How to solve redox reactions? Oxidation-reduction reactions (ORR): examples Oxidation-reduction reactions 9 chemistry.

Problem book on general and inorganic chemistry

2.2. Redox reactions

Look 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 for selecting odds using an electronic balance consists of the following steps:

a) write down the formulas of the reagents 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

Mn II¼ = Mn IV

b) compose equations for 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 Mn II- 2 e - = Mn IV

c) additional factors are selected for the equation of half-reactions so that the law of conservation of charge is satisfied for the reaction as a whole, for which the number of electrons accepted 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

Mn II- 2 e - = Mn IV 3

d) insert (using the found factors) stoichiometric coefficients into 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 reaction (if there are two such elements, then it is enough to equalize the number of atoms of one of them, and check for the second). The equation for the chemical reaction is obtained:

3 MnCO 3 + KClO 3 = 3 MnO 2 + KCl+ 3 CO 2

Example 3. Select the coefficients in the equation of the redox reaction

Fe 2 O 3 + CO ® Fe + CO 2

Solution

Fe 2 O 3 + 3 CO = 2 Fe +3 CO 2

Fe III + 3 e - = Fe 0 2

C II - 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. Select the coefficients in the equation of the redox reaction

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

Solution

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

FeII- e - = Fe III

- 11 e - 4

2S - I - 10 e - = 2S IV

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 classified as 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 per one formula unit of the substance.

Example 5. Select the coefficients in the oxidation-reduction reaction equation

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

Solution

2 (NH 4) 2 CrO 4 = 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 dismutation (disproportionation, autoxidation- self-healing), in which atoms of the same element in the reagent are oxidized and reduced, additional factors are first added to the right side of the equation, and then the coefficient for the reagent is found.

Example 6. Select the coefficients in the dismutation reaction equation

H2O2 ® H2O+O2

Solution

2 H 2 O 2 = 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 atoms of the same element of different reagents, as a result of their oxidation and reduction, receive the same oxidation state, additional factors are first added to the left side of the equation.

Example 7. Select the coefficients in the commutation reaction equation:

H 2 S + SO 2 = S + H 2 O

Solution

2H2S + SO2 = 3S + 2H2O

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 for selecting coefficients using 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 - acidic reaction medium, H2S - 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 - ), environment ( H+- more precisely, oxonium cation H3O+ ) and reducing agent ( H2S):

Cr2O72 - +H++H2S

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); This data is written down on the next two lines, the electron-ion equations for the reduction and oxidation half-reactions are drawn up, and additional factors are selected for the half-reaction equations:

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

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

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

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

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

K 2 Cr 2 O 7 + 4H 2 SO 4 + 3H 2 S = Cr 2 (SO 4) 3 + 7H 2 O + 3S ( t ) + K 2 SO 4

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

OxidizedAnd restored The oxidizing and reducing forms often differ in oxygen content (compare Cr2O72 - and Cr 3+ ). Therefore, when compiling half-reaction equations using the electron-ion balance method, they include the pairs H + / H 2 O (for an acidic medium) and OH - / H 2 O (for alkaline environment). If, when moving 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 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):

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

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

Lack of oxide ions in their original form (usually- in reduced) compared to the final form is compensated by the addition of water molecules (in an acidic environment) or hydroxide ions (in an alkaline environment):

acidic environment H 2 O = [ 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 equation of the redox reaction:

® MnSO 4 + H 2 O + Na 2 SO 4 + ¼

Solution

2 KMnO 4 + 3 H 2 SO 4 + 5 Na 2 SO 3 =

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 equation of the redox reaction:

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

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

SO 3 2 - + 2 OH - - 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 equation for the reduction half-reaction is:

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

and if in a slightly alkaline environment, then

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

Often, a weakly acidic and slightly alkaline medium is conventionally called neutral, and only water molecules are introduced into the half-reaction equations on the left. In this case, when composing the equation, you should (after selecting additional factors) write down an additional equation reflecting the formation of water from H + and OH ions - .

Example 10. Select the coefficients in the equation of the reaction occurring 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 = 2 MnO 2( t) + 3 Na 2 SO 4 + 2 KOH

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

MnO 4 - + 2 H 2 O + 3 e - = MnO 2( t) + 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 combining aqueous solutions of potassium permanganate and sodium sulfite, then it proceeds in a conditionally neutral (and in fact, slightly alkaline) environment due to the formation of potassium hydroxide. If the potassium permanganate solution is slightly acidified, the reaction will proceed in a weakly acidic (conditionally neutral) environment.

Example 11. Select the coefficients in the equation of the reaction occurring 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 = 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. Thus, from chemical practice it is known (and this must be remembered) that the permanganate ion in an acidic environment forms a manganese cation ( II) (pair MnO 4 - +H+/ Mn 2+ + H 2 O ), in a slightly alkaline environment- 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 oxidation states, i.e. unequal stability of specific forms in different environments of aqueous solution. All redox couples used in this section are given in problems 2.15 and 2.16.

Back forward

Attention! Slide previews are for informational purposes only and may not represent all the features of the presentation. If you are interested in this work, please download the full version.

Textbook: Rudzitis G.E., Feldman F.G. Chemistry: textbook for 9th grade of educational institutions / G.E. Rudzitis, F.G. Feldman. – 12th ed. – M.: Education, OJSC “Moscow Textbooks”, 2009. – 191 p.

Target: to form students’ understanding of redox processes and their mechanism

Expected results

Subject:

During the work, students

will acquire

the ability to analyze and objectively evaluate life situations related to chemistry, skills for safe handling of substances used in everyday life; the ability to analyze and plan environmentally friendly behavior in order to preserve health and the environment
the ability to establish connections between actually observed chemical phenomena and processes, explain the reasons for the diversity of substances, the dependence of the properties of substances on their structure;

master the scientific approach to drawing up the equation of redox reactions

Metasubject

During the work, students will be able

define concepts, create generalizations, establish analogies, classify, independently select grounds and criteria for classification, establish cause-and-effect relationships, build logical reasoning, inference (inductive, deductive and by analogy) and draw conclusions;
create, apply and transform signs and symbols, models and diagrams to solve educational and cognitive problems;
apply ecological thinking in cognitive, communicative, social practice and professional guidance

Personal

During the work, students will acquire

the foundations of ecological culture corresponding to the modern level of environmental thinking, experience of environmentally oriented reflective-evaluative and practical activities in life situations;

2.1. Chemical reaction. Conditions and signs of chemical reactions. Chemical equations.

2.2. Classification of chemical reactions according to changes in oxidation states of chemical elements

2.6. Redox reactions. Oxidizing agent and reducing agent.

Skills and activities tested by KIM GIA

Know/understand

chemical symbols: formulas of chemical substances, equations of chemical reactions
the most important chemical concepts: oxidation state, oxidizing agent and reducing agent, oxidation and reduction, main types of reactions in inorganic chemistry

1.2.1. characteristic features of the most important chemical concepts

1.2.2. about the existence of relationships between the most important chemical concepts

Compose

2.5.3. equations of chemical reactions.

