The reaction of a compound with a change in the oxidation state. Reactions without and with a change in the oxidation state. Self-help assignments

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

I. Reactions proceeding without a change in degree oxidation of elementsthat are part of the reactants. Such reactions are referred to as ion exchange reactions.

Na 2 CO 3 + H 2 SO 4 \u003d Na 2 SO 4 + CO 2 + H 2 O.

II. Reactions involving a change in the oxidation state of elements

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

5NaNO 2 + 2KMnO 4 + 3H 2 SO 4 \u003d 5NaNO 3 + 2MnSO 4 + K 2 SO 4 + 3H 2 O.

Oxidation state(oxidation) - a characteristic of the state of atoms of elements in a 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 were shifted towards a more electronegative element. Depending on the relative electronegativity of the elements forming the bond, the electron pair can be displaced to one of the atoms or symmetrically located relative to the nuclei of the atoms. Therefore, the oxidation state of the elements can be negative, positive or zero.

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

The oxidation state is indicated by +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 the element in simple substances is zero.

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

Has an oxidation state of +1 (excluding metal hydrides).

O has an oxidation state of –2 (excluding fluorides).

3. Elements of I, II and III groups 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 with a variable oxidation state, there is a concept of the highest and the lowest oxidation state.

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

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

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

Elements N, Cl: lowest oxidation state -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 the elements in a molecule is zero.

KNO 3; (+1) + X + 3 (-2) \u003d 0; X \u003d +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) \u003d -2; X \u003d +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 the elements changes are referred to as redox reactions / ОВР /. These reactions consist of oxidation and reduction processes.

By oxidationis called the process of giving up electrons by an element that is part of an atom, molecule or ion.

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 the 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 - are reducing agents. The reducing agent is oxidized.

Restorationthe process of attaching electrons by an element that is part of an atom, molecule or ion is called.

Cl 2 + 2e \u003d 2Cl -

Fe 3+ + e \u003d Fe 2+

Upon reduction, the oxidation state of the 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 - oxidizing agents. The oxidant is recovered.

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

To compose the equation of the redox reaction (ORR) in solutions, ionic electronic method (half-reaction method).

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

in an acidic medium - HOH and H + ions;

in a neutral environment - only NON;

in an alkaline medium - HOH and OH - ions.

When drawing up the OVR equations, you must adhere to a certain sequence:

1. Write a reaction scheme.

2. Identify the elements that have changed the oxidation state.

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

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

Note:

If the starting material contains fewer oxygen atoms than the products (Р РО 4 3-), then the lack of oxygen is supplied by the medium.

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

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

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

7. Sum up 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 oxygen test.

There are three types of redox reactions:

a) Intermolecular - reactions in which the oxidation state changes for the elements that make up various 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 in the elements that make up one molecule.

Redox processes. Compilation of redox reactions (ORR). A method for accounting for changes in the oxidation states of elements. OVR types. Ionic-electronic method of drawing up the OVR. The concept of the standard electrode potential. The use of standard redox potentials to clarify the fundamental possibility of the redox process.

Topic 4.2.1. Oxidation state

The oxidation state is a positive or negative number assigned to each atom in a compound and equal to the charge of the atom, provided that all chemical bonds in the compound are ionic. Since compounds with a purely ionic character chemical bond do not exist, the actual charges on the atoms never coincide with the oxidation states. Nevertheless, the use of oxidation states allows solving a number of chemical problems.

The oxidation state of an element in compounds is determined by the number of valence electrons involved in the formation of a chemical bond of a given element. But usually, to determine the oxidation states of elements, they do not paint the electronic configuration of valence electrons, but use a number of empirical rules:

1. The sum of the oxidation states of atoms in a particle is equal to its electric charge.

2. In simple substances (consisting of atoms of only one element), the oxidation state of the element is zero.

3. In binary compounds (consisting of atoms of two elements), the negative oxidation state is assigned to the atom with greater electronegativity. Usually the formulas of chemical compounds are written in such a way that the more electronegative atom is the second in the formula, although some formulas can be written differently:

Or (common notation), or.

4. In complex compounds, certain atoms are assigned constant oxidation states:

- fluorine always has an oxidation state of -1;

- metal elements usually have a positive oxidation state;

- hydrogen usually has an oxidation state of +1 (,), but in compounds with metals (hydrides) its oxidation state is -1:,;

- oxygen is characterized by an oxidation state of -2, but with a more electronegative fluorine -, and in peroxide compounds -,,, (sodium superoxide);

- the maximum positive oxidation state of an element usually coincides with the number of the group in which the element is located (Table 1).

Exceptions:

1) the maximum oxidation state is less than the group number: F, O, He, Ne, Ar, cobalt subgroup: Co (+ 2, + 3); Rh, Ir (+ 3, + 4, + 6), nickel subgroup: Ni (+2, rarely +4); Pd, Pt (+ 2, + 4, rarely +6);

2) the maximum oxidation state is higher than the group number: elements of the copper subgroup: Cu (+1, +2), Au (+1, +3).

- the lowest negative oxidation state of non-metallic elements is defined as the group number minus 8 (Table 4.1).

Table 4.1. Oxidation states of some elements

Element	Group number	Maximum positive oxidation state	The lowest negative oxidation state
Na
Al
N			5 – 8 = -3
S			6 – 8 = -2
Cl			7 – 8 = -1

Difficulties often arise in determining the oxidation states in complex compounds - salts, the formula of which contains several atoms, for which different oxidation states are possible. In this case, one cannot do without knowledge of the genetic relationship between the main classes not organic compounds, namely, knowledge of the formulas of acids, the derivatives of which are certain salts.

For example: determine the oxidation state of the elements in the compound Cr 2 (SO 4 ) 3 ... The student's reasoning in this case can be built in this way: Cr 2 (SO 4 ) 3 - This is a medium salt of sulfuric acid, in which the oxidation states of the elements are quite simple to arrange. IN Cr 2 (SO 4 ) 3 sulfur and oxygen have the same oxidation states, with the sulfate ion having a 2-: charge. Taking for easy to determine the oxidation state of chromium:. That is, this salt is chromium (III) sulfate:.

Topic 4.2.2. Redox processes

Redox reactions (ORR) are reactions that change the oxidation state of elements. The change in oxidation states occurs due to the transition of electrons from one particle to another.

