Reducing properties have ... Oxidation-reduction properties

Oxidation-reduction properties of individual atoms, as well as ions, are an important issue in modern chemistry. This material helps to explain the activity of elements and substances, to conduct a detailed comparison of the chemical properties of different atoms.

What is an oxidizer

Many tasks in chemistry, including the test questions of the unified state examination in grade 11, and the OGE in grade 9, are related to this concept. An oxidizer is considered to be atoms or ions, which in the process of chemical interaction take electrons from another ion or atom. If we analyze the oxidizing properties of atoms, we need a periodic system of Mendeleev. In periods lying in the table from left to right, the oxidizing ability of the atoms increases, that is, changes similarly to nonmetallic properties. In the main subgroups, a similar parameter decreases from top to bottom. Among the strongest simple substances with oxidative ability, fluoride is the leader. Such a term as "electronegativity", that is, the possibility of an atom accepting electrons in the case of chemical interaction, can be considered synonymous with oxidizing properties. Among complex substances that consist of two or more chemical elements, bright oxidants can be considered: potassium permanganate, potassium chlorate, ozone.

What is a reducing agent

The reducing properties of atoms are characteristic of simple substances exhibiting metallic properties. In the periodic table, the metal properties from left to right are weakened, and in the main subgroups (vertically) they increase. The essence of recovery in the recoil of electrons, which are located on the external energy level. The greater the number of electronic shells (levels), the easier it is to give "extra" electrons during the chemical interaction.

Excellent reactive properties have active (alkaline, alkaline-earth) metals. In addition, substances exhibiting similar parameters, we will select sulfur oxide (6), carbon monoxide. In order to acquire the maximum degree of oxidation, these compounds are forced to exhibit reducing properties.

The oxidation process

If during the chemical interaction the atom or ion gives electrons to another atom (ion), we are talking about the oxidation process. To analyze how the reducing properties and oxidizing ability change, a table of Mendeleev elements will be required, as well as a knowledge of the modern laws of physics.

Recovery process

Restorative processes assume the acceptance by ions or atoms of electrons from other atoms (ions) during direct chemical interaction. Excellent reducing agents are nitrite, alkali metal sulphites. The reducing properties in the system of elements vary in a manner similar to the metallic properties of simple substances.

The OVP parsing algorithm

In order that in the finished chemical reaction the student could arrange the coefficients, it is necessary to use a special algorithm. Oxidation-reduction properties help to solve a variety of computational problems in analytical, organic, general chemistry. We propose an order of parsing any reaction:

1. First, it is important to determine for each available element the degree of oxidation, using the rules.
2. Further, those atoms or ions that have changed their oxidation state are determined, will participate in the reaction.
3. Signs "minus" and "plus" indicate the number of free electrons taken and received during the chemical reaction.
4. Further, between the number of all electrons, the minimum common multiple is determined, that is, an integer that is divided without a remainder by the received and given electrons.
5. Then it is divided into electrons that participated in the chemical reaction.
6. Further, we determine which ions or atoms the reducing properties possess, and also determine the oxidants.
7. At the final stage, put the coefficients in the equation.

Applying the electronic balance method, we will arrange the coefficients in the given reaction scheme:

NaMnO ₄ + hydrogen sulphide + sulfuric acid = S + Mn SO ₄ + ... + ...

Algorithm for solving the problem

We will find out which substances must form after the interaction. Since the reaction already has an oxidizing agent (it will be manganese) and a reducing agent is defined (it will be sulfur), substances are formed in which oxidation states no longer change. Since the main reaction was between salt and strong oxygen-containing acid, one of the final substances would be water, and the second would be the sodium salt, more precisely sodium sulfate.

Let us now compose a scheme for the recoil and acceptance of electrons:

- Mn ⁺⁷ takes 5 e = Mn ^+2.

The second part of the scheme:

- S ^-2 gives 2e = S ⁰

We put the coefficients in the initial reaction, without forgetting to sum up all the sulfur atoms in the parts of the equation.

2NaMnO ₄ + 5H ₂ S + 3H ₂ SO ₄ = 5S + 2 MnSO ₄ + 8H ₂ O + Na ₂ SO ₄ .

The analysis of OBR with the participation of hydrogen peroxide

Applying the OVP parsing algorithm, we can formulate the equation of the proceeding reaction:

Hydrogen peroxide + sulfuric acid + potassium permanganate = Mn SO ₄ + oxygen + ... + ...

The degree of oxidation changed the oxygen ion (in hydrogen peroxide) and manganese cation in potassium permanganate. That is, the reducing agent, as well as the oxidizer, are present.

We will determine what kind of substances can still be obtained after the interaction. One of them will be water, which is obviously the reaction between acid and salt. Potassium did not form a new substance, the second product would be a potassium salt, namely sulphate, since the reaction was with sulfuric acid.

Scheme:

2O ^{- gives} 2 electrons and turns into O ₂ ⁰ 5

Mn ⁺⁷ takes 5 electrons and becomes the ion Mn ⁺² 2

We put the coefficients.

5H2O2 + 3H2SO4 + 2KMnO4 = 5O2 + 2MnSO4 + 8H2O + K2SO4

An example of IRS analysis involving potassium chromate

Using the electronic balance method, we will compose the equation with the coefficients:

FeCl ₂ + hydrochloric acid + potassium chromate = FeCl ₃ + CrCl ₃ + ... + ...

The degrees of oxidation changed iron (in iron II chloride) and the chromium ion in potassium dichromate.

Now we will try to find out what other substances are formed. One can be salt. Since potassium did not form any compound, therefore, the second product is the potassium salt, more precisely, chloride, because the reaction was with hydrochloric acid.

