These include reactions in which the reactants exchange electrons, while changing the oxidation state of the atoms of the elements that make up the reactants.

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, they play an extremely important role.

Oxidation is the process of donating electrons from an atom, molecule, or ion.

When an atom donates its electrons, it acquires a positive charge:

For example:

Al - 3e - \u003d Al 3+

H 2 - 2e - \u003d 2H +

When oxidized, the oxidation state rises.

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

2Cl - - 2e - \u003d Cl 2

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

Fe 2+ - e - \u003d Fe 3+

Reduction is the process of adding electrons to an atom, molecule, or ion.

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

For example:

Cl 2 + 2e- \u003d 2Cl -

S + 2e - \u003d S 2-

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

Fe 3+ + e- \u003d Fe 2+

or it can go to 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, while 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

Oxidizer + e - ↔ Reductant

Therefore, redox reactions are a unity of two opposite processes - oxidation and reduction

The most important reducing agents and oxidizing agents

Restorers

Oxidizers

Metals, hydrogen, coal Carbon monoxide(II) 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 SnCl 2, iron (II) sulfate FeSO 4, manganese (II) sulfate MnSO 4, chromium (III) sulfate Cr 2 (SO 4) 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 in 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 Ions of noble metals (Ag +, Au 3+, etc.) Iron(III) chloride FeCl 3 Hypochlorites, chlorates and perchlorates Royal vodka, a mixture of concentrated nitric and hydrofluoric acids Anode in electrolysis

Electronic balance method.

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

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

Al + O 2 \u003d Al 2 O 3

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

So, what is the electronic balance method? Balance is equality. Therefore, it is necessary to make the same number of electrons that one element gives and accepts another element in this reaction. Initially, this amount looks different, as can be seen from the different oxidation states of aluminum and oxygen:

Al 0 + O 2 0 \u003d Al 2 +3 O 3 -2

Aluminum donates electrons (acquires a positive oxidation state), and oxygen accepts electrons (acquires a negative oxidation state). To obtain an oxidation state of +3, 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- \u003d Al +3

O 2 0 + 4e- \u003d 2O -2

In order for the number of given and received electrons to equalize, 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 before the reducing agent (Al) we put the coefficient 4 we found, and before the oxidizing agent (O 2) - the coefficient we found 3, then the number of given and received electrons equalizes and becomes equal to 12. The electronic balance is achieved. It can be seen that a factor of 2 is required before the Al 2 O 3 reaction product. Now the redox reaction equation is equalized:

4Al + 3O 2 \u003d 2Al 2 O 3

All the advantages of the electron balance method are manifested 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 +7 oxidation state. 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 make an electronic balance diagram:

Mn +7 + 5e- = Mn +2

2Cl - - 2e- \u003d 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. A factor of 5 should be placed before Cl 2 (or 2 × 5 \u003d 10 before KCl), and a factor of 2 before KMnO 4. All other factors are tied to these two factors. This is much easier than simply brute force.

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 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 8 in front of H 2 SO 4 and H 2 O. We got the equation in final form.

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

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

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 formulated 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 side there are also 6 atoms; palladium 1 and 1; copper is also 1 and 1. This means that the equation is correct.

We rewrite this equation in ionic form:

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

And after the contraction of identical ions, we get

Cu + Pd 2+ = Cu 2+ + Pd

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

(using this reaction, chlorine is obtained in the laboratory).

We write the formulas of the initial and final substances of the reaction:

HCl + MnO 2 → Cl 2 + MnCl 2 + H 2 O

We show the change in the 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 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 2 electrons. This coefficient is already in the electronic equation:

2HCl + MnO 2 → Cl 2 + MnCl 2 + H 2 O

We find the coefficients for other reactants. It can be seen from the electronic equations that 2 moles of HCl account for 1 mole of MnO 2 . However, given that another 2 mol of acid is needed to bind the resulting doubly charged manganese ion, a factor of 4 should be set in front of the reducing agent. Then 2 mol of water will be obtained. The final equation is

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

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

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

Let's rewrite the equation in ionic form:

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

and after canceling identical ions in both parts of the equation, we get

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

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

Let's write the reaction scheme - the formulas of the starting and obtained 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 the 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 recoil and attachment of electrons:

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

5H 2 S + 2KMnO 4 + ZH 2 SO 4 \u003d 5S + 2MnSO 4 + K 2 SO 4 + 8H 2 O

The correctness of writing the equation is confirmed by counting the atoms of one element, such as oxygen; on the left side there are 2 4 + 3 4 = 20 and on the right side 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 initial substances and reaction products must be the same. Charges must also be conserved. The sum of the charges of the reactants must always be equal to the sum of the charges of the reaction products.

The electron-ion balance method is more versatile than the electron balance method and has an undeniable advantage in the selection of coefficients in many redox reactions, in particular, with the participation of organic compounds, in which even the procedure for determining oxidation states is very complicated.

OVR classification

There are three main types of redox reactions:

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

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

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

Consider examples of reactions of three types.