The process of loss of electrons by a particle is called oxidation, and the particle itself is oxidized. The process of attaching electrons by a particle is called restoration, while it itself is restored. That is, redox reactions are the unity of two opposite processes.

An oxidizing agent is a reagent that contains an element that lowers its oxidation state during ORR due to the addition of electrons. A reductant is a reagent that contains an element that increases its oxidation state by losing electrons.

For example:

reducing agent:

oxidizer:

reducing agent:

oxidizer:

Many redox reactions are accompanied by a color change in the solution.

For example:


purple	green	brown	colorless

Many redox reactions are widely used in practice.

BASIC TYPES

OXIDATION-REDUCTION REACTIONS

1) Intermolecular (outer-sphere electron transfer reactions) are reactions in which electronic transfer is carried out between different reagents, that is, the oxidizing agent and the reducing agent are part of different substances.

Ok-l voss-l

2) Intramolecular (reactions of inner-sphere electron transfer) - in these reactions, atoms different elements the same substance is an oxidizing agent and a reducing agent.

3) Self-oxidation reactions - self-recovery (disproportionation) - in these reactions, the oxidation state of the same element both increases and decreases.

Topic 4.2.3. Typical oxidants

1) Potassium tetraoxomanganate (VII) -

The oxidizing properties of the ion depend on the nature of the medium:

Acidic environment:

Neutral environment:

Alkaline medium:

2) Potassium dichromate -

Oxidizing properties also depend on the nature of the environment:

Acidic environment:

Neutral environment:

Alkaline medium:

3) Halogens.

4) Hydrogen in dilute acids.

5) Concentrated sulfuric acid

Sulfur reduction products depend on the nature of the reducing agent:

Low-active metal:

Medium activity metal:

Active metal:

6) Nitric acid

In nitric acid of any concentration, it is not protons that act as an oxidizing agent, but nitrogen, which has an oxidation state of +5. Therefore, hydrogen is never released in these reactions. Because nitrogen has a wide variety of oxidation states, it also has a wide range of reduction products. The reduction products of nitric acid depend on its concentration and the activity of the reducing agent.

When concentrated nitric acid interacts with metals, nitric oxide (IV) is usually released, and with non-metals - nitrogen oxide (II):

Interaction with metal:

Interaction with non-metal:

When diluted nitric acid interacts with metals, the products depend on the activity of the metal:

Low-active metal:

Active metal:

- active metal and very dilute acid:

7) Also used as oxidants PbO 2 , MnO 2 .

Topic 4.2.4. Typical reducing agents

one). Halide ions.

In a row, the reducing properties increase:

2). and its salts:

3). Ammonia and ammonium cation salts:

4). Derivatives:

IN aqueous solutions complexes easily turn into complexes:

5). All metals are capable, albeit to varying degrees, of exhibiting reducing properties.

6). Industry uses hydrogen, carbon (in the form of coal or coke) and CO .

Topic 4.2.5. Compounds capable of exhibiting both oxidizing and reducing properties

Some elements in an intermediate oxidation state have redox duality, i.e. with oxidants they are able to manifest themselves as reducing agents, and with reducing agents they behave as oxidizing agents.

NaNO 3; Na 2 SO 4; S; NH 2 OH; H 2 O 2 ... For example:

H 2 O 2 - reducing agent:

H 2 O 2 - oxidizer:

For example , H 2 O 2 can undergo disproportionation reactions:

Topic 4.2.3. Compilation of redox reactions

Two methods are used to compile an RIA:

1) electronic balance method:

This method is based on the use of oxidation states.

The oxidation state of manganese is reduced by 5 units,

in this case, the oxidation state of chlorine increases by 1 unit, but taking into account the resulting reaction product - a simple substance containing 2 moles of chlorine atoms - by 2 units.

Let us write this reasoning in the form of a balance and find the main coefficients using the concept of a common multiple for numbers showing increased no and lowering of oxidation states:

Let's arrange the obtained coefficients into the equation. Let us take into account that it is not only an oxidizing agent, but also binds the reaction products - manganese and potassium ions (the oxidation state in this case does not change), that is, the coefficient before will be greater than it follows from the balance.

The rest of the coefficients are found when calculating the balance of atoms, then from the balance of atoms we find the final coefficient before and from the balance of atoms we find the number of moles of water.

To check the correctness of the selected coefficients, we calculate the balance of moles of oxygen atoms. The final equation shows that out of 16 moles of acid taken for the reaction, 10 moles are spent on reduction, and 6 moles - on binding of the manganese (II) and potassium ions formed as a result of the reaction.

2) ion-electronic method (half-reaction method):

The oxidizing agent is contained in the ion.

In the partial equation of the reduction reaction for the balance of atoms, hydrogen cations must be added to the left side to bind oxygen atoms into water,

and for the balance of charges, add 5 moles of electrons to the same left side of the equation. We get:

The reducing agent is the ion, which contains.

In the particular equation of the oxidation reaction to balance the atoms, hydrogen cations must be added to the right side to bind the extra oxygen atoms to water, and for the balance of charges, add 2 moles of electrons to the same right side of the equation. We get:

Thus, we have two half-reactions:

For equalization, multiply the first half-reaction by 2, and the second by 5. Add the two half-reactions.

Complete ionic equation:

Reduce the same terms:

After reduction, the coefficients of the complete ionic equation can be transferred to the molecular equation.

Topic 4.2.4. Understanding the Standard Electrode Potential

The possibility of a redox reaction is judged by the values electrode potentials separate half-reactions.

If a metal plate is immersed in a solution containing ions of this metal, then a potential difference arises at the metal-solution interface, which is commonly called the electrode potential φ. Electrode potentials are determined experimentally. For standard conditions (the concentration of solutions is 1 mol / l, T \u003d 298 K), these potentials are called standard, denote φ 0. Standard electrode potentials are usually measured relative to a standard hydrogen electrode and are listed in look-up tables.

2H + + 2ē \u003d H 2 φ 0 \u003d 0.

The standard electrode potential is related to Gibbs free energy. For reaction under standard conditions:

ΔG \u003d - nFφ 0

F-Faraday constant (F \u003d 96500 C / mol), n is the number of transferred electrons.

The value of the electrode potential depends on the concentration of reagents and temperature. This relationship is expressed by the Nernst equation:

where φ is the value of the electrode potential, which depends on temperature and concentration.

