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 equation for the reaction 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 with 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, as 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 2 moles of water will be obtained. 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 compiled 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 the 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 thought of 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 turbidity 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 the 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 it is derived. Therefore, the method of half-reactions should be given preference and used when drawing up equations for all redox reactions occurring in aqueous solutions.

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 redox process equation, 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 electrical neutrality, 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.

1. How to define a redox reaction?

There are various classifications of chemical reactions. One of them includes those in which substances that interact with each other (or the substance itself) change the oxidation states of elements.

As an example, consider two reactions:

Zn 0 + 2Н +1 С1 -1 = Zn +2 Cl 2 -1 + Н 2 0 (1)
H +1 Cl -1 + K +1 O -2 H +1 = K +1 Cl -1 + H 2 +1 O -2 (2)

Reaction (1) involves zinc and hydrochloric acid. Zinc and hydrogen change their oxidation states, chlorine leaves its oxidation state unchanged:

Zn 0 - 2е = Zn 2+
2Н +1 + 2е = H 2 0
2Сl -1 = 2 Сl -1

And in reaction (2), ( neutralization reaction), chlorine, hydrogen, potassium, and oxygen do not change their oxidation states: Cl -1 = Cl -1, H +1 = H +1, K +1 = K +1, O -2 = O -2; Reaction (1) belongs to the redox type, and reaction (2) belongs to another type.

Chemical reactions that involve changesoxidation states of elements, are called redox.

In order to determine the redox reaction it is necessary to establish steppeno oxidation of elements on the left and right sides of the equation. To do this, you need to know how to determine the oxidation state of a particular element.

In the case of reaction (1), the elements Zn and H change their states, losing or gaining electrons. Zinc, giving up 2 electrons, goes into an ionic state - it becomes a Zn 2+ cation. In this case, the process occurs recovery and zinc is oxidized. Hydrogen gains 2 electrons, exhibits oxidative properties, itself in the reaction process is being restored.

2. Definitionoxidation states of elements.

Oxidation state of elements in its compounds is determined based on the position that the total total charge of the oxidation states of all elements of a given compound is equal to zero. For example, in the compound H 3 PO 4 oxidation states are +1 for hydrogen, +5 for phosphorus, and -2 for oxygen; Having composed a mathematical equation, we determine that in total number of particles(atoms or ions) will make up a charge equal to zero: (+1)x3+(+5)+(-2)x4 = 0

But in this example, the oxidation states of the elements are already specified. How can one determine the oxidation state of sulfur, for example, in the compound sodium thiosulfate Na 2 S 2 O 3, or manganese in the compound potassium permanganate- KMnO 4? To do this you need to know constant oxidation states of a number of elements. They have the following meanings:

1) Elements of group I of the periodic table (including hydrogen in combination with non-metals) +1;
2) Elements of group II of the periodic table +2;
3) Elements of group III of the periodic table +3;
4) Oxygen (except in combination with fluorine or peroxide compounds) -2;

Based on these constant values ​​of oxidation states (for sodium and oxygen), we determine oxidation state sulfur in the Na 2 S 2 O 3 compound. Since the total charge of all oxidation states of elements, the composition of which is reflected by a given compound formula, is equal to zero, then denoting the unknown charge on sulfur “ 2X"(since there are two sulfur atoms in the formula), we create the following mathematical equality:

(+1) x 2 + 2X+ (-2) x 3 = 0

Solving this equation for 2 x, we get

2X= (-1) x 2 + (+2) x 3
or
X = [(-2) + (+6)] : 2 = +2;

