Lesson in grade 9 on the topic:

"REDOX REACTIONS (ORD)"

TDC

Nurturing: create conditions for fostering activity and independence in the study of this topic, as well as the ability to work in a group, the ability to listen to your classmates.

Developing: continue the development of logical thinking, the ability to observe, analyze and compare, find causal relationships, draw conclusions, work with algorithms, form interest in the subject.

Educational :

1. to consolidate the concepts of “oxidation degree”, the processes of “oxidation”, “reduction”;
2. consolidate skills in compiling equations of redox reactions using the electronic balance method;
3. to teach to predict the products of redox reactions.

DURING THE CLASSES:

1. Organizing time.
2. Knowledge update.

1. What rules for determining the degree of atoms of chemical elements do you know? (slide 1)
2. Complete the task (slide 2)
3. Perform a self-test (slide 3)

1. Learning new material.

1. Complete the task (slide 4)

Determine what happens to the oxidation state of sulfur during the following transitions:

A) H 2 S → SO 2 → SO 3

B) SO 2 → H 2 SO 3 → Na 2 SO 3

What conclusion can be drawn after the execution of the second genetic chain?

What groups can chemical reactions be classified into according to changes in the oxidation state of atoms of chemical elements?

1. We check (slide 5).

1. We conclude: By changing the oxidation state of atoms of chemical elements involved in a chemical reaction, reactions are distinguished - with a change in CO and without a change in CO.

1. So, let's designate the topic of the lessonREDOX REACTIONS (ORD).

1. Write down the definition

OVR - reactions that occur with a change in the oxidation state of atoms,

Included in the reactants

1. Let's try to figure it out - what is the peculiarity of the processes of oxidation and reduction of elements during the formation of an ionic bond, using the example of a sodium fluoride molecule?

Look at the chart carefully and answer the questions:

1. What can be said about the completeness of the external level of fluorine and sodium atoms?

1. Which atom is easier to accept and which is easier to give up valence electrons in order to complete the outer level?

1. How can one formulate the definition of oxidation and reduction?

It is easier for the sodium atom to give up one electron before the completion of its external level (than to accept 7 ē up to eight, i.e. until completion), therefore, it gives its valence electron to the fluorine atom and helps it to complete its external level, while being a reducing agent, it oxidizes and increases its CO. It is easier for a fluorine atom, as a more electronegative element, to accept 1 electron to complete its external level, it takes a sodium electron, while being reduced, lowers its CO and is an oxidizing agent.

"Oxidizer as a notorious villain

Like a pirate, bandit, aggressor, Barmaley

Takes away electrons - and OK!

Damaged, restorer

He exclaims: “Here I am, help me!

Give me back my electrons!”

But no one helps and damage

Doesn't refund…”

1. Writing down definitions

The process of donating electrons from an atom is called oxidation.

An atom that donates electrons and increases its oxidation state is oxidized and is calledreducing agent.

The process of accepting electrons by an atom is calledrecovery.

An atom that accepts electrons and lowers its oxidation state is reduced and is called oxidizing agent.

1. ARRANGEMENT OF COEFFICIENTS IN OVR BY THE ELECTRONIC BALANCE METHOD

Many chemical reactions are equalized by a simple selection of coefficients.

But sometimes there are difficulties in the equations of redox reactions. To arrange the coefficients, the electronic balance method is used.

I suggest you lookANIMATION

Study the algorithm for compiling OVR equations using the electronic balance method (Appendix 1).

1. Anchoring

Arrange the coefficients in UHR

Al 2 O 3 + H 2 \u003d H 2 O+Al by electronic balance method, indicate the processes of oxidation (reduction), oxidizing agent (reducing agent), perform self-test.

1. Reflection

Answer the questions in the table "Questions to the student" (Appendix 2).

1. Summing up the lesson. DZ

1. Commented grading.
2. Homework: Complete the self-test (Appendix 3)

Preview:

To use the preview of presentations, create a Google account (account) and sign in: https://accounts.google.com

Slides captions:

Redox reactions (ORD)

Rules for calculating the oxidation state (CO) of elements:

Determine the oxidation states of atoms of chemical elements according to the formulas of their compounds: H 2 S, O 2, NH 3, HNO 3, Fe, K 2 Cr 2 O 7 Complete the task

1 -2 0 -3 +1 +1 +5 -2 H 2 S O 2 NH 3 HNO 3 0 +1 +7 -2 Fe K 2 Cr 2 O 7 Perform self test

