Ovr all the rules for the exam. How to solve problems C1 (30) at the exam in chemistry

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Line UMK Kuznetsova. Chemistry (10-11) (U)

Line UMK Kuznetsova. Chemistry (10-11) (B)

Line UMK N. E. Kuznetsova. Chemistry (10-11) (basic)

Organization of preparation for the exam in chemistry: redox reactions

How should the work in the classroom be organized so that students achieve good results in the exam?

The material was prepared based on the materials of the webinar "Organization of preparation for the exam in chemistry: redox reactions"

“We are considering the organization of training for the successful completion of tasks related to redox reactions. If we look at the specification and demo version, then such reactions are directly related to tasks #10 and #30, but this is the key topic of the school chemistry course. It touches on a variety of issues, a variety of properties of chemicals. It is very extensive,” emphasizes Lidia Asanova, webinar host, candidate of pedagogical sciences, author of teaching aids.

Task number 30, considering redox reactions, is a task of a high level of complexity. To receive the highest score (3) for its implementation, the student's answer must be:

determination of the degree of oxidation of elements that are an oxidizing agent and a reducing agent;
oxidizing agent and reducing agent (elements or substances);
the processes of oxidation and reduction, and on their basis the compiled electronic (electron-ionic) balance;
determination of missing substances in the reaction equation.

However, students often skip, do not place the coefficients, do not indicate the oxidizing agent and reducing agent, the degree of oxidation. How to organize the work in the lesson in order to achieve good results in the exam?

Particular attention in the textbook by O. S. Gabrielyan for grade 10, designed to study the subject in the amount of 3-4 hours a week, is given to applied topics: the manual covers issues related to chemistry in ecology, medicine, biology and culture. In the 11th grade, the course is completed and generalized.

1. Preparation for the exam should be carried out in the process of teaching a subject and preparation cannot be reduced only to training in performing tasks similar to tasks. examination work. Such “training” does not develop thinking, does not deepen understanding. But, by the way, in the examination task it is indicated that other wordings of the answer are allowed that do not distort its meaning. This means that creatively, with understanding approaching the solution of the task, you can get the highest score for completion, even if the answer is worded differently.

The main task of preparing for the exam is purposeful work on repetition, systematization and generalization of the studied material, on bringing it into the knowledge system key concepts chemistry course. Of course, experience in conducting a real chemical experiment is required.

2. There is a list of topics and concepts that students should not forget at all. Among them:

rules for determining the oxidation states of atoms (in simple substances, the oxidation state of elements is zero, the highest (maximum) oxidation state of elements of groups II-VII, as a rule, is equal to the number of the group in which the element is located in the periodic table, the lowest (minimum) oxidation state of metals is equal to zero, etc.);
the most important oxidizing and reducing agents, as well as the fact that the oxidation process is always accompanied by a reduction process;
redox duality;
types of OVR (intermolecular, intramolecular, co-proportionation reactions, disproportionation reactions (self-oxidation-self-healing)).

The table lists the types of oxidation- reducing reactions, factors influencing the course of reactions (photo pages). Examples are analyzed in detail, and, in addition, there are tasks on the topic "OVR" in the USE format.

For example:

“Using the electron balance method, write the equation for the chemical reaction:

N 2 O + KMnO 4 + ... = NO 2 + ... + K 2 SO 4 + H 2 O

Specify the oxidizing agent and reducing agent.

However, a variety of examples are given to work out the solution of problems. For example, in the manual “Chemistry. Deep level. Grade 11. Test papers" there are such:

“Based on the theory of redox processes, indicate the schemes of impossible reactions.

SO 2 + H 2 S → S + H 2 O

S + H 2 SO 4 → SO 2 + H 2 O

S + H 2 SO 4 → H 2 S + H 2 O

K 2 SO 3 + K 2 Cr 2 O 7 + H 2 SO 4 → K 2 SO 4 + K 2 CrO 4 + H 2 O

KMnO 4 + HCl → Cl2 + MnCl 2 + KCl + H 2 O

I 2 + SO 2 + H 2 O → HIO 3 + H 2 SO 4

Justify the answer. Convert the schemes of possible processes into reaction equations. Specify the oxidizing agent and reducing agent

"Compose the reaction equations in accordance with the scheme for changing the oxidation states of carbon atoms: C 0 → C - 4 → C -4 → C +4 → C +2 → C -2".

“Substances are given: carbon, nitric oxide (IV), sulfur oxide (IV), aqueous solution of potassium hydroxide. Write the equations for the four possible reactions between these substances, without repeating the pairs of reactants.

All this allows us to fully study the topic of redox reactions and work out the solution of a variety of problems.

*Since May 2017, the DROFA-VENTANA joint publishing group has been part of the Russian Textbook Corporation. The corporation also included the Astrel publishing house and the LECTA digital educational platform. Alexander Brychkin, a graduate of Financial Academy under the Government of the Russian Federation, candidate of economic sciences, head of innovative projects of the DROFA publishing house in the field of digital education (electronic forms of textbooks, Russian Electronic School, digital educational platform LECTA). Prior to joining the DROFA publishing house, he held the position of Vice President for Strategic Development and Investments of the EKSMO-AST publishing holding. Today, the Russian Textbook Publishing Corporation has the largest portfolio of textbooks included in the Federal List - 485 titles (approximately 40%, excluding textbooks for remedial school). The corporation's publishing houses own the sets of textbooks in physics, drawing, biology, chemistry, technology, geography, astronomy, most in demand by Russian schools - the areas of knowledge that are needed to develop the country's production potential. The corporation's portfolio includes textbooks and study guides for elementary school awarded the Presidential Prize in Education. These are textbooks and manuals on subject areas that are necessary for the development of the scientific, technical and industrial potential of Russia.

