How to determine whether it is giving or receiving electrons. Atoms or ions that gain electrons in this reaction are oxidizing agents, and those that donate electrons are reducing agents.

Chemical reactions that occur with a change in the degree of oxidation of the elements that make up the reactants are called redox.

Oxidation is the process of donating electrons from an atom, molecule, or ion. If an atom gives up its electrons, then it acquires a positive charge, for example:

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

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:

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

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

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

or it can go to a neutral atom:

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, and the reducing agent is oxidized.

It should be remembered that considering oxidation (reduction) as a process of donating (and accepting) electrons by atoms or ions does not always reflect the true situation, since in many cases there is not a complete transfer of electrons, but only a shift of the electron cloud from one atom to another.

However, for compiling the equations of redox reactions, it does not matter what kind of bond is formed in this case - ionic or covalent. Therefore, for simplicity, we will talk about the addition or donation of electrons, regardless of the type of bond.

Drawing up equations of redox reactions and selection of coefficients. When drawing up an equation for a redox reaction, it is necessary to determine the reducing agent, oxidizing agent, and the number of given and received electrons.

As a rule, the coefficients are selected using either the electron balance method or the electron-ion balance method (sometimes the latter is called the half-reaction method).

As an example of compiling equations of redox reactions, consider the process of pyrite oxidation with concentrated nitric acid:

First of all, let's define the products of the reaction. is a strong oxidizing agent, so sulfur will be oxidized to the maximum oxidation state and iron to, while it can be reduced to or. We will choose.

Where it will be located (on the left or right side), we do not yet know.

1. Let us first apply the method of electron-ion balance. This method considers the transition of electrons from one atom or ion to another, taking into account the nature of the medium (acidic, alkaline or neutral) in which the reaction takes place. - When compiling equations for the processes of oxidation and reduction, to equalize the number of hydrogen and oxygen atoms, one introduces (depending on the medium) either water molecules and hydrogen ions (if the medium is acidic), or water molecules and hydroxide ions (if the medium is alkaline). Accordingly, in the products obtained, on the right side of the electron-ionic equation, there will be hydrogen ions and water molecules (acidic medium) or hydroxide ions and water molecules (alkaline medium).

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

Consider for our case the oxidation half-reaction.

The molecule turns into an ion, completely dissociates into ions, we neglect hydrolysis) and two ions (dissociation):

In order to equalize oxygen, add 8 molecules to the left side and 16 ions to the right side (acidic environment!):

The charge on the left side is 0, the charge on the right side must therefore give up 15 electrons:

Consider now the reduction half-reaction of the nitrate ion:

It is necessary to subtract from the O atom. To do this, add 4 ions (acidic medium) to the left side, and 2 molecules to the right side

To equalize the charge to the left side (charge), add 3 electrons:

Finally we have:

Reducing both parts by we get the reduced ionic equation of the redox reaction:

Adding the appropriate number of ions to both sides of the equation, we find the molecular reaction equation:

Note that to determine the number of donated and received electrons, we never had to determine the oxidation state of the elements. In addition, we took into account the influence of the environment and automatically determined what is on the right side of the equation. What is certain is that this method is much more consistent with the chemical sense than the standard electron balance method, although the latter is somewhat easier to understand.

2. Equalize this reaction using the electronic balance method. The recovery process is described simply:

It is more difficult to draw up an oxidation scheme, since 2 elements are oxidized at once - Fe and S. You can attribute the oxidation state to iron to sulfur and take into account that there are two S atoms per 1 Fe atom:

It is possible, however, to dispense with determining the oxidation states and write down a scheme resembling scheme (7.1):

The right side has a charge of +15, the left side has a charge of 0, so it must give up 15 electrons. Write down the total balance:

5 molecules go to oxidation, and 3 more molecules are needed to form

To equalize hydrogen and oxygen, add 2 molecules to the right side:

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.

Consider, for example, the process of ethylene oxidation that occurs when it is passed through an aqueous solution of potassium permanganate. As a result, ethylene is oxidized to ethylene glycol, and permanganate is reduced to manganese (IV) oxide, in addition, as will be obvious from the final balance equation, potassium hydroxide is also formed on the right:

After carrying out the necessary reductions of similar terms, we write the equation in the final molecular form

Quantitative characteristics of redox reactions. A number of standard electrode potentials. The possibility of any redox reaction occurring under real conditions is due to a number of reasons: temperature, the nature of the oxidizing agent and reducing agent, the acidity of the medium, the concentration of substances involved in the reaction, etc.

