EducationSecondary education and schools

The degree of oxidation is what value? How to determine the degree of oxidation of elements?

Such a subject of the school curriculum as chemistry causes numerous difficulties for most modern schoolchildren, few can determine the degree of oxidation in compounds. The greatest difficulties for schoolchildren who study inorganic chemistry, that is, students of the main school (grades 8-9). Misunderstanding of the subject leads to the appearance of dislike among schoolchildren in this subject.

Teachers identify a number of reasons for such "dislike" of middle and senior students in chemistry: reluctance to understand complex chemical terms, inability to use algorithms to consider a particular process, problems with mathematical knowledge. The Ministry of Education of the Russian Federation introduced a major change in the content of the subject. In addition, "cut" and the number of hours for teaching chemistry. This negatively affected the quality of knowledge on the subject, the decrease of interest in the study of discipline.

What topics of the chemistry course are given most difficult for schoolchildren?

Under the new program in the course of the discipline "Chemistry" of the main school there are several serious topics included: periodic table of D. Mendeleyev's elements, classes of inorganic substances, ion exchange. It is most difficult for eighth-graders to determine the oxidation state of oxides.

Rules of the arrangement

First of all, students should be aware that oxides are complex two-element compounds, in which oxygen is included. An obligatory condition for the binary compound to belong to the class of oxides is the position of oxygen in the second compound in this compound.

To calculate such an indicator in any formulas of this class will be obtained only if the student has a certain algorithm.

Algorithm for acid oxides

To begin with, we note that the degree of oxidation is a numerical expression for the valence of the elements. Acidic oxides are formed by nonmetals or metals with a valence of four to seven, the second in such oxides necessarily oxygen.

In oxides, the oxygen valence always corresponds to two, it can be determined from the periodic table of DI Mendeleyev's elements. Such a typical nonmetal as oxygen, being in the 6th group of the main subgroup of the periodic table, takes two electrons to completely complete its external energy level. Nonmetals in compounds with oxygen most often exhibit a higher valence, which corresponds to the number of the group itself. It is important to recall that the degree of oxidation of chemical elements is an indicator that assumes a positive (negative) number.

The non-metal, at the beginning of the formula, has a positive oxidation state. Non-metal oxygen in oxides is stable, its index is -2. In order to verify the reliability of the distribution of values in acidic oxides, you will have to multiply all the numbers you put on the indices of a particular element. Calculations are considered reliable if the total of all the pros and cons of the delivered degrees is 0.

Compilation of two-element formulas

The degree of oxidation of atoms of elements gives a chance to create and record connections from two elements. When creating a formula, to begin with, both symbols are prescribed side by side, the oxygen is the second one. Above each of the recorded signs, the values of the degrees of oxidation are prescribed, then between the numbers found there is the number that will be divided into both digits without any remainder. This indicator must be separately divided by the numerical value of the degree of oxidation, obtaining indices for the first and second components of the two-element substance. The highest degree of oxidation is equal numerically to the value of the highest valence of a typical non-metal, identical to the number of the group where the nonmetal in the PS is.

Algorithm for setting numerical values in basic oxides

Similar compounds are oxides of typical metals. They in all compounds have an oxidation index of no more than +1 or +2. In order to understand what will be the degree of oxidation of the metal, we can use the periodic system. For metals of the main subgroups of the first group, this parameter is always constant, it is similar to the group number, that is, +1.

Metals of the main subgroup of the second group are also characterized by a stable degree of oxidation, in numerical terms +2. The degrees of oxidation of oxides in the total, taking into account their indices (numbers) should give zero, since a chemical molecule is considered to be neutral, devoid of charge, a particle.

The arrangement of degrees of oxidation in oxygen-containing acids

Acids are complex substances consisting of one or more hydrogen atoms, which are associated with some acidic residue. Given that oxidation states are digital indicators, some mathematical skills will be required to calculate them. Such an indicator for hydrogen (proton) in acids is always stable, is +1. Then you can indicate the degree of oxidation for a negative oxygen ion, it is also stable, -2.

Only after these actions, it is possible to calculate the oxidation state of the central component of the formula. As a specific sample, let us consider the determination of the degree of oxidation of elements in sulfuric acid H2SO4. Considering that in the molecule of a given complex substance contains two hydrogen protons, four oxygen atoms, we obtain an expression of this type + 2 + X-8 = 0. In order for the total to form zero, sulfur will have an oxidation state of +6

The arrangement of degrees of oxidation in salts

Salts are complex compounds consisting of metal ions and one or more acid residues. The procedure for determining the degrees of oxidation in each of the constituents in a complex salt is the same as in oxygen-containing acids. Given that the degree of oxidation of elements is a digital indicator, it is important to correctly identify the degree of oxidation of the metal.

