Redox reactions with changes in oxidation states. Drawing up redox equations

According to the change in the oxidation state, all chemical reactions can be divided into two types:

I. Reactions occurring without changing the degree of oxidation of the elements that make up the reactants. Such reactions are referred to as ion exchange reactions.

Na 2 CO 3 + H 2 SO 4 = Na 2 SO 4 + CO 2 + H 2 O.

II. Reactions that occur with a change in the oxidation state of elements,

included in the reactants. Such reactions are referred to as redox reactions.

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

Oxidation state(oxidation) - a characteristic of the state of the atoms of the elements in the composition of the molecule. It characterizes the uneven distribution of electrons between the atoms of elements and corresponds to the charge that an atom of an element would acquire if all the common electron pairs of its chemical bonds shifted towards a more electronegative element. Depending on the relative electronegativity of the elements that form a bond, an electron pair can be shifted to one of the atoms or symmetrically located relative to the nuclei of atoms. Therefore, the oxidation state of elements can be negative, positive, or zero.

Elements whose atoms accept electrons from other atoms have a negative oxidation state. Elements whose atoms donate their electrons to other atoms have a positive oxidation state. Atoms in the molecules of simple substances have a zero oxidation state, as well as if the substance is in the atomic state.

The oxidation state is denoted +1, +2.

Ion charge 1+, 2+.

The oxidation state of an element in a compound is determined by the rules:

1. The degree of oxidation of an element in simple substances is zero.

2. Some elements in almost all of their compounds exhibit a constant oxidation state. These elements include:

It has an oxidation state of +1 (with the exception of metal hydrides).

O has an oxidation state of -2 (with the exception of fluorides).

3. Elements of groups I, II and III of the main subgroups of the Periodic Table of Elements of D.I. Mendeleev have a constant oxidation state equal to the group number.

Elements Na, Ba, Al: oxidation state +1, +2, +3, respectively.

4. For elements that have a variable oxidation state, there is the concept of higher and lower oxidation states.

The highest oxidation state of an element is equal to the group number of the Periodic Table of Elements of D.I. Mendeleev, in which the element is located.

Elements N, Cl: the highest oxidation state is +5, +7, respectively.

The lowest oxidation state of an element is equal to the group number of the Periodic Table of Elements of D.I. Mendeleev, in which the element is located minus eight.

Elements N, Cl: the lowest oxidation state is -3, -1, respectively.

5. In single-element ions, the oxidation state of the element is equal to the charge of the ion.

Fe 3+ - the oxidation state is +3; S 2- - the oxidation state is -2.

6. The sum of the oxidation states of all atoms of elements in a molecule is zero.

KNO 3 ; (+1) + X+ 3 (-2) = 0; X = +5. The oxidation state of nitrogen is +5.

7. The sum of the oxidation states of all atoms of elements in an ion is equal to the charge of the ion.

SO 4 2- ; X+ 4 (-2) = -2; X= +6. The oxidation state of sulfur is +6.

8. In compounds consisting of two elements, the element that is written on the right always has the lowest oxidation state.

Reactions in which the oxidation state of elements changes are referred to as redox reactions /ORD/. These reactions consist of oxidation and reduction processes.

Oxidation The process of donating electrons by an element that is part of an atom, molecule or ion is called.

Al 0 - 3e \u003d Al 3+

H 2 - 2e \u003d 2H +

Fe 2+ - e \u003d Fe 3+

2Cl - - 2e \u003d Cl 2

When oxidized, the oxidation state of an element increases. A substance (atom, molecule, or ion) that contains an element that donates electrons is called a reducing agent. Al, H 2 , Fe 2+ , Cl - - reducing agents. The reducing agent is oxidized.

Recovery The process of adding electrons to an element that is part of an atom, molecule or ion is called.

Cl 2 + 2e \u003d 2Cl -

Fe 3+ + e \u003d Fe 2+

When reduced, the oxidation state of an element decreases. A substance (atom, molecule, or ion) that contains an element that accepts electrons is called an oxidizing agent. S, Fe 3+, Cl 2 are oxidizing agents. The oxidant is restored.

The total number of electrons in the system during a chemical reaction does not change. The number of electrons donated by the reducing agent is equal to the number of electrons attached by the oxidizing agent.

To compile the equation of the redox reaction (ORR) in solutions, the ion-electronic method (half-reaction method) is used.

OVR can occur in acidic, neutral or alkaline environments. The reaction equations take into account the possible participation of water molecules (HOH) and those contained in the solution, depending on the nature of the medium, an excess of H + or OH - ions:

in an acidic environment - HOH and H + ions;

in a neutral environment - only HOH;

in an alkaline environment - HOH and OH - ions.

When compiling the OVR equations, it is necessary to adhere to a certain sequence:

1. Write a reaction scheme.

2. Determine the elements that have changed their oxidation state.

3. Write a diagram in a short ion-molecular form: strong electrolytes in the form of ions, weak electrolytes in the form of molecules.

4. Compose equations for the processes of oxidation and reduction (equation of half-reactions). To do this, write down the elements that change the degree of oxidation in the form of real particles (ions, atoms, molecules) and equalize the number of each element in the left and right parts of the half-reaction.

Note:

If the original substance contains fewer oxygen atoms than the products (P PO 4 3-), then the lack of oxygen is supplied by the environment.

If the original substance contains more oxygen atoms than the products (SO 4 2-SO 2), then the released oxygen is bound by the medium.

