Lesson plan: redox reactions. Redox reactions
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"
LESSON OBJECTIVE: 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; oxidant; reducing agent; oxidation process; recovery process.
Equipment usedand reagents: solubility table, periodic table 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).
Lesson progress
I . Organizational moment
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 what 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, H2, H3PO4, NaHSO3, HNO3, Cu(NO2)2, NO2, Ba, Al.
For example: What will be the oxidation state of sulfur in sulfuric acid?
In molecules algebraic sum The oxidation states of elements taking into account the number of their atoms is 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 according to various characteristics (direction, number and composition of reacting and forming substances, use of a 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 - interaction 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. Final word teachers
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.
Literature used:
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 for the following compounds: 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 schemes:
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
Chemistry lesson on the topic “Oxidation-reduction reactions”
in 11th grade.
Prepared material
Angelica Svetlana Evgenievna,
first category chemistry teacher
MAOU secondary school No. 211, Novosibirsk
Target:
deepening the knowledge of students and preparing for Olympiads and the Unified State Exam.
Tasks:
Educational objectives:
To consolidate students' knowledge about redox reactions; consolidate students’ skills in drawing up equations of oxidation-reduction reactions
develop skills in drawing up equations of redox reactions
develop skills in identifying oxidizing and reducing agents
formation of a chemically literate personality, ready for life in a constantly changing environment, further education and self-education.
Developmental tasks:
contribute to the formation and development cognitive interest students to the subject
formation of skills to analyze, compare and generalize knowledge on the topic.
Educational tasks:
fostering a conscious need for knowledge;
fostering activity and independence when studying a given topic, the ability to work in a group, and the ability to listen to your classmates.
Lesson type: lesson - exercise.
Form organizations educational activities : individual and group.
Equipment : computer, multimedia projector, screen, document camera.
Teaching methods:
General method (partially search method).
Particular method (verbal – visual – practical).
Specific method (explanation with elements of conversation).
Lesson progress
Updating knowledge. Reproduction of previously acquired knowledge.
Organizational moment
Message of the topic, setting the topic and objectives of the lesson
Teacher.
What are redox reactions?
Any ORR is a set of processes of electron donation and addition.
What is the process of giving up electrons called?
The process of giving up electrons is called oxidation.
What are the particles that donate electrons called?
Particles (atoms, molecules or ions) that donate electrons are called restorers.
Teacher.
As a result of oxidation, the oxidation state of the reducing agent increases. Reducing agents can be particles in lower or intermediate oxidation states. The most important reducing agents are: all metals in the form of simple substances, especially active ones; C, CO, NH 3, PH 3, CH 4, SiH 4, H 2 S and sulfides, hydrogen halides and metal halides, metal hydrides, metal nitrides and phosphides.
What is the name of the process of adding electrons and the particles that accept electrons?
The process of adding electrons is called restoration. Particles that accept electrons are called oxidizing agents.
Teacher.
As a result of reduction, the oxidation state of the oxidizing agent decreases. Oxidizing agents can be particles in higher or intermediate oxidation states. The most important oxidizing agents: simple substances - non-metals, having high electronegativity (F 2, Cl 2, O 2), potassium permanganate, chromates and dichromates, nitric acid and nitrates, concentrated sulfuric acid, perchloric acid and perchlorates.
Operating with knowledge, mastering methods of activity in new conditions
Students take the Oxidation State TEST (Appendix 4.)
Generalization and systematization of knowledge and methods of action.
Teacher.
There are three types of redox reactions.
Intermolecular OVR - oxidizing agent and reducing agent are included in the composition various substances, For example:
Intramolecular OVR – an oxidizing agent and a reducing agent are part of one substance. It could be different elements, For example:
or one chemical element in different degrees oxidation, for example:
Disproportionation (auto-oxidation-self-healing)– the oxidizing agent and the reducing agent are the same element, which is in an intermediate oxidation state, for example:
To compile ORR equations, you can use the electronic balance method ( electronic circuits) or the electron-ion balance method. Let's consider one of the methods.
Electronic balance method:
Task 1. Create OVR equations using the electronic balance method, determine the type of OVR.
1. Zinc + potassium dichromate + sulfuric acid = zinc sulfate + chromium(III) sulfate + potassium sulfate + water.
Solution
Electronic balance:
2. Tin(II) sulfate + potassium permanganate + sulfuric acid = tin(IV) sulfate + manganese sulfate + potassium sulfate + water.
