Chemistry lesson on the topic "Redox Reactions" (Grade 9). Examples of redox reactions with solution. OVR: schemes Task book on general and inorganic chemistry

Target: development of skills and abilities in compiling equations of redox processes involving organic compounds.

Methods: story, work with a presentation, discussion, independent work, teamwork.

Teacher:

What are redox reactions in terms of the concept of "oxidation state of chemical elements"? (slide 2)

/ Redox reactions are reactions in which oxidation and reduction processes occur simultaneously and, as a rule, the oxidation states of elements change. /

Consider the process using the example of the interaction of acetaldehyde with concentrated sulfuric acid:

When compiling this equation, the electronic balance method is used. The method is based on a comparison of the oxidation states of atoms in the initial substances and reaction products. The main requirement for formulating equations by this method is that the number of given electrons must be equal to the number of received electrons.

Redox reactions are reactions in which electrons are transferred from one atom, molecule, or ion to another.

Oxidation is the process of donating electrons, which increases the oxidation state.

Reduction is the process of adding electrons, and the oxidation state decreases.

Atoms, molecules or ions that donate electrons are oxidized; are restorers.
Atoms, ions or molecules that accept electrons are reduced; are oxidizing agents.

Oxidation is always accompanied by reduction, reduction is associated with oxidation.

Redox reactions are the unity of two opposite processes: oxidation and reduction.

Independent work No. 2 on the instructive map: use the electronic balance method to find and set 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)

Teacher:

However, learning to find the coefficients in the OVR does not mean being able to compose them. It is necessary to know the behavior of substances in an OVR, to foresee the course of reactions, and to determine the composition of the resulting products depending on the reaction conditions.

In order to understand in which cases the elements behave as oxidizing agents, and in which cases as reducing agents, one must refer to the periodic system of D.I. Mendeleev. If we are talking about simple substances, then the reducing properties should be inherent in those elements that have a larger atomic radius compared to the rest 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 strongest reducing properties of them are alkali and alkaline earth metals located in the main subgroups of groups I and II (for example, sodium, potassium, calcium, etc.).

The most typical non-metals, which have a structure of the outer electron layer close to completion and a much 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 of groups VI-VII, for example, fluorine, chlorine, bromine, oxygen, sulfur, etc.

At the same time, it must be remembered that the division of simple substances into oxidizing and reducing agents is just as relative as the division into metals and non-metals. If non-metals enter an environment where a stronger oxidizing agent is present, then they can exhibit reducing properties. Elements in different oxidation states 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 the reducing agent can be an element in the lowest oxidation state. For example, in N -3 H 3 nitrogen in the -3 state can donate electrons, i.e. is a restorer.

Elements in intermediate positive oxidation states can both donate and accept electrons and are therefore capable of behaving as oxidizing or reducing agents depending on the conditions. For example, N +3 , S +4 . Getting into an environment with a strong oxidizing agent, they behave like reducing agents. Conversely, in a reducing environment, they behave as oxidizing agents.

According to the redox properties of the substance can be divided into three groups:

oxidizers

reducing agents

oxidizing agents - reducing agents

Independent work No. 3 on the instructive map: in which of the above schemes of reaction equations does MnO 2 exhibit the properties of an oxidizing agent, and in which - the properties of a reducing agent:

2MnO 2 + O 2 + 4KOH \u003d 2K 2 MnO 4 + 2H 2 O (MnO 2 is a reducing agent)

MnO 2 + 4HCI \u003d MnCI 2 + CI 2 + 2H 2 O (MnO 2 is an oxidizing agent)

The most important oxidizing agents and their reduction products

1. Sulfuric acid - H 2 SO 4 is an oxidizing agent

A) The 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 reduction product of a metal in the voltage series up to hydrogen is H2.

B) Consider another reaction - the interaction of zinc with concentrated H 2 SO 4 (slide 4)

Which atoms change their oxidation state? (zinc and sulfur)

Concentrated sulfuric acid (98%) contains 2% water, and the salt is obtained in solution. Sulfate ions are actually involved in the reaction. The recovery product is hydrogen sulfide.