5. Equalize the left and right parts of the equations by the number of charges. To do this, add or subtract the required number of electrons.

6. Select factors for the oxidation and reduction half-reactions so that the number of electrons during oxidation is equal to the number of electrons during reduction.

7. Summarize the half-reactions of oxidation and reduction, taking into account the found factors.

8. Write down the resulting ion-molecular equation in molecular form.

9. Carry out an oxygen test.

There are three types of redox reactions:

a) Intermolecular - reactions in which the oxidation state changes for the elements that make up different molecules.

2KMnO 4 + 5NaNO 2 + 3H 2 SO 4 2MnSO 4 + 5NaNO 3 + K 2 SO 4 + 3H 2 O

b) Intramolecular - reactions in which the oxidation state changes for the elements that make up one molecule.

1. How to determine the redox reaction?

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

As an example, consider two reactions:

Zn 0 + 2H +1 C1 -1 \u003d Zn +2 Cl 2 -1 + H 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 - 2e = Zn 2+
2H + 1 + 2e \u003d H 2 0
2Cl -1 \u003d 2 Cl -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 reaction, and reaction (2) belongs to another type.

Chemical reactions that are carried out with a changeoxidation states of elementsare 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. This requires knowing how to determine the oxidation state of an element.

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

2. Definitionoxidation states of elements.

The oxidation state of the 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 zero. For example, in the compound H 3 PO 4, the oxidation state of hydrogen is +1, phosphorus is +5, and oxygen is -2; Having made a mathematical equation, we determine that in the sum number of particles(atoms or ions) will have a charge equal to zero: (+1)x3+(+5)+(-2)x4 = 0

But in this example, the oxidation states of the elements are already set. How can one determine the degree of oxidation of sulfur, for example, in the compound sodium thiosulfate Na 2 S 2 O 3, or manganese in the compound potassium permanganate- KMnO 4 ? For 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 system (including hydrogen in combination with non-metals) +1;
2) Elements of group II of the periodic system +2;
3) Elements of group III of the periodic system +3;
4) Oxygen (except in combination with fluorine or in 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 the elements whose composition reflects this compound formula, is equal to zero, then denoting the unknown charge of sulfur " 2X”(since there are two sulfur atoms in the formula), we compose the following mathematical equation:

(+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 (+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-over" method to determine the 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 easy to establish that the oxidation state of titanium will be equal to (+3). Or, for example, in conjunction 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 in the formula of this compound (-4). Also, using the "criss-cross" method, it is not difficult to establish that if the following compound formula Cr 4 Si 3, then the degree of oxidation of chromium into it is (+3), and silicon (-4).
For salts, this is also not difficult. And it doesn't matter if it's given or medium salt or acid salt. In these cases, it is necessary to proceed from the salt-forming acid. For example, given salt sodium nitrate(NaNO3). It is known that it is a derivative of nitric acid (HNO 3), and in this compound the degree of nitrogen oxidation is (+5), therefore, in its salt - sodium nitrate, the degree of nitrogen oxidation is also (+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 (+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 donate electrons:

Cl +7 - e \u003d Cl +8
S +6 - e \u003d S +7

But the elements chlorine and sulfur cannot exist with such oxidation states. Similarly, only reducing agents are substances that have a minimum, as a rule, 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 would have to add electrons:

S-2 + e = S-3
J - + e \u003d J -2

But for sulfur and iodine, ions with such degrees of oxidation 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= Bi0

sn+2+2 e= Sn+4

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

SO 2 +4 + 2H 2 S -2 \u003d 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 and reducing agents 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 when the oxidizing atom and the reducing atom are in the same substance, for example:

N -3 H 4 N +5 O 3 \u003d 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 due to 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=Al3+

2Cl - - 2 e= Cl 2 0

Fe 2+ - e= Fe3+

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 acquires electrons, then such a process is called reduction, and this atom is an oxidizing agent, for example:

Ca 2+ + 2 e= Ca0

Cl 2 0 + 2 e= 2Cl -

Fe3+ + e= Fe 2+

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

Redox equations differ from molecular equations other reactions by the difficulty of selecting coefficients in front of the reactants and reaction products. For this use electronic balance method, or method of electron-ion equations(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 us 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

In order to make it necessary to take into account that amountparticles atoms in the left and right parts of the half-reactions should be the same

2J - - 2 e→ J 0 2
S+6+8 e→S-2

By setting the vertical line on the right of this half-reaction scheme, 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 values ​​of the coefficients:

2J - - 2 e→ J 0 2 |4
S+6+8 e→ S-2 |1

Let's sum up under this scheme half reactions horizontal line and summarize 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 that it is necessary. Substituting the obtained values ​​of the coefficients into the molecular equation, we bring it to this form:

8HJ + H 2 SO 4 \u003d 4J 2 + H 2 S + H 2 O

Having counted the number of hydrogen atoms in the left and right parts of the equation, we will make sure that the coefficient “4” in front of water needs to be corrected, we get the complete equation:

8HJ + H 2 SO 4 \u003d 4J 2 + H 2 S + 4H 2 O

This equation can be written using method of electronicion balance. In this case, there is no need to correct the coefficient in front of water molecules. The equation is compiled on the basis of the dissociation of ions of the compounds participating in the reaction: For example, dissociation of sulfuric acid leads to the formation of two hydrogen protons and a sulfate anion:

H 2 SO 4 ↔ 2H + + SO 4 2-

Similarly, the dissociation of hydrogen iodide and hydrogen sulfide can be written:

HJ ↔ H + + J -
H 2 S ↔ 2H + + S 2-

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

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

SO 4 -2 → S -2

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

SO 4 -2 → S -2 + 4H 2 O

Then, in the left part of the half-reaction, it is necessary to compensate for hydrogen atoms due to protons (since the reaction of the medium is acidic):

SO 4 2- + 8H + → S -2 + 4H 2 O

Having counted the number of passing electrons, we obtain a complete record of the equation in terms of half-reaction method:

SO 4 2- + 8H + + 8 e→ S -2 + 4H 2 O

Summing up 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- + 8Н + → 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 involved in the process is the same for both balance methods, but in the latter case, the number of protons and water molecules involved in the redox process is set “automatically”.

