2 Chemistry lesson in 8th grade on the topic “Oxidation-reduction reactions”
Annotation: A chemistry lesson on the topic “Oxidation-reduction reactions” is intended for 8th grade students. The lesson reveals the basic concepts of redox reactions: oxidation state, oxidizing agent, reducing agent, oxidation, reduction: the ability to compile redox records using the electronic balance method is developed.
Chemistry lesson in 8th grade on the topic
"Oxidation-reduction reactions"
THE PURPOSE OF THE LESSON: to form a system of knowledge about redox reactions, to teach how to make records of ORR using the electronic balance method.
LESSON OBJECTIVES:
Educational: consider the essence of redox processes, teach how to use “degrees of oxidation” to determine the processes of oxidation and reduction; teach students to equalize records of redox reactions using the electronic balance method.
Developmental: Improve the ability to make judgments about the type of chemical reaction by analyzing the degree of oxidation of atoms in substances; draw conclusions, work with algorithms, develop interest in the subject.
Educating: create a need for cognitive activity and value attitude towards knowledge; analyze the answers of your comrades, predict the result of work, evaluate your work; to cultivate a culture of communication through work in pairs “student-student”, “teacher-student”.
Lesson type: A lesson in learning new material.
Methods used in the lesson: Explanatory or lustrative.
Concepts introduced in the lesson: redox reactions; oxidizer; reducing agent; oxidation process; recovery process.
Equipment usedand reagents: solubility table, periodic table of D.I. Mendeleev, hydrochloric acid, sulfuric acid, zinc granules, magnesium shavings, copper sulfate solution, iron nail.
Form of work: individual, frontal.
Lesson time: (90 minutes, 2 lessons).
During the classes
I . Organizing time
II . Repetition of covered material
TEACHER: Guys, let's remember the previously studied material about the degree of oxidation, which we will need in the lesson.
Oral frontal survey:
What is electronegativity?
What is oxidation state?
Can the oxidation number of an element be zero? In which cases?
What oxidation state does oxygen most often exhibit in compounds?
Remember the exceptions.
What oxidation state do metals exhibit in polar and ionic compounds?
How is the oxidation state calculated using compound formulas?
The oxidation state of oxygen is almost always -2.
The oxidation state of hydrogen is almost always +1.
The oxidation state of metals is always positive and at its maximum value is almost always equal to the group number.
Oxidation state of free atoms and atoms in simple substances ah is always 0.
The total oxidation state of the atoms of all elements in a compound is 0.
TEACHER In order to consolidate the formulated rules, he invites students to calculate - to find the oxidation state of elements in simple substances and compounds:
S, H 2, H 3 PO 4, NaHSO 3, HNO 3, Cu(NO 2) 2, NO 2, Ba, Al.
For example: What will be the oxidation state of sulfur in sulfuric acid?
In molecules, the algebraic sum of the oxidation states of elements, taking into account the number of their atoms, is equal to 0.
H 2 +1 S x O 4 -2
(+1) * 2 +X *1 + (-2) . 4 = 0
X = + 6
H 2 +1 S +6 O 4 -2
III . Learning new material
TEACHER: Variety of classifications chemical reactions By various signs(direction, number and composition of reacting and forming substances, use of catalyst, thermal effect) can be supplemented with one more feature. This is a sign - a change in the oxidation state of atoms chemical elements, forming reacting substances.
On this basis, reactions are distinguished
Chemical reactions
Reactions that occur with a change in reactions that occur without a change in the oxidation state of elements. oxidation states of elements.
For example, in the reaction
1 +5 -2 +1 -1 +1 -1 +1 +5 -2
AgNO 3 + HCl AgCl + HNO 3 (student writes at the board)
The oxidation states of atoms of chemical elements did not change after the reaction. But in another reaction - the interaction of hydrochloric acid with zinc
2HCl + Zn ZnCl 2 + H 2 (student writes at the board)
the atoms of two elements, hydrogen and zinc, changed their oxidation states: hydrogen from +1 to 0, and zinc from 0 to +2. Therefore, in this reaction, each hydrogen atom received one electron
2H + 2eH2
and each zinc atom gave up two electrons
Zn - 2е Zn
TEACHER: What types of chemical reactions do you know?
WARNING: ORR includes all substitution reactions, as well as those compounding and decomposition reactions in which at least one simple substance.
TEACHER: Define OVR.
Chemical reactions that result in a change in the oxidation states of atoms of chemical elements or ions forming reacting substances are called redox reactions.
TEACHER: Guys, determine orally which of the proposed reactions is not redox:
1) 2Na + Cl 2 = 2NaCl
2) Na CL + AgNO 3 = NaNO 3 +AgCl↓
3) Zn + 2HCl = ZnCl 2 + H 2
4) S + O 2 = SO 2
STUDENTS: complete the task
TEACHER: As examples of OVR, we will demonstrate the following experience.
H 2 SO 4 + Mg MgSO 4 + H 2
Let us denote the oxidation state 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, magnesium and hydrogen, changed their oxidation states.
What happened to them?
Magnesium from a neutral atom turned into a conditional ion in the oxidation state +2, that is, it gave up 2e:
Mg 0 – 2е Mg +2
Write down in your notes:
Elements or substances that donate electrons are called reducing agents; during the reaction they oxidize.
The conditional H ion in the +1 oxidation state turned into a neutral atom, that is, each hydrogen atom received one electron.
2Н +1 +2е Н 2
Elements or substances that accept electrons are called oxidizing agents; during the reaction they are recovering.<Приложение 1>
These processes can be represented as a diagram:
Hydrochloric acid + magnesium magnesium sulfate + hydrogen
CuSO 4 + Fe (iron nail) = Fe SO 4 + Cu (nice red nail)
Fe 0 – 2 eFe +2
Cu +2 +2 eCu 0
The process of giving up electrons is called oxidation, and acceptance – restoration.
During the oxidation process, the oxidation state rises, in the process of recovery – goes down.
These processes are inextricably linked.
TEACHER: Let's complete the task according to the example described above.
Exercise: For redox reactions, indicate the oxidizing agent and the reducing agent, the processes of oxidation and reduction, and create electronic equations:
1) BaO + SO 2 = BaSO 3
2) CuCl 2 + Fe = FeCl 2 + Cu
3) Li + O 2 = Li 2 O 3
4) CuSO 4 + 2KOH = Cu(OH) 2 ↓ + K 2 SO 4
II part of the lesson (2nd lesson)
Electronic balance method as a way to compile OVR equations
Next, we will consider compiling equations for redox reactions using the electronic balance method. The electron balance method is based on the rule: the total number of electrons that the reducing agent gives up is always equal to total number electrons that the oxidizing agent gains.