Redox reactions. Corrosion of metals and methods of protection against it

Signs of redox reactions

The variety of classifications of chemical reactions according to various criteria (number and nature of reacting and formed substances, direction, phase composition, thermal effect, use of a catalyst) can be supplemented with one more feature. This sign is a change in the oxidation state of atoms chemical elements that form reactants.

For example, in the reaction

$(Ag)↖(+1)(N)↖(+5)(O_3)↖(-2)+(H)↖(+1)(Cl)↖(-1)=(Ag)↖(+1 )(Cl)↖(-1)+(H)↖(+1)(N)↖(+5)(O_3)↖(-2)$

the oxidation states of atoms of chemical elements did not change after the reaction. But in the reaction of the interaction of hydrochloric acid with zinc

$2(H)↖(+1)(Cl)↖(-1)+(Zn)↖(0)=(Zn)↖(+2)(Cl_2)↖(-1)+(H_2)↖(0) $

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:

$2H^(+)+2e↖(-)→H_2^0,$

and each zinc atom donated two electrons:

$(Zn)↖(0)-2e↖(-)→Zn^(+2).$

Chemical reactions, as a result of which there is a change in the oxidation states of atoms of chemical elements or ions that form reactants, are called redox reactions.

Oxidizing agent and reducing agent. Oxidation and reduction

Reduction is understood as the process of adding electrons to atoms, ions or molecules.

In this case, the degree of oxidation decreases.

For example, non-metal atoms can add electrons, thus turning into negative ions, i.e. recovering:

$(Cl^0+1ē)↙(\text"chlorine atom")→(Cl^(-1))↙(\text"chloride ion"),$

$(S^(0)+2ē)↙(\text"sulfur atom")→(S^(-2))↙(\text"chloride ion").$

Electrons can also attach to positive ions, turning them into atoms:

$(Cu^(+2)+2ē)↙(\text"copper(II) ion")→(Cu^0)↙(\text"copper atom"),$

$(Fe^(+3)+3ē)↙(\text"iron(III) ion")→(Fe^(0))↙(\text"iron atom").$

Positive ions can also accept electrons, in which the oxidation state decreases:

$(Fe^(+3)+1ē)↙(\text"iron(III) ion")→(Fe^(+2))↙(\text"iron(III) ion"),$

$(Sn^(+4)+2ē)↙(\text"tin(IV) ion")→(Sn^(+2))↙(\text"tin(II) ion").$

Atoms, ions or molecules that accept electrons are called oxidizing agents.

Oxidation is understood as the process of donating electrons by atoms, ions or molecules.

For example, metal atoms, losing electrons, turn into positive ions, i.e. oxidized:

$(Na^(0)-1ē)↙(\text"sodium atom")→(Na^(+1))↙(\text"sodium ion"),$

$(Al^(0)-3ē)↙(\text"aluminum atom")→(Al^(+3))↙(\text"aluminum ion").$

Negative ions can donate their electrons:

$(Cl^(-1)-1ē)↙(\text"chloride ion")→(Cl^(0))↙(\text"chlorine atom"),$

$(S^(-2)-2ē)↙(\text"sulphide ion")→(S^(0))↙(\text"sulfur atom").$

Some positive ions with lower oxidation states can also lose electrons:

$(Cu^(+1)-1ē)↙(\text"copper(I) ion")→(Cu^(+2))↙(\text"copper(II) ion"),$

$(Fe^(+2)-1ē)↙(\text"iron(II) ion")→(Fe^(+3))↙(\text"iron(III) ion").$

It can be noted that in this case the degree of oxidation increases.

Atoms, ions or molecules that donate electrons are called reducing agents.

Oxidation is always accompanied by reduction and vice versa, i.e. redox reactions are a unity of two opposite processes - oxidation and reduction. The scheme of the relationship between the change in oxidation states and the processes of oxidation and reduction can be represented as shown in the diagram below.

Knowing the formula of a substance and determining the oxidation states of the atoms of chemical elements in it, it is not difficult to predict what properties each element and the substance as a whole will exhibit. For example, nitrogen in nitric acid $H(N)↖(+5)O_3$ has a maximum oxidation state of $+5$, i.e. it has lost all electrons, so nitrogen and nitric acid will only exhibit oxidizing properties.

Nitrogen in ammonia $(N)↖(-3)(H_3)↖(+1)$ has a minimum oxidation state of $-3$, i.e. it will not be able to accept any more electrons, and therefore ammonia will only exhibit reducing properties.

Nitric oxide (II) $(N)↖(+2)(O)↖(-2)$. Nitrogen in this compound has an intermediate oxidation state and therefore can show as oxidative (for example, $N^(+2)+2ē→N^0$ or $N^(+2)+5ē→N^(-3)$ ) and restorative (for example, $N^(+2)-2ē→N^(+4)$) properties.

Electronic balance method

In redox reactions, the number of electrons donated by the reducing agent is equal to the number of electrons received by the oxidizing agent, i.e. respected electronic balance . The electron balance method is used to record the electronic equations for the processes of oxidation and reduction.