Let us analyze several specific cases of redox reactions that can be compiled by 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). 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\u003d CrO 2- 4 + 2H 2 O

We compose the second half-reaction in the already known way:
CrO 2 - + 4OH - -3 e\u003d CrO 4 2 - + 2H 2 O | 2
Br 0 2 + 2 e= Br - |3
__________
2CrO 2 - + 3Br 2 0 + 8OH - \u003d 2CrO 2- 4 + 6Br - + 4H 2 O

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

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

In a number of cases, non-dissociable substances simultaneously participate in the redox reaction. For example:

AsH 3 + HNO 3 \u003d 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\u003d 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 \u003d H 3 AsO 4 + 8NO 2 + 4H 2 O.

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

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

The redox process involves the atoms of copper, sulfur and nitrogen. When compiling the equation half-reaction method the following steps must be taken into account:

Cu + → Cu 2+
S 2- → S +6
N5+ → N+2

In this situation, it is necessary to combine the oxidation and reduction processes in one stage:

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

At which the redox half-reaction will take 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 \u003d 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, when glucose is oxidized 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 sheet half-reaction method The conversion of glucose takes into account the absence of its dissociation, 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\u003d 6CO 2 + 24H +

Half-reaction involving potassium permanganate will take the form:

MnO 4 - + 8H + + 5 e\u003d 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
MnO 4 - + 8H + + 5 e= Mn +2 + 4H 2 O | 24
___________________________________________________

5C 6 H 12 O 6 + 30H 2 O + 24MnО 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. Influence of the environment on the nature of the course of redox reactions.

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

For example, let's take Na 2 SO 3 as a reducing agent, 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
MnO4 - + 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 restored according to the scheme:

1) Acidic environment (H +) H 2 O 2 + 2H + + 2 e→ 2H2O

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

3) Alkaline medium (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= Fe3+ |2
H 2 O 2 + 2H + + 2 e\u003d 2H 2 O | 1
________________________________
2Fe 2+ + H 2 O 2 + 2H + → 2Fe 3+ + 2 H 2 O

However, when meeting with 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.

In the practical part of this topic, redox processes are considered, indicating only the initial 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+:

Fe3+ + e= Fe 2+

Chloride ion and potassium ion do not change their oxidation state in the reaction, and 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

Fe3+ + e═> Fe2+ |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, you can use the so-called "elevator system". Its principle is that any redox reaction can be represented as the movement of elevators in a multi-storey building in two mutually opposite directions. Moreover, the "floors" will be oxidation states relevant elements. Since any of the two half-reactions in a redox process is accompanied by either a decrease or an increase oxidation states of this or that element, then by simple reasoning one can assume about 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 sulfur states 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 be contrary to 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 case it should move in the opposite direction, i.e. be oxidative:

S 0 - 4 e=S+4

S 0 + 2 e=S-2

On the other hand, how likely is it that the recovery process will be carried out to state S +4 or to S +6? Since the reaction proceeds in an alkaline, and not in an acidic environment, its oxidizing ability is much lower, therefore the formation of the S +4 compound in this reaction is preferable than 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 | one

3S 0 + 6OH - \u003d 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 \u003d (A) + (B) + H 2 O

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

Phosphorus in the oxidation state (-3) will only exhibit 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 the state (-3) will reach its maximum oxidation state (+5).

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

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

P - 3 - 8 e= P+5 | one
N+ 5 + e= N+4 | eight

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

PH 3 + 8HNO 3 \u003d H 3 PO 4 + 8NO 2 + 4H 2 O

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Oxidation state

The oxidation state is the conditional charge of an atom in a molecule, calculated on the assumption that the molecule consists of ions and is generally electrically neutral.

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

The degree of oxidation is a formal concept; in some cases, the oxidation state does not coincide with the valency.

For example:

N2H4 (hydrazine)

nitrogen oxidation state - -2; nitrogen valence - 3.

Calculation of the degree of oxidation

To calculate the oxidation state of an element, the following provisions should be taken into account:

1. The oxidation states of atoms in simple substances are zero (Na 0; H2 0).

2. The algebraic sum of the oxidation states of all atoms that make up the molecule is always zero, and in a complex ion this sum is equal to the charge of the ion.

3. Atoms have a constant oxidation state: alkali metals (+1), alkaline earth metals (+2), hydrogen (+1) (except for hydrides NaH, CaH2, etc., where the oxidation state of hydrogen is -1), oxygen (-2) (except for F 2 -1 O +2 and peroxides containing the –O–O– group, in which the oxidation state of oxygen is -1).