1. Reactions of intermolecular oxidation-reduction are all the reactions we have already considered in this paragraph.
Let us consider a slightly more complicated case, when not all of the oxidizing agent can be consumed in the reaction, since part of it is involved in the usual, 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

Part of the particles NO 3 - participates in the reaction as an oxidizing agent, giving nitric oxide NO, and part of the NO 3 ions - unchanged passes into the copper compound Cu(NO 3) 2 . Let's make an electronic balance:

Cu 0 - 2e- \u003d Cu +2

N +5 + 3e- = N +2

We put the coefficient 3 found for copper in front of Cu and Cu(NO 3) 2 . But the coefficient 2 should be put 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 are not involved in the oxidation-reduction and are part of the Cu(NO 3) 2 product (NO 3 particles - here they are sometimes called the "ion - observer").

The remaining coefficients are selected without difficulty according to those already found:

3 Cu + 8HNO 3 \u003d 3 Cu (NO 3) 2 + 2 NO + 4H 2 O

2. Disproportionation reactions occur when molecules of the same substance are able to oxidize and reduce each other. This becomes possible if the substance contains in its composition atoms of any element in an intermediate oxidation state.

Therefore, the oxidation state can both decrease and increase. For example:

HN +3 O 2 \u003d 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 apply 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 \u003d 1 HNO 3 + 2 NO + H 2 O

Or, adding together moles of HNO 2:

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

Intramolecular oxidation-reduction reactions occur when oxidizing atoms and reducing atoms are adjacent in a molecule. Let's consider the decomposition of the KClO 3 berthollet salt 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- \u003d O 2 0

Here a difficulty arises - which of the two found coefficients 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 \u003d 2KCl + 3O 2

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

2KClO 3 \u003d 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 itself indicates, this method is based on the compilation of ionic equations for the oxidation process and the reduction process, followed by their summation into a general equation.
As an example, let's write an equation for the same reaction that was used in explaining the electron 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 the 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, two electrons must be subtracted from the left side of the circuit, after which the arrow can be replaced with an equal sign:

H 2 S - 2e - \u003d S + 2H +

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

The discoloration of the solution is associated with the transition of the MnO 4 - ion (it has a crimson color) to the Mn 2+ ion (practically colorless and only at a high concentration has a slightly pink color), which can be expressed by the scheme

MnO 4 - → Mn 2+

In an acidic solution, oxygen, which is part of the MnO 4 ions, together with hydrogen ions, eventually forms water. Therefore, the transition process is written as follows:

MnO 4 - + 8H + → Mn 2+ + 4H 2 O

In order 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 charge conservation condition, five electrons must be added to the left side of the scheme:

MnO 4 - + 8H + + 5e - \u003d 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 the general reaction equation, it is necessary to add the equations of half-reactions term by term, having previously equalized the numbers of given and received electrons. In this case, according to the rules for finding the smallest multiple, the corresponding factors are determined by which the equations of half-reactions are multiplied. Briefly, the entry is as follows:

And, having reduced 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 correct because the atoms and charges are equalized.

The half-reaction method is used to compose the reaction equation in ionic form. In order to move from it to an equation in molecular form, we proceed as follows: on the left side of the ionic equation, we select the corresponding cation for each anion, and an anion for each cation. 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, the formulation of the equations of 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 in comparison with the electron balance method is that that it uses not hypothetical ions, but real ones. In fact, there are no ions in the solution, but there are ions.

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

Writing separate ionic half-reaction equations is necessary to understand the chemical processes in a galvanic cell and during 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, it is not necessary 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 in the preparation of equations for all redox reactions occurring in aqueous solutions.

Chemical reactions that occur with a change in the oxidation states of elements are called redox reactions.

The main provisions of the theory of oxidation-reduction

1. The process of donating electrons by an atom or ion is called oxidation:

S 0 - 4e - ® S 4+ (oxidation)

An atom or ion that donates electrons is called a reducing agent (reductant): Zn 0 -2e - ® Zn 2+ (oxidation).

2. The process of adding electrons to an atom or ion is called recovery: S 6+ + 8e - ® S 2- (recovery).

Atoms or ions that accept electrons are called oxidizing agents (oxidizer): Cl - + e - ® Cl 0 (reduction).

The oxidizing agent is reduced during the reaction, and the reducing agent is oxidized. Oxidation is impossible without simultaneous reduction occurring with it, and vice versa, the reduction of one substance is impossible without the simultaneous oxidation of another.

3. In redox processes, the number of electrons given up in the oxidation process must always be equal to the number of electrons received in the reduction process.

Example:

Cu 2+ O 2- + H 2 0 \u003d Cu 0 + H 2 O 2-

oxidizer Cu 2+ +2e - ® Cu 0 reduction

reducing agent H 2 0 - 2e - ® 2H + oxidation

4. Equalization of the number of given and received electrons is carried out by selecting coefficients with a preliminary compilation of the electronic balance equation

Example:

Pb 2+ S 2- + HNO 3 ® S 0 + Pb 2+ (NO 3) 2 + N 2+ O 2- + H 2 O

Reducing agent S 2- - 2e - ® S 0 3 oxidation

oxidizer N 5+ + 3e - ® N 2+ 2 reduction

3PbS + 8HNO 3 ® 3S + 3Pb(NO 3) 2 + 2NO + 4H 2 O.