It can be difficult to take into account all these factors, but, remembering that any redox reaction proceeds with the transfer of electrons from the reducing agent to the oxidizing agent, it is possible to establish a criterion for the possibility of such a reaction.

The quantitative characteristics of redox processes are the normal redox potentials of oxidizing agents and reducing agents (or standard electrode potentials).

To understand the physicochemical meaning of such potentials, it is necessary to analyze the so-called electrochemical processes.

Chemical processes accompanied by the appearance of an electric current or caused by it are called electrochemical.

To understand the nature of electrochemical processes, we turn to the consideration of several fairly simple situations. Imagine a metal plate immersed in water. Under the action of polar water molecules, metal ions are detached from the surface of the plate and hydrated, they pass into the liquid phase. In this case, the latter becomes positively charged, and an excess of electrons appears on the metal plate. The further the process proceeds, the greater the charge of both the plate and the liquid phase becomes.

Due to the electrostatic attraction of solution cations and excess metal electrons, a so-called electric double layer appears at the phase boundary, which inhibits the further transition of metal ions into the liquid phase. Finally, the moment comes when an equilibrium is established between the solution and the metal plate, which can be expressed by the equation:

or taking into account the hydration of ions in solution:

The state of this equilibrium depends on the nature of the metal, the concentration of its ions in solution, on temperature and pressure.

When a metal is immersed not in water, but in a solution of a salt of this metal, the equilibrium shifts to the left in accordance with Le Chatelier's principle and the more, the higher the concentration of metal ions in the solution. Active metals, whose ions have a good ability to go into solution, will in this case be negatively charged, although to a lesser extent than in pure water.

Equilibrium (7.2) can be shifted to the right if electrons are removed from the metal in one way or another. This will dissolve the metal plate. On the contrary, if electrons are brought to a metal plate from the outside, then ions will precipitate from the solution on it.

As already noted, when a metal is immersed in a solution, a double electric layer is formed at the phase boundary. The potential difference that occurs between the metal and the surrounding liquid phase is called the electrode potential. This potential is a characteristic of the redox ability of the metal in the form of a solid phase.

Note that in an isolated metal atom (the state of a monatomic vapor that occurs at high temperatures and high degrees of rarefaction), the redox properties are characterized by a different value, called the ionization potential. The ionization potential is the energy required to detach an electron from an isolated atom.

The absolute value of the electrode potential cannot be measured directly. At the same time, it is not difficult to measure the electrode potential difference that occurs in a system consisting of two metal-solution pairs. Such pairs are called semi-elements. We agreed to determine the electrode potentials of metals with respect to the so-called standard hydrogen electrode, the potential of which is arbitrarily taken as zero. A standard hydrogen electrode consists of a specially prepared platinum plate immersed in an acid solution with a concentration of hydrogen ions and washed by a stream of hydrogen gas under pressure Pa, at a temperature

The emergence of a potential at a standard hydrogen electrode can be imagined as follows. Gaseous hydrogen, being adsorbed by platinum, passes into the atomic state:

Between atomic hydrogen formed on the surface of the plate, hydrogen ions in solution and platinum (electrons!) A state of dynamic equilibrium is realized:

The overall process is expressed by the equation:

Platinum does not take part in the redox process, but is only a carrier of atomic hydrogen.

If a metal plate immersed in a solution of its salt with a concentration of metal ions equal to 1 mol / l is connected to a standard hydrogen electrode, then a galvanic cell will be obtained. The electromotive force of this element (EMF), measured at and characterizes the standard electrode potential of the metal, usually denoted as.

Table 7.1 shows the values of the standard electrode potentials of some metals. The standard potentials of electrodes that act as reducing agents with respect to hydrogen have a “-” sign, and the “+” sign marks the standard potentials of electrodes that are oxidizing agents.

