Many substances have special properties, which in chemistry are usually called oxidizing or reducing.

Some chemical substances exhibit the properties of oxidizing agents, others - reducing agents, while some compounds can exhibit both properties simultaneously (for example, hydrogen peroxide H 2 O 2).

What are oxidizing and reducing agents, oxidation and reduction?

The redox properties of a substance are associated with the process of giving and receiving electrons by atoms, ions or molecules.

An oxidizing agent is a substance that accepts electrons during a reaction, i.e., is reduced; reducing agent - gives up electrons, i.e. oxidizes. The processes of transferring electrons from one substance to another are usually called redox reactions.

Compounds containing atoms of elements with the maximum oxidation state can only be oxidizing agents due to these atoms, because they have already given up all their valence electrons and are only able to accept electrons. The maximum oxidation state of an element's atom is equal to the number of the group in the periodic table to which the element belongs. Compounds containing atoms of elements with a minimum oxidation state can only serve as reducing agents, since they are only capable of donating electrons, because the outer energy level of such atoms is completed by eight electrons

Restorers		Oxidizing agents
Hydrogen, carbon Carbon monoxide Hydrogen sulfide Sulfur(IV) oxide Sulfurous acid and its salts Hydrogen halides Metal cations in lower degrees oxidation: Nitrous acid Hydrazine Cathode at electrolysis	SnCl 2, FeCl 2, MnSO 4, Cr 2 (SO 4) 3	Halogens Permanganates Manganates Manganese(IV) oxide Dichromats Nitric acid Sulfuric acid Lead(IV) oxide Hydrogen peroxide Mononasulfuric acid Dipersulfuric acids Metal cations in higher degrees oxidation: Potassium chlorate Anode during electrolysis	F 2 ; Cl2; I 2 ; Br 2 KMnO 4 K 2 Cr 2 O 7 K 2 CrO 4 H 2 SO 4 conc. PbO2 TlCl 3 , Au(CNS) 3

Compounds containing atoms of elements with intermediate oxidation states can be both oxidizing and reducing agents, depending on the partner with which they react and the reaction conditions. Thus, the typical oxidizing agent hydrogen peroxide, when interacting in an acidic environment with potassium permanganate, turns out to be a reducing agent:

5 H 2 O 2 + 2 KMnO 4 + 3 H 2 SO 4 = 2 MnSO 4 + K 2 SO 4 + 5 O 2 + 8 H 2 O,

and the typical reducing agent sodium sulfite oxidizes alkali metal sulfides:

Na 2 SO 3 + 2 Na 2 S+ 3 H 2 O = 3 S  + 6 NaOH.

In addition, reducing agents containing atoms in the lowest oxidation state can be oxidizing agents at the expense of another element. For example, a typical reducing agent ammonia can oxidize alkali metals at the expense of hydrogen atoms:

NH 3 + Na = NaH 2 N + 1/2 H 2.

Compilation of OVR equations

Redox reactions are described by reaction equations that reflect the amounts of substances that interact and the resulting products. To compile ORR equations, use or electronic balance method (scheme method), or electron-ion balance (half-reaction method).

The electronic balance method is more universal, since it allows one to establish stoichiometric ratios in ORR in any homo- and heterogeneous systems.

Electronic balance method‑ a method of finding coefficients in the equations of redox reactions, which considers the exchange of electrons between atoms of elements that change their oxidation state. The number of electrons given up by the reducing agent is equal to the number of electrons gained by the oxidizing agent.

The equation is compiled in several stages:

1. Write down the reaction scheme:

KMnO 4 + HCl → KCl + MnCl 2 + Cl 2 + H 2 O.

2. Place oxidation states above the signs of elements that change the oxidation state:

KMn +7 O 4 + HCl -1 → KCl + Mn +2 Cl 2 + Cl 2 0 + H 2 O.

3. Identify elements that change oxidation states and determine the number of electrons acquired by the oxidizing agent and given up by the reducing agent:

Mn +7 + 5ē → Mn +2.

2Cl -1 - 2ē → Cl 2 0.

4. Equalize the number of acquired and donated electrons, thereby establishing coefficients for compounds that contain elements that change the oxidation state:

Mn +7 + 5ē → Mn +2
2Cl -1 – 2ē → Cl 2 0

––––––––––––––––––––––––

2Mn +7 + 10Cl -1 → 2Mn +2 + 5Cl 2 0.

5. Select coefficients for the remaining participants in the reaction:

2KMn +7 O 4 + 16HCl -1 → 2KCl + 2Mn +2 Cl 2 + 5Cl 2 0 + 8H 2 O.

To select coefficients for equations of reactions occurring in aqueous solutions, the half-reaction method is preferable.

Firstly, it allows you to omit the operations of determining the oxidation state of elements.

Secondly, in the process of using it, an abbreviated ionic equation of the redox reaction is immediately obtained.

Thirdly, using the half-reaction equation, it is possible to establish the influence of the environment on the nature of the process.

In addition, when compiling an electron-ion balance, one operates with ions that actually exist in an aqueous solution, in contrast to the electron balance method, which deals with hypothetical particles such as Mn +7, Cr +6.

Electron-ion balance method (half-reaction method).

This method considers the transfer of electrons from one atom or ion to another, taking into account the nature of the medium (acidic, alkaline or neutral) in which the reaction occurs. When compiling equations for oxidation and reduction processes, to equalize the number of hydrogen and oxygen atoms, either water molecules and hydrogen ions are introduced (depending on the medium) (if the environment is acidic), or water molecules and hydroxide ions (if the environment is alkaline). Accordingly, in the resulting products, on the right side of the electron-ion equation there will be hydrogen ions and water molecules (acidic environment) or hydroxide ions and water molecules (alkaline environment).

That is, when writing electron-ion equations, one must proceed from the composition of the ions actually present in the solution . In addition, as in writing abbreviated ionic equations, substances that dissociate poorly, are poorly soluble, or are released as a gas should be written in molecular form.

Consider for example the following reaction:

H 2 O 2 + KMnO 4 + H 2 SO 4 → MnSO 4 + O 2 + H 2 O + K 2 SO 4.

When finding the stoichiometric coefficients of the equation of the redox process, the following operations must be performed.

1. Identify the oxidizing agent and reducing agent among the reacting substances. In our example, the oxidizing agent is KMnO 4, the reducing agent is H 2 O 2 and the products of their interaction are Mn 2+ and O 2.

2. Write out the half-reaction schemes:

H 2 O 2 → O 2 oxidation;

MnO → Mn 2+. recovery.

3. Equalize the schemes:

a) by element that changes the oxidation state (in our example this is not required);

b) by oxygen, adding it where needed in the form of water molecules if the reaction occurs in an acidic environment, and in the form of a hydroxide ion if the reaction occurs in an alkaline environment:

H 2 O 2 → O 2;

MnO → Mn 2+ + 4 H 2 O;

c) by hydrogen, adding it in the form of hydrogen ions if the reaction occurs in an acidic environment, and in the form of water molecules if the reaction occurs in an alkaline environment if:

H 2 O 2 → O 2 + 2 H +;

MnO+ 8 H + → Mn 2+ + 4 H 2 O;

d) by the total charge of the ions, adding or subtracting the required number of electrons:

H 2 O 2 - 2ē → O 2 + 2 H +;

MnO 4 - + 8 H + + 5 ē → Mn 2+ + 4H 2 O.

4. Taking into account the law of electroneutrality, equalize the number of donated and accepted electrons and sum up the left and right parts of the half-reactions separately:

H 2 O 2 - 2ē → O 2 + 2 H + | 2| 5

MnO+ 8 H + + 5 ē →Mn 2+ + 4 H 2 O | 5| 2

____________________________________________

5 H 2 O 2 + 2 MnO+ 16 H + = 5 O 2 + 10 H + + 2 Mn 2+ +8 H 2 O.

Reducing, we obtain the equation of this redox process in ionic form:

5 H 2 O 2 + 2 MnO+ 6 H + = 5 O 2 + 2 Mn 2+ +8 H 2 O.

5. Go to the molecular form of the equation, adding cations and anions that remain unchanged as a result of the reaction, that is, salt-forming ions (in our example, K + and SO 4 2- ions):

5 H 2 O 2 + 2 KMnO 4 + 3 H 2 SO 4 = 5 O 2 + 8 H 2 O + K 2 SO 4.

Let's consider another example - the process of pyrite oxidation with concentrated nitric acid.

