Reducing properties have Redox properties

2024 Author: Angel Austin | [email protected]. Last modified: 2024-01-02 17:39

The redox 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 oxidizing agent

Many tasks in chemistry, including test questions for the unified state exam in grade 11, and the OGE in grade 9, are associated with this concept. An oxidizing agent is considered to be atoms or ions that, in the process of chemical interaction, accept electrons from another ion or atom. If we analyze the oxidizing properties of atoms, we need the periodic system of Mendeleev. In periods located in the table from left to right, the oxidizing ability of atoms increases, that is, it changes similarly to non-metallic properties. In the main subgroups, this parameter decreases from top to bottom. Among the strongest simple substances with oxidizing ability, fluorine is in the lead. A term such as "electronegativity", that is, the ability of an atom to take in the case of a chemical interactionelectrons, can be considered synonymous with oxidizing properties. Among complex substances that consist of two or more chemical elements, bright oxidizing agents can be considered: potassium permanganate, potassium chlorate, ozone.

What is a reducing agent

The reducing properties of atoms are characteristic of simple substances that exhibit metallic properties. In the periodic table, in periods, the metallic properties weaken from left to right, and in the main subgroups (vertically) they increase. The essence of recovery is the return of electrons, which are located on the external energy level. The greater the number of electron shells (levels), the easier it is to give away "extra" electrons during a chemical interaction.

Active (alkaline, alkaline-earth) metals have excellent reducing properties. In addition, substances exhibiting similar parameters, we highlight sulfur oxide (6), carbon monoxide. In order to acquire the maximum oxidation state, these compounds are forced to exhibit reducing properties.

Oxidation process

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

Restoration process

Reduction processes involve the acceptance by ions of eitheratoms of electrons from other atoms (ions) during direct chemical interaction. Excellent reducing agents are nitrites, sulfites of alkali metals. The reducing properties in the system of elements change similarly to the metallic properties of simple substances.

OVR Parsing Algorithm

In order for the student to place the coefficients in the finished chemical reaction, it is necessary to use a special algorithm. Redox properties also help to solve various computational problems in analytical, organic, and general chemistry. We suggest the order of parsing any reaction:

1. First, it is important to determine the oxidation state of each available element using the rules.
2. Next, those atoms or ions that have changed their oxidation state are determined to participate in the reaction.
3. The minus and plus signs indicate the number of free electrons given and received during a chemical reaction.
4. Next, 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 the electrons involved in the chemical reaction.
6. Next, we determine which ions or atoms have reducing properties, and also determine oxidizers.
7. At the final stage put the coefficients in the equation.

Using the electronic balance method, let's place the coefficients in this reaction scheme:

NaMnO₄ + hydrogen sulfide + sulfuric acid=S + Mn SO_{4 +…+…}

Algorithm for solving the problem

Let's find out which substances should be formed after the interaction. Since there is already an oxidizing agent in the reaction (it will be manganese) and a reducing agent is defined (it will be sulfur), substances are formed in which the oxidation states no longer change. Since the main reaction proceeded between the s alt and a strong oxygen-containing acid, one of the final substances will be water, and the second will be sodium s alt, more precisely, sodium sulfate.

Now let's make a scheme for giving and receiving electrons:

- Mn⁺⁷ takes 5 e=Mn^+2.

Second part of the scheme:

- S^-2 gives2e=S⁰

We put the coefficients in the initial reaction, not forgetting to sum up all the sulfur atoms in the parts of the equation.

2NaMnO₄ + 5H₂S + 3H₂SO ₄ =5S + 2MnSO₄ + 8H₂O + Na₂SO _4.

Analysis of OVR involving hydrogen peroxide

Using the OVR parsing algorithm, we can compose an equation for the ongoing reaction:

hydrogen peroxide + sulfuric acid + potassium permagnanate=Mn SO_{4 + oxygen + …+…}

The oxidation states changed the oxygen ion (in hydrogen peroxide) and the manganese cation in potassium permanganate. That is, we have a reducing agent, as well as an oxidizing agent.

Let's determine what kind of substances can still be obtained after the interaction. One of them will be water, which is quite obviously a reaction between an acid and a s alt. Potassium did not form a newsubstances, the second product will be a potassium s alt, namely sulfate, since the reaction was with sulfuric acid.

Scheme:

2O ^{– donates} 2 electrons and turns into O ₂ ⁰ 5

Mn⁺⁷ takes 5 electrons and becomes Mn ion^{+2 2}

Set the coefficients.

5H₂O₂ + 3H₂SO₄ + 2KMnO₄=5O₂ + 2Mn SO₄ + 8H ₂O + K₂SO₄

Example of OVR analysis involving potassium chromate

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

FeCl₂ + hydrochloric acid + potassium chromate=FeCl₃+ CrCl_{3 + …+…}

Oxidation states changed iron (in ferric chloride II) and chromium ion in potassium dichromate.