Metals, arranged in ascending order of their standard electrode potentials, form the so-called electrochemical series of metal voltages:

A number of stresses characterize the chemical properties of metals:

1. The more negative the electrode potential of the metal, the greater its reducing ability.

2. Each metal is able to displace (restore) from salt solutions those metals that are in the electrochemical series of metal voltages after it.

3. All metals having a negative standard electrode potential, i.e. located in the electrochemical series of voltages of metals to the left of hydrogen, are able to displace it from acid solutions.

It should be noted that the presented series characterizes the behavior of metals and their salts only in aqueous solutions and at room temperature. In addition, it should be borne in mind that the standard electrode potentials indicated in the table take into account the features of the interaction of one or another ion with solvent molecules. This may violate some of the expected patterns in the arrangement of metals in the electrochemical series of metal voltages. For example, the electrochemical series of voltages of metals begins with lithium, while rubidium and potassium, which are more chemically active, are located to the right of lithium. This is due to the exceptionally high energy of the lithium ion hydration process compared to other alkali metal ions.

At the same time, Table 7.1 shows the standard redox potentials that were measured for non-metallic systems of the type (7.3) that are in equilibrium with respect to a normal hydrogen electrode.

The table shows the reduction half-reactions of the following general form:

As in the case of determining the value of metals, the values of non-metals are measured at a temperature of 25 ° C and at a concentration of all atomic and molecular species participating in the equilibrium equal to 1 mol/l.

Table 7.1. Standard redox potentials at 25 °C (298 K)

(see scan)

The algebraic value of the standard redox potential characterizes the oxidative activity of the corresponding oxidized form. Therefore, a comparison of the values of standard redox potentials allows us to answer the question: does this or that redox reaction proceed?

So, all half-reactions of oxidation of halide ions to free halogens

can be realized under standard conditions using lead (IV) oxide or potassium permanganate as an oxidizing agent. When using potassium dichromate, only reactions (7.5) and (7.6) can be carried out. Finally, the use of nitric acid as an oxidizing agent allows only the half-reaction with the participation of iodide ions (7.6).

Thus, a quantitative criterion for assessing the possibility of a particular redox reaction is the positive value of the difference between the standard redox potentials of the oxidation and reduction half-reactions.

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

During chemical reactions, the number and nature of bonds between interacting atoms can change, i.e. the oxidation states of atoms in molecules can change.

Reactions that change the oxidation state of atoms are called redox reactions.

Examples of redox reactions (abbreviated OVR):

The change in oxidation state is associated with the displacement or transfer of electrons. Regardless of whether electrons pass from one atom to another or are only partially pulled away by one of the atoms, we conventionally speak of giving and receiving electrons.

Processreturns electrons an atom or an ion is calledoxidation . Processaccession electrons is calledrestoration .

Substances whose atoms or ions donate electrons are called reducing agents . During the reaction, they are oxidized. Substances whose atoms or ions accept electrons are called oxidizers . During the reaction, they are restored.

The processes of oxidation and reduction are represented by electronic equations, which indicate the change in the oxidation state of the interacting atoms and the number of electrons donated by the reducing agent or accepted by the oxidizing agent.

Examples of equations expressing oxidation processes:

Equations expressing recovery processes:

A redox reaction is a single process in which oxidation and reduction occur simultaneously. The oxidation of one atom is always accompanied by the reduction of another and vice versa. Wherein general the number of electrons given up by the reducing agent is equal to the number of electrons attached by the oxidizing agent.

According to the law of equivalents the masses of the reactants are related to each other as the molar masses of their equivalents. The equivalent amount of a substance in an OVR depends on the number of electrons donated or attached to its atoms; the molar mass of the equivalent is calculated by the formula:

, (1)

where M is the molar mass of the substance, g/mol

M eq is the molar mass of the substance equivalent, g/mol

-number of donated or attached electrons

For example, in the reaction

a manganese atom attaches 5 electrons, so an equivalent amount
is 1/5 moth, and a sulfur atom donates 2 electrons and an equivalent amount
is 1/2 mole. The molar masses of the equivalents are respectively

Types of redox reactions

There are three types of chemical redox reactions: intermolecular, intramolecular, and self-oxidation-self-healing reactions. A separate group consists of electrochemical reactions.