5. When compiling the electronic balance equation, it is necessary to proceed from as many atoms or ions as they are included in the molecule of the original substance, and sometimes in the molecule of the reaction products

Example:

K 2 Cr 2 6+ O 7 + H 2 SO 4 + KJ - ® J 2 0 + Cr 2 3+ (SO 4) 3 + K 2 SO 4 + H 2 O

Oxidizer 2Cr 6+ + 6e - ® 2Cr 3+ 2 1 reduction

reducing agent 2J - - 2e - ® J 2 0 6 3 oxidation

6. Redox processes occur most often in the presence of an environment: neutral, acidic or alkaline.

Selection of coefficients in redox reactions

When choosing the coefficients, one must take into account the basic position: the number of electrons given up by reduction is equal to the number of electrons obtained by oxidation.

After identifying the oxidizing agent, reducing agent, a digital scheme for the transition of electrons (electron balance equation) is compiled to the corresponding reaction equality.

Example 1 Al + Cl 2 ® AlCl 3 , where Al – reducing agent, Cl 2 -oxidizing agent.

Electron transition scheme:

Al 0 - 3e - ® Al +3 3 1 oxidation

Cl 0 + e - ® Cl 1 1 3 reduction

It can be seen from this diagram that for one oxidized aluminum atom, three chlorine atoms are required to accept these three electrons (see second column). Therefore, for every aluminum atom, three chlorine atoms are needed, or for every two aluminum atoms, three chlorine molecules. We get the coefficients:

2Al + 3Cl 2 \u003d AlCl 3.

Example 2 N 3- H 3 + O 0 2 ® N 2+ O 2- + H 2 O, where O 2 is a typical oxidizing agent, and N 3- H 3 plays the role of a reducing agent.

We draw up a scheme (electronic balance):

N 3- - 5e - ® N +2 5 2 4 oxidation

O 0 + 2e - ® O -2 2 5 10 reduction

For 4 nitrogen atoms, 10 atoms or 5 oxygen molecules are required. We get the coefficients:

4NH 3 + 5O 2 \u003d 4NO + 6H 2 O.

Special cases of compiling the equalities of redox reactions

1. If in the reaction the number of electrons lost by the reducing agent and the number of electrons accepted by the oxidizing agent are even numbers, then when finding the coefficients, the number of electrons is divided by the common largest divisor.

Example:

H 2 SO 3 + HClO 3 ® H 2 SO 4 + HCl

Reductant S +4 - 2e - ® S +6 6 3 oxidation

oxidizing agentCl +5 + 6e - ® Cl - 2 1 reduction

The coefficients of the reducing agent and oxidizing agent will not be 2 and 6, but 1 and 3:

3H 2 SO 3 + 3HClO 3 \u003d 3H 2 SO 4 + HCl.

If the number of electrons lost by the reducing agent and gained by the oxidizing agent is odd, and as a result of the reaction an even number of atoms should be obtained, then the coefficients are doubled.

Example:

KJ - + KMn +7 O 4 + H 2 S +6 O 4 ® J o 2 + K 2 S +6 O 4 + Mn +2 SO 4 + H 2 O

Reducing agent J - -1e - ® J o 5 10 oxidation

The coefficients for the oxidizing agent and reducing agent will not be 1 and 5, but 2 and 10:

10KJ + 2KMnO 4 + 8H 2 SO 4 = 5J 2 + 6K 2 SO 4 + 2MnSO 4 + 8H 2 O.

2. Sometimes a reducing agent or an oxidizing agent is additionally consumed to bind the products formed as a result of the reaction.

Example:

HBr - + KMn +7 O 4 + HBr ®Br 0 2 + KBr - + Mn +2 Br 2 0 + H 2 O

Reducing agent Br - - e - ® Br 0 5 10 oxidation

oxidizing agent Mn +7 + 5e - ® Mn +2 1 2 reduction

In this reaction, ten HBr molecules react as reducing agents, and six HBr molecules are required to bind the resulting substances (salt formation):

10HBr + 2KMnO 4 + 6HBr = 5Br 2 + 2KBr + 2MnBr 2 + 8H 2 O.

3. Both positive and negative ions of the reducing agent molecule are oxidized simultaneously.

Example:

As 2 +3 S 3 -2 + HN +5 O 3 ® H 3 As +5 O 4 + H 2 S +6 O 4 + N +2 O + H 2 O

Here As +3 ions are oxidized to As 2 +3 ions and at the same time S -2 ions are oxidized to S +6 ions and N +5 anions are reduced to N +2.

