Study of the reaction rate of hydrogen peroxide decomposition in the presence of a catalyst by the gasometric method. Catalysts accelerating the decomposition of hydrogen peroxide
O.S.ZAYTSEV
EDUCATIONAL BOOK IN CHEMISTRY
FOR SECONDARY SCHOOL TEACHERS,
STUDENTS OF PEDAGOGICAL UNIVERSITIES AND SCHOOLCHILDREN OF GRADES 9–10,
DECIDED TO DEVOTE THEMSELVES TO CHEMISTRY AND NATURAL SCIENCE
TEXTBOOKTASK LABORATORY PRACTICESCIENTIFIC STORIES FOR READING
Continuation. See Nos. 4-14, 16-28, 30-34, 37-44, 47, 48/2002;
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23,
24, 25-26, 27-28, 29, 30, 31, 32, 35, 36, 37, 39, 41, 42, 43, 44, 46, 47/2003;
1, 2, 3, 4, 5, 7, 11, 13, 14, 16, 17, 20, 22/2004
§ 8.1 Redox reactions
(continuation)
TASKS AND QUESTIONS
1. Using the electron-ion method of selecting stoichiometric coefficients, compose the equations of redox reactions that proceed according to the following schemes (the water formula is not indicated):
Please note that among the compounds there are organic substances! Try to find coefficients using oxidation states or valences.
2.
Choose any two equations of electrode reactions:
Compose one summary equation from the two written equations of electrode processes. Name the oxidizing agent and reducing agent. Calculate the emf of the reaction, its G and the equilibrium constant. Make a conclusion about the direction of the shift in the equilibrium of this reaction.
If you forgot what to do, remember what was said above. You write out any two equations from this list. Look at the values of their electrode potentials and rewrite one of the equations in the opposite direction. What, why and why? Remember that the numbers of given and received electrons must be equal, multiply the coefficients by a certain number (which?) and sum both equations. The electrode potentials are also summed up, but you do not multiply them by the number of electrons involved in the process. A positive EMF value indicates the possibility of a reaction. For calculation G and the equilibrium constants, substitute the EMF value you calculated into the formulas that were derived earlier.
3. Is an aqueous solution of potassium permanganate stable? In another way, the question can be formulated as follows: will the permanganate ion react with water to form oxygen if
4. Oxidation by air oxygen in an aqueous solution is described by the equation:
O 2 + 4H + + 4 e\u003d 2H 2 O, E= 0.82 V.
Determine whether it is possible to oxidize the substances written on the right side of any equation of task 2 with air oxygen. Reducing agents are written on the right side of these equations. The teacher will give you the equation number.
You may find this task difficult to complete. This is the main flaw of your character - it seems to you that the task is impossible, and you immediately give up trying to solve it, although you have all the necessary knowledge. In this case, you should write the reaction equation between oxygen and hydrogen ions and the equation of interest to you. See which of the reactions has a higher ability to donate electrons (its potential should be more negative or less positive), rewrite its equation in the opposite direction, changing the sign of the electrode potential to the opposite, and sum it with another equation. A positive EMF value will indicate that the reaction is possible.
5.
Write the equation for the reaction between permanganate ion and hydrogen peroxide H 2 O 2 . In the reaction, Mn 2+ and O 2 are formed. What odds did you get?
And I got the following equation:
7H 2 O 2 + 2 + 6H + = 2Mn 2+ + 6O 2 + 10H 2 O.
Find a mistake if I made one, or explain why your coefficients are different. This task is designed to test your ingenuity and knowledge of the material of other sections of chemistry.
The reaction of a permanganate ion with hydrogen peroxide in an acid solution (sulfuric acid) can be represented by several equations with different coefficients, for example:
5H 2 O 2 + 2 + 6H + = 2Mn 2+ + 5O 2 + 8H 2 O,
7H 2 O 2 + 2 + 6H + = 2Mn 2+ + 6O 2 + 10H 2 O,
9H 2 O 2 + 2 + 6H + = 2Mn 2+ + 7O 2 + 12H 2 O.
Indicate the reason for this and write at least one more equation for the reaction of permanganate ion with hydrogen peroxide.