4. For elements, the positive oxidation state cannot exceed a value equal to the group number of the periodic system.

Examples:

V 2 +5 O 5 -2 ;Na 2 +1 B 4 +3 O 7 -2 ;K +1 Cl +7 O 4 -2 ;N -3 H 3 +1 ;K2 +1 H +1 P +5 O 4 -2 ;Na 2 +1 Cr 2 +6 O 7 -2

Reactions without and with a change in the oxidation state

There are two types of chemical reactions:

AReactions in which the oxidation state of the elements does not change:

Addition reactions

SO 2 + Na 2 O → Na 2 SO 3

Decomposition reactions

Cu(OH) 2 → CuO + H 2 O

Exchange reactions

AgNO 3 + KCl → AgCl + KNO 3

NaOH + HNO 3 → NaNO 3 + H 2 O

BReactions in which there is a change in the oxidation states of the atoms of the elements that make up the reacting compounds:

2Mg 0 + O 2 0 → 2Mg +2 O -2

2KCl +5 O 3 -2 →2KCl -1 + 3O 2 0

2KI -1 + Cl 2 0 → 2KCl -1 + I 2 0

Mn +4 O 2 + 4HCl -1 ® Mn +2 Cl 2 + Cl +1 2 0 + 2H 2 O

Such reactions are called redox reactions.

Redox reactions are reactions in which there is a change in the oxidation states of atoms. Redox reactions are very common. All combustion reactions are redox reactions.
The redox reaction consists of two processes that cannot proceed separately from each other. The process of increasing the oxidation state is called oxidation. Simultaneously with oxidation, reduction occurs, that is, the process of lowering the degree of oxidation.

Oxidation, reduction

Accordingly, two main participants are distinguished in redox reactions: an oxidizing agent and a reducing agent. The process of donating electrons is oxidation. When oxidized, the oxidation state rises. The oxidizing agent during the reaction lowers its oxidation state, recovering. Here it is necessary to distinguish between the chemical element-oxidizing agent and the substance-oxidizing agent.

N +5 - oxidizer; HN +5 O 3 and NaN +5 O 3 - oxidizing agents.
If we say that nitric acid and its salts are strong oxidizing agents, then by this we mean that the oxidizing agent is nitrogen atoms with an oxidation state of +5, and not the whole substance as a whole.
The second mandatory participant in the redox reaction is called the reducing agent. The process of adding electrons is reduction. When reduced, the oxidation state decreases.

The reducing agent increases its oxidation state by being oxidized during the course of the reaction. Just as in the case of an oxidizing agent, one should distinguish between a reducing agent and a reducing chemical element. Carrying out the reduction reaction of aldehyde to alcohol, we cannot simply take hydrogen with an oxidation state of -1, but take some hydride, lithium aluminum hydride is best.

H -1 - reducing agent; NaH -1 and LiAlH -1 4 - reducing agents.
In redox reactions, the complete transfer of electrons from the reducing agent to the oxidizing agent is extremely rare, since there are few compounds with ionic bonds. But when arranging the coefficients, we proceed from the assumption that such a transition does take place. This makes it possible to correctly determine the main coefficients in front of the formulas of the oxidizing agent and reducing agent.
5H 2 SO 3 + 2KMnO 4 \u003d 2H 2 SO 4 + 2MnSO 4 + K 2 SO 4 + 3H 2 O
S +4 – 2e → S +6 5 - reducing agent, oxidation
Mn +7 + 5e → Mn +2 2 - oxidizing agent, reduction

The atoms or ions that gain electrons in this reaction are oxidizing agents, and those that donate electrons are reducing agents.

Redox properties of a substance and the degree of oxidation of its constituent atoms

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. The minimum oxidation state for metal atoms is 0, for non-metals - (n–8) (where n is the group number in the periodic system). Compounds containing atoms of elements with an intermediate oxidation state can be both oxidizing and reducing agents, depending on the partner with which they interact and on the reaction conditions.

The most important reducing agents and oxidizing agents

Reclaimers:

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.

Hydrohalic acids and their salts.

Metal cations in lower oxidation states: SnCl 2, FeCl 2, MnSO 4, Cr 2 (SO 4) 3.

Nitrous acid HNO 2 ;

ammonia NH 3 ;

hydrazine NH 2 NH 2 ;

nitric oxide (II) (NO).

cathode in electrolysis.

Oxidizers

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).

Sulfuric acid (H 2 SO 4) conc.

Copper(II) oxide (CuO);

lead(IV) oxide (PbO 2);

silver oxide (Ag 2 O);

hydrogen peroxide (H 2 O 2).

Iron(III) chloride (FeCl 3).

Berthollet's salt (KClO 3).

Anode in electrolysis.

Each such half-reaction is characterized by a standard redox potential E 0 (dimension - volt, V). The larger E 0 , the stronger the oxidizing form as an oxidizing agent and the weaker the reduced form as a reducing agent, and vice versa.

The half-reaction was taken as the reference point of the potentials: 2H + + 2ē ® H 2, for which E 0 =0

For half-reactions M n+ + nē ® M 0 , E 0 is called the standard electrode potential. According to the magnitude of this potential, it is customary to arrange metals in a series of standard electrode potentials (a series of metal voltages):

A number of stresses characterize the chemical properties of metals:

1. The more to the left the metal is located in the series of voltages, the stronger its reducing ability and the weaker the oxidizing ability of its ion in solution (i.e., the easier it gives off electrons (oxidizes) and the more difficult it is for its ions to attach electrons back).