Depending on the activity of the metal, the products of the reduction of concentrated H 2 SO 4 are different: H 2 S, S, SO 2.

2. Another acid - nitric acid - is also an oxidizing agent due to the nitrate - NO 3 - ion. The oxidizing power of the nitrate ion is much higher than the H+ ion, and the hydrogen ion is not reduced to an atom, therefore, when nitric acid interacts with metals, hydrogen is never released, but various nitrogen compounds are formed. It depends on the concentration of the acid and the activity of the metal. Dilute nitric acid is reduced deeper than concentrated (for the same metal) (slide 6)

The diagrams indicate the products, the content of which is maximum among the possible products of acid reduction.

Gold and platinum do not react with HNO3, 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 it is difficult to obtain it in a free form. Therefore, in laboratories, potassium permanganate KMnO 4 is used as an oxidizing agent. Its oxidizing ability depends on the concentration of the solution, temperature and medium.

Creating a problem situation: I was preparing a solution of potassium permanganate (“potassium permanganate”) for the lesson, spilled a glass of solution and stained my favorite chemical coat. Suggest (after doing a laboratory experiment) a substance with which you can clean the dressing gown.

Oxidation-reduction reactions can take place in various media. Depending on the environment, the nature of the reaction between the same substances can change: the environment affects the change in the oxidation states of atoms.

Sulfuric acid is usually added to create an acidic environment. Salt and nitrogen are used less frequently, because. the first is capable of being oxidized, and the second is itself a strong oxidizing agent and can cause side processes. Potassium or sodium hydroxide is used to create an alkaline environment, and water is used to create a neutral one.

Laboratory experience: (TB rules)

Four numbered tubes are filled with 1-2 ml of dilute potassium permanganate solution. Add a few drops of sulfuric acid solution to the first tube, water to the second, potassium hydroxide to the third, leave the fourth tube as a control. Then pour the sodium sulfite solution into the first three test tubes, shaking gently. Note. How does the color of the solution change in each test tube. (slides 7, 8)

Results of the laboratory experiment:

Recovery products KMnO 4 (MnO 4) -:

in an acidic environment - Mn + 2 (salt), colorless solution;

in a neutral medium - MnO 2, brown precipitate;

in an alkaline medium - MnO 4 2-, green solution. (slide 9,)

For 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 + KOH → Na 2 SO 4 + K 2 MnO 4 + H 2 O

Select the coefficients 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 on their own)

You have done a laboratory experiment, suggest a substance with which you can clean a dressing gown.

Demo Experience:

Stains from a solution of potassium permanganate are quickly removed with a solution of hydrogen peroxide acidified with acetic acid:

2KMnO 4 + 9H 2 O2 + 6CH 3 COOH = 2Mn(CH 3 COO) 2 + 2CH 3 COOK + 7O 2 + 12H 2 O

Old spots of potassium permanganate contain manganese(IV) oxide, so another reaction will take place:

MnO 2 + 3H 2 O 2 + 2CH 3 COOH = Mn(CH 3 COO) 2 + 2O 2 + 4H 2 O (slide 12)

After removing stains, a piece of cloth must be washed with water.

Teacher:

The value of redox reactions

Purpose: To show students the importance of redox reactions in chemistry, technology, and everyday human life. Methods: work with presentation, discussion, independent work, collective work.

Within the framework of one lesson, it is impossible to consider the whole variety of redox reactions. But their importance in chemistry, technology, and everyday human life cannot be overestimated. Redox reactions underlie the production of metals and alloys, hydrogen and halogens, alkalis and drugs. The functioning of biological membranes, many natural processes are associated with redox reactions: metabolism, fermentation, respiration, photosynthesis. Without understanding the essence and mechanisms of the course of redox reactions, it is impossible to imagine the operation of chemical power sources (accumulators and batteries), the production of protective coatings, and the virtuoso 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 bleach, or whitewash, are used. Chlorine as a strong oxidizing agent is used to sterilize clean water and disinfect wastewater.