If you managed to explain the reason for such a strange phenomenon, explain the reason for the possibility of writing the following equations:
3H 2 O 2 + 2 + 6H + = 2Mn 2+ + 4O 2 + 6H 2 O,
H 2 O 2 + 2 + 6H + = 2Mn 2+ + 3O 2 + 4H 2 O.
Can reactions proceed according to these two equations?
Answer. The reaction of permanganate ions with hydrogen peroxide is superimposed by a parallel decomposition reaction of hydrogen peroxide:
2H 2 O 2 \u003d O 2 + 2H 2 O.
You can sum the basic reaction equation with an infinite number of this equation and get a lot of equations with different stoichiometric coefficients.
6. This task can serve as the topic of an essay or report.
Discuss the possibility of passing the reduction reaction of Fe 3+ ions with hydrogen peroxide in an aqueous solution:
2Fe 3+ + H 2 O 2 \u003d 2Fe 2+ + O 2 + 2H +.
Calculate the emf of the reaction, its G and the equilibrium constant, using the standard electrode potentials:
The study of the dependence of the reaction rate on the concentration of the components showed that with an increase in the concentration of Fe 3+ or H 2 O 2 individually, the reaction rate doubles. What is the kinetic equation for the reaction? Determine how the reaction rate will change with an increase in the concentration of Fe 3+ or H 2 O 2 three times. Predict how the reaction rate will change when the solution is diluted two or ten times with water.
The following reaction mechanism has been proposed:
H 2 O 2 \u003d H + H + (fast),
Fe 3+ + H = Fe 2+ + HO 2 (slow),
Fe 3+ + HO 2 = Fe 2+ + H + + O 2 (fast).
Prove that this mechanism does not contradict the above dependence of the rate on the concentrations of reactants. What is the limiting stage? What is its molecularity and what is its order? What is the general order of the reaction? Pay attention to the existence of such complex ions and molecules as H and HO 2 , and to the fact that two or even three particles are formed in each reaction. (Why are there no stages with the formation of one particle?)
7. Translate into Russian.
An important reaction type is the electron-transfer reaction, also known as the oxidation-reduction, or redox, reaction. In such a reaction one or more electrons appear to be transferred from one atom to another. Oxidation is a word originally meant combination with oxygen gas, but so many other reactions were seen to resemble reactions with oxygen that the term was eventually broadened to refer to any reaction in which a substance or species loses electrons. Reduction is a gain electrons. The term seems to have its origins in metallurgical terminology: the reduction of an ore to its metal. Reduction is just the opposite of oxidation. An oxidation cannot take place without its having a reduction coupled with it; that is, electrons cannot be lost unless something else gains them.
LABORATORY RESEARCH
The tasks offered to you, as it was before, are short research papers. Reactions that are important not only in chemistry but also in ecology were selected for the experiments. It is not necessary to complete all the experiments - choose the ones that interest you. It is desirable to work in small groups (2-3 people each). This reduces the time of the experiment, avoids mistakes and, most importantly, allows you to participate in scientific communication, which develops scientific speech.
1. Redox properties of hydrogen peroxide.
Hydrogen peroxide H 2 O 2 is the most important oxidizing agent that is used in everyday life, in technology, in the purification of water from organic contaminants. Hydrogen peroxide is an environmentally friendly oxidizing agent, because its decomposition products - oxygen and water - do not pollute the environment. The role of hydrogen peroxide and peroxide organic compounds in the processes of biological oxidation-reduction is known.
3–6% solutions of hydrogen peroxide for domestic and educational purposes are usually prepared from a 30% solution by dilution with water. Hydrogen peroxide decomposes during storage with the release of oxygen (do not store in tightly closed containers!). The lower the concentration of hydrogen peroxide, the more stable it is. To slow down decomposition, additives of phosphoric, salicylic acids and other substances are used. The salts of iron, copper, manganese and the catalase enzyme have a particularly strong effect on hydrogen peroxide.
A 3% solution of hydrogen peroxide in medicine is used for washing the mouth and gargling with stomatitis and sore throat.
30% hydrogen peroxide solution is called perhydrol. Perhydrol is not explosive. Getting on the skin, perhydrol causes burns, burning, itching and blistering, while the skin turns white. The burnt area should be rinsed quickly with water. Perhydrol in medicine is used to treat purulent wounds and to treat gums with stomatitis. In cosmetology, it is used to remove age spots on the skin of the face. Hydrogen peroxide stains on clothing cannot be removed. Hydrogen peroxide is used in the textile industry to bleach wool and silk, as well as furs.