2As +3 - 4e - ® 2As +5

reducing agents 3S -2 - 24e - ® 3S +6 oxidation

oxidant N +5 + 3e - ® N +2 reduction

In this reaction, for every three As 2 S 3 molecules, 28 HNO 3 molecules react. We check the correctness of the formulation of the reaction equations by counting the hydrogen and oxygen atoms in the right and left parts. Thus, we find that 4 more water molecules enter into the reaction, which must be assigned to the left side of the equation for its final recording:

3As 2 S 3 + 28HNO 3 + 4H 2 O = 6H 3 AsO 4 + 9H 2 SO 4 + 28NO

2As +3 –4e®2As +5 4

3S -2 -24e®3S + 24

Reducing agents 2As +3 + 3S -2 - 28e - ®2As +5 + 3S +6 3 oxidation

oxidizer N +5 + 3e - ®N +2 28 reduction

4. Reducing agent and oxidizing agent are ions of the same element, but included in different substances.

Example:

KJ - + KJ +5 O 3 + H 2 SO 4 ® J 0 2 + K 2 SO 4 + H 2 O

Reducing agent J - - e - ® J 0 5 oxidation

oxidizer J +5 + 5e - ®J 0 1 reduction

5KJ + KJO 3 + 3H 2 SO 4 = 3J 2 + 3K 2 SO 4 + 3H 2 O.

5. A reducing agent and an oxidizing agent are ions of the same element that are part of the same substance (self-oxidation - self-recovery).

Example:

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

Reductant N +3 - 2e - ® N +5 1 oxidation

oxidizer N +3 + e - ® N +2 2 reduction

Therefore, the reaction equality

Redox reactions - reactions occurring with a change in the oxidation state of elements.

Oxidation- the process of donating electrons by an atom

Recovery- the process of receiving electrons by an atom

Reducing agent an element that donates electrons

Oxidizing agent- an element that accepts electrons

For a visual, but simplified idea of ​​the reasons for the change in the charges of the elements, let's turn to the figures:

An atom is an electrically neutral particle. Therefore, the number of protons is equal to the number of electrons

If an element donates an electron, its charge changes. It becomes positively charged (if it receives, on the contrary, negatively)

That. the charge of an element is affected by the number of given or received electrons

I. Drawing up equations of redox reactions

1. Write down the reaction scheme

Na + Cl 2 -> NaCl

2. We arrange the oxidation states of the elements:

Na 0 + Cl 2 0 -> Na + Cl -

3. We write out the elements that have changed the oxidation state and determine the number of donated / received electrons:

Na 0 -1e -> Na +

Cl 2 + 2e -> 2Cl -

4. Find the least common multiple of the numbers of given and attached electrons:

That. we got the required coefficients

5. Arrange the coefficients:

2Na 0 + Cl 2 0 -> 2Na + Cl -

Oxidizers are particles (atoms, molecules, or ions) that accept electrons during a chemical reaction. In this case, the oxidation state of the oxidizing agent going down. At the same time, oxidizers are recovering.

Restorers are particles (atoms, molecules, or ions) that donate electrons during a chemical reaction. In this case, the oxidation state of the reducing agent rises. At the same time, restorers are oxidized.

Chemicals can be divided into typical oxidizers, typical reducing agents, and substances that may exhibit both oxidizing and reducing properties. Some substances practically do not show redox activity.

To typical oxidizers include:

• simple substances - non-metals with the strongest oxidizing properties (fluorine F 2, oxygen O 2, chlorine Cl 2);
• ionsmetals or non-metals with high positive (usually higher) oxidation states : acids (HN +5 O 3, HCl +7 O 4), salts (KN +5 O 3, KMn +7 O 4), oxides (S +6 O 3, Cr +6 O 3)
• compounds containing some metal cations having high oxidation states: Pb 4+ , ​​Fe 3+ , Au 3+ etc.

Typical reducing agents is usually:

• simple substances - metals(reducing abilities of metals are determined by a series of electrochemical activity);
• complex substances that contain atoms or ions of non-metals with a negative (usually lower) oxidation state: binary hydrogen compounds (H 2 S, HBr), salts of oxygen-free acids (K 2 S, NaI);
• some compounds containing cations with the lowest positive oxidation state(Sn 2+, Fe 2+, Cr 2+), which, donating electrons, can increase their oxidation state;
• compounds containing complex ions, consisting of non-metals with an intermediate positive oxidation state(S +4 O 3) 2–, (НР +3 O 3) 2– , in which elements can, by donating electrons, increase its positive oxidation state.

Most other substances can show both oxidizing and reducing properties.

Typical oxidizing and reducing agents are shown in the table.

In laboratory practice the most commonly used are the following oxidizers :

potassium permanganate (KMnO 4);

potassium dichromate (K 2 Cr 2 O 7);

nitric acid (HNO 3);

concentrated sulfuric acid (H 2 SO 4);

hydrogen peroxide (H 2 O 2);

oxides of manganese (IV) and lead (IV) (MnO 2 , PbO 2);

molten potassium nitrate (KNO 3) and melts of some other nitrates.

