A chemical element in a compound, calculated from the assumption that all bonds are ionic.

The oxidation states can have a positive, negative or zero value, therefore the algebraic sum of the oxidation states of elements in a molecule, taking into account the number of their atoms, is 0, and in an ion - the charge of the ion.

1. The oxidation states of metals in compounds are always positive.

2. The highest oxidation state corresponds to the group number periodic system, where this element is located (the exception is: Au+3(I group), Cu+2(II), from group VIII, the oxidation state +8 can only be in osmium Os and ruthenium Ru.

3. The oxidation states of non-metals depend on which atom it is connected to:

• if with a metal atom, then the oxidation state is negative;
• if with a non-metal atom, then the oxidation state can be both positive and negative. It depends on the electronegativity of the atoms of the elements.

4. The highest negative oxidation state of non-metals can be determined by subtracting from 8 the number of the group in which this element is located, i.e. the highest positive oxidation state is equal to the number of electrons on the outer layer, which corresponds to the group number.

5. The oxidation states of simple substances are 0, regardless of whether it is a metal or a non-metal.

Elements with constant oxidation states.

Element	Characteristic oxidation state	Exceptions
		Metal hydrides: LIH-1
		oxidation state called the conditional charge of the particle under the assumption that the bond is completely broken (has an ionic character). H- Cl = H + + Cl - , The bond in hydrochloric acid is covalent polar. The electron pair is more biased towards the atom Cl - , because it is more electronegative whole element. How to determine the degree of oxidation? Electronegativity is the ability of atoms to attract electrons from other elements. The oxidation state is indicated above the element: Br 2 0 , Na 0 , O +2 F 2 -1 ,K + Cl - etc. It can be negative and positive. The oxidation state of a simple substance (unbound, free state) is zero. The oxidation state of oxygen in most compounds is -2 (the exception is peroxides H 2 O 2, where it is -1 and compounds with fluorine - O +2 F 2 -1 , O 2 +1 F 2 -1 ). - Oxidation state a simple monatomic ion is equal to its charge: Na + , Ca +2 . Hydrogen in its compounds has an oxidation state of +1 (exceptions are hydrides - Na + H - and type connections C +4 H 4 -1 ). In metal-non-metal bonds, the atom that has the highest electronegativity has a negative oxidation state (electronegativity data are given on the Pauling scale): H + F - , Cu + Br - , Ca +2 (NO 3 ) - etc. Rules for determining the degree of oxidation in chemical compounds. Let's take a connection KMnO 4 , it is necessary to determine the oxidation state of the manganese atom. Reasoning: Potassium is an alkali metal in group I of the periodic table, and therefore has only a positive oxidation state of +1. Oxygen is known to have an oxidation state of -2 in most of its compounds. This substance is not a peroxide, which means it is no exception. Makes an equation: K+MnXO 4 -2 Let X- unknown to us the degree of oxidation of manganese. The number of potassium atoms is 1, manganese - 1, oxygen - 4. It is proved that the molecule as a whole is electrically neutral, so its total charge must be equal to zero. 1*(+1) + 1*(X) + 4(-2) = 0, X = +7, Hence, the oxidation state of manganese in potassium permanganate = +7. Let's take another example of an oxide Fe2O3. It is necessary to determine the oxidation state of the iron atom. Reasoning: Iron is a metal, oxygen is a non-metal, which means that it is oxygen that will be an oxidizing agent and have a negative charge. We know that oxygen has an oxidation state of -2. We consider the number of atoms: iron - 2 atoms, oxygen - 3. We make an equation where X- the oxidation state of the iron atom: 2*(X) + 3*(-2) = 0, Conclusion: the oxidation state of iron in this oxide is +3. Examples. Determine the oxidation states of all atoms in the molecule. 1. K2Cr2O7. Oxidation state K+1, oxygen O -2. Given indexes: O=(-2)×7=(-14), K=(+1)×2=(+2). Because the algebraic sum of the oxidation states of elements in a molecule, taking into account the number of their atoms, is 0, then the number of positive oxidation states is equal to the number of negative ones. Oxidation states K+O=(-14)+(+2)=(-12). It follows from this that the number of positive powers of the chromium atom is 12, but there are 2 atoms in the molecule, which means that there are (+12):2=(+6) per atom. Answer: K 2 + Cr 2 +6 O 7 -2. 2.(AsO 4) 3-. In this case, the sum of the oxidation states will no longer be equal to zero, but to the charge of the ion, i.e. - 3. Let's make an equation: x+4×(- 2)= - 3 . Answer: (As +5 O 4 -2) 3-.