Work with the presentation entry in a notebook.

Before giving examples of redox reactions with a solution, let us single out the main definitions associated with these transformations.

Those atoms or ions that, during the interaction, change their oxidation state with a decrease (accept electrons) are called oxidizing agents. Among the substances with such properties, strong inorganic acids can be noted: sulfuric, hydrochloric, nitric.

Oxidizer

Alkali metal permanganates and chromates are also strong oxidizing agents.

The oxidizing agent takes in the course of the reaction that it needs to complete the energy level (establishment of the completed configuration).

Reducing agent

Any redox reaction scheme involves the identification of a reducing agent. It includes ions or neutral atoms that can increase the oxidation state during the interaction (give electrons to other atoms).

Metal atoms can be cited as typical reducing agents.

Processes in OVR

What else is characterized by a change in the oxidation states of the starting substances.

Oxidation involves the process of giving off negative particles. Restoration involves taking them from other atoms (ions).

Parsing algorithm

Examples of redox reactions with a solution are offered in various reference materials designed to prepare high school students for graduate tests in chemistry.

In order to successfully cope with the tasks proposed in the OGE and the USE, it is important to know the algorithm for compiling and analyzing redox processes.

1. First of all, the charge values ​​\u200b\u200bof all the elements in the substances proposed in the scheme are put down.
2. Atoms (ions) are written out from the left side of the reaction, which, during the interaction, changed indicators.
3. With an increase in the degree of oxidation, the sign "-" is used, and with a decrease in "+".
4. Between the given and received electrons, the least common multiple is determined (the number by which they are divided without a remainder).
5. When dividing LCM into electrons, we obtain stereochemical coefficients.
6. We place them in front of the formulas in the equation.

The first example from the OGE

In the ninth grade, not all students know how to solve redox reactions. That is why they make many mistakes, do not get high scores for the OGE. The algorithm of actions is given above, now let's try to work it out on specific examples.

The peculiarity of the tasks concerning the placement of coefficients in the proposed reaction, issued to graduates of the main stage of education, is that both the left and right parts of the equation are given.

This greatly simplifies the task, since there is no need to independently invent interaction products, select the missing starting materials.

For example, it is proposed to use the electronic balance to identify the coefficients in the reaction:

At first glance, this reaction does not require stereochemical coefficients. But, in order to confirm his point of view, it is necessary for all elements to have charge numbers.

In binary compounds, which include copper oxide (2) and iron oxide (2), the sum of the oxidation states is zero, given that for oxygen it is -2, for copper and iron this indicator is +2. Simple substances do not give (do not accept) electrons, therefore they are characterized by a zero value of the oxidation state.

Let's make an electronic balance, showing the sign "+" and "-" the number of received and given in the course of the interaction of electrons.

Fe 0 -2e \u003d Fe 2+.

Since the number of electrons received and given away during the interaction is the same, it makes no sense to find the least common multiple, determine the stereochemical coefficients, put them in the proposed interaction scheme.

In order to get the maximum score for the task, it is necessary not only to write down examples of redox reactions with a solution, but also to write out the formula of the oxidizing agent (CuO) and reducing agent (Fe) separately.

The second example with the OGE

Let us give more examples of redox reactions with a solution that may be encountered by ninth-graders who have chosen chemistry as their final exam.

Suppose it is proposed to arrange the coefficients in the equation:

Na+HCl=NaCl+H2.

In order to cope with the task, it is first important to determine the indicators of oxidation states for each simple and complex substance. For sodium and hydrogen, they will be equal to zero, since they are simple substances.

In hydrochloric acid, hydrogen has a positive, and chlorine has a negative oxidation state. After placing the coefficients, we get the reaction with the coefficients.

