The Good Chemist's Guide: Redox Titration Experiment

Welcome chemists!!!

Here we are again! Ready to share our second experiment with you all.

After having learned about the different characteristics that can be found by looking at the labels of different substances containers, and after having studied properties of different elements and compounds, as shown in the following table.

... we have carried out a marvellous, an incredibly curious and interesting experiment... YES! We made colour disappear! You will learn about it now... It is not magic... It is Chemistry!

BACKGROUND INFORMATION

First of all... What is REDOX?

A Redox reaction is defined as that where an electron transfer occurs and at least one element chanes its oxidation state. In a redox reaction, there are alweays two coupled processes: one oxidation and one reduction:

. Oxidation is loss of electrons
. Reduction is gain of electrons.

The element which gets oxidised gives electrons to the other, reducing it. It is therefore the reductant. The one that takes the electrons and therefore gets reduced is called the oxidant.
In order to know what species is being reduced or oxidised, it is important to work out the oxidation state of the elements at both sides of the equation. The oxidation state of an element is defined as the hypothetical charge an atom would have if all its bonds to different atoms were 100% ionic.

A redox couple is the set composed by the oxidized and reduced forms of a determined
species. Ex: Zn2+/Zn0, Cu2+/Cu0.

BRIEF EXPERIMENT EXPLANATION

AIM:

To record the amount of aqueous potassium permanganate (0,05 M), required for the compound to stop reacting in 4 mL of commercial hydrogen peroxide and 4 mL of commercial sulphuric acid.

We will also determine which is the limiting reactant, as well as the excess. In order to work this out, we will do a series of calculations.

------ Please, find the images and videos on the following post ------

REDOX TITRATION MATERIALS

- Beaker
- Test tubes
- Burette
- Pipette
-Sulfuric acid (4M)
- H2O2
- Aqueous KMnO4 (0,05 M)

PROCESS

1. First of all, we fill up a 100mL burette with 80mL of aqueous KMnO4 (0,05 M). It is important that the tip of the burette under the stopcock is full of liquid, so that, when releasing the solution, we have control over the exact quantity that is being released. Therefore, adjust it to 0.

2. In a beaker, prepare a solution of 4mL of 4M H2SO4 and 4mL of H2O2. Place it under the burette.

3. Opening the burette's stopcock, start releasing the potassium permanganate aqueous solution into the solution in the beaker. Do the process slowly, almost drop by drop.

4. A reaction will take place. As the KMnO4 mixes with the solution, shaking the beaker with your hand, the pink/purple colour will start disappearing, being the solution colourless, due to the two compound's reaction.

Note: Potassium itself is colourless, it is MnO4(-2) what gave the solution the purple colour.

5. A moment will be reached in which the two solutions will stop reacting with each other, and the final solution will start adopting a pink colour. Record the exact volume of potassium permanganate released when the first drop of the KMnO4 solution stopped reacting in the solution, producing the light pink colour. This is why it is important to release the KMnO4 aqueous solution carefully.

OUR RESULTS: In our experiment, both solutions stopped reacting when we added 26,6 mL of aqueous KMnO4.

MATHEMATICAL CALCULATIONS
H2SO4
1 .- To balance de redox equation: H2O2 + KMnO4 ----------> O2 + MnSO4

Oxygen is being oxidised and manganese is being reduced. The balanced equatino would be as follows:

5H2O2 + 2KMnO4 + 3H2SO4 -----> 5O2 + 2MnSO4 + K2SO4 + H2O

We used 0.05 M KMnO4 and the reaction stopped when we poured 26.6 mL of potassium permanganate during titration. Therefore, we can calculate the number of moles: M = n/V ---> n = M x V --->
0.05 x 0.0266 = 0.00133 mol = 1.33 mmol of KMnO4

What we deduced previously from the experiment was that the reaction had stopped because there was an excess of KMnO4 and consequently, a lack of H2O2. By looking at the ratios of the balanced equation, we can say that:

nH2O2 = nKMnO4 x 5/2 --->1.33/2 x 5 = 3.325 mmol of H2O2.

If we have used 4mL of H2O2, we can calculate the concentration: M = n/V --> M = 3.325 mmol / 4mL = 0.83125 M H2O2

The concentration of H2O2 is measured in volumes. 'Volume' of O2 produced by a given volume of liquid H2O2. Commercial H2O2 has 10 volumes - 1:10
If we have 4mL of H2O2 we should obtain 40 mL of O2. We know that we obtain 3.325 mmol of O2 (ratio: 5H2O2 --> 5O2).

Conclusion

In this experiment, we have understood the theoretical contents learned in class by carrying out a practical experiment. We have studied the concepts of limiting reactant and we have learned how to determine the limiting reactant and the one in excess. What is more, we have learned how to use easy formulas and carry out a series of calculations in order to obtain the number of moles of potassium permanganate used, and we have compared our results with other students.

We consider that one off our strengths was that we released the potassium permanganate solution very slowly so that we knew exactly the precise volume of the solution that stopped reacting.

IMAGES