If the metal forming the salt is located in the main subgroup, its degree of oxidation will be stable, corresponding to the group number, is a positive value. If the salt contains a metal of a similar subgroup of the PS, exhibiting different valencies, the valence of the metal can be determined from the acid residue. Once the oxidation state of the metal is established, set the oxidation state of oxygen (-2), then calculate the degree of oxidation of the central element using the chemical equation.

As an example, let us consider the determination of the degrees of oxidation in elements in sodium nitrate (middle salt). NaNO3. The salt is formed by the metal of the main subgroup of group 1, therefore, the degree of oxidation of sodium will be +1. Oxygen in nitrates has a degree of oxidation of -2. To determine the numerical value of the degree of oxidation is the equation + 1 + X-6 = 0. Solving this equation, we get that X should be +5, this is the degree of nitrogen oxidation.

Basic terms in OVR

For oxidation, as well as the recovery process, there are special terms that schoolchildren are required to learn.

The degree of oxidation of an atom is its direct ability to attach (to give to others) electrons from some ions or atoms.

Oxidant is considered to be neutral atoms or charged ions, in the course of a chemical reaction they attach electrons to themselves.

The restorer will be uncharged atoms or charged ions, which lose their own electrons in the process of chemical interaction.

Oxidation is represented as a procedure for the release of electrons.

Recovery is associated with the adoption of additional electrons by an uncharged atom or ion.

Oxidation-reduction process is characterized by a reaction, during which the degree of oxidation of the atom necessarily changes. This definition allows us to understand how it is possible to determine whether the OVR reaction is.

Rules for parsing IAD

Using this algorithm, you can arrange the coefficients in any chemical reaction.

  1. First, you need to place the oxidation states in each chemical substance. Note that in a simple substance the degree of oxidation is zero, since there is no recoil (attachment) of negative particles. The rules for arranging the degrees of oxidation in binary and three-element substances were considered above.

  2. Then it is necessary to determine those atoms or ions in which oxidation states have changed in the course of the transformation that occurred.

  3. From the left side of the recorded equation, atoms or charged ions are separated, which have changed their oxidation states. This is necessary for balancing. The elements are always indicated by their values.

  4. Further, those atoms or ions that were formed during the reaction are written down, indicated by the sign + the number of electrons taken by the atom, - the number of negative particles given. If oxidation is reduced after the reaction process. This means that the electrons were taken by an atom (ion). With an increase in the degree of oxidation, the atom (ion) during the reaction gives off electrons.

  5. The smallest total number is divided first into the received ones, then to the electrons transferred in the process, the coefficients are obtained. The numbers found are the required stereochemical coefficients.

  6. Determine the oxidizer, reducing agent, the processes occurring during the reaction.

  7. The last stage will be the arrangement of the stereochemical coefficients in the reaction under consideration.

    Example of OBR

Consider the practical application of this algorithm to a specific chemical reaction.

Fe + CuSO4 = Cu + FeSO4

We calculate the parameters for all simple and complex substances.

Since Fe and Cu are simple substances, their degree of oxidation is 0. In CuSO4, then Cu + 2, then for oxygen-2, and for sulfur +6. In FeSO4: Fe +2, therefore, for O-2, according to calculations of S +6.

Now we are looking for elements that could change the indicators, in our situation they will be Fe and Cu.

Since after the reaction the value of the iron atom became +2, 2 electrons were released in the reaction. Copper changed its performance from +2 to 0, therefore, copper took 2 electrons. Now we determine the number of electrons taken and delivered by the iron atom and cation of bivalent copper. In the course of the transformation, two electrons are taken by the cation of bivalent copper, the same number of electrons is given up by the iron atom.

In this process, it makes no sense to determine the minimum common multiple, since an equal number of electrons is accepted and given during the transformation. Stereochemical coefficients will also correspond to unity. In the reaction, the properties of the reducing agent will exhibit iron, while oxidizing it. The cation of bivalent copper is reduced to pure copper, in the reaction it has the highest degree of oxidation.

Application of processes

Formulas for the degree of oxidation should be known to every student of grade 8-9, since this question is included in the OGE tasks. Any processes that occur with oxidative, restorative signs play an important role in our life. Without them, metabolic processes in the human body are impossible.

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