Determine what happens to the oxidation state of sulfur during the following transitions: A) H 2 S → SO 2 → SO 3 B) SO 2 → H 2 SO 3 → Na 2 SO 3 What conclusion can be drawn after the second genetic chain is completed? What groups can chemical reactions be classified into according to changes in the oxidation state of atoms of chemical elements? Complete the task

A) H 2 S -2 → S +4 O 2 → S +6 O 3 B) S +4 O 2 → H 2 S +4 O 3 → Na 2 S +4 O 3 In the first chain of transformations, sulfur increases its CO from (-2) to (+6). In the second chain, the oxidation state of sulfur does not change. Checking

Redox reactions (ORD) - reactions that occur with a change in the oxidation state of the atoms that make up the reactants Let's write the definition

Formation of an ionic bond, using the example of a sodium fluoride molecule

What can be said about the completeness of the external level of fluorine and sodium atoms? Which atom is easier to accept and which is easier to give up valence electrons in order to complete the outer level? How can one formulate the definition of oxidation and reduction? Answer the questions

Oxidation is the process of donating electrons from an atom. An oxidizing agent is an atom that accepts electrons and lowers its oxidation state, and is reduced during the reaction. The reducing agent is an atom that donates electrons and increases its oxidation state, in the course of the reaction it is oxidized. Recovery is the process of accepting electrons by an atom. Let's write down the definitions

1. Watch the animation. 2. Study the algorithm for compiling OVR equations using the electronic balance method (in a folder). ARRANGEMENT OF COEFFICIENTS IN OVR BY THE ELECTRONIC BALANCE METHOD

Arrange the coefficients in UCR Al 2 O 3 + H 2 \u003d H 2 O + Al using the electronic balance method, indicate the oxidation (reduction) processes, the oxidizing agent (reducing agent), perform a self-test. Anchoring

Answer the questions in the table "Questions to the student". Reflection

Preview:

Annex 2

Questions to the student

Date___________________Class______________________

Try to remember exactly what you heard in the lesson and answer the questions:

No. p / p	Questions	Answers
	What was the topic of the lesson?
	What was your goal in the lesson?
	How did your classmates work in class?
	How did you work in class?
	Today I found out...
	I was surprised...
	Now I can...
	I would like to…

Preview:

Annex 3

Test on the topic "REDOX REACTIONS"

Part "A" - choose one answer from the options

1. Redox reactions are called

A) Reactions that occur with a change in the oxidation state of the atoms that make up the reactants;

B) Reactions that proceed without changing the oxidation state of the atoms that make up the reactants;

C) Reactions between complex substances that exchange their constituents

2. An oxidizing agent is ...

A) An atom that donates electrons and lowers its oxidation state;

B) An atom that accepts electrons and lowers its oxidation state;

C) An atom that accepts electrons and increases its oxidation state;

D) An atom that donates electrons and increases its oxidation state

3. The recovery process is a process…

A) recoil of electrons;

B) Acceptance of electrons;

C) Increasing the degree of oxidation of the atom

4. This substance is only an oxidizing agent

A) H 2 S; B) H 2 SO 4; B) Na 2 SO 3; D) SO 2

5. This substance is only a reducing agent

A) NH3; B) HNO 3 ; B) NO 2; D) HNO 2

Part "B" - match(For example, A - 2)

1. Establish a correspondence between the half-reaction and the name of the process

2. Establish a correspondence between the equation of a chemical reaction and its type

A) 2H 2 + O 2 \u003d 2H 2 O	1) Expansions, OVR
B) 2CuO=2Cu+O 2	2) Connections, not OVR
C) Na 2 O + 2HCl \u003d 2NaCl + H 2 O	3) Exchange, no IAD
D) 4HNO 3 \u003d 4NO 2 + 2H 2 O + O 2	4) Connections, OVR

3. Establish a correspondence between the phosphorus atom in the formula of a substance and its redox properties that it can exhibit

Part "C" - solve the task

From the proposed reactions, select only OVR, determine the oxidation states of atoms, indicate the oxidizing agent, reducing agent, oxidation and reduction processes, arrange the coefficients using the electronic balance method:

NaOH + HCl \u003d NaCl + H 2 O

Fe (OH) 3 \u003d Fe 2 O 3 + H 2 O

Na + H 2 SO 4 \u003d Na 2 SO 4 + H 2

Synopsis of a lesson in chemistry in grade 9: "Oxidation-reduction reactions"

The purpose of the lesson:

Consider the essence of OVR, repeat the basic concepts of the degree of oxidation, oxidation and reduction.

