Redox of known volume and unknown concentration.

Redox titration reaction between Potassium Permanganate and Hydrogen peroxide solution

ABSTRACT
The aim of this experiment is to carry out titration.
According to Christian (2014), titration is a process in which a solution called the analyte reacts with a measured volume of reagent of known concentration. The reagent whose concentration is usually accurately known is dispensed from a burette into a conical flask containing another solution called the analyte and which is of known volume and unknown concentration.
To determine the concentration of the analyte accurately, the volume or the amount of the titrant needed to cause a colour change in the analyte is measured.
In this practical exercise, the reagent or titrant is potassium permanganate and is filled into the burette meanwhile hydrogen peroxide which is the analyte is contained in a conical flask. During titration, Manganate(VII) ion (MnO4-), oxidises hydrogen peroxide, (H2O2) to oxygen gas. The reaction is done with potassium manganate(VII) solution and hydrogen peroxide solution acidified with dilute sulphuric acid.
The manganate(VII) ions change to manganese(II) ions hence it is said to be reduced.
On the other hand, Hydrogen peroxide which is contained in the conical flask is seen to react with potassium permanganate causing the hydrogen peroxide to be oxidised.
The burette is filled to the required level using a funnel if necessary to avoid spillage and this will lead to a curved surface on the titrant which is caused by surface tension and which is referred to as the meniscus. The reading should be made taking into the consideration the point where the lower level of the meniscus ends.
The conical flask’s contents (analyte) are mixed regularly as the tap on the burette is opened and the reagent mixes with the analyte. A colour change is noticed every time the reagent is mixed with the analyte, however mixing the contents returns the colour to a transparent colour.
The tap is slowly turned and allowed to drip drop by drop until the colour of the analyte changes to a permanent pale pink colour, which is an indication that the analyte solution has been neutralised. When this constant colour change occurs, the tap is turned off completely and the volume of the reagent at this point is measured. This is called the endpoint of the titration and signals the completion of the reaction.
This change in colour is referred to as the end point and once this is determined, the amount of titrant (potassium permanganate) needed to react with the reagent (hydrogen peroxide) is calculated, given that the concentration of the titrant is of known concentration.
Another important aspect of titration usually associated with endpoint is a term referred to as the “equivalence point”. It is sometimes confused to be identical with the endpoint. According to Christian et al (2014), the equivalence point is the theoretical end of the titration where the number of moles (or equivalents) of titrant exactly equals the number of moles (or equivalents) of analyte.
Under normal laboratory conditions and with the experiment carried out without any deviations, the endpoint should exactly equal the equivalence point, but in the real world they are slightly different. The difference between the equivalence point and the endpoint is titration error (Christian et al, 2014).

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Introduction:
The aim of this report is to determine the endpoint of this titration experiment involving oxidation and reduction of Potassium Permanganate and Hydrogen peroxide called a redox reaction.

This is done by measuring the amount of the reagent (titrant) pipetted through the burette to completely neutralise the analyte and cause a colour change. When this happens, it is referred to as the endpoint of the titration. A colour indicator is used to visualise this endpoint.
Potassium permanganate is a very popular and widely used oxidising titrant; the reason being it acts as a self-indicator to determine the endpoint and it is also a very powerful oxidising agent, hence the choice of use in this experiment.
The definition of oxidation and reduction can be applied in various ways given that several parameters are taking into consideration such as oxidation numbers (states), loss or gain of electrons and losing or gaining Hydrogen.
According to Housecroft (2005), an oxidation-reduction reaction, as the name implies, is a reaction that involves the oxidation and reduction.
Oxidation usually refers to an increase in oxidation number, losing Hydrogen, losing one or more electrons or gaining oxygen.
Reduction on the other hand describes the exact opposite which is a decrease in oxidation number, gaining Hydrogen, gaining one or more electrons or losing oxygen. (Housecroft, 2005).

Oxidation and reduction reactions steps usually always complement each other whereby if there is a loss of electrons then there will be a gain of electrons by another element during the reaction.
Neither of the reactions occur in isolation as there will always be a reducing agent or reductant and alongside it, an oxidising agent or an oxidant during the titration process.(http://www.webassign.net/question_assets/ncsugenchem102labv1/lab_11/manual.html)
Taking into consideration the equation below:
Oxi1 + Red2 -; Red1 + Oxi2
Oxi1 is the oxidising agent and is reduced to Red1. Meanwhile Red2 is the reducing agent and is oxidised to Oxi2.
An oxidising agent will have a decrease in its oxidation number due to accepting electron(s):
Ma+ + ne- -; M(a-n)+
An example could be Fe3+ + e- -; Fe2+ wherebyt the oxidation number has decreased from +3 to +2.
On the other hand, a reducing agent will have an increase in its oxidation number due to giving up an electron:
Ma+ + -; M(a-n)+ + ne-
An example would be
2I- -; I2 + 2e-