Form of delivery: lesson using ICT, including paired, individual forms of organizing educational and cognitive activities of students.

Duration of training session: 45 minutes.

Using educational technologies: heuristic teaching method, collaborative learning

During the classes

I. Problematization, actualization, motivation – 10 min.

Frontal conversation

What are atoms and ions.
What is the difference?
What are electrons?
What is oxidation state?
How is oxidation number calculated?

On the board, students are asked to place the oxidation states in the following substances:

Сl 2 O 7, SO 3, H 3 PO 4, P 2 O 5, Na 2 CO 3, CuSO 4, Cl 2, HClO 4, K 2 Cr 2 O 7, Cr 2 (SO 4) 3, Al(NO 3) 3, CaSO 4,

NaMnO 4, MnCl 2, HNO 3, N 2, N 2 O, HNO 2, H 2 S, Ca 3 (PO 4) 2

II. Learning new material. Teacher's explanation. 15 minutes.

Basic concepts (slide 2):

Redox reactions- these are reactions in which the oxidation states of two elements change, one of which is a reducing agent and the other is an oxidizing agent

Reducing agent- this is the element that gives up electrons during the reaction and is itself oxidized

Oxidizer- this is the element that accepts electrons during the reaction and is itself reduced

Rules for composing redox equations(slide 3)

1. Write down the reaction equation (slide 4).

CuS+HNO 3 ->Cu(NO 3) 2 + S + NO+H 2 O

2. Let's arrange the oxidation states of all elements

Cu +2 S -2 +H +1 N +5 O -2 3 -> Cu +2 (N +5 O -2 3) -1 2 + S 0 + N +2 O -2 +H +1 2 O -2

3. Let’s highlight the elements that have changed their oxidation states

Cu +2 S -2 +H +1 N +5 O -2 3 -> Cu +2 (N +5 O -2 3) -1 2 + S 0 + N +2 O -2 +H +1 2 O -2

We see that as a result of the reaction, the oxidation states of two elements changed -

sulfur (S) changed completely (from – 2 before 0 )
nitrogen (N) changed partially (from +5 before +2 changed), some remained +5

4. Let’s write down those elements that have changed oxidation states and show the transition of electrons (slide 5.)

CuS -2 +HN +5 O 3 -> Cu(N +5 O 3) 2 + S 0 + N +2 O+H 2 O

S -2 - 2e S 0

5. Let’s compile an electronic balance and find the coefficients

6. Let’s substitute the coefficients found in the balance into the equation (coefficients are set for substances whose elements have changed their oxidation state) (slide 6).

CuS -2 +HN +5 O 3 -> Cu(N +5 O 3) 2 + 3 S0+ 2 N+2O+H2O

7. Let's deliver the missing coefficients using the equalization method

3CuS -2 +8HN +5 O 3 -> 3Cu(N +5 O 3) 2 + 3S 0 + 2N +2 O+4H 2 O

8. Using oxygen, let’s check the correctness of the equation (slide 7).

Before the reaction of oxygen 24 atoms = After the reaction of oxygen 24 atoms

9. Identify the oxidizing agent and the reducing agent and the processes - oxidation and reduction

S -2 (in CuS) is a reducing agent because donates electrons

N +5 (in HNO 3) is an oxidizing agent, because donates electrons

III. Consolidation of the studied material (25 min)

Students are asked to complete the task in pairs.

Task 1. 10 min. (slide 8)

Students are asked to create a reaction equation in accordance with the algorithm.

Mg+H 2 SO 4 -> MgSO 4 + H 2 S + H 2 O

Checking the job

4Mg 0 +5H 2 +1 S +6 O 4 -2 -> 4Mg +2 S +6 O 4 -2 + H 2 +1 S -2 + 4H 2 +1 O -2

Transition e –	Number of electrons	NOC	Odds
	2		4
			1

Task 2. 15 min. (slides 9, 10)

Students are asked to complete test(in pairs). The test items are checked and sorted out on the board.

Question No. 1

Which equation corresponds to a redox reaction?

CaCO 3 = CaO + CO 2
BaCl 2 + Na 2 SO 4 = BaSO 4 + 2NaCl
Zn + H 2 SO 4 = ZnSO 4 + H 2
Na 2 CO 3 + CO 2 + H 2 O = 2NaHCO 3

Question No. 2

In the reaction equation 2Al + 3Br 2 = 2AlBr 3 the coefficient in front of the reducing agent formula is equal to

Question No. 3

In the reaction equation 5Ca + 12HNO 3 = 5Ca(NO 3) 2 + N 2 + 6H 2 O the oxidizing agent is

Ca(NO3)2
HNO3
H2O

Question No. 4

Which of the proposed schemes will correspond to the reducer

S 0 > S -2
S +4 -> S +6
S -2 > S -2
S +6 -> S +4

Question No. 5

In the reaction equation 2SO 2 + O 2 -> 2 SO 3 sulfur

oxidizes
is being restored
neither oxidized nor reduced
both oxidizes and reduces

Question No. 6

Which element is the reducing agent in the reaction equation

2KClO 3 -> 2KCl + 3O 2

potassium
oxygen
hydrogen

Question No. 7

Scheme Br -1 -> Br +5 corresponds to element

oxidizing agent
restorer
both an oxidizing agent and a reducing agent

Question No. 8

Hydrochloric acid is the reducing agent in the reaction

PbO 2 + 4HCl = PbCl 2 + Cl 2 + 2H 2 O
Zn + 2HCl = ZnCl 2 + H 2
PbO + 2HCl = PbCl 2 + H 2 O
Na 2 CO 3 + 2HCl = 2NaCl+ CO 2 + H 2 O

Answers to test questions.

question number 1 2 3 4 5 6 7 8
answer 3 1 3 2 1 3 2 1

Homework: paragraph 5 ex. 6,7,8 p. 22 (textbook).

9.1. What are the chemical reactions?

Let us remember that we call any chemical phenomena in nature chemical reactions. During a chemical reaction, some chemical bonds are broken and others are formed. As a result of the reaction, other substances are obtained from some chemical substances (see Chapter 1).

While doing your homework for § 2.5, you became acquainted with the traditional selection of four main types of reactions from the entire set of chemical transformations, and then you also proposed their names: reactions of combination, decomposition, substitution and exchange.