After the explanation, students, under the guidance of the teacher, compose OVR equations according to the plans that the teacher made for this lesson <Приложение 2>.
Reminders are located on each student's desk.
TEACHER: Among the reactions we studied, redox reactions include:
Interaction metals with non-metals
2Mg + O 2 =2MgO
Oxidizing agent O 2 +4e 2O -2 1 reduction
2. Interaction metals with acid.
H 2 SO 4 + Mg = MgSO 4 + H 2
Reductant Mg 0 -2e Mg +2 2 oxidation
Oxidizing agent 2O -2 +4e O 2 0 1 reduction
3. Interaction metals with salt.
Cu SO 4 + Mg = MgSO 4 + Cu
Reductant Mg 0 -2e Mg +2 2 oxidation
Oxidizing agent Cu +2 +2e Cu 0 1 reduction
The reaction is dictated, one student independently draws up a reaction diagram at the board:
H 2 + O 2 → H 2 O
Let's determine which atoms of elements change their oxidation state.
(H 2 ° + O 2 ° → H 2 O 2).
Let's compose electronic equations for the processes of oxidation and reduction.
(H 2 ° -2e → 2H + – oxidation process,
O 2 ° +4e → 2O - ² - reduction process,
H 2 is a reducing agent, O 2 is an oxidizing agent)
Let's select the common dividend for given and received e and coefficients for electronic equations.
(∙2| Н 2 °-2е → 2Н + - oxidation process, the element is a reducing agent;
∙1| O 2 ° +4e → 2O - ² - reduction process, element – oxidizing agent).
Let's transfer these coefficients to the ORR equation and select coefficients in front of the formulas of other substances.
2 H 2 + O 2 → 2 H 2 O .
IV . Reinforcing the material learned
Exercises to consolidate the material:
Which nitrogen transformation scheme corresponds to this reaction equation?
4NH 3 +5O 2 → 4NO + 6H 2 O
1) N +3 → N +2 3) N +3 → N -3
2) N -3 → N -2 4) N -3 → N +2
2) Establish a correspondence between the change in the oxidation state of an atom sulfur and a scheme for the transformation of matter. Write down the numbers without spaces or commas.
TRANSFORMATION SCHEME
A) H 2 S + O 2 → SO 2 + H 2 O
B) H 2 SO 4 + Na → Na 2 SO 4 + H 2 S + H 2 O
B) SO 2 + Br 2 + H 2 O → H 2 SO 4 + HBr
CHANGE IN OXIDATION STATE
1) E +4 → E +6
2) E +6 → E -2
3) E +6 → E +4
4) E -2 → E +6
5) E -2 → E +4 answer (521)
3) Establish a correspondence between the transformation scheme and the change in oxidation state oxidizing agent in it.
TRANSFORMATION SCHEME
A) Cl 2 + K 2 MnO 4 → KMnO 4 + KCl
B) NH 4 Cl + KNO 3 → KCl + N 2 O + H 2 O
B) HI + FeCl 3 → FeCl 2 + HCl + I 2
CHANGE OF DEGREE
OXIDIZATION OXIDIZER
1) E +6 → E +7
2) E +5 → E +1
3) E +3 → E +2
4) E 0 → E -1
5) E -1 → E 0 answer (423)
V. Teacher's final words
Redox reactions represent the unity of two opposing processes: oxidation and reduction. In these reactions, the number of electrons given up by reducing agents is equal to the number of electrons added by oxidizing agents. The entire world around us can be considered as a giant chemical laboratory in which chemical reactions, mainly redox ones, take place every second.
VI . Reflection.
VIII . Homework:§ 43, exercise 1, 3, 7 pp. 234-235.
Used Books:
The oxidation state of simple substances is 0;
The degree of oxidation of metals in compounds is equal to
1. Gabrielyan O.S. "Chemistry. 8th grade: textbook. for general education institutions. –M. : Bustard, 2010.
Oxidation-reduction reactions. Khomchenko G.P., Sevastyanova K.I. - From Enlightenment, 1985.
MEMO FOR STUDENTS
Appendix No. 1
The most important reducing and oxidizing agents
|
Restorers |
Oxidizing agents |
|
|
Metals, N 2, coal, CO – carbon monoxide (II) H 2 S, SO 2, H 2 SO 3 and salts HJ, HBr, HCl SnCl 2, FeSO 4, MnSO 4, Cr2(SO4)3 HNO 2 - nitrous acid NH 3 – ammonia NO - nitric oxide (II) Aldehydes, alcohols, formic and oxalic acids, Cathode during electrolysis |
Halogens KMnO 4, K 2 MnO 4, MnO 2, K 2 Cr 2 O 7, K2CrO4 HNO 3 -nitric acid H 2 O 2 – hydrogen peroxide O 3 – ozone, O 2 H 2 SO 4 (conc.), H 2 S eO 4 CuO, Ag 2 O, PbO 2 Noble metal ions (Ag+, Au3+) FeCl3 Hypochlorites, chlorates and perchlorates "Aqua regia" Anode during electrolysis |
|
Appendix No. 2
Compilation algorithm chemical equations electronic balance method:
1.Draw up a reaction diagram.
2. Determine the oxidation states of elements in the reactants and reaction products.
Remember!
group number of these metals (forI - III groups).
The oxidation state of the oxygen atom in
connections is usually equal to - 2, except H 2 O 2 -1 and ОF 2.
The oxidation state of the hydrogen atom in
connections is usually +1, except MeH (hydrides).
Algebraic sum of oxidation states
elements in connections is 0.
3. Determine whether the reaction is redox or whether it proceeds without changing the oxidation states of the elements.
4. Underline the elements whose oxidation states change.
5. Compose electronic equations for oxidation and reduction processes.
6. Determine which element is oxidized (its oxidation state increases) and which element is reduced (its oxidation state decreases) during the reaction.
7. 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)
8. Define a reducing agent and an oxidizing agent.
9.Balance the number of electrons between the oxidizing agent and the reducing agent.
10. Determine the coefficients for the oxidizing agent and reducing agent, oxidation and reduction products.
11.Write down the coefficient before the formula of the substance that determines the solution environment.
12.Check the reaction equation.
Appendix 3
Independent work to test knowledge
Option 1
1. Indicate the oxidation state of elements in compounds whose formulas are IBr, TeCl 4, SeF e, NF 3, CS 2.
2. In the following reaction schemes, indicate the oxidation state of each element and arrange the coefficients using the electronic balance method:
1) F 2 + Xe → XeF 6 3) Na + Br 2 → NaBr
2) S + H 2 → H 2 S 4) N 2 + Mg → Mg 3 N 2
Option 2
1. Indicate the oxidation state of the elements in the compounds: H 2 S O 4, HCN, HN O 2, PC1 3
2. Complete the equations for the oxidation-reduction reactions:
1) CI 2 + Fe → 2) F 2 + I 2 → 3) Ca + C → 4) C + H 2 →
Indicate the oxidation states of the elements in the resulting products.