Therefore, the oxidation state of sulfur in the Na 2 S 2 O 3 compound is equal to (+2). But will it really always be necessary to use such an inconvenient method to determine the oxidation states of certain elements in compounds? Of course not always. For example, for binary compounds: oxides, sulfides, nitrides, etc., you can use the so-called “cross-on-cross” method to determine oxidation states. Let's say given compound formula:titanium oxide– Ti 2 O 3 . Using a simple mathematical analysis, based on the fact that the oxidation state of oxygen is known to us and is equal to (-2): Ti 2 O 3, it is not difficult to establish that the oxidation state of titanium will be equal to (+3). Or, for example, in connection methane CH 4 it is known that the oxidation state of hydrogen is (+1), then it is not difficult to determine the oxidation state of carbon. It will correspond to the formula of this compound (-4). Also, using the “cross-on-cross” method, it is not difficult to establish that if the following compound formula Cr 4 Si 3, then the oxidation state of chromium in it is (+3), and silicon (-4).
For salts this is also not difficult. Moreover, it does not matter whether it is given or medium salt or acid salt. In these cases, it is necessary to proceed from a salt-forming acid. For example, salt is given sodium nitrate(NaNO 3). It is known that it is a derivative of nitric acid (HNO 3), and in this compound the oxidation state of nitrogen is (+5), therefore, in its salt - sodium nitrate, the oxidation state of nitrogen is also equal to (+5). Sodium bicarbonate(NaHCO 3) is the acid salt of carbonic acid (H 2 CO 3). Just like in an acid, the oxidation state of carbon in this salt will be equal to (+4).

It should be noted that the oxidation states in compounds: metals and non-metals (when compiling electronic balance equations) are equal to zero: K 0, Ca 0, Al 0, H 2 0, Cl 2 0, N 2 0 As an example, we give the oxidation states of the most typical elements:

Only oxidizing agents are substances that have a maximum, usually positive, oxidation state, for example: KCl +7 O 4, H 2 S +6 O 4, K 2 Cr +6 O 4, HN +5 O 3, KMn +7 O 4 . This is easy to prove. If these compounds could be reducing agents, then in these states they would have to give up electrons:

Cl +7 – e = Cl +8
S +6 – e = S +7

But the elements chlorine and sulfur cannot exist with such oxidation states. Similarly, only reducing agents are substances that have a minimal, usually negative, oxidation state, for example: H 2 S -2, HJ -, N -3 H 3. In the process of redox reactions, such compounds cannot be oxidizing agents, since they we would have to add electrons:

S -2 + e = S -3
J - + e = J -2

But for sulfur and iodine, ions with such oxidation states are not typical. Elements with intermediate oxidation states, for example N +1, N +4, S +4, Cl +3, C +2, can exhibit both oxidizing and reducing properties.

3 . Types of redox reactions.

There are four types of redox reactions.

1) Intermolecular redox reactions.
The most common type of reaction. These reactions change oxidation stateselements in different molecules, for example:

2Bi +3 Cl 3 + 3Sn +2 Cl 2 = 2Bi 0 + 3Sn +4 Cl 4

Bi +3 - 3 e= Bi 0

Sn +2 + 2 e= Sn +4

2) A type of intermolecular redox reactions is the reaction proportionation, in which the oxidizing agent and the reducing agent are atoms of the same element: in this reaction, two atoms of one element with different oxidation states form one atom with a different oxidation state:

SO 2 +4 + 2H 2 S -2 = 3S 0 + 2H 2 O

S -2 - 2 e= S 0

S+4+4 e= S 0

3) Reactions disproportionation are carried out if the oxidizing agent and the reducing agent are atoms of the same element, or one atom of an element with one oxidation state forms a compound with two oxidation states:

N +4 O 2 + NaOH = NaN +5 O 3 + NaN +3 O 2 + H 2 O

N +4 - e= N +5

N +4 + e= N +3

4) Intramolecular redox reactions occur in cases where an oxidizing atom and a reducing atom are in the same substance, for example:

N -3 H 4 N +5 O 3 = N +1 2 O + 2H 2 O

2N -3 - 8 e=2N +1

2N +5 + 8 e= 2N +1

4 . The mechanism of redox reactions.

Redox reactions are carried out by the transfer of electrons from the atoms of one element to another. If an atom or molecule loses electrons, then this process is called oxidation, and this atom is a reducing agent, for example:

Al 0 - 3 e= Al 3+

2Cl - - 2 e= Cl 2 0

Fe 2+ - e= Fe 3+

In these examples, Al 0, Cl -, Fe 2+ are reducing agents, and the processes of their transformation into compounds Al 3+, Cl 2 0, Fe 3+ are called oxidative. If an atom or molecule gains electrons, then this process is called reduction, and this atom is an oxidizing agent, for example:

Ca 2+ + 2 e= Ca 0

Cl 2 0 + 2 e= 2Cl -

Fe 3+ + e= Fe 2+

Oxidizing agents, as a rule, are non-metals (S, Cl 2, F 2, O 2) or compounds of metals with the maximum oxidation state (Mn +7, Cr +6, Fe +3). Reducing agents are metals (K, Ca, Al) or compounds of non-metals with a minimum oxidation state (S -2, Cl -1, N -3, P -3);

Redox equations differ from molecular equations other reactions due to the complexity of selecting coefficients for reactants and reaction products. For this they use electronic balance method, or electron-ion equation method(sometimes the latter is called " half-reaction method"). As an example of compiling equations for redox reactions, consider a process in which concentrated sulfuric acid(H 2 SO 4) will react with hydrogen iodide (HJ):

H 2 SO 4 (conc.) + HJ → H 2 S + J 2 + H 2 O

First of all, let's establish that oxidation state iodine in hydrogen iodide is (-1), and sulfur in sulfuric acid: (+6). During the reaction, iodine (-1) will be oxidized to a molecular state, and sulfur (+6) will be reduced to the oxidation state (-2) - hydrogen sulfide:

J - → J 0 2
S +6 → S -2

To compose it is necessary to take into account that quantityparticles atoms in the left and right sides of the half-reactions should be the same

2J - - 2 e→ J 0 2
S +6 + 8 e→ S -2

By placing a vertical line on the right of this half-reaction diagram, we determine the reaction coefficients:

2J - - 2 e→ J 0 2 |8
S +6 + 8 e→ S -2 |2

Reducing by “2”, we get the final coefficient values:

2J - - 2 e→ J 0 2 |4
S +6 + 8 e→ S -2 |1

Let us summarize under this diagram half-reactions horizontal line and sum up what is involved in the reaction number of particles atoms:

2J - - 2 e→ J 0 2 |4
S +6 + 8 e→ S -2 |1
____________________
8J - + S +6 → 4 J 0 2 + S -2

After this it is necessary. Substituting the obtained values ​​of the coefficients into the molecular equation, we reduce it to this form:

8HJ + H 2 SO 4 = 4J 2 + H 2 S + H 2 O

By counting the number of hydrogen atoms on the left and right sides of the equation, we will be convinced of the need to correct the coefficient “4” before water, and we will obtain the complete equation:

8HJ + H 2 SO 4 = 4J 2 + H 2 S + 4H 2 O

This equation can be created using electronic methodion balance. In this case, there is no need to correct the coefficient in front of water molecules. The equation is based on the dissociation of ions of the compounds involved in the reaction: For example, dissociation of sulfuric acid results in the formation of two hydrogen protons and a sulfate anion:

H 2 SO 4 ↔ 2H + + SO 4 2-

The dissociation of hydrogen iodide and hydrogen sulfide can be written in a similar way:

HJ ↔ N + + J -
H 2 S ↔ 2Н + + S 2-

J 2 does not dissociate. H 2 O also practically does not dissociate. Composition half-reaction equations for iodine remains the same:

2J - - 2 e→ J 0 2
The half-reaction on sulfur atoms will take the following form:

SO 4 -2 → S -2

Since the right side of the half-reaction is missing four oxygen atoms, this amount must be balanced with water:

SO 4 -2 → S -2 + 4H 2 O

Then, on the left side of the half-reaction, it is necessary to compensate for the hydrogen atoms at the expense of protons (since the reaction of the medium is acidic):

SO 4 2- + 8H + → S -2 + 4H 2 O

By counting the number of electrons transferred, we obtain a complete representation of the equation according to half-reaction method:

SO 4 2- + 8H + + 8 e→ S -2 + 4H 2 O

Summing both half-reactions, we get electronic balance equation:

2J - - 2 e→ J 0 2 |8 4
SO 4 2- + 8H + + 8 e→ S -2 + 4H 2 O |2 1

8J - + SO 4 2- +8H + → 4J 2 0 + S 0 + 4H 2 O

From this entry it follows that the method electron-ion equation gives a more complete picture of the redox reaction than electronic balance method. The number of electrons participating in the process is the same for both methods of balance, but in the latter case, the number of protons and water molecules participating in the redox process is as if “automatically” established.