Examples of compound reactions:

C + O 2 = CO 2; (1)
Na 2 O + CO 2 = Na 2 CO 3; (2)
NH 3 + CO 2 + H 2 O = NH 4 HCO 3. (3)

Examples of decomposition reactions:

2Ag 2 O 4Ag + O 2; (4)
CaCO 3 CaO + CO 2; (5)
(NH 4) 2 Cr 2 O 7 N 2 + Cr 2 O 3 + 4H 2 O. (6)

Examples of substitution reactions:

CuSO 4 + Fe = FeSO 4 + Cu; (7)
2NaI + Cl 2 = 2NaCl + I 2; (8)
CaCO 3 + SiO 2 = CaSiO 3 + CO 2. (9)

Exchange reactions- chemical reactions in which starting substances seem to exchange their constituent parts.

Examples of exchange reactions:

Ba(OH) 2 + H 2 SO 4 = BaSO 4 + 2H 2 O; (10)
HCl + KNO 2 = KCl + HNO 2; (eleven)
AgNO 3 + NaCl = AgCl + NaNO 3. (12)

The traditional classification of chemical reactions does not cover all their diversity - in addition to the four main types of reactions, there are also many more complex reactions.
The identification of two other types of chemical reactions is based on the participation in them of two important non-chemical particles: electron and proton.
During some reactions, complete or partial transfer of electrons from one atom to another occurs. In this case, the oxidation states of the atoms of the elements that make up the starting substances change; of the examples given, these are reactions 1, 4, 6, 7 and 8. These reactions are called redox.

In another group of reactions, a hydrogen ion (H +), that is, a proton, passes from one reacting particle to another. Such reactions are called acid-base reactions or proton transfer reactions.

Among the examples given, such reactions are reactions 3, 10 and 11. By analogy with these reactions, redox reactions are sometimes called electron transfer reactions. You will become acquainted with OVR in § 2, and with KOR in the following chapters.

COMPOUNDING REACTIONS, DECOMPOSITION REACTIONS, SUBSTITUTION REACTIONS, EXCHANGE REACTIONS, REDOX REACTIONS, ACID-BASE REACTIONS.
Write down reaction equations corresponding to the following schemes:
a) HgO Hg + O 2 ( t); b) Li 2 O + SO 2 Li 2 SO 3; c) Cu(OH) 2 CuO + H 2 O ( t);
d) Al + I 2 AlI 3; e) CuCl 2 + Fe FeCl 2 + Cu; e) Mg + H 3 PO 4 Mg 3 (PO 4) 2 + H 2 ;
g) Al + O 2 Al 2 O 3 ( t); i) KClO 3 + P P 2 O 5 + KCl ( t); j) CuSO 4 + Al Al 2 (SO 4) 3 + Cu;
l) Fe + Cl 2 FeCl 3 ( t); m) NH 3 + O 2 N 2 + H 2 O ( t); m) H 2 SO 4 + CuO CuSO 4 + H 2 O.
Indicate the traditional type of reaction. Label redox and acid-base reactions. In redox reactions, indicate which atoms of elements change their oxidation states.

9.2. Redox reactions

Let's consider the redox reaction that occurs in blast furnaces during the industrial production of iron (more precisely, cast iron) from iron ore:

Fe 2 O 3 + 3CO = 2Fe + 3CO 2.

Let us determine the oxidation states of the atoms that make up both the starting substances and the reaction products


Fe2O3	+		=	2Fe	+

As you can see, the oxidation state of carbon atoms increased as a result of the reaction, the oxidation state of iron atoms decreased, and the oxidation state of oxygen atoms remained unchanged. Consequently, the carbon atoms in this reaction underwent oxidation, that is, they lost electrons ( oxidized), and the iron atoms – reduction, that is, they added electrons ( recovered) (see § 7.16). To characterize OVR, the concepts are used oxidizer And reducing agent.

Thus, in our reaction the oxidizing atoms are iron atoms, and the reducing atoms are carbon atoms.

In our reaction, the oxidizing agent is iron(III) oxide, and the reducing agent is carbon(II) monoxide.
In cases where oxidizing atoms and reducing atoms are part of the same substance (example: reaction 6 from the previous paragraph), the concepts of “oxidizing substance” and “reducing substance” are not used.
Thus, typical oxidizing agents are substances that contain atoms that tend to gain electrons (in whole or in part), lowering their oxidation state. Of the simple substances, these are primarily halogens and oxygen, and to a lesser extent sulfur and nitrogen. From complex substances - substances that contain atoms in higher oxidation states that are not inclined to form simple ions in these oxidation states: HNO 3 (N +V), KMnO 4 (Mn +VII), CrO 3 (Cr +VI), KClO 3 (Cl +V), KClO 4 (Cl +VII), etc.
Typical reducing agents are substances that contain atoms that tend to completely or partially donate electrons, increasing their oxidation state. Simple substances include hydrogen, alkali and alkaline earth metals, and aluminum. Of the complex substances - H 2 S and sulfides (S –II), SO 2 and sulfites (S +IV), iodides (I –I), CO (C +II), NH 3 (N –III), etc.
In general, almost all complex and many simple substances can exhibit both oxidizing and reducing properties. For example:
SO 2 + Cl 2 = S + Cl 2 O 2 (SO 2 is a strong reducing agent);
SO 2 + C = S + CO 2 (t) (SO 2 is a weak oxidizing agent);
C + O 2 = CO 2 (t) (C is a reducing agent);
C + 2Ca = Ca 2 C (t) (C is an oxidizing agent).
Let's return to the reaction we discussed at the beginning of this section.


Fe2O3	+		=	2Fe	+

Please note that as a result of the reaction, oxidizing atoms (Fe + III) turned into reducing atoms (Fe 0), and reducing atoms (C + II) turned into oxidizing atoms (C + IV). But CO 2 is a very weak oxidizing agent under any conditions, and iron, although it is a reducing agent, is under these conditions much weaker than CO. Therefore, the reaction products do not react with each other, and the reverse reaction does not occur. The given example is an illustration of the general principle that determines the direction of the flow of the OVR:

Redox reactions proceed in the direction of the formation of a weaker oxidizing agent and a weaker reducing agent.