The production of concentrated (90–98%) hydrogen peroxide solutions is constantly growing. Store such solutions in aluminum vessels with the addition of sodium pyrophosphate Na 4 P 2 O 7 . Concentrated solutions may decompose explosively. A concentrated solution of hydrogen peroxide on an oxide catalyst decomposes at 700 °C into water vapor and oxygen, which serves as an oxidizer for fuel in jet engines.
Hydrogen peroxide can exhibit both oxidizing and reducing properties.
The role of an oxidizing agent for hydrogen peroxide is more typical:
H 2 O 2 + 2H + + 2 e\u003d 2H 2 O,
for example in react:
2KI + H 2 O 2 + H 2 SO 4 \u003d I 2 + K 2 SO 4 + 2H 2 O.
Hydrogen peroxide as a reducing agent:
1) in an acid environment:
H 2 O 2 - 2 e\u003d O 2 + 2H +;
2) in the basic (alkaline) medium:
H 2 O 2 + 2OH - - 2 e\u003d O 2 + 2H 2 O.
Reaction examples:
1) in an acid environment:
2KMnO 4 + 5H 2 O 2 + 3H 2 SO 4 = K 2 SO 4 + 2MnSO 4 + 5O 2 + 8H 2 O;
2) in the main environment:
2KMnO 4 + H 2 O 2 + 2KOH \u003d 2K 2 MnO 4 + O 2 + 2H 2 O
The oxidizing properties of hydrogen peroxide are more pronounced in an acidic environment, while the reducing properties are more pronounced in an alkaline one.
1a. Decomposition of hydrogen peroxide.
Pour 2–3 ml of hydrogen peroxide solution into a test tube and heat the solution in a water bath. Gas release should begin. (What?) Prove experimentally that this is exactly the gas that you expected to receive.
Drop a grain of manganese dioxide into another test tube with a solution of hydrogen peroxide. Prove that the same gas is released.
Write the equation for the decomposition of hydrogen peroxide and separately the equations for receiving and returning electrons. What type of redox reaction is this?
Calculate the EMF of the reaction if:
Which of these two reactions has the greater ability to donate electrons and should be rewritten in the opposite direction? From the value of the EMF of the reaction, calculate G reactions and the equilibrium constant.
Compare results with G and equilibrium constant obtained from thermodynamic data:
Did your calculations match? If there is some discrepancy in the results, try to find the reasons.
1b. Detection of hydrogen peroxide.
To a diluted and acidified with sulfuric acid solution (2-3 ml) of potassium iodide, add a few drops of hydrogen peroxide solution. The solution will turn yellow-brown. When a few drops of starch solution are added to it, the color of the mixture instantly turns blue. Write the reaction equation (formed substances you know!).
Calculate the EMF of the reaction to make sure that the reaction is possible (select the reaction you need):
1c. Black lead sulfide and hydrogen peroxide.
The old masters painted their paintings with paints prepared on the basis of lead white, which included the white basic carbonate 2PbCO 3 Pb(OH) 2 . Over time, lead white turns black, and paints based on them change their color due to the action of hydrogen sulfide, and black lead sulfide PbS is formed. If the painting is carefully wiped with a dilute hydrogen peroxide solution, the lead sulfide turns into white lead sulfate PbSO 4 and the painting almost completely returns to its original appearance.
Pour 1–2 ml of a 0.1M solution of lead nitrate Pb (NO 3) 2 or lead acetate Pb (CH 3 COO) 2 into a test tube (sold in a pharmacy as a lead lotion). Add some hydrogen sulfide or sodium sulfide solution. Drain the solution from the resulting black precipitate and act on it with a solution of hydrogen peroxide. Write reaction equations.
All lead compounds are poisonous!
1g Preparation of a solution of hydrogen peroxide from hydroperite.
If you were unable to get a solution of hydrogen peroxide, then for laboratory work you can use hydroperite, the tablets of which can be bought at a pharmacy.