NO 3 - + 2ē + H 2 O \u003d NO 2 - + 2OH -, φ 0 \u003d - 0.01V

Let us take into account that \u003d \u003d 1 mol / L, pH + pOH \u003d 14, pH \u003d -lg, lg \u003d -lg - 14.

The electrode potential depends on the acidity of the pH medium. With acidification of the solution (with a decrease in pH), the oxidative function of NO 3 - will increase.

Topic 4.2.5. Direction of RVR flow

redox reactions

By the value of the standard electrode potential φ about, one can judge the reducing properties of the system: the more negative the value of φ about, the stronger the reducing properties, and the half-reaction proceeds more easily from right to left.

For example, let's compare the systems:

Li + + e ─ \u003d Li, φ 0 \u003d -3.045 B; Restorative

Ba 2+ + 2e ─ \u003d Ba, φ 0 \u003d - 2.91B activity of metals

Mg 2+ + 2e ─ \u003d Mg, φ 0 \u003d -2.363 B; falls as it increases

Zn 2+ + 2e - \u003d Zn, φ о \u003d -0.763 V standard

Fe 2+ + 2e ─ \u003d Fe, φ 0 \u003d -0.44 B; electrode potential φ about

Cd 2+ + 2e ─ \u003d Cd, φ 0 \u003d - 0.403 B;

Pd 2+ + 2e - \u003d Pd, φ о \u003d 0.987 V

Pt 2+ + 2e - \u003d Pt, φ о \u003d 1.188 V

Au 3+ + 3e ─ \u003d Au, φ 0 \u003d 1.50 B.

In the series of the above systems, a decreasing negative value of φ about corresponds to a drop in the restorative capacity of systems. Lithium has the highest reducing ability, that is, lithium is the most active of the metals presented, it loses its electrons most easily and goes into a positive oxidation state. The reducing activity of metals decreases in the series Li - Ba - Mg - Zn - Fe - Cd - Pd - Pt - Au.

According to the magnitude of the electrode potentials, N.N. Beketov placed metals in the so-called electrochemical series of metals, in which the electrode potential of the hydrogen electrode is taken as the reference point

Li Na K Mn Zn Cr Fe Co Ni H Cu Ag Pd Hg Pt Au

Metal activity decreases

1) Metals in the voltage series up to hydrogen (active metals for which φ 0 < 0), взаимодействуют с разбавленными кислотами с вытеснением водорода.

2) Each subsequent metal displaces the previous metals from its salt.

The larger the value of φ about, the stronger the oxidizing properties of the system, and the half-reaction proceeds more easily from left to right.

For example, let's compare the systems:

As can be seen from the values \u200b\u200bof standard electrode potentials F 2 is the strongest oxidizing agent, in the series F 2 - Cl 2 - Br 2 - I 2 the oxidizing properties of simple halogen substances decrease.

Comparing the values \u200b\u200bof standard electrode potentials of various systems, one can judge the direction of the redox reaction as a whole: a system with a more positive value of φ o is an oxidizing agent, and a system with a less positive value of a standard electrode potential is a reducing agent.

For example:

a) to obtain Br 2 by oxidizing Br ions - you can use Cl 2:

Cl 2 + 2e - \u003d 2Cl -, φ о \u003d 1.359 V

Br 2 + 2e - \u003d 2Br -, φ о \u003d 1.065 V

Overall reaction: Cl 2 + 2Br - \u003d Br 2 + 2Cl -

Complete reaction: Cl 2 + 2 КBr \u003d Br 2 + 2 КCl;

b) and to obtain F 2 by oxidizing F ions - you cannot use Cl 2:

F 2 + 2e - \u003d 2F -, φ о \u003d 2.870 V

Cl 2 + 2e - \u003d 2Cl -, φ о \u003d 1.359 V

The overall reaction: F 2 + 2 Cl - \u003d Cl 2 + 2F -, that is, the reaction Cl 2 + 2 KF \u003d cannot proceed.

You can also determine the direction of the flow and more complex redox reactions.

For example, let us answer the question: is it possible to reduce MnO 4 ions with Fe 3+ ions in an acidic medium? That is, does the reaction proceed:

MnO 4 - + H + + Fe 3+ \u003d Mn 2+ + Fe 2+ + H 2 O?

Main coeff.

MnO 4 - + 8H + + 5e - \u003d Mn 2+ + 4H 2 O, φ о 1 \u003d 1.505 V, 1

Since φ о 1\u003e φ о 2, the first half-reaction proceeds in the forward direction, and the second, relative to the first, proceeds in the opposite direction. Then, having equalized the number of electrons transferred in oxidation and reduction reactions, we obtain the following overall reaction:

In this reaction, the coefficients in front of all compounds are doubled in comparison with the coefficients obtained in the ionic equation, since the reaction products produced iron (III) sulfate having the formula Fe 2 (SO 4) 3 and containing 2 moles of Fe (III) atoms.

Practice 4.2. Redox reactions

1. Compilation of redox reactions by a method based on a change in the oxidation state of elements in a compound.

EXAMPLE 1.

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

KMn +7 O 4 - oxidizing agent: in an acidic medium Mn +7 → Mn +2, the oxidation state decreases by 5 units; Na 2 S +4 O 3 - reducing agent: S +4 → S +6, the oxidation state increases by 2 units. To put the coefficients in the reaction equation, we find the multiple for the numbers showing the increase and decrease in oxidation states:

For 2 moles of Mn (VII) atoms, 5 moles of S (IV) atoms are required:

2 Mn +7 + 5 S +4 \u003d 2 Mn +2 + 5 S +6 are the basic coefficients for an oxidizing agent and a reducing agent. Let's add the reaction products, substitute the main coefficients in the reaction equation, then calculate the balance of other elements: K, Na, S and H:

To check the correctness of the selected coefficients, we calculate the balance of moles of oxygen atoms. The sum of the coefficients in the redox reaction equation is 21.

EXAMPLE 2.