The first of the exam

How to supplement redox reactions? Examples with a solution found in the USE (Grade 11) involve the addition of gaps, as well as the placement of coefficients.

For example, you need to supplement the reaction with an electronic balance:

H 2 S+ HMnO 4 = S+ MnO 2 +…

Determine the reducing agent and oxidizing agent in the proposed scheme.

How to learn to compose redox reactions? The sample assumes the use of a specific algorithm.

First, in all substances given by the condition of the problem, it is necessary to set the oxidation states.

Next, you need to analyze which substance can become an unknown product in this process. Since an oxidizing agent is present here (manganese plays its role), a reducing agent (it is sulfur), the oxidation states do not change in the desired product, therefore, it is water.

Arguing about how to correctly solve redox reactions, we note that the next step will be to draw up an electronic ratio:

Mn +7 takes 3 e= Mn +4 ;

S -2 gives 2e= S 0 .

The manganese cation is a reducing agent, while the sulfur anion is a typical oxidizing agent. Since the smallest multiple between the received and given electrons will be 6, we get the coefficients: 2, 3.

The last step will be setting the coefficients in the original equation.

3H 2 S+ 2HMnO 4 = 3S+ 2MnO 2 + 4H 2 O.

The second sample of the OVR in the exam

How to write redox reactions correctly? Examples with a solution will help to work out the algorithm of actions.

It is proposed to use the electronic balance method to fill in the gaps in the reaction:

PH 3 + HMnO 4 = MnO 2 +…+…

We arrange the oxidation states of all elements. In this process, the oxidizing properties are manifested by manganese, which is part of the composition and the reducing agent should be phosphorus, changing its oxidation state to positive in phosphoric acid.

According to the assumption made, we obtain the reaction scheme, then we compose the electronic balance equation.

P -3 gives 8 e and turns into P +5 ;

Mn +7 takes 3e, going to Mn +4 .

The LCM will be 24, so phosphorus should have a stereometric coefficient of 3, and manganese -8.

We put the coefficients in the resulting process, we get:

3 PH 3 + 8 HMnO 4 = 8 MnO 2 + 4H 2 O+ 3 H 3 PO 4 .

The third example from the exam

Using the electron-ion balance, you need to compose a reaction, indicate the reducing agent and oxidizing agent.

KMnO 4 + MnSO 4 +…= MnO 2 +…+ H2SO 4 .

According to the algorithm, we place oxidation states for each element. Next, we determine those substances that are omitted in the right and left parts of the process. A reducing agent and an oxidizing agent are given here, so oxidation states do not change in the omitted compounds. The lost product will be water, and the starting compound will be potassium sulfate. We get the reaction scheme for which we will make an electronic balance.

Mn +2 -2 e= Mn +4 3 reducing agent;

Mn +7 +3e= Mn +4 2 oxidizing agent.

We write the coefficients in the equation, summing up the manganese atoms on the right side of the process, since it belongs to the disproportionation process.

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

Conclusion

Redox reactions are of particular importance for the functioning of living organisms. Examples of OVR are the processes of putrefaction, fermentation, nervous activity, respiration, and metabolism.

Oxidation and reduction are relevant for the metallurgical and chemical industries, thanks to such processes, metals can be restored from their compounds, protected from chemical corrosion, and processed.

To draw up a redox process in organic or it is necessary to use a certain algorithm of actions. First, in the proposed scheme, the oxidation states are arranged, then those elements that increased (lowered) the indicator are determined, and the electronic balance is recorded.

If you follow the sequence of actions proposed above, you can easily cope with the tasks offered in the tests.

In addition to the electronic balance method, the placement of coefficients is also possible by compiling half-reactions.