1. Intermolecular OVR are reactions in which the oxidizing agent and reducing agent are different substances:

2. Intramolecular OVR are reactions in which the oxidation states of different atoms of one molecule change:

3. Self-oxidation-self-healing reactions are reactions in which oxidation and reduction of atoms of the same element occur:

4. Electrochemical reactions are OVR, in which the processes of oxidation and reduction are spatially separated (occur on separate electrodes), and electrons are transferred from the reducing agent to the oxidizing agent through an external electrical circuit:

Redox reactions are usually complex, but knowing the formulas of the reactants and reaction products and being able to determine the oxidation states of atoms, one can easily place the coefficients in the equation of any OVR.

An oxidizing agent and a reducing agent are used to formulate a reaction in organic and inorganic chemistry. Consider the main characteristics of such interactions, identify the algorithm for compiling the equation and arranging the coefficients.

Definitions

An oxidizing agent is an atom or ion that, when interacting with other elements, accepts electrons. The process of accepting electrons is called reduction, and it is associated with a decrease in the oxidation state.

In the course of inorganic chemistry, two main methods for arranging coefficients are considered. The reducing agent and oxidizing agent in the reactions are determined by compiling the electronic balance or by the half-reaction method. Let us dwell in more detail on the first method of arranging the coefficients in the OVR.

Oxidation states

Before determining the oxidizing agent in the reaction, it is necessary to arrange the oxidation states of all elements in the substances involved in the transformation. It represents the charge of an atom of an element, calculated according to certain rules. In complex substances, the sum of all positive and negative oxidation states must be equal to zero. For metals of the main subgroups, it corresponds to valency and has a positive value.

For non-metals, which are located at the end in the formula, the degree is determined by subtracting the group number from eight and has a negative value.

For simple substances, it is equal to zero, since there is no process of accepting or giving off electrons.

For complex compounds consisting of several chemical elements, mathematical calculations are used to determine the oxidation states.

So, an oxidizing agent is an atom that, in the process of interaction, lowers its oxidation state, and a reducing agent, on the contrary, increases its value.

RIA examples

The main feature of tasks related to the arrangement of coefficients in redox reactions is the determination of missing substances and the compilation of their formulas. An oxidizing agent is an element that will accept electrons, but in addition to it, a reducing agent must also participate in the reaction, giving them away.

Here is a generalized algorithm that can be used to complete tasks offered to high school graduates at a unified state exam. Let's consider a few specific examples to understand that an oxidizing agent is not only an element in a complex substance, but also a simple substance.

First, you need to arrange for each element the values of the oxidation states, using certain rules.

Next, you need to analyze the elements that did not participate in the formation of substances, and draw up formulas for them. After all the gaps are eliminated, you can proceed to the process of compiling an electronic balance between the oxidizing agent and the reducing agent. The resulting coefficients are put into the equation, if necessary, adding them in front of those substances that are not included in the balance.

For example, using the electronic balance method, it is necessary to complete the proposed equation, place the necessary coefficients in front of the formulas.

H 2 O 2 + H 2 SO 4 + KMnO 4 \u003d MnSO 4 + O 2 + ... + ...

To begin with, for each, we determine the values of the oxidation states, we get

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

In the proposed scheme, they change for oxygen, as well as for manganese in potassium permanganate. Thus, we have found a reducing agent and an oxidizing agent. On the right side, there is no substance in which potassium would be, so instead of omissions, we will make up the formula for its sulfate.

The last step in this task will be the placement of the coefficients.

5H 2 O 2 + 3H 2 SO 4 + 2KMnO 4 \u003d 2Mn SO 4 + 5O 2 + 8H 2 O + K 2 SO 4

Acids, potassium permanganate, hydrogen peroxide can be considered as strong oxidizing agents. All metals exhibit reducing properties, turning into cations with a positive charge in reactions.

Conclusion

The processes concerning the acceptance and return of negative electrons occur not only in inorganic chemistry. Metabolism, which is carried out in living organisms, is a clear variant of the course of redox reactions in organic chemistry. This confirms the significance of the considered processes, their relevance for animate and inanimate nature.

FOUNDATIONS OF THEORETICAL CHEMISTRY

10. Redox reactions

Redox reactions in solutions.

Chemical reactions that occur with a change in the degree of oxidation of the elements that make up the reactants are called redox.