In chemistry, the terms "oxidation" and "reduction" mean reactions in which an atom or a group of atoms lose or, respectively, gain electrons. The oxidation state is a numerical value attributed to one or more atoms that characterizes the number of redistributed electrons and shows how these electrons are distributed between atoms during the reaction. Determining this quantity can be both a simple and quite complex procedure, depending on the atoms and the molecules consisting of them. Moreover, the atoms of some elements can have several oxidation states. Fortunately, there are simple unambiguous rules for determining the degree of oxidation, for the confident use of which it is enough to know the basics of chemistry and algebra.

Steps

Part 1

Determination of the degree of oxidation according to the laws of chemistry

Determine if the substance in question is elemental. The oxidation state of atoms outside a chemical compound is zero. This rule is true both for substances formed from individual free atoms, and for those that consist of two or polyatomic molecules of one element.

• For example, Al(s) and Cl 2 have an oxidation state of 0 because both are in a chemically uncombined elemental state.
• Please note that the allotropic form of sulfur S 8, or octasulfur, despite its atypical structure, is also characterized by a zero oxidation state.

1. Determine if the substance in question consists of ions. The oxidation state of ions is equal to their charge. This is true both for free ions and for those that are part of chemical compounds.

   • For example, the oxidation state of the Cl ion is -1.
   • The oxidation state of the Cl ion in the chemical compound NaCl is also -1. Since the Na ion, by definition, has a charge of +1, we conclude that the charge of the Cl ion is -1, and thus its oxidation state is -1.

2. Note that metal ions can have several oxidation states. Atoms of many metallic elements can be ionized to different extents. For example, the charge of ions of a metal such as iron (Fe) is +2 or +3. The charge of metal ions (and their degree of oxidation) can be determined by the charges of ions of other elements with which this metal is part of a chemical compound; in the text, this charge is indicated by Roman numerals: for example, iron (III) has an oxidation state of +3.

   • As an example, consider a compound containing an aluminum ion. The total charge of the AlCl 3 compound is zero. Since we know that Cl - ions have a charge of -1, and the compound contains 3 such ions, for the total neutrality of the substance in question, the Al ion must have a charge of +3. Thus, in this case, the oxidation state of aluminum is +3.

3. The oxidation state of oxygen is -2 (with some exceptions). In almost all cases, oxygen atoms have an oxidation state of -2. There are several exceptions to this rule:

   • If oxygen is in the elemental state (O 2 ), its oxidation state is 0, as is the case for other elemental substances.
   • If oxygen is included peroxides, its oxidation state is -1. Peroxides are a group of compounds containing a single oxygen-oxygen bond (ie the peroxide anion O 2 -2). For example, in the composition of the H 2 O 2 molecule (hydrogen peroxide), oxygen has a charge and an oxidation state of -1.
   • In combination with fluorine, oxygen has an oxidation state of +2, see the rule for fluorine below.

4. Hydrogen has an oxidation state of +1, with a few exceptions. As with oxygen, there are also exceptions. As a rule, the oxidation state of hydrogen is +1 (unless it is in the elemental state H 2). However, in compounds called hydrides, the oxidation state of hydrogen is -1.