1. Let's determine the oxidizing agent and the reducing agent among the reacting substances. In our example, the oxidizing agent is HNO 3, the reducing agent is FeS 2. Let us determine the reaction products. Nitric acid HNO 3 is a strong oxidizing agent, so sulfur will be oxidized to the maximum oxidation state S 6+, and iron to Fe 3+, while HNO 3 can be reduced to NO:

FeS 2 +HNO 3 → Fe(NO 3) 3 + H 2 SO 4 + NO.

2. Let's write out the half-reaction schemes

FeS 2 → Fe 3+ +SO oxidation;

NO→NO recovery.

3. We equalize the schemes:

FeS 2 + 8H 2 O - 15ē → Fe 3+ + 2SO + 16H +;

NO+4H + +3 ē → NO + 2H 2 O.

4. Taking into account the law of electrical neutrality, we equalize the number of donated and accepted electrons and sum up the left and right parts of the half-reactions separately:

FeS 2 + 8H 2 O - 15ē → Fe 3+ + 2SO+ 16H + | 15 | 1

NO+ 4H + +3 ē → NO + 2H 2 O | 3 | 5

FeS 2 + 8H 2 O +5NO+ 20H + =Fe 3+ +2SO+16H + + 5NO + 10H 2 O.

5. Reducing, we get the equation in ionic form:

FeS 2 + 5NO+ 4H + = Fe 3+ + 2SO + 5NO + 2H 2 O.

6. Let’s write the equation in molecular form, taking into account that some of the nitrate ions were not reduced, but participated in the exchange reaction, and some of the H + ions are present in the reaction products (H 2 SO 4):

Note that you never had to determine the oxidation state of the elements to determine the number of electrons given and received. In addition, we took into account the influence of the environment and automatically determined that H 2 O is on the right side of the equation. There is no doubt that this method is much more consistent with chemical meaning than the standard electronic balance method.

Chapter 10

Redox reactions.

Redox reactions – These are reactions that occur with a change in the oxidation states of the atoms of the elements that make up the molecules of the reacting substances:

2Mg + O 2  2MgO,

2KClO 3 2KCl + 3O 2.

Let us remind you that oxidation state – This is the conditional charge of an atom in a molecule, arising from the assumption that the electrons are not displaced, but are completely given to the atom of a more electronegative element.

The most electronegative elements in a compound have negative oxidation states, and the atoms of elements with less electronegativity have positive oxidation states.

Oxidation state is a formal concept; in some cases, the oxidation state of an element does not coincide with its valency.

To find the oxidation state of the atoms of the elements that make up the reacting substances, the following rules should be kept in mind:

1. The oxidation state of atoms of elements in molecules of simple substances is zero.

For example:

Mg 0 , Cu 0 .

2. The oxidation state of hydrogen atoms in compounds is usually +1.

For example: +1 +1

Exceptions: in hydrides (hydrogen compounds with metals), the oxidation state of hydrogen atoms is –1.

For example:

NaH –1.

3. The oxidation state of oxygen atoms in compounds is usually –2.

For example:

H 2 O –2, CaO –2.

Exceptions:

 The oxidation state of oxygen in oxygen fluoride (OF 2) is +2.

 the oxidation degree of oxygen in peroxides (H 2 O 2, Na 2 O 2) containing the –O–O– group is –1.

4. The oxidation state of metals in compounds is usually a positive value.

For example: +2

5. The oxidation state of non-metals can be both negative and positive.

For example: –1 +1

6. Amount c The oxidation state of all atoms in a molecule is zero.

Redox reactions are two interrelated processes - the oxidation process and the reduction process.

Oxidation process – is the process of giving up electrons by an atom, molecule or ion; in this case, the oxidation state increases, and the substance is a reducing agent:

– 2ē  2H + oxidation process,

Fe +2 – ē  Fe +3 oxidation process,

2J – – 2ē  oxidation process.

The reduction process is the process of adding electrons, while the oxidation state decreases, and the substance is an oxidizing agent:

+ 4ē  2O –2 reduction process,

Mn +7 + 5ē  Mn +2 reduction process,

Cu +2 +2ē  Cu 0 reduction process.

Oxidant – a substance that accepts electrons and is reduced in the process (the oxidation state of the element decreases).

Reducing agent – a substance that gives up electrons and is oxidized (the oxidation state of the element decreases).

It is possible to make a reasonable conclusion about the nature of the behavior of a substance in specific redox reactions based on the value of the redox potential, which is calculated from the value of the standard redox potential. However, in a number of cases, it is possible, without resorting to calculations, but knowing the general laws, to determine which substance will be an oxidizing agent and which will be a reducing agent, and make a conclusion about the nature of the redox reaction.

Typical reducing agents are:

 some simple substances:

metals: for example Na, Mg, Zn, Al, Fe,

non-metals: for example, H 2, C, S;

 some complex substances: for example, hydrogen sulfide (H 2 S) and sulfides (Na 2 S), sulfites (Na 2 SO 3), carbon monoxide (II) (CO), hydrogen halides (HJ, HBr, HCI) and salts of hydrohalic acids (KI, NaBr), ammonia (NH 3);

 metal cations in lower oxidation states: for example, SnCl 2, FeCl 2, MnSO 4, Cr 2 (SO 4) 3;

 cathode during electrolysis.

Typical oxidizing agents are:

 some simple substances - non-metals: for example, halogens (F 2, CI 2, Br 2, I 2), chalcogens (O 2, O 3, S);

 some complex substances: for example, nitric acid (HNO 3), sulfuric acid (H 2 SO 4 conc.), potassium premanganate (K 2 MnO 4), potassium bichromate (K 2 Cr 2 O 7), potassium chromate (K 2 CrO 4), manganese (IV) oxide (MnO 2), lead (IV) oxide (PbO 2), potassium chlorate (KCIO 3), hydrogen peroxide (H 2 O 2);

 anode during electrolysis.

When drawing up equations for redox reactions, it should be kept in mind that the number of electrons given up by the reducing agent is equal to the number of electrons accepted by the oxidizing agent.

There are two methods for composing equations for redox reactions - electron balance method and electron-ion method (half-reaction method) .

When compiling equations for redox reactions using the electronic balance method, a certain procedure must be followed. Let us consider the procedure for composing equations using this method using the example of the reaction between potassium permanganate and sodium sulfite in an acidic medium.

We write down the reaction scheme (indicate the reagents and reaction products):

We determine the oxidation state of atoms of elements that change its value:

7 + 4 + 2 + 6

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

3) We draw up an electronic balance diagram. To do this, we write down the chemical signs of the elements whose atoms change their oxidation state, and determine how many electrons the corresponding atoms or ions give up or gain.

We indicate the processes of oxidation and reduction, oxidizing agent and reducing agent.

We equalize the number of given and received electrons and, thus, determine the coefficients for the reducing agent and oxidizing agent (in this case they are equal to 5 and 2, respectively):

5 S +4 – 2 e- → S +6 oxidation process, reducing agent

2 Mn +7 + 5 e- → Mn +2 reduction process, oxidizing agent.

2KMnO 4 + 5Na 2 SO 3 + 8H 2 SO 4 = 2MnSO 4 + 5Na 2 SO 4 + K 2 SO 4 + 8H 2 O.

5) If hydrogen and oxygen do not change their oxidation states, then their number is calculated last and the required number of water molecules is added to the left or right side of the equation.

Redox reactions are divided into three types: intermolecular, intramolecular and self-oxidation – self-healing (disproportionation) reactions.

Reactions of intermolecular oxidation - reduction are called redox reactions in which the oxidizing and reducing agents are represented by molecules of different substances.

For example:

2Al + Fe 2 O 3 = 2Fe + Al 2 O 3,

Al 0 – 3e – → Al +3 oxidation, reducing agent,

Fe +3 +3e – → Fe 0 reduction, oxidizing agent.

In this reaction, the reducing agent (Al) and the oxidizing agent (Fe +3) are part of different molecules.

Intramolecular oxidation reactions – recovery reactions are called in which an oxidizing agent and a reducing agent are part of one molecule (and are represented either by different elements or by one element, but with different oxidation states):

2 KClO 3 = KCl + 3O 2

2 CI +5 + 6e – → CI –1 reduction, oxidizing agent

3 2O –2 – 4е – → oxidation, reducing agent

In this reaction, the reducing agent (O –2) and the oxidizing agent (CI +5) are part of one molecule and are represented by different elements.

In the reaction of thermal decomposition of ammonium nitrite, atoms of the same chemical element - nitrogen, which are part of one molecule - change their oxidation states:

NH 4 NO 2 = N 2 + 2H 2 O

N –3 – 3e – → N 0 reduction, oxidizing agent

N +3 + 3e – → N 0 oxidation, reducing agent.

Reactions of this type are often called reactions counter-proportionation .

Autoxidation reactions– self-healing(disproportionation) – These are reactions during which the same element with the same oxidation state both increases and decreases its oxidation state.