Oxidation

- is the process of donating electrons from an atom, molecule, or ion. If an atom gives up its electrons, then it acquires a positive charge: l - , gives up 1 electron, then it becomes a neutral atom:

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:

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

If an atom gains electrons, then it turns into a negatively charged ion:

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

or it can go to a neutral atom:

oxidizing agent
accepting electrons. restorer is an atom, molecule or ion, donating electrons.

Oxidizing agent

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

Determination of stoichiometric coefficients in the equations of redox reactions. When drawing up an equation for a redox reaction, it is necessary to determine the reducing agent, oxidizing agent, and the number of given and received electrons. As a rule, the coefficients are selected using either the method electronic balance

, either method electron-ion balance (sometimes the latter is called the method half reactions ).

As an example of compiling equations of redox reactions, consider the process of pyrite oxidation with concentrated nitric acid.

First of all, we define the products of the reaction.

HNO3 is a strong oxidizing agent, so sulfur will be oxidized to its maximum oxidation state S 6+, and iron - to Fe 3+, while HNO 3 can recover up to N0 or NO 2 . We will choose N O:

Where will be located

H2O (on the left or right side), we don't know yet.

1. Apply first electron-ion balance method

(half reactions). This method considers the transition of electrons from one atom or ion to another, taking into account the nature of the medium (acidic, alkaline or neutral) in which the reaction takes place.

When compiling equations for the processes of oxidation and reduction, to equalize the number of hydrogen and oxygen atoms, either water molecules and hydrogen ions are introduced (depending on the medium) (if the environment is acidic), or water molecules and hydroxide ions (if the medium is alkaline). Accordingly, in the products obtained, on the right side of the electron-ionic equation, there will be hydrogen ions and water molecules (acidic medium) or hydroxide ions and water molecules (alkaline medium).

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

Consider for our case the oxidation half-reaction. Molecule

FeS 2 turns into Fe ion 3+ (F e (N O 3) 3 completely dissociates into ions, hydrolysis is neglected) and two ions SO 4 2 - (dissociation of H 2 SO 4):

In order to equalize oxygen, add 8 H molecules to the left side

2 Oh, and to the right - 16 H ions+ (acid medium):

The charge on the left side is 0, the charge on the right side is +15, so

FeS 2 must donate 15 electrons:

Consider now the reduction half-reaction of the nitrate ion:

Must be taken away from

N O 3 2 O atoms. To do this, add 4 H ions to the left side 1+ (acidic environment), and to the right - 2 H molecules 2 O:

To equalize the charge to the left side (charge

+3) add 3 electrons:

Finally we have:

Reducing both parts by 16N

+ and 8Н 2 Oh, we get the reduced ionic equation of the redox reaction:

By adding the appropriate number of ions to both sides of the equation

NO 3 - and H+ we find the molecular reaction equation:

Please note that you never had to determine the oxidation state of the elements to determine the number of given and received electrons. In addition, we took into account the influence of the environment and automatically determined that H

2 O is on the right side of the equation. There is no doubt that this method much more consistent with the chemical sense than the standard electron balance method, although the latter is somewhat easier to understand.

2. We equalize this reaction by the method electronic balance . The recovery process is described:

It is more difficult to draw up an oxidation scheme, since two elements are oxidized at once -

Fe and S. It is possible to attribute to iron the oxidation state 2+, to sulfur 1- and take into account that there are two S atoms per Fe atom:

However, it is possible to do without determining the oxidation states and write down a scheme resembling the scheme

The right side has a charge of +15, the left side has a charge of 0, so

FeS 2 must donate 15 electrons. Write down the total balance:

five HNO molecules

3 going to be oxidized FeS2, and three more molecules HNO3 necessary for education Fe (N O 3) 3:

To equalize hydrogen and oxygen, add two H molecules to the right side

2 O:

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, involving organic compounds, in which even the procedure for determining the oxidation states is very complicated.

