Supplement over the topic of the OSR. An abstract lesson on the theme "Redox reaction" (grade 11). Vi. VSP

The motto of the lesson: "Someone loses, and someone finds ..."

Objectives lesson:
Training:
consolidate the concept of "oxidation degree", "oxidation" processes, "recovery";
consolidate the skills in the preparation of equations of redox reactions by the electronic balance method;
Teach to predict the products of oxidation reactions.
Developing:
Continue the development of logical thinking, skills to observe, analyze and compare, to find causal relationships, draw conclusions, work with algorithms, to form interest in the subject.
Educational:
Form scientific worldview of students; improve labor skills;
To teach listening to the teacher and your classmates, be attentive to yourself and others, evaluate yourself and others, to talk.

I. Organizational moment

The topic of the lesson is declared, the relevance of this topic and its connection with life is justified. Redox processes belong to the most common chemical reactions and are of great importance in theory and practice. The metabolic processes of metabolic processes flowing in the living organism, rotting and fermentation, photosynthesis are related. Redox processes accompany the cyphans of substances in the closure. They can be observed in combustion of fuel, in the processes of metal corrosion, with electrolysis and metal smelting. With their help, rubber, acids and other valuable products are obtained.
Redox reactions underlie the conversion of the energy of interacting chemicals into electrical energy in electroplating and fuel cells. Humanity has long been used by the OSR, at first, without understanding their essence. Only by the beginning of the 20th century was created an electronic theory of redox processes. The lesson will have to recall the basic provisions of this theory, as well as learn how to make the equations of chemical reactions occurring in solutions, and find out what the mechanism of such reactions depends.
II. Repetition and generalization of the material studied earlier
1. The degree of oxidation.
The organization of the conversation aimed at updating the reference knowledge about the degree of oxidation and the rules of its definition, on the following issues:
- What is electric monitance?
- What is the degree of oxidation?
- Can the degree of oxidation of the element be equal to zero? In which cases?
- What degree of oxidation most often shows oxygen in connections?
- Remember exceptions.
- What degree of oxidation is metals in polar and ionic connections?
Following the conversation, the rules for determining the degrees of oxidation are formulated
To secure the formulated rules, it is proposed to determine the degree of oxidation of elements in connections:
H2SO4, H2, H2SO3, HCIO4, VA, KMNO4, AI2 (SO4) 3, HNO3, BA (NO3) 2, HCN, K4, NH3, (HN4) 2SO4.
This task with selective responses is used for an oral front survey.
2. Oxidation and recovery processes. Redox reactions.
During the conversation, there is an actualization of knowledge of redox processes.
Specify the type of chemical reaction to the right. If necessary, lay out the coefficients. If S.O. Elements before and after the reaction are changed, then you write the word "yes", if you do not change, then write the word "no".
I Option:
HG + S → HG s
Nano3 → Nano2 + O2
Cuso4 + NaOH → Na 2SO4 + Cu (OH) 2
II Option:
Al (OH) 3 → Al 2O3 + H2O
H2O + P2O5 → H3PO4
Fe + HCl → FECL2 + H2
All types of work are checked together with the class. The board remains the equations of chemical reactions, and then the class is invited to answer the questions:
1) In all cases, there is a change in the degrees of oxidation of chemical elements? (not).
2) Does this depend on the type of chemical reactions in the number of reagents and reaction products? (not).
Questions are offered:
- What is the name of the recovery process?
- How does the degree of oxidation of the element change when recovering?
- What is oxidation?
- How does the degree of oxidation of the element be changed during oxidation?
- Give the definition of the concepts of the "oxidizing agent" and "reducing agent".
From a modern point of view, the change in the degree of oxidation is associated with pulling or moving electrons. Therefore, along with the above, it is possible to give another definition: these are such reactions in which the transition of electricity from one atoms, molecules or ions to the other.
We conclude: "What is the essence of the OSR?"
Redox reactions are of the unity of two opposite processes - oxidation and recovery. In these reactions, the number of electrons given by the reducing agents is equal to the number of electrons connected by oxidizing agents. At the same time, no matter whether the electrons are transmitted from one atom to another completely or only partially, they are delayed to one of the atoms, it is conventionally indicated only on the return or attachment of electrons. That is why the motto of the lesson is chosen: "Someone loses, and someone finds ..."
3. Functions of compounds in the OSR.
1. Consider the degree of oxidation of the elements, prove that these substances exhibit the properties of oxidants.
CL2, HCLO4, H2SO4, KMNO4, SO2
2. Consider the degrees of elements oxidation, prove that these substances exhibit the properties of reducing agents:
HCl, NH3, H2S, K, SO2
As a result of this work, students form rules for determining the function of the compound in the OSR:
1. If the element exhibits the highest degree of oxidation in the connection, then this connection can only be an oxidizer.
2. If the element exhibits the lowest degree of oxidation in connection, this connection can be a reducing agent
Solving problematic issues:
- Can one and the same substance be as an oxidizing agent and a reducing agent?
- Can one and the same element show the properties of both the oxidizing agent and the reducing agent?
Formulation of the third rule.
3. If the element exhibits an intermediate degree of oxidation in connection, this compound can be both a reducing agent and oxidizing agent.

III. The placement of coefficients in the OSR equations by the electronic balance method.

Studying the skills for determining the degree of oxidation, the preparation of patterns of redox reactions by the method of electronic balance (work at the board and in notebooks) with the development of reasoning and analysis skills through the comments of the responses to students.
By the method of electronic balance, select the coefficients in the schemes of redox reactions and specify the process of oxidation and recovery:
K2Cr2O7 + H2SO + H2SO4 → K2SO4 + CR2 (SO4) 3 + S + H2O

H2S + K2CR2O7 + H2SO4 → S + CR2 (SO4) 3 + K2SO4 + H2O

K2CR2O7 + HCl → CL2 + KCL + CRCL3 + H2O

H2O2 + KMNO4 + H2SO4 → O2 + K2SO4 + MNSO4 + H2O

Questions from part with (C1) Kimov Ege:

Nano2 + KMNO4 + H2SO4 → Nano3 + MNSO4 + ... + ...

Nano3 + Nai + H2SO4 → NO + I2 + ... + ...

KMNO4 + Na2SO3 + H2SO4 → MNSO4 + ... + ... + ...

Check - frontal survey, clarification of signs of redox reactions.
Questions from part in (B2) Kimov Ege:
Install the correspondence between the reaction equation and the change in the degree of oxidizing the oxidant in this reaction:

A) S02 + N02 \u003d S03 + NO 1) -1 → 0
B) 2nh3 + 2NA \u003d 2NANH2 + H2 2) 0 → -2
C) 4n02 + 02 + 2H20 \u003d 4HN03 3) +4 → +2
D) 4nh3 + 6no \u003d 5n2 + 6N20 4) +1 → 0
5) +2 → 0
6) 0 → - 1

Reaction equation Changes in the degree of oxidizing the oxidant

A) 2NH3 + 2NA \u003d 2NANH2 + H2 1) -1 → 0
B) H2S + 2NA \u003d Na2S + H2 2) 0 → - 1
4NH3 + 6NO \u003d 5N2 + 6N20 3) + 2 → 0
D) 2H2S + 302 \u003d 2S02 + 2N20 4) + 1 → 0
5) +4 → +2
6) 0→ -2
Install the correspondence between the reaction equation and substance that is a reducing agent in this reaction.
Recovery equation
A) NO + N02 + H20 \u003d 2HN02 1) N02
B) SO2 + 2H2S \u003d 3S + 2H20 2) H2S
BR2 + S02 + 2H20 \u003d 2HBR + H2SO4 3) BR2
D) 2ki + v2 \u003d 2kvg + i2 4) s02
5) No.
6) KI.
IV. Stage of consolidation of knowledge (ends with a test).
Test
1) What is the lowest degree of sulfur oxidation?
a) -6; b) -4; at 2; d) 0; e) +6.

2) What is the degree of phosphorus oxidation in the MG3P2 connection?
a) +3; b) +5; c) 0; d) -2; d) -3.

3) What elements have a constant oxidation degree +1?
a) hydrogen; b) lithium; c) copper;
d) magnesium; e) selenium.

4) What is the highest degree of oxidation of manganese?
a) -1; b) 0; c) +7; d) +4; e) +6.

5) What is the degree of chlorine oxidation in the Ca (CLO) 2 connection?
a) +2; b) +1; c) 0; d) -1; D 2.

6) Which of the following substances can only be oxidizing agents?
a) NH3; b) br2; c) KCLO3; d) Fe; e) hno3.

7) What is the name of the process below and how many electrons is involved in it?

a) recovery, 1e; b) oxidation, 2e;
c) recovery, 2e; d) oxidation, 1e.

8) Which of the listed substances can be oxidizing agents and reducing agents? Perhaps several answer options.
a) SO2; b) na; c) h2; d) K2Cr2O7; e) hno2.