For example, the reaction of interaction of aluminum with copper (II) chloride is described by the scheme:

$(Cu)↖(+2)(Cl_2)↖(-1)+(Al)↖(0)→(Al)↖(+3)(Cl_3)↖(-1)+(Cu)↖(0) ,$

and the electronic equations will look like:

$(Cu^(+2)+2ē→Cu^0)↙(\text"oxidizing agent")↖(\text"reducing agent")|3,$

$(Al^(0)-3ē→Al^(+3))↙(\text"oxidizing agent")↖(\text"reducing agent")|2.$

The molecular equation for this reaction is:

$3CuCl_2+2Al=2AlCl_3+3Cu$.

We will show how, using the electron balance method, one can arrange the coefficients in the equation of a complex redox reaction. It is known that the first rule of a number of stresses of metals on the interaction of metals with acid solutions does not apply to concentrated sulfuric acid and nitric acid of any concentration.

In contrast to hydrochloric acid, in which hydrogen cations were the oxidizing agent for metal atoms, in sulfuric and nitric acids, sulfur and nitrogen atoms from sulfate ions and nitrate ions are oxidizing agents. Therefore, $H_2SO_4$(conc.) and $HNO_3$(any concentration) interact with metals standing in the series of voltages both before and after hydrogen, while being reduced to $SO_2$, $NO$, etc. For example, when interacting with dilute nitric acid with copper, copper (II) nitrate, nitric oxide (II) and water are obtained. Let's write down the formulas of the initial and final substances, indicating the oxidation states:

$(H)↖(+1)(N)↖(+5)(O_3)↖(-2)+(Cu)↖(0)→(Cu)↖(+2)((N)↖(+5 )(O_3)↖(-2))_(2)+(N)↖(+2)(O)↖(-2)+(H_2)↖(+1)(O)↖(-2).$

We emphasize the signs of chemical elements that have changed their oxidation states:

$H(N)↙(-)↖(+5)O_(3)+(Cu)↙(=)↖(0)→(Cu)↙(=)↖(+2)(NO_3)_2+(N) ↙(-)↖(+2)O+H_2O.$

Let us compose electronic equations, i.e. reflect the processes of recoil and attachment of electrons:

$(N^(+5)+3ē→N^(+2))↙(\text"oxidant")↖(\text"recovery")|2,$

$(Cu^(0)-2ē→Cu^(+2))↙(\text"reducing agent")↖(\text"oxidation")|3.$

We put the coefficient $3$ in front of $Cu^0$ and in front of the copper (II) nitrate formula, in which $Cu^(+2)$, since copper occurs only once with such values of oxidation states. We put the coefficient $2$ only before the formula of a substance with $N^(+2)$, since this value of the oxidation state for nitrogen occurs only once in the reaction scheme, but we do not write the coefficient before $HNO_3$, because $N^(+ 5)$ occurs again in the formula $Cu(NO_3)_2$. Our entry looks like:

$HNO_3+3Cu→3Cu(NO_3)_2+2NO+H_2O.$

Now let's equalize the number of nitrogen atoms. After the reaction, it is equal to $3·2=6$ from $Cu(NO_3)_2$ and two more atoms from $2NO$, for a total of $8$.

Therefore, before $HNO_3$ we write the coefficient $8$:

$8HNO_3+3Cu→3Cu(NO_3)_2+2NO+H_2O.$

and equalize the number of hydrogen atoms:

$8HNO_3+3Cu→3Cu(NO_3)_2+2NO+4H_2O.$

Let's check the correctness of the arrangement of the coefficients by counting the number of oxygen atoms before and after the reaction: $24$ atoms before the reaction and $24$ atoms after the reaction. The coefficients are placed correctly, so let's replace the arrow in the equation with an equal sign:

$8HNO_3+3Cu=3Cu(NO_3)_2+2NO+4H_2O.$

Corrosion of metals

When metals interact with substances environment on their surface, compounds are formed that have completely different properties than the metals themselves. AT ordinary life we often repeat the words "rust", "rust", seeing a brownish-yellow coating on products made of iron and its alloys. Rusting is a special case of corrosion.

Corrosion is a process of spontaneous destruction of metals under the influence of the external environment.

However, almost all metals undergo destruction, as a result of which many of their properties deteriorate (or are completely lost): strength, ductility, gloss decrease, electrical conductivity decreases, and friction between moving machine parts increases, the dimensions of parts change, etc.

Corrosion of metals is solid and local.

The first is not as dangerous as the second; its manifestations can be taken into account when designing structures and apparatuses. Local corrosion is much more dangerous, although metal losses here can be small. One of its most dangerous types is point. It consists in the formation of through lesions, i.e. point cavities - pittings, while the strength of individual sections decreases, the reliability of structures, apparatus, structures decreases.

Corrosion of metals causes great economic harm. Mankind suffers huge material losses as a result of the destruction of pipelines, machine parts, ships, bridges, and various equipment.

Corrosion leads to a decrease in the reliability of metal structures. Taking into account the possible destruction, it is necessary to overestimate the strength of some products (for example, aircraft parts, turbine blades), and therefore increase the consumption of metal, which requires additional economic costs.

Corrosion leads to production downtime due to the replacement of failed equipment, to the loss of raw materials and products as a result of the destruction of gas, oil and water pipelines. It is impossible not to take into account the damage to nature, and hence to human health, caused by the leakage of oil products and other chemicals. Corrosion can lead to contamination of products, and consequently, to a decrease in its quality. The costs of compensating for losses associated with corrosion are enormous. They make up $30%$ of annual metal production worldwide.

From all that has been said, it follows that a very important problem is to find ways to protect metals and alloys from corrosion. They are very varied. But for their selection it is necessary to know and take into account the chemical essence of corrosion processes.

By chemical nature corrosion is a redox process. Depending on the environment in which it occurs, there are several types of corrosion.