Add and equalize the redox reaction:

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

KMn +7 O 4 - oxidizing agent: in a neutral medium Mn +7 → Mn +4, the oxidation state decreases by 3 units; Na 2 S +4 O 3 - reducing agent: S +4 → S +6, the oxidation state increases by 2 units. To put the coefficients in the reaction equation, we find the multiple for the numbers showing the increase and decrease in oxidation states:

For 2 moles of Mn (VII) atoms, 3 moles of S (IV) atoms are required:

2 Mn +7 + 3 S +4 \u003d 2 Mn +4 + 3 S +6 - these are the main coefficients for an oxidizing agent and a reducing agent. Let's add the reaction products, substitute the main coefficients into the reaction equation, then calculate the balance of other elements: K, Na and H:

To check the correctness of the selected coefficients, we calculate the balance of moles of oxygen atoms. The sum of the coefficients in the equation of the redox reaction is 13.

EXAMPLE 3

Add and equalize the redox reaction:

KMnO 4 + Na 2 SO 3 + KOH → K 2 MnO 4 + ...

KMn +7 O 4 - oxidizing agent: in an alkaline medium Mn +7 → Mn +6, the oxidation state decreases by 1 unit; Na 2 S +4 O 3 - reducing agent: S +4 → S +6, the oxidation state increases by 2 units. To put the coefficients in the reaction equation, we find the multiple for the numbers showing the increase and decrease in oxidation states:

For 2 moles of Mn (VII) atoms, 1 mole of S (IV) atoms is required:

2 Mn +7 + S +4 \u003d 2 Mn +6 + S +6 are the main coefficients for an oxidizing agent and a reducing agent. Let's add the reaction products, substitute the main coefficients into the reaction equation, then calculate the balance of other elements: K, Na and H:

To check the correctness of the selected coefficients, we calculate the balance of moles of oxygen atoms.

The sum of the coefficients in the equation of the redox reaction is 9.

EXAMPLE 4

Add and equalize the redox reaction:

K 2 Cr 2 O 7 + Na 2 SO 3 + H 2 SO 4 → Cr 2 (SO 4) 3 + ...

K 2 Cr 2 +6 O 7 - oxidizing agent: 2Cr +6 → 2Cr +3, the oxidation state decreases by 6 units; Na 2 S +4 O 3 - reducing agent: S +4 → S +6, the oxidation state increases by 2 units. To put the coefficients in the reaction equation, we find the multiple for the numbers showing the increase and decrease in oxidation states:

For 2 moles of Cr (VI) atoms, 3 moles of S (IV) atoms are required:

2 Cr +6 + 3 S +4 \u003d 2 Cr +3 + 3 S +6 are the main coefficients for an oxidizing agent and a reducing agent. Let's add the reaction products, substitute the main coefficients in the reaction equation, then calculate the balance of other elements: K, Na, S and H:

To check the correctness of the selected coefficients, we calculate the balance of moles of oxygen atoms. The sum of the coefficients in the redox reaction equation is 17.

EXAMPLE 5

The sum of the coefficients in the equation of the redox reaction

K 2 MnO 4 + FeSO 4 + H 2 SO 4 → MnSO 4 + ...

K 2 Mn +6 O 4 - oxidizing agent: in an acidic medium Mn +6 → Mn +2, the oxidation state decreases by 4 units; Fe +2 SO 4 - reducing agent: Fe +2 → Fe +3, the oxidation state increases by 1 unit. To put the coefficients in the reaction equation, we find the multiple for the numbers showing the increase and decrease in oxidation states:

For 1 mole of Mn (VII) atoms, 4 moles of Fe (II) atoms are required:

Mn +6 + 4 Fe +2 \u003d Mn +2 + 4 Fe +3 are the main coefficients for an oxidizing agent and a reducing agent. Let's add the reaction products, substitute the main coefficients in the reaction equation, then calculate the balance of other elements: K, S and H:

To check the correctness of the selected coefficients, we calculate the balance of moles of oxygen atoms. The sum of the coefficients in the redox reaction equation is 17.

2. Compilation of redox reactions by the electronic balance method

EXAMPLE 6

If an acidic solution of potassium tetraoxomanganate (VII) is used as an oxidizing agent:

then the reducer can be the system:

Fe 3+ + e - \u003d Fe 2+, φ o \u003d 0.771 V

Co 3+ + e - \u003d Co 2+, φ о \u003d 1.808 V

By the value of the standard redox potential φ about, one can judge the redox properties of the system. A system with a more positive value of φ o is an oxidizing agent, and a system with a less positive value of the standard redox potential φ o is a reducing agent. Therefore, for the system MnO 4 - + 8H + + 5e - \u003d Mn 2+ + 4H 2 O, φ о \u003d 1.505 V, the system Fe 3+ + e - \u003d Fe 2+, φ о \u003d 0.771 V.

EXAMPLE 7

Rh 3+ + 3e - \u003d Rh, φ о \u003d 0.8 V

Bi 3+ + 3e - \u003d Bi, φ о \u003d 0.317 V

Ni 2+ + 2e - \u003d Ni, φ о \u003d -0.250 V

2H + + 2e - \u003d H 2, φ о \u003d 0.0 V

which metal can dissolve in hydrochloric acid?

By the value of the standard electrode potential φ about, one can judge the redox properties of the system. A system with a more positive φ o value is an oxidizing agent, and a system with a less positive value of the standard electrode potential is a reducing agent. In hydrochloric acid (HCl), H + cations are an oxidizing agent, accept electrons and are reduced to H 2, for this reaction φ о \u003d 0 V. Therefore, only that metal dissolves in HCl that can be a reducing agent under these conditions, that is, for which φ about< 0, а именно никель:

Ni + 2 HCl \u003d NiCl 2 + H 2

EXAMPLE 8

Based on the values \u200b\u200bof standard electrode potentials of half-reactions:

Zn 2+ + 2e - \u003d Zn, φ о \u003d -0.763 V

Cd 2+ + 2e - \u003d Cd, φ о \u003d -0.403 V

which metal is the most active?

The more active the metal, the greater its reducing properties. The reduction properties of the system can be judged by the value of the standard redox potential φ о: the more negative the value of φ о, the stronger the reducing properties of the system, and the half-reaction proceeds more easily from right to left. Consequently, zinc has the highest reducibility, that is, zinc is the most active of the metals presented.

EXAMPLE 9

If an acidic solution of iron (III) chloride is used as an oxidizing agent:

which system can be a reducer:

I 2 + 2e - \u003d 2I -, φ о \u003d 0.536 V

Br 2 + 2e - \u003d 2Br -, φ о \u003d 1.065 V

Pb 4+ + 2e - \u003d Pb 2+, φ о \u003d 1.694 V?