   • For example, in H 2 O, the oxidation state of hydrogen is +1, since the oxygen atom has a charge of -2, and two +1 charges are needed for overall neutrality. However, in the composition of sodium hydride, the oxidation state of hydrogen is already -1, since the Na ion carries a charge of +1, and for total electroneutrality, the charge of the hydrogen atom (and thus its oxidation state) must be -1.

5. Fluorine Always has an oxidation state of -1. As already noted, the degree of oxidation of some elements (metal ions, oxygen atoms in peroxides, and so on) can vary depending on a number of factors. The oxidation state of fluorine, however, is invariably -1. This is explained by the fact that this element has the highest electronegativity - in other words, fluorine atoms are the least willing to part with their own electrons and most actively attract other people's electrons. Thus, their charge remains unchanged.

6. The sum of the oxidation states in a compound is equal to its charge. The oxidation states of all the atoms that make up a chemical compound, in total, should give the charge of this compound. For example, if a compound is neutral, the sum of the oxidation states of all its atoms must be zero; if the compound is a polyatomic ion with a charge of -1, the sum of the oxidation states is -1, and so on.

   • This good method checks - if the sum of the oxidation states is not equal to the total charge of the compound, then you made a mistake somewhere.

   Part 2

   Determining the oxidation state without using the laws of chemistry

   1. Find atoms that do not have strict rules regarding oxidation state. In relation to some elements, there are no firmly established rules for finding the degree of oxidation. If an atom does not fall under any of the rules listed above, and you do not know its charge (for example, the atom is part of a complex, and its charge is not indicated), you can determine the oxidation state of such an atom by elimination. First, determine the charge of all other atoms of the compound, and then from the known total charge of the compound, calculate the oxidation state of this atom.

      • For example, in the Na 2 SO 4 compound, the charge of the sulfur atom (S) is unknown - we only know that it is not zero, since sulfur is not in the elementary state. This connection is a good example to illustrate algebraic method determining the degree of oxidation.

   2. Find the oxidation states of the rest of the elements in the compound. Using the rules described above, determine the oxidation states of the remaining atoms of the compound. Don't forget about the exceptions to the rule in the case of O, H, and so on.

      • For Na 2 SO 4 , using our rules, we find that the charge (and hence the oxidation state) of the Na ion is +1, and for each of the oxygen atoms it is -2.
   3. In compounds, the sum of all oxidation states must equal the charge. For example, if the compound is a diatomic ion, the sum of the oxidation states of the atoms must be equal to the total ionic charge.
   4. It is very useful to be able to use the periodic table of Mendeleev and know where the metallic and non-metallic elements are located in it.
   5. The oxidation state of atoms in the elementary form is always zero. The oxidation state of a single ion is equal to its charge. Elements of group 1A of the periodic table, such as hydrogen, lithium, sodium, in elemental form have an oxidation state of +1; the oxidation state of group 2A metals, such as magnesium and calcium, in its elemental form is +2. Oxygen and hydrogen, depending on the species chemical bond, may have 2 different meanings degree of oxidation.

M.A.AKHMETOV

Lecture notes
in general chemistry

Continuation. For the beginning, see№ 8, 12, 13, 20, 23, 25-26, 40/2004

Chapter 5
redox
reactions

5.1. Determination of the degree of oxidation

Redox reactions are reactions that involve the transfer of electrons from one atom to another. The transition of electrons is judged by changes in the oxidation states of atoms. If the oxidation state of an atom changes, then its electronic environment also changes. There are two ways to determine the oxidation states of atoms: first– by gross formula , second– according to the structural formula .
When determining the oxidation states of atoms in the first way, the rule is used: the sum of the oxidation states of all the atoms that form the particle is equal to the charge of the particle . For a molecule, this sum is equal to zero, and for an ion, its charge.
As an illustration, let us determine the oxidation state of atoms in sodium thiosulfate Na 2 S 2 O 3 by the first method. Among the elements that form a particle, oxygen is the most electronegative - it will accept electrons. Since oxygen is in the main subgroup of group VI, it lacks two electrons to complete the electron layer. Therefore, the oxygen atom will accept two electrons and acquire an oxidation state of -2. The most electropositive atom is sodium, which has only one electron in the outer electronic level (sodium will give it away). These considerations, taking into account the sodium thiosulfate formula, allow us to draw up the equation:

2 (+1) + 2X + 3 (–2) = 0,

the solution of which will give the value of the oxidation state of the sulfur atom (+2).
It is possible to determine the oxidation states of atoms in complex ions. Let's take an anion as an example. In it, the most electronegative oxygen atom accepts two electrons and has an oxidation state of -2. The oxidation state of the chromium atom is determined from the equation:

2X + 7 (–2) = –2

and is equal to +6.
The second way to find the oxidation states of atoms - according to the structural formula - is based on the definition: oxidation state – this is the conditional integer charge that would be on an atom if all of its polar covalent bonds became ionic. Depicting the structural formula of sodium thiosulfate

determine the oxidation states of its atoms.
Sodium atoms connected by single bonds to more electronegative oxygen atoms will naturally give them their outer electrons, each acquiring an oxidation state of +1. Oxygen atoms that have two bonds each with more electropositive atoms will conditionally accept two electrons each and will have an oxidation state of -2. It can be seen from the structural formula that the compound contains two sulfur atoms in different environments. One of the S atoms is connected only by a double bond to the other S atom, and its oxidation state is zero. The second sulfur atom has four bonds to the three more electronegative oxygen atoms and therefore has an oxidation state of +4.
The average oxidation state of sulfur atoms, as in determining it by the first method, is +2 ((+4+0)/2).
The oxygen atom does not always have an oxidation state of -2. For example, in its combination with fluorine atoms, it has a positive oxidation state. In peroxides, the oxidation state of each oxygen atom is , in superoxides it is only , and in ozonides it is even . Also, at the sulfur atom, the oxidation state can be equal to -1, for example, in disulfides. In some oxides, for example, Fe 3 O 4 and Pb 3 O 4 , the oxidation states of atoms are determined based on the fact that these oxides are mixed: Fe 2 O 3 FeO and PbO 2 2PbO, respectively.

5.2. Writing Equations
redox reactions

The selection of coefficients in the equations of redox reactions is carried out by compiling the electronic balance. The selection method, which boils down to counting the number of atoms on the right and left sides of the equation, does not always guarantee correct definition coefficients. Thus, in the three equations below for the oxidation of triethylamine with nitric acid, there are equal numbers of carbon, hydrogen, oxygen, and nitrogen atoms on the left and right sides, but only one of them is realized:

4 (C 2 H 5) 3N + 36HNO 3 \u003d 24CO 2 + 48H 2 O + 6NO 2 + 17N 2,

2 (C 2 H 5) 3N + 78HNO 3 \u003d 12CO 2 + 54H 2 O + 78NO 2 + N 2,

(C 2 H 5) 3 N + 11HNO 3 \u003d 6CO 2 + 13H 2 O + 4NO 2 + 4N 2.

The theory of the redox process involves the transfer of electrons from the atoms of the reducing agent to the atoms of the oxidizing agent. According to the law of conservation of matter, the total number of electrons donated by the reducing agent is equal to the total number of electrons received by the oxidizing agent. This simple idea guides the formulation of equations for redox reactions. The task is to select the proportionality coefficients at which the electronic balance is achieved.
Let us analyze an example of the oxidation of an ethylbenzene molecule with potassium permanganate in an acidic medium when heated. We write the reaction equation and indicate the oxidation states of those atoms that changed it, and we will determine their oxidation states in the molecules of ethylbenzene and benzoic acid using the appropriate structural formulas:

A carbon atom directly bonded to the benzene ring will change its oxidation state from -2 to +3 (donate 5 electrons). The carbon atom of the methyl group will change its oxidation state from -3 to +4 in carbon dioxide (donate 7 electrons). In total, the ethylbenzene molecule will donate 12 electrons. The manganese atom will change its oxidation state from +7 to +2 (accept 5 electrons). In this case, we have the equation:

12X = 5y,

whose minimal positive integer solutions are X = 5, at = 12.
The selection of coefficients in the equations by the disproportionation reaction by the electronic balance method must be carried out on their right side. As an example, let's analyze the disproportionation of Berthollet salt (without a catalyst):

From the changes in the oxidation states of atoms during the reaction, it follows that he received 6 electrons, and allegedly gave 2 electrons.
Then

(KCl) \u003d 3 (KClO 4).

