Semireaction method: algorithm

Many chemical processes pass with a changeoxidative degrees of atoms, which form reacting compounds. Writing the equations of oxidation-reduction reactions is often accompanied by difficulty in the arrangement of the coefficients before each formula of substances. For these purposes, techniques related to the electron or electron-ion balance of the charge distribution have been developed. The second method of constructing equations is described in detail in the article.

Semireaction method, essence

It is also called the electron-ion balancedistribution factor coefficients. A method is based on the exchange of negatively charged particles between anions or cations in dissolved media with different values of the hydrogen index.

In the reactions of the electrolytes of the oxidative andions of negative or positive charge participate in the reduction process. The equations of the molecular ion type, based on the half-reaction method, clearly demonstrate the essence of any process.

To form a balance, use a specialthe designation of strong-link electrolytes as ionic particles, and weak compounds, gases and sediments in the form of undissociated molecules. In the composition of the scheme it is necessary to indicate particles in which the degree of their oxidation varies. To determine the dissolution medium in the balance, acid (H⁺), alkaline (OH^-) and neutral (H₂O) conditions.

For what use?

In OVR, the half-reaction method is aimed at writing ionic equations separately for the oxidative and reducing processes. The final balance will be their summation.

Stages of implementation

The half-reaction method has its own peculiarities of writing. The algorithm includes the following steps:

- The first thing to do is write down the formulas for all the reactants. For example:

H₂S + KMnO₄ + HCl

- Then it is necessary to establish a function, from a chemical point of view, of each component process. In this reaction KMnO₄acts as an oxidizing agent, H₂S is a reducing agent, and HCl determines the acid medium.

- The third stage needs to be written from a new lineformulas of ionic reacting compounds with a strong electrolyte potential, whose atoms exhibit a change in the degrees of their oxidation. In this interaction, MnO₄^-acts as an oxidizing agent, H₂S is a reducing reagent, and H⁺ or an oxonium cation H₃O⁺ determines the acid medium. Gaseous, solid or weak electrolytic compounds are expressed as whole molecular formulas.

Knowing the original components, try to determine,which of the oxidizing and reducing reagents will be the reduced and oxidized form, respectively. Sometimes final substances are already set in conditions, which facilitates the work. In the following equations, the transition H₂S (hydrogen sulphide) in S (sulfur), and the anion MnO₄^- in the cation Mn²⁺.

To balance the atomic particles in the left and right sections, the hydrogen cation H⁺ or molecular water. To the alkaline solution, hydroxide ions OH^- or H₂O.

MnO₄^-→ Mn²⁺

In solution, the oxygen atom from manganate ions together with H⁺ form water molecules. To equalize the number of elements, the equation is written as: 8H⁺+ MnO₄^- → 4H₂O + Mn²⁺.

Then electric balancing is performed. To do this, consider the total amount of charges in the left area, it turns out to be +7, and then on the right side, it's +2. To balance the process to the starting materials, five negative particles are added: 8H⁺+ MnO₄^- + 5e^-→ 4H₂O + Mn²⁺. A half reaction recovery is obtained.

Now, the oxidation process follows the number of atoms. For this, hydrogen cations are added to the right-hand side: H₂S → 2H⁺ + S.

After the charge is equalized: H₂S -2e^- → 2H⁺ + S. It is seen that two negative particles are taken from the initial compounds. A half reaction of the oxidizing process is obtained.

Write both equations in a column and aligngiven and received charges. By the rule of determining the smallest multiple, each factor is chosen for each half reaction. The oxidizing and reducing equation is multiplied by it.

Now it is possible to summarize the two balances by adding the left and right sides to each other and reducing the number of electron particles.

8H⁺+ MnO₄^- + 5e^-→ 4H₂O + Mn²⁺ | 2

H₂S -2e^- → 2H⁺ + S | 5

16H⁺ + 2MnO₄^- + 5H₂S → 8H₂O + 2Mn²⁺ + 10H⁺ + 5S

In the resulting equation, the number H⁺ reduce by 10: 6H⁺ + 2MnO₄^- + 5H₂S → 8H₂O + 2Mn²⁺ + 5S.

