1. Redox titrations
Pharma.analytical chemistry II
Dr.Jehad M Diab, faculty of pharmacy
Damascus University
2. REDOX titrations
A volumetric method of analysis which
relies on oxidation or reduction of the
analyte using redox indicators or
potentiometry.
This unit covers:
changes in solution potential during a
titration
basic calculations
methods of sample preparation and
common titrants
Example applications Dr.Jehad diab
7. Dr.Jehad diab
A more complex example
Determine of Eeq for the following reaction:
Mno4- +5 Fe2+ 8H+ = Mn2+ + 5Fe3++ 4H2O
e eq= 0.77+0.059/1 × log [Fe3+] / [Fe2+] (1)
e eq=1.51+0.059/5 × log [Mno4-][H+]8/[Mn2+] (2) × 5
1+2:
6eeq= 0.77+5 × 1.51+0.059 × log [Fe3+][MnO4- ][H+]8/[Fe2+][Mn2+]
[Fe2+]=5[MnO4-] , [Fe3+]=5[Mn2+] ,
[Fe3+]/[Fe2+]= [Mn2+] / [MnO4-]
6eeq = 0.77+5×1.51+.059 log [Mn2+][MnO4-][H+]8/[MnO4-][Mn2+]
8. 6eeq = 0.77+5×1.51+.059 log [Mn2+][MnO4-][H+]8/[MnO4-][Mn2+]
6eeq=0.77+5×1.51+0.059 log[H+]8
eeq=(0.77+5×1.51)/6 + 0.059/6 log[H+]8
eeq=(0.77+5×1.51)/6 - 0.079 pH
If [H+]=1M (pH= 0):
eeq=(0.77+5×1.51)/6 + 0.059/6 log[1]8 =1.39 v
So, the potential at equivalence point is dependent
on [H+]. If [H+]=1M we can calculate from the
following equation:
eeq=(n1e0ox + n2e0Red ) / n1+n2 Dr.J.Diab
22. Ce4+ + Fe2+ Ce3++ Fe3+
E = e0 – 0.059 log [Ce3+]/ [Ce4+]
(200 x0.05/ 210)
1.70 -
(10ml x0.1M/210ml)
1.64 v
volt
Or E =1.70-0.059log 100%/10%=1.64 v
1.66 V
Dr.Jehad diab
23.
24. Homework
You are titrating 50 ml 0.1M of Co2+ solution
wuth0.1 M Ce4+ titrant.E0Co=0.85 v ,E0Ce=1.70.
What is the potential for the titration
system after addition of:
a. 0 ml of Ce4+
b. 25 ml of Ce4+
c. 50 ml of Ce4+
d. 75 ml of Ce4+
Dr.Jehad diab
25. Ce4+ + Co2+ Ce3++ Co3+
Equivalent point volume:
CCe4+ VCe4+ = CCo2+ V Co2+
0.1 x VCe4+ = 50 x0.1 ==> VCe4+ = 50 ml
a) Addition of 0 ml of Ce4+ ,absence of Co3+ ,E can
not be calculated
b) Addition of 25 ml of Ce4+ before the equivalent
point; excess of Co2+
E= 0.85+0.059/1xlog([25x0.1)/75/(50 x0.1-25x0.1)/75=0.85v
c)Addition of 50 ml of Ce4+;at the equivalent point:
E= (e0Co+ e0Ce)/2= (0.85 +1.70)/2= 1.275 v
Dr.Jehad diab
26. c)Addition of 75 ml of Ce4+;after the equivalent point
excess of Ce4+
E= e0Ce+0.059/1xlog[Ce4+]/[Ce3+]
E =1.70+0.059/1xlog(75x0.1-50x0.1)/125 /(50x0.1)/125= 1.682 v
Dr.Jehad diab
29. Redox indicators
Specific indicators:
Starch:
Starch +I2 <--> blue complex
It is an easy to detect and rapid indicator. This
explains why iodine is a common titrant even
though it is a weak oxidant.
KSCN:
Fe3++ SCN- ( Indicator) → FeSCN2+(red complex)
Determination of Fe3+ with Ti3+ in presence of
SCN-
Fe3+ with Ti3+ --> Fe2+ with Ti4+
when [Fe3+] decreases then color of red complex
disappears which indicates the end point.