9) What is the name of the process below and how many electrons involved in it?

a) recovery, 8e; b) oxidation, 4e;
c) oxidation, 8e; d) recovery, 4e.

10) Which of the following substances can only be recovers? Perhaps several answer options.
a) H2S; b) kmno4; c) SO2; d) nh3; e) Na.

Answers. 1 - in; 2 - d; 3 - b, g; 4 - in; 5 B; 6 - d; 7 - b; 8 - a, in, d; 9 - a; 10 - a, g, d.
V. Deeperation and expansion of knowledge (lecture part of the lesson)
The value of redox reactions
Redox reactions accompany many processes carried out in industry and in various areas of life: gas burning in a gas stove, cooking, washing, cleaning household items, manufacture of shoes, perfume, textile products ...
Whether we light the match, whether bizarre fireworks burn in the sky - all these are redox processes.
For the purposes of bleaching and disinfection, they use the oxidative properties of such most well-known means as hydrogen peroxide, potassium permanganate, chlorine and chlorine, or lecherous, lime.
If it is required to oxidize from the surface of the product any easily destructive substance, hydrogen peroxide is used. It serves to whiten silk, feathers, fur. With it, the vintage paintings also restore. Due to the harmability for the body, hydrogen peroxide is used in the food industry for whitening chocolate, scars and shells in the production of sausages.
The disinfecting effect of potassium permanganate is also based on its oxidative properties.
Chlorine as a strong oxidizing agent is used to sterilize pure water and disinfecting wastewater. Chlorine destroys many paints, on which it is based on the use of paper and tissues. Chlorine, or Bellen, lime is one of the most common oxidizers both in everyday life and in production scale.
In nature, redox reactions are extremely common. They play a large role in biochemical processes: breathing, metabolism, nervous activity of man and animals. The manifestation of various vital functions of the body is associated with the cost of energy, which our organism gets from food as a result of redox reactions.
Vi. Summarizing.

Estimates for the lesson and is given homework:
A. Determine the degrees of oxidation of elements by formulas:
HNO2, FE2 (SO4) 3, NH3, NH4CL, KCLO3, VA (NO3) 2, NSLO4
B. Plant the coefficients by the method of electronic balance:
KMNO4 + Na2SO3 + H2O → MnO2 + Na2 SO4 + KOH
S. KMNO4 + Na2SO3 + KOH → ... + K2 MnO4 + ...

Literature:

Gabrielyan O.S. Chemistry-8. M.: Drop, 2002;
Gabrielyan O.S., Voskoboinikova N.P., Yashukova A.V. Teacher's desk book. 8th grade. M.: Drop, 2002;
Small children's encyclopedia. Chemistry. M.: Russian Encyclopedic Association, 2001; Encyclopedia for "Avanta +" children. Chemistry. T. 17. M.: Avanta +, 2001;
Homchenko G.P., Sevastyanova K.I. Redox reactions. M.: Enlightenment, 1989.
V.A. Shelonians. Signal models and tasks: Redox reactions. Ooipkro, Omsk- 2002
A.G. Kulman. General Chemistry, Moscow-1989.
Full text material Abstract lesson for grade 8 "Redox reactions" Look in the downloaded file.
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1 Support. The theme "oxidative reaction reactions of the OSR" is oxidizing as a statement of reactions due to the redistribution of electrons between the interacting chemical particles, as a result of which the degrees of oxidation of the atoms included in their composition are changed. The degree of oxidation of the element in the compound is: a) a conditional charge attributed to the atom when assuming that all connections are built on ion type; B) a charge that would arise on the atom if the electronic pairs of which it is associated with other atoms would be shifted to a more electronegative atom. !!! The value of the degree of oxidation is set above the symbol of the chemical element. The rules for determining the degrees of oxidation: 1) the sum of the degrees of oxidation of all atoms in the compound is zero (the principle of electronutrality). 2) The sum of the degrees of the oxidation of elements in the ion is equal to the charge of the ion. 3) The degree of oxidation of the element in a simple substance is zero. 4) The degree of oxidation of a single-charge ion is equal to the charge of the ion. 5) Hydrogen in non-metallal compounds has a degree of oxidation +1 (except for boron and silicon), with metals, boron and silicon, the degree of hydrogen oxidation is 1. 6) oxygen in oxides, as a rule, has a degree of oxidation 2. in peroxides its degree of oxidation equal to 1 (H 2 O 2, Na 2 O 2), in a compound with fluorine (+2) of 2, in superoxides (1/2), in ozonides (1/3). 7) The highest (positive) degree of oxidation of the element is equal to the number of the periodic system group in which the element is located. Exceptions: elements 1B of group (CU, AG, AU) and VIII B groups (except osmia), oxygen, fluorine. 8) lower (negative) The degree of oxidation is characteristic of non-metals and is equal to the number of the periodic system of minus 8. The responses include reactions: * replacement, * compounds, * decomposition. The following types of OPR are distinguished: * Intermolecular varying degrees of oxidation of atoms in different molecules, * intramolecular oxidizer and the reducing agent are in the same substance (most often it is a thermal decomposition reaction), * disproportionation (dismutation) or self-examination of the oxidizing agent and reducing agent The atoms of the same element in the connection are performed.

2 Exercise 1. Determine the degrees of oxidation of elements in compounds: phosphorus: HRO 3, H 3 PO 3, H 3 PO 4, H 4 P 2 O 7, CA 3 (PO 4) 2, pH 3, pH 4 +, RO 3 . Serious: H 2 S, FES, FES 2, AS 2 S 3, H 2 SO 3, H 2 SO 4, Na 2 S 2 O 3, SO 4, AG 2 S, H 2 SO 5, SO 2, K 2 SO 3. Nitrogen: N 2 O, NO 2, N 2, NH 3, Ca 3 N 2, N 2 H 4, NH 4 NO 3, CH 3 NH 2, C 6 H 5 NO 2, C 6 H 5 NH 2, NO 2 NO 3. Oxygen: K 2 O, KO 3, H 2 O 2, O 3, O 2, OF 2. Carbon: Co, CO 2, CH 4, CH 3 Soam, C 2 H 5 , CH 3 PIN 3, NSO 3 H2CO 3, CH 2 O. Manganese: MNSO 4, MNO 2, K 2 MNO 4, KMNO 4, MN 2 O 7, MNO 4. Chrome: Cr 2 O 3, K 2 CR 2 O 7, K 2 CRO 4, NA 2 CRO 2, NA 3, CR 2 (SO 4) 3. Exercise 2. Which of the listed phenomena is a redox processing process? Option 1 Option 2 Option 3 1 Wheelpage of silver objects Eating ozone in the air during a thunderstorm translating of a Negaine lime into hare 2 Rusting of iron in the wet air Synthesis Ammonia Electrolysis of the melt of the table salt 3 combustion of gasoline in the engine combustion of wood absorption of moisture p 2 o 5 4 Gas release when Pumping of chalk 5 Gas release when dissolving zinc in hydrochloric acid Milk sinusia violation of gaseous products When calculating the ammonia, warming solution when diluting sulfuric acid burning of a candle Any ARP bilateral process consisting of semi-formation of oxidation and recovery. Restorener (E) The degree of oxidation increases. ("Gave the electron, acquired oxygen, oxidized") Oxidizer (+ E) the degree of oxidation decreases. (Took, recovered) // Restorener is the one who gives the electrons, he himself gives the robber to the villainoomyokuslitel //. In the preparation of the ORV equations, compliance with the two most important rules:

3 1. ELECTRONICAL RULES: The number of electrons given to the wist, should be equal to the number of electrons adopted in the semi-reaction of recovery. 2. The rule of consistency of charges: the sum of all charges in the left side of the equation is equal to the sum of all charges in the right part of the equation. Currently, 2 methods of finding stoichiometric coefficients are most often used: 1. Electronic balance method. Used to describe heterogeneous processes. 2. Semi-formation method or ion electronic method. Used to describe reactions flowing in aqueous solutions where electron percene processes and ion exchange are combined. The second method has a number of indisputable advantages: * There is no need to determine the degrees of oxidation of atoms of individual elements, which is especially essential in the case of reactions occurring with the participation of organic substances. * Reaction products are easily defined in the equalization process. * Water formulas, acids or alkali indicate the medium and are given to correctly determine the reaction products. In the process of adjustment, they can move from one part of the equation to another and even disappear. It should be noted that for the successful operation of the OSR, the pH of the medium often has to regulate. In this case, in oxidally reduction pairs of reactant substances, auxiliary substances are introduced, creating the necessary medium: for pH 7 sulfuric acid, for pH 7 sodium hydroxides or potassium. In aqueous solutions in oxidally reduction processes, water ions are actively involved: H + and it will remember the following rules: * Recovery: one oxygen atom, leaving the oxidant particle, two ions are spent in an acidic medium and one water molecule is formed; In a neutral and alkaline medium, one water molecule is spent and two ions are formed. * Oxidation: one oxygen atom connecting to the recovery particle is spent in an acidic and neutral medium one water molecule and two ions n + is formed; In an alkaline medium, two ions are spent and one water molecule is formed. For example: Table 1. PH Restoration Oxidation< 7 MnO 4 + 8H + + 5e Mn H 2 O SO H 2 O 2e SO 4 + 2H + = 7 MnO 4 +2Н 2 О + 3e MnO 2 + 4ОН SO H 2 O 2e SO 4 + 2H +