The redox properties of substances can only be compared under identical conditions. In some cases, this comparison can be made quantitatively.
While doing your homework for the first paragraph of this chapter, you became convinced that it is quite difficult to select coefficients in some reaction equations (especially ORR). To simplify this task in the case of redox reactions, the following two methods are used:
A) electronic balance method And
b) electron-ion balance method.
You will learn the electron balance method now, and the electron-ion balance method is usually studied in higher education institutions.
Both of these methods are based on the fact that electrons in chemical reactions neither disappear nor appear anywhere, that is, the number of electrons accepted by atoms is equal to the number of electrons given up by other atoms.
The number of given and accepted electrons in the electron balance method is determined by the change in the oxidation state of atoms. When using this method, it is necessary to know the composition of both the starting substances and the reaction products.
Let's look at the application of the electronic balance method using examples.

Example 1. Let's create an equation for the reaction of iron with chlorine. It is known that the product of this reaction is iron(III) chloride. Let's write down the reaction scheme:

Fe + Cl 2 FeCl 3 .

Let us determine the oxidation states of the atoms of all elements that make up the substances participating in the reaction:

Iron atoms give up electrons, and chlorine molecules accept them. Let us express these processes electronic equations:
Fe – 3 e– = Fe +III,
Cl2+2 e –= 2Cl –I.

In order for the number of electrons given to be equal to the number of electrons received, the first electronic equation must be multiplied by two, and the second by three:

Fe – 3 e– = Fe +III,
Cl2+2 e– = 2Cl –I

2Fe – 6 e– = 2Fe +III,
3Cl 2 + 6 e– = 6Cl –I.

By introducing coefficients 2 and 3 into the reaction scheme, we obtain the reaction equation:
2Fe + 3Cl 2 = 2FeCl 3.

Example 2. Let's create an equation for the combustion reaction of white phosphorus in excess chlorine. It is known that phosphorus(V) chloride is formed under these conditions:

				+V –I
P 4	+	Cl2		PCl 5.

White phosphorus molecules give up electrons (oxidize), and chlorine molecules accept them (reduce):

P 4 – 20 e– = 4P +V
Cl2+2 e– = 2Cl –I

1
10

2
20

P 4 – 20 e– = 4P +V
Cl2+2 e– = 2Cl –I

P 4 – 20 e– = 4P +V
10Cl 2 + 20 e– = 20Cl –I

The initially obtained factors (2 and 20) had a common divisor, by which (like future coefficients in the reaction equation) they were divided. Reaction equation:

P4 + 10Cl2 = 4PCl5.

Example 3. Let's create an equation for the reaction that occurs when iron(II) sulfide is roasted in oxygen.

Reaction scheme:

				+III –II		+IV –II
	+	O2			+

In this case, both iron(II) and sulfur(–II) atoms are oxidized. The composition of iron(II) sulfide contains atoms of these elements in a 1:1 ratio (see the indices in the simplest formula).
Electronic balance:

4	Fe+II – e– = Fe +III S–II–6 e– = S +IV	In total they give 7 e –
7	O 2 + 4e – = 2O –II

Reaction equation: 4FeS + 7O 2 = 2Fe 2 O 3 + 4SO 2.

Example 4. Let's create an equation for the reaction that occurs when iron(II) disulfide (pyrite) is roasted in oxygen.

Reaction scheme:

				+III –II		+IV –II
	+	O2			+

As in the previous example, both iron(II) atoms and sulfur atoms are also oxidized here, but with an oxidation state of I. The atoms of these elements are included in the composition of pyrite in a ratio of 1:2 (see the indices in the simplest formula). It is in this regard that the iron and sulfur atoms react, which is taken into account when compiling the electronic balance:

	Fe+III – e– = Fe +III 2S–I – 10 e– = 2S +IV	In total they give 11 e –
	O2+4 e– = 2O –II

Reaction equation: 4FeS 2 + 11O 2 = 2Fe 2 O 3 + 8SO 2.

There are also more complex cases of ODD, some of which you will become familiar with while doing your homework.

OXIDIZING ATOM, REDUCING ATOM, OXIDIZING SUBSTANCE, REDUCING SUBSTANCE, ELECTRONIC BALANCE METHOD, ELECTRONIC EQUATIONS.
1. Compile an electronic balance for each OVR equation given in the text of § 1 of this chapter.
2. Make up equations for the ORRs that you discovered while completing the task for § 1 of this chapter. This time, use the electronic balance method to set the odds. 3.Using the electron balance method, create reaction equations corresponding to the following schemes: a) Na + I 2 NaI;
b) Na + O 2 Na 2 O 2;
c) Na 2 O 2 + Na Na 2 O;
d) Al + Br 2 AlBr 3;
e) Fe + O 2 Fe 3 O 4 ( t);
e) Fe 3 O 4 + H 2 FeO + H 2 O ( t);
g) FeO + O 2 Fe 2 O 3 ( t);
i) Fe 2 O 3 + CO Fe + CO 2 ( t);
j) Cr + O 2 Cr 2 O 3 ( t);
l) CrO 3 + NH 3 Cr 2 O 3 + H 2 O + N 2 ( t);
l) Mn 2 O 7 + NH 3 MnO 2 + N 2 + H 2 O;
m) MnO 2 + H 2 Mn + H 2 O ( t);
n) MnS + O 2 MnO 2 + SO 2 ( t)
p) PbO 2 + CO Pb + CO 2 ( t);
c) Cu 2 O + Cu 2 S Cu + SO 2 ( t);
t) CuS + O 2 Cu 2 O +SO 2 ( t);
y) Pb 3 O 4 + H 2 Pb + H 2 O ( t).

9.3. Exothermic reactions. Enthalpy

Why do chemical reactions occur?
To answer this question, let us remember why individual atoms combine into molecules, why an ionic crystal is formed from isolated ions, and why the principle of least energy applies when the electron shell of an atom is formed. The answer to all these questions is the same: because it is energetically beneficial. This means that during such processes energy is released. It would seem that chemical reactions should occur for the same reason. Indeed, many reactions can be carried out, during which energy is released. Energy is released, usually in the form of heat.

If during an exothermic reaction the heat does not have time to be removed, then the reaction system heats up.
For example, in the methane combustion reaction

CH 4 (g) + 2O 2 (g) = CO 2 (g) + 2H 2 O (g)

so much heat is released that methane is used as fuel.
The fact that this reaction releases heat can be reflected in the reaction equation:

CH 4 (g) + 2O 2 (g) = CO 2 (g) + 2H 2 O (g) + Q.

This is the so called thermochemical equation. Here the symbol "+ Q" means that when methane is burned, heat is released. This heat is called thermal effect of reaction.
Where does the released heat come from?
You know that during chemical reactions chemical bonds are broken and formed. In this case, the bonds between carbon and hydrogen atoms in CH 4 molecules, as well as between oxygen atoms in O 2 molecules, are broken. In this case, new bonds are formed: between carbon and oxygen atoms in CO 2 molecules and between oxygen and hydrogen atoms in H 2 O molecules. To break bonds, you need to expend energy (see “bond energy”, “atomization energy”), and when forming bonds, energy is released. Obviously, if the “new” bonds are stronger than the “old” ones, then more energy will be released than absorbed. The difference between the released and absorbed energy is the thermal effect of the reaction.
Thermal effect (amount of heat) is measured in kilojoules, for example:

2H 2 (g) + O 2 (g) = 2H 2 O (g) + 484 kJ.