Therefore, it is necessary to put a coefficient 3 in front of potassium perchlorate KClO 4:

4KClO 3 \u003d KCl + 3KClO 4.

5.3. Electrolysis

The decomposition of an electrolyte (in solution or melt) when an electric current passes through it is called electrolysis .
The instrumentation of the electrolysis process boils down to the fact that two electrodes connected to a current source are lowered into a vessel with an electrolyte solution or melt (Fig. 5.1).

A negatively charged electrode is called cathode (cations are attracted to it), and a positively charged electrode - anode (it attracts anions). The electrical circuit is closed due to redox processes taking place on the electrodes. At the cathode, cations are reduced, and at the anode, anions are oxidized.
Let's start the consideration of the process with the simplest case - melt electrolysis. In the electrolysis of melts at the cathode metal cations are reduced to pure metal, and at the anode simple anions are oxidized to a simple substance, for example:

2Cl - - 2 e\u003d Cl 2,

S 2– – 2 e= S.

If the anion has a complex structure, then in this case a process takes place that requires the least amount of energy. If the salt is resistant to heat and the element atom in the anion is in the highest oxidation state, then oxygen is usually oxidized to a simple substance:

– 2e\u003d SO 3 + 1 / 2O 2.

If an element atom is in an intermediate oxidation state, then it is most likely that in this case, not oxygen will be oxidized, but an atom of another element in the anion, for example:

– e= NO 2 .

Electrolysis in solutions is more complicated in terms of determining the products. This is due to the appearance of another component - water. Metals with standard electrode potentials from –1.67 V (Al) and lower (located to the left of manganese in the series of metal voltages) are usually not reduced from aqueous solutions. In such systems, hydrogen is released at the cathode. This is primarily due to the fact that these metals (including magnesium and aluminum without a protective oxide film) react with water. But this does not mean at all that electrode processes of the type

Na + + e= Na

do not occur in aqueous solutions. One of the ways to obtain metallic sodium is the electrolysis of an aqueous solution of NaCl (brine). The secret of this process lies in the use of a mercury cathode. The reduced sodium atoms are absorbed by the mercury layer, which protects them from contact with water. The subsequent separation into components of the resulting sodium amalgam (amalgam is an alloy, one of the components of which is mercury) is achieved by rectification. The released mercury is then returned to the working cycle.
The impossibility of obtaining metals interacting with water by electrolysis of aqueous solutions of the corresponding electrolytes is also evidenced by the following reasoning. Let calcium be reduced during the electrolysis of an aqueous solution at the cathode:

Ca 2+ + 2 e= Ca.

The metal, having recovered, will react with water:

Ca + 2H 2 O \u003d Ca (OH) 2 + H 2.

Consequently, instead of metal, hydrogen will be released at the cathode.
Metals with standard electrode potentials in the range from –1.05 V to 0 V (located in the electrochemical series between aluminum and hydrogen) are reduced from aqueous solutions in parallel with hydrogen. The ratio of products (metal and hydrogen) is determined by the concentration of the solution, its acidity and some other factors (the presence of other, especially complex, salts in the solution; the material from which the electrode is made). The higher the salt concentration, the greater the proportion of the released metal. The more acidic the environment, the more likely hydrogen will be released. Metals with positive standard electrodes
potentials (located in the series of voltages of metals to the right of hydrogen) are released during the electrolysis of solutions in the first place. For example:

Ag + + e= Ag.