4\u003e 7 MNO 4 + E MNO 4 SO 3 + 2OH 2E SO 4 + H 2 O Exercise 3. Make an equation of oxidation or recovery semi-formation equations, taking into account the acidity of the medium: Table 2. Acosal medium pH< 7 Нейтральная рн = 7 Щелочная рн > 7 1 NO 3 NO 2 NO 2 NO 3 CRO 2 CRO 4 2 MNO 4 MN 2+ MNO 4 MNO 2 AL ALO 2 3 CR 3+ CR 2 O 7 SO 3 SO 4 NO 3 N 2 Exercise 4. Make the recovery semi-reaction equations nitrution (NO 3) in an acidic medium to: a) NO 2, b) NO, c) N 2 O, D) NN 4 +. Exercise 5. The transition schemes are given: 1) SO 2 S, 2) Caco 3 Ca (HCO 3) 2, 3) CR 2 O 3 CRO 3, 4) H 2 O 2 O 2, 5) CR 2 O 7 CRO 4 What process corresponds to each transition: a) oxidative, b) reducing, c) exchange? Algorithm for selecting coefficients in the reaction equations by the electron balance method by the method of an electronically balance using the following chemical reaction equation: H 2 O 2 + K 2 CR 2 O 7 + H 2 SO 4 \u003d 1. Write the schemes of two incomplete semi-formations: the oxidant transition to its reduced shape The reducing agent in its oxidized form: CR 2 O 7 CR 3+ Oxidizer H 2 O 2 O 2 Restorener 2. Take a material balance. To do this, equalize the number of atoms of all elements, except for oxygen and hydrogen: CR 2 O 7 2CR 3+ H 2 O 2 O 2 After that, the number of oxygen and hydrogen atoms are equalized depending on the medium in which the reaction proceeds. CR 2 O H + 2CR H 2 O H 2 O 2 O 2 + 2H + 3. Test balance of charges. The total charge on the left and right is equalized by adding or subtracting electrons in the left side of the schemes. CR 2 O H + + 6E 2CR H 2 O H 2 O 2 2E O 2 + 2H + 4. To semoreactivity schemes, select the coefficients so that the number of removable electrons is equal to the number of accepted. CR 2 O H + + 6E 2CR H 2 O 1 H 2 O 2 2E O 2 + 2H + 3

5 5. Fold the sections of half-formations, taking into account the selected coefficients: CR 2 OH + + 3H 2 O 2 2CR H 2 O + 3O 2 + 6H + 6. Reduce "Similar" Members: Cr 2 OH + + 3H 2 O 2 2CR H 2 O + 3O 2 7. To each ion, the counterions in the desired amount taking into account the starting materials. Exactly the same ions and in the same amount added to the right-hand part of the equation: Cr 2 O H + + 3H 2 O 2 2CR H 2 O + 3O 2 2K + 4SO 4 2K + + 4SO 4 8. Write the formulas of substances in molecular form. On the right side of the equation, the ions that give uni-soluble or smallssaging substances are connected. The remaining ions are combined arbitrarily 3H 2 O 2 + K 2 CR 2 O 7 + 4H 2 SO 4 \u003d CR 2 (SO 4) 3 + 3O 2 + K 2 SO H 2 O Exercise 6. Make the equations of oxidizing agent reactions between potassium permanganate and sulfite Potassium in acidic, neutral and alkaline media using the oxidation and recovery and recovery semi-processing table 1. Transition metals in the lowest oxidation (Ions Sn 2 +, Fe 2+, Cu +, Hg 2 2+, etc.), interacting with oxidizing agents, can increase their degree of oxidation, for example: 5FeCl 2 + KMNO HCl \u003d 5 FECL 3 + MnCl 2 + KCl + 4 H 2 O Exercise 7. Extract the schemes of similar reactions; Stoichiometric coefficients Select the ion electronic method. a) FESO 4 + HNO 3 (conc.) \u003d Fe (NO 3) 3 + NO 2 + b) SNCl 2 + CL 2 \u003d SNCl 4 in an acidic medium chromat goes into dichromat: 2K 2 CRO 4 + H 2 SO 4 \u003d K 2 CR 2 O 7 + K 2 SO 4 + H 2 O Ne further restoration of the ion dichromate to CR 3+: CR 2 OH + + 6E 2CR H 2 O Exercise 8. Make the equations of oxidizing agent reactions and dispel the coefficients by half-reaction. a) k 2 Cr 2 O 7 + SO 2 + H 2 SO 4 \u003d CR 2 (SO 4) 3 + K 2 SO 4 + H 2 O b) k 2 Cr 2 O 7 + KNO 2 + H 2 SO 4 \u003d Kno 3 + c) K 2 CR 2 O 7 + Ki + H 2 SO 4 \u003d i 2 + in oxidizing-substitutative reactions The hydrogen peroxide can be like an oxidizing agent: H 2 O 2 + 2H + + 2E 2H 2 O

6 H 2 O 2 + 2E 2OH and reducing agent: H 2 O 2 + 2OH 2E O 2 + 2H 2 OH 2 O 2 2E 2 + 2H + Exercise 9. Pick up the stoichiometric coefficients in the equations of reactions leaking from The participation of hydrogen peroxide. Specify, in what of these hydrogen peroxide is an oxidizer, and in which reducing agent? a) H 2 O 2 + PBS \u003d PBSO H 2 O b) H 2 O 2 + NIS + CH 3 COOH \u003d S + Ni (CH 3 COO) 2 + H 2 O c) H 2 O 2 + KMNO 4 + H 2 SO 4 \u003d O 2 + MNSO 4 + K 2 SO 4 + H 2 O is a very strong oxidizer is ammonium persulfate (NH 4) 2 S 2 O 8. In the preparation of equations, it can be considered that persulphates decomposes, highlighting atomic oxygen playing a role Oxidizer: (NH 4) 2 S 2 O 8 + H 2 O \u003d (NH 4) 2 SO 4 + H 2 SO 4 + [O] S 2 O 8 + 2E \u003d 2SO 4 Exercise 10. Separate the coefficients in the following Equations of the oxidizing agent process involving ammonium persulfate, potassium and sodium: a) (NH 4) 2 S 2 O 8 + Mn (NO 3) 2 + H 2 O \u003d HMNO 4 + (NH 4) 2 SO 4 + H 2 SO 4 b) K 2 S 2 O 8 + Mn (NO 3) 2 + H 2 O \u003d HMNO 4 + HNO 3 + KHSO 4 c) Na 2 S 2 O 8 + CRCl 3 + NaOH \u003d Na 2 SO 4 + Na 2 Cro 4 +? The equivalent number of oxidizing agent (reducing agent) is a quantity of its amount, which, restoring (oxidizing), attachs (releases) 1 mol of electrons. The molar mass of the oxidant equivalent (reducing agent) is equal to the molar mass divided into an equivalent number: ME (x) \u003d m (x) / z, g / mol Exercise 11. Calculate the equivalent number and molar mass of equivalent of sulfuric acid in the proposed reactions: a) Zn + H 2 SO 4 (RSS) \u003d ZNSO 4 + H 2 b) 2HBr + H 2 SO 4 (conc.) \u003d Br 2 + SO H 2 O c) 8Hi + H 2 SO 4 (conc.) \u003d 4i 2 + H 2 S + 4 H 2 O Exercise 12. Schemes of redox reactions are proposed. By the method of ionically electronic balance, arrange stoichiometric coefficients, calculate the equivalent number and molar masses of the oxidant equivalent and the reducing agent: 1. Kmno 4 + No + H 2 SO 4 \u003d HNO 3 + MNSO 4 + K 2 SO 4 + KNO 3 + H 2 O 2. PBO 2 + CR (NO 3) 3 + H 2 O \u003d Pb (NO 3) 2 + H 2 Cr 2 O 7 3. FESO 4 + K 2 CR 2 O 7 + H 2 SO 4 \u003d CR 2 (SO 4) 3 + K 2 SO 4 + Fe 2 (SO 4) 3 + H 2 O 4. KNO 2 + K 2 CRO 4 + KOH + H 2 O \u003d KNO 3 + K 3 5. KMNO 4 + P + H 2 SO 4 \u003d H 3 PO 4 + K 2 SO 4 + MNSO 4