This notation means that 484 kilojoules of heat will be released if two moles of hydrogen react with one mole of oxygen to produce two moles of gaseous water (water vapor).

Thus, in thermochemical equations, the coefficients are numerically equal to the amounts of substance of the reactants and reaction products.

What determines the thermal effect of each specific reaction?
The thermal effect of the reaction depends
a) on the aggregative states of the starting substances and reaction products,
b) on temperature and
c) on whether the chemical transformation occurs at constant volume or at constant pressure.
The dependence of the thermal effect of a reaction on the state of aggregation of substances is due to the fact that the processes of transition from one state of aggregation to another (like some other physical processes) are accompanied by the release or absorption of heat. This can also be expressed by a thermochemical equation. Example – thermochemical equation for condensation of water vapor:

H 2 O (g) = H 2 O (l) + Q.

In thermochemical equations, and, if necessary, in ordinary chemical equations, the aggregative states of substances are indicated using letter indices:
(d) – gas,
(g) – liquid,
(t) or (cr) – solid or crystalline substance.
The dependence of the thermal effect on temperature is associated with differences in heat capacities starting materials and reaction products.
Since the volume of the system always increases as a result of an exothermic reaction at constant pressure, part of the energy is spent on doing work to increase the volume, and the heat released will be less than if the same reaction occurs at a constant volume.
Thermal effects of reactions are usually calculated for reactions occurring at constant volume at 25 °C and are indicated by the symbol Q o.
If energy is released only in the form of heat, and a chemical reaction proceeds at a constant volume, then the thermal effect of the reaction ( Q V) is equal to the change internal energy(D U) substances participating in the reaction, but with the opposite sign:

Q V = – U.

The internal energy of a body is understood as the total energy of intermolecular interactions, chemical bonds, the ionization energy of all electrons, the bond energy of nucleons in nuclei, and all other known and unknown types of energy “stored” by this body. The “–” sign is due to the fact that when heat is released, the internal energy decreases. That is

U= – Q V .

If the reaction occurs at constant pressure, then the volume of the system can change. Doing work to increase the volume also takes part of the internal energy. In this case

U = –(QP+A) = –(QP+PV),

Where Q p– the thermal effect of a reaction occurring at constant pressure. From here

Q P = – U–PV .

A value equal to U+PV got the name enthalpy change and denoted by D H.

H=U+PV.

Hence

Q P = – H.

Thus, as heat is released, the enthalpy of the system decreases. Hence the old name for this quantity: “heat content”.
Unlike the thermal effect, a change in enthalpy characterizes a reaction regardless of whether it occurs at constant volume or constant pressure. Thermochemical equations written using enthalpy change are called thermochemical equations in thermodynamic form. In this case, the value of the enthalpy change under standard conditions (25 °C, 101.3 kPa) is given, denoted H o. For example:
2H 2 (g) + O 2 (g) = 2H 2 O (g) H o= – 484 kJ;
CaO (cr) + H 2 O (l) = Ca(OH) 2 (cr) H o= – 65 kJ.

Dependence of the amount of heat released in the reaction ( Q) from the thermal effect of the reaction ( Q o) and the amount of substance ( n B) one of the participants in the reaction (substance B - the starting substance or reaction product) is expressed by the equation:

Here B is the amount of substance B, specified by the coefficient in front of the formula of substance B in the thermochemical equation.

Task

Determine the amount of hydrogen substance burned in oxygen if 1694 kJ of heat was released.

Solution

2H 2 (g) + O 2 (g) = 2H 2 O (g) + 484 kJ.

Q = 1694 kJ, 6. The thermal effect of the reaction between crystalline aluminum and gaseous chlorine is 1408 kJ. Write the thermochemical equation for this reaction and determine the mass of aluminum required to produce 2816 kJ of heat using this reaction.
7. Determine the amount of heat released during combustion of 1 kg of coal containing 90% graphite in air, if the thermal effect of the combustion reaction of graphite in oxygen is 394 kJ.

9.4. Endothermic reactions. Entropy

In addition to exothermic reactions, reactions are possible in which heat is absorbed, and if it is not supplied, the reaction system is cooled. Such reactions are called endothermic.

The thermal effect of such reactions is negative. For example:
CaCO 3 (cr) = CaO (cr) + CO 2 (g) – Q,
2HgO (cr) = 2Hg (l) + O 2 (g) – Q,
2AgBr (cr) = 2Ag (cr) + Br 2 (g) – Q.

Thus, the energy released during the formation of bonds in the products of these and similar reactions is less than the energy required to break bonds in the starting substances.
What is the reason for the occurrence of such reactions, since they are energetically unfavorable?
Since such reactions are possible, it means that there is some factor unknown to us that is the reason for their occurrence. Let's try to find it.

Let's take two flasks and fill one of them with nitrogen (colorless gas) and the other with nitrogen dioxide (brown gas) so that both the pressure and temperature in the flasks are the same. It is known that these substances do not react chemically with each other. Let's tightly connect the flasks with their necks and install them vertically, so that the flask with heavier nitrogen dioxide is at the bottom (Fig. 9.1). After some time, we will see that brown nitrogen dioxide gradually spreads into the upper flask, and colorless nitrogen penetrates into the lower one. As a result, the gases mix, and the color of the contents of the flasks becomes the same.
What causes gases to mix?
Chaotic thermal motion of molecules.
The above experience shows that a process can occur spontaneously, without any of our (external) influence, the thermal effect of which is zero. But it is really equal to zero, because in this case there is no chemical interaction (chemical bonds are not broken or formed), and the intermolecular interaction in gases is negligible and practically the same.
The observed phenomenon is a special case of the manifestation of a universal law of Nature, according to which systems consisting of a large number of particles always tend to the greatest disorder.
The measure of such disorder is a physical quantity called entropy.

Thus,

the MORE ORDER, the LESS ENTROPY,
the LESS ORDER, the MORE ENTROPY.