2. Each metal is able to displace from salt solutions those metals that are in the series of voltages to the right of it, i.e. restores ions of subsequent metals into electrically neutral atoms, giving up electrons and turning into ions itself.

3. Only metals standing in the series of voltages to the left of hydrogen (H) are able to displace it from acid solutions (for example, Zn, Fe, Pb, but not Cu, Hg, Ag).

Galvanic cells

Every two metals, being immersed in solutions of their salts, which communicate with each other through a siphon filled with electrolyte, form a galvanic cell. Plates of metals immersed in solutions are called element electrodes.

If you connect the outer ends of the electrodes (the poles of the element) with a wire, then from the metal, which has a lower potential value, electrons begin to move to the metal, which has a larger one (for example, from Zn to Pb). The escape of electrons disturbs the balance that exists between the metal and its ions in solution, and causes a new number of ions to pass into the solution - the metal gradually dissolves. At the same time, electrons passing to another metal discharge the ions in solution near its surface - the metal is released from the solution. The electrode at which oxidation occurs is called the anode. The electrode where the reduction takes place is called the cathode. In a lead-zinc cell, the zinc electrode is the anode and the lead electrode is the cathode.

Thus, in a closed galvanic cell, an interaction takes place between a metal and a salt solution of another metal, which are not in direct contact with each other. Atoms of the first metal, by donating electrons, turn into ions, and ions of the second metal, by gaining electrons, turn into atoms. The first metal displaces the second from its salt solution. For example, during the operation of a galvanic cell composed of zinc and lead, immersed respectively in solutions of Zn (NO 3) 2 and Pb (NO 3) 2, the following processes occur at the electrodes:

Zn – 2ē → Zn 2+

Pb 2+ + 2ē → Pb

Summing up both processes, we obtain the equation Zn + Pb 2+ → Pb + Zn 2+ , which expresses the reaction occurring in the element in ionic form. The molecular equation for the same reaction will be:

Zn + Pb(NO 3) 2 → Pb + Zn(NO 3) 2

The electromotive force of a galvanic cell is equal to the potential difference between its two electrodes. When determining it, the smaller one is always subtracted from the larger potential. For example, the electromotive force (EMF) of the considered element is:

It will have such a value provided that the metals are immersed in solutions in which the concentration of ions is 1 g-ion / l. At other concentrations of solutions, the values ​​of the electrode potentials will be somewhat different. They can be calculated using the formula:

E \u003d E 0 + (0.058 / n) lgC

where E is the desired potential of the metal (in volts)

E 0 - its normal potential

n - valency of metal ions

C - concentration of ions in solution (g-ion / l)

Example

Find the electromotive force of the element (emf) formed by a zinc electrode immersed in a 0.1 M solution of Zn (NO 3) 2 and a lead electrode immersed in a 2 M solution of Pb (NO 3) 2.

Solution

We calculate the potential of the zinc electrode:

E Zn \u003d -0.76 + (0.058 / 2) lg 0.1 \u003d -0.76 + 0.029 (-1) \u003d -0.79 v

We calculate the potential of the lead electrode:

E Pb \u003d -0.13 + (0.058 / 2) lg 2 \u003d -0.13 + 0.029 0.3010 \u003d -0.12 v

Find the electromotive force of the element:

E. d. s. = -0.12 - (-0.79) = 0.67v

Electrolysis

by electrolysis The process of decomposition of a substance by electric current is called.

The essence of electrolysis lies in the fact that when current is passed through an electrolyte solution (or molten electrolyte), positively charged ions move to the cathode, and negatively charged ions move to the anode. Having reached the electrodes, the ions are discharged, as a result of which the components of the dissolved electrolyte or hydrogen and oxygen are released from the water at the electrodes.

To convert different ions into neutral atoms or groups of atoms, different voltages of electric current are required. Some ions lose their charges more easily, others more difficult. The degree of ease with which metal ions discharge (attach electrons) is determined by the position of the metals in the voltage series. The more to the left the metal is in the series of voltages, the greater its negative potential (or less positive potential), the more difficult, other things being equal, its ions are discharged (the ions Аu 3+, Ag + are most easily discharged; the most difficult are Li +, Rb +, K +).

If there are ions of several metals in the solution at the same time, then the ions of the metal with the lower negative potential (or higher positive potential) are discharged first. For example, from a solution containing Zn 2+ and Cu 2+ ions, metallic copper is first released. But the value of the potential of the metal also depends on the concentration of its ions in solution; the ease of discharge of ions of each metal also changes in the same way depending on their concentration: an increase in concentration facilitates the discharge of ions, a decrease makes it difficult. Therefore, during the electrolysis of a solution containing ions of several metals, it may happen that the release of a more active metal will occur earlier than the release of a less active one (if the concentration of the ions of the first metal is significant, and the second is very low).

In aqueous solutions of salts, in addition to salt ions, there are always water ions (H + and OH -). Of these, hydrogen ions will discharge more easily than all metal ions preceding hydrogen in the voltage series. However, due to the negligible concentration of hydrogen ions during the electrolysis of all salts, except for the salts of the most active metals, metal is released at the cathode, and not hydrogen. Only during the electrolysis of salts of sodium, calcium and other metals up to and including aluminum, hydrogen ions are discharged and hydrogen is released.