For example: 0 -1 +1

Cl 2 + H 2 O = HCI + HCIO

CI 0 + 1e – → CI –1 reduction, oxidizing agent

CI 0 – 1e – → CI +1 oxidation, reducing agent.

Disproportionation reactions are possible when the element in the starting substance has an intermediate oxidation state.

The properties of simple substances can be predicted by the position of the atoms of their elements in the periodic table of elements D.I. Mendeleev. Thus, all metals in redox reactions will be reducing agents. Metal cations can also be oxidizing agents. Nonmetals in the form of simple substances can be both oxidizing and reducing agents (excluding fluorine and inert gases).

The oxidizing ability of nonmetals increases in a period from left to right, and in a group - from bottom to top.

Reducing abilities, on the contrary, decrease from left to right and from bottom to top for both metals and non-metals.

If the redox reaction of metals occurs in solution, then to determine the reducing ability, use range of standard electrode potentials (metal activity series). In this series, metals are arranged as the reducing ability of their atoms decreases and the oxidizing ability of their cations increases ( see table 9 applications ).

The most active metals, standing in the series of standard electrode potentials up to magnesium, can react with water, displacing hydrogen from it.

For example:

Ca + 2H 2 O = Ca(OH) 2 + H 2

When interacting metals with salt solutions, it should be borne in mind that each more active metal (not interacting with water) is capable of displacing (reducing) the metal behind it from the solution of its salt.

Thus, iron atoms can reduce copper cations from a solution of copper sulfate (CuSO 4):

Fe + CuSO 4 = Cu + FeSO 4

Fe 0 – 2e – = Fe +2 oxidation, reducing agent

Cu +2 + 2e – = Cu 0 reduction, oxidizing agent.

In this reaction, iron (Fe) is placed in the activity series before copper (Cu) and is the more active reducing agent.

The reaction of, for example, silver with a solution of zinc chloride will be impossible, since silver is located in a series of standard electrode potentials to the right of zinc and is a less active reducing agent.

All metals that are in the activity series before hydrogen can displace hydrogen from solutions of ordinary acids, that is, reduce it:

Zn + 2HCl = ZnCI 2 + H 2

Zn 0 – 2e – = Zn +2 oxidation, reducing agent

2H + + 2e – → reduction, oxidizing agent.

Metals that are in the activity series after hydrogen will not reduce hydrogen from solutions of ordinary acids.

To determine whether there may be oxidizing agent or reducing agent complex substance, it is necessary to find the oxidation state of the elements that make it up. Elements found in highest oxidation state , can only lower it by accepting electrons. Hence, substances whose molecules contain atoms of elements in the highest oxidation state will only be oxidizing agents .

For example, HNO 3, KMnO 4, H 2 SO 4 in redox reactions will act only as an oxidizing agent. The oxidation states of nitrogen (N +5), manganese (Mn +7) and sulfur (S +6) in these compounds have maximum values ​​(coincide with the group number of a given element).

If elements in compounds have a lower oxidation state, then they can only increase it by donating electrons. At the same time, such substances containing elements in the lowest oxidation state will act only as a reducing agent .

For example, ammonia, hydrogen sulfide and hydrogen chloride (NH 3, H 2 S, HCI) will only be reducing agents, since the oxidation states of nitrogen (N –3), sulfur (S –2) and chlorine (Cl –1) are the lowest for these elements .

Substances that contain elements with intermediate oxidation states can be both oxidizing and reducing agents, depending on the specific reaction. Thus, they can exhibit redox duality.

Such substances include, for example, hydrogen peroxide (H 2 O 2), an aqueous solution of sulfur (IV) oxide (sulfurous acid), sulfites, etc. Such substances, depending on environmental conditions and the presence of stronger oxidizing agents (reducing agents), may exhibit in some cases, oxidizing properties, and in others, reducing properties.

As is known, many elements have a variable oxidation state, being part of various compounds. For example, sulfur in the compounds H 2 S, H 2 SO 3, H 2 SO 4 and sulfur S in the free state have oxidation states of –2, +4, +6 and 0, respectively. Sulfur belongs to the elements r-electronic family, its valence electrons are located on the last s- And r-sublevels (...3 s 3r). A sulfur atom with an oxidation state of – 2 valence sublevels is fully completed. Therefore, a sulfur atom with a minimum oxidation state (–2) can only donate electrons (oxidize) and be only a reducing agent. A sulfur atom with an oxidation state of +6 has lost all its valence electrons and in this state can only accept electrons (be reduced). Therefore, the sulfur atom with the maximum oxidation state (+6) can only be an oxidizing agent.

Sulfur atoms with intermediate oxidation states (0, +4) can both lose and gain electrons, that is, they can be both reducing agents and oxidizing agents.

Similar reasoning is valid when considering the redox properties of atoms of other elements.

The nature of the redox reaction is influenced by the concentration of substances, the solution environment and the strength of the oxidizing agent and reducing agent. Thus, concentrated and dilute nitric acid react differently with active and low-active metals. The depth of nitrogen reduction (N+5) of nitric acid (oxidizing agent) will be determined by the activity of the metal (reducing agent) and the concentration (dilution) of the acid.

4HNO 3(conc.) + Cu = Cu(NO 3) 2 + 2NO 2 + 2H 2 O,

8HNO 3(dil.) + 3Cu = 3Cu(NO 3) 2 + 2NO + 4H 2 O,

10HNO 3(conc.) + 4Мg = 4Mg(NO 3) 2 + N 2 O + 5H 2 O,

10HNO 3 (dil.) + 4Mg = 4Mg(NO 3) 2 + NH 4 NO 3 + 3H 2 O.

The reaction of the environment has a significant influence on the course of redox processes.

If potassium permanganate (KMnO4) is used as an oxidizing agent, then depending on the reaction of the solution medium, Mn +7 will be reduced in different ways:

in an acidic environment (up to Mn +2) the reduction product will be a salt, for example, MnSO 4,

in a neutral environment (up to Mn +4) the reduction product will be MnO 2 or MnO(OH) 2,

in an alkaline environment (up to Mn +6) the reduction product will be manganate, for example, K 2 MnO 4.

For example, when reducing a solution of potassium permanganate with sodium sulfite, depending on the reaction of the medium, the corresponding products will be obtained:

sourWednesday –

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

neutralWednesday –

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

alkalineWednesday –

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

The temperature of the system also affects the course of the redox reaction. Thus, the products of the interaction of chlorine with an alkali solution will be different depending on temperature conditions.

When chlorine interacts with cold lye solution the reaction proceeds with the formation of chloride and hypochlorite:

Cl 2 + KOH → KCI + KCIO + H 2 O

CI 0 + 1e – → CI –1 reduction, oxidizing agent

CI 0 – 1e – → CI +1 oxidation, reducing agent.

If you take hot concentrated KOH solution, then as a result of interaction with chlorine we obtain chloride and chlorate:

0 t° -1 +5

3CI 2 + 6KOH → 5KCI + KCIO 3 + 3H 2 O

5 │ CI 0 + 1e – → CI –1 reduction, oxidizing agent

1 │ CI 0 – 5e – → CI +5 oxidation, reducing agent.

10.1. Questions for self-control on the topic

1. What reactions are called redox?

2. What is the oxidation state of an atom? How is it determined?

3. What is the oxidation state of atoms in simple substances?

4. What is the sum of the oxidation states of all atoms in a molecule?

5. What process is called the oxidation process?

6. What substances are called oxidizing agents?

7. How does the oxidation state of an oxidizing agent change in redox reactions?

8. Give examples of substances that are only oxidizing agents in redox reactions.

9. What process is called the recovery process?

10. Define the concept of “reducing agent”.

11. How does the oxidation state of a reducing agent change in redox reactions?

12. What substances can only be reducing agents?

13. Which element is an oxidizing agent in the reaction of dilute sulfuric acid with metals?

14. Which element is an oxidizing agent in the interaction of concentrated sulfuric acid with metals?

15. What function does nitric acid perform in redox reactions?

16. What compounds can be formed as a result of the reduction of nitric acid in reactions with metals?

17. Which element is the oxidizing agent in concentrated, dilute and very dilute nitric acid?

18. What role can hydrogen peroxide play in redox reactions?

19. How are all redox reactions classified?

10.2. Tests for self-testing of theory knowledge on the topic “Oxidation-reduction reactions”

Option #1

1) CuSO 4 + Zn = ZnSO 4 + Cu,

2) CaCO 3 + CO 2 + H 2 O = Ca(HCO 3) 2,

3) SO 3 + H 2 O = H 2 SO 4,

4) FeCl 3 + 3NaOH = Fe(OH) 3 + 3NaCl,

5) NaHCO 3 + NaOH = Na 2 CO 3 + H 2 O.