7 6. KCLO 3 + KJ + H 2 SO 4 \u003d KCL + J 2 + K 2 SO 4 7. KMNO 4 + SO 2 + H 2 O \u003d H 2 SO 4 + MNSO 4 + K 2 SO 4 8. KJ + H 2 SO 4 \u003d J 2 + KHSO 4 + H 2 S + H 2 O 9. KMNO 4 + Na 2 SO 3 + H 2 SO 4 \u003d k 2 SO 4 + MNSO 4 + H 2 O 10. BI 2 (SO 4) 3 + Cl 2 + NaOH \u003d Nabio 3 + NaCl + H 2 O 11. K 2 Cr 2 O 7 + Na 2 SO 3 + H 2 SO 4 \u003d Na 2 SO 4 + K 2 SO 4 + CR 2 (SO 4) 3 + H 2 O 12. CUS + HNO 3 \u003d CU (NO 3) 2 + H 2 SO 4 + NO 2 + H 2 O 13. CUS + HNO 3 \u003d CU (NO 3) 2 + S + NO + H 2 O 14. KMNO 4 + H 2 O 2 + H 2 SO 4 \u003d O 2 + MNSO 4 + K 2 SO 4 + H 2 O 15. Si + K 2 Cr 2 O 7 + H 2 SO 4 \u003d CR 2 (SO 4) 3 + H 2 SiO 3 + K 2 SO 4 + H 2 O 16. KMNO 4 + KJ + H 2 SO 4 \u003d MNSO 4 + K 2 SO 4 + J 2 + H 2 O 17. AU + H 2 SEO 4 \u003d AU 2 (SEO 4) 3 + SEO 2 + H 2 O 18. K 2 Cr 2 O 7 + H 2 S + H 2 SO 4 \u003d Cr 2 (SO 4) 3 + K 2 SO4 + S + H 2 O 19. MNO + PBO 2 + HNO 3 \u003d HMNO 4 + PB (NO 3) 2 + H 2 O 20. P + HNO 3 + H 2 O \u003d H 3 PO 4 + NO 21. N 2 H 4 + KMNO 4 + H 2 SO 4 \u003d N 2 + K 2 SO 4 + MNSO 4 + H 2 O 22. K 2 CR 2 O 7 + HCl \u003d KCl + CrCl 3 + Cl 2 + H 2 O 23. NAH + K 2 CR 2 O 7 + H 2 SO 4 \u003d Na 2 SO 4 + Cr 2 (SO 4) 3 + K 2 SO 4 + H 2 + H 2 O 24. Naaso 2 + J 2 + NaOH \u003d Na 3 ASO 4 + Naj + H 2 O 25. Nabio 3 + MNSO 4 + H 2 SO 4 \u003d HmnO4 + Bi 2 (SO 4) 3 + Na 2 SO 4 + H 2 O 26. NA 2 FeO 4 + MNSO 4 + H 2 SO 4 \u003d HMNO 4 + Fe 2 (SO 4) 3 + Na 2 SO 4 + H 2 O 27. Zn + H 3 ASO 3 + HCl \u003d ZnCl 2 + ASH 3 + H 2 O 28. MNO 2 + KNO 2 + H 2 SO 4 \u003d MNSO 4 + KNO 3 + H 2 O 29. KJO 3 + KJ + H 2 SO 4 \u003d J 2 + K 2 SO 4 + H 2 O 30. Br 2 + Cl 2 + KOH \u003d KCL + KBRO 3 + H 2 O Evhr equilibrium constant helps to judge not only about the direction, but also about the depth of the process. For any OSR, the equilibrium constant can be calculated if the oxidative reductive potentials of oxidation semi-formations and recovery are known: LGK \u003d (E 0 0 OX E Red) 0.059 N, where to the equilibrium constant of the oxidative reducing reaction, E 0 (OX) and E 0 (RED ) Normal oxidizer and reducing agent potentials, N Number of ions participating in oxidation semicrees or recovery (equivalent number). Knowing a constant equilibrium, you can calculate the completeness of the reaction flow, without resorting to the experiment. Suppose it is necessary to calculate the reaction depth: Sn + Pb (CH 3 Coo) 2 Pb + Sn (CH 3 COO) 2. Find in the directory the values \u200b\u200bof the standard semi-precious potentials: E 0 (Pb / Pb 2+) \u003d 0.126 B; E 0 (Sn / Sn 2+) \u003d 0.136 B [0.126 (0.136)] 2 0.059 LGK \u003d 0, 339 k \u003d / \u003d 10 0.339 \u003d 2.2

8 This means that the equilibrium in the system under consideration will come when the concentration of lead ions in the solution will be 2.2 times less than the concentration of tin ions. That is, 1 mol lead ions should have 2.2 mol tin ions. Consequently, the reaction proceeds reversible. (SN 2) 2.2 100 (2.2 1) 69% Exercise 13. Calculate equilibrium constant for reaction: 5FeCl 2 + KMNO 4 + 4H 2 SO 4 \u003d Fe 2 (SO 4) 3 + 3FeCl 3 + MNSO 4 + KCL + 4H 2 O, if the standard half-absorption potentials are: E 0 (MnO4 + 8H +) / (Mn H 2 O) \u003d 1.52 b; E 0 (Fe 2+ / Fe 3+) \u003d 0.77 B. The value of the reaction has a large influence on the value of the equilibrium constant. There is a rule to create the reaction medium necessary for the optimal flow of the process: if the hydrogen cations are accumulated as a result of the OSR, it is created by an alkaline medium, and if the anions are hydroxyl acid. Exercise 14. In which direction is the equilibrium, with an increase in the pH of the solution: Na 2 SO 3 + Br 2 + H 2 O Na 2 SO HBr 3 K 2 MNO H 2 O MNO 2 2 2KMNO KOH? The equilibrium constant allows you to predict the possibility of dissolving the substance. Consider whether the dissolution of copper sulfide is possible in nitric acid? 3CUS + 2 HNO HNO 3 3 S + 3 CU (NO 3) NO + 4 H 2 O 3CUS + 2NO H + 3 S + 3 CU NO + 4H 2 OE 0 1 (NO 3 + 4H + / NO + 2H 2 O ) \u003d 0.96 B Calculate the oxidizing agent reaction potential (E 0 2) CUS CU 2+ + S, S 2E S. E 0 2 \u003d E 0 (S / S) + 0.059 / 2 LG 1 /, where 0 (S / S) \u003d 0.51V at \u003d 1 mol / l, the concentration over the precipitate of copper sulfide can be calculated from the size of the solubility. PR (CUS) \u003d 3, therefore, \u003d PR (CUS) / \u003d 3, then E 0 2 \u003d 0.51 + 0.059 / 2 LG 1/3, \u003d +0.63 in the number of ions participating in an oxidizing agent reaction equal to 6 , therefore, LG K \u003d (0.96 0.63) 6 / 0.059 \u003d 33 and K \u003d 33, that is, CUS is well soluble in nitric acid. The dependence of oxidizing plant potentials on the pH of the medium can be calculated by the formula of Nernst, taking into account the concentration of hydrogen ions:

9 E E 0 0.059 [OX] [H] LG N M, where m coefficient at a concentration of hydrogen ions in the half-reaction equation, for example: MnO 4 + 5e + 8 H + Mn 2+ 4 H 2 O; _, 059 [MnO4] [H] E E (MnO4 / Mn) LG. 2 5 [Mn] Changing the concentration of hydrogen ions, can be corrected (reduce or increase) oxidizing agent potential. This makes it possible to use one or another oxidizing agent and selectively. Exercise 15. Calculate the oxidizing agent potential of the SO 4 / SO 3 system, if the solution contains 0.001 mol / l of SO 4 ions, 0.05 mol / l of SO 3 ions, 2.9 mol / l of hydrogen ions, and the standard oxidizing agent of the H 2 system potential SO 3 + H 2 O SO H + is equal to 0.20 V. The equilibrium of most oxidizing agent reactions can be displaced by changing the pH of the medium. This is especially true of reactions that have the difference in the potentials of the oxidizing agent and the reducing agent is small. Consider, for example, is it possible to interact in chloridion with dichromation in an acidic environment? To do this, we will form ion-electronic equations of semoretakes: CR 2 O H + + 6E 2 CR H 2 O; E 0 \u003d 1.33 B; Cl 2 + 2e 2 Cl; E 0 \u003d 1.36 B. Since the potential of the second half-reaction is higher than the first, under standard conditions, the reaction does not go. However, if a single-concentrated solvent is added to a single-axis solution of potassium dichromate, a solution of hydrochloric acid, a chlorine isolation reaction begins :, 059 [CR2O7] [H] E (CR2O7 [/ 2CR) E (CR2O7 / 2 CR) LG [CR ] At the concentrations of dichromation 1 mol / l, and the concentration of hydrochloric acid 3 mol / l We obtain the following potential value :, 059 3 E (CR 2O7 / CR) 1.33 LG 1.39 V. 6 1 Thus, increasing the concentration of hydrogen ions It was possible to react in the right direction. Find what is the concentration of hydrogen ions in the CR 2 O H + + 6E 2 CR H 2 O system, if the oxidizing agent potential is 1.33, and the concentrations of dichromate and chromium (3) ions are equal, respectively, 1 and 10 6 mol / l?