At the anode, either ions of acid residues or hydroxide ions of water can be discharged. If the ions of acidic residues do not contain oxygen (Cl -, S 2-, CN -, etc.), then it is usually these ions that are discharged, and not hydroxyl ions, which lose their charge much more difficult, and Cl 2, S and t are released at the anode .d. On the contrary, if the salt of an oxygen-containing acid or the acid itself is subjected to electrolysis, then the hydroxyl ions are discharged, and not the ions of oxygen residues. The neutral OH groups formed during the discharge of hydroxyl ions immediately decompose according to the equation:

4OH → 2H 2 O + O 2

As a result, oxygen is released at the anode.

Electrolysis of nickel chloride solution NiCl 2

The solution contains Ni 2+ and Cl - ions, as well as an insignificant concentration of H + and OH - ions. When current is passed, Ni 2+ ions move to the cathode, and Cl - ions move to the anode. Taking two electrons from the cathode, Ni 2+ ions turn into neutral atoms that are released from the solution. The cathode is gradually covered with nickel.

Chlorine ions, reaching the anode, donate electrons to it and turn into chlorine atoms, which, when combined in pairs, form chlorine molecules. Chlorine is released at the anode.

Thus, at the cathode recovery process, at the anode - oxidation process.

Electrolysis of potassium iodide solution KI

Potassium iodide is in solution in the form of K + and I - ions. When current is passed, K + ions move to the cathode, I - ions move to the anode. But since potassium is in the series of voltages much to the left of hydrogen, it is not potassium ions that are discharged at the cathode, but hydrogen ions of water. The resulting hydrogen atoms are combined into H 2 molecules, and thus hydrogen is released at the cathode.

As the hydrogen ions are discharged, more and more water molecules dissociate, as a result of which hydroxide ions (released from the water molecule) accumulate at the cathode, as well as K + ions, which continuously move towards the cathode. A solution of KOH is formed.

At the anode, iodine is released, because I - ions are discharged more easily than water hydroxyl ions.

Electrolysis of potassium sulfate solution

The solution contains ions K + , SO 4 2- and ions H + and OH - from water. Since K + ions are more difficult to discharge than H + ions, and SO 4 2- ions than OH - ions, when an electric current is passed, hydrogen ions will be discharged at the cathode, and hydroxyl groups will be discharged at the anode, that is, in fact, there will be water electrolysis. At the same time, due to the discharge of hydrogen and hydroxide ions of water and the continuous movement of K + ions to the cathode, and SO 4 2- ions to the anode, an alkali solution (KOH) is formed at the cathode, and a sulfuric acid solution is formed at the anode.

Electrolysis of a copper sulfate solution with a copper anode

Electrolysis proceeds in a special way when the anode is made of the same metal whose salt is in solution. In this case, no ions are discharged at the anode, but the anode itself gradually dissolves, sending ions into the solution and giving electrons to the current source.

The whole process is reduced to the release of copper at the cathode and the gradual dissolution of the anode. The amount of CuSO 4 in the solution remains unchanged.

Laws of electrolysis (M. Faraday)

1. The weight of the substance released during electrolysis is proportional to the amount of electricity flowing through the solution and practically does not depend on other factors.

2. Equal amounts of electricity emit during electrolysis from various chemical compounds equivalent amounts of substances.

3. To isolate one gram equivalent of any substance from an electrolyte solution, 96,500 coulombs of electricity must be passed through the solution.

m (x) = ((I t) / F) (M (x) / n)

where m (x) is the amount of reduced or oxidized substance (g);

I - the strength of the transmitted current (a);

t is the electrolysis time (s);

M(x) - molar mass;

n is the number of electrons acquired or given away in redox reactions;

F - Faraday's constant (96500 cou/mol).

Based on this formula, you can make a number of calculations related to the electrolysis process, for example:

1. Calculate the amounts of substances released or decomposed by a certain amount of electricity;

2. Find the current strength by the amount of the released substance and the time spent on its release;

3. Establish how long it will take to release a certain amount of a substance at a given current strength.

Example 1

How many grams of copper will be released on the cathode when a current of 5 amperes is passed through a solution of copper sulfate СuSO 4 for 10 minutes?

Solution

Determine the amount of electricity flowing through the solution:

Q = It,

where I is the current strength in amperes;

t is the time in seconds.

Q=5A 600s=3000coulombs

The equivalent of copper (at. mass 63.54) is 63.54: 2 \u003d 31.77. Therefore, 96500 coulombs emit 31.77 g of copper. Desired amount of copper:

m = (31.77 3000) / 96500 » 0.98 g

Example 2

How long does it take to pass a current of 10 amperes through an acid solution to obtain 5.6 liters of hydrogen (at n.a.)?

Solution

We find the amount of electricity that must pass through the solution in order to release 5.6 liters of hydrogen from it. Since 1 g-eq. hydrogen occupies at n. y. volume is 11.2 liters, then the desired amount of electricity

Q = (96500 5.6) / 11.2 = 48250 coulombs

Let's determine the current passage time:

t = Q / I = 48250 / 10 = 4825 s = 1 h 20 min 25 s

Example 3

When current was passed through a solution of silver salt on the cathode, 10 min. 1 g of silver. Determine the strength of the current.

Solution

1 g-eq. silver is equal to 107.9 g. To isolate 1 g of silver, 96500: 107.9 = 894 coulombs must pass through the solution. Hence the current

I \u003d 894 / (10 60) "1.5A

Example 4

Find the equivalent of tin, if at a current of 2.5 amperes from a solution of SnCl 2 in 30 minutes. 2.77 g of tin is released.