Types of corrosion

The most common types of corrosion are chemical and electrochemical.

I. Chemical corrosion takes place in a non-conductive medium. This type of corrosion manifests itself in the case of the interaction of metals with dry gases or non-electrolyte liquids (gasoline, kerosene, etc.). Parts and assemblies of engines, gas turbines, rocket launchers are exposed to such destruction. Chemical corrosion is often observed during the processing of metals at high temperatures.

For example:

$2(Fe)↖(0)+3(S)↖(+4)O_2+3(O_2)↖(0)→↖(t)(Fe_2)↖(+3)((S)↖(+6) (O_4)↖(-2))_3,$

$2(Fe)↖(0)+3(Cl_3)↖(0)→2(Fe)↖(+3)(Cl_3)↖(-1),$

$2(Zn)↖(0)+(O_2)↖(0)→2(Zn)↖(+2)(O)↖(-2).$

Most metals are oxidized by atmospheric oxygen, forming oxide films on the surface. If this film is strong, dense, well bonded to the metal, then it protects the metal from destruction. Such protective films appear in $Zn, Al, Cr, Ni, Pb, Sn, Nb, Ta$, etc. In iron, it is loose, porous, easily separated from the surface, and therefore is not able to protect the metal from further destruction.

II. Electrochemical corrosion occurs in a conductive medium (in an electrolyte with the appearance of an electric current inside the system). As a rule, metals and alloys are heterogeneous and contain inclusions of various impurities. When they come into contact with electrolytes, some parts of the surface begin to play the role of an anode (donate electrons), while others act as a cathode (accept electrons).

Let us consider the destruction of an iron sample in the presence of a tin impurity.

On iron, as a more active metal, in contact with the electrolyte, the processes of oxidation (dissolution) of the metal and the transition of its cations to the electrolyte occur:

$(Fe)↖(0)-2e=Fe^(2+)$ (anode).

Depending on the electrolyte environment, various processes can take place on the cathode. In one case, gas evolution ($Н_2$) will be observed. In the other, the formation of rust, consisting mainly of $Fe_2O_3·nH_2O$.

So, electrochemical corrosion is a redox reaction that occurs in media that conduct current (unlike chemical corrosion). The process occurs when two metals come into contact or on the surface of a metal containing inclusions that are less active conductors (it may also be a non-metal).

At the anode (a more active metal), metal atoms are oxidized to form cations (dissolution).

At the cathode (a less active conductor), hydrogen ions or oxygen molecules are reduced to form $H_2$ or $OH^-$ hydroxide ions, respectively.

Hydrogen cations and dissolved oxygen are the most important oxidizing agents causing electrochemical corrosion.

The corrosion rate is the greater, the more the metals (metal and impurities) differ in their activity (for metals, the farther apart they are located in a series of voltages). Corrosion increases significantly with increasing temperature.

The electrolyte can be sea water, river water, condensed moisture and, of course, well-known electrolytes - solutions of salts, alkalis, acids.

You obviously remember that in winter, technical salt (sodium chloride, sometimes calcium chloride) is used to remove snow and ice from sidewalks. The resulting solutions flow into sewer pipelines, thereby creating a favorable environment for electrochemical corrosion of underground utilities.

Corrosion protection methods

Already in the design of metal structures and their manufacture, corrosion protection measures are provided:

1.Surface grinding products so that moisture does not linger on them.

2.Application of alloyed alloys containing special additives: chromium, nickel, which, when high temperature form a stable oxide layer on the metal surface (for example, $Cr_2O_3$). Alloy steels are well-known - stainless steels, from which household items (knives, forks, spoons), machine parts, tools are made.

3. Application of protective coatings. Consider their types.

BUT. non-metallic- non-oxidizing oils, special varnishes, paints, enamels. True, they are short-lived, but they are cheap.

B. Chemical- artificially created surface films: oxide, nitride, silicide, polymer, etc. For example, all small arms and parts of many precision instruments are burnished - this is the process of obtaining the thinnest film of iron oxides on the surface of a steel product. The resulting artificial oxide film is very strong (mainly of the composition $(Fe)↖(+2)(Fe_2)↖(+3)O_4$ and gives the product a beautiful black color and blue tint. Polymer coatings are made from polyethylene, polyvinyl chloride, polyamide resins. They are applied in two ways: a heated product is placed in a polymer powder, which melts and welds to the metal, or the metal surface is treated with a polymer solution in a low-boiling solvent, which quickly evaporates, and the polymer film remains on the product.

AT. metal- this is a coating with other metals, on the surface of which stable protective films are formed under the action of oxidizing agents. Applying chromium to the surface - chromium plating, nickel - nickel plating, zinc - galvanizing, tin - tinning, etc. A chemically passive metal - gold, silver, copper - can also serve as a coating.

4. Electrochemical methods of protection.

BUT. Protective (anodic)- a piece of a more active metal (protector) is attached to the protected metal structure, which serves as an anode and is destroyed in the presence of an electrolyte. Magnesium, aluminum, zinc are used as a protector in the protection of ship hulls, pipelines, cables and other steel products.

B. cathodic- the metal structure is connected to the cathode of an external current source, which excludes the possibility of its anode destruction.

5. Special treatment of electrolyte or other media in which the protected metal structure is located.

BUT. The introduction of substances-inhibitors that slow down corrosion.

It is known that Damascus craftsmen used solutions of sulfuric acid with the addition of brewer's yeast, flour, and starch to remove scale and rust. These impurities were among the first inhibitors. They did not allow the acid to act on the weapon metal, as a result, only scale and rust were dissolved. Ural gunsmiths used for these purposes "pickling soups" - solutions of sulfuric acid with the addition of flour bran.