3. Sodium iodide + potassium permanganate + potassium hydroxide = iodine + potassium manganate + sodium hydroxide.
4. Sulfur + potassium chlorate + water = chlorine + potassium sulfate + sulfuric acid.
5. Potassium iodide + potassium permanganate + sulfuric acid = manganese(II) sulfate + iodine + potassium sulfate + water.
6. Iron(II) sulfate + potassium dichromate + sulfuric acid = iron(III) sulfate + chromium(III) sulfate + potassium sulfate + water.
7. Ammonium nitrate = nitric oxide (I) + water.
Answers to exercises in task 1
Task 3. Compose OVR equations.
2. Manganese(IV) oxide + oxygen + potassium hydroxide = potassium manganate +.......................
3. Iron(II) sulfate + bromine + sulfuric acid = .......................
4. Potassium iodide + iron(III) sulfate = ....................... .
5. Hydrogen bromide + potassium permanganate = ..................................
6. Hydrogen chloride + chromium(VI) oxide = chromium(III) chloride + .......................
7. Ammonia + bromine = .......................
8. Copper(I) oxide + nitric acid = nitric oxide(II) + .......................
9. Potassium sulfide + potassium manganate + water = sulfur + .......................
10. Nitric oxide (IV) + potassium permanganate + water = .......................
11. Potassium iodide + potassium dichromate + sulfuric acid = ..................................
Answers to exercises task 3
Definition and explanation homework.
Appendix 1.
Reducers:
Carbon(II) monoxide (CO)
Hydrogen sulfide (H2S)
sulfur oxide (IV) (SO2)
sulfurous acid H2SO3 and its salts
Hydrohalic acids and their salts
Metal cations in lower oxidation states: SnCl2, FeCl2, MnSO4, Cr2(SO4)3
Nitrous acid HNO2
ammonia NH3
hydrazine NH2NH2
nitric oxide (II) (NO)
Cathode during electrolysis
Oxidizing agents:
Halogens
Potassium permanganate(KMnO4)
Potassium manganate (K2MnO4)
manganese(IV) oxide (MnO2)
Potassium dichromate (K2Cr2O7)
potassium chromate (K2CrO4)
Nitric acid (HNO3)
Sulfuric acid (H2SO4) conc.
Copper(II) oxide (CuO)
lead(IV) oxide (PbO2)
silver oxide (Ag2O)
hydrogen peroxide (H2O2)
Iron(III) chloride (FeCl3)
Berthollet's salt (KClO3)
Anode during electrolysis.
Appendix 2.
Rules for determining the degree of oxidation
The oxidation state of atoms of simple substances is zero.
The sum of the oxidation states of atoms in a complex substance (in a molecule) is zero.
Oxidation state of atoms alkali metals +1.
The oxidation state of alkaline earth metal atoms is +2.
The oxidation state of boron and aluminum atoms is +3.
The oxidation state of hydrogen atoms is +1 (in hydrides of alkali and alkaline earth metals –1).
The oxidation state of oxygen atoms is –2 (in peroxides –1).
Appendix 3.
Memo
possible oxidation states of elements
Manganese: +2, +3, +4, +6, +7.
Chromium: +2, +3, +6.
Iron: +2, +3, +6.
Nitrogen: -3, 0, +1, +2, +4, +5.
Sulfur: -2, 0, +4, +6.
Phosphorus: -3, 0, +3, +5.
Chlorine: -1, 0, +1, + 3, +5, +7.
Metals with higher oxidation states form acidic oxides.
Potassium permanganate: KMnO 4.
It is a strong oxidizing agent. It easily oxidizes many organic matter, converts iron(2) salts into iron(3) salts, sulfurous acid into sulfuric acid and releases chlorine from hydrochloric acid.
When entering into chemical reactions, the MnO 4 - ion can be reduced to varying degrees:
In an acidic environment (pH
In a neutral environment (pH=7) to MnO 2.
In an alkaline environment (pH>7) up to MnO 4 2-
Hydrogen peroxide.
The oxidation state of the element oxygen in hydrogen peroxide is
1, i.e. has an intermediate value between the oxidation state of the oxygen element in water (-2) and in molecular oxygen (0). Therefore, hydrogen peroxide exhibits redox duality.