By the value of the standard redox potential φ about, one can judge the redox properties of the system. A system with a more positive value of φ o is an oxidizing agent, and a system with a less positive value of the standard redox potential is a reducing agent. Therefore, for the system Fe 3+ + e - \u003d Fe 2+, φ о \u003d 0.771 V, the reducing agent can be the system I 2 + 2e - \u003d 2I -, φ о \u003d 0.536 V.

Main coeff.

Fe 3+ + e - \u003d Fe 2+, φ о 1 \u003d 0.771 V 2

I 2 + 2e - \u003d 2I -, φ о 2 \u003d 0.536 V 1

Since φ о 1\u003e

2 Fe 3+ + 2I - \u003d 2 Fe 2+ + I 2

By adding ions of the opposite sign, we get the complete equation:

2 FeCl 3 + 2 KI \u003d 2 FeCl 2 + 2 KCl + I 2

EXAMPLE 10

Is it possible to reduce MnO 4 ions with Fe 3+ ions in an acidic environment?

Let's write the question in the form of a reaction equation:

MnO 4 - + H + + Fe 3+ \u003d Mn 2+ + Fe 2+ + H 2 O.

Let us select suitable half-reactions from the reference table and give their standard electrode potentials:

Main coeff.

MnO 4 - + 8H + + 5e - \u003d Mn 2+ + 4H 2 O, φ о 1 \u003d 1.505 V, 1

Fe 3+ + e - \u003d Fe 2+, φ о 2 \u003d 0.771 V 5

MnO 4 - + 8H + + 5 Fe 3+ \u003d Mn 2+ + 5Fe 2+ + 4H 2 O

That is, it is possible to reduce MnO 4 ions with Fe 3+ ions in an acidic medium. The complete reaction is:

In this reaction, the coefficients in front of all compounds are doubled in comparison with the coefficients obtained in the ionic equation, since iron (III) sulfate, having the formula Fe 2 (SO 4) 3, was obtained in the reaction products.

TASKS FOR INDEPENDENT SOLUTION

1. Determine the oxidation state of the elements in the compounds:

H 3 PO 4 , K 3 PO 4 , N 2 O 5 , NH 3 , Cl 2 , KCl, KClO 3 , Ca(ClO 4 ) 2 , NH 4 Cl, HNO 2 , Li, Li 3 N, Mg 3 N 2 , NF 3 , N 2 , NH 4 NO 3 , H 2 O, H 2 O 2 , KOH, KH, K 2O 2 , BaO, BaO 2 , OF 2 , F 2 , NF 3 , Na 2 S, FeS, FeS 2 , NaHS, Na 2 SO 4 , NaHSO 4 , SO 2 , SOCl 2 , SO 2 Cl 2 , MnO 2 , Mn(OH) 2 , KMnO 4 , K 2 MnO 4 , Cr, Cr(OH) 2 , Cr(OH) 3 , K 2 CrO 4 , K 2 Cr 2 O 7 , (NH 4 ) 2 Cr 2 O 7 , K 3 [ Al(OH) 6 ], Na 2 [ Zn(OH) 4 ], K 2 [ ZnCl 4 ], H 2 SO 3 , FeSO 3 , Fe 2 (SO 3 ) 3 , H 3 PO 4 , Cu 3 PO 4 , Cu 3 (PO 4 ) 2 , Na 2 SiO 3 , MnSiO 3 , PbSO 4 , Al 2 (SO 4 ) 3 , Fe 2 (SO 4 ) 3 , NH 4 Cl, (NH 4 ) 2 SO 4 , Cr 2 (SO 4 ) 3 , CrSO 4 , NiSO 4 , [ Zn(OH 2 ) 6 ] SO 4 , Fe(NO 3 ) 2 , Fe(NO 3 ) 3 , PbCO 3 , Bi 2 (CO 3 ) 3 , Ag 2 S, Hg 2 S, HgS, Fe 2 S 3 , FeS, SnSO 4 .

2. Indicate the oxidizing agent and the reducing agent, draw up schemes for changing the oxidation states, add and arrange the coefficients in the reaction equation:

a. MnO 2 + HCl (conc) →

b. KMnO 4 + H 2 S + H 2 SO 4 →

in. FeCl 3 + SnCl 2 →

d. KMnO 4 + H 2 O 2 + H 2 SO 4 → O 2

d. Br 2 + KOH →

i.e. Zn + HNO 3 → NH 4 NO 3 + ...

g. Cu + HNO 3 → NO 2 + ...

h. K 2 MnO 4 + FeSO 4 + H 2 SO 4 →

and. K 2 Cr 2 O 7 + (NH 4) 2 S + H 2 O → Cr (OH) 3 +… + NH 3 +…

k. H 2 S + Cl 2 →

l. K 2 Cr 2 O 7 + HCl → CrCl 3 + ...

m. FeCl 3 + H 2 S →

n. KMnO 4 + NaNO 2 + H 2 SO 4 →

about. Cl 2 + KOH →

a) Based on the standard values \u200b\u200bof the electrode potentials, arrange the metals in order of enhancement of the reducing properties:

Ba 2+ + 2e ─ \u003d Ba, φ 0 \u003d -2.91 B;

Au 3+ + 3e ─ \u003d Au, φ 0 \u003d 1.50 B;

Fe 2+ + 2e ─ \u003d Fe, φ 0 \u003d -0.44 B.

What happens when an iron plate is immersed in an AuCl 3 solution

b) Based on standard values \u200b\u200bof electrode potentials of half-reactions

MnO 4 - + 8H + + 5e - \u003d Mn 2+ + 4H 2 O, φ о \u003d 1.505 V,

Pb 4+ + 2e - \u003d Pb 2+, φ о \u003d 1.694 V

give a reasonable answer to the question - is it possible to oxidize Mn 2+ ions with Pb 4+ ions? Give the overall reaction, indicate the oxidizing agent and the reducing agent.

c) Based on the standard values \u200b\u200bof the electrode half-reaction potentials, give a reasonable answer to the question - is it possible to oxidize Fe 2+ ions using Pb 4+ ions? Give the overall reaction, indicate the oxidizing agent and the reducing agent.

d) Based on the standard values \u200b\u200bof the electrode potentials, arrange the metals in the order of enhancement of the reducing properties:

Mg 2+ + 2e ─ \u003d Mg

Cd 2+ + 2e ─ \u003d Cd

Сu 2+ + 2e ─ \u003d Cu

What happens when a copper strip is immersed in a cadmium chloride solution?