10 Verification Tasks 1. Group independent work 1. Elementary ions are given: F, H +, H, CU +, CU 2+, Fe 2+, S, S 2, SN 2+, MG 2+, Mn 2+, Cl . Which of them are able to exercise: a) only the function of the oxidant; b) only the function of the reducing agent; c) dual function? 2. Danations: NO 2, HNO 3, SO 2, H 2 S 2 O 7, MNO 2, HBRO, CL 2 O 7, CRO 3, K 2 MNO 4, H 2 SO 5, H 2 O 2, NH 3 , N 2 H 4, H 2, HI. Which of them are able to exercise: a) only the function of the oxidant; b) only the function of the reducing agent; c) dual function? 3. DAYSEDIONIONES: SO 4, NO 3, NO 2, NH 4 +, NO 2, CLO, CLO 4, MNO 4, MNO 4, ALH 4. Which of them are able to exercise: a) only the function of the oxidant; b) only the function of the reducing agent; c) dual function? 4. Compounds are given: KCLO 2, HCl, HNO 2, KNO 3, H 2 S, CLO 2, H 3 PO 3, H 3 PO 4, MNO 2, BR 2. Which of them are capable of disproportionation reactions? 5. Done incomplete half-share schemes: * MNO 4 .. \u003d MNO 4; * CR 2 O H + \u003d 2Cr H 2 O; * 2no H + \u003d N 2 O + 5 H 2 O; * Pt 0 + .. \u003d Pt +4: * H 2 O OH \u003d O H 2 O, * SO OH \u003d SO 4 + H 2 O. What process, oxidation or recovery reflects each diagram? Specify the number of sent or received in each electron scheme. 6. Dana Schemes of semoretosis. Determine the type of process: oxidation or recovery. Extract the reaction schemes if the process proceeds in an acidic environment: * NO 3 NO; * SO 4 H 2 S; * MNO 2 MNO 4; * CR 3+ CR 2 O Using the half-formation method, add the right parts of the equations of redox processes: * KMNO 4 + K 2 S + H 2 SO 4 K 2 SO 4 + * KMNO 4 + K 2 S + H 2 OK 2 SO 4 + * KMNO 4 + K 2 S + KOH K 2 SO 4 + * K 2 Cr 2 O 7 + Ki + H 2 SO 4 I 2 + CR 2 (SO 4) 3 +

11 3. Testbacks 1. The oxidation of the atom oxidation in the compound is a) the number of its valence electrons, b) the conditional charge provided that all connections are ionic. C) the number of electrons missing until the external layer is completed. D) the number of electronic pairs binding an atom with neighboring atoms. 2. Which of the data of elementary ions is able to show only the oxidizing agent function? A) N +, b) n, c) i, d) Cu + 3. Which of the elementary ions is able to show only the reducing agent function? A) Ca 2+, b) Fe 2+, c) H +, D) AU 4. Which of the data of complex ions is able to show only the oxidizing agent function? A) Cro 4, b) NH + 4. C) ALH 4, D) S 2 O 3 5. Which of these complex ions is able to show only the reducing agent function? A) MNO 4, B) PO 3 4, B) 4, D) SiO 4 6. Which compounds have a dual function? A) H 4 P 2 O 7, b) NH 4 NO 3, C) Na 2 Cr 2 O 7, D) KCLO 4 7. Which of the above compounds is capable of reaction of disproportionation? A) KCLO 4, B) BR 2, C) KMNO 4, D) NH 3 8. In which compound chlorine shows the degree of oxidation +1? A) CL 2 O, B) CH 3 CL, C) CaCl 2, D) SOCL 2 9. In what compound the degree of carbon oxidation is zero? A) CH 3 CH 2 OH, B) CH 3 COOH, C) (CH 3) 2 CO, D) CH 3 CH among these processes indicate the oxidative processes. A) H 2 O 2 H 2 O, B) MNO 4 MNO 4, C) NH + 4 NO 3, D) H 2 O 2 O among these processes, specify the rehabilitation processes. A) H 2 O 2 H 2 O, B) MNO 4 MNO 4, C) NH + 4 NO 3, D) H 2 O 2 O What schemes do not reflect the flow rate of the OSR? A) CR 2 O 7 + H 2 O 2Cro 4 + 2H +, b) zn + 2h + zn 2+ + h 2, c) CO 3 + H 2 O + CO 2 2HCO 3, D) Fe 2+ + NO 3 + 2H + Fe 3+ + NO 2 + H 2 O 13. Which of the processes relate to ASP? A) ozone formation during a thunderstorm, b) milk sinusia, c) firing pyrite (FES 2) in the production of sulfuric acid, d) sedimentation of suspended impurities when adding Al 2 (SO 4) to the wastewater (SO 4) in which environment prohibit According to the scheme: MNO 4 MNO 2? A) acid, b) alkaline, c) neutral, d) medium does not play a substantial role 15. In which environment, the process of restoring permanganate ion is proceeding according to the scheme: MNO 4 Mn 2+? A) acid, b) alkaline, c) neutral, d) medium does not play a significant role

12 16. In which environment, the process of recovery of permanganate ion is proceeding according to the scheme: MNO 4 MNO 4? A) acid, b) alkaline, c) neutral, d) medium does not play a substantial role 17. What substances can not be released in the interaction of dilute nitric acid with active metals? A) NO 2, B) H 2, C) N 2, D) NO 18. What function does hydrogen peroxide in the oxidatively reduced process performs if the reaction products are the molecular oxygen? A) oxidizing agent, b) reducing agent, c) of the reaction medium, d) of the solvent 19. What function does hydrogen peroxide in the oxidative regenerative process performs if the reaction products are water? A) solvent, b) reducing agent, c) of the reaction medium, d) oxidizing agent 20. What is the equivalence factor of the chemical particle in the process of formation? A) the smallest common to the number of given and received electrons, b) the value of the inverse number of removable electrons, c) of the inverse number of adopted electrons, d) the value inverse the smallest shared multiple for the number of given and received electrons. 21. What is the equivalence factor of the chemical particle in the recovery process? A) the smallest common to the number of given and received electrons, b) the value of the inverse number of removable electrons, c) of the inverse number of adopted electrons, d) the value inverse the smallest shared multiple for the number of given and received electrons. 22. What is the name of oxidative reaction reactions, during which the atoms of the same element are the oxidizing agent, and the reducing agent? A) self-sender self-examination reaction. B) dismutation reactions. C) intramolecular reactions. D) disproportionation reactions. 23. Which of the proposed transform schemes correspond to intramolecular oxidative reducing reactions? A) NH 4 NO 3 N 2 O + H 2 O b) Cl 2 + NaOH Na 2 CLO 3 + NaCl c) S + NaOH Na 2 SO 3 + Na 2 S g) (NH 4) 2 Cr 2 O 7 n 2 + Cr 2 O 3 + H 2 O 24. Rate the correctness of the following judgments: 1) 1) Hydrogen in non-metal compounds has a degree of oxidation +1 (except for boron and silicon), with metals, boron and silicon, the degree of hydrogen oxidation is equal to 1. 2) oxygen in oxides, as a rule, has a degree of oxidation 2. In peroxides, its degree of oxidation is 1 (H 2 O 2, Na 2 O 2), in a compound with fluorine (+2) of 2, in superoxides (1/2 ), in Ozonides (1/3). A) both judgments are true. B) Both judgments are incorrect. C) just the first judgment is true. D) only the second judgment is true.