If peroxide serves as an oxidizing agent, then it is reduced to water H 2 O.
If peroxide serves as a reducing agent, then it is oxidized to molecular oxygen-O 2.
Chromate and dichromate salts.
Chromates (colored bright yellow) in an acidic environment turn into dichromates (orange), dichromates in an alkaline environment turn into chromates.
Chromates and dichromates are strong oxidizing agents and in the equations of redox reactions they change the oxidation state from +6 to +3.
Chlorine compounds.
HClO-hypochlorous acid (hypochlorite salts)
HClO 2 -chloride (chlorite salts)
HClO 3 -chlorate (chlorate salts)
HClO 4 -chlorine (perchlorate salts)
When halogens interact with alkalis, hypochlorites are formed in a cold solution, and chlorates are formed in a hot solution (for example, potassium chlorate or Berthollet salt-KClO 3).
Concentrated nitric acid
If concentrated nitric acid is taken as a starting material for ORR with other substances, as a result of the reaction it is reduced to nitric oxide NO 2
Appendix 4.
TEST “Oxidation states”
Option 1.
1 .
An ion containing 16 protons and 18 electrons has a charge
1) +4 2) -2 3) +2 4) -4
2. The ion has an eight-electron outer shell
1) P 3+ 2) S 2- 3) C1 5+ 4) Fe 2+
3. Ca 2+ and
1) K + 2) Ne 0 3) Ba 2+ 4) F -
4. Electronic configurationIs 2 2 s 2 2 p 6 corresponds to an ion
1) A1 3+ 2) Fe 3+ 3) Zn 2+ 4) Cr 3+
Option 2.
1. The ion has a two-electron outer shell
1) S 6+ 2) S 2- 3) Br 5+ 4) Sn 4+
2. Electronic configuration Is 2 2s 2 2p 6 3s 2 3p 6 corresponds to the ion
1) C l - 2) N 3- 3) Br - 4) O 2-
3. Same electronic structure have particles
1) Na 0 and Na + 2) Na 0 and K 0 3) Na + and F - 4) Cr 2+ and Cr 3+
4. The Al 3+ ion has the following electronic configuration:
1) 1s 2 2s 2 2p 6 2) 1s 2 2s 2 2p 6 3s 1 3) 1s 2 2s 2 2p 6 3s 2 3p 1 4) Is 2 2s 2 2p 6 3s 2 3p 6 4s 1
Answers :
Option 1.
Option2.
Appendix 5.
Homework
Task. Copper-based alloys are called bronzes. Beryllium bronze rings – exact copy gold. They do not differ from gold ones either in color or in weight and, suspended on a thread, when they hit the glass they emit melodious sound. In short, a fake cannot be detected by eye, ear, or tooth. Suggest ways to identify a fake: in your own kitchen, in a chemical laboratory (2 ways). Write down the reaction equations and name their characteristics.
Answer.
In the kitchen. Heat the “golden” ring on a gas stove; the copper oxidizes in air to black copper (II) oxide CuO (that is, the bronze ring darkens when heated).
In the laboratory. Dissolve the ring in nitric acids e. High-grade gold does not dissolve in nitric acid, but copper, which is part of bronze, interacts with HNO 3. Signs: solution blue color, release of brown "fox tail" gas.
Cu + 4HNO 3 conc. = Cu(NO 3) 2 + 2NO 2 + 2H 2 O
Gold does not dissolve in conc. H 2 SO 4, but copper dissolves when heated:
Сu + 2H 2 SO 4 conc. = CuSO 4 + SO 2 + 2H 2 O
Signs: blue solution, gas evolution.
Analysis of the lesson conducted
The lesson was taught in 11th grade. The set goal - deepening the knowledge of students and preparing for Olympiads and the Unified State Exam - was achieved. Students were given reminders necessary for a more complete understanding of the topic and used when doing homework.
The main problems that students encountered in mastering the content educational material on the topic “Oxidation-reduction reactions”, related to the compilation of ORR by the electronic balance method.
Using an algorithm compiled by the teacher together with the students, it was possible to correct the basic steps for writing an OVR and avoid basic mistakes.
Consider the diagrams of reaction equations below. What is their significant difference? Did the oxidation states of the elements change in these reactions?
