Pharmaceutical Analytical Chemistry PHCM223 Lecture 9
REDOX REACTIONS (II)
Dr. Nesrine El Gohary
10th lecture
2
• Identify some important oxidizing agents.
• Define Iodimetric and Iodometric titrations.
• Apply redox titrations.
Learning outcomes
Common Oxidizing agents used as titrants
Potassium Permanganate
KMNO4
Potassium dichromate
K2Cr2O7
Cerium (IV) Ce4+
Iodine I2
Cerium (IV) Ce4+
Like permanganate it is a powerful oxidizing agent.
Its potential depends on the acid in which the reaction takes place. It is 1.44 V on using H2SO4 and 1.70 V in perchloric acid (HClO4).
It not used in basic solutions since it is precipitated.
It is used in acidic medium where it is reduced to colorless Ce3+ ion. Ce4+ + e Ce3+
It can be used in the same titrations as permanganate but the oxidation of chloride is slow so could be used with HCl solutions.
The salt of cerium, ammonium hexanitrocerate, (NH4)2Ce(NO3)6 is a primary standard material.
yellow colorless
The yellow color or Ce4+ at the endpoint is not clear to be used as a self indicator so ferroin is used as an indicator with Ce 4+ titrations.
Iodine I2 Iodine is a weak oxidizing agent ((E°= 0.536V).
It is used to titrate only strong reducing agents, thus this increases its selectivity where it is possible to titrate strong reducing agents in the presence of weak ones.
Titrations performed with I2 are called Iodimetric titrations
These titrations are performed in neutral or mildly alkaline (pH8) to weakly acid solutions.
If the pH is too alkaline: I2 will disproportionate (undergo oxidation and reduction reaction at the same time) to hypoiodate and iodide
I2 + 2OH- IO- + I- + H2O
If the pH is too acidic: Starch the indicator used in Iodimetric titrations is hydrolyzed.
I2
Reducing agent + Starch
Start point: Colorless End point: Blue color
Iodine I2 Cont.
Iodine has a low solubility in water, so the actual titrant is I3- .
I3- is prepared by dissolving iodine in concentrated solutions
of potassium iodide.
I2+I- I3-
Triiodide
Although Pure iodine is available but its solution should be standardized using As2O3.
It should be standardized because it is highly volatile.
Iodometric Titration
Sample Oxidizing agent + Excess of I-
I2 is liberated in an amount equivalent to the sample.
Na2S2O3
Reducing agent
The liberated Iodine (I2) is titrated against a reducing agent which is sodium thiosulphate.
Example: Determination of dichromate (Cr2O7
2-) Cr2O7
2- + 6I- + 14H+ 2Cr3+ + 3I2 + 7H2O 1mole of Cr2O7
2- produces 3 moles of I2
I2 + 2S2O32- 2I- + S4O6
2-
1 moles of I2 reacts with 2 moles of S2O32-
1 mole of Cr2O72- is equivalent to 6 moles of S2O3
2-
Thiosulphate (S2O32- ) is oxidized in this reaction to tetrathionate (S4O6
2-) and Iodine (I2) is reduced to iodide (I-).
Iodometric Titration Cont.
Start point : large amount of I2
Near the end point: small amount of I2
Near the end point: small amount of I2 after addition of starch
The end point: no I2 is present
Starch is used as an indicator for this titration but it is added near the endpoint, when the color of the solution is pale yellow due to a small amount of I2 remaining.
Iodometric Titration Cont.
Starch is not added from the beginning of titration for two reasons:
1. Iodine starch complex is only slowly dissociated and so in the presence of a large amount of iodine at the beginning of titration, it will be difficult to break that complex.
2- Iodometric titration usually occur in acid medium which leads to the hydrolysis of starch.
Iodide is not used directly as a titrant mainly because of the lack of convenient visual indicator system, as well as the low speed of the reaction.
Notes on Iodometric titrations:
Thiosulphate (S2O32- ) is usually used as a titrant for Iodine (I2) because it reacts
quantitatively with Iodine and it is oxidized to tetrathionate (S4O62-) . It is not directly used
as a titrant for strong oxidizing agents because it could be oxidized to higher oxidation states such as sulphate (SO4
2-) and the reaction doesn’t proceed quantitatively.
PROBLEMS
A solution of Na2S2O3 is standardized iodometrically against 0.1262 g of high purity KBrO3, requiring 44.97 ml Na2S2O3. What is the molarity of the Na2S2O3?
Example 1:
KBrO3 0.1262 g + Excess of I-
Na2S2O3
44.97 ml M??
I2
BrO3- + 6I- + 6H+ Br- + 3I2 + 3H2O
3I2 + 6S2O32- 6I- + 3S4O6
2-
What is the relation between the sample BrO3
- and the titrant S2O32- ?