Consider, for example, the process of ethylene oxidation that occurs when it is passed through an aqueous solution of potassium permanganate. As a result, ethylene is oxidized to ethylene glycol HO-

CH 2 - CH 2 -OH, and permanganate is reduced to manganese (IV) oxide, in addition, as will be obvious from the final balance equation, potassium hydroxide is also formed on the right:

After carrying out the necessary reductions of similar terms, we write the equation in the final molecular form

Standard potentials of redox reactions.
The possibility of any redox reaction occurring under real conditions is due to a number of reasons: temperature, the nature of the oxidizing agent and reducing agent, the acidity of the medium, the concentration of substances involved in the reaction, etc. It can be difficult to take into account all these factors, but, remembering that any redox reaction proceeds with the transfer of electrons from the reducing agent to the oxidizing agent, it is possible to establish a criterion for the possibility of such a reaction.

The quantitative characteristics of redox processes are normal redox potentials of oxidizing and reducing agents (or standard potentials electrodes).

To understand the physicochemical meaning of such potentials, it is necessary to analyze the so-called electrochemical processes.

Chemical processes accompanied by the appearance of an electric current or caused by it are called electrochemical.

, both plates and liquid phase.

Due to the electrostatic attraction of solution cations and excess metal electrons, a so-called electric double layer appears at the phase boundary, which inhibits the further transition of metal ions into the liquid phase. Finally, there comes a moment when an equilibrium is established between the solution and the metal plate, which can be expressed by the equation:

or taking into account the hydration of ions in solution:

The state of this equilibrium depends on the nature of the metal, the concentration of its ions in solution, on temperature and

pressure.

The equilibrium can be shifted to the right if electrons are removed from the metal in one way or another. This will dissolve the metal plate. On the contrary, if electrons are brought to a metal plate from the outside, then ions will be deposited on it

from solution.

When a metal is immersed in a solution, a double electric layer is formed at the phase boundary. The potential difference that occurs between the metal and the surrounding liquid phase is called the electrode potential. This potential is a characteristic of the redox ability of the metal in the form of a solid phase.

In an isolated metal atom (the state of a monatomic vapor that occurs at high temperatures and high degrees of rarefaction), the redox properties are characterized by a different quantity called the ionization potential. The ionization potential is the energy required to detach an electron from an isolated atom.

The absolute value of the electrode potential cannot be measured directly. At the same time, it is not difficult to measure the electrode potential difference that occurs in a system consisting of two metal-solution pairs. Such couples are called half elements . We agreed to determine the electrode potentials of metals with respect to the so-called standard hydrogen electrode, the potential of which is arbitrarily taken as zero. A standard hydrogen electrode consists of a specially prepared platinum plate immersed in an acid solution with a hydrogen ion concentration of 1 mol/l and washed by a hydrogen gas jet at a pressure of 10

5 Pa, at 25 °C. A number of standard electrode potentials.
If a metal plate immersed in a solution of its salt with a concentration of metal ions equal to 1 mol / l is connected to a standard hydrogen electrode, then a galvanic cell will be obtained. The electromotive force of this element (EMF), measured at 25 ° C, characterizes standard electrode potential of a metal, commonly referred to as E°.

The standard potentials of electrodes that act as reducing agents with respect to hydrogen have the “-” sign, and the “+” sign has the standard potentials of electrodes that are oxidizing agents.

Metals, arranged in ascending order of their standard electrode potentials, form the so-called electrochemical voltage series of metals :Li, Rb, K, wa Sr, Ca, Na, Mg, Al, Mn, Zn, Cr, Fe, Cd, Co, Ni, Sn, Pb, H, Sb, Bi, Cu, Hg, Ag, Pd, Pt, Au.

A number of stresses characterize the chemical properties of metals:

1. The more negative the electrode potential of the metal, the greater its reducing ability.

2. Each metal is able to displace (restore) from salt solutions those metals that are in the electrochemical series of metal voltages after it.

3. All metals that have a negative standard electrode potential, that is, those that are in the electrochemical series of metal voltages to the left of hydrogen, are able to displace it from acid solutions.

As in the case of determining the E° value of metals, the E° values of non-metals are measured at a temperature of 25 ° C and at a concentration of all atomic and molecular particles involved in the equilibrium equal to 1 mol/l.

The algebraic value of the standard redox potential characterizes the oxidative activity of the corresponding oxidized form. So Comparison of the values of standard redox potentials allows answering the question: does this or that redox reaction take place?

A quantitative criterion for assessing the possibility of a particular redox reaction occurring is the positive value of the difference between the standard redox potentials of the half-reactions of oxidation and reduction.