13 25. Intending the correctness of the following judgments: 1) the conditional charge attributed to the atom when the assumption is assumed that all connections are built on ion type; 2) a charge that would arise on the atom if the electronic pairs, which it is associated with other atoms, would be shifted to a more electronegative atom. A) both judgments are true. B) Both judgments are incorrect. B) Tolcopher judgment is true. D) only the second judgment is true. 26. What rules must be observed in the selection of coefficients in oxidative reducing reactions? A) lever rule. B) the rule of constancy amount of charges. C) additivity rule. D) Electronic Balance Rule. 27. Which of the proposed transform schemes correspond to the oxidative reaction reactions of disproportionation? A) NH 4 NO 3 N 2 O + H 2 O b) Cl 2 + NaOH Na 2 CLO 3 + NaCl c) S + NaOH Na 2 SO 3 + Na 2 S g) (NH 4) 2 Cr 2 O 7 N 2 + Cr 2 O 3 + H 2 O 28. The degree of oxidation is +4 sulfur exhibits in compounds: a) MgSB) SO 2 V) K 2 SO 3 g) S which of the ions only exhibits reducing properties? A) JO 3 b) jo c) j2 g) 30. What is the erroneous reaction? A) H 2 SO 4 + 2AG AG 2 SO 4 + H 2 b) 2H 2 S + 4AG + O 2 2AGS + 2H 2 O c) 2H 2 SO 4 + 2AG AG 2 SO 4 + SO 2 + 2H 2 O g) 2AGNO 3 + K 2 SO 4 AG 2 SO 4 + 2KNO 3 Positioning of coefficients in the schemes of redox reactions occurring in aqueous solutions, ion-electronic method 31. KMNO 4 + NO + H 2 SO 4 \u003d HNO 3 + MNSO 4 + K 2 SO 4 + KNO 3 + H 2 O 32. PBO 2 + CR (NO 3) 3 + H 2 O \u003d PB (NO 3) 2 + H 2 Cr 2 O FESO 4 + K 2 Cr 2 O 7 + H 2 SO 4 \u003d CR 2 (SO 4) 3 + K 2 SO 4 + Fe 2 (SO 4) 3 + H 2 O 34. KNO 2 + K 2 CRO 4 + KOH + H 2 O \u003d KNO 3 + K 3 35. KMNO 4 + P + H 2 SO 4 \u003d H 3 PO 4 + K 2 SO 4 + MNSO KCLO 3 + KJ + H 2 SO 4 \u003d KCL + J 2 + K 2 SO KMNO 4 + SO 2 + H 2 O \u003d H 2 SO 4 + MNSO 4 + K 2 SO KJ + H 2 SO 4 \u003d J 2 + KHSO 4 + H 2 S + H 2 O 39. KMNO 4 + K 2 SO 3 + H 2 SO 4 \u003d K 2 SO 4 + MNSO 4 + H 2 O 40. BI 2 (SO 4) 3 + Cl 2 + NaOH \u003d Nabio 3 + NaCl + H 2 O 41. K 2 CR 2 O 7 + Na 2 SO 3 + H 2 SO 4 \u003d Na 2 SO 4 + K 2 SO 4 + CR 2 (SO 4) 3 + H 2 O 42. CUS + HNO 3 \u003d CU (NO 3) 2 + H 2 SO 4 + NO 2 + H 2 O 43 . CUS + HNO 3 \u003d CU (NO 3) 2 + S + NO + H 2 O

14 44. KMNO 4 + H 2 O 2 + H 2 SO 4 \u003d O 2 + MNSO 4 + K 2 SO 4 + H 2 O 45. Si + K 2 Cr 2 O 7 + H 2 SO 4 \u003d CR 2 (SO 4) 3 + H 2 SiO 3 + K 2 SO 4 + H 2 O 46. KMNO 4 + KJ + H 2 SO 4 \u003d MNSO 4 + K 2 SO 4 + J 2 + H 2 O 47. AU + H 2 SEO 4 \u003d Au 2 (SEO 4) 3 + SEO 2 + H 2 O 48. K 2 Cr 2 O 7 + H 2 S + H 2 SO 4 \u003d CR 2 (SO 4) 3 + K 2 SO4 + S + H 2 O 49. MNO + PBO 2 + HNO 3 \u003d HMNO 4 + PB (NO 3) 2 + H 2 O 50. P + HNO 3 + H 2 O \u003d H 3 PO 4 + NO 51. N 2 H 4 + KMNO 4 + H 2 SO 4 \u003d N 2 + K 2 SO 4 + MNSO 4 + H 2 O 52. K 2 Cr 2 O 7 + HCl \u003d KCl + CRCl 3 + Cl 2 + H 2 O 53. NAH + K 2 CR 2 O 7 + H 2 SO 4 \u003d Na 2 SO 4 + CR 2 (SO 4) 3 + K 2 SO 4 + H 2 + H 2 O 54. Naaso 2 + J 2 + NaOH \u003d Na 3 ASO 4 + Naj + H 2 O 55. Nabio 3 + MNSO 4 + H 2 SO 4 \u003d HmnO4 + Bi 2 (SO 4) 3 + Na 2 SO 4 + H 2 O 56. Na 2 Feo 4 + MNSO 4 + H 2 SO 4 \u003d HMNO 4 + Fe 2 (SO 4) 3 + Na 2 SO 4 + H 2 O 57. Zn + H 3 ASO 3 + HCl \u003d ZnCl 2 + ASH 3 + H 2 O 58. MNO 2 + KNO 2 + H 2 SO 4 \u003d MNSO 4 + KNO 3 + H 2 O 59. KJO 3 + KJ + H 2 SO 4 \u003d J 2 + K 2 SO 4 + H 2 O 60 . Br 2 + Cl 2 + Koh \u003d KCl + KBRO 3 + H 2 O

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Theme of the lesson: someone loses, and someone finds.

(Redox reactions)

The purpose of the lesson: Systematize, expand, deepen and update the knowledge of students on redox reactions based on interdimensional relations.

Tasks lesson

1) Training - expand and secure:

- knowledge of the degree of oxidation, oxidizers and reducing agents;

- skills solving cognitive tasks;

- the ability to apply the knowledge gained in everyday life.

2) Educational - to bring up a sense of responsible attitude to health; Careful attitude towards nature.

3) Developing - Development of skills and skills that contribute to an integrated approach to solving cognitive tasks.

Type of lesson: Lesson for the formation and improvement of knowledge

Equipment:

Multimedia projector, computer, interactive SMART board. Solutions of substances: KMNO4, H2SO4, KOH, K2SO3 necessary for conducting a demonstration experiment (OPRin various environments)

DURING THE CLASSES

Organization of lesson(1 min)

Hello guys. Your performance and activity in the lesson will help us expand the boundaries of knowledge, and will make all of us more sensitive and attentive to nature around us.

Setting the goal and actualization of knowledge (2 min)

The world around us is great and diverse. Life surrounds us everywhere. It is evidenced by the buzz of insects, chirping birds, rustling of small animals. It exists like in ice polar zones (Slide number 1), So in hot deserts (Slide number 2),. We encounter it everywhere, starting with the sunken sea consecrated sun and ending with the highest depths of the ocean. Under our legs, the inconspicuous number of microorganisms work, making the soil fertile and suitable for plant growth (Slide number 3), Which, in turn, are needed by other forms of life. The Earth is saturated in such an abundance that it will shock our imagination. In short, life is the greatest value of our planet and its most unique distinctive feature. The world around us is a giant chemical laboratory in which thousands of reactions occur every second, mostly redox, and as long as they exist, these reactions, while there are conditions for their flow, possibly all the magnificence around us, life itself is possible.

The topic of our today's lesson:

(Slide number 4) Someone loses, and someone finds

(Oxidative - Recovery Reactions)

Nature is the best and objective teacher.

when solving the most difficult issuesscience.

V. V. Dokuchaev

III. Fastening the existing knowledge of students, expansion and formation of new ones.(10 min)

1. Blitz poll:

(If there is an interactive SMART Notebook interactive whiteboard in the office, the front conversation takes place; function: Question-answer, curtain, if not, then with the help of the presentation: the question and answer with the animation effect

1.1. What reactions are called redox?

OSR is reactions flowing with a change in the degree of oxidation of atoms that are part of the reacting substances.

1.2. Redox reaction is represented by the scheme

a) Zno + HCl \u003d ZnCl2 + H2O

b) C2H4 + BR2 \u003d C2H4BR2

c) NaHCO3 \u003d Na2CO3 + H2O + CO2

1.3. Give the degree of oxidation of the atom.

The degree of oxidation is the conditional charge of the atoms of the chemical element in a compound calculated on the basis of the assumption that all compounds (and ionic and covalently - polar) consist only of ions.

1.4. What is the basis of the electronic balance method?

This method is based on comparing the degree of oxidation of atoms in the initial substances and reaction products.

1.5. What are the atoms, molecules and ions, which give electrons?

Atoms, molecules or ions that give electrons are called reducing agents. During the reaction they are oxidized

1.6. What is the restoration properties depend on?

The lower the degree of oxidation of the element, the smaller it is electrothy, the stronger the rehabilitation properties.

1.7. What are the numbers, molecules and ions that attach electrons?

Atoms, molecules or ions connecting electrons are called oxidizing agents. During the reaction they are restored

1.8. What does oxidative properties depend on?

The higher the degree of oxidation of the element and its more electronegility, the stronger the oxidative properties.

2. Expanding knowledge of students:

Teacher: The manifestation of redox properties affects such a factor as the stability of a molecule or ion. The stronger the particle, to a lesser extent, it shows redox properties. For example, nitrogen electronenence is sufficiently high, but in its molecule a triple relationship (N = N), molecule is very stable, nitrogen is chemically passive. Or HCLO is a stronger oxidizer than HCLO4, since chlorothic acid is less resistant to the compound solution than the chlorine.

For substances having elements atoms in an intermediate degree of oxidation, there is a manifestation of both oxidative and restorative properties.