Let's look at several specific cases of redox reactions that can be compiled using the method electron-ion balance. Some redox processes are carried out with the participation of an alkaline environment, for example:

KCrO 2 + Br 2 + KOH → KBr + K 2 CrO 4 +H 2 O

In this reaction, the reducing agent is chromite ion (CrO 2 -), which is oxidized to chromate ion (CrO -2 4). The oxidizing agent - bromine (Br 0 2) is reduced to bromide ion (Br -):
CrO 2 - → CrO 4 2-
Br 0 2 → 2 Br -

Since the reaction occurs in an alkaline medium, the first half-reaction must be composed taking into account hydroxide ions (OH -):
CrO 2 - + 4OH - - 3 e= CrO 2- 4 + 2H 2 O

We compose the second half-reaction in a well-known way:
CrO 2 - + 4OH - -3 e= CrO 4 2 - + 2H 2 O |2
Br 0 2 + 2 e= Br - |3
__________
2CrO 2 - + 3Br 2 0 + 8OH - = 2CrO 2- 4 + 6Br - + 4H 2 O

After this it is necessary to finally assign coefficients in the reaction equation and completely molecular equation this redox process will take the form:

2KCrO 2 + 3Br 2 + 8KOH = 2K 2 CrO 4 + 6KBr + 4H 2 O.

In some cases, non-dissociable substances also participate in the redox reaction. For example:

AsH 3 + HNO 3 = H 3 AsO 4 + NO 2 + 4H 2 O

Then half-reaction method is compiled taking into account this process:

AsH 3 + 4H 2 O – 8 e= AsO 4 3- + 11H + |1
NO 3 + 2H + + e= NO 2 + H 2 O |8

AsH 3 + 8NO 3 + 4H 2 O + 2H + = AsO 4 3- + 8NO 2 + 11H + O

Molecular equation will take the form:

AsH 3 + 8HNO 3 = H 3 AsO 4 + 8NO 2 + 4H 2 O.

Redox reactions are sometimes accompanied by the simultaneous process of oxidation-reduction of several substances. For example, in a reaction with copper sulfide it interacts concentrated nitric acid:

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

The redox process involves atoms of copper, sulfur and nitrogen. When composing the equation half-reaction method It is necessary to take into account the stages of this process:

Cu+ → Cu2+
S 2- → S +6
N 5+ → N +2

In this situation, it is necessary to combine oxidative and reduction processes into one stage:

2Cu + - 2 e→ 2Cu 2+ | 10 e
S 2- - 8 e→ S 6+
_______________________
N 5+ + 3 e→ N 2+ | 3 e

In which the redox half-reaction takes the form:

2Cu + - 2 e→ 2Cu 2+
S 2- - 8 e→ S 6+ 3 ( recovery processes)
_______________________
N 5+ + 3 e→ N 2+ 10 (oxidation process)
_____________________________________

6Cu + + 3S 2- + 10N 5+ → 6Cu 2+ + 3S 6+ + 10N 2+

Eventually molecular reaction equation will take the form:

3Cu 2 S + 22HNO 3 = 6Cu(NO 3) 2 + 3H 2 SO 4 + 10NO + 8H 2 O.

Particular attention should be paid to redox reactions involving organic substances. For example, during the oxidation of glucose potassium permanganate in an acidic environment the following reaction occurs:

C 6 H 12 O 6 + KMnO 4 + H 2 SO 4 > CO 2 + MnSO 4 + K 2 SO 4 + H 2 O

When drawing up a balance half-reaction method transformation of glucose, the absence of its dissociation is taken into account, but the correction of the number of hydrogen atoms is carried out due to protons and water molecules:

C 6 H 12 O 6 + 6H 2 O - 24 e= 6CO 2 + 24H +

Half-reaction involving potassium permanganate will take the form:

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

As a result, we obtain the following scheme of the redox process:

C 6 H 12 O 6 + 6H 2 O - 24 e= 6CO 2 + 24H + | 5
MnО 4 - +8H + + 5 e= Mn +2 + 4H 2 O |24
___________________________________________________

5C 6 H 12 O 6 + 30H 2 O + 24MnO 4 - + 192H + = 30CO 2 + 120H + + 24Mn 2+ + 96H 2 O

By reducing the number of protons and water molecules on the left and right sides half-reactions, we get the final molecular equation:

5C 6 H 12 O 6 + 24KMnO 4 + 36H 2 SO 4 = 30CO 2 + 24MnSO 4 + 12K 2 SO 4 + 66H 2 O

5. The influence of the environment on the nature of redox reactions.

Depending on the environment (excess H +, neutral, excess OH -), the nature of the reaction between the same substances may change. Typically used to create an acidic environment sulfuric acid(H 2 SO 4), Nitric acid(HNO 3), hydrochloric acid (HCl), sodium hydroxide (NaOH) or potassium hydroxide (KOH) is used as an OH medium. For example, we will show how the environment affects potassium permanganate(КMnO 4) . and its reaction products:

For example, let’s take Na 2 SO 3 as a reducing agent and KMnO 4 as an oxidizing agent.