Equations of connection between entropy ( S) and other quantities are studied in physics and physical chemistry courses. Entropy unit [ S] = 1 J/K.
Entropy increases when a substance is heated and decreases when it cools. It increases especially strongly during the transition of a substance from solid to liquid and from liquid to gaseous state.
What happened in our experience?
When two different gases were mixed, the degree of disorder increased. Consequently, the entropy of the system has increased. With zero thermal effect, this was the reason for the spontaneous occurrence of the process.
If we now want to separate the mixed gases, then we will have to do work , that is, to expend energy for this. Spontaneously (due to thermal movement), mixed gases will never separate!
So, we have discovered two factors that determine the possibility of many processes, including chemical reactions:
1) the system’s desire to minimize energy ( energy factor) And
2) the system’s desire for maximum entropy ( entropy factor).
Let us now see how various combinations of these two factors affect the possibility of chemical reactions occurring.
1. If, as a result of the proposed reaction, the energy of the reaction products turns out to be less than the energy of the starting substances, and the entropy is greater (“downhill to greater disorder”), then such a reaction can and will proceed exothermic.
2. If, as a result of the proposed reaction, the energy of the reaction products turns out to be greater than the energy of the starting substances, and the entropy is less (“uphill to greater order”), then such a reaction does not proceed.
3. If in the proposed reaction the energy and entropy factors act in different directions (“downhill, but to greater order” or “uphill, but to greater disorder”), then without special calculations it is impossible to say anything about the possibility of such a reaction occurring ("who will win"). Think about which of these cases are endothermic reactions.
The possibility of a chemical reaction occurring can be assessed by calculating the change during the reaction of a physical quantity that depends on both the change in enthalpy and the change in entropy in this reaction. This physical quantity is called Gibbs energy(in honor of the 19th century American physical chemist Josiah Willard Gibbs).

G= H–T S

Condition for spontaneous reaction:

G< 0.

At low temperatures, the factor determining the possibility of a reaction occurring is largely the energy factor, and at high temperatures it is the entropy factor. From the above equation, in particular, it is clear why decomposition reactions that do not occur at room temperature (entropy increases) begin to occur at elevated temperatures.

ENDOTHERMIC REACTION, ENTROPY, ENERGY FACTOR, ENTROPY FACTOR, GIBBS ENERGY.
1.Give examples of endothermic processes known to you.
2.Why is the entropy of a sodium chloride crystal less than the entropy of the melt obtained from this crystal?
3. Thermal effect of the reaction of copper reduction from its oxide with carbon

2CuO (cr) + C (graphite) = 2Cu (cr) + CO 2 (g)

is –46 kJ. Write down the thermochemical equation and calculate how much energy is needed to produce 1 kg of copper from this reaction.
4. When calcining calcium carbonate, 300 kJ of heat was expended. At the same time, according to the reaction

CaCO 3 (cr) = CaO (cr) + CO 2 (g) – 179 kJ

24.6 liters of carbon dioxide were formed. Determine how much heat was wasted uselessly. How many grams of calcium oxide were formed?
5.When magnesium nitrate is calcined, magnesium oxide, nitrogen dioxide gas and oxygen are formed. The thermal effect of the reaction is –510 kJ. Make up a thermochemical equation and determine how much heat is absorbed if 4.48 liters of oxygen are released. What is the mass of decomposed magnesium nitrate?

Target: development of skills in drawing up equations of redox processes involving organic compounds.

Methods: story, work with presentation, discussion, independent work, group work.

Teacher:

What are oxidation-reduction reactions from the point of view of the concept of “degree of oxidation of chemical elements”? (slide 2)

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

Let's 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 comparing the oxidation states of atoms in the starting materials and reaction products. The main requirement when composing equations using this method is that the number of electrons given must be equal to the number of electrons received.

Oxidation-reduction reactions are reactions in which electrons transfer from one atom, molecule or ion to another.

Oxidation is the process of losing electrons and increasing the oxidation state.

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

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

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

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

Independent work No. 2 according to the instruction card: using the electronic balance method, find and put the coefficients in the following redox reaction scheme:

MnO 2 + H 2 SO 4 → MnSO 4 + O 2 + H 2 O (2MnO 2 + 2H 2 SO 4 → 2MnSO 4 + O 2 +2H 2 O)

Teacher:

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

In order to understand in which cases elements behave as oxidizing agents, and in which - as reducing agents, you need to turn to the periodic table of D.I. Mendeleev. If we are talking about simple substances, then reducing properties should be inherent in those elements that have a larger atomic radius compared to others and a small (1 - 3) number of electrons at the external energy level. Therefore, they can give them away relatively easily. These are mostly metals. The most powerful reducing properties of them are the alkali and alkaline earth metals located in the main subgroups of groups I and II (for example, sodium, potassium, calcium, etc.).

The most typical nonmetals, which have a close to complete structure of the outer electron layer and a significantly smaller atomic radius compared to metals of the same period, quite easily accept electrons and behave as oxidizing agents in redox reactions. The most powerful oxidizing agents are light elements of the main subgroups VI – VII groups, for example fluorine, chlorine, bromine, oxygen, sulfur, etc.

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

Elements in intermediate positive oxidation states can both donate and accept electrons and are therefore able to behave as oxidizing or reducing agents depending on conditions. For example, N +3, S +4. When placed in an environment with a strong oxidizing agent, they behave as reducing agents. And, conversely, in a reducing environment they behave as oxidizing agents.

Based on their redox properties, substances can be divided into three groups:

oxidizing agents

reducing agents

oxidizing agents - reducing agents

Independent work No. 3 on the instruction card: in which of the given reaction equation schemes MnO 2 exhibits the properties of an oxidizing agent, and in which - the properties of a reducing agent:

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

MnO 2 + 4HCI = 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) Equation for the interaction of zinc with dilute H 2 SO 4 (slide 3)

Which ion is the oxidizing agent in this reaction? (H+)

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

B) Let's consider another reaction - the interaction of zinc with concentrated H 2 SO 4 (slide 4)

Which atoms change oxidation state? (zinc and sulfur)

Concentrated sulfuric acid (98%) contains 2% water, and the salt is obtained in solution. The reaction actually involves sulfate ions. The reduction product is hydrogen sulfide.

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

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

The diagrams indicate products whose content is the highest among possible acid reduction products

Gold and platinum do not react with HNO3, but these metals dissolve in “regia vodka” - a mixture of concentrated hydrochloric and nitric acids in a 3: 1 ratio.

Au + 3HCI (conc.) + HNO 3 (conc.) = AuCI 3 + NO + 2H 2 O

3. The most powerful oxidizing agent among simple substances is fluorine. But it is too active and difficult to obtain in free form. Therefore, in laboratories, potassium permanganate KMnO 4 is used as an oxidizing agent. Its oxidizing ability depends on the concentration of the solution, temperature and environment.