2. Guided by the structure of atoms, determine under what number the formula of the ion is indicated, which can only be an oxidizing agent:

1) Mn
, 2) NO 3– , 3) ​​Br – , 4) S 2– , 5) NO 2– ?

3. What number is the formula of the substance that is the most powerful reducing agent from among those given below:

1) NO 3–, 2) Cu, 3) Fe, 4) Ca, 5) S?

4. What number indicates the amount of substance KMnO 4, in moles, which interacts with 10 mol of Na 2 SO 3 in the reaction represented by the following scheme:

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

1) 4, 2) 2, 3) 5, 4) 3, 5) 1.

5. What number is given for the disproportionation reaction (auto-oxidation - self-healing)?

1) 2H 2 S + H 2 SO 3 = 3S + 3H 2 O,

2) 4KClO 3 = KCl + 3KClO 4,

3) 2F 2 + 2H 2 O = 4HF + O 2.

4) 2Au 2 O 3 = 4Au + 3O 2,

5) 2KClO 3 = 2KCl + 3O 2.

Option No. 2

1. What number is the equation for the redox reaction?

1) 4KClO 3 = KCl + 3KClO 4,

2) CaCO 3 = CaO + CO 2,

3) CO 2 + Na 2 O = Na 2 CO 3,

4) CuOHCl + HCl = CuCl 2 + H 2 O,

5) Pb(NO 3) 2 + Na 2 SO 4 = PbSO 4 + 2NaNO 3.

2. What number is the formula of a substance that can only be a reducing agent:

1) SO 2, 2) NaClO, 3) KI, 4) NaNO 2, 5) Na 2 SO 3?

3. What number is the formula of the substance that is the most powerful oxidizing agent from among those given:

1) I 2, 2) S, 3) F 2, 4) O 2, 5) Br 2?

4. What number is the volume of hydrogen in liters under normal conditions that can be obtained from 9 g of Al as a result of the following redox reaction:

2Al + 6H 2 O = 2Al(OH) 3 + 3H 2

1) 67,2, 2) 44,8, 3) 33,6, 4) 22,4, 5) 11,2?

5. What number is given for the scheme of the redox reaction that occurs at pH > 7?

1) I 2 + H 2 O → HI + HIO,

2) FeSO 4 + HIO 3 + … → I 2 + Fe(SO 4) 3 + …,

3) KMnO 4 + NaNO 2 + … → MnSO 4 + …,

4) KMnO 4 + NaNO 2 + … → K 2 MnO 4 + …,

5) CrCl 3 + KMnO 4 + … → K 2 Cr 2 O 7 + MnO(OH) 2 + ….

Option No. 3

1. What number is the equation for the redox reaction?

1) H 2 SO 4 + Mg → MgSO 4 + H 2 ,

2) CuSO 4 + 2NaOH →Cu(OH) 2 + Na 2 SO 4,

3) SO 3 + K 2 O → K 2 SO 4,

4) CO 2 + H 2 O → H 2 CO 3,

5) H 2 SO 4 + 2KOH → K 2 SO 4 + 2H 2 O.

2. Guided by the structure of the atom, determine what number is the formula of the ion that can be a reducing agent:

1) Ag + , 2) A l3+ , 3) ​​C l7+ , 4) Sn 2+ , 5) Zn 2+ ?

3. What number is the recovery process listed under?

1) NO 2– → NO 3–, 2) S 2– → S 0, 3) Mn 2+ → MnO 2,

4) 2I – → I 2, 5)
→ 2Cl – .

4. Under what number is the mass of reacted iron given, if as a result of the reaction represented by the following scheme:

Fe + HNO 3 → Fe(NO 3) 3 + NO + H 2 O

11.2 L of NO (no) formed?

1) 2,8, 2) 7, 3) 14, 4) 56, 5) 28.

5. What number is given in the diagram of the reaction of self-oxidation-self-reduction (dismutation)?

1) HI + H 2 SO 4 → I 2 + H 2 S + H 2 O,

2) FeCl 2 + SnCl 4 → FeCl 3 + SnCl 2,

3) HNO 2 → NO + NO 2 + H 2 O,

4) KClO 3 → KCl + O 2,

5) Hg(NO 3) 2 → HgO + NO 2 + O 2.

Answers to test tasks can be found on page

10.3. Questions and exercises for independent

work on studying the topic.

1. Indicate the number or sum of conventional numbers under which the schemes of redox reactions are located:

1) MgCO 3 + HCl  MgCl 2 + CO 2 + H 2 O,

2) FeO + P  Fe + P 2 O 5,

4) H 2 O 2  H3O + O 2, 8) KOH + CO 2  KHCO 3.

2. Indicate the number or sum of conventional numbers under which the redox processes are located:

1) electrolysis of sodium chloride solution,

2) pyrite firing,

3) hydrolysis of sodium carbonate solution,

4) lime slaking.

3. Indicate the number or sum of conventional numbers under which the names of groups of substances characterized by an increase in oxidizing properties are located:

1) chlorine, bromine, fluorine,

2) carbon, nitrogen oxygen,

3) hydrogen, sulfur, oxygen,

4) bromine, fluorine, chlorine.

4. Which of the substances – chlorine, sulfur, aluminum, oxygen– is it a stronger reducing agent? In your answer, indicate the molar mass of the selected compound.

5. Indicate the number or sum of conventional numbers under which only oxidizing agents are located:

1) K 2 MnO 4, 2) KMnO 4, 4) MnO 3, 8) MnO 2,

16) K 2 Cr 2 O 7, 32) K 2 SO 3.

6. Indicate the number or sum of conventional numbers under which the formulas of substances with redox duality are located:

1) KI, 2) H 2 O 2, 4) Al, 8) SO 2, 16) K 2 Cr 2 O 7, 32) H 2.

7. Which of the connections – iron oxide(III), chromium oxide(III), sulfur oxide(IV), nitric oxide(II), nitric oxide(V) – can it only be an oxidizing agent? In your answer, indicate the molar mass of the selected compound.

8. Indicate the number or sum of conventional numbers under which the formulas of substances that have an oxidation state of oxygen are located - 2:

1) H 2 O, Na 2 O, Cl 2 O, 2) HPO 3, Fe 2 O 3, SO 3,

4) OF 2, Ba(OH) 2, Al 2 O 3, 8) BaO 2, Fe 3 O 4, SiO 2.

9. Which of the following compounds can only be an oxidizing agent: sodium nitrite, sulfurous acid, hydrogen sulfide, nitric acid? In your answer, indicate the molar mass of the selected compound.

10. Which of the following nitrogen compounds is NH 3; HNO3; HNO2; NO 2 – can it only be an oxidizing agent? In your answer, write down the relative molecular weight of the selected compound.

11. Under what number, among the names of substances listed below, is the most powerful oxidizing agent indicated?

1) concentrated nitric acid,

2) oxygen,

3) electric current at the anode during electrolysis,

12. Which of the following nitrogen compounds is HNO 3; NH3; HNO2; NO – can it only be a reducing agent? In your answer, write down the molar mass of the selected compound.

13. Which of the compounds is Na 2 S; K 2 Cr 2 O 7 ; KMnO4; NaNO2; KClO 4 - can be both an oxidizing agent and a reducing agent, depending on the reaction conditions? In your answer, write down the molar mass of the selected compound.

14. Indicate the number or sum of conventional numbers, where ions that can be reducing agents are indicated:

1) (MnO 4) 2–, 2) (CrO 4) –2, 4) Fe +2, 8) Sn +4, 16) (ClO 4) –.

15. Indicate the number or sum of conventional numbers under which only oxidizing agents are located:

1) K 2 MnO 4 , 2) HNO 3 , 4) MnO 3 , 8) MnO 2 , 16) K 2 CrO 4 , 32) H 2 O 2 .

16. Indicate the number or sum of conventional numbers, under which only the names of substances are located, between which redox reactions cannot occur:

1) carbon and sulfuric acid,

2) sulfuric acid and sodium sulfate,

4) hydrogen sulfide and hydrogen iodide,

8) sulfur oxide (IV) and hydrogen sulfide.

17. Indicate the number or sum of conventional numbers under which the oxidation processes are located:

1) S +6  S –2, 2) Mn +2  Mn +7, 4) S –2  S +4,

8) Mn +6  Mn +4, 16) O 2  2O –2, 32) S +4  S +6.

18. Indicate the number or sum of conditional numbers under which the recovery processes are located:

1) 2I –1  I 2, 2) 2N +3  N 2, 4) S –2  S +4,

8) Mn +6  Mn +2, 16) Fe +3  Fe 0, 32) S 0  S +6.