To reducers that apply in laboratory practice relate:

• magnesium (Mg), aluminum (Al), zinc (Zn) and other active metals;
• hydrogen (H 2) and carbon (C);
• potassium iodide (KI);
• sodium sulfide (Na 2 S) and hydrogen sulfide (H 2 S);
• sodium sulfite (Na 2 SO 3);
• tin chloride (SnCl 2).

Classification of redox reactions

Redox reactions are usually divided into four types: intermolecular, intramolecular, disproportionation reactions (self-oxidation-self-reduction), and counter-disproportionation reactions.

Intermolecular reactions proceed with a change in the degree of oxidation different elements from different reagents. At the same time, they form various products of oxidation and reduction .

2Al0 + Fe +3 2 O 3 → Al +3 2 O 3 + 2Fe 0,

C 0 + 4HN +5 O 3 (conc) = C +4 O 2 + 4N +4 O 2 + 2H 2 O.

Intramolecular reactions are reactions in which different elements from one reagent move into different products such as:

(N -3 H 4) 2 Cr+6 2 O 7 → N 2 0 + Cr +3 2 O 3 + 4 H 2 O,

2 NaN +5 O -2 3 → 2 NaN +3 O 2 + O 0 2 .

Disproportionation reactions (self-oxidation-self-healing) - these are reactions in which the oxidizing agent and reducing agent - the same element of the same reagent, which goes into different products:

3Br 2 + 6 KOH → 5KBr + KBrO 3 + 3 H 2 O,

Reproportionation (proportionation, counterdisproportionation ) are reactions in which an oxidizing agent and a reducing agent are the same element, Which one of different reagents goes into one product. Reaction inverse to disproportionation.

2H 2 S -2 + S + 4 O 2 \u003d 3S + 2H 2 O

Basic rules for compiling redox reactions

Redox reactions are accompanied by oxidation and reduction processes:

Oxidation is the process of donating electrons by a reducing agent.

Recovery is the process of adding electrons to an oxidizing agent.

Oxidizing agent recovering, and the reducing agent oxidized .

In redox reactions, the electronic balance: The number of electrons that the reducing agent donates is equal to the number of electrons that the oxidizing agent receives. If the balance is drawn up incorrectly, you will not be able to draw up complex OVRs.

Several methods for compiling redox reactions (ORRs) are used: the electron balance method, the electron-ion balance method (half-reaction method), and others.

Consider in detail electronic balance method .

It is quite easy to “recognize” OVR - it is enough to arrange the oxidation states in all compounds and determine that the atoms change the oxidation state:

K + 2 S -2 + 2K + Mn +7 O -2 4 = 2K + 2 Mn +6 O -2 4 + S 0

We write out separately the atoms of the elements that change the oxidation state, in the state BEFORE the reaction and AFTER the reaction.

The oxidation state is changed by manganese and sulfur atoms:

S -2 -2e = S 0

Mn +7 + 1e = Mn +6

Manganese absorbs 1 electron, sulfur donates 2 electrons. At the same time, it is necessary to comply electronic balance. Therefore, it is necessary to double the number of manganese atoms, and leave the number of sulfur atoms unchanged. We indicate the balance coefficients both before the reagents and before the products!

Scheme for compiling OVR equations using the electronic balance method:

Attention! There can be several oxidizing or reducing agents in a reaction. The balance must be drawn up so that the TOTAL number of given and received electrons is the same.

General patterns of redox reactions

The products of redox reactions often depend on process conditions. Consider the main factors affecting the course of redox reactions.

The most obvious determining factor is reaction solution medium - . As a rule (but not necessarily), the substance defining the medium is listed among the reagents. The following options are possible:

• oxidative activity intensifies in a more acidic environment and the oxidant is reduced more deeply(for example, potassium permanganate, KMnO 4, where Mn +7 is reduced to Mn +2 in an acidic environment, and to Mn +6 in an alkaline environment);
• oxidative activity intensifies in a more alkaline environment, and the oxidizing agent is reduced more deeply (for example, potassium nitrate KNO 3, where N +5, when interacting with a reducing agent in an alkaline medium, is reduced to N -3);
• or the oxidizing agent is practically not subject to changes in the environment.

The reaction medium makes it possible to determine the composition and form of existence of the remaining OVR products. The basic principle is that products are formed that do not interact with reagents!

Note! E If the solution medium is acidic, then bases and basic oxides cannot be present among the reaction products, because they interact with acid. Conversely, in an alkaline medium, the formation of acid and acid oxide is excluded. This is one of the most common, and most gross mistakes.

Also, the direction of OVR flow is affected by the nature of the reactants. for example, during the interaction of nitric acid HNO 3 with reducing agents, a pattern is observed - the greater the activity of the reducing agent, the more nitrogen N + 5 is reduced.

With an increase temperature most OVRs tend to be more intense and deeper.

In heterogeneous reactions, the composition of products is often influenced by fineness of the solid . For example, powdered zinc with nitric acid forms one product, while granular zinc forms completely different products. The greater the degree of grinding of the reagent, the greater its activity, usually.