Examples of the use of modern inhibitors: during transportation and storage, hydrochloric acid is perfectly "tamed" by butylamine derivatives, and sulfuric acid - by nitric acid, volatile diethylamine is injected into various containers. Note that inhibitors act only on the metal, making it passive in relation to the medium, for example, to an acid solution. More than $5$ thousand corrosion inhibitors are known to science.

B. Removal of oxygen dissolved in water (deaeration). This process is used in the preparation of water entering boiler plants.

Part I

Task number 30 at the exam in chemistry is devoted to the topic "Oxidation - reduction reactions." Previously, this type of task was included in version of the exam number C1.

The meaning of task 30: it is necessary to place the coefficients in the reaction equation using the electronic balance method. Usually, only the left side of the equation is given in the condition of the problem, the student must independently add the right side.

The complete solution of the problem is estimated at 3 points. One point is given for the determination of the oxidizing agent and reducing agent, another one - directly for the construction of the electronic balance, the last - for the correct arrangement of the coefficients in the reaction equation. Note: at the USE-2018, the maximum score for solving task 30 will be 2 points.

In my opinion, the hardest part of this process is the first step. Not everyone is able to correctly predict the result of a reaction. If the products of interaction are indicated correctly, all subsequent stages are already a matter of technology.

First step: remember oxidation states

We must start with the concept element's oxidation state. If you are unfamiliar with this term, refer to the "Oxidation state" section in the Chemistry Handbook. You must learn to confidently determine the oxidation states of all elements in inorganic compounds and even in the simplest organic matter. Without a 100% understanding of this topic, it is pointless to move on.

step two: oxidizing agents and reducing agents. Redox reactions

I want to remind you that everything chemical reactions in nature can be divided into two types: redox and proceeding without changing oxidation states.

In the course of OVR (this is the reduction we will use below for redox reactions), some elements change their oxidation states.

An element whose oxidation state going down, is called oxidizing agent.
An element whose oxidation state rises, is called reducing agent.

The oxidizing agent is reduced during the reaction.
The reducing agent is oxidized during the reaction.

Example 1. Consider the reaction of sulfur with fluorine:

S + 3F 2 = SF 6 .

List the oxidation states of all the elements. We see that the oxidation state of sulfur increases (from 0 to +6), and the oxidation state of fluorine decreases (from 0 to -1). Conclusion: S - reducing agent, F 2 - oxidizing agent. During the process, sulfur is oxidized and fluorine is reduced.

Example 2. Let's discuss the reaction of manganese (IV) oxide with hydrochloric acid:

MnO 2 + 4HCl \u003d MnCl 2 + Cl 2 + 2H 2 O.

During the reaction, the oxidation state of manganese decreases (from +4 to +2), and the oxidation state of chlorine increases (from -1 to 0). Conclusion: manganese (in the composition of MnO 2) is an oxidizing agent, chlorine (in the composition of HCl is a reducing agent). Chlorine is oxidized, manganese is reduced.

Please note that in the last example, not all chlorine atoms have changed their oxidation state. This did not affect our conclusions in any way.

Example 3. Thermal decomposition of ammonium dichromate:

(NH 4) 2 Cr 2 O 7 \u003d Cr 2 O 3 + N 2 + 4H 2 O.

We see that both the oxidizing agent and the reducing agent are part of the same "molecule": chromium changes its oxidation state from +6 to +3 (i.e., it is an oxidizing agent), and nitrogen - from -3 to 0 (hence, nitrogen - reducing agent).

Example 4. The interaction of nitrogen dioxide with an aqueous solution of alkali:

2NO 2 + 2NaOH \u003d NaNO 3 + NaNO 2 + H 2 O.

Having arranged the oxidation states (I hope you do this without difficulty!), We find a strange picture: the oxidation state of only one element changes - nitrogen. Some of the N atoms increase their oxidation state (from +4 to +5), some decrease (from +4 to +3). In fact, there is nothing strange in this! In this process, N(+4) is both an oxidizing agent and a reducing agent.

Let's talk a little about the classification of redox reactions. Let me remind you that all OVRs are divided into three types:

1) intermolecular OVR (the oxidizing agent and the reducing agent are in the composition of different molecules);
2) intramolecular OVR (oxidizing agent and reducing agent are in the same molecule);
3) disproportionation reactions (oxidizing agent and reducing agent are atoms of the same element with the same initial oxidation state in the composition of one molecule).

I think that, based on these definitions, you can easily understand that the reactions from examples 1 and 2 are intermolecular OVR, the decomposition of ammonium dichromate is an example of intramolecular OVR, and the interaction of NO 2 with alkali is an example of a disproportionation reaction.

Step Three: we begin to master the method of electronic balance

To test how well you have mastered the previous material, I will ask you a simple question: "Can you give an example of a reaction in which oxidation occurs, but there is no reduction, or, conversely, there is oxidation, but there is no reduction?"

The correct answer is: "No, you can't!"

Indeed, let the oxidation state of element X increase during the reaction. This means that X donates electrons. But to whom? After all, electrons cannot simply evaporate, disappear without a trace! There is some other element Y whose atoms will accept these electrons. The electrons have a negative charge, hence the oxidation state of Y will decrease.

Conclusion: if there is a reducing agent X, then there will definitely be an oxidizing agent Y! Moreover, the number of electrons donated by one element will be exactly equal to the number of electrons received by another element.