At the anode, during the electrolysis of aqueous solutions, all simple anions are oxidized, with the exception of fluoride. For example:

2I - - 2 e= I 2 .

Fluorine cannot be obtained by electrolysis of aqueous solutions, because it reacts with water:

F 2 + H 2 O \u003d 2HF + 1 / 2O 2.

If the salt undergoing electrolysis contains a complex anion in which the heteroatom (not oxygen) is in the highest oxidation state, then oxygen is formed at the anode, i.e. water decomposes:

H 2 O - 2 e= 2H + + 1/2O 2 .

The complex anion itself can also serve as a source of oxygen:

– 2e\u003d SO 3 + 1 / 2O 2.

The resulting acid anhydride will immediately react with water:

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

When a heteroatom is in an intermediate oxidation state, it is oxidized, not the oxygen atom. An example of such a process is the oxidation of a sulfite ion under the action of an electric current:

The resulting sulfuric anhydride SO 3 immediately reacts with water.
Anions of carboxylic acids are decarboxylated as a result of electrolysis, forming hydrocarbons:

2R-COO - - 2 e= R–R + 2CO 2 .

5.4. The direction of the oxidation
recovery processes
and the influence of the acidity of the environment on it

The redox or standard electrode potentials serve as a measure of the redox ability of substances in aqueous solutions. Let us determine, for example, whether the Fe 3+ iron cation can oxidize halogen anions to KCl, KBr, and KI. Knowing the standard electrode potentials ( 0), it is possible to calculate the electromotive force (EMF) of the process. It is defined as the difference between such potentials of the oxidizing agent and the reducing agent, and the reaction proceeds at a positive EMF value:

Table 5.1

Determination of the possibility of leakage
redox processes
based on standard electrode potentials

Tab. 5.1 shows that only one of the investigated processes is possible. Indeed, of all the above potassium halides, only KI reacts with iron trichloride:

2FeCl 3 + 2KI = 2FeCl 2 + I 2 + 2KCl.

There is another simple way to determine the direction of the process. If we write two equations of the half-reactions of the process one under the other so that the standard electrode potential of the upper half-reaction is less than the lower one, then the letter Z written between them (Fig. 5.2) will indicate with its ends the directions of the stages of the allowed process (rule Z).

From the same substances, by changing the pH of the medium, different products can be obtained. For example, the permanganate anion is reduced in an acidic medium to form a manganese(II) compound:

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

In a neutral environment, manganese dioxide MnO 2 is formed:

2KMnO 4 + 3Na 2 SO 3 + H 2 O \u003d 2KOH + 2MnO 2 + 3Na 2 SO 4.

In an alkaline environment, the permanganate anion is reduced to the manganate anion:

2KMnO 4 + Na 2 SO 3 + 2KOH \u003d 2K 2 MnO 4 + Na 2 SO 4 + H 2 O.

5.5. Exercises

1. Determine the oxidation states of atoms in the following compounds: BaO 2 , CsO 2 , RbO 3 , F 2 O 2 , LiH, F 2 , C 2 H 5 OH, toluene, benzaldehyde, acetic acid.

One of the basic concepts in chemistry, widely used in the preparation of equations of redox reactions, is oxidation state atoms.

For practical purposes (when compiling equations of redox reactions), it is convenient to represent the charges on atoms in molecules with polar bonds as integers equal to the charges that would arise on the atoms if the valence electrons were completely transferred to more electronegative atoms, i.e. e. if the bonds were completely ionic. Such charge values ​​are called oxidation states. The oxidation state of any element in a simple substance is always 0.

in molecules complex substances some elements always have a constant oxidation state. Most elements are characterized by variable oxidation states, which differ both in sign and magnitude, depending on the composition of the molecule.

Often the oxidation state is equal to the valency and differs from it only in sign. But there are compounds in which the oxidation state of the element is not equal to its valency. As already noted, in simple substances, the oxidation state of an element is always zero, regardless of its valency. The table compares the valencies and oxidation states of some elements in various compounds.