In the first equation, the oxidation states of the elements did not change, but in the second they changed - for copper and iron.
The second reaction is a redox reaction.
Reactions that result in changes in the oxidation states of the elements that make up the reactants and reaction products are called oxidation-reduction reactions (ORR).
COMPILATION OF EQUATIONS FOR REDOX REACTIONS.
There are two methods for composing redox reactions - the electron balance method and the half-reaction method. Here we will look at the electronic balance method.
In this method, the oxidation states of atoms in the starting substances and in the reaction products are compared, and we are guided by the rule: the number of electrons donated by the reducing agent must be equal to the number of electrons gained by the oxidizing agent.
To create an equation, you need to know the formulas of the reactants and reaction products. Let's look at this method with an example.
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Arrange the coefficients in the reaction, the scheme of which is: HCl + MnO 2 = Cl 2 + MnCl 2 + H 2 O |
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Algorithm for setting coefficients |
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1. We indicate the oxidation states of chemical elements. Chemical elements in which the oxidation states have changed are emphasized. |
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2. We compose electronic equations in which we indicate the number of given and received electrons. Behind the vertical line we put the number of electrons transferred during the oxidation and reduction processes. Find the least common multiple (shown in the red circle). We divide this number by the number of electrons moved and get the coefficients (shown in the blue circle). This means that before manganese there will be a coefficient of -1, which we do not write, and before Cl 2 there will also be -1. |
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Let's consider a more complex equation: |
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H 2 S + KMnO 4 + H 2 SO 4 = S + MnSO 4 + K 2 SO 4 + H 2 O |
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We arrange the oxidation states of chemical elements: |
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The electronic equations will take the following form Before sulfur with oxidation states -2 and 0 we put a coefficient of 5, before manganese compounds -2, we equalize the number of atoms of other chemical elements and obtain the final reaction equation |
Basic principles of the theory of redox reactions
1. Oxidation called process of losing electrons by an atom, molecule, or ion.
For example :
Al – 3e - = Al 3+
Fe 2+ - e - = Fe 3+
H 2 – 2e - = 2H +
2Cl - - 2e - = Cl 2
During oxidation, the oxidation state increases.
2. Recovery called process of gaining electrons by an atom, molecule, or ion.
For example:
S + 2е - = S 2-
WITH l 2 + 2е- = 2Сl -
Fe 3+ + e - = Fe 2+
During reduction, the oxidation state decreases.
3. Atoms, molecules or ions that donate electrons are called restorers . During the reactionthey oxidize.
Atoms, molecules or ions that gain electrons are called oxidizing agents . During the reactionthey are recovering.
Since atoms, molecules and ions are part of certain substances, these substances are called accordingly restorers or oxidizing agents.
4. Redox reactions represent the unity of two opposing processes - oxidation and reduction.
The number of electrons given up by the reducing agent is equal to the number of electrons gained by the oxidizing agent.
EXERCISES
Simulator No. 1 Oxidation-reduction reactions
Simulator No. 2 Electronic balance method
Simulator No. 3 Test “Oxidation-reduction reactions”
ASSIGNMENT TASKS
No. 1. Determine the oxidation state of atoms of chemical elements using the formulas of their compounds: H 2 S, O 2, NH 3, HNO 3, Fe, K 2 Cr 2 O 7
No. 2. Determine what happens to the oxidation state of sulfur during the following transitions:
A) H 2 S → SO 2 → SO 3
B ) SO 2 → H 2 SO 3 → Na 2 SO 3
What conclusion can be drawn after completing the second genetic chain?
What groups can chemical reactions be classified into based on changes in the oxidation state of atoms of chemical elements?
No. 3. Arrange the coefficients in CHR using the electronic balance method, indicate the processes of oxidation (reduction), oxidizing agent (reducing agent); write down the reactions in complete and ionic form:
A) Zn + HCl = H 2 + ZnCl 2
B) Fe + CuSO 4 = FeSO 4 + Cu
No. 4. Given diagrams of reaction equations:
СuS + HNO 3 (diluted ) = Cu(NO 3) 2 + S + NO + H 2 O
K + H 2 O = KOH + H 2
Arrange the coefficients in the reactions using the electronic balance method.
Indicate the substance - an oxidizing agent and a substance - a reducing agent.