It is on this fact that electronic balance method used in problem C1.

Let's start learning this method with examples.

Example 4

C + HNO 3 \u003d CO 2 + NO 2 + H 2 O

electronic balance method.

Solution. Let's start by determining oxidation states (do it yourself!). We see that during the process, two elements change their oxidation states: C (from 0 to +4) and N (from +5 to +4).

Obviously, carbon is a reducing agent (is oxidized), and nitrogen (+5) (as part of nitric acid) is an oxidizing agent (reduced). By the way, if you correctly identified the oxidizing agent and the intel, you are already guaranteed 1 point for problem N 30!

Now the fun begins. Let's write the so-called. oxidation and reduction half-reactions:

A carbon atom leaves 4 electrons, a nitrogen atom takes 1 e. The number of given electrons is not equal to the number of received ones. This is bad! The situation needs to be corrected.

We "multiply" the first half-reaction by 1, and the second by 4.

C(0) - 4e	=	C(+4)	(1)
N(+5) + 1e	=	N(+4)	(4)

Now everything is fine: for one carbon atom (donating 4 e) there are 4 nitrogen atoms (each of which accepts one e). The number of given electrons is equal to the number of received ones!

What we have just written, in fact, is called electronic balance. If you write this balance correctly on the real exam in chemistry, you are guaranteed another 1 point for task C1.

The last stage: it remains to transfer the obtained coefficients into the reaction equation. We don’t change anything before the formulas C and CO 2 (since the coefficient 1 is not set in the equation), we put four before the formulas HNO 3 and NO 2 (because the number of nitrogen atoms in the left and right parts of the equation should be 4) :

C + 4HNO 3 \u003d CO 2 + 4NO 2 + H 2 O.

It remains to make the last check: we see that the number of nitrogen atoms is the same on the left and right, the same applies to C atoms, but there are still problems with hydrogen and oxygen. But everything is easy to fix: we put the coefficient 2 in front of the formula H 2 O and we get the final answer:

C + 4HNO 3 \u003d CO 2 + 4NO 2 + 2H 2 O.

That's all! The problem is solved, the coefficients are placed, and we get one more point for the correct equation. Outcome: 3 points for a perfectly solved problem 30. Congratulations!

Example 5. Arrange the coefficients in the reaction equation

NaI + H 2 SO 4 = Na 2 SO 4 + H 2 S + I 2 + H 2 O

electronic balance method.

Solution. List the oxidation states of all the elements. We see that during the process, two elements change their oxidation states: S (from +6 to -2) and I (from -1 to 0).

Sulfur (+6) (in the composition of sulfuric acid) is an oxidizing agent, and iodine (-1) in the composition of NaI is a reducing agent. During the reaction, I(-1) is oxidized, S(+6) is reduced.

Write down the oxidation and reduction half-reactions:

pay attention to important point: There are two atoms in the iodine molecule. "Half" of the molecule cannot take part in the reaction, so in the corresponding equation we write not I, but I 2 .

We "multiply" the first half-reaction by 4, and the second by 1.

2I(-1) - 2e	=	I 2 (0)	(4)
S(+6) + 8e	=	S(-2)	(1)

The balance is built, for 8 given electrons there are 8 received ones.

We transfer the coefficients to the reaction equation. Before the formula I 2 we put 4, before the formula H 2 S - we mean the coefficient 1 - this, I think, is obvious.

NaI + H 2 SO 4 = Na 2 SO 4 + H 2 S + 4I 2 + H 2 O

But further questions may arise. First, it would be wrong to put a four in front of the NaI formula. Indeed, already in the oxidation half-reaction itself, the symbol I is preceded by a coefficient of 2. Therefore, on the left side of the equation, one should write not 4, but 8!

8NaI + H 2 SO 4 \u003d Na 2 SO 4 + H 2 S + 4I 2 + H 2 O

Secondly, often in such a situation, graduates put a factor of 1 in front of the sulfuric acid formula. They argue as follows: "In the reduction half-reaction, a coefficient of 1 was found, this coefficient refers to S, which means that the formula of sulfuric acid should be preceded by a unit."

These arguments are wrong! Not all sulfur atoms changed the oxidation state, some of them (as part of Na 2 SO 4) retained the +6 oxidation state. These atoms are not taken into account in the electronic balance and the coefficient 1 has nothing to do with them.

All this, however, will not prevent us from completing the decision. It is only important to understand that in further reasoning we are no longer relying on the electronic balance, but simply on common sense. So, I remind you that the coefficients in front of H 2 S, NaI and I 2 are "frozen", they cannot be changed. But the rest - you can and should.

On the left side of the equation there are 8 sodium atoms (as part of NaI), on the right - so far only 2 atoms. We put a factor of 4 in front of the sodium sulfate formula:

8NaI + H 2 SO 4 \u003d 4Na 2 SO 4 + H 2 S + 4I 2 + H 2 O.

Only now it is possible to equalize the number of atoms S. There are 5 of them on the right, therefore, before the sulfuric acid formula, you need to put a factor of 5:

8NaI + 5H 2 SO 4 \u003d 4Na 2 SO 4 + H 2 S + 4I 2 + H 2 O.

The last problem: hydrogen and oxygen. Well, I think you yourself guessed that there is not enough coefficient 4 in front of the water formula on the right side:

8NaI + 5H 2 SO 4 \u003d 4Na 2 SO 4 + H 2 S + 4I 2 + 4H 2 O.

Once again, we carefully check everything. Yes everything is correct! Problem solved, we got our legitimate 3 points.

