Oxidation - reduction (Redox) reactions are the 2nd group of titrimetric analysis we shall think under volumetric analyses. The others, these as precipitation titration and complexometric titration we said will be encountered at subsequent levels of your programme.
All the basic laws involved in acid - base titrations are as well applicable in redox reactions. Redox reactions are though rather slow such that the experiments might involve heating the reactions mixture or using catalysts, to hasten the reaction. The standard reagents used in
Redox reactions are oxidizing or reducing agents. As was done for acid - base reactions, the concept of molar concentrations is as well applicable to Redox reactions (we might hence refresh ourselves through the mole concept conversed in module 1 of this course).
The amount of the oxidizing agent and that of the reducing agent titrated are narrated exactly as in the ratio following in the stoichiometry in the balanced equation of the reaction at the end point. This means that
The amount in moles dm-3 of oxidizing agent = amount in moles dm-3 of oxidizing agent.
In our treatment of Redox titrations, we shall concern ourselves by only 2 different oxidizing agents. Such are potassium permanganate and the other is iodine solution. But 1st let us deal through potassium permanganate, solution.
Standardization of a solution of potassium permanganate by a standard reductant acid.
Potassium permanganate (KMnO4):
The most extensively utilized oxidizing agent in redox titrations is potassium permanganate. It is found valuable in both acidic and basic media. It might as well be utilized in neutral solution. It though acts best in acid medium. Potassium permanganate in both solid and aqueous form is purple in colour so that addition of an organic indicator to detect the end point is meaningless since the transform in colour of the organic dye indicator will not be distinctly noticed as it will be masked via the colour of the potassium permanganate. Therefore, a KMnO4 act is a self indicator. The 1st drop of excess KMnO4 produces a distinct colour transform.
Finally, since of its intense coloration, it is often extremely difficult to read the meniscus of the KMnO4 solution in a burette. The practice therefore is to examine the surface of the solution. Potassium permanganate cannot he used as a primary standard since it is difficult to gain pure. Even in solutions, it is decomposed through light and traces of organic matter. Therefore, the potassium permanganate solution intended for redox titrations must be standardized prior to use.
From the traits of potassium permanganate described hence, it would be obvious to us that before we can utilize potassium permanganate as an analytical reagent, its actual concentration must be determined since it can't be attained pure.
In practice, it is standardized via use of a number of standard reagents. Some of these are ammonium iron (11) sulphate, sodium ethanedioate and ethane dioic acid. In this particular laboratory exercise, we shall make use of a standard solution of ethanedioic acid (oxalic acid) a hydrated dibasic acid, which can be obtained in high purity. Oxalic acid is a good reducing agent even though it is an acid.
Since the best performance of KMnO4 the in the acid medium, if the contents of the conical flask turns brown, it means insufficient acid catalyst was added, addition of more acid would revert the solution to colorless. Not all acids act as catalysts for permanganate reaction although. The only acid suitable for employ is dilute sulphuric acid. Hydrochloric acid will be oxidized to chlorine.
2MnO-4(aq) + 10Cl- (aq) +16H+(aq) → 2Mn2+ (aq) + 5Cl2 + 8H2O(I)
While nitric acid would compete through the permanganate ion for the reducing agent since itself is a strong oxidizing agent like the KMnO4.
Oxalic acid solution
Pipette (25cm3 )
Measuring a cylinder
Conical flask (25cm3 )
1. Rinse the burette (50.0cm3) twice through a few cm3 of KMnO4 solution. Fill the burette by the KMnO4 solution above the Zero mark and drain to this mark, making sure the burette tip is full and the air is expelled within the body of the KmnO4 solution.
2. Pipette out 25.0cm3 of the Oxalic acid solution to a 250cm3 conical flask, add about 15cm3 of 2M H2SO4 and heat to between 60-80°c.
3. Titrate the hot mixture by KMnO4 solution until it holds a permanent faint pink colour.
4. Repeat the titration using fresh sample each time until at least 2 titrations are within 0.5cm3 accuracy.
5. Record the titre values using the format in earlier experiments and discover the average titre value.
6. From your consequences determine.
a. The number of moles of oxalic acid utilized in the titration.
b. The number of moles of permanganate utilized in the titration.
c. The concentration in moldm-3 of permanganate that is molarity.
d. The concentration in gdm-3 (Mn = 55, K = 39.0 =16, C=12, H = 1,N =14).
Volume of KMnO4
Treatments of Results:
We should be familiar through the procedure in treating consequences that we have attained from performing a volumetric (titrimetric) analysis. The procedure for Redox reactions is essentially similar to acid / base titration. The major difficulty is the perceived difficulty in balancing the redox equation. The subsequent steps can be utilized as guidelines:
Step 1: The equation of the reaction
MNO- 4 (aq) + 5C2O42- (aq) + 16H+ aq → 2Mn 2+(aq) + 10CO2 (g) 10CO2 (g) 8H2O(1)
From the equation, the mole ratio between the reactants is,
5moles C2O42-(aq) is = 2mol MnO4-(aq)
Step 2. We shall suppose the followings:
1. The potassium permanganate solution was prepared through dissolving between 1.4 and 1.6g of the solid in 1dm3 of solution
2. The standardized oxalic acid solution was prepared via dissolving 0.84g of solid in 250cm3 of solution.
3. The average titre value obtained for KMnO4 is Vcm3.
Step 3. The concentration of Na2C2O4 in moldm-3 = 4 x 0.84 / 90
= 0.037 moldm-3
No of moles utilized in the titration = Molarity x volume/ 1000
= 0.0009 moles of Na2C2O4
Step 4. Find the number of moles of KMnO4
From the equation of reaction
5 moles C2O42- = 2mol MnO4-
0.0009 mole C2O42- = 2/5 x 0.0009mole MnO4-
=0.0037 mole MnO4-
Step 5. Find the amount in moles dm-3 of the KMnO4 solution
Moles dm-3 No of mole x 100 / Vm
Moles dm -3 0.000376 x 1000 / Vm
0.37/ Vm = Xmolesdm-3
Step 6.Find the mass Concentration of the KMnO4
Mass concentration = Con. in moldm-3 x molar Mass KMnO4
= X moldm-3 x 158g mol-1
= 158Xgdm -3
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