Equipment and reagents: A set of test tubes, solutions: CuSO4 , H2SO4, NaOH, H2O, Na2SO3.

The course of the lesson in chemistry in grade 9

Organizing time.

Today in the lesson we will continue familiarization with redox reactions, we will consolidate the knowledge acquired in previous classes, get acquainted with the oxidation-reduction reactions, find out what role the environment has on the course of redox processes. OVR are among the most common chemical reactions and are of great importance in theory and practice. OM processes accompany the cycles of substances in nature, they are associated with metabolic processes occurring in a living organism, decay, fermentation, and photosynthesis. They can be observed during the combustion of fuel, in the process of smelting metals, during electrolysis, in corrosion processes. (slides 1-7).

The topic of redox reactions is not new, students were asked to repeat some concepts and skills. Question for the class? What is the degree of oxidation? (Without this concept and the ability to arrange the oxidation state of chemical elements, it is not possible to consider this topic.) Students are invited to determine the oxidation state in the following compounds: KCIO3, N2, K2Cr2O7, P2O5, KH, HNO3. Check their assignments with the notes on the board. In all cases, there is a change in the degree of oxidation. To do this, we will carry out laboratory work (on the tables are instructions for performing experiments, briefing on tb).

Conduct experiments :1. CuSO4 + 2NaOH= Na2SO4 + Cu(OH)2

CuSO4 + Fe= CuFeSO4

Arrange with make notes. Conclusion: not all reactions are classified as OVR. (slide 8).

What is the essence of the OVR? (slide 9).

OVR is a unity of two opposite processes of oxidation and reduction. In these reactions, the number of electrons donated by the reducing agent is equal to the number of electrons attached by the oxidizing agent. The reducing agent increases its degree of oxidation, the oxidizing agent lowers it. (It was not by chance that the motto of the lesson was chosen.) Consider a chemical reaction (it is of great importance from an environmental point of view, since it allows you to collect accidentally spilled mercury under normal conditions.

H g0 + 2Fe+3Cl3-=2Fe+2Cl2-1 + Hg+2Cl2-1

Hg0 - 2ē → Hg+2

Fe+3+ē→ Fe+2

Students are asked to solve a problem. How does the environment affect the behavior of the same oxidizing agent, for example: KMnO4

Laboratory work 2 is being carried out according to the options:

2KMnO4+ 5Na2SO3 + 3H2SO4 = 2MnSO4 + 5Na2SO4 + K2SO4 + 3H2O

2KMnO4+ Na2SO3 2KOH= 2K2Mn04+Na2SO4 H2O

2KMnO4 +3Na2SO3 +H2O= 2KOH +3Na2SO4+ 2MnO2

Conclusion: the environment affects the oxidizing properties of substances. (Slide 10)

KMnO4 in an acidic medium - Mn + 2 - colorless solution.

In a neutral environment -MnO2 - brown precipitate,

In an alkaline environment, -MnO4-2 is green.

Depending on the pH of the solution, KMnO4 oxidizes various substances, being reduced to Mn compounds of different oxidation states.

The lesson is summed up. Ratings are given.

Reflection.

The class expresses their opinion about the work in the lesson.

Homework

Download presentation for the lesson in chemistry: "Redox reactions"

The lesson discusses the essence of redox reactions, their difference from ion exchange reactions. Changes in the oxidation states of the oxidizing agent and reducing agent are explained. The concept of electronic balance is introduced.

Topic: Redox reactions

Lesson: Redox reactions

Consider the reaction of magnesium with oxygen. We write the equation for this reaction and arrange the values ​​of the oxidation states of the atoms of the elements:

As can be seen, magnesium and oxygen atoms in the composition of the initial substances and reaction products have different values ​​of oxidation states. Let us write down the schemes of oxidation and reduction processes occurring with magnesium and oxygen atoms.

Before the reaction, the magnesium atoms had an oxidation state equal to zero, after the reaction - +2. Thus, the magnesium atom lost 2 electrons:

Magnesium donates electrons and oxidizes itself, which means that it is a reducing agent.

Before the reaction, the oxidation state of oxygen was zero, and after the reaction it became -2. Thus, the oxygen atom has attached 2 electrons to itself:

Oxygen accepts electrons and is itself reduced, which means that it is an oxidizing agent.

We write the general scheme of oxidation and reduction:

The number of given electrons is equal to the number of received ones. The electronic balance is maintained.

IN redox reactions processes of oxidation and reduction occur, which means that the oxidation states of chemical elements change. This is a hallmark redox reactions.

Redox reactions are reactions in which chemical elements change their oxidation state.