19. Indicate the number or sum of conditional numbers under which the recovery processes are located:

1) C 0  CO 2, 2) Fe +2  Fe +3,

4) (SO 3) 2–  (SO 4) 2–, 8) MnO 2  Mn +2.

20. Indicate the number or sum of conditional numbers under which the recovery processes are located:

1) Mn +2  MnO 2, 2) (IO 3) –  (IO 4) –,

4) (NO 2) –  (NO 3) –, 8) MnO 2  Mn +2.

21. Indicate the number or sum of the conventional numbers under which the ions that are reducing agents are located.

1) Ca +2, 2) Al +3, 4) K +, 8) S –2, 16) Zn +2, 32) (SO 3) 2–.

22. What number is the formula of a substance, when interacting with which hydrogen acts as an oxidizing agent?

1) O 2, 2) Na, 3) S, 4) FeO.

23. What number is the equation of the reaction in which the reducing properties of the chloride ion are manifested?

1) MnO 2 + 4HCl = MnCl 2 + Cl 2 + 2H 2 O,

2) CuO + 2HCl = CuCl 2 + H 2 O,

3) Zn + 2HCl = ZnCl 2 + H 2,

4) AgNO 3 + HCl = AgCl + HNO 3.

24. When interacting with which of the indicated substances - O 2, NaOH, H 2 S - does sulfur (IV) oxide exhibit the properties of an oxidizing agent? Write the equation for the corresponding reaction and indicate in your answer the sum of the coefficients of the starting substances.

25. Indicate the number or sum of the conventional numbers under which the disproportionation reaction schemes are located:

1) NH 4 NO 3  N 2 O + H 2 O, 2) NH 4 NO 2  N 2 + H 2 O,

4) KClO 3  KClO 4 + KCl, 8) KClO 3  KCl + O 2.

26. Draw up an electron balance diagram and indicate how much potassium permanganate is involved in the reaction with ten moles of sulfur (IV) oxide. The reaction proceeds according to the scheme:

KMnO 4 + SO 2  MnSO 4 + K 2 SO 4 + SO 3.

27. Draw up an electronic balance diagram and indicate how much potassium sulfide reacts with six moles of potassium permanganate in the reaction:

K 2 S + KMnO 4 + H 2 O  MnO 2 + S + KOH.

28. Draw up an electronic balance diagram and indicate how much potassium permanganate reacts with ten moles of iron (II) sulfate in the reaction:

KMnO 4 + FeSO 4 + H 2 SO 4  MnSO 4 + Fe 2 (SO 4) 3 + K 2 SO 4 + H 2 O.

29. Draw up an electron balance diagram and indicate how much potassium chromite (KCrO2) reacts with six moles of bromine in the reaction:

KCrO 2 + Br 2 + KOH  K 2 CrO 4 + KBr + H 2 O.

30. Draw up an electronic balance diagram and indicate how much manganese (IV) oxide reacts with six moles of lead (IV) oxide in the reaction:

MnO 2 + PbO 2 + HNO 3  HMnO 4 + Pb(NO 3) 2 + H 2 O.

31. Write down the reaction equation:

KMnO 4 + NaI + H 2 SO4  I 2 + K 2 SO 4 + MnSO 4 + Na 2 SO 4 + H 2 O.

32. Write down the reaction equation:

KMnO 4 + NaNO 2 + H 2 O  MnO 2 + NaNO 3 + KOH.

In your answer, indicate the sum of the stoichiometric coefficients in the reaction equation.

33. Write down the reaction equation:

K 2 Cr 2 O 7 + HCl conc.  KCl + CrCl 3 + Cl 2 + H 2 O.

In your answer, indicate the sum of the stoichiometric coefficients in the reaction equation.

34. Draw up an electron balance diagram and indicate how much sodium nitrite (NaNO2) reacts with four moles of potassium permanganate in the reaction:

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

35. Draw up an electronic balance diagram and indicate how much hydrogen sulfide reacts with six moles of potassium permanganate in the reaction:

KMnO 4 + H 2 S + H 2 SO 4  S + MnSO 4 + K 2 SO 4 + H 2 O.

36. What amount of iron substance in moles will be oxidized by oxygen with a volume of 33.6 liters (n.s.) in the reaction proceeding according to the scheme below?

Fe + H 2 O + O 2  Fe(OH) 3.

37. Which of the following metals – Zn, Rb, Ag, Fe, Mg – is not soluble in dilute sulfuric acid? In your answer, indicate the relative atomic mass of this metal.

38. Which of the following metals – Zn, Rb, Ag, Fe, Mg – is not soluble in concentrated sulfuric acid? In your answer, indicate the serial number of the element in the periodic table D.I. Mendeleev.

39. Indicate the number or sum of conventional numbers under which metals are located that are passivated in concentrated solutions of oxidizing acids.

1) Zn, 2) Cu, 4) Au, 8) Fe, 16) Mg, 32) Cr.

40. Indicate the number or sum of conventional numbers under which the chemical symbols of metals are located that do not displace hydrogen from a dilute solution of sulfuric acid, but displace mercury from solutions of Hg 2+ salts:

1) Fe, 2) Zn, 4) Au, 8) Ag, 16) Cu.

41. Under what number are the chemical symbols of metals indicated, each of which does not react with nitric acid?

1) Zn, Ag; 2) Pt, Au; 3) Cu, Zn; 4) Ag, Hg.

42. What number is indicated for the method of producing chlorine in industry?

1) electrolysis of sodium chloride solution;

2) the effect of manganese oxide (1V) on hydrochloric acid;

3) thermal decomposition of natural chlorine compounds;

4) the effect of fluorine on chlorides.

43. What number is the chemical formula of the gas that is predominantly released when a concentrated solution of nitric acid acts on copper?

1) N 2, 2) NO 2, 3) NO, 4) H 2.

44. Under what number are the formulas of the reaction products of hydrogen sulfide combustion in air with a lack of oxygen indicated?

1) SO 2 + H 2 O, 2) S + H 2 O,

3) SO 3 + H 2 O, 4) SO 2 + H 2.

Indicate the number of the correct answer.

45. Write an equation for the reaction between concentrated sulfuric acid and copper. In your answer, indicate the sum of the coefficients in the reaction equation.

10.4. Answers to self-test tasks

knowledge of theory on the topic.

"Redox reactions"

Option #1		Option No. 2		Option No. 3
5oxidatively Document Increases 4) the oxidation state of iron decreases Oxidatively-restorative reaction connection occurs between: 1) hydrogen chloride and... potassium bichromate K2Cr2O7 can perform in oxidatively-restorative reactions function: 1) both an oxidizing agent and... “Preparation of reaction equations in molecular and ionic forms. Calculation problems for calculating the mass fraction of a substance in a solution.” Target Document ... oxidatively-restorative reactions, practicing practical skills in drawing up equations oxidatively-restorative reactions electronic balance method. Theory. Oxidatively-restorative are called reactions ...

These include reactions in which reacting substances exchange electrons, thereby changing the oxidation states of the atoms of the elements that make up the reacting substances.

For example:

Zn + 2H + → Zn 2+ + H 2 ,

FeS 2 + 8HNO 3 (conc) = Fe(NO 3) 3 + 5NO + 2H 2 SO 4 + 2H 2 O,

The vast majority of chemical reactions are redox reactions; they play an extremely important role.

Oxidation is the process of losing electrons by an atom, molecule or ion.

If an atom gives up its electrons, it acquires a positive charge:

For example:

Al - 3e - = Al 3+

H 2 - 2e - = 2H +

During oxidation, the oxidation state increases.

If a negatively charged ion (charge -1), for example Cl -, gives up 1 electron, then it becomes a neutral atom:

2Cl - - 2e - = Cl 2

If a positively charged ion or atom gives up electrons, then the magnitude of its positive charge increases according to the number of electrons given up:

Fe 2+ - e - = Fe 3+

Reduction is the process of gaining electrons by an atom, molecule or ion.

If an atom gains electrons, it becomes a negatively charged ion:

For example:

Сl 2 + 2е- = 2Сl -

S + 2е - = S 2-

If a positively charged ion accepts electrons, its charge decreases:

Fe 3+ + e- = Fe 2+

or it can go into a neutral atom:

Fe 2+ + 2e- = Fe 0

An oxidizing agent is an atom, molecule, or ion that accepts electrons. A reducing agent is an atom, molecule, or ion that donates electrons.