37. Potential Indicator
End point is determined by measuring the potential of
indicator electrode against reference electrode and plotting
the potential against the volume of titrant.
(Pt for E( mv) and
glass electrode for
pH)
Dr.Jehad diab
38. Potentiometric titration اﻟﻣﻌﺎﯾرة اﻟﻛﻣوﻧﯾﺔ
It is possible to monitor
the course of a titration
using potentiometric
measurements. The Pt
electrode, for example,
is appropriate for
monitoring an redox
titration and determining
an end point in lieu
of an indicator. The
procedure has been
called a potentiometric
titration. The end point
occurs when the easured
potential undergoes a
sharp change—when all
the oxd. or red. in the
titration vessel is reacted Dr.Jehad diab
55. Dr.Jehad diab
Primary applications of Cr2O7-2
- Determination Fe2+
- Indirect determination of oxidizing agents;
A known excess of fe2+ is added to the
sample which is oxidant such as MnO4- and
the excess of fe2+ is back titrated with
Cr2O7-2
- Ethanol (C2H5OH)
Reactions:
6Fe2+ + Cr2O72- + 14H+ → 6Fe3+ + 2Cr3+ + 7H2O
3C2H5OH + 2Cr2O72- + 16H+ → 4Cr3+ + 3CH3COOH + 11H2O
56. of titrant
Moles of Fe2+ = 6 moles of Cr2O7-2
W Fe/ Mw Fe = 6 CCr2O7-2 x VCr2O7-2 (L)
0.2464/56 = 6 CCr2O7-2 x 39.31/1000
CCr2O7-2 = 0.01871 M
Dr.Jehad diab
59. MnO4- at pH < 1(H2SO4 is used ,HCl can not be used
because MnO4- oxidize 2Cl- to Cl2)
MnO4- + 8H+ + 5 e- → Mn2+ + 4 H2O
(E0 = 1.51 V strong acidic med)
MnO4- + 2H2O +3e- →MnO2 +4OH- (E0=0.59 v weak basic med)
MnO4- + 4H+ +3e- →MnO2 +2H2O (E0=1.69 v weak acidic to neutral )
MnO4- + e- →MnO4-2 (Eo=0.56 v strong basic med)
Standardization of MnO4-
Dr.Jehad diab
61. Common titrants
Oxidizing titrants
Cerium (IV) (Ce4+),E0=1.44 v and 1.70 v in 1M
H2SO4 and 1M HClO4 respectively
- Commonly used in place of KMnO4
- Works best in acidic solution
- Can be used in most applications in
previous table
- Used to analyze some organic compounds
- Color change not distinct to be its own
indicator ,Ferroin indicator is used
Yellow colorless
Dr.Jehad diab
62. Common titrants Dr.Jehad diab
Oxidizing titrants
I2 (Iodimetry)
I2 + 2e- →2I-
I2 + I- → I3- (I- added to increase solubility of I2)
I3- + 2 e- ------> 3 I- , E0 =0.536 V
Week oxidizing agents
Unstable
Needed to be re-standardized
2- →starch
I2 + 2S2O3 S4O62- + 2I-
( end point: disappearance of blue color)
Less popular than Ce4+, MnO4- ,Cr2O72-
used for the determination of strong reductants
63. Applications of Iodine in Redox
Titrations(Iodimetry) اﻟﻣﻘﯾﺎس اﻟﯾودي
I3- + 2 e- → 3 I- , I2 + 2e- →2I-
Iodimetric titrations carried out in weak acidic to weak basic
medium because in strong basic I2 + OH- → OI-+H+
And 3OI-disproportionate to IO3- +2I- this resulted in error by
disturbing the stoichiometry of the reaction. In strong acidic
medium starch decomposes.