1 mmol BrO3- ≡ 6 mmol S2O3
2-
No. of millimoles of BrO3- = (Wt/Mwt)
= 126.2 mg/167.01 = 0.755
0.755 ≡ ??? No. of millimoles of S2O3
2- consumed = 4.53
MV = 4.53
M x 44.97 = 4.53 M of S2O3
2- = 0.1007 M
Example 2:
A 0.7120 g Specimen of iron ore was brought into solution and passed through a Jones reductor. Titration of Fe(II) produced required 39.21 ml of 0.02086 M KMnO4. Express the results of this analysis in terms of percent: (a) Fe and (b) Fe2O3
0.712 g sample
Fe2+
KMnO4 0.02086 M
39.21 ml
% Fe ??
% Fe2O3 ??
MnO4- + 5Fe2+ + 8H+ Mn2+ + 5Fe3+ + 4H2O
1 millimole of MnO4- ≡ 5 millimoles of Fe2+
(39.21 x 0.02086) ≡ ???? millimoles of MnO4
-
Millimoles of Fe2+ = 4.089 millimole
Wt = 4.089 At.wt
Wt of Fe = 228.35 mg = 0.22835 g
% of Fe = 0.22835 x 100 = 32.07 % 0.712
No of millimoles of Fe2O3 are ½ that of Fe Millimoles of Fe2O3 = 4.089/2 = 2.0445
Wt of Fe2O3 = 2.0445 x 159.69 = 326.48 mg = 0.32648 g
At. Wt of Fe = 55.845 g/mol
% of Fe2O3 = 0.32648 x 100 = 45.85 % 0.712
Example 3: The organic matter in a 0.9280 g sample of burn ointment was eliminated by ashing, following which the solid residue of ZnO was dissolved in acid. Treatment with (NH4)2C2O4 resulted in the formation of the sparingly soluble ZnC2O4. The solid was filtered, washed and then redissolved in dilute acid. The liberated H2C2O4 required 37.81 ml of 0.01508 M KMnO4. Calculate the percentage of ZnO in the medication.
2MnO4- + 16H+ +10e 2Mn2+ + 8H2O
5(COOH)2 10CO2 + 10e + 10H+
_______________________________________________________________________ 2MnO4
- + 5(COOH)2 + 6H+ 2Mn2+ + 10CO2 + 8H2O
0.9280 g sample
ZnO ZnC2O4 H2C2O4
KMnO4 0.01508 M
37.81 ml
% of ZnO ???
2 moles of MnO4- ≡ 5 moles of (COOH)2 ≡ 5 moles of ZnO
(0.01508×37.81×10-3) moles of MnO4- ≡ ???moles of ZnO
No of moles of ZnO = 0.01508×37.81×10-3 x 5 =1.425×10-3 moles 2
% of ZnO = (No. of moles × M.Wt/Wt. of sample)×100 = (1.425×10-3×81.39/0.928)×100 =12.5%
Example 4: The tetraethyl lead [Pb(C2H5)4] in a 25.0 ml sample of aviation gasoline was shaken with 15.0 ml of 0.02095 M I2. The reaction is : Pb(C2H5)4 + I2 Pb(C2H5)3I + C2H5I After the reaction was complete, the unused I2, was titrated with 6.09 ml of 0.03465 M Na2S2O3. Calculate the weight in milligrams of Pb(C2H5)4 ( 323.4 g/mol) in each liter of gasoline.
Pb(C2H5)4 + 15 ml of 0.02095 M I2
Unreacted I2
Na2S2O3 0.03465 M
6.09 ml
Conc. Of Pb(C2H5)4 in mg/l ???
25.0 ml sample
First determine the volume of unreacted I2
3I2 + 6S2O32- 6I- + 3S4O6
2-
1 millimole of I2 ≡ 2 millimoles of S2O3
2-
???? ≡ (6.09 x 0.03465) millimoles of S2O3
2-
Millimoles of I2 = 0.1055 millimole
MV = 0.1055 millimole 0.02095 x V = 0.1055 millimole
V of unreacted I2 = 5.04 ml
Example 4 cont.:
V of I2 reacted with the sample = 15- 5.04 = 9.96 ml
Pb(C2H5)4 + I2 Pb(C2H5)3I + C2H5I
1 millimole of Pb(C2H5)4 ≡ 1 millimole of I2
(MV)Tetraethyl lead = (MV)Iodine
(? X 25.0 ml) = (0.02095 x 9.96)
Concentration of Pb(C2H5)4 = 8.34963×10-3 Mol/L
N.B. :The weight in milligrams of Pb(C2H5)4 in each liter of gasoline is the Conc. in part per millions, thus:
Conc. In ppm = M×M.Wt×1000=8.34963×10-3×323.4×1000 =2700.27 mg/L
Determine the volume of I2 that reacted with the sample
• Try to solve problems 28- 44 , Chapter 18 , Skoog , pages 467-469.
• Kindly notice that , you can refer to the equations given in slide 3 in lecture 9, if not mentioned in the problem.
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Further practice