The oxidation state of an atom (element) in a compound, this is the conditional charge calculated assuming that the compound consists only of ions. When determining the degree of oxidation, it is conditionally assumed that the valence electrons in the compound pass to more electronegative atoms, and therefore the compounds consist of positively and negatively charged ions. In reality, in most cases, there is not a complete return of electrons, but only a displacement of an electron pair from one atom to another. Then another definition can be given: The oxidation state is that electric charge, which would arise on an atom if the electron pairs by which it is connected with other atoms in the compound passed to more electronegative atoms, and the electron pairs binding the same atoms were divided between them.

When calculating the oxidation states, the series is used simple rules:

1 . The oxidation state of elements in simple substances, both monatomic and molecular, is zero (Fe 0, O 2 0).

2 . The oxidation state of an element in the form of a monatomic ion is equal to the charge of this ion (Na +1, Ca +2, S -2).

3 . In compounds with a covalent polar bond, a negative charge refers to a more electronegative atom, and a positive charge to a less electronegative atom, and the oxidation states of the elements take on the following values:

The oxidation state of fluorine in compounds is always -1;

The oxidation state of oxygen in compounds is -2 (); with the exception of peroxides, where it is formally equal to -1 (), oxygen fluoride, where it is equal to +2 (), as well as superoxides and ozonides, in which the oxidation state of oxygen is -1/2;

The oxidation state of hydrogen in compounds is +1 (), with the exception of metal hydrides, where it is -1 ( );

For alkaline and alkaline earth elements, the oxidation state is +1 and +2, respectively.

Most elements can exhibit variable oxidation states.

4 . The algebraic sum of the oxidation states in a neutral molecule is zero, in a complex ion it is the charge of the ion.

For elements with a variable oxidation state, its value is easy to calculate, knowing the formula of the compound and using rule No. 4. For example, it is necessary to determine the oxidation state of phosphorus in phosphoric acid H 3 PO 4 . Since oxygen has CO \u003d -2, and hydrogen has CO \u003d +1, then for a zero sum in phosphorus, the oxidation state should be equal to +5:

For example, in NH 4 Cl, the sum of the oxidation states of all hydrogen atoms is 4 × (+1), and the oxidation state of chlorine is -1, therefore, the oxidation state of nitrogen should be equal to -3. In the sulfate ion SO 4 2–, the sum of the oxidation states of the four oxygen atoms is -8, so sulfur must have an oxidation state of +6 so that the total charge of the ion is -2.

The concept of the degree of oxidation for most compounds is conditional, because does not reflect the real effective charge of the atom, but this concept is very widely used in chemistry.

The maximum, and for non-metals and the minimum, oxidation state has a periodic dependence on the serial number in PSCE D.I. Mendeleev, which is due to the electronic structure of the atom.