With regards to titration techniques, there are mainly four types of titration in the various industries.
They are namely Acid base titration, Oxidation-Reduction titration (redox), Precipitation titration and Complexometric titration. (Painter, 2018).
The most common one being the acid/base titrations which are mainly used for determining the concentration of analytes which are usually of unknown concentration. They are also usually used for standardising acids and bases and will either require the use of a colour indicator as in the practical experiment carried out in our scenario or the use of a pH meter. (Bartleson, 2017)
Redox titration which is another very popular titration method is redox reaction involving the titrant and the analyte whereby there is a loss or gain of electrons between both solutions. The gain in electrons is referred to as reduction and loss of electrons if an oxidation reaction.
Considering that titration has different methods as highlighted above, it is used by many industries due to its versatility to analyse chemical compounds. Examples of the milieus in which it has been used according to Highland J (2018) include:
Environmental studies whereby it is used to analyse precipitation and to quantify the amount the degree of contamination in natural rainwater or snow.
In Wastewater analysis by being a key mechanism in analysing the amount of impurities in waste water and which will require filtering (Highland J, 2018)
In the Nutrition or food industry, it is used to carry out quantitative analysis of acidic juice and water. This process modifies the acidity of the components of the products in order to meet the needs of those with varying or special dietary requirements.
It has also been used and is currently being implemented in the wine industry in varying the acidity of wine to maintain the quality of wine desired. This is important as quality of wine is usually determined by its acidic balance.
Finally, it is also implemented in the Pharmacological industry whereby various titration techniques are used to ensure quality control in production of drugs.

Materials and Methods
Despite following the methods outlined in the practical sheet, there were still a few deviations from the practical schedule as the results obtained were not as expected hence the titration had to be carried out four times to obtain at least three titres with a difference of no more than 0.2cm3.
The highest difference in titre readings between two titrations was 0.8cm3 which was a high percent error.
This was potentially caused because of random and systematic errors which affected the results obtained.
According to Christian G et al (2014), every measurement or experiment always has a random error present during measurement, which could be because of some level of imprecision associated with it. This usually results in random results being obtained. An example of a random error could be the ability to accurately read the burette volume of the titrant taking into consideration the lower end of the meniscus.
On the other hand Christian G et al (2014) also discusses systematic errors which are more likely to be due to the ability to bias a result consistently in a particular direction. An example could be improperly rinsing and drying the burette before and after use.
Common sources of error which could have affected the accuracy of the results could be: that the beakers had not been properly rinsed after use and hence had residue from other elements hence causing contamination.
Another source of error could have been not running some solution through the tip of the burette to fill the burette tip completely, making sure there are no air bubbles and that the level of the solution falls to or below the zero mark.
Given that the experiment had to be carried out four times due to a few deviations as the results obtained were not as expected. This was repeated to obtain titres with a difference of no more than 0.2cm3.
It could have come from the failure to stop the drip of KMn04 early enough given each titration was done by different members of the team during the experiment. This could therefore have produced an inaccurate volume of KMn04 used in the reaction based on the burette reading.
Another mistake could have come across as a result of inaccurate readings of the burette or conical flasks.

Results

My results from the titration can be seen in the table below:

Table 1: Data for the Titration of 50.0 mL of 0.100 M Fe2+ with 0.100 M Ce4+

Titration 1 Titration 2 Titration 3 Titration 4
Final vol/cm3 24.4cm3 23.9cm3 23.1cm3 23.3cm3
Initial vol/cm3 0cm3 0cm3 0cm3 0cm3
Titre/cm3 24.4cm3 23.9cm3 23.1cm3 23.3cm3
Mean Titre

The results to be used in the calculations will be the mean titre of the experiment.
This can be calculated by adding up the titres of titration 3 and titration 4 as they are within 0.2cm3 of each other and then dividing by 4.
Before doing this, it is necessary to demonstrate how the initial titre readings where obtained by using the formula below and getting the figures from Table 1 above:
Titration 3:
Titre = final burette reading – initial burette reading
= 23.1cm3 – 0 cm3
= 23.1cm3
Titration 4:
Titre = final burette reading – initial burette reading
= 23.3cm3 – 0 cm3
= 23.3cm3

The mean titre therefore is calculated as follows:
Mean titre = Titration 3 + Titration 4
——————————————-
4
= 23.1cm3 + 23.3cm3
——————————–
4
Mean titre = 23.2cm3
Discussion:

1. Using the two half-equations determine the oxidation numbers of all elements in reactants and products and so confirm, with reasons, the identities of the oxidant and reductant.