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

Topic: Redox reactions

Lesson: Redox reactions

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

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

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

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

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

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

We write the general scheme of oxidation and reduction:

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

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

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

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

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

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

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

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

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

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

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

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

C -8e \u003d C - oxidation process

O + 2e \u003d O - recovery process

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

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

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

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

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

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

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

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

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

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

Additional web resources

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

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

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

Homework

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

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

Task book on general and inorganic chemistry

2.2. Redox reactions

See tasks >>>

Theoretical part

Redox reactions include chemical reactions that are accompanied by a change in the oxidation states of elements. In the equations of such reactions, the selection of coefficients is carried out by compiling electronic balance. The method of selecting coefficients using the electronic balance consists of the following steps:

a) write down the formulas of the reactants and products, and then find the elements that increase and decrease their oxidation states, and write them out separately:

MnCO 3 + KClO 3 ® MnO2+ KCl + CO2

Cl V¼ = Cl - I

MnII¼ = MnIV

b) compose the equations of half-reactions of reduction and oxidation, observing the laws of conservation of the number of atoms and charge in each half-reaction:

half reaction recovery Cl V + 6 e - = Cl - I

half reaction oxidation MnII- 2 e - = MnIV

c) select additional factors for the equation of half-reactions so that the law of conservation of charge is fulfilled for the reaction as a whole, for which the number of accepted electrons in the reduction half-reactions is made equal to the number of electrons donated in the oxidation half-reaction:

Cl V + 6 e - = Cl - I 1

MnII- 2 e - = Mn IV 3

d) put down (according to the factors found) stoichiometric coefficients in the reaction scheme (coefficient 1 is omitted):

3 MnCO 3 + KClO 3 = 3 MNO 2 + KCl+CO2

d) equalize the number of atoms of those elements that do not change their oxidation state during the course of the reaction (if there are two such elements, then it is enough to equalize the number of atoms of one of them, and check the second one). Get the equation of the chemical reaction:

3 MnCO 3 + KClO 3 = 3 MNO 2 + KCl+ 3CO2

Example 3. Fit Coefficients in Redox Equation

Fe 2 O 3 + CO ® Fe + CO2

Solution

Fe 2 O 3 + 3 CO \u003d 2 Fe + 3 CO 2

FeIII + 3 e - = Fe 0 2

CII - 2 e - = C IV 3

With the simultaneous oxidation (or reduction) of atoms of two elements of one substance, the calculation is carried out for one formula unit of this substance.

Example 4 Fit Coefficients in Redox Equation

Fe(S ) 2 + O 2 = Fe 2 O 3 + SO 2

Solution

4 Fe(S ) 2 + 11 O 2 = 2 Fe 2 O 3 + 8 SO 2

FeII- e - = FeIII

- 11 e - 4

2S - I - 10 e - = 2SIV

O 2 0 + 4 e - = 2O - II + 4 e - 11

In examples 3 and 4, the functions of the oxidizing and reducing agent are divided between different substances, Fe 2 O 3 and O 2 - oxidizing agents, CO and Fe(S)2 - reducing agents; such reactions are intermolecular redox reactions.

When intramolecular oxidation-reduction, when in the same substance the atoms of one element are oxidized, and the atoms of another element are reduced, the calculation is carried out for one formula unit of the substance.

Example 5 Find the coefficients in the equation of the redox reaction

(NH 4) 2 CrO 4 ® Cr 2 O 3 + N 2 + H 2 O + NH 3

Solution

2 (NH 4) 2 CrO 4 \u003d Cr 2 O 3 + N 2 +5 H 2 O + 2 NH 3

Cr VI + 3 e - = Cr III 2

2N - III - 6 e - = N 2 0 1

For reactions dismutations (disproportionation, autoxidation- self-healing), in which the atoms of the same element in the reagent are oxidized and reduced, additional factors are put down first on the right side of the equation, and then the coefficient for the reagent is found.

Example 6. Fit Coefficients in Dismutation Reaction Equation

H2O2 ® H 2 O + O 2

Solution

2 H 2 O 2 \u003d 2 H 2 O + O 2

O - I+ e - =O - II 2

2O - I - 2 e - = O 2 0 1

For the commutation reaction ( synproportionation), in which the atoms of the same element of different reagents, as a result of their oxidation and reduction, receive the same oxidation state, additional factors are put down first on the left side of the equation.