Electrolysis of solutions.

The combination of redox reactions that occur on electrodes in electrolyte solutions or melts when an electric current is passed through them is called electrolysis.

At the cathode of the current source, the process of transferring electrons to cations from a solution or melt occurs, therefore the cathode is the "reductor". At the anode, electrons are donated by anions, therefore the anode is the "oxidizer".

During electrolysis, competing processes can occur both at the anode and at the cathode.

When electrolysis is carried out using an inert (non-consumable) anode (for example, graphite or platinum), as a rule, two oxidative and two reduction processes are competing:

at the anode - oxidation of anions and hydroxide ions,

at the cathode - reduction of cations and hydrogen ions.

When electrolysis is carried out using an active (consumable) anode, the process becomes more complicated and the competing reactions on the electrodes are:

at the anode - oxidation of anions and hydroxide ions, anodic dissolution of the metal - the material of the anode;

at the cathode - the reduction of the salt cation and hydrogen ions, the reduction of metal cations obtained by dissolving the anode.

When choosing the most probable process at the anode and cathode, one should proceed from the position that the reaction that requires the least energy consumption will proceed. In addition, to select the most probable process at the anode and cathode during the electrolysis of salt solutions with an inert electrode, the following rules are used:

The following products can form at the anode: a) during the electrolysis of solutions containing anions F - , SO 4 2- , N About 3 - , RO 4 3 - , as well as alkali solutions, oxygen is released; b) during the oxidation of anions C l - , V r - , I-chlorine, bromine, iodine are released, respectively;c) during the oxidation of anions of organic acids, the process occurs:

2. In the electrolysis of salt solutions containing ions located in a series of voltages to the left of Al

3+ , hydrogen is released at the cathode; if the ion is located in the voltage series to the right of hydrogen, then metal is released at the cathode.

3. During the electrolysis of salt solutions containing ions located in a series of voltages between

Al + and H + , competing processes of both cation reduction and hydrogen evolution can occur at the cathode.

Let us consider as an example the electrolysis of an aqueous solution of copper chloride on inert electrodes. Cu ions are present in solution.

2+ and 2Cl - , which, under the influence of electric current, are directed to the corresponding electrodes:

Metallic copper is released at the cathode, and chlorine gas is released at the anode.

If in the considered example of solution electrolysis

CuCl 2 take a copper plate as an anode, then copper is released at the cathode, and at the anode, where oxidation processes occur, instead of discharging C ions l - and the release of chlorine proceeds the oxidation of the anode (copper). In this case, the dissolution of the anode itself occurs, and in the form of Cu ions itgoes into solution. Electrolysis CuCl 2 with a soluble anode can be written as follows:

The electrolysis of salt solutions with a soluble anode is reduced to the oxidation of the anode material (its dissolution) and is accompanied by the transfer of metal from the anode to the cathode. This property is widely used in the refining (purification) of metals from contamination.

Electrolysis of melts. To obtain highly active metals (sodium, aluminum, magnesium, calcium, etc.), which easily interact with water, electrolysis of molten salts or oxides is used:

If an electric current is passed through an aqueous solution of an active metal salt and an oxygen-containing acid, then neither the metal cations nor the ions of the acid residue are discharged. Hydrogen is released at the cathode

and on anode - oxygen, and electrolysis is reduced to the electrolytic decomposition of water.

Electrolysis of electrolyte solutions is energetically more profitable than melts, since electrolytes - salts and alkalis - melt at very high temperatures.

Faraday's law of electrolysis.

The dependence of the amount of a substance formed under the action of an electric current on time, current strength and the nature of the electrolyte can be established on the basis of a generalized Faraday's law :

where t - the mass of the substance formed during electrolysis (g); E - equivalent mass of a substance (g / mol); M is the molar mass of the substance (g/mol); P- the number of given or received electrons;

I - current strength (A); t- process duration(with); F - Faraday's constant,characterizing the amount of electricity required to release 1 equivalent mass of a substance(F= 96,500 C/mol = 26.8 Ah/mol).

Portal for the student. Self-training

Types of redox reactions

Definitions

Oxidation states

RIA examples

Conclusion

FOUNDATIONS OF THEORETICAL CHEMISTRY

10. Redox reactions

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