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 of the unified state exam in the 11th grade, and the OGE in the 9th grade, are associated with this concept. An oxidizing agent 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 the 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 leads. Such a term as "electronegativity", that is, the ability of an atom to accept electrons in the case of chemical interaction, can be considered a synonym for oxidative properties. Among the complex substances, which 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 exhibiting metallic properties. In the periodic table, in the periods, the metallic properties weaken from left to right, and in the main subgroups (vertically) they increase. The essence of recovery is in the return of electrons, which are located at the external energy level. The larger the number of electron shells (levels), the easier it is to give off "extra" electrons during chemical interaction.
Active (alkaline, alkaline-earth) metals possess excellent reducing properties. In addition, substances exhibiting similar parameters, we allocate sulfur oxide (6), carbon monoxide. In order to obtain the maximum degree of oxidation, these compounds are forced to exhibit reducing properties.
Oxidation process
If during a chemical interaction an atom or an ion gives electrons to another atom (ion), we are talking about the oxidation process. To analyze how the reducing properties and oxidizing ability change, you will need a periodic table of elements, as well as knowledge of modern laws of physics.
Recovery process
Restorative processes involve the adoption by ions or atoms of electrons from other atoms (ions) during direct chemical interaction. Excellent reducing agents are nitrites, alkali metal sulfites. 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 set the coefficients in the finished chemical reaction, it is necessary to use a special algorithm. Redox properties also help to solve various design problems in analytical, organic, and general chemistry. We offer the procedure for parsing any reaction:
- First, it is important to determine the oxidation state of each element using the rules.
- Next, those atoms or ions that have changed their oxidation state are determined and will participate in the reaction.
- The minus and plus signs indicate the number of free electrons given and received during the chemical reaction.
- Then, between the number of all electrons, the minimum common multiple is determined, that is, an integer that can be divided into received and given electrons without a remainder.
- Then it is divided into electrons involved in the chemical reaction.
- Next, we determine which ions or atoms have reducing properties and also determine oxidizing agents.
- At the final stage, we put the coefficients in the equation.
Using the electronic balance method, we arrange the coefficients in this reaction scheme:
NaMnO 4 + hydrogen sulfide + sulfuric acid = S + Mn SO 4 + ... + ...
The algorithm for solving the problem
We will find out which substances should be formed 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 the oxidation state does not change anymore. Since the main reaction took place between the salt and a strong oxygen-containing acid, one of the final substances will be water, and the second - sodium salt, more precisely, sodium sulfate.
We now compose a scheme of 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 summarize all the sulfur atoms in the parts of the equation.
2NaMnO 4 + 5H 2 S + 3H 2 SO 4 = 5S + 2MnSO 4 + 8H 2 O + Na 2 SO 4 .
Analysis of OVR with the participation of hydrogen peroxide
Using the algorithm for parsing OVR, 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.
We determine what kind of substances can still be obtained after the interaction. One of them will be water, which, quite obviously, presents the reaction between acid and salt. Potassium did not form a new substance, the second product will be a potassium salt, namely sulfate, since the reaction was with sulfuric acid.
Scheme:
2O - gives away 2 electrons and turns into O 2 0 5
Mn +7 takes 5 electrons and becomes an ion Mn +2 2
We put the coefficients.
5H 2 O 2 + 3H 2 SO 4 + 2KMnO 4 = 5O 2 + 2Mn SO 4 + 8H 2 O + K 2 SO 4
An example of an OVR analysis involving potassium chromate
Using the electronic balance method, we compose an equation with the coefficients:
FeCl 2 + hydrochloric acid + potassium chromate = FeCl 3 + CrCl 3 + ... + ...
The oxidation states changed iron (in iron chloride II) and chromium ion in potassium dichromate.
Now we will try to find out what other substances are formed. One may be salt. Since potassium did not form any compound, therefore, the second product will be a potassium salt, more precisely, chloride, because the reaction took place with hydrochloric acid.
Let's make a diagram:
Fe +2 gives e = Fe +3 6 reducing agent,
2Cr +6 takes 6 e = 2Cr +3 1 an oxidizing agent.