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

Oxidation is always accompanied by reduction, and vice versa, reduction is always associated with oxidation, which can be expressed by the equations:

Reducing agent - e - ↔ Oxidizing agent

Oxidizing agent + e - ↔ Reducing agent

Therefore, redox reactions represent the unity of two opposite processes - oxidation and reduction

The most important reducing and oxidizing agents

Restorers

Oxidizing agents

Metals, hydrogen, coal Carbon(II) monoxide CO Hydrogen sulfide H 2 S, sulfur oxide (IV) SO 2, sulfurous acid H 2 SO 3 and its salts Hydroiodic acid HI, hydrobromic acid HBr, hydrochloric acid HCl Tin(II) chloride SnCl2, iron(II) sulfate FeSO4, manganese(II) sulfate MnSO4, chromium(III) sulfate Cr2 (SO4) 3 Nitrous acid HNO 2, ammonia NH 3, hydrazine N 2 H 4, nitric oxide (II) NO Phosphorous acid H 3 PO 3 Aldehydes, alcohols, formic and oxalic acids, glucose Cathode during electrolysis

Halogens Potassium permanganate KMnO 4, potassium manganate K 2 MnO 4, manganese(IV) oxide MnO 2 Potassium dichromate K 2 Cr 2 O 7 , potassium chromate K 2 CrO 4 Nitric acid HNO 3 Oxygen O 2, ozone O 3, hydrogen peroxide H 2 O 2 Sulfuric acid H 2 SO 4 (conc.), selenic acid H 2 SeO 4 Copper(II) oxide CuO, silver(I) oxide Ag 2 O, lead(IV) oxide PbO 2 Noble metal ions (Ag +, Au 3+, etc.) Iron(III) chloride FeCl 3 Hypochlorites, chlorates and perchlorates Aqua regia, a mixture of concentrated nitric and hydrofluoric acids Anode during electrolysis

Electronic balance method.

To equalize OVR, several methods are used, of which we will now consider one - the electronic balance method.

Let's write the reaction equation between aluminum and oxygen:

Al + O 2 = Al 2 O 3

Don't be fooled by the simplicity of this equation. Our task is to understand a method that in the future will allow you to equalize much more complex reactions.

So, what is the electronic balance method? Balance is equality. Therefore, the number of electrons that one element gives up and the other element accepts in a given reaction should be made equal. Initially, this amount looks different, as can be seen from the different oxidation states of aluminum and oxygen:

Al 0 + O 2 0 = Al 2 +3 O 3 -2

Aluminum gives up electrons (acquires a positive oxidation state), and oxygen accepts electrons (acquires a negative oxidation state). To obtain the +3 oxidation state, an aluminum atom must give up 3 electrons. An oxygen molecule, in order to turn into oxygen atoms with an oxidation state of -2, must accept 4 electrons:

Al 0 - 3e- = Al +3

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

In order for the number of given and received electrons to be equal, the first equation must be multiplied by 4, and the second by 3. To do this, it is enough to move the numbers of given and received electrons against the top and bottom lines as shown in the diagram above.

If now in the equation we put the coefficient 4 we found in front of the reducing agent (Al), and the coefficient 3 we found in front of the oxidizing agent (O 2), then the number of given and received electrons is equalized and becomes equal to 12. Electronic balance has been achieved. It can be seen that a coefficient of 2 is required before the reaction product Al 2 O 3. Now the equation of the redox reaction is equalized:

4Al + 3O 2 = 2Al 2 O 3

All the advantages of the electronic balance method appear in more complex cases than the oxidation of aluminum with oxygen.

For example, the well-known “potassium permanganate” - potassium permanganate KMnO 4 - is a strong oxidizing agent due to the Mn atom in the oxidation state +7. Even the chlorine anion Cl – gives it an electron, turning into a chlorine atom. This is sometimes used to produce chlorine gas in the laboratory:

K + Mn +7 O 4 -2 + K + Cl - + H 2 SO 4 = Cl 2 0 + Mn +2 SO 4 + K 2 SO 4 + H 2 O

Let's create an electronic balance diagram:

Mn +7 + 5e- = Mn +2

2Cl - - 2e- = Cl 2 0

Two and five are the main coefficients of the equation, thanks to which it is possible to easily select all other coefficients. Before Cl 2 you should put a coefficient of 5 (or 2 × 5 = 10 before KСl), and before KMnO 4 - a coefficient of 2. All other coefficients are tied to these two coefficients. This is much easier than acting by simply crunching numbers.

2 KMnO 4 + 10KCl + 8H 2 SO 4 = 5 Cl 2 + 2MnSO 4 + 6K 2 SO 4 + 8H 2 O

To equalize the number of K atoms (12 atoms on the left), it is necessary to put a coefficient of 6 in front of K 2 SO 4 on the right side of the equation. Finally, to equalize oxygen and hydrogen, it is enough to put a coefficient of 8 in front of H 2 SO 4 and H 2 O. We get the equation in its final form.

The electronic balance method, as we see, does not exclude the usual selection of coefficients in the equations of redox reactions, but can significantly facilitate such selection.

Drawing up an equation for the reaction of copper with a solution of palladium (II) nitrate. Let us write down the formulas of the initial and final substances of the reaction and show the changes in oxidation states:

from which it follows that for a reducing agent and an oxidizing agent, the coefficients are equal to 1. The final reaction equation is:

Cu + Pd(NO 3) 2 = Cu(NO 3) 2 + Pd

As you can see, electrons do not appear in the overall reaction equation.

To check the correctness of the equation, we count the number of atoms of each element in its right and left sides. For example, on the right side there are 6 oxygen atoms, on the left there are also 6 atoms; palladium 1 and 1; copper is also 1 and 1. This means that the equation is written correctly.

Let's rewrite this equation in ionic form:

Cu + Pd 2+ + 2NO 3 - = Cu 2+ + 2NO 3 - + Pd

And after the reduction of identical ions we get

Cu + Pd 2+ = Cu 2+ + Pd

Drawing up a reaction equation for the interaction of manganese (IV) oxide with concentrated hydrochloric acid

(chlorine is produced using this reaction in the laboratory).

Let's write down the formulas of the starting and final substances of the reaction:

HCl + MnO 2 → Cl 2 + MnCl 2 + H 2 O

Let us show the change in oxidation states of atoms before and after the reaction:

This reaction is redox, since the oxidation states of chlorine and manganese atoms change. HCl is a reducing agent, MnO 2 is an oxidizing agent. We compose electronic equations:

and find the coefficients for the reducing agent and the oxidizing agent. They are respectively equal to 2 and 1. The coefficient 2 (and not 1) is set because 2 chlorine atoms with an oxidation state of -1 give up 2 electrons. This coefficient is already in the electronic equation:

2HCl + MnO 2 → Cl 2 + MnCl 2 + H 2 O

We find coefficients for other reacting substances. From the electronic equations it is clear that for 2 mol of HCl there is 1 mol of MnO 2. However, taking into account that another 2 moles of acid are needed to bind the resulting doubly charged manganese ion, a coefficient of 4 should be placed in front of the reducing agent. Then you will get 2 moles of water. The final equation is

4HCl + MnO2 = Cl2 + MnCl2 + 2H2O

Checking the correctness of writing the equation can be limited to counting the number of atoms of one element, for example chlorine: on the left side there are 4 and on the right side 2 + 2 = 4.

Since the electron balance method depicts reaction equations in molecular form, after compilation and verification they should be written in ionic form.

Let's rewrite the equation in ionic form:

4Н + + 4Сl - + МnО 2 = Сl 2 + Мn 2 + + 2Сl - + 2Н 2 О

and after canceling identical ions on both sides of the equation we get

4H + + 2Cl - + MnO 2 = Cl 2 + Mn 2 + + 2H 2 O

Drawing up a reaction equation for the interaction of hydrogen sulfide with an acidified solution of potassium permanganate.

Let's write the reaction scheme - the formulas of the starting and resulting substances:

H 2 S + KMnO 4 + H 2 SO 4 → S + MnSO 4 + K 2 SO 4 + H 2 O

Then we show the change in oxidation states of atoms before and after the reaction:

The oxidation states of sulfur and manganese atoms change (H 2 S is a reducing agent, KMnO 4 is an oxidizing agent). We compose electronic equations, i.e. We depict the processes of electron loss and gain:

And finally, we find the coefficients for the oxidizing agent and the reducing agent, and then for the other reactants. From the electronic equations it is clear that we need to take 5 mol H 2 S and 2 mol KMnO 4, then we get 5 mol S atoms and 2 mol MnSO 4. In addition, from a comparison of the atoms on the left and right sides of the equation, we find that 1 mol K 2 SO 4 and 8 mol of water are also formed. The final reaction equation will be

5Н 2 S + 2КМnО 4 + ЗН 2 SO 4 = 5S + 2МnSO 4 + К 2 SO 4 + 8Н 2 О

The correctness of writing the equation is confirmed by counting the atoms of one element, for example oxygen; on the left side there are 2 4 + 3 4 = 20 and on the right side there are 2 4 + 4 + 8 = 20.