Option 3
1. Indicate the oxidation state in compounds whose formulas are XeF 4, CC 1 4, PC1 b, SnS 2.
2. Write the reaction equations: a) dissolution of magnesium in a solution of sulfuric acid; b) interaction of sodium bromide solution with chlorine. Which element is oxidized and which is reduced?
Option 4
1. Make up formulas following connections: a) lithium nitride (a compound of lithium with nitrogen); b) aluminum sulfide (compound of aluminum with sulfur); c) phosphorus fluoride, in which the electropositive element exhibits the maximum degree of oxidation.
2. Write the equations for the reactions: a) magnesium iodide with bromine; b) dissolving magnesium in a solution of hydrobromic acid. Indicate what is an oxidizing agent and what is a reducing agent in each case.
Option 5
1. Make up formulas for the following compounds: a) fluorine with xenon; b) beryllium with carbon, in which the electropositive element exhibits the maximum oxidation state.
2. Arrange the coefficients using the electronic balance method in the following diagrams:
1) KI + Cu(N ABOUT 3 ) 2 → CuI + I 2 +KN ABOUT 3
2) MnS + HN ABOUT 3 ( conc. .) → MnS ABOUT 4 + N ABOUT 2 +H 2 ABOUT
Option 6
1. Indicate the oxidation states of each element in compounds whose formulas are Na 2 S O 3, KSO 3, NaCIO, Na 2 Cr O 4, N H 4 ClO 4, BaMn O 4.
2. Write the equations for the reactions: a) lithium iodide with chlorine; b) lithium with hydrochloric acid. Enter the oxidation states of all elements and coefficients using the electronic balance method.
Option 7
1. Calculate the oxidation states of manganese, chromium and nitrogen in compounds whose formulas are KMnO 4, Na 2 Cr 2 O 7, NH 4 N O 3.
2. Indicate the oxidation states of each element and arrange the coefficients using the electronic balance method in the following diagrams:
2) H 2 S O 3 + I 2 + H 2 O → H 2 S O 4 + HI
Option 8
1. What is the oxidation state of carbon in carbon monoxide (IV) and does it change
Lesson topic: Oxidation-reduction reactions.
The purpose of the lesson: Summarize, systematize and expand students' knowledge about redox reactions, the most important oxidizing agents and their reduction products.
Tasks:
Strengthen the ability to determine the oxidation states of elements, oxidizing agent and reducing agent, and arranging coefficients using the electronic balance method.
Improve the ability to determine the redox properties of substances, predict reaction products depending on the activity of metals, the concentration of acids and the reaction of the solution environment.
Develop the ability to compose equations for chemical reactions occurring in various environments using the example of manganese compounds.
Show the diversity and importance of OVR in nature and Everyday life.
Continue preparing for the Unified State Exam in Chemistry.
During the classes
1. Organizational moment
Good afternoon Have a good mood!
The topic of our lesson: “Oxidation-reduction reactions” (Presentation, slide 1)
Redox reactions are among the most common chemical reactions and have great value in theory and practice. The most important processes on the planet are associated with this type of chemical reactions. Humanity has been using OVRs for a long time, without initially understanding their essence. Only at the beginning of the 20th century was the electronic theory of oxidation - recovery processes. During the lesson you will remember the basic principles of this theory, the electronic balance method, learn how to write equations for chemical reactions occurring in solutions, and find out what the mechanism of such reactions depends on.
2. Repetition and generalization of previously studied material
The topic of OVR is not new for you; it runs like a red thread throughout the entire chemistry course. Therefore, I propose to review some concepts and skills on this topic.
The first question is: “What is the oxidation state?” Without this concept and the ability to arrange the oxidation states of chemical elements, it is not possible to consider this topic.
/ Oxidation state is the conditional charge of an atom of a chemical element in a compound, calculated on the basis of the assumption that all compounds consist only of ions. The oxidation state can be positive, negative or zero, depending on the nature of the compounds involved./
Some elements have constant oxidation states, others have variable ones.
For example, elements with a constant positive oxidation state include alkali metals: Li +1, Na +1, K +1, Rb +1, Cs +1, Fr +1, the following elements of group II of the periodic table: Be +2, Mg + 2, Ca +2, Sr +2, Ba +2, Ra +2, Zn +2, as well as the element III A of group - A1 +3 and some others. Metals in compounds always have a positive oxidation state.
Among non-metals, F has a constant negative oxidation state (-1).
In simple substances formed by atoms of metals or non-metals, the oxidation states of the elements are zero, for example: Na°, Al°, Fe°, H 2 0, O 2 0, F 2 0, Cl 2 0, Br 2 0.
Hydrogen is characterized by oxidation states: +1 (H 2 0), -1 (NaH).
Oxygen is characterized by oxidation states: -2 (H 2 0), -1 (H 2 O 2), +2 (OF 2).
It should be remembered that in general the molecule is electrically neutral, therefore in any molecule the algebraic sum of oxidation states is equal to zero, and in a complex ion - the charge of the ion.
For example, let's calculate the oxidation state of chromium in potassium dichromate K 2 Cr 2 O 7 .
The oxidation state of potassium is +1, oxygen -2.
Let's count the number of negative charges: 7 (-2) = -14
The number of positive charges should be + 14. Potassium has two positive charges, therefore chromium has 12.
Since there are two chromium atoms in the formula, we divide 12 by two: 12: 2 = 6.
6 is the oxidation state of chromium.
Verification: algebraic sum of positive and negative powers oxidation of elements is zero, the molecule is electrically neutral.
Independent work No. 1 according to the instruction card: using the information provided, calculate the oxidation states of the elements in the compounds: MnO 2, H 2 SO 4, K 2 SO 3, H 2 S, KMnO 4.
What are oxidation-reduction reactions from the point of view of the concept of “degree of oxidation of chemical elements”? (slide 2)
/ Redox reactions- these are reactions in which oxidation and reduction processes simultaneously occur and, as a rule, the oxidation states of elements change./
Let's consider the process using the example of the interaction of zinc with dilute sulfuric acid:
When compiling this equation, the electronic balance method is used. The method is based on comparing the oxidation states of atoms in the starting materials and reaction products. The main requirement when composing equations using this method is that the number of electrons given must be equal to the number of electrons received.
Oxidation-reduction reactions are reactions in which electrons transfer from one atom, molecule or ion to another.