The lesson examines the essence of redox reactions and their difference from ion exchange reactions. The changes in the oxidation states of the oxidizing agent and the reducing agent are explained. The concept of electronic balance is introduced.
Topic: Redox reactions
Lesson: Redox Reactions
Consider the reaction of magnesium with oxygen. Let's write down the equation of this reaction and arrange the values of the oxidation states of the atoms of the elements:
As can be seen, the magnesium and oxygen atoms in the starting materials and reaction products have different meanings oxidation states. Let us write down diagrams of the oxidation and reduction processes occurring with magnesium and oxygen atoms.
Before the reaction, magnesium atoms had an oxidation state of zero, after the reaction - +2. Thus, the magnesium atom has lost 2 electrons:
Magnesium donates electrons and is itself oxidized, which means it is a reducing agent.
Before the reaction, the oxidation state of oxygen was zero, and after the reaction it became -2. Thus, the oxygen atom added 2 electrons to itself:
Oxygen accepts electrons and is itself reduced, which means it is an oxidizing agent.
Let's write down the general scheme of oxidation and reduction:
The number of electrons given is equal to the number of electrons received. Electronic balance is maintained.
IN redox reactions processes of oxidation and reduction occur, which means the oxidation states of chemical elements change. This hallmark redox reactions.
Redox reactions are reactions in which chemical elements change their oxidation state
Let's look at specific examples of how to distinguish a redox reaction from other reactions.
1. NaOH + HCl = NaCl + H 2 O
In order to say whether a reaction is redox, it is necessary to assign the values of the oxidation states of atoms of chemical elements.
1-2+1 +1-1 +1 -1 +1 -2
1. NaOH + HCl = NaCl + H 2 O
Please note that the oxidation states of all chemical elements to the left and right of the equal sign remain unchanged. This means that this reaction is not redox.
4 +1 0 +4 -2 +1 -2
2. CH 4 + 2O 2 = CO 2 + 2H 2 O
As a result of this reaction, the oxidation states of carbon and oxygen changed. Moreover, carbon increased its oxidation state, and oxygen decreased. Let's write down the oxidation and reduction schemes:
C -8e = C - oxidation process
О +2е = О - recovery process
So that the number of electrons given is equal to the number of electrons received, i.e. complied with electronic balance, it is necessary to multiply the second half-reaction by a factor of 4:
C -8e = C - reducing agent, oxidizes
O +2e = O 4 oxidizing agent, reduced
During the reaction, the oxidizing agent accepts electrons, lowering its oxidation state, and it is reduced.
The reducing agent gives up electrons during the reaction, increasing its oxidation state, it is oxidized.
1. Mikityuk A.D. Collection of problems and exercises in chemistry. 8-11 grades / A.D. Mikityuk. - M.: Publishing house. "Exam", 2009. (p.67)
2. Orzhekovsky P.A. Chemistry: 9th grade: textbook. for general education establishment / P.A. Orzhekovsky, L.M. Meshcheryakova, L.S. Pontak. - M.: AST: Astrel, 2007. (§22)
3. Rudzitis G.E. Chemistry: inorganic. chemistry. Organ. chemistry: textbook. for 9th grade. / G.E. Rudzitis, F.G. Feldman. - M.: Education, OJSC “Moscow Textbooks”, 2009. (§5)
4. Khomchenko I.D. Collection of problems and exercises in chemistry for high school. - M.: RIA “New Wave”: Publisher Umerenkov, 2008. (p.54-55)
5. Encyclopedia for children. Volume 17. Chemistry / Chapter. ed. V.A. Volodin, Ved. scientific ed. I. Leenson. - M.: Avanta+, 2003. (p. 70-77)
Additional web resources
1. Single collection of digital educational resources(video experiments on the topic) ().
2. A unified collection of digital educational resources (interactive tasks on the topic) ().
3. Electronic version magazine "Chemistry and Life" ().
Homework
1. No. 10.40 - 10.42 from the “Collection of problems and exercises in chemistry for high school” by I.G. Khomchenko, 2nd ed., 2008
2. Participation in the reaction of simple substances is a sure sign of a redox reaction. Explain why. Write the equations for the reactions of compound, substitution and decomposition involving oxygen O 2 .
Alexandrova Anfisa Mikhailovna
Chemistry teacher
Municipal educational institution "Privolzhskaya average" secondary school» Volzhsky district RME
Topic: “Oxidation-reduction reactions”
Lesson type: lesson – generalization and repetition of material with a combination of frontal, paired and individual work by students.