H2S + I2 → S + 2I- + 2H+
H3AsO3 +I2+ H2O →H3AsO4 +2I- +2H++ 2e
SO32- + I2+H2O →SO42- + 2I- + 2H+
Sn2+ + I2 → Sn4+ + 2 I-
AsO33- + I2 + H2O →AsO43- +2 I- +3H+
N2H4 + 2I2 → N2 +4H+ +4I- Dr.Jehad diab
64. Iodimetric determination of hydroquinone by
back titration
Excess of standard II2 is added and excess is back
2
titrated with Na2S2O3
I2 + 2S2O32- → S4O62- + 2I-
Dr.Jehad diab
67. Oxidizing titrants
Potassium iodate:KIO3 ﯾودات اﻟﺑوﺗﺎﺳﯾوم
IO3- + 5I- (excess) + 6H+(0.1-1M) → 3I2 + 3H2O
IO3- + 2I2 + 10Cl- + 6H+(3M) → 5ICl2- + 3H2O
IO3- + 2I- + 6Cl- + 6H+(3M) → 3ICl2- + 3H2O
In intermediate step iodate is converted to I2
;I2 In presence of CHCl3 resulted in violet
color. in final step iodine converted to ICl2-
and violet color disappear which match the
end point.
Dr.Jehad diab
68. Application:
Determination of iodine(I2) and iodide(I-) in an
aqueous mixture
the iodine in an aliquot of a mixture is determined
by standard sodium thiosulfate solution in
presence of starch as indicator . the concentration
of iodine plus iodide is then determined with
standard potassium iodate solution in strong HCl
in presence of chloroform as indicator.
I2 + 2S2O32- → S4O62- + 2I-
IO3- + 2I2 + 10Cl- + 6H+(3M) → 5ICl2- + 3H2O
IO3- + 2I- + 6Cl- + 6H+(3M) → 3ICl2- + 3H2O
Dr.Jehad diab
79. The titration’s end point is signaled when the solution
changes from the yellow color of the products to the
brown color of the Karl Fisher Reagent. Dr.Jehad diab
80. Example: A 1.0120 g mineral sample was crushed
and dissolved in acid solution. This sample was
passed through a Jones reactor (Fe3+ converted
to Fe2+). Titration of the Fe(II) required 23.29 ml
of 0.01992 M KMnO4. What is the %Fe in the
sample?
wFe 2
2
moles of Fe 56 5
moles of MnO4 0.02329 0.01992 1
2+
wFe2 0.12990 g
2+ 0.12990 100
% Fe2 12.8320%
1.0120 Dr.Jehad diab
81. Example:
A 0.1165 g of primary standard Cu was
dissolved and then treated with an excess of
kI. Reaction:
2Cu2+ + 4I- CU2I2(s) +I2
Calculate the normality of Na2S2O3 soln. if
36.24 ml were needed titrate the librated I2
I2 + 2S2O32- → S4O62- + 2I- , 2Cu2+ ≡ I ≡2e-
2
0.1165
N 0.03624
63.54
N 0.10125 Dr.Jehad diab
82. Dr.Jehad diab
Example:
The amount of ascorbic acid, C6H8O6, in orange juice was
determined by oxidizing the ascorbic acid to
dehydroascorbic acid, C6H6O6, with a known excess of I3–,
and back titrating the excess I3– with Na2S2O3. A 5.00-mL
sample of filtered orange juice was treated with 50.00 mL
of excess 0.01023 M I3–. After the oxidation was complete,
13.82 mL of 0.07203 M Na2S2O3 was needed to reach the
starch indicator end point. Report the concentration of
ascorbic acid in milligrams per 100 mL.
C6H8O6 + I3- C6H6O6 + 3I- + 2H+ (I3- =I2)
I3- + 2S2O32- 3I- + S4O62-
86. Problem: Calculate the normality of the solution
produced by dissolving 2.064 g of primary
standard K2Cr2O7 in sufficient water to give 500
ml.
N=2.064 g / (294/6)=0.421g/0.500L=0.0842 N
Problem: calculate the molarity of the I2 solution
that is 0.04N with respect to the following reaction:
I2 + H2S →2I- +2H++S
M=0.04/2=0.02 M
Problem: calculate the molarity of the
KIO3solution that is 0.04 N with respect to the
following reaction:
IO3-+2I-+6H++6Cl- →3ICl2-+3H2O
M=0.04/4=0.01 M Dr.Jehad diab