Oxidation is the process of losing electrons and increasing the oxidation state.
Reduction is the process of adding electrons, and the oxidation state decreases.
Atoms, molecules or ions that donate electrons become oxidized; are reducing agents.
Atoms, ions, or molecules that accept electrons are reduced; are oxidizing agents.
Oxidation is always accompanied by reduction; reduction is associated with oxidation.
Oxidation-reduction reactions are the unity of two opposite processes: oxidation and reduction.
Independent work No. 2 according to the instruction card: using the electronic balance method, find and put the coefficients in the following redox reaction scheme:
MnO 2 + H 2 SO 4 → MnSO 4 + O 2 + H 2 O (2MnO 2 + 2H 2 SO 4 → 2MnSO 4 + O 2 +2H 2 O)
However, learning to find coefficients in OVR does not mean being able to compile them. It is necessary to know the behavior of substances in the reaction reaction, to provide for the course of reactions, to determine the composition of the resulting products depending on the reaction conditions.
In order to understand in which cases elements behave as oxidizing agents, and in which - as reducing agents, you need to turn to periodic table D.I. Mendeleev. If we are talking about simple substances, then reducing properties should be inherent in those elements that have a larger atomic radius compared to others and a small (1 - 3) number of electrons at the external energy level. Therefore, they can give them away relatively easily. These are mostly metals. The most powerful reducing properties of them are the alkali and alkaline earth metals located in the main subgroups of groups I and II (for example, sodium, potassium, calcium, etc.).
The most typical nonmetals, which have a close to complete structure of the outer electron layer and a significantly smaller atomic radius compared to metals of the same period, quite easily accept electrons and behave as oxidizing agents in redox reactions. The most powerful oxidizing agents are the light elements of the main subgroups VI – VII groups, for example fluorine, chlorine, bromine, oxygen, sulfur, etc.
At the same time, we must remember that the division of simple substances into oxidizing agents and reducing agents is as relative as the division into metals and non-metals. If non-metals enter an environment where a stronger oxidizing agent is present, they can exhibit reducing properties. Items in different degrees oxidations can behave differently.
If an element has its highest oxidation state, then it can only be an oxidizing agent. For example, in HN +5 O 3 nitrogen in the + 5 state can only be an oxidizing agent and accept electrons.
Only an element in the lowest oxidation state can be a reducing agent. For example, in N -3 H 3 nitrogen in the -3 state can donate electrons, i.e. is a reducing agent.
Elements in intermediate positive oxidation states can both donate and accept electrons and are therefore able to behave as oxidizing or reducing agents depending on conditions. For example, N +3, S +4. When placed in an environment with a strong oxidizing agent, they behave as reducing agents. And, conversely, in a reducing environment they behave as oxidizing agents.
Based on their redox properties, substances can be divided into three groups:
oxidizing agents
reducing agents
oxidizing agents - reducing agents
Independent work No. 3 on the instruction card: in which of the given reaction equation schemes MnO 2 exhibits the properties of an oxidizing agent, and in which - the properties of a reducing agent:
2MnO 2 + O 2 + 4KOH = 2K 2 MnO 4 + 2H 2 O (MnO 2 – reducing agent)
MnO 2 + 4HCI = MnCI 2 + CI 2 + 2H 2 O (MnO 2 - oxidizer)
3. Deepening and expanding knowledge
The most important oxidizing agents and their reduction products
1. Sulfuric acid - H 2 SO 4 is an oxidizing agent
A) Equation for the interaction of zinc with dilute H 2 SO 4 (slide 3)
Which ion is the oxidizing agent in this reaction? (H+)
The product of reduction by the metal in the voltage series up to hydrogen is H 2 .
B) Let's consider another reaction - the interaction of zinc with concentrated H 2 SO 4 (slide 4)
Which atoms change oxidation state? (zinc and sulfur)
Concentrated sulfuric acid (98%) contains 2% water, and the salt is obtained in solution. The reaction actually involves sulfate ions. The reduction product is hydrogen sulfide.
Depending on the activity of the metal, the reduction products of concentrated H 2 SO 4 are different: H 2 S, S, SO 2.
2. Another acid – nitric – is also an oxidizing agent due to the nitrate ion NO 3 - . The oxidizing capacity of the nitrate ion is significantly higher than the H + ion, and the hydrogen ion is not reduced to an atom, therefore, when interacting nitric acids When interacting with metals, hydrogen is never released, but various nitrogen compounds are formed. This depends on the acid concentration and the activity of the metal. Dilute nitric acid is reduced more deeply than concentrated (for the same metal) (slide 6)
The diagrams indicate products whose content is the highest among possible products acid reduction
Gold and platinum do not react with HNO 3, but these metals dissolve in “aqua regia” - a mixture of concentrated hydrochloric and nitric acids in a 3: 1 ratio.
Au + 3HCI (conc.) + HNO 3 (conc.) = AuCI 3 + NO + 2H 2 O
3. The most powerful oxidizing agent among simple substances is fluorine. But it is too active and difficult to obtain in free form. Therefore, in laboratories they use potassium permanganate KMnO 4 . Its oxidizing ability depends on the concentration of the solution, temperature and environment.
Creating a problem situation: I was preparing a solution of potassium permanganate (“potassium permanganate”) for a lesson, spilled a glass with the solution and stained my favorite chemistry coat. Suggest (after performing a laboratory experiment) a substance that can be used to clean the robe.
Oxidation–reduction reactions can occur in various environments. Depending on the environment, the nature of the reaction between the same substances may change: the environment affects the change in the oxidation states of atoms.
Typically, sulfuric acid is added to create an acidic environment. Hydrochloric and nitrogen are used less frequently, because the first is capable of oxidizing, and the second is itself a strong oxidizing agent and can cause side processes. To create an alkaline environment, potassium or sodium hydroxide is used, and water is used to create a neutral environment.
Laboratory experience:(TB rules)
1-2 ml of a diluted solution of potassium permanganate is poured into four numbered test tubes. Add a few drops of sulfuric acid solution to the first test tube, water to the second, potassium hydroxide to the third, and leave the fourth test tube as a control. Then pour sodium sulfite solution into the first three test tubes, shaking gently. Check. How does the color of the solution change in each test tube? (slides 7, 8)
Results of laboratory experiment:
Reduction products KMnO 4 (MnO 4 -):
V acidic environment– Mn +2 (salt), colorless solution;
in a neutral environment – MnO 2, brown precipitate;
in an alkaline medium - MnO 4 2-, green solution. (slide 9,)
To the reaction schemes:
KMnO 4 + Na 2 SO 3 + H 2 SO 4 → MnSO 4 + Na 2 SO 4 + K 2 SO 4 + H 2 O
KMnO 4 + Na 2 SO 3 + H 2 O→ MnO 2 ↓ + Na 2 SO 4 + KOH
KMnO 4 + Na 2 SO 3 + TOOH→ Na 2 SO 4 + K 2 MnO 4 + H 2 O
Select odds using the electronic balance method. Specify the oxidizing agent and reducing agent (slide 10)
(The task is multi-level: strong students write down the reaction products independently)
You have done a laboratory experiment, suggest a substance that can be used to clean the gown.