In an acidic environment:

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

SO 3 2- + H 2 O - 2 e→ SO 4 2- + 2H + |5
MnO 4 - + 8H + + 5 e→ Mn 2+ + 4H 2 O |2
________________________________________________
5SO 3 2- + 2MnO 4 - + 6H + → 5SO 4 2- + 2Mn 2+ + 3H 2 O

In neutral (or slightly alkaline):

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

SO 3 2- + H 2 O - 2 e→ SO 4 2- + 2H + |3
MnO 4 - + 2H 2 O + 3 e→ MnO 2 + 4OH |2
_____________________________________
3SO 3 2- + 2 MnO 4 - + H 2 O → 3SO 4 2- + 2MnO 2 + 2OH

In a highly alkaline environment:

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

SO 3 2- + 2 OH - - 2 e→ SO 4 2- + H 2 O |1
MnO 4 - + e→ MnO 4 2 |2
____________________________________

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

Hydrogen peroxide(H 2 O 2) depending on the environment is reduced according to the scheme:

1) Acidic environment (H +) H 2 O 2 + 2H + + 2 e→ 2H 2 O

2) Neutral medium (H 2 O) H 2 O 2 + 2 e→ 2OH

3) Alkaline environment (OH -) H 2 O 2 + 2 e→ 2OH

Hydrogen peroxide(H 2 O 2) acts as an oxidizing agent:

2FeSO 4 + H 2 O 2 + H 2 SO 4 → Fe 2 (SO 4) 3 + 2H 2 O

Fe 2+ - e= Fe 3+ |2
H 2 O 2 + 2H + + 2 e= 2H 2 O |1
________________________________
2Fe 2+ + H 2 O 2 + 2H + → 2Fe 3+ + 2 H 2 O

However, when encountering very strong oxidizing agents (KMnO 4) Hydrogen peroxide(H 2 O 2) acts as a reducing agent:

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

H 2 O 2 – 2 e→ O 2 + 2H + |5
MnO 4 - + 8H + + 5 e→ Mn 2+ + 4H 2 O |2
_________________________________
5H 2 O + 2 MnO 4 - + 6H + → 5O 2 + 2Mn 2+ + 8H 2 O

6. Determination of products of redox reactions.

The practical part of this topic discusses redox processes, indicating only the starting reagents. Reaction products usually need to be determined. For example, the reaction involves ferric chloride(FeCl 3) and potassium iodide(KJ):

FeCl 3 + KJ = A + B + C

required to install compound formulas A, B, C, formed as a result of the redox process.

The initial oxidation states of the reagents are as follows: Fe 3+, Cl -, K +, J -. It is easy to assume that Fe 3+, being an oxidizing agent (has a maximum oxidation state), can only reduce its oxidation state to Fe 2+:

Fe 3+ + e= Fe 2+

The chloride ion and potassium ion do not change their oxidation state in the reaction, but the iodide ion can only increase its oxidation state, i.e. go to state J 2 0:

2J - - 2 e= J 2 0

As a result of the reaction, in addition to the redox process, there will be exchange reaction between FeCl 3 and KJ, but taking into account the change in oxidation states, the reaction is not determined according to this scheme:

FeCl 3 + KJ = FeJ 3 + KCl,

but will take the form

FeCl 3 + KJ = FeJ 2 + KCl,

where the product C is the compound J 2 0:

FeCl 3 + 6KJ = 2FeJ 2 + 6KJ + J 2

Fe 3+ + e═> Fe 2+ |2

2J - - 2 e═> J 2 0 |1

________________________________

2Fe +3 + 2J - = 2Fe 2+ + J 2 0

In the future, when determining the products of the redox process, the so-called “elevator system” can be used. Its principle is that any redox reaction can be represented as the movement of elevators in a multi-story building in two mutually opposite directions. Moreover, the “floors” will be oxidation states corresponding elements. Since any of the two half-reactions in the redox process is accompanied by either a decrease or an increase oxidation states of one or another element, then by simple reasoning we can assume their possible oxidation states in the resulting reaction products.

