in peroxides O= -1).
 The O.N. of hydrogen = +1 (except in metallic hydride H=-1)
 The algebraic sum of the positive and negative O.N. of the
atoms of the formula (O.N of a compound) = zero.
 The algebraic sum of the positive and negative O.N. of the
atoms in a polyatomic ion (radical )= the charge of the ion.
5

5. Oxidation Numbers of Some Substances
Substance Oxidation Numbers
NaCl
H2
NH3
H2O2
LiH
K2CrO4
SO4
2-
KClO3
Na = +1, Cl = -1
H = 0
N = -3, H = +1
H = +1, O = -1
Li = +1, H = -1
K = +1, Cr = +6, O= -2
O = -2, S = +6
K = +1, Cl = +5, O= -2
6

6. For manganese
Species Mn Mn2+ Mn3+ MnO2 MnO4
2 MnO4

O.N. 0 +2 +3 +4 +6 +7
For nitrogen
Species NH3 N2H4 NH2OH N2 N2O NO HNO2 HNO3
O.N. –3 –2 –1 0 +1 +2 +3 +5
Oxidation states of manganese and nitrogen
in different species
7

7. O-2
H+1
H+1
N-3
H+1
H+1
H+1
C-4
H+1
H+1
H+1
H+1
O.N in covalent compounds as NH3, CH4 and H2O
are as follows:
8

8. Eqivalent weights of oxidants & reductant
a. Method of limited application (old definition of eq. wts).
b. Ion-electron method (By the number of electrons lost or
gained by one mole of the substance in a half reaction).
c. Oxidation number method
Is the weight of the substance that reacts with or contains
1 g of available hydrogen or 8 g of available oxygen.
2KMnO4 + acid  K2O + 2MnO + 5O
2KMnO4 → 5O, i.e Eq. W = 2x KMnO4 /10 or KMnO4 / 5 .
2KMnO4 + alkaline  K2O + 2MnO2 +3O Eq. W = MW/3
K2Cr2O7  K2O + Cr2O3 + 3O Eq. W = MW/6
acid
K Mn O4
Mn S O4
-8
+7
+1 +2 -8
+6
KMnO4 = MW/5
9

9. Al + HCl = AlCl3 +
H2
0 +1 -1 +3 -1
0
Oxidation: ( Al = Al3+ + 3e- )  2 = 6 e-
Reduction: ( 2 H+ + 2e = H2 )  3 = 6 e-
+ 3+
Balancing Redox Reactions
Using the Half-Reaction Method
10

10.  Balance each half reaction:
MnO4
 + 5é Mn2+
C2O4
2 2 CO2 + 2é
2 MnO4
 + 5 C2O4
2 2 Mn2+ + 10 CO2 + 8 H2O
 Balance oxygen atoms by adding water
2 MnO4
 + 5 C2O4
2 + 16 H+ 2 Mn2+ + 10 CO2 + 8 H2O
 Balance hydrogen atoms by adding H+
 Make number of electrons gained in each reaction equal
2 MnO4
 + 5 C2O4
2 2 Mn2+ + 10 CO2
2 MnO4
 + 5 C2O4
2 + 16 H+ 2 Mn2+ + 10 CO2 + 8 H2O
Balancing Redox Reactions
Using the Half-Reaction Method
11

11. Balancing Redox Reactions
Using the Half-Reaction Method
 Assign oxidation numbers
 Identify substance oxidized
 Identify substance reduced
 Write 1/2 reactions (eliminate spectator ions, if
necessary, adding appropriate number of
electrons to balance charge
 Balance electrons gained & lost
 Add 1/2 reactions to get overall reaction
 Add spectator ions (if necessary) and finish the
balancing
12

12. Cl2 + 2Fe2+  2Fe3+ + 2Cl-
Ce4+ + Fe2+ → Ce3+ + Fe3+
Cr2O7
2 + Fe2+ +  Cr3+ + Fe3+ +
KClO3 + Fe2+ + → Fe3+ + KCl +
MnO4
- + Fe2+ +  Mn2+ + Fe3+
Cl2 + Fe2+ 
Cl2 + Fe2+  Fe3+ + 2Cl-
KClO3 + 6Fe2+ + → 6Fe3+ + KCl +
KClO3 + 6Fe2+ + 6H+ → 6Fe3+ + KCl + 3H2O
Cr2O7
2 + 6Fe2+ + 14H+  2Cr3+ + 6Fe3+ + 7H2O
MnO4
- + 5Fe2+ + 8H+  Mn2+ + 5Fe3++ 4H2O
Examples
13