We check the correctness of the composition of the ionicbalance by counting the number of oxygen atoms before and after the arrow, which is 8. It is also necessary to compare the charges of the final and initial parts of the balance: (+6) + (-2) = +4. If everything is the same, then it is compiled correctly.

The half-reaction method ends with a transition fromion recording to the molecular equation. For each anionic and cationic particle of the left side of the balance, the ion opposite in charge is selected. Then they are transferred to the right side, in the same amount. Now the ions can be combined into whole molecules.

6H⁺ + 2MnO₄^- + 5H₂S → 8H₂O + 2Mn²⁺ + 5S

6Cl^- + 2K⁺ → 6Cl^- + 2K⁺

H₂S + KMnO₄ + 6HCl → 8H₂O + 2MnCl₂ + 5S + 2KCl.

To apply the half-reaction method, whose algorithm reduces to writing a molecular equation, one can, along with writing electronic type balances.

Determination of oxidants

Such a role belongs to ionic, atomic ormolecular particles that accept negatively charged electrons. Oxidizing substances undergo restoration in reactions. They have an electronic defect, which can easily be replenished. Such processes include redox half-reactions.

Not all substances have the ability to attach electrons. To strong oxidizing reagents include:

halogenated representatives;
acid type nitric, selenium and sulfuric;
Potassium permanganate, dichromate, manganate, chromate;
manganese and lead tetravalent oxides;
silver and gold are ionic;
oxygen gas compounds;
copper bivalent and silver monovalent oxides;
chlorine-containing salt components;
royal vodka;
hydrogen peroxide.

Determination of reducing agents

Such a role belongs to ionic, atomic ormolecular particles that give off a negative charge. In reactions, the reducing substances undergo an oxidative action when electrons are split off.

Restorative properties possess:

representatives of many metals;
sulfur of the tetravalent compound and hydrogen sulphide;
halogenated acids;
iron, chromium and manganese sulfates;
tin bivalent chloride;
nitrogen-containing reagents such as nitric acid, divalent oxide, ammonia and hydrazine;
Natural carbon and its oxide are divalent;
hydrogen molecules;
acid phosphorous.

Advantages of the electron-ion method

To write oxidation-reduction reactions, the half-reaction method is used more often than the balance of the electron species.

method of half reactions in an alkaline medium

This is due to the advantages electron-ion method:

At the time of writing, the equations consider the real ions and compounds that exist in the solution.
You can initially not have information about the resulting substances, they are determined at the final stages.
Data on the oxidative degree are not always necessary.
Thanks to the method, it is possible to find out the number of electrons that participate in half-reactions, as the hydrogen index of the solution changes.
By the reduced equations of the ionic species, the peculiarities of the processes and the structure of the resulting substances are studied.

Half reactions in acidic solution

Carrying out calculations with an excess of hydrogenions obeys the basic algorithm. The method of half reactions in an acid medium begins with the recording of the constituents of any process. Then they are expressed in the form of equations of the ionic form with observance of the balance of the atomic and electronic charge. The processes of oxidative and reducing nature are recorded separately.

To equalize the atomic oxygen in the direction of the reactions with its excess, hydrogen cations are added. Number of H⁺ should be enough to get molecular water. In the direction of lack of oxygen, H₂O.

Then balance the hydrogen atoms and electrons.

Summarize the parts of the equations before and after the arrow with the distribution of the coefficients.

Reduction of identical ions andmolecules. To the already recorded reagents, the addition of the missing anionic and cationic species is performed in the total equation. Their number after and before the arrow should coincide.

The OVR equation (the half-reaction method) is considered to be satisfied when writing a ready-made expression of the molecular form. Each component must have a certain multiplier.

Examples for acidic media

The interaction of sodium nitrite with acidchlorinated leads to the production of sodium nitrate and hydrochloric acid. To arrange the coefficients, the half-reaction method is used, examples of writing equations are associated with indicating the acidic medium.