Hydroperite is a complex compound of hydrogen peroxide with carbamide (urea) NH 2 CONH 2 H 2 O 2 . When dissolved in water, a solution of hydrogen peroxide and carbamide NH 2 CONH 2 is obtained. A solution of hydroperite is used instead of a solution of hydrogen peroxide as an antiseptic and for dyeing hair. To rinse the mouth and throat, dissolve 1 tablet in a glass of water (0.25% hydrogen peroxide solution). One tablet of hydroperite weighs 1.5 g and corresponds to 15 ml
(1 tablespoon) 3% hydrogen peroxide solution.
Calculate how many hydroperite tablets should be dissolved in 100 ml of water to obtain approximately 1% hydrogen peroxide solution. What volume of oxygen (N.O.) can be obtained from one tablet of hydroperite?
Empirically determine how many milliliters of oxygen can be obtained from one tablet of hydroperite. Propose the design of the device and assemble it. Bring the volume of released oxygen to normal conditions. To get more accurate calculation results, you can take into account the vapor pressure of water over the solution, which at room temperature (20 ° C) is approximately 2300 Pa.
And natural resources
Department of Chemistry and Ecology
STUDY OF THE REACTION RATE OF DECOMPOSITION
HYDROGEN PEROXIDE IN THE PRESENCE OF A CATALYST
GAS METHOD.
in the discipline "Physical and colloidal chemistry"
for the specialty 060301.65 − Pharmacy
Velikiy Novgorod
1 The purpose of the work……………………………………………………………………..3
2 Basic theoretical provisions…………………………………………….3
4 Experimental part………………………………………………………4
4.1 Decomposition of hydrogen peroxide in the presence of manganese dioxide MnO2 ………..…………………………………………………………………….4
4.2 Decomposition of hydrogen peroxide in the presence of a catalyst at temperature T2 .................................................................. ................................................. ................6
5 Requirements for the content of the report……………………………………………..6
6 Exemplary control questions and tasks……...……………………………7
1 OBJECTIVES OF THE WORK
1. Determine the rate constant, reaction order, half-life at temperature T1.
2. Construct a graph of the dependence of the amount of released O2 on time, determine graphically the half-life.
3. Determine the activation energy of the reaction, calculate the temperature coefficient of the reaction rate.
2 MAIN THEORETICAL PROVISIONS
The use of hydrogen peroxide in many technological processes, medicine and agriculture is based on its oxidizing properties. The process of H2O2 decomposition in aqueous solutions proceeds spontaneously and can be represented by the equation:
H2O2®H2O +1/2 O2
The process can be accelerated with a catalyst. These can be anions and cations, such as CuSO4 (homogeneous catalysis). Solid catalysts (coal, metals, salts, and metal oxides) also have an accelerating effect on the decomposition of H2O2. The course of a heterogeneous catalytic reaction of H2O2 decomposition is affected by the pH of the medium, the state of the surface, and catalytic poisons, for example, C2H5OH, CO, HCN, H2S.
In the cells of plants, animals, and humans, catalytic decomposition of hydrogen peroxide is also carried out. The process is carried out under the action of catalase and peroxidase enzymes, which, unlike non-biological catalysts, have exceptionally high catalytic activity and specificity of action.
The decomposition of H2O2 is accompanied by the release of O2. The volume of released oxygen is proportional to the amount of decomposed hydrogen peroxide. The paper uses the gasometric method.
3 SAFETY REQUIREMENTS
When performing this laboratory work, it is necessary to follow the general rules for working in a chemical laboratory.
4 EXPERIMENTAL
4.1 Decomposition of hydrogen peroxide in the presence of manganese dioxideMNO2 .
Before starting the experiment, it is necessary to prepare a catalyst: grease a small piece of a glass rod with BF glue or starch paste. It is necessary to grease only the end with glue, pour a little MnO2 powder on the watch glass, touch the end of the stick to the powder so that a small amount of MnO2 remains on the glass. Glue Dry for several minutes (1-2 min). The pressure inside the system for collecting H2O2 must be brought to atmospheric pressure: open the stopper of the reaction tube, use a balancing bottle to set the water level in the burette to zero.
The scheme of the device for measuring the rate of decomposition of H2O2 is shown in Fig. 1.
water
 |
test tube with H2O2
Gif" width="10">.gif" width="10"> catalyst
Fig.1 - Instrument for studying the kinetics of H2O2 decomposition.