So, in examples 4 and 5, we discussed in detail algorithm for solving problem C1 (30). In your solution of a real examination problem, the following points must be present:

1) oxidation states of ALL elements;
2) an indication of the oxidizing agent and reducing agent;
3) scheme of electronic balance;
4) final reaction equation with coefficients.

A few comments about the algorithm.

1. The oxidation states of all elements on the left and right sides of the equation should be indicated. Everyone, not just the oxidizer and reductant!

2. The oxidizing agent and reducing agent must be clearly and clearly indicated: the element X (+...) in the composition ... is an oxidizing agent, is reduced; element Y(...) in the composition... is a reducing agent, oxidized. Not everyone will be able to decipher the inscription in small underline "approx. in-sya" under the formula of sulfuric acid as "sulfur (+6) in the composition of sulfuric acid - an oxidizing agent, is being reduced."

Don't be sorry for the letters! You do not give an ad in the newspaper: "Sd. room. from Sun."

3. The electronic balance scheme is just a scheme: two half-reactions and corresponding coefficients.

4. No one needs detailed explanations of exactly how you placed the coefficients in the equation on the exam. It is only necessary that all the numbers are correct, and the entry itself is made in legible handwriting. Be sure to double check yourself!

And once again about the assessment of task C1 at the exam in chemistry:

1) determination of the oxidizing agent (oxidizing agents) and reducing agent (reducing agents) - 1 point;
2) electronic balance scheme with correct coefficients - 1 point;
3) the main reaction equation with all coefficients - 1 point.

Outcome: 3 points for complete solution task no. 30.

Note: I remind you once again that at the Unified State Exam-2018, the maximum score for solving problem N 30 will be 2 points.

I'm sure you understand what the idea behind the electronic balance method is. We understood in general terms how the solution of example N 30 is built. In principle, everything is not so complicated!

Unfortunately, the following problem arises at the real exam in chemistry: the reaction equation itself is not completely given. That is, the left side of the equation is present, but the right side either has nothing at all, or the formula of one substance is indicated. You will have to complete the equation yourself, relying on your knowledge, and only then start placing the coefficients.

This can be very difficult. There are no universal recipes for writing equations. In the next part, we will discuss this issue in more detail and look at more complex examples.

How to solve problems C1 (36) at the exam in chemistry. Part I

Task N 36 at the Unified State Examination in Chemistry is devoted to the topic "Oxidation - Reduction Reactions". Previously, a task of this type was included in the USE version under the number C1.

The meaning of task C1: it is necessary to arrange the coefficients in the reaction equation using the electronic balance method. Usually, only the left side of the equation is given in the condition of the problem, the student must independently add the right side.

First step: remember oxidation states

We must start with the concept element's oxidation state. If you are unfamiliar with this term, refer to the "Oxidation state" section in the Chemistry Handbook. You must learn to confidently determine the oxidation states of all elements in inorganic compounds and even in the simplest organic substances. Without a 100% understanding of this topic, it is pointless to move on.

Step two: oxidizing agents and reducing agents. Redox reactions

I want to remind you that all chemical reactions in nature can be divided into two types: redox and those occurring without changing oxidation states.

In the course of OVR (this is the reduction we will use below for redox reactions), some elements change their oxidation states.

Example 1. Consider the reaction of sulfur with fluorine:

S + 3F 2 = SF 6 .

Example 2. Let's discuss the reaction of manganese (IV) oxide with hydrochloric acid:

MnO 2 + 4HCl \u003d MnCl 2 + Cl 2 + 2H 2 O.

Please note that in the last example, not all chlorine atoms have changed their oxidation state. This did not affect our conclusions in any way.

Example 3. Thermal decomposition of ammonium dichromate:

(NH 4) 2 Cr 2 O 7 \u003d Cr 2 O 3 + N 2 + 4H 2 O.

Example 4. The interaction of nitrogen dioxide with an aqueous solution of alkali:

2NO 2 + 2NaOH \u003d NaNO 3 + NaNO 2 + H 2 O.

Let's talk a little about the classification of redox reactions. Let me remind you that all OVRs are divided into three types:

1) intermolecular OVR (the oxidizing agent and the reducing agent are in the composition of different molecules);
2) intramolecular OVR (oxidizing agent and reducing agent are in the same molecule);
3) disproportionation reactions (oxidizing agent and reducing agent are atoms of the same element with the same initial oxidation state in the composition of one molecule).

Step three: start mastering the electronic balance method

The correct answer is: "No, you can't!"

It is on this fact that electronic balance method used in problem C1.

Let's start learning this method with examples.

Example 4

C + HNO 3 \u003d CO 2 + NO 2 + H 2 O

electronic balance method.

Solution. Let's start by determining oxidation states (do it yourself!). We see that during the process, two elements change their oxidation states: C (from 0 to +4) and N (from +5 to +4).

Now the fun begins. Let's write the so-called. oxidation and reduction half-reactions:

A carbon atom leaves 4 electrons, a nitrogen atom takes 1 e. The number of given electrons is not equal to the number of received ones. This is bad! The situation needs to be corrected.

We "multiply" the first half-reaction by 1, and the second by 4.

C(0) - 4e	=	C(+4)	(1)
N(+5) + 1e	=	N(+4)	(4)

Now everything is fine: for one carbon atom (donating 4 e) there are 4 nitrogen atoms (each of which accepts one e). The number of given electrons is equal to the number of received ones!

What we have just written, in fact, is called electronic balance. If you write this balance correctly on the real exam in chemistry, you are guaranteed another 1 point for task C1.

C + 4HNO 3 \u003d CO 2 + 4NO 2 + H 2 O.