13. BrO3
 + 5Br  + 6H+→ 3Br2 + 3H2O
SnCl2 + 2HgCl2 → Hg2Cl2  + SnCl4
3As2O3 + 2BrO3
 → As2O5 + 2Br 
K2S2O8 + 2Fe2+ + 2H+ → 2Fe3+ + 2KHSO4
I2 + 2Na2S2O3  2NaI + Na2S4O6
2KMnO4 + 16HCl → 2 KCl + 2MnCl2 + 5Cl2 + 8 H2O
2MnO2 + 2AsO3
3- + 4H+  2Mn2+ + 2AsO4
3- + 2H2O
BrO3
 + Br  + → 3Br2 +
Examples
14

14. Nernest Equation for Electrode Potential (E)
Et = Electrode potential at temperature t.
E = Standard electrode potential (constant depend
on the system)
R = Gas constant
T = Absolute Temp. (t°C + 273)
F = Faraday constant (96500 Coulombs)
Loge = ln (natural logarithm = 2.303 log)
n = Valency of the ion
[Mn+]=Molar concentration of metal ions in solution
Et = Eo + log [Mn+]
RT
nF
E25 °C = Eo + log [Mn+]
0.0591
n
17

15. Standard Oxidation Potential (Eo
)
E25 °C = Eo + log [Oxidized] / [Reduced]
0.0591
n
The higher E°, the stronger the oxidizing power of its oxidant
and the weaker the reducing power of its reduced form.
The most powerful oxidizing agents are those at the top
(higher +ve) and the most powerful reducing agents are at
the bottom (higher – ve).
22

16. If any two redox systems are combined, the stronger oxidizing
agent gains electrons from the stronger reducing agent with the
formation of weaker reducing and oxidizing agents.
Cl2/Cl- (E° = + 1.36 V) & Fe3+/Fe2+ (E°= + 0.77 V)
Cl2 + 2Fe2+  2Fe3+ + 2Cl-
strong oxid. + strong red. → weak oxid. + weak red.
agent agent agent agent
23 Standard Oxidation Potential (Eo
)

17. Oxidation Potentials: Electrochemical Series
25

18. The potential of MnO4
/Mn2+ varies with the ratio
[MnO4
]/[Mn2+].
If ferrous is titrated with MnO4
 in presence of Cl ,
chloride will interfere by reaction with MnO4
 and gives
higher results.
Factors Affecting Oxidation Potential
1. Common Ion
E25 °C = Eo + log [Oxid] /[Red]
0.0591
n
Zimmermann’s Reagent (MnSO4, H3PO4 and H2SO4)
 MnSO4 has a common ion (Mn2+) with the reductant that
lowers the potential of MnO4
/Mn2+ system:
 Phosphoric acid lowers the potential of Fe3+/Fe2+ system
by complexation with Fe3+ as [Fe(PO4)2]3.
 Sulphuric acid is used for acidification.
26

19. 2. Effect of pH
The oxidation potential of an oxidizing agent containing
oxygen increases by increasing acidity and vice versa.
 Potassium permenganate:
MnO4
 + 8H+ + 5e  Mn2+ + 4H2O
E = Eo + log
0.0591
5
MnO4
/Mn2+
[MnO4
][H+]8
[Mn2+]
 Potassium dichromate:
Cr2O7
2 + 14H+ +6e  2Cr3+ + 7H2O
E = Eo + log
0.0591
6
Cr2O7
2/Cr3+
[Cr2O7
2][H+]14
[Cr3+]
27 Factors Affecting Oxidation Potential

20. As5+ or Sb5+ can oxidize I  I2 or I2 oxidizes As3+ and
Sb3+  As5+ or Sb5+ depending on the pH of the medium:
 In presence of much acid,
The Eº of As5+/As3+ or Sb5+/Sb3+  (it oxidizes I  I2)
As2O5 + 4I + 4H+ → As2O3 + 2I2 + 2H2O
Sb2O5 + 4I + 4H+ → Sb2O3 + 2I2 + 2H2O
 if Eº is reduced by decreasing [H+] (in presence of
sodium bicarbonate), arsenite is oxidized with iodine.
28
Arsenate/ arsenite system:
 Arsenate /arsenite & antimonate/antimonite
 In slightly acid or neutral medium,
The Eº of As5+/As3+ or Sb5+/Sb3+  (I2 oxidizes As2O3 and
Sb2O3  As2O5 and Sb2O5)
Factors Affecting Oxidation Potential
HCO3
-
H+
AsO3
3- + I2 + H2O
AsO4
3- + 2I- + 2H+