e) Based on the standard values \u200b\u200bof the electrode half-reaction potentials

Ir 3+ + 3e - \u003d Ir,

NO 3 - + 4H + + 3e - \u003d NO + 2H 2 O,

give a reasonable answer to the question - does iridium dissolve in nitric acid? Give the total reaction, indicate the oxidizing agent and the reducing agent

f) Based on the standard values \u200b\u200bof the electrode potentials, arrange the halogens in the order of increasing their oxidizing properties:

Cl 2 + 2e ─ \u003d 2Cl ─ φ 0 \u003d 1.359 B;

Br 2 + 2e ─ \u003d 2Br ─ φ 0 \u003d 1.065 B;

I 2 + 2e ─ \u003d 2I ─ φ 0 \u003d 0.536 B;

F 2 + 2e ─ \u003d 2F ─ φ 0 \u003d 2.87 B.

Prove whether it is possible to use the oxidation reaction of Br ions with chlorine Cl 2 to obtain bromine?

g) Based on the standard values \u200b\u200bof the electrode half-reaction potentials

Fe 3+ + e - \u003d Fe 2+, φ o \u003d 0.771 V,

Br 2 + 2e - \u003d 2Br -, φ о \u003d 1.065 V

give a reasonable answer to the question - is it possible to oxidize Fe 2+ ions with Br 2? Give the overall reaction, indicate the oxidizing agent and the reducing agent.

h) Based on the standard values \u200b\u200bof the electrode potentials, arrange the metals in the order of enhancement of the reducing properties:

Zn 2+ + 2e - \u003d Zn, φ о \u003d - 0.763 V

Hg 2+ + 2e - \u003d Hg, φ о \u003d 0.850 V

Cd 2+ + 2e - \u003d Cd, φ о \u003d - 0.403 V.

What happens when a cadmium plate is immersed in a zinc chloride solution?

Classification chemical reactions in inorganic and organic chemistry carried out on the basis of various classifying features, information about which is given in the table below.

By changing the oxidation state of elements

The first sign of classification is based on the change in the oxidation state of the elements that form reagents and products.
a) redox
b) without changing the oxidation state
Redox are called reactions accompanied by a change in oxidation states chemical elementsincluded in the reagents. To redox in inorganic chemistry includes all substitution reactions and those decomposition reactions and compounds in which at least one simple substance is involved. All exchange reactions belong to the reactions that proceed without changing the oxidation states of the elements that form the reactants and reaction products.

By the number and composition of reagents and products

Chemical reactions are classified by the nature of the process, that is, by the number and composition of reagents and products.

Compound reactions are called chemical reactions as a result of which complex molecules are obtained from several simpler ones, for example:
4Li + O 2 \u003d 2Li 2 O

Decomposition reactions are called chemical reactions, as a result of which simple molecules are obtained from more complex ones, for example:
CaCO 3 \u003d CaO + CO 2

Decomposition reactions can be viewed as the inverse of compound.

Substitution reactions chemical reactions are called, as a result of which an atom or group of atoms in a molecule of a substance is replaced by another atom or group of atoms, for example:
Fe + 2HCl \u003d FeCl 2 + H 2 \u206d

Their distinguishing feature is the interaction of a simple substance with a complex one. Such reactions also exist in organic chemistry.
However, the concept of "substitution" in organic matter is broader than in inorganic chemistry. If in the molecule of the starting substance any atom or functional group is replaced by another atom or group, these are also substitution reactions, although from the point of view of inorganic chemistry the process looks like an exchange reaction.
- exchange (including neutralization).
Exchange reactions are called chemical reactions that occur without changing the oxidation states of the elements and lead to the exchange of the constituent parts of the reagents, for example:
AgNO 3 + KBr \u003d AgBr + KNO 3

If possible, flow in the opposite direction

If possible, flow in the opposite direction - reversible and irreversible.

Reversible are called chemical reactions occurring at a given temperature in two opposite directions simultaneously with comparable rates. When writing the equations of such reactions, the equal sign is replaced with oppositely directed arrows. The simplest example reversible reaction is the synthesis of ammonia by the interaction of nitrogen and hydrogen:

N 2 + 3H 2 ↔2NH 3

Irreversible are called reactions that proceed only in the forward direction, as a result of which products are formed that do not interact with each other. Irreversible include chemical reactions, as a result of which low-dissociated compounds are formed, release a large number energy, as well as those in which the final products leave the reaction sphere in a gaseous form or in the form of a precipitate, for example:

HCl + NaOH \u003d NaCl + H2O

2Ca + O 2 \u003d 2CaO

BaBr 2 + Na 2 SO 4 \u003d BaSO 4 ↓ + 2NaBr

Thermal effect

Exothermic are called chemical reactions with the release of heat. Conventional designation of the change in enthalpy (heat content) ΔH, and the heat effect of the reaction Q. For exothermic reactions, Q\u003e 0, and ΔH< 0.

Endothermic are called chemical reactions that take place with heat absorption. For endothermic reactions Q< 0, а ΔH > 0.

Compound reactions will generally be exothermic and decomposition reactions will be endothermic. A rare exception is the reaction of nitrogen with oxygen - endothermic:
N2 + О2 → 2NO - Q

By phase

Homogeneous are called reactions that take place in a homogeneous medium (homogeneous substances, in one phase, for example, r-g, reactions in solutions).

Heterogeneous are called reactions that take place in a heterogeneous medium, on the contact surface of reactants in different phases, for example, solid and gaseous, liquid and gaseous, in two immiscible liquids.

By using the catalyst

A catalyst is a substance that accelerates a chemical reaction.

Catalytic reactions proceed only in the presence of a catalyst (including enzymatic).

Non-catalytic reactions go in the absence of a catalyst.

By the type of disconnection

Homolytic and heterolytic reactions are distinguished according to the type of chemical bond breaking in the parent molecule.

Homolytic are called reactions in which, as a result of the breaking of bonds, particles are formed that have an unpaired electron - free radicals.