NaNO₂ + HClO₃ → NaNO₃ + HCl

ClO₃^- + 6H⁺ + 6e^- → 3H₂O + Cl^- | 1

NO₂^- + H₂O-2e^- → NO₃^- + 2H⁺ | 3

ClO₃^- + 6H⁺ + 3H₂O + 3NO₂^-→ 3H₂O + Cl^- + 3NO₃^- + 6H⁺

ClO₃^- + 3NO₂^-→ Cl^- + 3NO₃^-

3Na⁺ + H⁺ → 3Na⁺ + H⁺

3NaNO₂ + HClO₃ → 3NaNO₃ + HCl.

In this process, nitrate is used to produce nitratesodium, and hydrochloric acid is formed from chloric acid. The oxidative degree of nitrogen varies from +3 to +5, and the charge of chlorine +5 becomes -1. Both products do not form a precipitate.

Half-reactions for alkaline medium

Carrying out calculations with an excess of hydroxideions corresponds to calculations for acidic solutions. The method of half-reactions in an alkaline medium also begins with the expression of the constituent parts of the process in the form of ionic equations. Differences are observed during the alignment of the number of atomic oxygen. Thus, molecular water is added to the reaction with its excess, and the hydroxide anions are added to the opposite part.

The coefficient in front of the molecule H₂O shows the difference in the amount of oxygen after and before the arrow, and for OH ions^- it's doubled. During oxidation, the reagent acting as a reducing agent takes O atoms away from the hydroxyl anions.

The half-reaction method ends with the remaining stages of the algorithm, which coincide with processes that have an acid excess. The end result is the equation of the molecular type.

Examples for alkaline medium

When iodine is mixed with sodium hydroxideformed sodium iodide and iodate, a water molecule. To obtain the balance of the process, the half-reaction method is used. Examples for alkaline solutions have their own specific features related to the equalization of atomic oxygen.

NaOH + I₂→ NaI + NaIO₃ + H₂O

I + e^- → I^- | 5

6OH^- + I - 5e^- → I^-+ 3H₂O + IO₃^- | 1

I + 5I + 6OH^- → 3H₂O + 5I^- + IO₃^-

6Na⁺ → Na⁺ + 5Na⁺

6NaOH + 3I₂→ 5NaI + NaIO₃ + 3H₂O.

The result of the reaction is the disappearance of violet staining of molecular iodine. There is a change in the degree of oxidation of this element from 0 to -1 and +5 with the formation of iodide and sodium iodate.

Reactions in a neutral medium

Usually, these are the processes that occur during the hydrolysis of salts with the formation of a weakly acidic (with a hydrogen index of 6 to 7) or a slightly alkaline solution (with a pH of 7 to 8).

The method of half reactions in a neutral medium is recorded in several variants.

The first method does not take into account salt hydrolysis. The medium is assumed to be neutral, and molecular water is attributed to the left of the arrow. In this case, one half-reaction is taken as acidic and the other half as alkaline.

The second method is suitable for processes in which an approximate value of the hydrogen index can be set. Then the reactions for the ion-electron method are considered in an alkaline or acidic solution.

Example with a neutral medium

When hydrogen sulfide is combined with sodium dichromate in water, a precipitate of sulfur, sodium and chromium of trivalent hydroxides is obtained. This is a typical reaction for a neutral solution.

Na₂Cr₂O₇ + H₂S + H2O → NaOH + S + Cr (OH)₃

H₂S-2e^- → S + H⁺ | 3

7H₂O + Cr₂O₇^2- + 6e^- → 8OH^-+ 2Cr (OH)₃ | 1

7H₂O + 3H₂S + Cr₂O₇^2- → 3H⁺ + 3S + 2Cr (OH)₃ + 8OH^-. Hydrogen cations and hydroxide-anions, when combined, form 6 water molecules. They can be removed in the right and left parts, leaving an excess in front of the arrow.

H₂O + 3H₂S + Cr₂O₇^2- → 3S + 2Cr (OH)₃ + 2OH^-

2Na⁺ → 2Na⁺

Na₂Cr₂O₇ + 3H₂S + H₂O → 2NaOH + 3S + 2Cr (OH)₃

At the end of the reaction, a precipitate is formed from the hydroxidechromium blue and yellow sulfur in an alkaline solution with sodium hydroxide. The oxidative degree of the element S with -2 becomes 0, and the chromium charge with +6 turns into +3.