Such a subject of the school curriculum as chemistry causes numerous difficulties for the majority of modern schoolchildren; few can determine the degree of oxidation in compounds. The greatest difficulties are among students who study inorganic chemistry, that is, students in a primary school (grades 8–9). Misunderstanding of the subject leads to hostility among schoolchildren to this subject.
Teachers identify a number of reasons for such “dislike” of middle and high school students for chemistry: reluctance to understand complex chemical terms, inability to use algorithms to consider a specific process, problems with mathematical knowledge. The Ministry of Education of the Russian Federation made a major change in the content of the subject. In addition, the number of hours for teaching chemistry was “cut”. This negatively affected the quality of knowledge in the subject, a decrease in interest in the study of discipline.
What are the hardest topics for students in chemistry courses?
According to the new program, several serious topics are included in the course of the subject “Chemistry” of the basic school: the periodic table of the elements of D. I. Mendeleev, classes of inorganic substances, ion exchange. Eighth graders are most difficult to determine the degree of oxidation of oxides.
Placement rules
First of all, students should know that oxides are complex two-element compounds, which include oxygen. An obligatory condition for the binary compound to belong to the class of oxides is the location of oxygen second in this compound.
It will be possible to calculate such an indicator in any formulas of this class only if the student has a certain algorithm.
Algorithm for Acid Oxides
To begin with, we note that oxidation states are numerical expressions of the valency of elements. Acid oxides are formed by non-metals or metals with valencies from four to seven, the second in such oxides is necessarily oxygen.
In oxides, the oxygen valency always corresponds to two; it can be determined from the periodic table of the elements of D. I. Mendeleev. Such a typical non-metal 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. Non-metals in compounds with oxygen most often exhibit higher valency, 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 assuming a positive (negative) number.
The non-metal at the beginning of the formula has a positive oxidation state. Nonmetal oxygen is stable in oxides, its rate is -2. In order to check the accuracy of the arrangement of values in acidic oxides, you will have to multiply all the numbers you set by the indices of a particular element. The calculations are considered reliable if the total result of all the pros and cons of the degrees set is 0.
Drawing up two-element formulas
The oxidation state of the atoms of the elements gives a chance to create and record compounds of two elements. When creating the formula, for a start, both symbols are written side by side, they must put oxygen in the second. The oxidation state values are written above each of the recorded signs, then the number is found between the numbers found, which will be divisible by both numbers without any remainder. This indicator must be divided separately by the numerical value of the degree of oxidation, getting the indices for the first and second components of a two-element substance. The highest oxidation state is numerically equal to the value of the higher valency of a typical non-metal, identical to the number of the group where the non-metal in the PS is.
Algorithm for setting numerical values in basic oxides
Such compounds are considered to be oxides of typical metals. In all compounds they have an oxidation state index of not more than +1 or +2. In order to understand what will be the degree of oxidation of the metal, you 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 oxidation state, in digital terms +2. The oxidation states of oxides in total, taking into account their indices (numbers), should give zero, since a chemical molecule is considered to be a neutral, chargeless, particle.
The arrangement of oxidation states in oxygen-containing acids
Acids are complex substances consisting of one or more hydrogen atoms that are bound to some kind of acid residue. Considering 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. Next, you can specify the oxidation state for the 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, we consider the determination of the degree of oxidation of elements in sulfuric acid H2SO4. Given that the molecule of this complex substance contains two hydrogen protons, 4 oxygen atoms, we obtain an expression of this form + 2 + X-8 = 0. In order for a total to form zero, sulfur will have an oxidation state of +6
The degree of oxidation in salts
Salts are complex compounds consisting of metal ions and one or more acid residues. The method for determining the oxidation states of each of the components 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 indicate the degree of oxidation of the metal.
If the metal forming the salt is located in the main subgroup, its oxidation state will be stable, corresponds to the group number, and is a positive value. If the salt contains a metal of a similar subgroup of PS exhibiting different valencies, the valency of the metal can be determined by the acid residue. After the oxidation state of the metal has been established, the oxidation state of oxygen is set (-2), then the oxidation state of the central element is calculated using the chemical equation.
As an example, we consider the determination of the oxidation states of 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 sodium oxidation will be +1. In oxygen in nitrates, the oxidation state is -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.
Key Terms in OVR
For the oxidative as well as the reduction process, there are special terms that students must learn.
The degree of oxidation of an atom is its direct ability to attach to itself (give away to others) electrons from some ions or atoms.
Oxidizing agents are neutral atoms or charged ions that attach electrons to themselves during a chemical reaction.
The reductant will be uncharged atoms or charged ions, which in the process of chemical interaction lose their own electrons.
Oxidation is represented as an electron recoil procedure.
Recovery is associated with the adoption of additional electrons by an uncharged atom or ion.
The redox process is characterized by a reaction during which the degree of oxidation of an atom necessarily changes. This definition allows you to understand how you can determine whether the reaction is an OVR.
RIA Analysis Rules
Using this algorithm, you can arrange the coefficients in any chemical reaction.
First you need to arrange the degree of oxidation in each chemical substance. Note that in a simple substance, the oxidation state is zero, since there is no recoil (adherence) of negative particles. The rules for the arrangement of oxidation states in binary and three-element substances were considered by us above.
Then it is necessary to determine those atoms or ions for which the oxidation state has changed during the course of the transformation.
From the left side of the written equation, atoms or charged ions are distinguished that have changed their oxidation states. This is necessary to balance. Above the elements must indicate their values.
Next, those atoms or ions that are formed during the reaction are recorded, indicated by the sign + the number of electrons received by the atom, the number of negative particles delivered. If, after the interaction process, the oxidation state decreases. This means that the electrons were taken by an atom (ion). With an increase in the oxidation state, an atom (ion) gives up electrons during the reaction.
The smallest total number is first divided by received, then by the electrons given in the process, and coefficients are obtained. The numbers found will be the desired stereochemical coefficients.
The oxidizing agent, reducing agent, processes occurring during the reaction are determined.
The final step will be the arrangement of stereochemical coefficients in the reaction under consideration.
OVR example
Consider the practical application of this algorithm for a specific chemical reaction.
Fe + CuSO4 = Cu + FeSO4
We calculate indicators for all simple and complex substances.
Since Fe and Cu are simple substances, their oxidation state is 0. In CuSO4, then Cu + 2, then oxygen has 2, and sulfur has +6. In FeSO4: Fe +2, therefore, for O-2, according to the 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 given in the reaction. Copper changed its performance from +2 to 0, therefore, copper took 2 electrons. Now we determine the number of received and given electrons by the iron atom and the cation of divalent copper. During the conversion, two electrons were taken by a cation of divalent copper, the same number of electrons were given by an iron atom.
In this process, it makes no sense to determine the minimum total multiple, since an equal number of electrons is received and given during the conversion. Stereochemical coefficients will also correspond to unity. In the reaction, the properties of the reducing agent will be shown by iron, while it is oxidized. The divalent copper cation is reduced to pure copper, in the reaction it has a high oxidation state.
Process application
Formulas for the degree of oxidation should be known to every student in grades 8–9, since this issue is included in the tasks of the OGE. Any processes that occur with oxidizing, reducing signs play an important role in our life. Without them, metabolic processes in the human body are impossible.