2 ml of a 3% solution of H2O2 are measured with a pipette or measuring cylinder, poured into test tube 1. If the experiment is carried out at room temperature, prepare a stopwatch, a table for recording experimental data, Dip the catalyst applied on a piece of glass rod into the test tube. Close the reaction vessel with a stopper. The volume of oxygen released is recorded first after 30 s, then the interval can be increased to 1 min.
As the liquid level in the burette decreases, the equalizing bottle is lowered so that the liquid level in the burette and the bottle does not change, the level difference is minimal.
The reaction is considered complete when the liquid level in the buret stops falling.
The volume of oxygen corresponding to the complete decomposition of H2O2 -V¥ can be obtained if the reaction vessel is placed in a glass of hot water. After cooling the tube to room temperature. After that, the volume of O2 corresponding to the complete decomposition of H2O2 is determined.
Table - Experimental data
Assuming that the order of the reaction is first, the reaction rate constant is calculated from the first order kinetic equation:
Based on the results of the experiment, the average value of the reaction rate constant is calculated.
The half-life of hydrogen peroxide is calculated from the equation:
t0.5 = 0.693/k using the average rate constant.
Determine the rate constant and the half-life graphically, using the dependence Vt= f (t) and ln(V¥ - Vt) = f (t), which are shown in Fig. 2 and Fig. 3. Compare the results obtained by two methods - analytical and graphic.
V¥https://pandia.ru/text/80/128/images/image032_11.gif" width="211" height="12">.gif" width="616" height="64">
t, min t, min
Rice. 2 – Dependence Vt = f(t) Fig.3 – Dependence ln(V¥ – Vt) = f(t)
4.2 Decomposition of hydrogen peroxide in the presence of a catalyst at temperature T2
The experiment is repeated by placing the reaction vessel in a water bath or a glass of water at a temperature of T2 (as directed by the teacher). The data is entered into the table:
Knowing the rate constants k1 and k2 at two different temperatures, we can calculate the activation energy Ea using the Arrhenius equation:
Ea = 
In addition, you can calculate the temperature coefficient using the van't Hoff rule:
k2/k1 = γ ∆t/10
5 REQUIREMENTS FOR THE CONTENT OF THE REPORT
The report must contain:
1. purpose of the work;
2. results of measuring the volume of oxygen released during the decomposition of peroxide;
3. calculation of the reaction rate constant and the half-life (half-life) of hydrogen peroxide;
4. graph of dependence Vt = f(t) and the results of graphical determination of the half-life of hydrogen peroxide;
5. graph of dependence ln(V¥ – Vt) = f(t) to determine the reaction rate constant;
6. results of measurements of the volume of oxygen released during the decomposition of peroxide at an elevated temperature and calculation of the reaction rate constant;
7. calculation of the activation energy according to the Arrhenius equation and calculation of the temperature coefficient of the reaction rate according to the van't Hoff rule;
8. conclusions.
6 EXAMPLE QUESTIONS AND QUESTIONS
1. The reaction rate constant depends on:
a) the nature of the reagents;
b) temperature;
c) concentrations of reagents;
d) the time elapsed since the start of the reaction.
2. Order of reaction
a) formal value;
b) is determined only experimentally;
c) can be calculated theoretically;
d) is equal to the sum of the exponents p + q, in the equation υ = k · CAp · CBq.
3. Activation energy of a chemical reaction
a) the excess energy compared to the average energy of the molecules, necessary for the collision between the molecules to become active;
b) depends on the nature of the reagents;
c) measured in J/mol;
d) increases with the introduction of a catalyst into the system.
4. The half-life of a certain radioactive isotope is 30 days. Calculate the time after which the amount of the isotope will be 10% of the original.
5. The reaction of the first order at a certain temperature proceeds by 25% in 30 minutes. Calculate the half-life of the starting material.
6. How many times will the reaction rate increase with an increase in temperature by 40K, if the temperature coefficient of the reaction rate is 3?
7. With an increase in temperature by 40K, the rate of a certain reaction increased by 39.06 times. Determine the temperature coefficient of the reaction rate.
The decomposition of hydrogen peroxide under the action of a catalyst and in the presence of liquid soap is one of the most beautiful chemical experiments. In Russian literature, this experiment does not have a specific name, in English sources it is called Elephant's Toothpaste, which in free translation sounds like Elephant Toothpaste.