Demonstration experience:
Stains from a solution of potassium permanganate are quickly removed with a solution of hydrogen peroxide, acidified acetic acid:
2KMnO 4 + 9H 2 O 2 + 6CH 3 COOH = 2Mn(CH 3 COO) 2 + 2CH 3 COOK + 7O 2 + 12H 2 O
Old potassium permanganate stains contain manganese(IV) oxide, so another reaction will occur:
MnO 2 + 3H 2 O 2 + 2CH 3 COOH = Mn(CH 3 COO) 2 + 2O 2 + 4H 2 O (slide 12)
After removing stains, the piece of fabric must be rinsed with water.
The importance of redox reactions
It is impossible to consider all the variety of redox reactions within the framework of one lesson. But their importance in chemistry, technology, and everyday human life cannot be overestimated.
Student: Redox reactions underlie the production of metals and alloys, hydrogen and halogens, alkalis and medicines.
The functioning of biological membranes and many natural processes are associated with redox reactions: metabolism, fermentation, respiration, photosynthesis. Without understanding the essence and mechanisms of redox reactions, it is impossible to imagine the operation of chemical power sources (accumulators and batteries), the production of protective coatings, and the masterful processing of metal surfaces of products.
For the purposes of bleaching and disinfection, the oxidizing properties of such well-known agents as hydrogen peroxide, potassium permanganate, chlorine and chlorine, or bleach, are used.
Chlorine as a strong oxidizing agent is used for sterilization clean water and wastewater disinfection.
4. Consolidation of the studied material
Test :
Concentrated H 2 SO 4 at ordinary temperature passivates:
Concentrated HNO 3 does not react with metal:
Dilute HNO 3 with active metals is reduced to:
Which reduction product of KMnO 4 is missing: 2KMnO 4 + 3K 2 SO 3 + H 2 O = + 3K 2 SO 4 + 2KOH
In an acidic environment, KMnO 4 is reduced to:
(mutual checking of tests in pairs)
5. Homework
Using the diagrams given in the lesson, complete the reaction equations and arrange the coefficients in them using the electronic balance method:
AI + H 2 SO 4 (conc.) →
Ag + HNO 3 (conc.) →
KBr + KMnO 4 + H 2 SO 4 → …….. + Br 2 + K 2 SO 4 + H 2 O (slide 13)
6.Summing up the lesson
Instructional card
I . Repetition and generalization of previously studied material
Exercise 1: Calculate the oxidation states of elements in compounds:
MnO 2 , H 2 SO 4 , K 2 SO 3 , H 2 S, KMnO 4 .
Task 2: Using the electronic balance method, find and put the coefficients in the following redox reaction scheme:
MnO 2 +H 2 SO 4 → MnSO 4 + O 2 +H 2 O
Task 3: In which of the following reaction equation schemes does MnO 2 exhibit the properties of an oxidizing agent, and in which does it exhibit the properties of a reducing agent:
A) 2 MnO 2 + O 2 + 4 KOH = 2 K 2 MnO 4 + 2 H 2 O B) MnO 2 + 4 HCI = MnCI 2 + C.I. 2 + 2 H 2 O
II . Deepening and expanding knowledge:
Laboratory experience: (follow safety regulations)
1-2 ml of a diluted solution of potassium permanganate is poured into four numbered test tubes. Add a few drops of sulfuric acid solution to the first test tube, water to the second, potassium hydroxide to the third, and leave the fourth test tube as a control. Then pour sodium sulfite solution into the first three test tubes, shaking gently.
Note how the color of the solution changes in each test tube:
1 test tube -
2 test tubes –
3 test tube –
4 tube - control
Exercise: To the reaction schemes:
KMnO 4 +Na 2 SO 3 +H 2 SO 4 →MnSO 4 +Na 2 SO 4 +K 2 SO 4 +H 2 O
KMnO 4 +Na 2 SO 3 +H 2 O→MnO 2 ↓+Na 2 SO 4 + KOH
KMnO 4 +Na 2 SO 3 + TOOH →Na 2 SO 4 + K 2 MnO 4 +H 2 O
Select odds using the electronic balance method. Specify the oxidizing agent and the reducing agent.
III . Reinforcing the material learned
Test:
1.In an acidic environmentKMnO 4 restored to:
A) salt Mn +2 B) MnO 2 C) K 2 MnO 4
2. ConcentratedH 2 SO 4 Passivates at normal temperature:
A) Zn B) Cu C) AI
3. ConcentratedHNO 3 does not react with metal:
A) Ca B) Au C) Mg
4.DilutedHNO 3 with active metals it is reduced to:
A)NO B) N 2 C) N 2 O
5. Which recovery productKMnO 4 missing:
2KMnO 4 + 3K 2 SO 3 + H 2 O = + 3K 2 SO 4 + 2KOH
A) MnO 2 B) 2MnSO 4 C) K 2 MnO 4
Test score (based on peer review results)
IV . Homework
Using the diagrams given in the lesson, complete the reaction equations and place the coefficients in them:
1. AI + H 2 SO 4 (conc.) →
2. Ag + HNO 3 (conc.) →
3. KBr + KMnO 4 + H 2 SO 4 → …….. + Br 2 + K 2 SO 4 + H 2 O
Oxidation state
Redox properties of substances
Types of oxidation-reduction reactions
Direction of redox reactions
Redox reactions include those that are accompanied by the movement of electrons from one particle to another. When considering the patterns of redox reactions, the concept of oxidation degree is used.
1. Oxidation state
Concept oxidation states introduced to characterize the state of elements in connections. The oxidation state means the conventional charge of an atom in a compound, calculated based on the assumption that the compound consists of ions. The oxidation state is indicated by an Arabic numeral with a plus sign when electrons are displaced from a given atom to another atom and by a number with a minus sign when electrons are displaced in the opposite direction. A number with a “+” or “-“ sign is placed above the element symbol. Oxidation number indicates the oxidation state of an atom and is simply convenient form to account for electron transfer: it should not be considered as either the effective charge of an atom in the molecule (for example, in a LiF molecule, the effective charges of Li and F are respectively + 0.89 and -0.89, while the oxidation states are +1 and -1), nor as the valency of the element (for example, in the compounds CH 4, CH 3 OH, HCOOH, CO 2, the valence of carbon is 4, and the oxidation states are respectively -4, -2, +2, +4). Numerical values Valence and oxidation states can coincide in absolute value only when compounds with an ionic structure are formed.