21. E (I2/2I) system increases by the addition of HgCl2 since it
complexes with iodide ions.
Hg2+ + 4I  [HgI4]2 (low dissociation complex)
3. Effect of Complexing Agents
• E (Fe3+/Fe2+) is reduced by the addition of F or PO4
3 due to
the formation of the stable complexes [FeF6]3 and
[Fe(PO4)2]3 respectively. Thus, ferric ions, in presence of F or
PO4
3 cannot oxidize iodide although Eo(Fe3+/Fe2+) = 0.77
while Eo(I2/2I ) = 0.54.
Iodine: I2 + 2e  2I E =Eo + log
0.0591
2
I2/I
[I2]
[I]2
Ferric: Fe3++e  Fe2+
E =Eo + log
0.0591
1
Fe3+/Fe2+
[Fe3+]
[Fe2+]
29 Factors Affecting Oxidation Potential

22. Addition of Zn2+ salts which precipitates ferrocyanide:
[Fe(CN)6]4- + Zn2+  Zn2 [Fe(CN)6]
The oxidation potential of ferri/ferrocyanide system in
presence of Zn2+ ions can oxidize iodide to iodine,
although the oxidation potential of I2/2I- system is higher.
4. Effect of Precipitating Agents
Ferricyanide: [Fe(CN)6]3 + e  [Fe(CN)6]4
E = Eo + log
0.0591
1
Ferri/Ferro
[[Fe(CN)6]3]
[[Fe(CN)6]4]
30 Factors Affecting Oxidation Potential

23. In presence of tartarate or citrate I2 oxidizes Cu+ to cu+2 as
these anions form stable complexes with Cu+2
Cu+2/Cu+ system E°= 0.15
I2/2I system E°= 0.54
Expected
Actually
2Cu+2 + 4I → 2CuI2
The instability of CuI2 results in formation of Cu2I2 ppt
which  Cu+ → ↑ E° of Cu+2/Cu+ system to become
able to oxidize I → I2
unstable
31
I2 oxidize Cu+ → Cu+2
The reverse occurs due to instability of CuI2
→ Cu2I2  + I2  S2O3
-2
We add KSCN… Why?
Factors Affecting Oxidation Potential
4. Effect of Precipitating Agents

24. It is the plot of potential (E, volts) versus the volume (ml) of
titrant.
Redox Titration Curve
Example: Titration of 100 ml
0.1 N Ferrous sulphate ≠ 0.1 N ceric sulphate.
Ce4+ + Fe2+ (sample)→ Ce3+ + Fe3+
The change in potential during titration can be either
measured or calculated using Nernest equation as follows:
32
 At the beginning (0.00 mL of Ce+4 added, inital point):
E = Eo + 0.0592/n log [Fe3+]/[Fe2+]
No Ce+4 present; minimal, unknown [Fe+3]; thus, insufficient
information to calculate E

25. 36 Redox Titration Curve

26.  After adding 10 mL Ce4+: E= 0.77+ 0.0591/1 log 10/90 =0.69 V
 After adding 50 mL Ce4+: E =0.77 + 0.0591/1 log 50/50 = 0.77 V
 After adding 90 mL Ce4+:E=0.77 + 0.0591/1 log 90/10 = 0.81 V
 After adding 99 mL Ce4+: E = 0.77 + 0.0591/1 log 99/1 = 0.87 V
 After adding 99.9 mL Ce4+: E=0.77+0.0591 log 99.9/0.1= 0.93 V
After adding 100 mL Ce4+: the two potentials are identical
33 Redox Titration Curve

27. E = EO
1 + 0.0591/1 log [Fe3+]/[Fe2+]
E = EO
2 + 0.0591/1 log [Ce4+]/[Ce3+]
 Summation of two equations:
2 E = Eo
1 + Eo
2 + 0.0591 log
[Fe3+][Ce4+]
[Fe2+][Ce3+]
 At the end point: Fe2+ = Ce4+ , and Fe3+ = Ce3+
2 E = Eo
1 + Eo
2
E = ( Eo
1 + Eo
2 ) / 2 E = 0.77 + 1.45 = 1.10 V
After adding 101 ml Ce4+: E=1.45+0.0591/1 log1/100= 1.33 V
After adding 110 mL Ce4+: E=1.45+0.0591 log 10/100=1.39 V
34 Redox Titration Curve