For the experiment, you need 30-50% hydrogen peroxide (perhydrol), some liquid detergent (which gives a lot of stable foam) and a catalyst for the decomposition of hydrogen peroxide. When the peroxide comes into contact with the catalyst, oxygen is actively evolved, and due to the presence of liquid soap, a lot of foam is formed (for a short time). Dyes are often added to the mixture, which color the foam in different colors. The stream of foam that escapes from the opening of the flask or cylinder often really resembles "elephant toothpaste".
As a catalyst for the decomposition of hydrogen peroxide, various substances and mixtures can be used, for example: copper ammonia, potassium iodide, and even a suspension of yeast.
In the past, I have already conducted this experiment, but I did not bother to look at my old notes, as a result, the first experiment did not work out. I took 7.5 g of copper sulfate, added 30 ml of concentrated ammonia solution to it, mixed thoroughly. The solution was poured into a liter flask, added 50 ml of Gala liquid dish detergent and 80 ml of distillate, mixed again. 100 ml of perhydrol taken from the refrigerator was poured into the flask. Or rather, I tried to pour it: a violent reaction began, as a result, I did not have time to add about 1/3 of the perhydrol. Rapid decomposition of peroxide took place, but the experiment was disgusting: there was little foam.
Then I looked at the old records. It turned out that last time I took completely different amounts of substances:
"In a 300 ml conical flask, pour 10-20 ml of liquid Gala dish detergent (or any similar detergent). In another flask, dissolve 3-4 grams of copper sulfate in an excess of strong ammonia solution (add ammonia until the copper sulfate is completely dissolves.) Blue copper (II) ammonia is formed:
CuSO 4 + 6NH 3 + 2H 2 O \u003d (OH) 2 + (NH 4) 2 SO 4
Pour the copper ammonia solution into the detergent bottle and mix well. Place the flask on the table and quickly add 50-100 ml of a 30-50% hydrogen peroxide solution to it. There will be a strong release of gas. A fountain of foam will hit from the flask. The entire space around the flask will be filled with a large lump of foam in a few seconds. Steam will rise from the foam - the decomposition reaction of hydrogen peroxide proceeds with the release of heat. In our experiments, the height and width of the resulting foam was about 60 cm.
In other words, less copper ammonia and liquid soap had to be taken.
For the second experiment, I took a 300 ml conical flask, 2 g of copper sulphate, to which I added 20% concentrated ammonia solution. After dissolving the copper sulfate, add 20 ml of liquid dish detergent Gala, mix. I took 70 ml of hydrogen peroxide, but managed to add only 50 ml - active foam formation began.
The foam that came out of the flask really looked like toothpaste squeezed out of a tube. Due to copper ammonia, blue-colored stripes stretched along the foam. The experiment turned out well, but the decomposition was slow and took more than 2.5 minutes.
I remembered that I conducted the experiment described in the article
Hydrogen peroxide (peroxide) is a colorless syrupy liquid with a density that hardens at -. This is a very fragile substance that can decompose with an explosion into water and oxygen, and a large amount of heat is released:
Aqueous solutions of hydrogen peroxide are more stable; in a cool place they can be stored for quite a long time. Perhydrol - a solution that goes on sale - contains. It, as well as in highly concentrated solutions of hydrogen peroxide, contains stabilizing additives.
The decomposition of hydrogen peroxide is accelerated by catalysts. If, for example, a little manganese dioxide is thrown into a solution of hydrogen peroxide, then a violent reaction occurs and oxygen is released. Catalysts that promote the decomposition of hydrogen peroxide include copper, iron, manganese, as well as ions of these metals. Already traces of these metals can cause decay.
Hydrogen peroxide is formed as an intermediate product during the combustion of hydrogen, but due to the high temperature of the hydrogen flame, it immediately decomposes into water and oxygen.
Rice. 108. Scheme of the structure of the molecule. The angle is close to , the angle is close to . Link lengths: .
However, if a hydrogen flame is directed at a piece of ice, traces of hydrogen peroxide can be found in the resulting water.
Hydrogen peroxide is also obtained by the action of atomic hydrogen on oxygen.