When determining the degree of oxidation, the following rules are used:
Atoms of elements that are in a free state or in the form of molecules of simple substances have an oxidation state of zero, for example Fe, Cu, H 2, N 2, etc.
The oxidation state of an element in the form of a monoatomic ion in a compound having an ionic structure is equal to the charge of this ion,
1 -1 +2 -2 +3 -1
for example, NaCl, Cu S, AlF 3.
Hydrogen in most compounds has an oxidation state of +1, with the exception of metal hydrides (NaH, LiH), in which the oxidation state of hydrogen is -1.
The most common oxidation state of oxygen in compounds is -2, with the exception of peroxides (Na 2 O 2, H 2 O 2), in which the oxidation state of oxygen is –1 and F 2 O, in which the oxidation state of oxygen is +2.
For elements with a variable oxidation state, its value can be calculated by knowing the formula of the compound and taking into account that the algebraic sum of the oxidation states of all elements in a neutral molecule is zero. In a complex ion, this sum is equal to the charge of the ion. For example, the oxidation state of the chlorine atom in the HClO 4 molecule, calculated based on the total charge of the molecule = 0, where x is the oxidation state of the chlorine atom), is +7. The oxidation state of the sulfur atom in the (SO 4) 2- [x + 4(-2) = -2] ion is +6.
2. Redox properties of substances
Any redox reaction consists of oxidation and reduction processes. Oxidation - is the process of donating electrons by an atom, ion, or molecule of a reactant. Substances that give their electrons during the reaction and are oxidized are called restorers.
Reduction is the process of an atom accepting electrons ion or reagent molecule.
Substances that accept electrons and are reduced in the process are called oxidizing agents.
Oxidation-reduction reactions always occur as a single process called redox reaction. For example, when metallic zinc interacts with copper ions reducing agent(Zn) donates its electrons oxidizing agent– copper ions (Cu 2+):
Zn + Cu 2+ Zn 2+ + Cu
Copper is released on the surface of the zinc, and zinc ions go into solution.
The redox properties of elements are related to the structure of their atoms and are determined by their position in the periodic system D.I. Mendeleev. The reducing ability of the element is due to the weak bond of valence electrons with the nucleus. Metal atoms containing a small number of electrons at the outer energy level are prone to losing them, i.e. are easily oxidized, playing the role of reducing agents. The most powerful reducing agents are the most active metals.
The criterion for the redox activity of elements can be their value relative electronegativity: the higher it is, the more pronounced the oxidizing ability of the element, and the lower it is, the more pronounced its reducing activity is. Nonmetal atoms (for example, F, O) have high value electron affinity and relative electronegativity, they readily accept electrons, i.e. are oxidizing agents.
The redox properties of an element depend on the degree of its oxidation. For the same element there are different lower, higher and intermediate oxidation states.
As an example, consider sulfur S and its compounds H 2 S, SO 2 and SO 3. The relationship between the electronic structure of the sulfur atom and its redox properties in these compounds is clearly presented in Table 3.1.
In the H 2 S molecule, the sulfur atom has a stable octet configuration of the outer energy level 3s 2 3p 6 and therefore can no longer add electrons, but can give them away.
The state of an atom in which it can no longer accept electrons is called the lowest oxidation state.
In the lowest oxidation state, the atom loses its oxidizing ability and can only be a reducing agent.
Table 1.
|
Substance formula |
Electronic formula |
Redox properties |
|
|
|
1s 2 2s 2 2p 6 3s 2 3p 6 |
|
|
|
|
1s 2 2s 2 2p 6 3s 2 3p 4 |
oxidizer |
reducing agent |
|
|
1s 2 2s 2 2p 6 3s 2 3p o |
oxidizer |
|
|
|
1s 2 2s 2 2p 6 3s o 3p 0 |
oxidizer |
|
–2
;
- 6
;
- 8
reducing agent
+
2
–4
;
-
6
+
4 
;
+
6
-2
reducing agent
+
2
;
+ 6
;
+
8
In the SO 3 molecule, all the outer electrons of the sulfur atom are shifted to the oxygen atoms. Therefore, in this case, the sulfur atom can only accept electrons, exhibiting oxidizing properties.
The state of an atom in which it has given up all its valence electrons is called the highest oxidation state. An atom in the highest oxidation state can only be an oxidizing agent.
In the SO 2 molecule and elemental sulfur S, the sulfur atom is located in intermediate oxidation states, i.e., having valence electrons, the atom can give them away, but without having a complete R - sublevel, can also accept electrons until its completion.
An atom of an element with an intermediate oxidation state can exhibit both oxidizing and reducing properties, which is determined by its role in a particular reaction.
For example, the role of the sulfite anion SO 
in the following reactions is different:
5Na 2 SO 3 + 2KMnO 4 + 3H 2 SO 4 2MnSO 4 + 5Na 2 SO 4 + K 2 SO 4 + 3H 2 O (1)
H 2 SO 3 + 2 H 2 S 3 S + 3 H 2 O (2)
In reaction (1), the sulfite anion SO 
in the presence of a strong oxidizing agent, KMnO 4 plays the role of a reducing agent; in reaction (2) sulfite anion SO 
- an oxidizing agent, since H 2 S can only exhibit reducing properties.
Thus, among complex substances restorers can be:
1. Simple substances whose atoms have low ionization energies and electronegativity (in particular, metals).
2. Complex substances containing atoms in lower oxidation states:
H Cl,H 2 S,N H 3
Na 2 S O3, Fe Cl2, Sn(NO 3) 2 .
Oxidizing agents can be:
1. Simple substances whose atoms have high values of electron affinity and electronegativity - nonmetals.
2. Complex substances containing atoms in higher oxidation states: +7 +6 +7
K Mn O 4 , K 2 Cr 2 O 7, HClO 4.
3. Complex substances containing atoms in intermediate oxidation states:
Na 2 S O3, Mn O2, Mn SO4.
Lesson motto: “Someone loses, but someone finds...”
Lesson objectives:
Educational:
consolidate the concepts of “oxidation state”, processes of “oxidation”, “reduction”;
consolidate skills in drawing up equations of redox reactions using the electronic balance method;
teach to predict the products of redox reactions.
Educational:
Continue development logical thinking, skills to observe, analyze and compare, find cause-and-effect relationships, draw conclusions, work with algorithms, and develop interest in the subject.