Now let's try to find out what other substances are formed. One can be s alt. Since potassium did not form any compound, therefore, the second product will be a potassium s alt, more precisely, chloride, because the reaction took place with hydrochloric acid.

Let's make a diagram:

Fe⁺² gives e= Fe^{+3 6 reducer,}

2Cr⁺⁶ accepts 6 e=2Cr ^{+31 oxidizer.}

Put the coefficients in the initial reaction:

6K₂Cr₂O₇ + FeCl₂+ 14HCl=7H₂O + 6FeCl₃ + 2CrCl_{3 + 2KCl}

ExampleOVR analysis involving potassium iodide

Armed with the rules, let's make an equation:

potassium permanganate + sulfuric acid + potassium iodide…manganese sulfate + iodine +…+…

Oxidation states changed manganese and iodine. That is, a reducing agent and an oxidizing agent are present.

Now let's find out what we end up with. The compound will be with potassium, that is, we will get potassium sulfate.

Recovery processes occur in iodine ions.

Let's draw up an electron transfer scheme:

- Mn⁺⁷ accepts 5 e=Mn^{+2 2 is an oxidant,}

- 2I^- give away 2 e=I₂^{0 5 is a reducing agent.}

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

210KI + KMnO₄ + 8H₂SO₄ =2MnSO ₄ + 5I₂ + 6K₂SO₄ + 8H 2_O

Example of analysis of OVR involving sodium sulfite

Using the classical method, we will compose an equation for the circuit:

- sulfuric acid + KMnO_{4 + sodium sulfite… sodium sulfate + manganese sulfate +…+…}

After interaction we get sodium s alt, water.

Let's make a diagram:

- Mn⁺⁷ takes 5 e=Mn^{+2 2,}

- S⁺⁴ gives 2 e=S^{+6 5.}

Arrange the coefficients in the reaction under consideration, do not forget to add the sulfur atoms when arranging the coefficients.

3H₂SO₄ + 2KMnO₄ + 5Na₂ SO₃ =K₂SO₄ + 2MnSO₄ + 5Na₂ SO₄ + 3H_2O.

Example of analysis of OVR involving nitrogen

Let's do the following task. Using the algorithm, we will compose the complete reaction equation:

- manganese nitrate + nitric acid + PbO₂=HMnO₄+Pb(NO₃₎ 2₊

Let's analyze what substance is still formed. Since the reaction took place between a strong oxidizing agent and s alt, it means that the substance will be water.

Show the change in the number of electrons:

- Mn⁺² gives away 5 e=Mn^{+7 2 exhibits the properties of a reducing agent,}

- Pb⁺⁴ takes 2 e=Pb^{+2 5 oxidizer.}

3. We arrange the coefficients in the initial reaction, be sure to add up all the nitrogen available on the left side of the original equation:

- 2Mn(NO₃)₂ + 6HNO₃ + 5PbO 2 _=2HMnO4 _{+ 5Pb(NO}3₎2 _{+ 2H} 2_O.

This reaction does not exhibit the reducing properties of nitrogen.

Second redox reaction with nitrogen:

Zn + sulfuric acid + HNO₃=ZnSO_{4 + NO+…}

- Zn⁰ give away 2 e=Zn^{+23 will be a restorer,}

N⁺⁵accepts 3 e=N^{+2 2 is an oxidizer.}

Arrange the coefficients in a given reaction:

3Zn + 3H₂SO₄ + 2HNO₃ =3ZnSO ₄ + 2NO + 4H_2O.

The importance of redox reactions

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

Practical application of RIA

In order to preserve the surface of the metal, you need to know that active metals have restorative properties, so you can cover the surface with a layer of a more active element, while slowing down the process of chemical corrosion. Due to the presence of redox properties, drinking water is purified and disinfected. No problem can be solved without correctly placing the coefficients in the equation. In order to avoid mistakes, it is important to have an understanding of all redoxparameters.

Protection against chemical corrosion

Corrosion is a particular problem for human life and activity. As a result of this chemical transformation, the destruction of the metal occurs, the parts of the car, machine tools lose their operational characteristics. In order to correct such a problem, tread protection is used, metal is coated with a layer of varnish or paint, and anti-corrosion alloys are used. For example, an iron surface is covered with a layer of active metal - aluminum.

Conclusion

Various recovery reactions occur in the human body, ensure the normal functioning of the digestive system. Such basic life processes as fermentation, decay, respiration are also associated with restorative properties. All living beings on our planet have similar abilities. Without reactions with the return and acceptance of electrons, mining, industrial production of ammonia, alkalis, and acids is impossible. In analytical chemistry, all methods of volumetric analysis are based precisely on redox processes. The fight against such an unpleasant phenomenon as chemical corrosion is also based on the knowledge of these processes.