Consider specific examples of how to distinguish a redox reaction from other reactions.

1. NaOH + HCl \u003d NaCl + H 2 O

In order to say whether a reaction is redox, it is necessary to arrange the values ​​of the oxidation states of the atoms of chemical elements.

1-2+1 +1-1 +1 -1 +1 -2

1. NaOH + HCl \u003d NaCl + H 2 O

Please note that the oxidation states of all chemical elements to the left and right of the equal sign remained unchanged. This means that this reaction is not a redox reaction.

4 +1 0 +4 -2 +1 -2

2. CH 4 + 2O 2 \u003d CO 2 + 2H 2 O

As a result of this reaction, the oxidation states of carbon and oxygen have changed. Moreover, carbon increased its oxidation state, and oxygen lowered it. Let's write down the oxidation and reduction schemes:

C -8e \u003d C - oxidation process

O + 2e \u003d O - recovery process

So that the number of given electrons is equal to the number of received ones, i.e. respected electronic balance, it is necessary to multiply the second half-reaction by a factor of 4:

C -8e \u003d C - reducing agent, oxidized

O + 2e \u003d O 4 oxidizing agent, reduced

The oxidizing agent accepts electrons during the reaction, lowering its oxidation state, it is reduced.

The reducing agent donates electrons during the reaction, increasing its oxidation state, it is oxidized.

1. Mikityuk A.D. Collection of tasks and exercises in chemistry. Grades 8-11 / A.D. Mikityuk. - M.: Ed. "Exam", 2009. (p. 67)

2. Orzhekovsky P.A. Chemistry: 9th grade: textbook. for general inst. / P.A. Orzhekovsky, L.M. Meshcheryakova, L.S. Pontak. - M.: AST: Astrel, 2007. (§22)

3. Rudzitis G.E. Chemistry: inorgan. chemistry. Organ. chemistry: textbook. for 9 cells. / G.E. Rudzitis, F.G. Feldman. - M.: Enlightenment, JSC "Moscow textbooks", 2009. (§ 5)

4. Khomchenko I.D. Collection of problems and exercises in chemistry for high school. - M .: RIA "New Wave": Publisher Umerenkov, 2008. (p. 54-55)

5. Encyclopedia for children. Volume 17. Chemistry / Chapter. ed. V.A. Volodin, leading. scientific ed. I. Leenson. - M.: Avanta +, 2003. (p. 70-77)

Additional web resources

1. A single collection of digital educational resources (video experiences on the topic) ().

2. A single collection of digital educational resources (interactive tasks on the topic) ().

3. Electronic version of the journal "Chemistry and Life" ().

Homework

1. No. 10.40 - 10.42 from the "Collection of tasks and exercises in chemistry for high school" I.G. Khomchenko, 2nd ed., 2008

2. Participation in the reaction of simple substances is a sure sign of a redox reaction. Explain why. Write the equations for the reactions of connection, substitution and decomposition with the participation of oxygen O 2.

The topic of the lesson is "Redox Reactions".

Goals:

Tutorial: pto acquaint students with a new classification of chemical reactions on the basis of changes in the oxidation states of elements - with redox reactions (ORD).Form the concept of oxidizing - restorativereactions, as chemical reactions on the basis of changes in the oxidation state of elements. Define the terms oxidizing agent and reducing agent. Describe the unity and continuity of the processes of oxidation and reduction, to teach students to arrange the coefficients using the electronic balance method.

Educational: pTo continue the formation of skills to compose equations of chemical reactions. To contribute to the expansion of the horizons of students, the development of skills and abilities to apply the knowledge gained to explain the phenomena of the world around.Continue the development of logical thinking, the ability to analyze and compare.Improve practical skills in working with laboratory equipment and chemical reagents; to supplement students' knowledge of the rules of work in the chemistry room. Develop the ability to observe, draw conclusions.

Educational: with contribute to the formation of a culture of interpersonal communication on the example of the ability to listen to each other, ask questions to each other, analyze the answers of comrades, predict the result of work, evaluate their work.To form a scientific outlook of students, to improve labor skills.

Lesson type: learning new material.

Didactic goals:create conditions for understanding and understanding the block of new educational information.

Lesson form: lesson - discussion with elements of problem-based learning.

Teaching methods:explanatory - illustrative, problematic, partially - search.

During the classes

Organizing time.