Lesson type- explanatory and illustrative.
Methods and methodological techniques. Verbal-visual and demonstration-practical. Independent work to find the correct answers, discussion of the selected answer, laboratory experiment, followed by writing reaction equations, discussion of the results of the work.
Target: deepening knowledge on compiling OVR equations using the electronic balance method.
Lesson objectives:
Educational: repeat the basic concepts about the processes of oxidation and reduction, the degree of oxidation, oxidizing agents and reducing agents, consider the essence of redox reactions, develop skills in drawing up equations of chemical reactions occurring in different environments electronic balance method.
Educational: contribute to the formation and development of students' cognitive interest in the subject, promote the development of students' speech, the formation of the ability to analyze, compare, and generalize knowledge on the topic.
Educational: fostering a conscious need for knowledge, improving the ability to listen to the opinions of each member of the team.
Reagents: solutions of potassium permanganate, sulfuric acid, sodium sulfite, water.
Equipment: pipettes, test tubes.
Lesson plan:
I. Updating knowledge.
V. Homework.
VI. Reflection and summing up.
Lesson motto: “Someone loses, but someone finds...”
I. Updating knowledge.
Conversation on previously studied material.
1) What reactions are called redox?
Oxidation-reduction reactions are reactions in which electrons transfer from one atom, molecule or ion to another.
2) What is the oxidation process?
Oxidation is the process of losing electrons and increasing the oxidation state.
3) What process is called recovery?
Reduction is the process of adding electrons, and the oxidation state decreases.
4) What are the names of particles that donate electrons?
Atoms, molecules or ions that donate electrons become oxidized; are reducing agents.
5) What are the particles that accept electrons called?
Atoms, ions, or molecules that accept electrons are reduced; are oxidizing agents.
6) What is “oxidation state”?
The oxidation state is the conditional charge of an atom in a molecule, calculated under the assumption that the molecule consists of ions and is generally electrically neutral (the conditional charge of an atom that we assign to it in the case of accepting or losing electrons).
7) What method of composing the equation of redox reactions do you know? What rule underlies this method?
Independent work of students at the board using cards (with further discussion).
1. Determine the valence and oxidation states of elements in following connections:
CH 4, Cl 2, CO 2, NH 3, C 2 H 4, CH 3 COOH, V 2 O 5, Na 2 B 4 O 7, KClO 4, K 2 HPO 4, Na 2 Cr 2 O 7.
Answer: You can use Appendix 1 to complete the task.
IV I I IV II III I IV I IV I IV II II I V II
C -4 H +1 4, Cl 0 2, C +4 O -2 2, N -3 H +1 3, C -2 2 H +1 4, C -3 H +1 3 C +3 O -2 O -2 H +1, V +5 2 O -2 5 ,
I VII II I I V II I VI II
K +1 Cl +7 O -2 4 , K +1 2 H +1 P +5 O -2 4 , Na +1 2 Cr +6 2 O -2 7 .
2. In which of the reaction equations below does MnO 2 exhibit the properties of an oxidizing agent, and in which does it exhibit the properties of a reducing agent?
A ) 2MnO 2 + 2H 2 SO 4 2MnSO 4 + O 2 + 2H 2 O;
b ) 2MnO 2 + O 2 + 4KOH 2K 2 MnO 4 + 2H 2 O;
V ) MnO 2 + H 2 = MnO + H 2 O;
G ) 2MnO 2 + 3NaBiO 3 + 6HNO 3 = 2HMnO 4 + 3BiONO 3 + 3NaNO 3 + 2H 2 O
Answer:
The oxidizing agent accepts electrons and the oxidation state decreases, which means that in cases A And V MnO 2 is an oxidizing agent. The reducing agent gives up electrons and the oxidation state increases, which means that in cases b And G MnO 2 is a reducing agent.
II. Motivation and goal setting.
Redox reactions are extremely common. They are associated, for example, with the processes of respiration and metabolism 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 and acids are obtained, as well as many other valuable products. Redox reactions underlie the conversion of chemical energy into electrical energy in galvanic and fuel cells.
Problem: I was preparing a solution of potassium permanganate (“potassium permanganate”) for the 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.
III. Development and expansion of knowledge.
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.