We rewrite the equation in ionic form:

5H 2 S + 2MnO 4 - + 6H + = 5S + 2Mn 2+ + 8H 2 O

It is known that a correctly written reaction equation is an expression of the law of conservation of mass of substances. Therefore, the number of the same atoms in the starting materials and reaction products must be the same. The charges must also be conserved. The sum of the charges of the starting substances must always be equal to the sum of the charges of the reaction products.

The electron-ion balance method is more universal compared to the electronic balance method and has an undeniable advantage in selecting coefficients in many redox reactions, in particular, involving organic compounds, in which even the procedure for determining oxidation states is very complex.

OVR classification

There are three main types of redox reactions:

1) Intermolecular oxidation-reduction reactions
(when the oxidizing agent and the reducing agent are different substances);

2) Disproportionation reactions
(when the same substance can serve as an oxidizing agent and a reducing agent);

3) Intramolecular oxidation-reduction reactions
(when one part of the molecule acts as an oxidizing agent, and the other as a reducing agent).>

Let's look at examples of three types of reactions.

1. Intermolecular oxidation-reduction reactions are all the reactions we have already discussed in this paragraph.
Let's consider a slightly more complex case, when not all of the oxidizing agent can be consumed in the reaction, since part of it is involved in an ordinary, non-redox exchange reaction:

Cu 0 + H + N +5 O 3 -2 = Cu +2 (N +5 O 3 -2) 2 + N +2 O -2 + H 2 O

Some NO 3 - particles participate in the reaction as an oxidizing agent, producing nitric oxide NO, and some NO 3 - ions pass unchanged into the copper compound Cu(NO 3) 2. Let's create an electronic balance:

Cu 0 - 2e- = Cu +2

N +5 + 3e- = N +2

Let us put the coefficient 3 found for copper in front of Cu and Cu(NO 3) 2. But coefficient 2 should be placed only in front of NO, because all the nitrogen present in it participated in the redox reaction. It would be a mistake to put a factor of 2 in front of HNO 3, because this substance also includes those nitrogen atoms that do not participate in oxidation-reduction and are part of the product Cu(NO 3) 2 (NO 3 particles - here sometimes called “ion” -observer").

The remaining coefficients can be easily selected using those already found:

3 Cu + 8HNO 3 = 3 Cu(NO 3) 2 + 2 NO + 4H 2 O

2. Disproportionation reactions occur when molecules of the same substance are capable of oxidizing and reducing each other. This becomes possible if the substance contains atoms of any element in an intermediate oxidation state.

Consequently, the oxidation state can either decrease or increase. For example:

HN +3 O 2 = HN +5 O 3 + N +2 O + H 2 O

This reaction can be represented as a reaction between HNO 2 and HNO 2 as an oxidizing agent and a reducing agent and using the electron balance method:

HN +3 O 2 + HN +3 O 2 = HN +5 O3 + N +2 O + H 2 O

N +3 - 2e- = N +5

N +3 + e- = N +2

We get the equation:

2HNO 2 + 1HNO 2 = 1 HNO 3 + 2 NO + H 2 O

Or, adding the moles of HNO 2 together:

3HNO2 = HNO3 + 2NO + H2O

Intramolecular oxidation-reduction reactions occur when oxidizing atoms and reducing atoms are adjacent in a molecule. Let's consider the decomposition of Berthollet salt KClO 3 when heated:

KCl +5 O 3 -2 = KCl - + O 2 0

This equation also obeys the electronic balance requirement:

Cl +5 + 6e- = Cl -

2O -2 - 2e- = O 2 0

Here a difficulty arises - which of the two coefficients found should be put in front of KClO 3 - after all, this molecule contains both an oxidizing agent and a reducing agent?

In such cases, the found coefficients are placed in front of the products:

KClO 3 = 2KCl + 3O 2

Now it is clear that KClO 3 must be preceded by a factor of 2.

2KClO 3 = 2KCl + 3O 2

The intramolecular reaction of decomposition of berthollet salt when heated is used in the production of oxygen in the laboratory.

Half-reaction method

As the name suggests, this method is based on drawing up ionic equations for the oxidation process and the reduction process and then summing them into an overall equation.
As an example, let's create an equation for the same reaction that was used to explain the electronic balance method.
When hydrogen sulfide H 2 S is passed through an acidified solution of potassium permanganate KMnO 4, the crimson color disappears and the solution becomes cloudy.
Experience shows that cloudiness of the solution occurs as a result of the formation of elemental sulfur, i.e. process flow:

H 2 S → S + 2H +

This scheme is equalized by the number of atoms. To equalize by the number of charges, you need to subtract two electrons from the left side of the diagram, after which you can replace the arrow with an equal sign:

H 2 S - 2е - = S + 2H +

This is the first half-reaction - the process of oxidation of the reducing agent H 2 S.

Discoloration of the solution is associated with the transition of the MnO 4 - ion (it has a crimson color) into the Mn 2+ ion (almost colorless and only at high concentrations it has a faint pink color), which can be expressed by the diagram

MnO 4 - → Mn 2+

In an acidic solution, oxygen, which is part of the MnO 4 ions, together with hydrogen ions ultimately forms water. Therefore, we write the transition process like this:

MnO 4 - + 8H + → Mn 2+ + 4H 2 O

To replace the arrow with an equal sign, the charges must also be equalized. Since the initial substances have seven positive charges (7+), and the final substances have two positive charges (2+), then to fulfill the condition of conservation of charges, five electrons must be added to the left side of the diagram:

MnO 4 - + 8H + + 5e - = Mn 2+ + 4H 2 O

This is the second half-reaction - the process of reduction of the oxidizing agent, i.e. permanganate ion

To compile a general reaction equation, it is necessary to add the half-reaction equations term by term, having previously equalized the numbers of electrons given and received. In this case, according to the rules for finding the smallest multiple, the corresponding factors are determined by which the half-reaction equations are multiplied. The abbreviated form is as follows:

And, reducing by 10H +, we finally get

5H 2 S + 2MnO 4 - + 6H + = 5S + 2Mn 2+ + 8H 2 O

We check the correctness of the equation compiled in ionic form: the number of oxygen atoms on the left side is 8, on the right side 8; number of charges: on the left side (2-)+(6+) = 4+, on the right side 2(2+) = 4+. The equation is written correctly, since the atoms and charges are equal.

Using the half-reaction method, the reaction equation is compiled in ionic form. To move from it to an equation in molecular form, we do this: on the left side of the ionic equation, we select the corresponding cation for each anion, and for each cation - an anion. Then we write the same ions in the same number on the right side of the equation, after which we combine the ions into molecules:

Thus, compiling equations for redox reactions using the half-reaction method leads to the same result as the electron balance method.

Let's compare both methods. The advantage of the half-reaction method compared to the electronic balance method is that. that it uses not hypothetical ions, but actually existing ones. In fact, there are no ions in a solution, but there are ions.

With the half-reaction method, it is not necessary to know the oxidation state of atoms.

Writing individual ionic half-reaction equations is necessary to understand the chemical processes in a galvanic cell and in electrolysis. With this method, the role of the environment as an active participant in the entire process is visible. Finally, when using the half-reaction method, you do not need to know all the resulting substances; they appear in the reaction equation when deriving it. Therefore, the method of half-reactions should be given preference and used when drawing up equations for all redox reactions occurring in aqueous solutions.

According to their function in redox processes, their participants are divided into oxidizing agents and reducing agents.

Oxidizing agents are atoms, molecules or ions that accept electrons from other atoms. The oxidation state of the oxidizing agent decreases.

Restorers– atoms, molecules or ions that donate electrons to other atoms. The oxidation state of the reducing agent increases. During redox reaction, the oxidizing agent is reduced, the reducing agent is oxidized, and both processes occur simultaneously.

Accordingly, oxidizing agents and reducing agents interact in such proportions that the numbers of electrons accepted and given up are the same.

The specific manifestation of oxidizing or reducing properties by atoms of various elements depends on many factors. The most important of them include the position of the element in the periodic table, the oxidation state of the element in a given substance, the special properties of other participants in the reaction (the nature of the medium for solutions, the concentration of reagents, temperature, stereochemical properties of complex particles, etc.)

Oxidizing agents.

Oxidizing agents can be both simple and complex substances. Let's try to determine what factors determine the oxidative (and reductive) properties of substances.