Creating a problem situation: I was preparing a solution of potassium permanganate (“potassium permanganate”) for the lesson, spilled a glass with the solution and stained my favorite chemical coat. Suggest (after performing a laboratory experiment) a substance that can be used to clean the robe.

Oxidation–reduction reactions can occur in various environments. Depending on the environment, the nature of the reaction between the same substances may change: the environment affects the change in the oxidation states of atoms.

Typically, sulfuric acid is added to create an acidic environment. Hydrochloric and nitrogen are used less frequently, because the first is capable of oxidizing, and the second is itself a strong oxidizing agent and can cause side processes. To create an alkaline environment, potassium or sodium hydroxide is used, and water is used to create a neutral environment.

Laboratory experience: (TB regulations)

1-2 ml of a diluted solution of potassium permanganate is poured into four numbered test tubes. Add a few drops of sulfuric acid solution to the first test tube, water to the second, potassium hydroxide to the third, and leave the fourth test tube as a control. Then pour sodium sulfite solution into the first three test tubes, shaking gently. Check. How does the color of the solution change in each test tube? (slides 7, 8)

Results of laboratory experiment:

Reduction products KMnO 4 (MnO 4) -:

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. (slide 9,)

To the reaction schemes:

KMnO 4 + Na 2 SO 3 + H 2 SO 4 → MnSO 4 + Na 2 SO 4 + K 2 SO 4 + H 2 O

KMnO 4 + Na 2 SO 3 + H 2 O → MnO 2 ↓ + Na 2 SO 4 + KOH

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

Select odds using the electronic balance method. Specify the oxidizing agent and reducing agent (slide 10)

(The task is multi-level: strong students write down the reaction products independently)

You have done a laboratory experiment, suggest a substance that can be used to clean the gown.

Demonstration experience:

Stains from a solution of potassium permanganate are quickly removed with a solution of hydrogen peroxide acidified 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 potassium permanganate stains contain manganese(IV) oxide, so another reaction will occur:

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

After removing stains, the piece of fabric must be rinsed with water.

Teacher:

The importance of redox reactions

Goal: Show students the importance of redox reactions in chemistry, technology, and everyday human life. Methods: presentation, discussion, independent work, group work.

It is impossible to consider all the variety of redox reactions within the framework of one lesson. But their importance in chemistry, technology, and everyday human life cannot be overestimated. Redox reactions underlie the production of metals and alloys, hydrogen and halogens, alkalis and drugs. The functioning of biological membranes and many natural processes are associated with redox reactions: metabolism, fermentation, respiration, photosynthesis. Without understanding the essence and mechanisms of redox reactions, it is impossible to imagine the operation of chemical power sources (accumulators and batteries), the production of protective coatings, and the masterful processing of metal surfaces of products. For the purposes of bleaching and disinfection, the oxidizing properties of such well-known agents as hydrogen peroxide, potassium permanganate, chlorine and chlorine, or bleach, are used. Chlorine, as a strong oxidizing agent, is used to sterilize clean water and disinfect wastewater.

Working with the presentation, writing in a notebook.

Reactions during which the elements that make up the reacting substances change the oxidation state are called oxidation-reduction (ORR).

Oxidation state. To characterize the state of elements in compounds, the concept of oxidation state was introduced. The oxidation state (s.o.) is a conditional charge that is assigned to an atom under the assumption that all bonds in the molecule or ion are extremely polarized. The oxidation state of an element within a molecule of a substance or ion is defined as the number of electrons displaced from an atom of a given element (positive oxidation state) or to an atom of a given element (negative oxidation state). To calculate the oxidation state of an element in a compound, one should proceed from the following provisions (rules):

1. The oxidation state of elements in simple substances, in metals in the elemental state, in compounds with non-polar bonds is equal to zero. Examples of such compounds are N 2 0, H 2 0, Cl 2 0, I 2 0, Mg 0, Fe 0, etc.

2. In complex substances, elements with higher electronegativity have a negative oxidation state.

Since during the formation of a chemical bond, electrons are displaced to the atoms of more electronegative elements, the latter have a negative oxidation state in compounds.

O -2 Cl O -2 N + Element EO

In some cases, the oxidation state of an element numerically coincides with the valence (B) of the element in a given compound, as, for example, in HClO 4.

The examples below show that the oxidation state and valence of an element can vary numerically:

N ≡ N В (N)=3; s.o.(N)=0

H + C -2 O -2 H +

EO (C) = 2.5 V(C) = 4 s.o.(C) = -2

EO (O) = 3.5 V (O) = 2 s.o. (O) = -2

EO (N) = 2.1 V(N) = 1 s.o.(N) = +1

3. There are higher, lower and intermediate oxidation states.

Highest oxidation state– this is its greatest positive value. The highest oxidation state is usually equal to the group number (N) of the periodic table in which the element is found. For example, for elements of period III it is equal to: Na +2, Mg +2, AI +3, Si +4, P +5, S +6, CI +7. The exceptions are fluorine, oxygen, helium, neon, argon, as well as elements of the cobalt and nickel subgroup: their highest oxidation state is expressed by a number whose value is lower than the number of the group to which they belong. Elements of the copper subgroup, on the contrary, have a highest oxidation state greater than one, although they belong to group I.

Lowest degree oxidation is determined by the number of electrons missing to the stable state of the atom ns 2 nр 6. The lowest oxidation state for non-metals is (N-8), where N is the number of the group of the periodic table in which the element is located. For example, for non-metals of the III period it is equal to: Si -4, P -3, S -2, CI ˉ. The lowest oxidation state for metals is its lowest possible positive value. For example, manganese has the following oxidation states: Mn +2, Mn +4, Mn +6, Mn +7; d.o.=+2 is the lowest oxidation state for manganese.

All other occurring oxidation states of an element are called intermediate. For example, for sulfur, the oxidation state of +4 is intermediate.

4. A number of elements exhibit a constant oxidation state in complex compounds:

a) alkali metals – (+1);

b) metals of the second group of both subgroups (except for Нg) – (+2); mercury can exhibit oxidation states (+1) and (+2);

c) metals of the third group, the main subgroup – (+3), with the exception of Tl, which can exhibit oxidation states (+1) and (+3);

e) H +, except for metal hydrides (NaH, CaH 2, etc.), where its oxidation state is (-1);

f) O -2, with the exception of peroxides of elements (H 2 O 2, CaO 2, etc.), where the oxidation state of oxygen is (-1), superoxides of elements

(KO 2, NaO 2, etc.), in which its oxidation state is – ½, fluoride

oxygen ОF 2.

5. Most elements can exhibit varying degrees of oxidation in compounds. When determining their oxidation state, they use the rule according to which the sum of the oxidation states of elements in electrically neutral molecules is equal to zero, and in complex ions - the charge of these ions.