Example 7 Select the coefficients in the commutation reaction equation:

H 2 S + SO 2 \u003d S + H 2 O

Solution

2 H 2 S + SO 2 \u003d 3 S + 2H 2 O

S - II - 2 e - = S 0 2

SIV+4 e - = S 0 1

To select coefficients in the equations of redox reactions occurring in an aqueous solution with the participation of ions, the method is used electron-ion balance. The method of selection of coefficients using the electron-ion balance consists of the following steps:

a) write down the formulas of the reagents of this redox reaction

K 2 Cr 2 O 7 + H 2 SO 4 + H 2 S

and establish the chemical function of each of them (here K2Cr2O7 - oxidizing agent, H 2 SO 4 - acid reaction medium, H 2 S - reducing agent);

b) write down (on the next line) the formulas of the reagents in ionic form, indicating only those ions (for strong electrolytes), molecules (for weak electrolytes and gases) and formula units (for solids) that will take part in the reaction as an oxidizing agent ( Cr2O72 - ), environments ( H+- more precisely, the oxonium cation H3O+ ) and reducing agent ( H2S):

Cr2O72 - + H + + H 2 S

c) determine the reduced formula of the oxidizing agent and the oxidized form of the reducing agent, which must be known or specified (for example, here the dichromate ion passes chromium cations ( III), and hydrogen sulfide - into sulfur); these data are written on the next two lines, the electron-ion equations of the reduction and oxidation half-reactions are compiled, and additional factors are selected for the half-reaction equations:

half reaction reduction of Cr 2 O 7 2 - + 14 H + + 6 e - \u003d 2 Cr 3+ + 7 H 2 O 1

half reaction H 2 S oxidation - 2 e - = S(t) + 2H + 3

d) by summing up the equations of half-reactions, they compose the ionic equation of this reaction, i.e. supplement entry (b):

Cr2O72 - + 8 H + + 3 H 2 S = 2 Cr 3+ + 7 H 2 O + 3 S ( T )

d) on the basis of the ionic equation make up the molecular equation of this reaction, i.e. supplement the entry (a), and the formulas of cations and anions that are absent in the ionic equation are grouped into formulas of additional products ( K2SO4):

K 2 Cr 2 O 7 + 4H 2 SO 4 + 3H 2 S \u003d Cr 2 (SO 4) 3 + 7H 2 O + 3S ( m) + K 2 SO 4

f) check the selected coefficients by the number of atoms of elements in the left and right parts of the equation (usually it is enough to check only the number of oxygen atoms).

oxidizedAnd restored forms of oxidizing and reducing agent often differ in oxygen content (compare Cr2O72 - and Cr3+ ). Therefore, when compiling half-reaction equations using the electron-ion balance method, they include H + / H 2 O pairs (for an acidic environment) and OH - / H 2 O (for an alkaline environment). If during the transition from one form to another, the original form (usually - oxidized) loses its oxide ions (shown below in square brackets), then the latter, since they do not exist in a free form, must be combined with hydrogen cations in an acidic environment, and in an alkaline environment - with water molecules, which leads to the formation of water molecules (in an acidic environment) and hydroxide ions (in an alkaline environment):

acid environment[ O2 - ] + 2 H + = H 2 O

alkaline environment [ O 2 - ] + H 2 O \u003d 2 OH -

Lack of oxide ions in their original form (more often- reduced) in comparison with the final form is compensated by the addition of water molecules (in an acid medium) or hydroxide ions (in an alkaline medium):

acidic environment H 2 O \u003d [ O 2 - ] + 2 H +

alkaline environment2 OH - = [ O 2 - ] + H 2 O

Example 8 Select the coefficients using the electron-ion balance method in the redox reaction equation:

® MnSO 4 + H 2 O + Na 2 SO 4 + ¼

Solution

2 KMnO 4 + 3 H 2 SO 4 + 5 Na 2 SO 3 \u003d

2 MnSO 4 + 3 H 2 O + 5 Na 2 SO 4 + + K 2 SO 4

2 MnO 4 - + 6 H + + 5 SO 3 2 - = 2 Mn 2+ + 3 H 2 O + 5 SO 4 2 -

MnO4 - + 8H + + 5 e - = Mn 2+ + 4 H 2 O2

SO 3 2 - + H2O - 2 e - = SO 4 2 - + 2 H + 5

Example 9. Select the coefficients using the electron-ion balance method in the redox reaction equation:

Na 2 SO 3 + KOH + KMnO 4 ® Na 2 SO 4 + H 2 O + K 2 MnO 4

Solution

Na 2 SO 3 + 2 KOH + 2 KMnO 4 = Na 2 SO 4 + H 2 O + 2 K 2 MnO 4

SO 3 2 - + 2OH - + 2 MnO 4 - = SO 4 2 - + H 2 O + 2 MnO 4 2 -

MnO4 - + 1 e - = MnO 4 2 - 2

SO 3 2 - + 2OH - - 2 e - = SO 4 2 - + H 2 O 1

If the permanganate ion is used as an oxidizing agent in a weakly acidic environment, then the reduction half-reaction equation is:

MnO4 - + 4 H + + 3 e - = MnO 2( m) + 2 H 2 O

and if in a weakly alkaline medium, then

MNO 4 - + 2 H 2 O + 3 e - = MnO 2( m) + 4 OH -

Often, a weakly acidic and weakly alkaline medium is conditionally called neutral, while only water molecules are introduced into the half-reaction equations on the left. In this case, when compiling the equation, one should (after selecting additional factors) write an additional equation that reflects the formation of water from H + and OH ions - .

Example 10. Select the coefficients in the equation for the reaction taking place in a neutral medium:

KMnO 4 + H 2 O + Na 2 SO 3 ® Mn ABOUT 2( t) + Na 2 SO 4 ¼

Solution

2 KMnO 4 + H 2 O + 3 Na 2 SO 3 \u003d 2 MnO 2( t) + 3 Na 2 SO 4 + 2 KOH

MnO4 - + H 2 O + 3 SO 3 2 - = 2 MnO 2( m) + 3 SO 4 2 - + 2 OH -

MNO 4 - + 2 H 2 O + 3 e - = MnO 2( m) + 4 OH -

SO 3 2 - + H2O - 2 e - = SO 4 2 - +2H+

8OH - + 6 H + = 6 H 2 O + 2 OH -

Thus, if the reaction from example 10 is carried out by simply draining aqueous solutions of potassium permanganate and sodium sulfite, then it proceeds in a conditionally neutral (and in fact, in a slightly alkaline) environment due to the formation of potassium hydroxide. If the solution of potassium permanganate is slightly acidified, then the reaction will proceed in a weakly acidic (conditionally neutral) medium.

Example 11. Select the coefficients in the equation for the reaction taking place in a weakly acidic environment:

KMnO 4 + H 2 SO 4 + Na 2 SO 3 ® Mn ABOUT 2( t) + H 2 O + Na 2 SO 4 + ¼

Solution

2KMnO 4 + H 2 SO 4 + 3Na 2 SO 3 \u003d 2Mn O 2( T ) + H 2 O + 3Na 2 SO 4 + K 2 SO 4