The oxidizing ability of simple substances can be judged by the values ​​of relative electronegativity ( χ ). This concept reflects the ability of an atom to shift electron density towards itself from other atoms, i.e. is actually a measure of the oxidizing power of simple substances. Indeed, the strongest oxidizing properties are exhibited by active nonmetals with maximum electronegativity values. So, fluorineF 2 exhibits only oxidizing properties because it matters the most χ , equal to 4.1 (on the Allred-Rochow scale). The second place is occupied by oxygen O 2, for it χ = 3.5, ozone O 3 exhibits even stronger oxidizing properties. The third place is taken by nitrogen ( χ =3.07), but its oxidizing properties appear only at high temperatures, since the nitrogen molecule N 2 has very high strength, because atoms are connected by a triple bond. Chlorine and bromine have fairly strong oxidizing properties.

On the other hand, the minimum values ​​of electronegativity are inherent in metals ( χ = 0.8-1.6). This means that the intrinsic electrons of metal atoms are held very loosely and can easily move to atoms with higher electronegativity. Metal atoms to the zero degree can exhibit only restorative properties and cannot accept electrons. The most pronounced reducing properties are exhibited by metals of groups IA and IIA.

Redox properties of complex substances

The criterion for the oxidizing ability of atoms can be the degree of oxidation. The maximum oxidation state corresponds to the transfer of all valence electrons to other atoms. Such an atom can no longer give away electrons, but can only accept them. Thus, in maximum oxidation state, an element can exhibit only oxidizing properties A. However, it should be noted that the maximum degree of oxidation does not automatically mean the manifestation of pronounced oxidizing properties. In order for the properties of a strong oxidizing agent to be realized, the particle must be unstable, maximally asymmetrical, with an uneven distribution of electron density. Thus, in dilute solutions the sulfate ion SO 4 2- containing a sulfur atom in the maximum oxidation state +6 , does not exhibit oxidizing properties at all, since it has a highly symmetric tetrahedral structure. Whereas in concentrated solutions of sulfuric acid, a significant proportion of particles are in the form of undissociated molecules and HSO 4 - ions, which have an asymmetric structure with an uneven distribution of electron density. As a consequence of this, concentrated sulfuric acid, especially when heated, is a very strong oxidizing agent.

On the other hand, the minimum oxidation state of an element means that the nonmetal atom has accepted the maximum possible number of electrons into valence sublevels and can no longer accept electrons. Hence,

non-metal atoms in the minimum oxidation state can exhibit only reducing properties.

It may be recalled that the minimum oxidation state of a non-metal is equal to the group number –8. As in the case of sulfuric acid, to realize the reducing properties it is not enough to have only a minimum oxidation state. An example is nitrogen in the –3 oxidation state. The highly symmetrical ammonium ion NH 4 + is an extremely weak reducing agent in solution. The ammonia molecule, which has less symmetry, exhibits fairly strong reducing properties when heated. The reduction reaction from oxides can be given:

3FeO+ 2NH 3 = 3Fe+3H 2 O+N 2.

As for simple substances with intermediate values ​​of electronegativity ( χ = 1.9 – 2.6), then for non-metals one can expect the implementation of both oxidizing and reducing properties. Such substances include hydrogenH2, carbonC, phosphorusP, sulfurS, iodineI2 and other non-metals of average activity. Naturally, metals simple substances are excluded from this category because cannot accept electrons.

These substances, when interacting with active oxidizing agents, exhibit the properties of reducing agents, and when reacting with reducing agents, they exhibit the properties of oxidizing agents. As an example, we give the reactions of sulfur:

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

S+O 2 =SO 2 Fe+S=FeS

As you can see, in the first reaction sulfur is a reducing agent, and in the second it is an oxidizing agent.

Complex substances containing atoms in intermediate oxidation states will also exhibit the properties of both oxidizing and reducing agents. There are a lot of such substances, so we will name only the most common ones. These are sulfur compounds (+4): in an acidic environment SO 2, and in an alkaline and neutral environment SO 3 2- and HSO 3 -. If these compounds participate in the reaction as reducing agents, then they will be oxidized to sulfur +6 (in the gas phase to SO 3, and in solution to SO 4 2-. If sulfur compounds (+4) react with active reducing agents, then reduction occurs to elemental sulfur, or even hydrogen sulfide.

SO 2 + 4HI=S+ 2I 2 +2H 2 O

Many nitrogen compounds also exhibit redox duality. The behavior of NO 2 nitrite ions is of particular interest - . When they are oxidized, nitrate ion NO 3 is formed - , and upon reduction gaseous nitrogen monoxide NO. Example: 2NaNO 2 + 2NaI+2H 2 SO 4 =I 2 +NO+ 2Na 2 SO 4 +2H 2 O.

Let's look at another example, this time taking hydrogen peroxide, in which the oxidation state of oxygen is (-1). If oxidation of this substance takes place, the degree of oxygen will increase to 0, and the release of hydrogen gas will be observed:

H 2 O 2 +Cl 2 = 2HCl+O 2.

In oxidation reactions, the oxidation state of oxygen in peroxides is reduced to (-2), which corresponds to either water H 2 O or hydroxide ion OH - . As an example, we give a reaction often used in restoration work, in which black lead sulfide, under the action of a dilute solution of hydrogen peroxide, is converted into white sulfate: PbS (black) + 4H 2 O 2 = PbSO 4 (white) + 4H 2 O.

Thus, to complete the introductory part, we present the main oxidizing agents, reducing agents and substances that can exhibit both oxidizing and reducing properties.

Oxidizing agents:F 2 , O 2 , O 3 , Cl 2 , Br 2 , HNO 3 , H 2 SO 4 (conc.), KMnO 4 , K 2 Cr 2 O 7 , PbO 2 , NaBiO 3 , Fe 3+ ions in aqueous solution ,Cu 2+ ,Ag + .

Restorers:H 2 S, (S 2-), HI (I -), HBr (Br -), HCl (weak), NH 3 (at high temperatures), ions in aqueous solution Fe 2+, Cr 2+, Sn 2+ etc.

Substances with dual properties:H 2 ,C,P,As,S,I 2 ,CO,H 2 O 2 ,Na 2 O 2 ,NaNO 2 ,SO 2 (SO 3 2-) and, formally, almost all substances containing atoms with intermediate degree of oxidation.

Drawing up equations for redox reactions.

There are several ways to compose OVR equations. Typically used

a) electronic balance method,

b) electron-ion balance method.

Both methods are based on finding such quantitative relationships between the oxidizing agent and the reducing agent, at which equality of received and given electrons is observed.

The electronic balance method is more universal, although less visual. It is based on calculating the change in the oxidation states of the oxidizing and reducing atoms in the initial and final substances. When working with this method, it is convenient to follow this algorithm.

The molecular diagram of the redox reaction is written down,

The oxidation states of atoms (usually those that change it) are calculated

The oxidizing agent and the reducing agent are determined,

The number of electrons accepted by the oxidizing agent and the number of electrons given up by the reducing agent are established,

The coefficients are found, when multiplied by which the numbers of given and received electrons are equalized,

Coefficients are selected for other participants in the reaction.

Let's consider the oxidation reaction of hydrogen sulfide.

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

In this reaction, sulfur (-2) is the reducing agent and molecular oxygen is the oxidizing agent. Then we create an electronic balance.

S -2 -6e - →S +4 2 - multiplication factor for the reducing agent

O 2 +4e - →2O -2 3 - multiplication factor for the oxidizing agent

We write the formulas of substances taking into account multiplication coefficients

2H 2 S+ 3O 2 = 2SO 2 +2H 2 O

Let's consider another case - the decomposition of aluminum nitrate Al(NO 3) 3. In this substance, nitrogen atoms have the highest oxidation state (+5), and oxygen atoms have the lowest (-2). It follows that nitrogen will be an oxidizing agent, and oxygen will be a reducing agent. We draw up an electronic balance, knowing that all nitrogen is reduced to nitrogen dioxide, and oxygen is oxidized to molecular oxygen. Taking into account the numbers of atoms, we write:

3N +5 +3e - → 3N +4 4

2O -2 -4e - →O 2 o 3

then the decomposition equation will be written as follows: 4Al(NO 3) 3 = Al 2 O 3 + 12NO 2 + 3O 2.

Method electronic balance usually used to determine coefficients in ORR occurring in heterogeneous systems containing solids or gases.

For reactions occurring in solutions, it is usually used electron-ion balance method, which takes into account the influence of various factors on the composition of the final products.

This method takes into account: a) the acidity of the medium, b) the concentration of reacting substances, c) the actual state of the reacting particles in solution, d) the influence of temperature, etc. In addition, for this method there is no need to use the oxidation state.