Usually, to create an acidic environment, add sulfuric acid. 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. Note how the color of the solution changes in each test tube.
Results of laboratory experiment:
KMnO reduction products 4 (MnO 4 -):
in an acidic environment – Mn +2 (salt), colorless solution;
in a neutral environment – MnO 2, brown precipitate;
in an alkaline medium - MnO 4 2-, green solution.
Exercise . Reaction diagrams:
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 + KOH → K 2 MnO 4 + Na 2 SO 4 + H 2 O
The task is multi-level: strong students write down the reaction products on their own:
KMnO 4 + Na 2 SO 3 + H 2 SO 4 →
KMnO 4 + Na 2 SO 3 + H 2 O →
KMnO 4 + Na 2 SO 3 + KOH →
Select the coefficients using the electronic balance method using the algorithm (Appendix 1). Specify the oxidizing agent and the reducing agent.
Answer:
2KMnO 4 + 5Na 2 SO 3 + 3H 2 SO 4 → 2MnSO 4 + 5Na 2 SO 4 + K 2 SO 4 + 3H 2 O
2KMnO 4 + 3Na 2 SO 3 + H 2 O → 2MnO 2 ↓ + 3Na 2 SO 4 + 2KOH
2KMnO 4 + Na 2 SO 3 + 2KOH → 2K 2 MnO 4 + Na 2 SO 4 + H 2 O
You have done a laboratory experiment, suggest a substance that can be used to clean the gown.
The following diagrams show the reaction products. Specify the reagents, write reaction equations using the electron balance method:
(students work in pairs)
a) KI + KMnO 4 + . . . ->MnSO 4 + I 2 + K 2 SO 4 + H2O
Answer: because as a result of the reaction, Mn +2 is obtained, therefore the process takes place in an acidic environment with the participation of sulfuric acid and potassium sulfate is formed.
10KI + 2 KMnO 4 + 8H 2 SO 4 = 2MnSO 4 + 5I 2 + 6K 2 SO 4 +8H 2 O
2I -1 -2e -> I 2 0 5 - oxidation, reducing agent
Mn +7 + 5e -> Mn +2 2- reduction, oxidizing agent
b ) NaI + KMnO 4 + . . . -> I 2 + K 2 MnO 4 + NaOH
Answer: because as a result of the reaction K 2 MnO 4 is obtained, therefore the process takes place in an alkaline environment with the participation of potassium hydroxide
2NaI + 2 KMnO 4 + 2KOH = I 2 + 2K 2 MnO 4 + 2NaOH
2I -1 -2e -> I 2 0 1- oxidation, reducing agent
Mn +7 + 1e -> Mn +6 2- reduction, oxidizing agent
V ) . . . + KMnO 4 + H 2 O -> NaNO 3 + MnO 2 + KOH
Answer: in this reaction the oxidizing agent KMnO 4 is known, it is easy to assume that sodium nitrite, where N +3, is reduced to nitrate:
3 NaNO 2 + 2 KMnO 4 + H 2 O = 3NaNO 3 + 2MnO 2 + 2KOH,
N +3 – 2e -> N +5 3 - oxidation, reducing agent
Mn +7 + 3e -> Mn +4 2 - reduction, oxidizing agent
In addition to potassium permanganate, other substances also have oxidizing properties. You can get acquainted with them in Appendix 2.
1) H2SO4 (diluted), oxidizing agent H +1
The product of reduction by a metal in the voltage series up to hydrogen is H2.
For example,
H 2 SO 4 (diluted) + Zn -> ZnSO 4 + H 2,
H 2 SO 4 (diluted) + Cu does not react.
2) H 2 SO 4 (concentrated), oxidizing agent S +6
Depending on the activity of the metal, the reduction products of concentrated H 2 SO 4 different: H 2 S; S; SO 2 . The reduction product also depends on the concentration acids (Table 18, page 250 of the textbook).
3) HNO 3, oxidizing agent N +5 (Table 18 p. 250 of the textbook).
Concentrated HNO 3 passivates metals such as Fe, Cr, Al, which is associated with the formation of a thin but very dense oxide film on the surface of these metals.
Au and Pt 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.
For example:
Au + 3HCl (conc.) + HNO 3 (conc.) = AuCl 3 + NO + 2H 2 O.