H2O2(aq) ? O2(g) + 2H+(aq) + 2e-
colourless colourless
MnO4-(aq) +8H+(aq) + 5e- ? Mn2+(aq) + 4H2O(l)
deep purple pale pink

For H2O2
H = +1
O = -1
O normally has an Oxidation number of -2.(except in peroxides where it is -1 and +1 in O2F2, +2 in OF2.)
For O2 = 0

For Mn2+
Mn = +2
For MnO4-

MnO4- -; Mn2+
+7 -2 +2 0
+5 +2 0

MnO4- is the oxidant hence has been reduced to Mn2+ because the oxidation number of Manganese has reduced from +5 to +2
H2O2 is the reductant hence has been oxidised or decomposed to O2 because the oxidation number of Oxygen has increased from -1 to 0

O:
H2O2(aq) ? O2(g)
Oxidised

R:
MnO4- ? Mn2+
Reduced

2. Then balance for electrons transferred and so determine the overall balanced reaction equation. From this determine the amount of H2O2 reacting with 1 mol of KMnO4.

Combine these redox couples into two half-reactions: one for the oxidation, and one for the reduction.
O:
H2O2 ? O2
-1 0
R:
MnO4- + ? Mn2+
+5 2+

R:
Balance oxygen by adding H20 to right side
MnO4- ? Mn2+ + 4H20
Add H+ to balance hydrogens on the left side
MnO4- + 8H+ ? Mn2+ + 4H2O
-1 +8 +2 0
Charges are not balanced hence add 5 electrons
MnO4- + 8H+ + 5e- ? Mn2+ + 4H2O hence
MnO4-(aq) +8H+(aq) + 5e- ? Mn2+(aq) + 4H2O(l)

K MnO4- ? K+ + MnO4-

It is worth noting here that Potassium is a spectator ion

Finally, a final check is carried out to confirm that the equation is balanced by ensuring it contains the same number of atoms on both sides of the equation.

O:
Balance Hydrogen by adding H+ to right side
H2O2 ? O2 + 2H+
+2

Charges are not balanced hence Add 2 electrons to right side
H2O2 ? O2 + 2H+ + 2e-

Therefore
H2O2(aq) ? O2(g) + 2H+(aq) + 2e-

H2O2(aq) ? O2(g) + 2H+(aq) + 2e- – First balanced equation
MnO4-(aq) +8H+(aq) + 5e- ? Mn2+(aq) + 4H2O(l) – Second balanced equation

Balanced equation: H2O2(aq) + MnO4-(aq) + 3e- + 6H+ ? O2(g) + Mn2+ (aq) + 4H2O(l)

3. Calculate the amount, n(MnO4-) used in the titration.

From the reaction equation:
H2O2(aq) + MnO4-(aq) + 3e- + 6H+ ? O2(g) + Mn2+ (aq) + 4H2O(l)

H2O2 + MnO4- = 1:1
For every 1 mole of MnO4- we need 1mole of H2O2

1mole of H2O2 reacts with 1 mole of MnO4-

Moles of MnO4- = conc x volume of titre
= 0.02 M x 23.2cm3
1000
= 0.00046 moles

4. Using the value from 3. above and the result from 2. above calculate the amount of H2O2 reacted in the titration.

From the reaction equation:
H2O2(aq) + MnO4-(aq) + 3e- + 6H+ ? O2(g) + Mn2+ (aq) + 4H2O(l)

H2O2 + MnO4- = 1:1
For every 1 mole of MnO4- we need 1mole of H2O2

It can be deduced that 1mole of H2O2 reacts with 1 mole of MnO4-
Consequently the exact concentration of H2O2 can be worked out as follows:
25ml of KMnO4- = 23.2ml of H2O2
Hence 0.00046 moles of MnO4- reacts with 0.00046 moles of H2O2

5. Calculate the concentration of the H2O2 solution.

C = n x 1000
V
= 0.00046 moles x 1000
25
= 0.0184M of H2O2 in 23.2 ml of MnO4-

Conclusion:
In the second part of the lab, redox reactions occurred when oxygen was presented to the solution. This happened when the solution was stirred by allowing the oxygen from the air to enter the solution. By transferring electrons, there were signs that reactions occurred through color changes and precipitates.

References

1. Housecroft C. (2005). Inorganic Chemistry Gosport: Ashford Colour Press Ltd

2. Bartleson B (2017) Purpose of Titration. Updated April 24, 2017
https://sciencing.com/purpose-titration-5406434.html
3. Painter T(2018) Types of Titration Updated April 30, 2018). https://sciencing.com/types-titration-14630.html

4. Christian G et al (2014). Analytical Chemistry. United States of America: Library of Congress Cataloging-In-Publication Data.

5. https://sciencing.com/list-5772040-titration-used-industry-.html

By James Highland; Updated January 10, 2018 Where is titration used in the industry

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