As an example, let's calculate the oxidation state of phosphorus in orthophosphoric acid H 3 PO 4. The sum of all oxidation states in a compound must be equal to zero, so we denote the oxidation state of phosphorus by X and, multiplying the known oxidation states of hydrogen (+1) and oxygen (-2) by the number of their atoms in the compound, we create the equation: (+1)* 3+X+(-2)*4 = 0, of which X = +5.

Let's calculate the oxidation state of chromium in the dichromate ion (Cr 2 O 7) 2-.

The sum of all oxidation states in a complex ion must be equal to (-2), so let’s denote the oxidation state of chromium by X and create the equation 2X + (-2)*7 = -2, from which X = +6.

The concept of oxidation state for most compounds is conditional, because does not reflect the real effective charge of the atom. In simple ionic compounds, the oxidation state of their constituent elements is equal to the electric charge, since during the formation of these compounds there is an almost complete transfer of electrons from one

1 -1 +2 -1 +3 -1

atom to another: NaI, MgCI 2, AIF 3. For a compound with a polar covalent bond, the actual effective charge is less than the oxidation number, but this concept is very widely used in chemistry.

The main provisions of the theory of OVR:

1. Oxidation is the process of giving up electrons by an atom, molecule or ion. Particles that donate electrons are called reducing agents; during the reaction they are oxidized, forming an oxidation product. In this case, the elements involved in oxidation increase their oxidation state. For example:

AI – 3e -  AI 3+

H 2 – 2e -  2H +

Fe 2+ - e -  Fe 3+

2. Recovery is the process of adding electrons to an atom, molecule or ion. Particles that gain electrons are called oxidizing agents; during the reaction they are reduced to form a reduction product. In this case, the elements participating in the reduction reduce their oxidation state. For example:

S + 2e -  S 2-

CI 2 + 2e -  2 CI ˉ

Fe 3+ + e -  Fe 2+

3. Substances containing reducing or oxidizing particles are respectively called reducing agents or oxidizing agents. For example, FeCI 2 is a reducing agent due to Fe 2+, and FeCI 3 is an oxidizing agent due to Fe 3+.

4. Oxidation is always accompanied by reduction and, conversely, reduction is always associated with oxidation. Thus, ORR represents the unity of two opposite processes - oxidation and reduction

5. The number of electrons donated by the reducing agent is equal to the number of electrons accepted by the oxidizing agent.

Drawing up equations of redox reactions. Two methods of composing equations for OVR are based on the last rule:

1. Electronic balance method.

Here, the number of electrons gained and lost is calculated based on the oxidation states of the elements before and after the reaction. Let's look at the simplest example:

Na0+Cl  Na + Cl

2Na 0 – eˉ  Na + - oxidation

1 Cl 2 + 2eˉ  2 Cl - recovery

2 Na + Cl 2 = 2Na + + 2Cl

2 Na + Cl 2 = 2 NaCl

This method is used if the reaction does not occur in solution (in the gas phase, thermal decomposition reaction, etc.).

2. Ion-electronic method (half-reaction method).

This method takes into account the solution environment and gives an idea of the nature of the particles that actually exist and interact in solutions. Let's look at it in more detail.

Algorithm for selecting coefficients using the ion-electronic method:

1. Draw up a molecular diagram of the reaction indicating the starting materials and reaction products.

2. Draw up a complete ionic-molecular reaction scheme, writing weak electrolytes, sparingly soluble, insoluble and gaseous substances in molecular form, and strong electrolytes in ionic form.

3. Having excluded from the ion-molecular scheme ions that do not change as a result of the reaction (without taking into account their quantity), rewrite the scheme in a brief ion-molecular form.

4. Identify the elements that change their oxidation state as a result of the reaction; find the oxidizing agent, reducing agent, reduction products, oxidation.

5. Draw up diagrams of half-reactions of oxidation and reduction, for this:

a) indicate the reducing agent and oxidation product, oxidizing agent and reduction product;

b) equalize the number of atoms of each element in the left and right sides of the half-reactions (perform a balance by element) in the sequence: element changing the oxidation state, oxygen, other elements; It should be remembered that in aqueous solutions, H 2 O molecules, H + or OH – ions can participate in reactions, depending on the nature of the medium:

c) equalize the total number of charges in both parts of the half-reactions; To do this, add or subtract the required number of electrons on the left side of the half-reactions (charge balance).

6. Find the least common multiple (LCM) for the number of electrons given and received.

7. Find the main coefficients for each half-reaction. To do this, divide the number obtained in step 6 (LCM) by the number of electrons appearing in this half-reaction.

8. Multiply the half-reactions by the obtained main coefficients, add them together: the left side with the left, the right side with the right (get the ionic-molecular equation of the reaction). If necessary, “bring similar” ions taking into account the interaction between hydrogen ions and hydroxide ions: H + +OH ˉ= H 2 O.

9. Arrange the coefficients in the molecular equation of the reaction.

10. Conduct a check for particles that are not involved in the ORR, excluded from the complete ion-molecular scheme (item 3). If necessary, coefficients for them are found by selection.

11. Perform final oxygen check.

1. Acidic environment.

Molecular reaction scheme:

KMnO 4 + NaNO 2 + H 2 SO 4  MnSO 4 + NaNO 3 + H 2 O + K 2 SO 4

Full ion-molecular reaction scheme:

K + +MnO +Na++NO +2H++SO  Mn 2+ + SO + Na + + NO + H 2 O + 2K + +SO .

Brief ion-molecular reaction scheme:

MnO +NO +2H +  Mn 2+ + NO +H2O

ok product ok product ok

During the reaction, the oxidation state of Mn decreases from +7 to +2 (manganese is reduced), therefore, MnO – oxidizing agent; Mn 2+ – reduction product. The oxidation degree of nitrogen increases from +3 to +5 (nitrogen is oxidized), therefore NO – reducing agent, NO – oxidation product.

Half reaction equations:

2MnO + 8 H+ + 5e -  Mn 2+ + 4 H 2 O- recovery process

10 +7 +(-5) = +2

5 NO + H 2 O– 2e -  NO + 2 H+ - oxidation process

2MnO + 16H + + 5NO + 5H 2 O = 2Mn 2+ +8H 2 O + 5NO + 1OH + (complete ion-molecular equation).

In the overall equation, we exclude the number of identical particles located on both the left and right sides of the equality (we present similar ones). In this case, these are H + and H 2 O ions.

The short ion-molecular equation will be

2MnO + 6H + + 5NO  2Mn 2+ + 3H 2 O + 5NO .