C + 4HNO 3 \u003d CO 2 + 4NO 2 + 2H 2 O.

That's all! The problem is solved, the coefficients are placed, and we get one more point for the correct equation. Outcome: 3 points for a perfectly solved problem C 1. Congratulations on that!

Example 5. Arrange the coefficients in the reaction equation

NaI + H 2 SO 4 = Na 2 SO 4 + H 2 S + I 2 + H 2 O

electronic balance method.

Solution. List the oxidation states of all the elements. We see that during the process, two elements change their oxidation states: S (from +6 to -2) and I (from -1 to 0).

Sulfur (+6) (in the composition of sulfuric acid) is an oxidizing agent, and iodine (-1) in the composition of NaI is a reducing agent. During the reaction, I(-1) is oxidized, S(+6) is reduced.

Write down the oxidation and reduction half-reactions:

Pay attention to an important point: there are two atoms in the iodine molecule. "Half" of the molecule cannot take part in the reaction, so in the corresponding equation we write not I, but I 2 .

We "multiply" the first half-reaction by 4, and the second by 1.

2I(-1) - 2e	=	I 2 (0)	(4)
S(+6) + 8e	=	S(-2)	(1)

The balance is built, for 8 given electrons there are 8 received ones.

We transfer the coefficients to the reaction equation. Before the formula I 2 we put 4, before the formula H 2 S - we mean the coefficient 1 - this, I think, is obvious.

NaI + H 2 SO 4 = Na 2 SO 4 + H 2 S + 4I 2 + H 2 O

8NaI + H 2 SO 4 \u003d Na 2 SO 4 + H 2 S + 4I 2 + H 2 O

On the left side of the equation there are 8 sodium atoms (as part of NaI), on the right - so far only 2 atoms. We put a factor of 4 in front of the sodium sulfate formula:

8NaI + H 2 SO 4 \u003d 4Na 2 SO 4 + H 2 S + 4I 2 + H 2 O.

Only now it is possible to equalize the number of atoms S. There are 5 of them on the right, therefore, before the sulfuric acid formula, you need to put a factor of 5:

8NaI + 5H 2 SO 4 \u003d 4Na 2 SO 4 + H 2 S + 4I 2 + H 2 O.

The last problem: hydrogen and oxygen. Well, I think you yourself guessed that there is not enough coefficient 4 in front of the water formula on the right side:

8NaI + 5H 2 SO 4 \u003d 4Na 2 SO 4 + H 2 S + 4I 2 + 4H 2 O.

Once again, we carefully check everything. Yes everything is correct! Problem solved, we got our legitimate 3 points.

So, in examples 4 and 5, we discussed in detail algorithm for solving problem C1. In your solution of a real examination problem, the following points must be present:

1) oxidation states of ALL elements;
2) an indication of the oxidizing agent and reducing agent;
3) scheme of electronic balance;
4) final reaction equation with coefficients.

A few comments about the algorithm.

1. The oxidation states of all elements on the left and right sides of the equation should be indicated. Everyone, not just the oxidizer and reductant!

Don't be sorry for the letters! You do not give an ad in the newspaper: "Sd. room. from Sun."

3. The electronic balance scheme is just a scheme: two half-reactions and corresponding coefficients.

And once again about the assessment of task C1 at the exam in chemistry:

1) determination of the oxidizing agent (oxidizing agents) and reducing agent (reducing agents) - 1 point;
2) electronic balance scheme with correct coefficients - 1 point;
3) the main reaction equation with all coefficients - 1 point.

Outcome: 3 points for the complete solution of problem N 36.

I'm sure you understand what the idea behind the electronic balance method is. Understood in general terms how the solution of example C1 is built. In principle, everything is not so difficult!

This can be very difficult. There are no universal recipes for writing equations. In the next part, we will discuss this issue in more detail and look at more complex examples.

Redox reactions

For a job well done, you'll get 2 points. Approximately 10-15 minutes.

To complete task 30 in chemistry, you must:

know what is
be able to write equations of redox reactions

Tasks for training

To complete the task, use the following list of substances: potassium permanganate, potassium bicarbonate, sodium sulfite, barium sulfate, potassium hydroxide. Permissible use aqueous solutions substances.

From the proposed list of substances, select substances between which a redox reaction is possible, and write down the equation for this reaction. Make an electronic balance, indicate the oxidizing agent and reducing agent.

Solution

Use the following list of substances: sulfur oxide (IV), potassium chloride, sodium sulfate, barium permanganate, aluminum hydroxide. The use of aqueous solutions is acceptable.

Solution
Use the following list of substances: sodium sulfide, potassium chloride, sulfuric acid, potassium permanganate, lithium hydroxide. The use of aqueous solutions is acceptable.

From the proposed list, select substances between which a redox reaction is possible. Write down the equation for this reaction. Make an electronic balance, indicate the oxidizing agent and reducing agent.

Solution
Use the following list of substances: potassium dichromate, lithium chloride, sodium orthophosphate, potassium chloride, potassium sulfite. The use of aqueous solutions is acceptable.

From the proposed list, select substances between which a redox reaction is possible. Write down the equation for this reaction. Make an electronic balance, indicate the oxidizing agent and reducing agent.

Solution
Use the following list of substances: silver nitrate, ammonium chloride, phosphine, rubidium acetate, zinc oxide. The use of aqueous solutions is acceptable.

From the proposed list, select substances between which a redox reaction is possible. Write down the equation for this reaction. Make an electronic balance, indicate the oxidizing agent and reducing agent.