Journey to the past:

Teacher: In the III century BC. a monument was built on the island of Rhodes in the form of a huge statue of Helios (among the Greeks, the God of the Sun). The grandiose idea and perfection of execution of the Colossus of Rhodes - one of the wonders of the world - amazed everyone who saw it. (showing the colossus on the slide). We do not know for sure what the statue looked like, but it is known that it was made of bronze and reached a height of about 33 m. The statue was created by the sculptor Haret, it took 12 years to build. The bronze shell was attached to the iron frame. The hollow statue began to be built from the bottom and, as it grew, it was filled with stones to make it more stable. About 50 years after the completion of construction, the Colossus collapsed. During the earthquake, he broke at the level of his knees. Scientists believe the reason for the fragility of this miracle was the corrosion of the metal, and the basis of the corrosion process is redox reactions.Write in your notebook the topic of the lesson: “Oxidation- restorative reactions."

So, today in the lesson we will get acquainted with redox reactions and find out what is the difference between exchange reactions and redox reactions. We will learn how to determine the oxidizing agent and reducing agent in the reactions. We will learn how to draw up diagrams of the processes of giving and receiving electrons.

Knowledge update.

To begin with, let's remember what an oxidation state is and how the oxidation state is determined in simple and complex substances.

The oxidation state is the conditional charge of an atom in a compound. The oxidation state coincides with the valency, but unlike the valence, the oxidation state is negative.

Rules for determining oxidation states:

1. For free atoms and for simple substances, the oxidation state is 0:

Na, H 2 , N 2 , S, Al, F 2 .

2. Metals in all compounds have a positive oxidation state (its maximum value is equal to the group number):

a) for metals of the main subgroup of group I +1;

b) for metals of the main subgroup of group II +2;

c) aluminum has +3.

3. In compounds, oxygen has an oxidation state of -2

(exceptionO +2 F 2 and peroxides:H 2 O 2 -1 ; K 2 O 2 -1 ).

4. In compounds with non-metals, hydrogen has an oxidation state of +1, and with metals -1.

5. In compounds, the sum of the oxidation states of all atoms is 0.

H +1 Cl -1 H 2 +1 S -2 H 2 +1 S +6 O 4 -2

1 - 1 = 0 (2 1) - 2 = 0 (1 2) + 6 - (2 4) = 0

Exploring a new topic.

In the 8th grade, you got acquainted with the reactions of combination, decomposition, substitution and exchange.This classification of chemical reactions is based on the number and composition of the initial and resulting substances. Let's consider chemical reactions from the point of view of oxidation (donation of electrons) and reduction (addition of electrons) of the atoms of elements. Above the signs of chemical elements we put down their oxidation states.

Did the oxidation states of the elements change in these reactions?

In the first equation, the oxidation states of the elements did not change, while in the second they did - for copper and iron.

The second reaction is redox.

Reactions, as a result of which the oxidation states of the elements that make up the reactants and reaction products change, are called redox reactions ( ).

In redox reactions, electrons are transferred from one atom, molecule, or ion to another. The process of donating electrons is calledoxidation .

H 2 0 - 2ē 2H + 2Br - - 2ēBr 2 0 S -2 - 2ēS 0

The process of adding electrons is calledrecovery :

Mn +4 + 2ē Mn +2 S 0 + 2ē S -2 Cr +6 +3ē Cr +3

Atoms or ions that donate electrons in this reaction areoxidizers , and which donate electrons -reducing agents .

Drawing up equations of redox reactions.

There are two methods for compiling redox reactions - the electron balance method and the half-reaction method. Here we will look at .
In this method, the oxidation states of atoms in the starting substances and in the reaction products are compared, while being guided by the rule: the number of electrons donated by the reducing agent must be equal to the number of electrons attached by the oxidizing agent.
To draw up an equation, you need to know the formulas of the reactants and reaction products. Let's look at this method with an example.

Algorithm for compiling OVR equations using the electronic balance method:

Draw up a reaction scheme.

Al + HCl → AlCl 3 + H 2

Determine the oxidation states of the elements in the reactants and reaction products.

Al 0 + H +1 Cl -1 → Al +3 Cl 3 -1 + H 2 0

Determine whether the reaction is redox or it proceeds without changing the oxidation states of the elements.

This reaction is OVR

Underline the elements whose oxidation states change.

Al 0 + H +1 Cl -1 → Al +3 Cl 3 -1 + H 2 0

Determine which element is oxidized (its oxidation state goes up) and which element is reduced (its oxidation state goes down) during the reaction.