Heterolytic are called reactions proceeding through the formation of ionic particles - cations and anions.

homolytic (equal gap, each atom receives 1 electron)
heterolytic (unequal break - one gets a pair of electrons)

Radical (chain) are called chemical reactions involving radicals, for example:

CH 4 + Cl 2 hv → CH 3 Cl + HCl

Ionic called chemical reactions involving ions, for example:

KCl + AgNO 3 \u003d KNO 3 + AgCl ↓

Heterolytic reactions of organic compounds with electrophiles are called electrophilic - particles that carry a whole or fractional positive charge. They are classified into electrophilic substitution and electrophilic addition reactions, for example:

C 6 H 6 + Cl 2 FeCl3 → C 6 H 5 Cl + HCl

H 2 C \u003d CH 2 + Br 2 → BrCH 2 –CH 2 Br

Nucleophilic refers to heterolytic reactions of organic compounds with nucleophiles - particles that carry a whole or fractional negative charge. They are classified into nucleophilic substitution and nucleophilic addition reactions, for example:

CH 3 Br + NaOH → CH 3 OH + NaBr

CH 3 C (O) H + C 2 H 5 OH → CH 3 CH (OC 2 H 5) 2 + H 2 O

Classification of organic reactions

Classification organic reactions given in the table:

DEFINITION

Oxidation state is a quantitative assessment of the state of an atom of a chemical element in a compound, based on its electronegativity.

It takes both positive and negative values. To indicate the oxidation state of an element in a compound, you need to put an Arabic numeral above its symbol with the corresponding sign ("+" or "-").

It should be remembered that the oxidation state is a quantity that has no physical meaning, since it does not reflect the real charge of an atom. However, this concept is widely used in chemistry.

Oxidation state table of chemical elements

The maximum positive and minimum negative oxidation states can be determined using the Periodic Table of D.I. Mendeleev. They are equal to the number of the group in which the element is located and the difference between the value of the "highest" oxidation state and the number 8, respectively.

If we consider chemical compounds more specifically, then in substances with non-polar bonds, the oxidation state of the elements is zero (N 2, H 2, Cl 2).

The oxidation state of metals in the elementary state is zero, since the distribution of electron density in them is uniform.

In simple ionic compounds, the oxidation state of their constituent elements is equal to the electric charge, since during the formation of these compounds, an almost complete transition of electrons from one atom to another occurs: Na +1 I -1, Mg +2 Cl -1 2, Al +3 F - 1 3, Zr +4 Br -1 4.

When determining the oxidation state of elements in compounds with polar covalent bonds, the values \u200b\u200bof their electronegativities are compared. Since during the formation of a chemical bond, electrons are displaced to atoms of more electronegative elements, the latter have a negative oxidation state in the compounds.

There are elements for which only one value of the oxidation state is characteristic (fluorine, metals of groups IA and IIA, etc.). Fluorine, which has the highest electronegativity value, in compounds always has a constant negative oxidation state (-1).

Alkaline and alkaline earth elements, which are characterized by a relatively low value of electronegativity, always have a positive oxidation state equal to (+1) and (+2), respectively.

However, there are also such chemical elements, which are characterized by several values \u200b\u200bof the oxidation state (sulfur - (-2), 0, (+2), (+4), (+6), etc.).

In order to make it easier to remember how many and which oxidation states are characteristic for a particular chemical element, tables of oxidation states of chemical elements are used, which look like this:

Serial number	Russian / English name	Chemical symbol	Oxidation state
	Hydrogen / Hydrogen
	Helium / Helium
	Lithium / Lithium
	Beryllium / Beryllium
			(-1), 0, (+1), (+2), (+3)
	Carbon / Carbon		(-4), (-3), (-2), (-1), 0, (+2), (+4)
	Nitrogen / Nitrogen		(-3), (-2), (-1), 0, (+1), (+2), (+3), (+4), (+5)
	Oxygen / Oxygen		(-2), (-1), 0, (+1), (+2)
	Fluorine / Fluorine

	Sodium / Sodium
	Magnesium / Magnesium
	Aluminum / Aluminum
	Silicon / Silicon		(-4), 0, (+2), (+4)
	Phosphorus		(-3), 0, (+3), (+5)
	Sulfur / Sulfur		(-2), 0, (+4), (+6)
	Chlorine / Chlorine		(-1), 0, (+1), (+3), (+5), (+7), rarely (+2) and (+4)
	Argon / Argon
	Potassium
	Calcium / Calcium
	Scandium / Scandium
	Titanium / Titanium		(+2), (+3), (+4)
	Vanadium / Vanadium		(+2), (+3), (+4), (+5)
	Chromium / Chromium		(+2), (+3), (+6)
	Manganese / Manganese		(+2), (+3), (+4), (+6), (+7)
	Iron / Iron		(+2), (+3), rarely (+4) and (+6)
	Cobalt / Cobalt		(+2), (+3), rarely (+4)
	Nickel / Nickel		(+2), rarely (+1), (+3) and (+4)
	Copper / Copper		+1, +2, rarely (+3)