Educational:
To form the scientific worldview of students; improve work skills;
teach to listen to the teacher and your classmates, to be attentive to yourself and others, to evaluate yourself and others, and to have a conversation.
I. Organizational moment
The topic of the lesson is announced, the relevance of this topic and its connection with life is justified. Redox processes are among the most common chemical reactions and are of great importance in theory and practice. They are associated with metabolic processes occurring in a living organism, rotting and fermentation, photosynthesis. Redox processes accompany the cycles of substances in nature. They can be observed during fuel combustion, in metal corrosion processes, during electrolysis and smelting of metals. With their help, alkalis, acids and other valuable products.
Redox reactions underlie the energy conversion of interacting chemical substances into electrical energy in galvanic and fuel cells. Humanity has been using OVRs for a long time, without initially understanding their essence. Only at the beginning of the 20th century was the electronic theory of redox processes created. During the lesson you will remember the main provisions of this theory, as well as learn how to draw up equations for chemical reactions occurring in solutions, and find out what the mechanism of such reactions depends on.
II. Repetition and generalization of previously studied material
1. Oxidation state.
Organizing a conversation aimed at updating background knowledge on the degree of oxidation and the rules for its determination, on the following issues:
- What is electronegativity?
- What is the oxidation state?
- Can the oxidation state of an element be zero? In which cases?
- What oxidation state does oxygen most often exhibit in compounds?
- Remember the exceptions.
- What oxidation state do metals exhibit in polar and ionic compounds?
Based on the results of the conversation, rules for determining oxidation states are formulated
To consolidate the formulated rules, it is proposed to determine the oxidation state of elements in compounds:
H2SO4, H2, H2SO3, HCIO4, Ba, KMnO4, AI2(SO4)3, HNO3, Ba(NO3)2, HCN, K4, NH3, (HN4)2SO4.
This task with selective answers is used for oral frontal questioning.
2. Oxidation and reduction processes. Redox reactions.
During the conversation, knowledge about redox processes is updated.
Indicate the type of chemical reaction on the right. Adjust the coefficients as necessary. If s.o. elements before and after the reaction change, then write the word “yes” on the left; if they do not change, then write the word “no”.
Option I:
Hg + S → Hg S
NaNO3 →NaNO2 + O2
CuSO4 + NaOH →Na2SO4 + Cu(OH)2
Option II:
Al(OH)3 → Al2O3 + H2O
H2O + P2O5 → H3PO4
Fe + HCl → FeCl2 + H2
All types of work are checked together with the class. The equations of chemical reactions remain on the board, and then the class is asked to answer the questions:
1) Do oxidation states of chemical elements change in all cases? (No).
2) Does it depend on the type of chemical reactions in terms of the number of reactants and reaction products? (No).
Suggested questions:
- What is the recovery process called?
- How does the oxidation state of an element change during reduction?
- What is oxidation?
- How does the oxidation state of an element change during oxidation?
- Define the terms “oxidizing agent” and “reducing agent”.
WITH modern point In view, the change in oxidation state is associated with the withdrawal or movement of electrons. Therefore, along with the above, another definition can be given: these are reactions in which electrons transfer from one atom, molecule or ion to another.
We conclude: “What is the essence of OVR?”
Redox reactions represent the unity of two opposing processes - oxidation and reduction. In these reactions, the number of electrons given up by the reducing agents is equal to the number of electrons gained by the oxidizing agents. At the same time, regardless of whether electrons move from one atom to another completely or only partially, or are attracted to one of the atoms, we conventionally speak only of the release or addition of electrons. That is why the motto of the lesson was chosen: “Someone loses, and someone finds...”
3. Functions of connections in OVR.
1. Having calculated the oxidation states of elements, prove that these substances exhibit the properties of oxidizing agents.
Cl2, HClO4, H2SO4, KMnO4, SO2
2. Calculate the oxidation states of the elements, 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 connection function in the OVR:
1. If an element exhibits the highest oxidation state in a compound, then this compound can only be an oxidizing agent.
2. If an element exhibits a lower oxidation state in a compound, then this compound can be a reducing agent
Solving problematic issues:
- Can the same substance be both an oxidizing agent and a reducing agent?
- Can one and the same element exhibit the properties of both an oxidizing agent and a reducing agent?
Formulating the third rule.
3. If an element exhibits an intermediate oxidation state in a compound, then this compound can be both a reducing agent and an oxidizing agent.
III. Arrangement of coefficients in OVR equations using the electronic balance method.
Practicing skills in determining the degree of oxidation, drawing up diagrams of redox reactions using the electronic balance method (work at the board and in notebooks) with the development of reasoning and analysis skills through student comments on answers.
Using the electronic balance method, select the coefficients in the redox reaction schemes and indicate the process of oxidation and reduction:
K2Cr2O7 + H2S + 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 C (C1) of the Unified State Exam KIM:
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 B (B2) of the Unified State Exam KIM:
Establish a correspondence between the reaction equation and the change in the oxidation state of the oxidizing agent in this reaction:
A) S02 + N02 = S03+NO 1) -1 → 0
B) 2NH3 + 2Na = 2NaNH2 + H2 2) 0 → -2
B) 4N02 + 02 + 2H20 = 4HN03 3) +4 → +2
D) 4NH3 + 6NO = 5N2 + 6H20 4) +1 → 0
5) +2 → 0
6) 0 → - 1
Reaction equation Change in oxidation state of oxidizing agent
A) 2NH3 + 2Na = 2NaNH2 + H2 1) -1 → 0
B) H2S + 2Na = Na2S + H2 2) 0 → - 1
4NH3 + 6NO = 5N2 + 6H20 3) +2→ 0
D) 2H2S + 302 = 2S02 + 2H20 4) + 1 → 0
5) +4 → +2
6) 0→ -2
Establish a correspondence between the equation of the reaction and the substance that is the reducing agent in this reaction
Reaction Equation Reductant
A) NO + N02 + H20 = 2HN02 1) N02
B) SO2 + 2H2S = 3S + 2H20 2) H2S
Br2 + S02 + 2H20 = 2HBr + H2SO4 3) Br2
D) 2KI + Br2 = 2KVg + I2 4) S02
5) NO
6) KI
IV. Stage of consolidation of knowledge (ends with a test).
Test
1) What is the lowest oxidation state of sulfur?
a) –6; b) –4; at 2; d) 0; e) +6.
2) What is the oxidation state of phosphorus in the Mg3P2 compound?
a) +3; b) +5; c) 0; d) –2; e) –3.
3) Which elements have a constant oxidation state of +1?
a) Hydrogen; b) lithium; c) copper;
d) magnesium; d) selenium.
4) What is it equal to? highest degree manganese oxidation?
a) –1; b) 0; c) +7; d) +4; e) +6.