Consider the most typical laboratory oxidizers.

Basic schemes of redox reactions

Scheme for the recovery of permanganates

Permanganates contain a powerful oxidizing agent - manganese in oxidation state +7. Salts of manganese +7 color the solution in Violet Colour.

Permanganates, depending on the medium of the reaction solution, are reduced in different ways.

AT acidic environment recovery is deeper Mn2+. Manganese oxide in the +2 oxidation state exhibits basic properties, therefore, in acidic environment salt is formed. Salts of manganese +2 colorless. AT neutral solution manganese is recovered to the degree of oxidation +4 , with education amphoteric oxide MnO 2 — brown sediment insoluble in acids and alkalis. AT alkaline environment, manganese is reduced minimally - to the nearest oxidation states +6 . Manganese compounds +6 exhibit acidic properties, in an alkaline medium they form salts - manganates. Manganates give the solution green coloring .

Consider the interaction of potassium permanganate KMnO 4 with potassium sulfide in acidic, neutral and alkaline media. In these reactions, the oxidation product of the sulfide ion is S 0 .

5 K 2 S + 2 KMnO 4 + 8 H 2 SO 4 \u003d 5 S + 2 MnSO 4 + 6 K 2 SO 4 + 8 H 2 O,

3 K 2 S + 2 KMnO 4 + 4 H 2 O = 2 MnO 2 ↓ + 3 S↓ + 8 KOH,

A common mistake in this reaction is an indication of the interaction of sulfur and alkali in the reaction products. However, sulfur interacts with alkali under rather harsh conditions (elevated temperature), which does not correspond to the conditions for this reaction. Under normal conditions, it will be correct to indicate exactly molecular sulfur and alkali separately, and not the products of their interaction.

K 2 S + 2 KMnO 4 - (KOH) \u003d 2 K 2 MnO 4 + S ↓

Difficulties also arise when compiling this reaction. The fact is that in this case, writing the molecule of the medium (KOH or other alkali) in the reagents is not required to equalize the reaction. Alkali takes part in the reaction and determines the product of the reduction of potassium permanganate, but the reactants and products are equalized even without its participation. This seemingly paradox can be easily resolved if we remember that a chemical reaction is just a conditional notation that does not indicate every ongoing process, but is just a reflection of the sum of all processes. How to determine it yourself? If you act according to the classical scheme - balance-balance coefficients - metal equalization, then you will see that metals are equalized by balance coefficients, and the presence of alkali on the left side of the reaction equation will be superfluous.

Permanganates oxidize:

• nonmetals with a negative oxidation state to simple substances (with oxidation state 0), exceptions — phosphorus, arsenic - up to +5 ;
• nonmetals with an intermediate oxidation state to the highest degree of oxidation;
• active metals stable positive the degree of oxidation of the metal.

KMnO 4 + NeMe (lowest d.d.) = NeMe 0 + other products

KMnO 4 + NeMe (intermediate s.o.) = NeMe (higher s.o.) + other products

KMnO 4 + Me 0 = Me (stable s.d.) + other products

KMnO 4 + P -3, As -3 = P +5, As +5 + other products

Chromate/Bichromate Recovery Scheme

A feature of chromium with valence VI is that it forms 2 types of salts in aqueous solutions: chromates and bichromates, depending on the solution medium. Active metal chromates (for example, K 2 CrO 4) are salts that are stable in alkaline environment. Dichromates (bichromates) of active metals (for example, K 2 Cr 2 O 7) - salts, stable in an acidic environment .

Chromium(VI) compounds are reduced to chromium(III) compounds . Chromium compounds Cr +3 are amphoteric, and depending on the medium of the solution, they exist in solution in various forms: in an acidic medium in the form salts(amphoteric compounds form salts when interacting with acids), in a neutral medium - insoluble amphoteric chromium (III) hydroxide Cr(OH) 3 , and in an alkaline environment, chromium (III) compounds form complex salt, For example, potassium hexahydroxochromate (III) K 3 .

Chromium VI compounds oxidize:

• nonmetals in a negative oxidation state to simple substances (with oxidation state 0), exceptions — phosphorus, arsenic - up to +5;
• nonmetals in an intermediate oxidation state to the highest degree of oxidation;
• active metals from simple substances (oxidation point 0) to compounds with stable positive the degree of oxidation of the metal.

Chromate/bichromate + neMe (negative d.d.) = neMe 0 + other products

Chromate/bichromate + NeMe (intermediate positive s.d.) = NeMe (highest s.d.) + other products

Chromate / bichromate + Me 0 \u003d Me (stable s.d.) + other products

Chromate/dichromate + P, As (negative d.d.) = P, As +5 + other products

Decomposition of nitrates

Nitrate salts contain nitrogen in oxidation state +5 - strong oxidizing agent. Such nitrogen can oxidize oxygen (O -2). This happens when nitrates are heated. In this case, in most cases, oxygen is oxidized to the oxidation state 0, i.e. before molecular oxygen O2 .