3. Fastening. Express survey

3.1. Restorers include:

AL, CL2, HBR, O3, KMNO4,

3.2. Oxidizers include

H2 SO4, O2, H2, MG, K2MNO4

3.3. And oxidative and reducing properties may have

CO, NANO3, HNO2, CU2O, H2SO3

3.4. Stronger oxidants are

a) hno3 or hno2?

b) S or SO2?

c) Cu, Cu2o or Cuo?

4. Individual knowledge control

4.1. Test control of knowledge available

	The redox reactions do not relate to the reaction, shown by the scheme: A) NO + O2 ® NO2 B) C2H4 + BR2 ® C2H4BR2 C) Caco3 ®CAO + CO2 D) KNO3 ® KNO2 + O2	Redox reactions include the reaction, shown by the scheme: A) H2O + CAO ® CA (OH) 2 B) H2O + P2O5 ® HPO3 C) NaHCO3 ® Na2CO3 + H2O + CO2 D) CH4 + CL2 ®ch3Cl + HCl
	The degree of oxidation (-3) has nitrogen in the connection: A) hno2 b) hno3 C) nh4no3 g) nano2	The degree of oxidation (-1) has a sulfur in connection: A) fes b) feso4
	The oxidation process takes place in the case when:	The recovery process takes place in the case when: A) neutral atoms turn into negatively charged ions B) neutral atoms turn into positively charged ions C) positive charge ion increases D) negative ion charge increases
	Only oxidative properties exhibits: A) s0 b) ci + 7 c) n20 g) n + 3	And oxidative and restorative properties show: A) Mn + 7 b) Cu0 c) Cu + 2 g) Cu + 1
	Recoveful properties of simple substances formed by the elements of the second period from left to right: A) decrease B) enhanced C) change periodically D) do not change	The oxidative properties of simple substances formed by the elements of the seventh group, the main subgroup with increasing the charge of the kernel: A) decrease B) enhanced C) change periodically D) do not change

4.2. Differentiated pupils survey at the board (the remaining students at this time work independently)

KMNO4 + KOH + K2SO3 \u003d K2MNO4 + K2SO4 + H2O

KMNO4 + H2SO4 + K2SO3 \u003d MNSO4 + K2SO4 + H2O

KMNO4 \u003d K2MNO4 + MNO2 + O2

SO2 + H2S \u003d S + H2O

C4N10 + O2 \u003d CH3SOON + H2O

4.3. Expansion of students' knowledge

Features of redox reactions in organic

You can use the average arithmetic degrees of oxidation of hydrogen atoms

C410 / 4N10 + 1 + O20 \u003d C20H4 + 1O2-2 + H2 + 1O-2

4C10 / 4 -10 E \u003d 4C0 2 - Restore

O20 +4 E \u003d 2O-2 5 - Oxidizer

5. Introduction of new knowledge:

5.1. Classification of redox reactions.

Students (with the help of the teacher) determine the type of those OVR, the equations of the reactions of which are written on the board and in the notebooks of students.

There are two types of oxidative reaction reactions:

Intermolecular oh

The intermolecular OSR includes reactions in which the oxidizing agent and the reducing agent are in different substances.

Question: Determine the intermolecular oh?

Intramolecular oh

Intramolecular ASPs include reactions in which the oxidizing agent and the reducing agent are in one substance.

Question: Determine intramolecular oh?

In a separate group, the reactions of disproportionation and comparison can be distinguished

Disproportionation reactions include intramolecular oh, the flow of which is accompanied by simultaneously increasing and decrease in the degree of oxidation of atoms of the same element (compared to the initial degree of oxidation).

Question: What kind of reactions can be attributed to disproportionation reactions?

Comporting reactions include intermolecular AUROs, in which the oxidizing agent and the reducing agent is the atom of the same chemical element, but in different degrees of oxidation.

Question: Which of the reactions can be attributed to the reactions of comparison?

5.2. Conditions of the OSR, depending on the reaction of the medium.

Demonstration experiment "The influence of the environment on the conditions of the flow

Potassium permanganate solution in 2 flasks. Add sulfuric acid to the first solution, in the second - alkali, then add potassium sulfite to each solution and mix.

Question: Look carefully on the reaction equations and determine which flask is alkaline, and in which acid medium?

5.3. Transferring knowledge available in a non-standard situation.

Problem Situation:

Which of the shown schemes reflects the actual chemical phenomenon, and what is the erroneous?

(if there is an interactive whiteboard in the office, the problem situation is permitted with its help)

HCLO3 \u003d HCLO2 + HCl

HCLO3 \u003d HCLO4 + HCl

Conclusion: The second HSR is possible, as there is an oxidizing agent and a reducing agent.

As the scale of chemical production is expanded, unfortunately, the number of accidents associated with the emissions of harmful substances is growing. Phenol can get into the water - causing poisoning - vomiting and pain in the opposite region. Offer an effective way to clean water from phenol.

The newspaper "Izvestia" there was a note "Wedding nuts" - instructive from a chemical point of view. The author writes: "Berylliye bronze rings is an exact copy of the gold. They do not differ from the latter neither in color, nor by weight and, suspended on the thread, when they hit the glass, they make a characteristic, sincerely - melodious sound. In short, the fake will not detect either on the eye, neither taste or tooth. " Offer ways with which you can distinguish a fake?

Vi. VSP

Everyone knows the seven wonders of the world. Redox reactions underlie "seven wonders of alive and inanimate nature"

Photosynthesis, breath, rotting,fermentation, corrosion, electrolysis, burning.

Conversation with students about the meaning of photosynthesis and his role in nature

The first cry of the child, creates the first breath, the beginning of a new life. Breathing is characteristic of most living organisms, it is simply inseparable from life.

Due to the processes of rotting, the volumes of substances are carried out in nature. Pinged bacteria, translating the organic substance into an inorganic, as if the cycle of life begin.

The fermentation can be carried out under the action of yeast, the meaning of which everyone knows, it is enough to dwell on the bread maker ...

Everyone knows about the harmful effect of corrosion, but it is impossible to underestimate its meaning, I will dwell only on one fact. With deep antiquity there is a method of converting iron to steel, through rusting. Circassians in the Caucasus have buried strip iron into the ground, and, doubting it in 10-15 years, they shot their sabers from it, which could even reflect the rifle barrel, the shield of the enemy. After digging, the rusty iron together with the organic substances was heated in the mountains, and then cooled with water - hardened.

Slide number 10.

The gilding of objects is known for a long time, since the gilded products are very beautiful. Before the electrolysis and galvanotechnics were not invented, the products made of metals were golden: they were applied to the tough amalgam of gold (alloy it with mercury); Then he glue hot; At the same time, mercury evaporated, and the gold remained. But mercury pairs are very poisonous, so, for example, with the gilding of the domes of St. Isaac's Cathedral in St. Petersburg, 60 workers died from mercury poisoning.

Slides number 11, 12, 13

It was very difficult for our ancestors, those who were responsible for the safety and maintenance of the whole tribe, and those who only have learned to mine. With fire, a lot is connected: it is the warmth of a native focus, soothing the flame of the candle, cooking food, the song by the fire ... But it is impossible to joke with fire, it is necessary to carefully and carefully handle it, because its force is not only creating, but also the destructive, capable To destroy all living things.

VII. Definition and explanationhome tasks

1. The value of the OSR in the nature and life of a person (creative mini-messages of students).

2. Differentiated task:

Estip the coefficients by the method of electronic balance, determine the oxidizing agent and reducing agent in the following schemes

Level 1:

Zn + H2SO4 (CON) \u003d ZNSO4 + H2O + S

2 level

KmnO4 + HCl \u003d CL2 + KCL + MNCL2 +?

3 level

FESO4 + KMNO4 + H2SO4 \u003d FE2 (SO4) 3 +? +? +?

C2H4 + KMNO4 + H2O \u003d C2H6O2 + MNO2 +?

Working lessons (abstract lessons)

Basic general education

Line Ukk O. S. Gabrielyan. Chemistry (8-9)

Attention! The site administration site is not responsible for the maintenance of methodological developments, as well as for the compliance of the development of GEF.

References:

1. Desktop book of chemistry teacher. 8th grade. O.S. Gabrielyan, N.P.Vomkocynikova, A.V. Yashukova (M.: Drop). 2003
2. Efu Chemistry grade 8. O.S. Gabrielyan, (M.: Drop).
3. Workbook to the textbook O.S. Gabrielyan Chemistry Grade 8. O.S. Gabrielyan, A.S. Sweets (M.: Drop-2013).

Objectives lesson:

• training: introduce students with a new classification of chemical reactions on the basis of changes in the degrees of oxidizes of elements - redox reactions, repeat the concepts of "oxidizing", "reducing agent", "oxidation", "restoration";
• developing: Continue the development of logical thinking, the formation of interest in the subject, using modern technologies in learning.
• educational: To form a scientific worldview of students, the formation of a culture of interpersonal communication: to evaluate its work ..