As an example, consider a reaction in which sulfur reacts with concentrated sodium hydroxide solution ( NaOH):

S + NaOH(conc) = (A) + (B) + H 2 O

Since in this reaction changes will occur only with the oxidation states of sulfur, for clarity we will draw up a diagram of its possible states:

Compounds (A) and (B) cannot simultaneously be the states of sulfur S +4 and S +6, since in this case the process would occur only with the release of electrons, i.e. would be restorative:

S 0 - 4 e= S +4

S 0 - 6 e= S +6

But this would contradict the principle of redox processes. Then it should be assumed that in one case the process should proceed with the release of electrons, and in the other it should move in the opposite direction, i.e. be oxidizing:

S 0 - 4 e= S +4

S0+2 e= S -2

On the other hand, how likely is it that the recovery process will be carried out to the state S +4 or to S +6? Since the reaction occurs in an alkaline rather than an acidic environment, its oxidative potential is much lower, therefore the formation of the S +4 compound in this reaction is preferable to S +6. Therefore, the final reaction will take the form:

4S + 6NaOH(conc) = Na 2 SO 3 + 2Na 2 S + 3H 2 O

S 0 +2 e= S - 2 | 4 | 2

S 0 + 6OH - - 4 e= SO 3 2 - + 3H 2 O | 2 | 1

3S 0 + 6OH - = 2S - 2 + SO 3 2 - + 3H 2 O

As another example, consider the following reaction between phosphine and concentrated nitric acid(HNO3) :

PH 3 + HNO 3 = (A) + (B) + H 2 O

In this case, we have changing oxidation states of phosphorus and nitrogen. For clarity, we present state diagrams of their oxidation states.

Phosphorus in the state of oxidation state (-3) will exhibit only reducing properties, so in the reaction it will increase its oxidation state. Nitric acid itself is a strong oxidizing agent and creates an acidic environment, so phosphorus from a state of (-3) will reach its maximum oxidation state (+5).

In contrast, nitrogen will lower its oxidation state. In reactions of this type, usually to a state of (+4).

Further, it is not difficult to assume that phosphorus in the (+5) state, being a product (A), can only be orthophosphoric acid H 3 PO 4, since the reaction medium is strongly acidic. Nitrogen in such cases usually takes on the oxidation state (+2) or (+4), more often (+4). Therefore, the product (B) will be Nitric oxide NO2. All that remains is to solve this equation using the balance method:

P - 3 – 8 e= P +5 | 1
N+ 5 + e= N +4 | 8

P - 3 + 8N +5 = P +5 + 8N +4

PH 3 + 8HNO 3 = H 3 PO 4 + 8NO 2 + 4H 2 O

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Oxidation-reduction reactions (ORR)- reactions accompanied by the addition or loss of electrons, or redistribution of electron density on atoms (change in oxidation state).

Stages of OVR

Oxidation- donation of electrons by atoms, molecules or ions. As a result, the oxidation state increases. Reducing agents give up electrons.

Recovery- addition of electrons. As a result, the oxidation state decreases. Oxidizing agents accept electrons.

OVR- a coupled process: if there is reduction, then there is oxidation.

OVR rules

Equivalent exchange of electrons and atomic balance.

Acidic environment

In an acidic environment, the released oxide ions bind with protons to form water molecules; the missing oxide ions are supplied by water molecules, then protons are released from them.

Where there are not enough oxygen atoms, we write as many water molecules as there are not enough oxide ions.

Sulfur in potassium sulfite has an oxidation state of +4, manganese in potassium permanganate has an oxidation state of +7, sulfuric acid is the reaction medium.
Managanese in the highest oxidation state is an oxidizing agent, therefore, potassium sulfite is a reducing agent.

Note: +4 is an intermediate oxidation state for sulfur, so it can act as both a reducing agent and an oxidizing agent. With strong oxidizing agents (permanganate, dichromate), sulfite is a reducing agent (oxidized to sulfate); with strong reducing agents (halogenides, chalcogenides), sulfite is an oxidizing agent (reduced to sulfur or sulfide).