Element	Oxidation State Values ​​and Compound Examples
F	–1 (HF, KF)
O	–2 (H 2 O, CaO, CO 2); –1 (H 2 O 2); +2 (of 2)
N	–3 (NH3); –2(N 2 H 4); –1 (NH 2 OH); +1 (N 2 O); +2 (NO); +3 (N 2 O 3 , HNO 2); +4 (NO 2); +5 (N 2 O 5, HNO 3)
Cl	–1 (HCl, NaCl); +1 (NaClO); +3 (NaClO2); +5 (NaClO 3); +7 (Cl 2 O 7, NaClO 4)
Br	–1 (KBr); +1 (BrF); +3 (BrF 3); +5 (KBrO 3)
I	–1 (HI); +1 (ICl); +3 (ICl 3); +5 (I 2 O 5); +7 (IO 3 F, K 5 IO 6)
C	–4 (CH4); +2 (CO); +4 (CO 2 , CCl 4)
Si	–4 (Ca 2 Si); +2 (SiO); +4 (SiO 2 , H 2 SiO 3 , SiF 4)
H	–1 (LiH); +1 (H 2 O, HCl)
S	–2 (H 2 S, FeS); +2 (Na 2 S 2 O 3); +3 (Na 2 S 2 O 4); +4 (SO 2 , Na 2 SO 3 , SF 4); +6 (SO 3 , H 2 SO 4 , SF 6)
Se, Te	–2 (H 2 Se, H 2 Te); +2 (SeCl 2 , TeCl 2); +4 (SeO 2 , TeO 2); +6 (H 2 SeO 4 , H 2 TeO 4)
P	–3 (PH 3); +1 (H3PO2); +3 (H3PO3); +5 (P 2 O 5 , H 3 PO 4)
As, Sb	–3 (GaAs, Zn 3 Sb 2); +3 (AsCl 3 , Sb 2 O 3); +5 (H 3 AsO 4 , SbCl 5)
Li, Na, K	+1 (NaCl)
Be, Mg, Ca	+2 (MgO, CaCO 3)
Al	+3 (Al 2 O 3 , AlCl 3)
Cr	+2 (CrCl2); +3 (Cr 2 O 3 , Cr 2 (SO 4) 3); +4 (CrO2); +6 (K 2 CrO 4 , K 2 Cr 2 O 7)
Mn	+2 (MnSO4); +3 (Mn 2 (SO 4) 3); +4 (MnO2); +6 (K2MnO4); +7 (KMnO 4)
Fe	+2 (FeO, FeSO 4); +3 (Fe 2 O 3, FeCl 3); +4 (Na 2 FeO 3)
Cu	+1 (Cu 2 O); +2 (CuO, CuSO 4 , Cu 2 (OH) 2 CO 3)
Ag	+1 (AgNO3)
Au	+1 (AuCl); +3 (AuCl 3 , KAuCl 4)
Zn	+2 (ZnO, ZnSO4)
hg	+1 (Hg 2 Cl 2); +2 (HgO, HgCl 2)
sn	+2 (SnO); +4 (SnO 2 , SnCl 4)
Pb	+2 (PbO, PbSO 4); +4 (PbO2)

In chemical reactions, the rule of conservation of the algebraic sum of the oxidation states of all atoms must be fulfilled. In the full equation chemical reaction oxidation and reduction processes must exactly compensate each other. Although the degree of oxidation, as noted above, is a rather formal concept, it is used in chemistry for the following purposes: firstly, to draw up equations for redox reactions, and secondly, to predict redox properties of the elements in the connection.

Many elements are characterized by several values ​​of oxidation states, and by calculating its oxidation state, redox properties can be predicted: an element in the highest negative oxidation state can only donate electrons (oxidize) and be a reducing agent, in the highest positive oxidation state it can only accept electrons (reduce ) and be an oxidizing agent, in intermediate oxidation states - both oxidized and reduced.

Oxidation-reduction is a single, interconnected process. Oxidation corresponds to an increase in the oxidation state of the element, and recovery - its reduction.

Many manuals adhere to the interpretation of oxidation as the loss of electrons, and reduction as their addition. This approach, proposed by the Russian scientist Pisarzhevsky (1916), is applicable to electrochemical processes on electrodes and refers to the discharge (charging) of ions and molecules.

However, the explanation of the change in oxidation states as processes of detachment and addition of electrons is generally incorrect. It can be applied to some simple ions like

Cl - - ®Cl 0 .

To change the oxidation state of atoms in complex ions of the type

CrO 4 2 - ®Cr +3

a decrease in the positive oxidation state of chromium from +6 to +3 corresponds to a smaller real increase in the positive charge (on Cr in CrO 4 2 - real charge "+0.2 electron charge, and on Cr +3 - from +2 to +1.5 in different compounds).

The charge transfer from the reducing agent to the oxidizing agent, which is equal to the change in the degree of oxidation, occurs in this case with the participation of other particles, for example, H + ions:

CrO 4 2 - + 8H + + 3 ®Cr +3 + 4H 2 O.

The entry presented is titled half reactions .

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