	Gallium / Gallium		(+3), rarely (+2)
	Germanium / Germanium		(-4), (+2), (+4)
	Arsenic / Arsenic		(-3), (+3), (+5), rarely (+2)
	Selenium / Selenium		(-2), (+4), (+6), rarely (+2)
	Bromine / Bromine		(-1), (+1), (+5), rarely (+3), (+4)
	Krypton / Krypton
	Rubidium / Rubidium
	Strontium / Strontium
	Yttrium / Yttrium
	Zirconium / Zirconium		(+4), rarely (+2) and (+3)
	Niobium / Niobium		(+3), (+5), rarely (+2) and (+4)
	Molybdenum / Molybdenum		(+3), (+6), rarely (+2), (+3) and (+5)
	Technetium
	Ruthenium / Ruthenium		(+3), (+4), (+8), rarely (+2), (+6) and (+7)
	Rhodium / Rhodium		(+4), rarely (+2), (+3) and (+6)
	Palladium / Palladium		(+2), (+4), rarely (+6)
	Silver / Silver		(+1), rarely (+2) and (+3)
	Cadmium / Cadmium		(+2), rarely (+1)
	Indium / Indium		(+3), rarely (+1) and (+2)
	Tin / Tin		(+2), (+4)
	Antimony / Antimony		(-3), (+3), (+5), rarely (+4)
	Tellurium / Tellurium		(-2), (+4), (+6), rarely (+2)
			(-1), (+1), (+5), (+7), rarely (+3), (+4)
	Xenon / Xenon
	Cesium / Cesium
	Barium / Barium
	Lanthanum / Lanthanum
	Cerium / Cerium		(+3), (+4)
	Praseodymium / Praseodymium
	Neodymium / Neodymium		(+3), (+4)
	Promethium / Promethium
	Samarium / Samarium		(+3), rarely (+2)
	Europium / Europium		(+3), rarely (+2)
	Gadolinium / Gadolinium
	Terbium / Terbium		(+3), (+4)
	Dysprosium / Dysprosium
	Holmium / Holmium
	Erbium / Erbium
	Thulium / Thulium		(+3), rarely (+2)
	Ytterbium / Ytterbium		(+3), rarely (+2)
	Lutetium / Lutetium
	Hafnium / Hafnium
	Tantalum / Tantalum		(+5), rarely (+3), (+4)
	Tungsten / Tungsten		(+6), rarely (+2), (+3), (+4) and (+5)
	Rhenium / Rhenium		(+2), (+4), (+6), (+7), rarely (-1), (+1), (+3), (+5)
	Osmium / Osmium		(+3), (+4), (+6), (+8), rarely (+2)
	Iridium / Iridium		(+3), (+4), (+6), rarely (+1) and (+2)
	Platinum / Platinum		(+2), (+4), (+6), rarely (+1) and (+3)
	Gold / Gold		(+1), (+3), rarely (+2)
	Mercury / Mercury		(+1), (+2)
	Thallium / Thallium		(+1), (+3), rarely (+2)
	Lead / Lead		(+2), (+4)
	Bismuth		(+3), rarely (+3), (+2), (+4) and (+5)
	Polonium / Polonium		(+2), (+4), rarely (-2) and (+6)
	Astatine / Astatine
	Radon / Radon
	Francium / Francium
	Radium / Radium
	Actinium / Actinium
	Thorium / Thorium
	Proactinium / Protactinium
	Uranium / Uranium		(+3), (+4), (+6), rarely (+2) and (+5)

Examples of problem solving

EXAMPLE 1

Answer We will alternately determine the oxidation state of phosphorus in each of the proposed transformation schemes, and then choose correct option answer.

The oxidation state of phosphorus in phosphine is (-3), and in orthophosphoric acid - (+5). Change in the oxidation state of phosphorus: +3 → +5, i.e. first answer option.
The oxidation state of a chemical element in a simple substance is zero. The oxidation state of phosphorus in the oxide of the composition P 2 O 5 is (+5). Change in the oxidation state of phosphorus: 0 → +5, i.e. third answer option.
The oxidation state of phosphorus in the acid of the composition HPO 3 is (+5), and H 3 PO 2 is (+1). Change in the oxidation state of phosphorus: +5 → +1, i.e. fifth answer option.

EXAMPLE 2

The task

The oxidation state (-3) carbon has in the compound: a) CH 3 Cl; b) C 2 H 2; c) HCOH; d) C 2 H 6.

Decision

In order to give the correct answer to the question posed, we will alternately determine the oxidation state of carbon in each of the proposed compounds.

a) the oxidation state of hydrogen is (+1), and chlorine is (-1). Let's take the oxidation state of carbon as "x":

x + 3 × 1 + (-1) \u003d 0;

The answer is wrong.

b) the oxidation state of hydrogen is (+1). Let's take the oxidation state of carbon for "y":

2 × y + 2 × 1 \u003d 0;

The answer is wrong.

c) the oxidation state of hydrogen is (+1), and oxygen is (-2). Let's take the oxidation state of carbon as "z":

1 + z + (-2) +1 \u003d 0:

The answer is wrong.

d) the oxidation state of hydrogen is (+1). Let's take the oxidation state of carbon for "a":

2 × a + 6 × 1 \u003d 0;

Correct answer.

Answer

Option (d)

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 \u003d Na 2 SO 3

Decomposition reactions

Cu (OH) 2 \u003d  CuO + H 2 O

Exchange reactions

AgNO 3 + KCl \u003d AgCl + KNO 3

NaOH + HNO 3 \u003d 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 and the transfer of electrons from one compound to another:

2Mg 0 + O 2 0 \u003d 2Mg +2 O -2

2KI -1 + Cl 2 0 \u003d 2KCl -1 + I 2 0

Mn +4 O 2 + 4HCl -1 \u003d Mn +2 Cl 2 + Cl 2 0 + 2H 2 O

These reactions are called redox reactions.

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 the compound have negative oxidation states, while the atoms of elements with less electronegativity are positive.

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

for example:

N 2 H 4 (hydrazine)

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

Calculation of the oxidation state

To calculate the oxidation state of an element, consider the following:

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

2. The algebraic sum of the oxidation states of all the 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. The following atoms have a constant oxidation state in compounds with atoms of other elements: alkali metals (+1), alkaline earth metals (+2), fluorine

(-1), hydrogen (+1) (except for metal hydrides Na + H -, Ca 2+ H 2 - and others, where the oxidation state of hydrogen is -1), oxygen (-2) (except 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 of:

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

Oxidation, reduction

In redox reactions, electrons from one atom, molecule or ion are transferred to another. The process of donating electrons is oxidation. With oxidation, the oxidation state increases:

H 2 0 - 2ē \u003d 2H + + 1 / 2О 2

S -2 - 2ē \u003d S 0

Al 0 - 3ē \u003d Al +3

Fe +2 - ē \u003d Fe +3

2Br - - 2ē \u003d Br 2 0

Electron attachment process - reduction: Reduction lowers the oxidation state.

Mn +4 + 2ē \u003d Mn +2

S 0 + 2ē \u003d S -2

Cr +6 + 3ē \u003d Cr +3

Cl 2 0 + 2ē \u003d 2Cl -

O 2 0 + 4ē \u003d 2O -2

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

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

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

Compounds containing atoms of elements with the maximum oxidation state can only be oxidizing agents due to these atoms, because they have already given up all their valence electrons and are only able to accept electrons. The maximum oxidation state of an atom of an element is equal to the number of the group in the periodic table to which this 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 with eight electrons. The minimum oxidation state for metal atoms is 0, for non-metals - (n – 8) (where n is the group number in 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 and oxidizing agents

Reducing agents

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 the lowest oxidation states: SnCl 2, FeCl 2, MnSO 4, Cr 2 (SO4) 3.

Nitrous acid HNO 2;

ammonia NH 3;

hydrazine NH 2 NH 2;

nitric oxide (II) (NO).