Let's make a scheme:

Fe ⁺² gives e = Fe ⁺³ 6 reducing agent,

2Cr ^{+ 6} takes 6 e = 2Cr ⁺³ 1 oxidizer.

We put the coefficients in the initial reaction:

6K ₂ Cr ₂ O ₇ + FeCl ₂ + 14HCl = 7H ₂ O + 6FeCl ₃ + 2CrCl ₃ + 2KCl

An example of IRS analysis involving potassium iodide

Armed with the rules, we will compose the equation:

Potassium permanganate + sulfuric acid + potassium iodide ... manganese sulfate + iodine + ... + ...

Degrees of oxidation changed manganese and iodine. That is, a reducing agent and an oxidizing agent are present.

Now we will find out what will eventually form in us. The compound will be in potassium, that is, we get potassium sulfate.

Restorative processes occur in iodine ions.

Let's make the scheme of electron transfer:

- Mn ⁺⁷ takes 5 e = Mn ⁺² 2 is an oxidizer,

- 2I ^- gives back 2 e = I ₂ ⁰ 5 is a reducing agent.

We arrange the coefficients in the initial reaction, do not forget to sum up all the sulfur atoms in this equation.

210KI + KMnO ₄ + 8H ₂ SO ₄ = 2MnSO ₄ + 5I ₂ + 6K ₂ SO ₄ + 8H ₂ O

An example of IRS analysis involving sodium sulphite

Using the classical method, let's compose the equation for the circuit:

- sulfuric acid + KMnO ₄ + sodium sulphite ... sodium sulfate + manganese sulphate + ... + ...

After the interaction, we get a sodium salt, water.

Let's make a scheme:

- Mn ⁺⁷ takes 5 e = Mn ⁺² 2,

- S ⁺⁴ gives 2 e = S ⁺⁶ 5.

We place the coefficients in the reaction under consideration, we do not forget to add sulfur atoms when the coefficients are arranged.

3H ₂ SO ₄ + 2KMnO ₄ + 5Na ₂ SO ₃ = K ₂ SO ₄ + 2 MnSO ₄ + 5Na ₂ SO ₄ + 3H ₂ O.

An example of an OBD analysis involving nitrogen

Perform the following task. Using the algorithm, we will compose the complete equation of the reaction:

- manganese nitrate + nitric acid + PbO ₂ = HMnO ₄ + Pb (NO ₃ ) ₂ +

Let us analyze what substance is still being formed. Since the reaction took place between a strong oxidizer and salt, it means that the substance will be water.

Let us show the change in the number of electrons:

- Mn ⁺² gives 5 e = Mn ⁺⁷ 2 shows the properties of the reducing agent,

- Pb ⁺⁴ takes 2 e = Pb ⁺² 5 with an oxidizer.

3. We arrange the coefficients in the initial reaction, we must add all the nitrogen present in the left part of the original equation:

- 2Mn (NO3) ₂ + 6HNO3 + 5PbO2 = 2HMnO4 + 5Pb (NO3) ₂ + 2H2O.

In this reaction, the reducing properties of nitrogen do not appear.

The second sample of oxidation-reduction reaction with nitrogen:

Zn + sulfuric acid + HNO ₃ = ZnSO ₄ + NO + ...

- Zn ⁰ gives 2 e = Zn ⁺² 3 is a reducing agent,

N ⁺⁵ takes 3 e = N ⁺² 2 is an oxidizing agent.

We place the coefficients in the given reaction:

3Zn + 3H2SO4 + 2HNO3 = 3ZnSO4 + 2NO + 4H2O.

Significance of oxidation-reduction reactions

The most famous reductive reactions are photosynthesis, characteristic of plants. How do the reducing properties change? The process takes place in the biosphere, leading to an increase in energy through an external source. It is this energy that mankind uses for its needs. Among the examples of oxidation and reduction reactions associated with chemical elements, the transformations of nitrogen, carbon, and oxygen compounds are of particular importance. Thanks to photosynthesis, the earth's atmosphere has such a composition, which is necessary for the development of living organisms. Thanks to photosynthesis, the amount of carbon dioxide in the air shell does not increase, the surface of the Earth does not overheat. The plant not only develops by means of an oxidation-reduction reaction, but also forms such substances necessary for a person, as oxygen, glucose. Without this chemical reaction, a complete cycle of substances in nature is impossible, as well as the existence of organic life.

Practical application of OVR

In order to preserve the surface of the metal, it is necessary to know that active metals have reducing properties, so it is possible to coat the surface with a layer of a more active element, while slowing down the process of chemical corrosion. Due to the presence of oxidation-reduction properties, purification and disinfection of drinking water is carried out. No problem can be solved without setting the coefficients in the equation correctly. In order to avoid mistakes, it is important to have an idea of all oxidation-reduction parameters.

Protection against chemical corrosion

A particular problem for human life and activity is corrosion. As a result of this chemical transformation is the destruction of metal, lose their performance characteristics of the car parts, machine tools. In order to correct a similar problem, tread protection is used, the coating of the metal with a layer of varnish or paint, the use of anticorrosive alloys. For example, the iron surface is covered with a layer of active metal - aluminum.

Conclusion

A variety of restorative reactions occur in the human body, ensure normal functioning of the digestive system. Such basic processes of vital activity, such as fermentation, putrefaction, respiration, are also associated with restorative properties. Possess such opportunities all living beings on our planet. Without reactions with the return and acceptance of electrons, it is impossible to extract minerals, industrial production of ammonia, alkalis, acids. In analytical chemistry, all methods of volumetric analysis are based on oxidation-reduction processes. The fight against such an unpleasant phenomenon as chemical corrosion, is also based on the knowledge of these processes.