In industry, hydrogen peroxide is obtained mainly by electrochemical methods, for example, anodic oxidation of solutions of sulfuric acid or ammonium hydrosulfate, followed by hydrolysis of the resulting peroxysulfuric acid (see § 132). The processes taking place in this case can be represented by a diagram:
In hydrogen peroxide, hydrogen atoms are covalently bonded to oxygen atoms, between which a simple bond also occurs. The structure of hydrogen peroxide can be expressed by the following structural formula: H-O-O-H.
Molecules have significant polarity, which is a consequence of their spatial structure (Fig. 106).
In a hydrogen peroxide molecule, the bonds between hydrogen and oxygen atoms are polar (due to the displacement of common electrons towards oxygen). Therefore, in an aqueous solution, under the influence of polar water molecules, hydrogen peroxide can split off hydrogen ions, that is, it has acidic properties. Hydrogen peroxide is a very weak dibasic acid in an aqueous solution; it decomposes, albeit to a small extent, into ions:
Dissociation on the second stage
practically does not flow. It is suppressed by the presence of water - a substance that dissociates to form hydrogen ions to a greater extent than hydrogen peroxide. However, when hydrogen ions are bound (for example, when alkali is introduced into a solution), dissociation in the second stage occurs.
Hydrogen peroxide reacts directly with some bases to form salts.
So, under the action of hydrogen peroxide on an aqueous solution of barium hydroxide, a precipitate of the barium salt of hydrogen peroxide precipitates:
Salts of hydrogen peroxide are called peroxides or peroxides. They consist of positively charged metal ions and negatively charged ions, the electronic structure of which can be represented by the diagram:
The degree of oxidation of oxygen in hydrogen peroxide is -1, i.e., it has an intermediate value between the degree of oxidation of oxygen in water and in molecular oxygen (0). Therefore, hydrogen peroxide has the properties of both an oxidizing agent and a reducing agent, i.e., it exhibits redox duality. Nevertheless, oxidizing properties are more characteristic of it, since the standard potential of the electrochemical system
in which it acts as an oxidizing agent, is 1.776 V, while the standard potential of the electrochemical system
in which hydrogen peroxide is a reducing agent, is 0.682 V. In other words, hydrogen peroxide can oxidize substances that do not exceed 1.776 V, and restore only those that are more than 0.682 V. According to Table. 18 (on page 277) you can see that the first group includes many more substances.
Examples of reactions in which it serves as an oxidizing agent are the oxidation of potassium nitrite
and isolation of iodine from potassium iodide:
It is used for bleaching fabrics and furs, used in medicine (3% solution - a disinfectant), in the food industry (for food preservation), in agriculture for dressing seeds, as well as in the production of a number of organic compounds, polymers, porous materials. As a strong oxidizing agent, hydrogen peroxide is used in rocket technology.
Hydrogen peroxide is also used to renew old oil paintings that have darkened over time due to the conversion of white lead into black lead sulfide under the influence of traces of hydrogen sulfide in the air. When such paintings are washed with hydrogen peroxide, lead sulfide is oxidized to white lead sulfate:
Purpose and objectives 1. Purpose: To find out which products contain catalysts that accelerate the decomposition of hydrogen peroxide, and which do not. 2. Tasks: o Find out what a catalyst is o Conduct an experiment with hydrogen peroxide and find out which products are a catalyst. 1. Purpose: To find out which products contain catalysts that accelerate the decomposition of hydrogen peroxide, and which do not. 2. Tasks: o Find out what a catalyst is o Conduct an experiment with hydrogen peroxide and find out which products are a catalyst.
What products are catalysts? 1. We took a hematogen, dripped hydrogen peroxide and saw that oxygen is released, therefore. hydrogen peroxide decomposes. 2. We also took other foods, such as raw meat, raw potatoes, beets, bread, garlic, banana, cocoa, and found that they also contain a catalyst.
Conclusion In the course of the work, we found out that the products containing catalysts for the decomposition of hydrogen peroxide are: hematogen, raw meat, raw potatoes, beets, bread, garlic, banana, cocoa. They are not: apple, tea leaves, cookies, orange / tangerine, sausage, smoked meat, ketchup, honey, chocolate candy. We also learned what a catalyst is and how to conduct this experiment.