Al 0 → Al +3 oxidized

H +1 → H 2 0 recovering

On the left side of the diagram, use arrows to indicate the oxidation process (displacement of electrons from an atom of an element) and the reduction process (displacement of electrons to an atom of an element)

Al 0 – 3 ē →Al +3 oxidation process

2 H +1 + 2 ē →H 2 0 recovery process

Determine the reducing agent and oxidizing agent.

Al 0 – 3 ē →Al +3 reducing agent

2 H +1 + 2 ē →H 2 0 oxidizer

Balance the number of electrons between the oxidizing agent and reducing agent.

Al 0 – 3 → Al +3

2H +1 + 2 ē → H 2 0

Determine the coefficients for the oxidizing agent and reducing agent, oxidation and reduction products.

Al 0 – 3 → Al +3

x 2

2H +1 + 2 ē → H 2 0

x 3

Place the coefficients in front of the formulas of the oxidizing agent and reducing agent.

2 Al+ 6 HCl → 2 AlCl 3 + 3 H 2

Check the reaction equation.

Let's count the number of atoms on the right and left, if there is an equal number of them, we have equalized the equation.

Consolidation.

№1. Determine the oxidation state of atoms of chemical elements according to the formulas of their compounds:H 2 S, O 2 , NH 3 , HNO 3 , Fe, K 2 Cr 2 O 7

№2. Determine what happens to the oxidation state of sulfur during the following transitions:H 2 S → SO 2 → SO 3

№3. Arrange the coefficients in UCR using the electronic balance method, indicate the oxidation (reduction) processes, the oxidizing agent (reducing agent); write the reactions in full and ionic form:

A) Zn + HCl = H 2 + ZnCl 2

B) Fe + CuSO 4 = FeSO 4 + Cu

№4. Are givenschemeequationsreactions:
WITHuS + HNO 3 ( diluted) = Cu(NO 3 ) 2 + S + NO + H 2 O

K+H 2 O=KOH+H 2
Arrange the coefficients in the reactions using the electronic balance method.
Name the oxidizing agent and the reducing agent.

Homework: p 1, exercise 1, 6 p 7.

Reactions in which the elements that make up the reacting substances change their oxidation state are called redox reactions (ORD).

The degree of oxidation. To characterize the state of elements in compounds, the concept of the degree of oxidation has been introduced. The oxidation state (s.o.) is the conditional charge that is assigned to an atom under the assumption that all bonds in a molecule or ion are polarized to the limit. The oxidation state of an element in a molecule of a substance or ion is defined as the number of electrons displaced from an atom of a given element (positive oxidation state) or to an atom of a given element (negative oxidation state). To calculate the oxidation state of an element in a compound, one should proceed from the following provisions (rules):

1. The oxidation state of elements in simple substances, in metals in the elemental state, in compounds with non-polar bonds is equal to zero. Examples of such compounds are N 2 0 , H 2 0 , Cl 2 0 , I 2 0 , Mg 0 , Fe 0 , etc.

2. In complex substances, elements with higher electronegativity have a negative oxidation state.

Since, during the formation of a chemical bond, electrons are displaced to atoms of more electronegative elements, the latter have a negative oxidation state in compounds.

O -2 Cl O -2 H + Element EO

In some cases, the oxidation state of an element numerically coincides with the valence (B) of the element in a given compound, as, for example, in HClO 4 .

The examples below show that the oxidation state and valence of an element can differ numerically:

N ≡ N B (N)=3; s.d.(N)=0

H + C -2 O -2 H +

EO (C) = 2.5 V (C) = 4 s.o. (C) = -2

EO (O) \u003d 3.5 V (O) \u003d 2 s.o. (O) \u003d -2

EO(N) = 2.1 V(N) = 1 s.o.(N) = +1

3. There are higher, lower and intermediate oxidation states.

Highest oxidation state is its largest positive value. The highest oxidation state, as a rule, is equal to the group number (N) of the periodic system in which the element is located. For example, for elements of period III, it is equal to: Na +2, Mg +2, AI +3, Si +4, P +5, S +6, CI +7. The exceptions are fluorine, oxygen, helium, neon, argon, as well as elements of the cobalt and nickel subgroups: their highest oxidation state is expressed by a number whose value is lower than the number of the group to which they belong. For elements of the copper subgroup, on the contrary, the highest oxidation state is greater than one, although they belong to group I.

lowest degree oxidation is determined by the number of electrons that do not reach the stable state of the atom ns 2 np 6 . The lowest oxidation state for non-metals is (N-8), where N is the group number of the periodic system in which the element is located. For example, for non-metals of the III period, it is equal to: Si -4, P -3, S -2, CI ˉ. The lowest oxidation state for metals is its smallest positive value possible. For example, manganese has the following oxidation states: Mn +2, Mn +4, Mn +6, Mn +7; s.o.=+2 is the lowest oxidation state for manganese.