Solution

The amount of electricity passed through the solution in 30 minutes.

Q \u003d 2.5 30 60 \u003d 4500 coulombs

Since to isolate 1 g-eq. required 96,500 pendants, then the equivalent of tin.

E Sn \u003d (2.77 96500) / 4500 \u003d 59.4

Corrosion

Before leaving the discussion of electrochemistry, let's apply what we have learned to the study of one very important problem - corrosion metals. Corrosion is caused by redox reactions, in which a metal, as a result of interaction with any substance from its environment, turns into an undesirable compound.

One of the most widely known corrosion processes is the rusting of iron. From an economic point of view, this is a very important process. According to available estimates, 20% of the iron produced annually in the United States goes to replace iron products that have become unusable due to rusting.

It is known that oxygen is involved in the rusting of iron; iron does not oxidize in water in the absence of oxygen. Water also takes part in the rusting process; iron does not corrode in oxygenated oil unless it contains traces of water. Rusting is accelerated by a number of factors, such as the pH of the medium, the presence of salts in it, the contact of iron with metal, which is more difficult to oxidize than iron, and also under the influence of mechanical stresses.

Corrosion of iron is basically an electrochemical process. Some parts of the surface of iron serve as an anode on which it is oxidized:

Fe (solid) → Fe 2+ (aqueous) + 2e - Eº oxide \u003d 0.44 V

The resulting electrons move through the metal to other parts of the surface, which play the role of the cathode. Oxygen reduction takes place on them:

O 2 (g.) + 4H + (aq.) + 4e - → 2H 2 O (l.) Eº restore \u003d 1.23 V

Note that H + ions are involved in the process of O 2 reduction. If the concentration of H + decreases (ie, with an increase in pH), the recovery of O 2 becomes more difficult. It has been observed that iron in contact with a solution whose pH is above 9-10 does not corrode. During the corrosion process, the Fe 2+ ions formed on the anode are oxidized to Fe 3+ . Fe 3+ ions form hydrated iron oxide (III), which is called rust:

4Fe 2+ (aq.) + O 2 (g.) + 4H 2 O (l.) +2 X H 2 O (l.) → 2Fe 2 O 3 . x H 2 O ( tv.) + 8H + (aq.)

Since the role of the cathode is usually played by that part of the surface that is best provided with an influx of oxygen, rust most often appears in these areas. If you carefully examine a shovel that has stood for some time in open, moist air with dirt adhering to the blade, you will notice that depressions have formed on the surface of the metal under the dirt, and rust has appeared everywhere where O 2 could penetrate.

Motorists often encounter increased corrosion in the presence of salts in areas where salt is abundantly sprinkled on roads in winter to combat icing. The influence of salts is explained by the fact that the ions they form create the electrolyte necessary for the emergence of a closed electrical circuit.

The presence of anodic and cathodic sites on the iron surface leads to the creation of two different chemical environments on it. They can arise due to the presence of impurities or defects in the crystal lattice (apparently due to stresses inside the metal). In places where there are such impurities or defects, the microscopic environment of a particular iron atom can cause some increase or decrease in its oxidation state compared to normal positions in the crystal lattice. Therefore, such places can play the role of anodes or cathodes. Ultra-pure iron, in which the number of such defects is minimized, corrodes much less than ordinary iron.

Iron is often coated with paint or some other metal, such as tin, zinc, or chromium, to protect its surface from corrosion. The so-called "tinplate" is obtained by covering sheet iron with a thin layer of tin. Tin protects iron only as long as the protective layer remains intact. As soon as it is damaged, air and moisture begin to affect the iron; tin even accelerates the corrosion of iron because it serves as a cathode in the electrochemical process of corrosion. Comparison of the oxidation potentials of iron and tin shows that iron is oxidized more easily than tin:

Fe (solid) → Fe 2+ (water) + 2e - Eº oxide \u003d 0.44 V

Sn (tv.) → Sn 2+ (water) + 2e - Eº oxide \u003d 0.14 V

Therefore, iron serves in this case as an anode and is oxidized.

"Galvanized" (galvanized) iron is obtained by coating the iron with a thin layer of zinc. Zinc protects iron from corrosion even after the integrity of the coating is broken. In this case, iron plays the role of a cathode during corrosion, because zinc is oxidized more easily than iron:

Zn (solid) → Zn 2+ (water) + 2e - Eº oxide \u003d 0.76 V

Therefore, zinc plays the role of an anode and corrodes instead of iron. Such protection of the metal, in which it plays the role of a cathode in the process of electrochemical corrosion, is called cathodic protection. Pipes laid underground often protect against corrosion by making them the cathode of an electrochemical cell. To do this, blocks of some active metal, most often magnesium, are buried in the ground along the pipeline, and they are connected with wire to pipes. In wet soil, the active metal acts as an anode, and the iron pipe receives cathodic protection.

While our discussion is focused on iron, it is not the only metal subject to corrosion. At the same time, it may seem strange that an aluminum can, carelessly left in the open air, corrodes immeasurably more slowly than an iron can. Judging by the standard oxidation potentials of aluminum (Eº oxide = 1.66 V) and iron (Eº oxide = 0.44 V), it should be expected that the corrosion of aluminum should occur much faster. The slow corrosion of aluminum is explained by the fact that a thin dense oxide film forms on its surface, which protects the metal underneath from further corrosion. Magnesium, which has a high oxidation potential, is protected from corrosion due to the formation of the same oxide film. Unfortunately, the oxide film on the surface of iron has a too loose structure and is not capable of creating reliable protection. However, a good protective oxide film is formed on the surface of iron-chromium alloys. Such alloys are called stainless steel.