28. 20 40 60 80 100 120 140
0
Potential
(V)
Ce4+ (mL)
35 Redox Titration Curve

29. Detection of End Point in Redox Titrations
1. Self Indicator (No Indicator)
When the titrant solution is coloured (KMnO4):
KMnO4 (violet) + Fe2+ + H+  Mn2+ (colourless) + Fe2+.
The disappearance of the violet colour of KMnO4 is due to its
reduction to the colourless Mn2+.
When all the reducing sample (Fe2+) has been oxidized
(equivalence point), the first drop excess of MnO4
 colours the
solution a distinct pink.
37

30. In Titration of Fe2+ by Cr2O7
2
Cr2O7
2 + 3Fe2+ + 14H+  2Cr3+ + 3Fe2+ + 7H2O
The reaction proceeds until all Fe2+ is converted into Fe3+
Fe2+ +Ferricyanide (indicator) Fe 2+ ferricyanide (blue)].
Fe 2+ + [Fe(CN)6]3 → Fe3[Fe(CN)6]2.
The end point is reached when the first drop fails to give a blue
colouration with the indicator (on plate)
2. External Indicator
38 Detection of End Point in Redox Titrations

31. 3. Internal Redox Indicator
Redox indicators are compounds which have different
colours in the oxidized and reduced forms.
Inox + n e Inox
They change colour when the oxidation potential of the titrated
solution reaches a definite value:
E = E° + 0.0591/n log [InOX]/[Inred]
When [Inox] = [Inred] , E = E°
Indicator colours may be detected when:
[Inox]/[Inred] = 1/10 or 10/1
hence, Indicator range: E = E°In ± 0.0591/n
39 Detection of End Point in Redox Titrations

32. 4. Irreversible Redox Indicators
N N
(CH3)2N SO3Na + Br2 + 2 H2O
Methyl orange
(CH3)2N NO + SO3Na + 4 HBr
ON
In acid solutions, methyl orange is red.
Addition of strong oxidants (Br2) would destroy the indicator
and thus it changes irreversibly to pale yellow colour
Some highly coloured organic compounds that
undergo irreversible oxidation or reduction
Methyl Orange
42 Detection of End Point in Redox Titrations

33. Importance of Redox Reactions in Biological
systems and Pharmaceutical Applications
 Oxidizing agents act on human tissues by one of the
following mechanisms:
1. Germicides through liberation of oxygen in the tissues such as
hydrogen peroxide
2. Bactericidal against anaerobes as KMnO4 (for wounds) or Cl2
(for dental therapy) through denaturing the proteins by direct
oxidation.
 Oxidation can have deleterious effect on the cells and tissues
of human body because it produces reactive oxygen and
nitrogen species, such as OH, NO, NO2
 and alkoxyl radicals
RO which damage cells and other body components.
 Antioxidants (reducing agents) can counteract the effect of
the reactive oxygen and nitrogen species and protect the
other compounds in the body from oxidation.
44

34. Importance of Redox Reactions in Biological
systems and Pharmaceutical Applications
 Typical antioxidants include vitamin A, C and E; minerals
such as selenium; herbs such as rosemary.
 Antioxidants act by one or more of the following
mechanisms:
1. Reaction with the generated free radicales can counteract
the effect of the reactive oxygen and nitrogen species
2. Binding the metal ions needed for catalyzing the
formation of reactive radicales
3. Repairing the oxidative damage to the biomolecules or
induction of the enzymes that catalyze the repair
mechanisms.
45

35. Redox reactions are included not only in inorganic or
organic drug analysis but also in the metabolic pathways of
drugs.
Catabolic pathways: oxidation of C-atoms of carbohydrate
or lipid or protein
Anabolic pathways: reduction of C-atoms of carbohydrate
or lipid or protein
Phase-I Metabolic Pathways (Functionalization
reactions)
I- Oxidative reactions
A. Oxidation of aromatic compounds → hydroxy aromatic
compounds
B. Oxidation of olefins → Epioxides
C. Oxidative dealkylation
Redox Reactions and Biological Action
46

36. RH
(Parent
drug)
P-450-(Fe3+)
RH─P-450-(Fe3+) RH─P-450-(Fe2+)
NADPH + flavoprotein reductase
H2O
ROH
(oxidized product)
RH─P-450-(Fe2+)
O2
-
Redox Reactions in Biological System
48