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 an OVR analysis involving potassium iodide
Armed with the rules, we make the equation:
potassium permanganate + sulfuric acid + potassium iodide ... manganese sulfate + iodine + ... + ...
The oxidation states changed manganese and iodine. That is, a reducing agent and an oxidizing agent are present.
Now we will find out what will form in the end. The compound will be in potassium, that is, we get potassium sulfate.
Recovery processes occur in iodine ions.
We compose the electron transfer scheme:
- Mn +7 takes 5 e = Mn +2 2 is an oxidizing agent,
- 2I - gives 2 e = I 2 0 5 is a reducing agent.
We put the coefficients in the initial reaction, do not forget to summarize 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 a SAR analysis involving sodium sulfite
Using the classical method, we compose the equation for the circuit:
- sulfuric acid + KMnO 4 + sodium sulfite ... sodium sulfate + manganese sulfate + ... + ...
After the interaction, we obtain a sodium salt, water.
Let's make a diagram:
- Mn +7 takes 5 e = Mn +2 2,
- S +4 gives 2 e = S +6 5.
We arrange the coefficients in the reaction under consideration, do not forget to add sulfur atoms when arranging the coefficients.
3H 2 SO 4 + 2KMnO 4 + 5Na 2 SO 3 = K 2 SO 4 + 2MnSO 4 + 5Na 2 SO 4 + 3H 2 O.
An example of an analysis of nitrogen-containing reactions
We perform the following task. Using the algorithm, we make up the complete reaction equation:
- 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 oxidizing agent and salt, it means that the substance will be water.
We show the change in the number of electrons:
- Mn +2 gives 5 e = Mn +7 2 shows the properties of a reducing agent,
- Pb +4 takes 2 e = Pb +2 5 as an oxidizing agent.
3. We arrange the coefficients in the initial reaction, be sure to add all the nitrogen available on the left side of the original equation:
- 2Mn (NO 3 ) 2 + 6HNO 3 + 5PbO 2 = 2HMnO 4 + 5Pb (NO 3 ) 2 + 2H 2 O.
In this reaction, the reducing properties of nitrogen are not manifested.
The second sample of the redox reaction with nitrogen:
Zn + sulfuric acid + HNO 3 = ZnSO 4 + NO + ...
- Zn 0 gives 2 e = Zn +2 3 will be a reducing agent,
N +5 takes 3 e = N +2 2 is an oxidizing agent.
We put the coefficients in a given reaction:
3Zn + 3H 2 SO 4 + 2HNO 3 = 3ZnSO 4 + 2NO + 4H 2 O.
The significance of redox reactions
The most famous reduction reactions are photosynthesis characteristic of plants. How change recovery properties? 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 examples of oxidizing and reducing reactions associated with chemical elements, transformations of nitrogen, carbon, and oxygen compounds are of particular importance. Thanks to photosynthesis, the terrestrial atmosphere has the 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 such substances necessary for humans as oxygen, glucose. Without this chemical reaction, a full circulation of substances in nature, as well as the existence of organic life, is impossible.
The practical application of OVR
In order to preserve the metal surface, it is necessary to know that active metals have reducing properties, therefore it is possible to 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 cleaned and disinfected. No problem can be solved without arranging the coefficients correctly in the equation. In order to avoid errors, it is important to have an idea of all the redox parameters.
Chemical corrosion protection
A particular problem for life and human activity is corrosion. As a result of this chemical transformation, the destruction of the metal occurs, the automobile parts and machine tools lose their operational characteristics. In order to fix this problem, tread protection, metal coating with a layer of varnish or paint, the use of anti-corrosion alloys are used. For example, the iron surface is covered with a layer of active metal - aluminum.
Conclusion
A variety of recovery reactions occur in the human body, ensure the normal functioning of the digestive system. Such basic vital processes as fermentation, decay, respiration, are also associated with regenerative properties. All living beings on our planet have similar capabilities. Without reactions with the return and acceptance of electrons, the extraction of minerals, the industrial production of ammonia, alkalis, and acids are 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 knowledge of these processes.