Sulfur goes from oxidation state +4 to +6 - sulfite is oxidized to sulfate. Manganese goes from oxidation state +7 to +2 (acidic environment) - the permanganate ion is reduced to Mn 2+.

2. Compose half-reactions. Equalizing manganese: 4 oxide ions are released from permanganate, which are bound by hydrogen ions (acidic medium) into water molecules. Thus, 4 oxide ions bind to 8 protons in 4 water molecules.

In other words, there are 4 oxygen missing on the right side of the equation, so we write 4 water molecules, and 8 protons on the left side of the equation.

Seven minus two is plus five electrons. You can equalize by total charge: on the left side of the equation there are eight protons minus one permanganate = 7+, on the right side there is manganese with a charge of 2+, water is electrically neutral. Seven minus two is plus five electrons. Everything is equalized.

Equating the sulfur: the missing oxide ion on the left side of the equation is supplied by a water molecule, which subsequently releases two protons on the right side.
On the left the charge is 2-, on the right it is 0 (-2+2). Minus two electrons.

Multiply the upper half-reaction by 2, the lower half-reaction by 5.

We reduce protons and water.

Sulfate ions bind to potassium and manganese ions.

Alkaline environment

In an alkaline environment, the released oxide ions are bound by water molecules, forming hydroxide ions (OH - groups). The missing oxide ions are supplied by hydroxo groups, which must be taken twice as much.

Where there are not enough oxide ions, we write hydroxo groups 2 times more than what is missing, on the other hand - water.

Example. Using the electron balance method, create a reaction equation, determine the oxidizing agent and reducing agent:

Determine the degree of oxidation:

Bismuth (III) with strong oxidizing agents (for example, Cl 2) in an alkaline environment exhibits reducing properties (oxidizes to bismuth V):

Since on the left side of the equation there are not enough 3 oxygens for balance, we write 6 hydroxo groups, and on the right - 3 waters.

The final reaction equation is:

Neutral environment

In a neutral environment, the released oxide ions are bound by water molecules to form hydroxide ions (OH - groups). The missing oxide ions are supplied by water molecules. H + ions are released from them.

Using the electron balance method, create a reaction equation, determine the oxidizing agent and reducing agent:

1. Determine the oxidation state: sulfur in potassium persulfate has an oxidation state of +7 (it is an oxidizing agent, because it has the highest oxidation state), bromine in potassium bromide has an oxidation state of -1 (it is a reducing agent, because it has the lowest oxidation state), water is the reaction medium.

Sulfur goes from oxidation state +7 to +6 - persulfate is reduced to sulfate. Bromine goes from oxidation state -1 to 0 - bromide ion is oxidized to bromine.

2. Compose half-reactions. We equalize sulfur (coefficient 2 before sulfate). Oxygen Eq.
On the left side there is a charge of 2-, on the right side there is a charge of 4-, 2 electrons are attached, so we write +2

We equalize the bromine (coefficient 2 before the bromide ion). On the left side the charge is 2-, on the right side the charge is 0, 2 electrons are given, so we write -2

3. Summary equation of electronic balance.

4. Final reaction equation: Sulfate ions combine with potassium ions to form potassium sulfate, a factor of 2 before KBr and before K2SO4. Water turned out to be unnecessary - put it in square brackets.

OVR classification

1. Oxidizing agent and reducing agent- different substances
2. Self-oxidizing agents, self-reducing agents (disproportionation, dismutation). An element in an intermediate oxidation state.
3. Oxidizing agent or reducing agent - medium for the process
4. Intramolecular oxidation-reduction. The same substance contains an oxidizing agent and a reducing agent.
   Solid-phase, high-temperature reactions.

Quantitative characteristics of ORR

Standard redox potential, E 0- electrode potential relative to standard hydrogen potential. More about.

To undergo ORR, it is necessary that the potential difference be greater than zero, that is, the potential of the oxidizing agent must be greater than the potential of the reducing agent:

,

For example:

The lower the potential, the stronger the reducing agent; the higher the potential, the stronger the oxidizing agent.
Oxidizing properties are stronger in an acidic environment, while reducing properties are stronger in an alkaline environment.

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