All other oxidation states of an element are called intermediate. For example, for sulfur, the oxidation state of +4 is intermediate.

4. A number of elements exhibit a constant oxidation state in complex compounds:

a) alkali metals - (+1);

b) metals of the second group of both subgroups (with the exception of Hg) - (+2); mercury can exhibit oxidation states (+1) and (+2);

c) metals of the third group, the main subgroup - (+3), with the exception of Tl, which can exhibit oxidation states (+1) and (+3);

e) H + , except for metal hydrides (NaH, CaH 2, etc.), where its oxidation state is (-1);

f) O -2, with the exception of peroxides of elements (H 2 O 2, CaO 2, etc.), where the oxidation state of oxygen is (-1), superoxides of elements

(KO 2, NaO 2, etc.), in which its oxidation state is - ½, fluoride

oxygen ОF 2 .

5. Most elements can show different oxidation states in compounds. When determining their oxidation state, they use the rule according to which the sum of the oxidation states of elements in electrically neutral molecules is zero, and in complex ions it is the charge of these ions.

As an example, let's calculate the oxidation state of phosphorus in orthophosphoric acid H 3 PO 4 . The sum of all oxidation states in the compound should be equal to zero, so let's denote the oxidation state of phosphorus as X and, multiplying the known oxidation states of hydrogen (+1) and oxygen (-2) by the number of their atoms in the compound, we will write the equation: (+1) * 3+X+(-2)*4 = 0, of which X = +5.

Let's calculate the oxidation state of chromium in the dichromate ion (Cr 2 O 7) 2-.

The sum of all oxidation states in a complex ion should be equal to (-2), therefore, we denote the oxidation state of chromium through X, we compose the equation 2X + (-2) * 7 \u003d -2, from which X \u003d +6.

The concept of the degree of oxidation for most compounds is conditional, because does not reflect the real effective charge of the atom. In simple ionic compounds, the oxidation state of their constituent elements is equal to the electric charge, since during the formation of these compounds, an almost complete transfer of electrons from one

1 -1 +2 -1 +3 -1

atom to another: NaI , MgCI 2 , AIF 3 . For a compound with a polar covalent bond, the actual effective charge is less than the oxidation state, but this concept is very widely used in chemistry.

The main provisions of the OVR theory:

1. Oxidation The process of donating electrons by an atom, molecule, or ion is called. Particles that donate electrons are called reducing agents; during the reaction, they are oxidized, forming an oxidation product. In this case, the elements involved in the oxidation increase their oxidation state. For example:

AI - 3e -  AI 3+

H 2 - 2e -  2H +

Fe 2+ - e -  Fe 3+

2. Recovery The process of adding electrons to an atom, molecule, or ion is called. Particles that accept electrons are called oxidizers; during the reaction they are reduced, forming a reduction product. In this case, the elements involved in the reduction lower their oxidation state. For example:

S + 2e -  S 2-

CI 2 + 2e -  2 CI ˉ

Fe 3+ + e -  Fe 2+

3. Substances containing reducing or oxidizing particles are respectively called reducing agents or oxidizing agents. For example, FeCI 2 is a reducing agent due to Fe 2+ , and FeCI 3 is an oxidizing agent due to Fe 3+ .

4. Oxidation is always accompanied by reduction and, conversely, reduction is always associated with oxidation. Thus, OVR is a unity of two opposite processes - oxidation and reduction

5. The number of electrons donated by the reducing agent is equal to the number of electrons received by the oxidizing agent.

Drawing up equations of redox reactions. Two methods for compiling equations for the OVR are based on the last rule:

1. Electronic balance method.

Here, the number of added and lost electrons is calculated based on the values ​​of the oxidation states of the elements before and after the reaction. Let's look at the simplest example:

Na0 + Cl Na + Cl

2Na 0 – eˉ  Na + - oxidation

1 Cl 2 + 2eˉ  2 Cl - recovery

2 Na + Cl 2 = 2Na + + 2Cl

2 Na + Cl 2 = 2NaCl

This method is used if the reaction does not proceed in solution (in the gas phase, thermal decomposition reactions, etc.).

2. Ion-electronic method (half-reaction method).

This method takes into account the environment of the solution, gives an idea of ​​the nature of the particles actually existing and interacting in solutions. Let's dwell on it in more detail.