Depending on the type of metal that forms the salt, various products are formed during the thermal (temperature) decomposition of nitrates: if metal active(in the series of electrochemical activity are to magnesium), then nitrogen is reduced to an oxidation state of +3, and upon decomposition nitrite salts and molecular oxygen are formed .

for example:

2NaNO 3 → 2NaNO 2 + O 2 .

Active metals occur in nature in the form of salts (KCl, NaCl).

If a metal is in the electrochemical activity series to the right of magnesium and to the left of copper (including magnesium and copper) , then the decomposition produces metal oxide in a stable oxidation state, nitric oxide (IV)(brown gas) and oxygen. Metal oxide also forms during decomposition lithium nitrate .

for example, decomposition zinc nitrate:

2Zn(NO 3) 2 → 2ZnO + 4NO 2 + O 2.

Metals of medium activity are most often found in nature in the form of oxides (Fe 2 O 3, Al 2 O 3, etc.).

ions metals, located in the series of electrochemical activity to the right of copper are strong oxidizing agents. At decomposition of nitrates they, like N +5, participate in the oxidation of oxygen, and are reduced to simple substances, i.e. metal is formed and gases are released nitric oxide (IV) and oxygen .

for example, decomposition silver nitrate:

2AgNO 3 → 2Ag + 2NO 2 + O 2 .

Inactive metals occur in nature in the form of simple substances.

Some exceptions!

Decomposition ammonium nitrate :

In the ammonium nitrate molecule there is both an oxidizing agent and a reducing agent: nitrogen in the -3 oxidation state exhibits only reducing properties, nitrogen in the +5 oxidation state only oxidizing.

When heated, ammonium nitrate decomposing. At temperatures up to 270 o C, nitric oxide (I)("laughing gas") and water:

NH 4 NO 3 → N 2 O + 2H 2 O

This is an example of a reaction counterdisproportionation .

The resulting oxidation state of nitrogen is the arithmetic mean of the oxidation state of nitrogen atoms in the original molecule.

At a higher temperature, nitric oxide (I) decomposes into simple substances - nitrogen and oxygen:

2NH 4 NO 3 → 2N 2 + O 2 + 4H 2 O

At decomposition ammonium nitrite NH4NO2 counter-disproportionation also occurs.

The resulting oxidation state of nitrogen is also equal to the arithmetic mean of the oxidation states of the initial nitrogen atoms - the oxidizing agent N +3 and the reducing agent N -3

NH 4 NO 2 → N 2 + 2H 2 O

Thermal decomposition manganese(II) nitrate accompanied by metal oxidation:

Mn(NO 3) 2 \u003d MnO 2 + 2NO 2

Iron(II) nitrate at low temperatures it decomposes to iron oxide (II), when heated, iron is oxidized to an oxidation state of +3:

2Fe(NO 3) 2 → 2FeO + 4NO 2 + O 2 at 60°C
4Fe(NO 3) 2 → 2Fe 2 O 3 + 8NO 2 + O 2 at >60°C

Nickel(II) nitrate decomposes to nitrite when heated.

Oxidizing properties of nitric acid

Nitric acid HNO 3 when interacting with metals is practically never forms hydrogen , unlike most mineral acids.

This is due to the fact that the acid contains a very strong oxidizing agent - nitrogen in the +5 oxidation state. When interacting with reducing agents - metals, various products of nitrogen reduction are formed.

Nitric acid + metal \u003d metal salt + nitrogen reduction product + H 2 O

Nitric acid can be converted into nitric oxide (IV) NO 2 (N +4); nitric oxide (II) NO (N +2); nitric oxide (I) N 2 O ("laughing gas"); molecular nitrogen N 2 ; ammonium nitrate NH 4 NO 3. As a rule, a mixture of products is formed with a predominance of one of them. Nitrogen is reduced in this case to oxidation states from +4 to −3. The depth of recovery depends primarily by nature reducing agent and from the concentration of nitric acid . This is how the rule works: the lower the concentration of the acid and the higher the activity of the metal, the more electrons nitrogen receives, and the more reduced products are formed.

Some patterns will allow you to correctly determine the main product of nitric acid reduction by metals in the reaction:

• under action very dilute nitric acid on the metals usually formed ammonium nitrate NH 4 NO 3 ;

for example, interaction of zinc with very dilute nitric acid:

4Zn + 10HNO 3 = 4Zn(NO 3) 2 + NH 4 NO 3 + 3H 2 O

• concentrated nitric acid in the cold passivates some metals - chromium Cr, aluminum Al and iron Fe . When the solution is heated or diluted, the reaction proceeds;

metal passivation - this is the transfer of the metal surface to an inactive state due to the formation of thin layers of inert compounds on the metal surface, in this case mainly metal oxides, which do not react with concentrated nitric acid

• Nitric acid does not react with platinum subgroup metals — gold Au, platinum Pt, and palladium Pd;

• when interacting concentrated acid with inactive metals and metals of medium activity nitric acid is reduced to nitric oxide (IV) NO 2 ;

for example, oxidation of copper with concentrated nitric acid:

Cu + 4HNO 3 \u003d Cu (NO 3) 2 + 2NO 2 + 2H 2 O

• when interacting concentrated nitric acid with active metals formed nitrogen oxide (I) N 2 O ;

for example, oxidation sodium concentrated nitric acid:

Na + 10HNO 3 \u003d 8NaNO 3 + N 2 O + 5H 2 O

• when interacting dilute nitric acid with inactive metals (in the activity series to the right of hydrogen) the acid is reduced to nitric oxide (II) NO ;
• when interacting dilute nitric acid with intermediate activity metals either nitric oxide (II) NO, or nitric oxide N 2 O, or molecular nitrogen N 2 - depending on additional factors (metal activity, metal grinding degree, acid dilution degree, temperature).
• when interacting dilute nitric acid with active metals formed molecular nitrogen N 2 .

For an approximate determination of the products of the reduction of nitric acid in the interaction with different metals, I propose to use the pendulum principle. The main factors that shift the position of the pendulum are the concentration of the acid and the activity of the metal. To simplify, we use 3 types of acid concentrations: concentrated (more than 30%), dilute (30% or less), very dilute (less than 5%). We divide metals by activity into active (before aluminum), medium activity (from aluminum to hydrogen) and inactive (after hydrogen). The products of the reduction of nitric acid are arranged in descending order of the degree of oxidation:

NO2; NO; N 2 O; N 2 ; NH4NO3

The more active the metal, the more we move to the right. The greater the concentration or the lower the dilution of the acid, the more we shift to the left.

for example , concentrated acid and inactive metal copper Cu interact. Therefore, we shift to the extreme left position, nitric oxide (IV), copper nitrate and water are formed.

The interaction of metals with sulfuric acid

Dilute sulfuric acid interacts with metals like a normal mineral acid. Those. interacts with metals that are located in a series of electrochemical voltages up to hydrogen. The oxidizing agent here is H + ions, which are reduced to molecular hydrogen H 2. In this case, metals are oxidized, as a rule, to minimum degree of oxidation.

for example:

Fe + H 2 SO 4 (razb) \u003d FeSO 4 + H 2

interacts with metals standing in a series of voltages both before and after hydrogen.

H 2 SO 4 (conc) + metal \u003d metal salt + sulfur reduction product (SO 2, S, H 2 S) + water

When concentrated sulfuric acid interacts with metals, a metal salt (in a stable oxidation state), water and a sulfur reduction product are formed - sulfur dioxide S +4 O 2, molecular sulfur S or hydrogen sulfide H 2 S -2, depending on the degree of concentration, activity of the metal, its degree of grinding, temperature, etc. When concentrated sulfuric acid interacts with metals, molecular hydrogen is not formed!

The basic principles of the interaction of concentrated sulfuric acid with metals:

1. concentrated sulfuric acid passivates aluminium, chrome, iron at room temperature, or in the cold;

2. concentrated sulfuric acid does not interact with gold, platinum and palladium ;

3. With inactive metals concentrated sulfuric acid recovers to sulfur oxide (IV).

for example, copper is oxidized with concentrated sulfuric acid:

Cu 0 + 2H 2 S +6 O 4 (conc) = Cu +2 SO 4 + S +4 O 2 + 2H 2 O

4. When interacting with active metals and zinc concentrated sulfuric acid formssulfur S or hydrogen sulfide H 2 S 2- (depending on temperature, degree of grinding and activity of the metal).

for example , interaction of concentrated sulfuric acid with zinc:

8Na 0 + 5H 2 S +6 O 4(conc) → 4Na 2 + SO 4 + H 2 S — 2 + 4H2O

Hydrogen peroxide

Hydrogen peroxide H 2 O 2 contains oxygen in the -1 oxidation state. Such oxygen can both increase and decrease the oxidation state. Thus, hydrogen peroxide exhibits both oxidizing and reducing properties.

When interacting with reducing agents, hydrogen peroxide exhibits the properties of an oxidizing agent, and is reduced to an oxidation state of -2. As a rule, the product of reduction of hydrogen peroxide is water or hydroxide ion, depending on the reaction conditions. For example:

S +4 O 2 + H 2 O 2 -1 → H 2 S +6 O 4 -2

When interacting with oxidizing agents, peroxide is oxidized to molecular oxygen (oxidation state 0): O 2 . for example :

2KMn +7 O 4 + 5H 2 O 2 -1 + 3H 2 SO 4 → 5O 2 0 + 2Mn +2 SO 4 + K 2 SO 4 + 8H 2 O

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

Some chemicals exhibit the properties of oxidizing agents, others - reducing agents, while some compounds can exhibit both properties at the same time (for example, hydrogen peroxide H 2 O 2).

What is an oxidizing agent and a reducing agent, 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., is oxidized. The processes of transferring electrons from one substance to another are usually called redox reactions.

Compounds containing atoms of elements with a maximum degree of oxidation 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 atom of an element 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 external energy level of such atoms is completed by eight electrons