2 MnO 4 - + 2 H + + 3 SO 3 2 - = 2 MnO 2( t) + H 2 O + 3 SO 4 2 -

MnO4 - +4H + + 3 e - = Mn O 2( t ) + 2 H 2 O2

SO 3 2 - + H2O - 2 e - = SO 4 2 - + 2 H + 3

Forms of existence of oxidizing agents and reducing agents before and after the reaction, i.e. their oxidized and reduced forms are called redox couples. So, it is known from chemical practice (and this needs to be remembered) that the permanganate ion in an acidic medium forms a manganese cation ( II ) (pair MNO 4 - + H + / Mn 2+ + H 2 O ), in a weakly alkaline medium- manganese(IV) oxide (pair MNO 4 - +H+ ¤ Mn O 2 (t) + H 2 O or MNO 4 - + H 2 O = Mn O 2(t) + OH - ). The composition of oxidized and reduced forms is determined, therefore, by the chemical properties of a given element in various degrees of oxidation, i.e. unequal stability of specific forms in various media of an aqueous solution. All redox pairs used in this section are given in problems 2.15 and 2.16.

What is an OVR? Examples of such reactions can be found not only in inorganic, but also in organic chemistry. In the article we will give definitions to the main terms used in the analysis of such interactions. In addition, we will give some OVR, examples and solutions of chemical equations that will help you understand the algorithm of actions.

Basic definitions

But first, let's recall the basic definitions that will help you understand the process:

• An oxidizing agent is an atom or ion capable of accepting electrons in the process of interaction. In the form of serious oxidizing agents are mineral acids, potassium permanganate.
• A reducing agent is an ion or atom that donates valence electrons to other elements.
• The process of attaching free electrons is called oxidation, and recoil is called reduction.

Action algorithm

How to parse OVR equations? The examples offered to school graduates involve the placement of coefficients by electronic balance. Here is the procedure:

1. First, it is necessary to set for all elements the values ​​of the oxidation states in simple and complex substances participating in the proposed chemical transformation.
2. Next, those elements that have changed the digital value are selected.
3. The signs "+" and "-" indicate the received and given electrons, their number.
4. Further, the least common multiple is determined between them, the coefficients are determined.
5. The resulting numbers are put into the reaction equation.

First example

How do I complete an OVR-related task? The examples offered at the final exams in grade 9 do not imply the addition of formulas of substances. Guys, as a rule, need to determine the coefficients and substances that have changed valency values.

Consider those OVR (reactions), examples of which are offered to graduates of the 11th grade. Schoolchildren must independently supplement the equation with substances and only after that, use the electronic balance to arrange the coefficients:

H 2 O 2 + H 2 SO 4 + KMnO 4 \u003d Mn SO 4 + O 2 + ... + ...

To begin with, we will arrange the oxidation states in each compound. So, in hydrogen peroxide for the first element, it corresponds to +1 , at oxygen -1 . In sulfuric acid, the following indicators: +1, +6, -2 (summing up to zero). Oxygen is a simple substance, so it has a zero oxidation state.

The electronic balance for this interaction has the following form:

• Mn +7 takes 5 e = Mn+2 2, is an oxidizing agent;
• 2I- gives 2e = I 2 0 5, acts as a reducing agent.

At the final stage of this task, we will arrange the coefficients in the finished scheme and get:

2KMnO 4 + 8H 2 SO 4 + 10KI = 2MnSO 4 + 5I 2 + 6K 2 SO 4 + 8H 2 O .

Conclusion

These processes have found serious application in chemical analysis. With their help, you can open and separate various ions, carry out the oxidimetry method.

A variety of physical and chemical methods of analysis are based on OVR. The theory of acidic and basic interactions explains the kinetics of the ongoing processes and makes it possible to carry out quantitative calculations using equations.

In order for schoolchildren who have chosen chemistry for the final exam to successfully pass these tests, it is necessary to work out the algorithm for equalizing the OVR based on electronic balance. Teachers work out with their students the method of arranging the coefficients, using a variety of examples from inorganic and organic chemistry.

Tasks related to determining the oxidation states of chemical elements in simple and complex substances, as well as drawing up a balance between accepted and given electrons, are an obligatory element of examination tests at the main, general level of education. Only in the case of successful completion of such tasks, we can talk about the effective development of the school course in inorganic chemistry, and also count on getting a high mark on the OGE, the Unified State Examination.