Means of education:

• Electronic application to the textbook "Chemistry 8th grade". O.S. Gabrielyan, (M.: Drop).
• Interactive textbook "Visual Chemistry. Chemistry. 8-9 class. " Moscow: Exam Media LLC 2011-2013

Tutorial: Ef. Gabrielyan O.S. Chemistry. Grade 8: - M.: Drop, 2015

During the classes

1. Organizational stage

Preparation of students to work in the lesson. Rules of work and TB in a smart class when working with laptops

2. Actualization of knowledge of students

BUT) Recall all the classification of chemical reactions and signs that are based on each classification. Reiteration. "Types of chemical reactions" (according to the training center 2)

Work on literature 1:

1. According to the type and composition of reacting and formed substances, there are reactions:

a) compounds;
b) decomposition;
c) substitution;
d) exchange (including the neutralization reaction).

2. According to the aggregate state of substances (phase) distinguish reactions:

a) homogeneous;
b) heterogeneous.

3. The thermal effect of the reaction is divided into:

a) exothermic (including combustion reactions);
b) endothermic.

4. On the use of the catalyst, reactions are isolated:

a) catalytic (including enzymatic);
b) noncatalithic.

5. In the direction distinguish reactions:

a) reversible;
b) irreversible.

B) Perform a complete characteristic of the sulfur oxide synthesis reaction (6) from sulfur oxide (4) and oxygen:

3. Assimplement of new Effound Knowledge

BUT) Recall what S.O. And how it changes with XP. (Repetition followed by testing for learning 2.)

B) explanation of the material on Efu p. 263-265.

IN)Work on electronic EFC application.

D) work on literature 2

4. Primary consolidation of knowledge

BUT)Students perform the task. Electronic application

With difficulty using page 264-265 ef.

B) Performing a task by electronic application, finding the oxidizing agent, reducing agent, transition of electrons, work at the board.

Topic: "Redox reactions".

Objectives lesson:

Consider the essence of redox processes, to teach the use of "degrees of oxidation" to determine the processes of oxidation and recovery.

teach students equalizing recording redox reactions by electronic balance.

Improve the ability to express a judgment on the type of chemical reaction, analyzing the degree of oxidation of atoms in substances;

draw conclusions, work with algorithms, form interest in the subject.

Form scientific worldview of students; improve labor skills;

to teach listening to the teacher and your classmates, be attentive to yourself and others, evaluate yourself and others, to talk.

Equipment and reagents: salonic acid, sulfuric acid, zinc in granules, magnesium chips, copper sulfate solution, iron nail.

During the classes

Organizing time.

The concept of oxidative reducing reactions

Let's remember the classification of chemical reactions that you know.

In terms of the number and composition of reagents and products,

On thermal effect

Towards,

Catalyst participation.

There is another classification based on changing or preserving degrees of oxidation of atoms of chemical elements forming reagents and reaction products. On this basis distinguish reactions

Chemical reactions

Reactions flowing with a change in the reaction flowing without changing the degree

degree of oxidation of elements, oxidation

the substances involved in the reaction (OVR)

The teacher asks students to remember

What is called the degree of oxidation (C.O.) and how is it calculated by the formulas of the compounds?

Degree of oxidation the conditional charge of atoms in the chemical compound calculated on the basis of the assumption is that this compound consists of simple ions.

The degree of oxygen oxidation is almost always equal to -2.

The degree of hydrogen oxidation is almost always equal to +1.

The degree of metal oxidation is always positive and in the maximum value is almost always equal to the number of the group.

The degree of oxidation of free atoms and atoms in simple substances is always equal to 0.

The total degree of oxidation of atoms of all elements in the compound is 0.

Here the teacher offers students verbally to calculate - to find the degree of oxidation of the elements.

What will be the degree of oxidation of sulfur and phosphorus

In molecules, the algebraic sum of the degrees of the oxidation of elements, taking into account the number of their atoms equal to 0.

H. 2 +1 S. x. O. 4 -2 N. 3 RO 4

(+1) . 2 + x + (-2) . 4 = 0

X \u003d +6.

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

What types of chemical reactions do you know?

Students answer.

The OSR includes all reactions of substitution, as well as those reactions of compound and decomposition in which participatesat least one simple substance .

Give the definition of the OSR.

OSR is the reaction, during which the degrees of oxidation change.

And one more definition. " Chemical reactions that proceed with changes in the degrees of oxidation of atoms in the molecules of reactant substances are called Oxidative and recovery. "

Why are these reactions so called?

Consider examples of such chemical reactions.

As examples of the ASP, the teacher demonstrates the next experience.

H. 2 SO.4 + MG. = MgSO. 4 + H. 2

Denote the degree of oxidation of all elements in the formulas of substances - reagents and products of this reaction:

As can be seen from the reaction equation, the atoms of two elements of magnesium and hydrogen changed their oxidation degrees.

What happened to them?

Magnesium from the neutral atom turned into a conditional ion to the degree of oxidation +2, that is, I gave 2ee:

MG. 0 - 2E. MG. +2

Write down in your abstract:

Elements or substances that give electrons are calledreducing agents; During the reaction, theyoxidize .

The conditional ion H in the degree of oxidation is +1 turned into a neutral atom, that is, each hydrogen atom received one electron.

2N +1 + 2EN 2

Elements or substances taking electrons are calledoxidifiers ; During the reaction, theyrestore .

These processes can be represented as a scheme:

Salonic acid + magnesium magnesium sulfate + hydrogen

Cuso. 4 + FE. (iron nail) \u003dFE.SO. 4 + Cu. (Beautiful red nail)

FE. 0 – 2 e. FE. +2

Cu. +2 +2 e. Cu. 0

Someone loses, and someone finds ...

Electron recoil process is calledoxidation , and adoption - restoration.

In the process of oxidation, the degree of oxidationrises , in the recovery process -falls.

These processes are inextricably linked.

3. Method of electronic balance as a method for compiling ASP equations

Next, consider the compilation of equations of redox reactions by the electronic balance. The coefficients are then placed. The basis of the e-balance method is the rule:the total number of electrons that gives the reducing agent is always equal to the total number of electrons that the oxidizing agent attachs.

After explanation, students under the guidance of the teacher constitute the ORP equations according to the plans that the teacher compiled to this lesson (see Appendix). Plans are at each student on the desk.

Teacher: Among the reactions studied by us are redox reactions:

Interaction metals with non-metals .

2 MG. + O. 2 =2 MGO.

Best-L. MG. 0 -2 e. --- MG. +2 2 oxidation

OK Isl- O. 2 +4 e. ---2 O. -2 1 recovery

2. Interaction metals with acid.

H. 2 SO.4 + MG. = MgSO. 4 + H. 2

Best-L. MG. 0 -2 e. --- MG. +2 2 oxidation

OK Isl-F O. -2 +4 e. --- O. 2 0 1 recovery

3. Interactionmetals with salt.

Cu. SO.4 + MG. = MgSO. 4 + Cu.

Best-L. MG. 0 -2 e. --- MG. +2 2 oxidation

OK Isl- Cu. +2 +2 e. --- Cu. 0 1 recovery

The reaction is dictated, one student independently makes up the reaction scheme at the board:

H. 2 + O. 2 → H. 2 O.

We define the atoms of which elements change the degree of oxidation.

(H.2 ° +. O.2 ° → H.2 O. 2 ).

Make electronic equations of oxidation and recovery processes.

(H.2 ° -2. e. → 2 H. + - the process of oxidation,

O.2 ° +4. e. → 2 O. - ² - recovery process,

H 2 - reducing agent, o2 - Oxidizer)

We will select the general divider for the given and adopted E and coefficients for electronic equations.

(∙ 2 | H 2 ° -2E → 2N + - process of oxidation, element - reducing agent;

∙1| O.2 ° +4. e. → 2 O. - ² - Recovery process, element - oxidizing agent).

We transfer these coefficients to the OSR equation and select the coefficients in front of the formulas of other substances.

2 H. 2 + O. 2 → 2 H. 2 O. .

The plan for compiling the equations of the OSR

and electronic balance to them

1. Record the reaction scheme.

2. Determine the atoms of which elements change the degree of oxidation.

3. Make the electronic equations of oxidation and recovery processes.

4. Choose a common divide for given and received electrons and coefficients for electronic equations.

5. To transfer these coefficients to the OSR equation and select coefficients in front of the formulas of other substances.

We conclude : "What is the essence of the OSR?"

Redox reactions are the unity of two opposite processes: oxidation and recovery. In these reactions, the number of electrons given by the reducing agents is equal to the number of electrons connected by oxidizing agents. At the same time, no matter whether the electrons are transmitted from one atom to another completely or only partially, they are delayed to one of the atoms, it is conventionally indicated on the return or addition of electrons.

Homework.

§ 43, UPR.1, 3, 7