On the basis of changes in the oxidation states of the atoms that make up the reactants, chemical reactions are divided into two types.

1) Reactions that proceed without changing the oxidation states of atoms.

For example:

2+4-2 t +2 -2 +4 -2
CaCO 3 \u003d CaO + CO 2

In this reaction, the oxidation state of each of the atoms remained unchanged.

2) Reactions occurring with a change in the oxidation states of atoms.

For example:

0 +2 -1 0 +2 -1
Zn + CuCl 2 \u003d Cu + ZnCl 2

In this reaction, the oxidation states of the zinc and copper atoms changed.

Redox reactions are the most common chemical reactions.

In practice, a redox reaction is the addition or loss of electrons. Some atoms (ions, molecules) donate or receive electrons from others.

Oxidation.

The process of donating electrons from an atom, ion, or molecule is called oxidation.

When an electron is donated, the oxidation state of an atom increases.

A substance whose atoms, ions, or molecules donate electrons is called reducing agent.

In our example, atoms in the 0 oxidation state have moved into atoms with the +2 oxidation state. That is, an oxidation process has taken place. In this case, the zinc atom, which gave up two electrons, is a reducing agent (it increased the oxidation state from 0 to +2).

The oxidation process is recorded by an electronic equation, which indicates the change in the oxidation state of atoms and the number of electrons donated by the reducing agent.

For example:

0 +2 0
Zn - 2e - = Zn (oxidation, Zn - reducing agent).

Recovery.

The process of adding electrons is called restoration.

When electrons are added, the oxidation state of an atom decreases.

A substance whose atoms, ions, or molecules acquire electrons is called oxidizing agent.

In our example, the transition of copper atoms with an oxidation state of +2 to atoms with an oxidation state of 0 is a reduction process. At the same time, a copper atom with an oxidation state of +2, accepting two electrons, lowers the oxidation state from +2 to 0 and is an oxidizing agent.

The oxidation process is also written by an electronic equation:

2 0 0
Cu + 2e - = Cu (reduction, Cu is an oxidizing agent).

The reduction process and the oxidation process are inseparable and proceed simultaneously.

0 +2 0 +2
Zn + CuCl 2 \u003d Cu + ZnCl 2
reducing agent oxidizing agent
oxidized reduced

Calculation of the degree of oxidation

Summary

1. The formation of personnel is one of the most significant areas of work of the personnel manager.

2. In order to provide the organization with the necessary human resources, it is important to develop an adequate situation in the external environment and the technology of activity, the structure of the company; calculate the need for staff.

3. To develop recruitment programs, it is necessary to analyze the personnel situation in the region, develop procedures for attracting and evaluating candidates, and carry out adaptation measures to include new employees in the organization.

test questions

1. What groups of factors should be taken into account when creating an organizational structure?
2. What stages of organization design can be distinguished?
3. Explain the concept of “qualitative assessment of staffing needs”.
4. Describe the concept of "additional need for staff."
5. What is the purpose of the analysis of the personnel situation in the region?
6. What is the purpose of performance analysis?
7. What are the stages of performance analysis?
8. Explain what a professiogram is?
9. What environmental factors influence the recruitment process?
10. Describe the sources of internal and external recruitment.
11. How to evaluate the quality of a set?
12. What methods are used to evaluate candidates?
13. What competitive recruitment paradigms do you know?
14. Name the stages of adaptation of an employee in an organization.

To calculate the oxidation state of an element, the following provisions should be taken into account:

1. The oxidation states of atoms in simple substances are equal to zero (Na 0; H 2 0).

2. The algebraic sum of the oxidation states of all atoms that make up the molecule is always zero, and in a complex ion this sum is equal to the charge of the ion.

3. Atoms have a constant oxidation state: alkali metals (+1), alkaline earth metals (+2), hydrogen (+1) (except for hydrides NaH, CaH 2, etc., where the oxidation state of hydrogen is -1), oxygen (-2 ) (except for F 2 -1 O +2 and peroxides containing the –O–O– group, in which the oxidation state of oxygen is -1).

4. For elements, the positive oxidation state cannot exceed a value equal to the group number of the periodic system.

Examples:

V 2 +5 O 5 -2; Na 2 +1 B 4 +3 O 7 -2; K +1 Cl +7 O 4 -2; N -3 H 3 +1; K 2 +1 H +1 P +5 O 4 -2; Na 2 +1 Cr 2 +6 O 7 -2

There are two types of chemical reactions:

A Reactions in which the oxidation state of the elements does not change:

Addition reactions

SO 2 + Na 2 O Na 2 SO 3

Decomposition reactions

Cu(OH) 2 - t CuO + H 2 O

Exchange reactions

AgNO 3 + KCl AgCl + KNO 3

NaOH + HNO 3 NaNO 3 + H 2 O

B Reactions in which there is a change in the oxidation states of the atoms of the elements that make up the reacting compounds:

2Mg 0 + O 2 0 2Mg +2 O -2

2KCl +5 O 3 -2 - t 2KCl -1 + 3O 2 0

2KI -1 + Cl 2 0 2KCl -1 + I 2 0

Mn +4 O 2 + 4HCl -1 Mn +2 Cl 2 + Cl 2 0 + 2H 2 O

Such reactions are called redox.