4) K 2 C r 2 O 7 in an acidic environment is reduced to Cr 3+
in a neutral environment to Cr 2 O 3
in an alkaline environment up to CrO 4 2-
Redox reactions in organic chemistry is associated either with the formation of oxygen bonds or with the elimination of hydrogen.
Rule for the formation of bonds: - OH → -1e
O → -2е
abstraction of 1 atom H → -1e
I V. Consolidation of the studied material.
To reinforce the material covered, I offer test assignments.
Option 1
1. Which non-metal is a strong oxidizing agent?
1) fluorine 2) sulfur 3) ozone 4) silicon
2. The degree of oxidation of sulfur in potassium sulfate is equal to
1)+6 2)+4 3)0 4)-2
3. In which of the following reactions does the chlorine atom act as a reducing agent?
1) Cu + Cl 2 = CuCl 2
2) HCl + NaOH = NaCl + H 2 O
3) HCl + MnO 2 = MnCl 2 + Cl 2 + H 2 O
4) Cl 2 + H 2 = HCl
5) +2 → 0
6) 0 → - 1
5. Using the electronic balance method, create a reaction equation:
PbS + H 2 O 2 →PbSO 4 + H 2 O
6. Using the electron balance method, create a reaction equation:
KBr + KMnO 4 + H 2 SO 4 → …….. + Br 2 + K 2 SO 4 + H 2 O
Identify the oxidizing agent and the reducing agent.
Answer: 1-1; 2-1; 3-3; 4-A3, B4, B2, G5.
Option 2
1. In which of the following compounds the sulfur atom is in the oxidation state +6
1) FeSO 3 2) S 3) SO 2 4) K 2 SO 4
2. Which element is reduced in the reaction Fe 2 O 3 + CO = Fe + CO 2
1) iron 2) oxygen 3) carbon
3. Select the reaction equation in which the element carbon is the oxidizing agent.
1)2 C + O 2 = 2CO
2) CO 2 + 2Mg = 2MgO + C
3) CH 4 + 2O 2 = CO 2 + 2H 2 O
4) C + 2H 2 SO 4 = CO 2 + 2H 2 O + 2SO 2
4. Establish a correspondence between the reaction equation and the change in the oxidation state of the oxidizing agent in this reaction:
Reaction equation Change in oxidation state of oxidizing agent
A) S O 2 + N O 2 = S O 3 +NO 1) -1 → 0
B) 2NH 3 + 2Na = 2NaNH 2 + H 2 2) 0 → -2
B) 4N O 2 + O 2 + 2H 2 O = 4HN O 3 3) +4 → +2
D) 4NH 3 + 6NO = 5N 2 + 6H 2 O 4) +1 → 0
5) +2 → 0
6) 0 → - 1
5. Using the electronic balance method, create an equation for the reaction:
NaNO 2 + NH 4 Cl → NaCl + 2H 2 O + N 2
Identify the oxidizing agent and the reducing agent.
6. Using the electronic balance method, create a reaction equation:
KI+H 2 SO 4 + NaNO 2 → …… + K 2 SO 4 +Na 2 SO 4 + NO + H 2 O
Identify the oxidizing agent and the reducing agent.
Answer: 1-4; 2-1; 3-2; 4-A3, B4, B2, G5.
V. Homework.
1. Complete the reaction equations and arrange the coefficients using the electronic balance method:
1. K 2 Cr 2 O 7 + KNO 2 + …….→ KNO 3 + Cr 2 (SO 4 ) 3 + …..+H 2 O
2. C 6 H 5 -CH 3 + KMnO 4 + H 2 SO 4 → C 6 H 5 COOH +….+….+…..
3. C 2 H 5 OH + K 2 Cr 2 O 7 + H 2 SO 4 → CH 3 COOH +….+….+…..
4.Na 2 SO 3 + K 2 Cr 2 O 7 + H 2 SO 4 → ….+….+….+…..
2. Make up an equation for the oxidation of formaldehyde with a solution of potassium permanganate acidified with sulfuric acid, taking into account that formaldehyde is oxidized to CO 2, select the coefficients using the electronic balance method. 2
connections is usually equal to - 2, except H 2 O 2 -1 and ОF 2.
4. The oxidation state of the hydrogen atom in
connections is usually +1, except MeH (hydrides).
5.Algebraic sum of oxidation states
elements in connections is 0.
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