Algorithm for selection of coefficients by ion-electronic method:

1. Draw up a molecular scheme of the reaction indicating the starting materials and reaction products.

2. Draw up a complete ion-molecular scheme of the reaction, writing weak electrolytes, poorly soluble, insoluble and gaseous substances in molecular form, and strong electrolytes in ionic form.

3. Having excluded from the ion-molecular scheme ions that do not change as a result of the reaction (without taking into account their number), rewrite the scheme in a short ion-molecular form.

4. Mark the elements that change the oxidation state as a result of the reaction; find the oxidizing agent, reducing agent, reduction products, oxidation.

5. Draw up schemes of half-reactions of oxidation and reduction, for this:

a) indicate the reducing agent and the oxidation product, the oxidizing agent and the reduction product;

b) equalize the number of atoms of each element in the left and right parts of the half-reactions (perform a balance by elements) in the sequence: an element that changes the oxidation state, oxygen, other elements; it should be remembered that in aqueous solutions, H 2 O molecules, H + or OH ions, depending on the nature of the medium, can participate in reactions:

c) equalize the total number of charges in both parts of the half-reactions; to do this, add or subtract the required number of electrons on the left side of the half-reactions (charge balance).

6. Find the least common multiple (LCM) for the number of given and received electrons.

7. Find the main coefficients for each half-reaction. To do this, the number (NOC) obtained in paragraph 6 is divided by the number of electrons appearing in this half-reaction.

8. Multiply the half-reactions by the obtained basic coefficients, add them together: the left side on the left, the right side on the right (obtain the ion-molecular reaction equation). If necessary, “bring similar” ions, taking into account the interaction between hydrogen ions and hydroxide ions: H + +OH ˉ= H 2 O.

9. Arrange the coefficients in the molecular reaction equation.

10. Carry out a check for particles that do not participate in the OVR and are excluded from the complete ion-molecular scheme (item 3). If necessary, the coefficients for them are found by selection.

11. Perform a final oxygen test.

1. Acidic environment.

Molecular scheme of the reaction:

KMnO 4 + NaNO 2 + H 2 SO 4  MnSO 4 + NaNO 3 + H 2 O + K 2 SO 4

Full ion-molecular scheme of the reaction:

K+MnO + Na + + NO +2H++ SO  Mn2+ + SO + Na + + NO + H2O + 2K + +SO .

Brief ion-molecular scheme of the reaction:

MNO + NO + 2H +  Mn 2+ + NO + H2O

ok-l v-l product v-niya product ok-iya

During the reaction, the oxidation state of Mn decreases from +7 to +2 (manganese is reduced), therefore, MnO - oxidizing agent; Mn 2+ is a reduction product. The oxidation state of nitrogen rises from +3 to +5 (nitrogen is oxidized), therefore, NO – reducing agent, NO is an oxidation product.

Half reaction equations:

2MNO + 8 H+ + 5e -  Mn 2+ + 4 H 2 O- recovery process

10 +7 +(-5) = +2

5 NO + H 2 O– 2e -  NO + 2 H+ - oxidation process

2MnO + 16H + + 5NO + 5H 2 O = 2Mn 2+ + 8H 2 O + 5NO + 1OH + (complete ionic-molecular equation).

In the total equation, we exclude the number of identical particles that are both on the left and on the right sides of the equation (we give similar ones). In this case, these are H + and H 2 O ions.

A brief ion-molecular equation will have the form

2MnO + 6H + + 5NO  2Mn 2+ + 3H 2 O + 5NO .

In molecular form, the equation is

2KMnO 4 + 5 NaNO 2 + 3 H 2 SO 4 \u003d 2MnSO 4 + 5NaNO 3 + 3H 2 O + K 2 SO 4.

Let's check the balance for the particles that did not participate in the OVR:

K + (2 = 2), Na + (5 = 5), SO (3 = 3). Oxygen balance: 30 = 30.

2. Neutral environment.

Molecular scheme of the reaction:

KMnO 4 + NaNO 2 + H 2 O  MnO 2 + NaNO3 + KOH

Ionic-molecular scheme of the reaction:

K++MnO + Na + + NO + H 2 O  MnO 2 + Na + + NO +K++OH

Brief ion-molecular scheme:

MNO + NO + H 2 O  MnO 2 + NO +OH-

ok-l v-l product v-niya product ok-iya

Half reaction equations:

2MnO + 2H 2 O+ 3eˉ MnO 2 +4OH - recovery process

6 -1 +(-3) = -4

3 NO +H 2 O– 2eˉ NO + 2H + - oxidation process