5) What is the oxidation state of chlorine in the compound Ca(ClO)2?
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 presented below and how many electrons are involved in it?
a) restoration, 1e; b) oxidation, 2e;
c) restoration, 2e; d) oxidation, 1f.
8) Which of the following substances can be both oxidizing and reducing agents? There are several possible answers.
a) SO2; b) Na; c) H2; d) K2Cr2O7; e) HNO2.
9) What is the name of the process presented below and how many electrons are involved in it?
a) restoration, 8e; b) oxidation, 4f;
c) oxidation, 8f; d) restoration, 4f.
10) Which of the following substances can only be reducing agents? There are several possible answers.
a) H2S; b) KMnO4; c) SO2; d) NH3; e) Na.
Answers. 1 – in; 2 – d; 3 – b, d; 4 – in; 5 B; 6 – d; 7 – b; 8 – a, c, d; 9 – a; 10 – a, d, d.
V. Deepening and expanding knowledge (Lecture part of the lesson)
The importance of redox reactions
Redox reactions accompany many processes carried out in industry and in various fields everyday life: burning gas in a gas stove, cooking, washing, cleaning household items, making shoes, perfumes, textiles...
Whether we light a match, whether fancy fireworks are burning in the sky - all these are redox processes.
For the purposes of bleaching and disinfection, the oxidizing properties of such well-known agents as hydrogen peroxide, potassium permanganate, chlorine and chlorine, or bleach, are used.
If it is necessary to oxidize any easily degradable substance from the surface of the product, hydrogen peroxide is used. It is used to bleach silk, feathers, and fur. It is also used to restore ancient paintings. Due to its harmlessness to the body, hydrogen peroxide is used in the food industry for bleaching chocolate, scars and casings in the production of sausages.
The disinfectant effect of potassium permanganate is also based on its oxidizing properties.
Chlorine, as a strong oxidizing agent, is used to sterilize clean water and disinfect wastewater. Chlorine destroys many paints, which is the basis for its use in bleaching paper and fabrics. Chloric, or bleaching, lime is one of the most common oxidizing agents both in everyday life and on an industrial scale.
In nature, redox reactions are extremely common. They play an important role in biochemical processes: respiration, metabolism, nervous activity humans and animals. Manifestation of various vital functions The body is associated with the expenditure of energy that our body receives from food as a result of redox reactions.
VI. Summarizing.
Grades for the lesson are given and homework:
A. Determine the oxidation states of elements using the formulas:
HNO2, Fe2(SO4)3, NH3, NH4Cl, KClO3, Ва(NO3)2, НClО4
B. Arrange the coefficients using the electronic balance method:
KMnO4 +Na2SO3+H2O → MnO2+ Na2 SO4+ KOH
C. KMnO4 + Na2SO3+ KOH → … + K2 MnO4 + …
Literature:
Gabrielyan O.S. Chemistry-8. M.: Bustard, 2002;
Gabrielyan O.S., Voskoboynikova N.P., Yashukova A.V. Teacher's handbook. 8th grade. M.: Bustard, 2002;
Small children's encyclopedia. Chemistry. M.: Russian Encyclopedic Partnership, 2001; Encyclopedia for children "Avanta+". Chemistry. T. 17. M.: Avanta+, 2001;
Khomchenko G.P., Sevastyanova K.I. Redox reactions. M.: Education, 1989.
V.A. Shelontsev. Iconic models and problems: redox reactions. OOIPKRO, Omsk - 2002
A.G. Kuhlman. General chemistry, Moscow-1989.
For the full text of the material Lesson notes for grade 8 “Oxidation-reduction reactions,” see the downloadable file.
The page contains a fragment.
Lesson developments (lesson notes)
Basics general education
Line UMK O. S. Gabrielyan. Chemistry (8-9)
Attention! The site administration is not responsible for the content of methodological developments, as well as for the compliance of the development with the Federal State Educational Standard.
References:
- Chemistry teacher's handbook. 8th grade. O.S. Gabrielyan, N.P. Voskoboynikova, A.V. Yashukova (M.: Bustard). 2003
- EFU Chemistry 8th grade. O.S. Gabrielyan, (M.: Bustard).
- Workbook to the textbook O.S. Gabrielyan Chemistry 8th grade. O.S. Gabrielyan, A.S. Sladkov (M.: Bustard-2013).
Lesson Objectives:
- educational: introduce students to new classification chemical reactions based on changes in the oxidation states of elements - redox reactions, repeat the concepts of “oxidizing agent”, “reducing agent”, “oxidation”, “reduction”;
- developing: continue the development of logical thinking, the formation of interest in the subject, using modern technologies in teaching.
- educational: to form the scientific worldview of students, the formation of culture interpersonal communication: evaluate your work..
Means of education:
- Electronic supplement to the textbook “Chemistry 8th grade”. O.S. Gabrielyan, (M.: Bustard).
- Interactive tutorial“VISUAL CHEMISTRY. Chemistry. 8-9 grade." Moscow: Exam-Media LLC 2011-2013
Textbook: EFU Gabrielyan O.S. Chemistry.8th grade: – M.: Bustard, 2015
During the classes
1. Organizational stage
Preparing students for work in the classroom. Rules of work and safety in the smart class when working with laptops
2. Updating students’ knowledge
A) Let's remember all the classifications of chemical reactions known to you and the characteristics that underlie each classification. Repetition. "Types of Chemical Reactions" (according to learning tool 2)
Literature work 1:
1. According to the type and composition of the reacting and resulting substances, there are reactions:
a) connections;
b) decomposition;
c) substitution;
d) exchange (including the neutralization reaction).
2. By state of aggregation substances (phase) reactions are distinguished:
a) homogeneous;
b) heterogeneous.
3. According to the thermal effect, reactions are divided into:
a) exothermic (including combustion reactions);
b) endothermic.
4. Based on the use of a catalyst, reactions are distinguished:
a) catalytic (including enzymatic);
b) non-catalytic.
5. Reactions are distinguished by direction:
a) reversible;
b) irreversible.
B) Give full description reactions for the synthesis of sulfur oxide (6) from sulfur oxide (4) and oxygen:
3. Assimilation of new knowledge on EFU
A) Let's remember what S.O. is. and how it changes with XP. (Repetition followed by testing using learning tool 2.)
B) Explanation of material on EFU pp. 263–265.
IN) Work on the EFU electronic application.
D) Work on literature 2
4. Primary consolidation of knowledge
A) Students complete the task. ELECTRONIC APPLICATION
If there is any difficulty, use pages 264-265 of the EFU.
B) Completing a task on an electronic application, finding an oxidizing agent, a reducing agent, electron transfer, working at the board.
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