Semireaction method: algorithm

Many chemical processes pass with a change in the oxidative degrees of atoms, which form reactive 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 electronic or electron-ion balance of the charge distribution have been developed. The second method of constructing equations is described in detail in the article.

Semi-reaction method, essence

It is also called the electron-ion balance of the distribution of coefficient factors. 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 oxidizing and reducing electrolytes, ions with a negative or positive charge participate. The molecular-ionic equations, based on the half-reaction method, clearly demonstrate the essence of any process.

To form a balance, a special designation of strong link electrolytes is used 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 oxidation varies. To determine the dissolution medium in the balance are designated acid (H ⁺ ), alkaline (OH ^- ) and neutral (H ₂ O) conditions.

For what use?

In OVR, the half-reaction method is aimed at writing ionic equations separately for 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:

H ₂ S + KMnO ₄ + HCl

- Then it is necessary to establish the function, from a chemical point of view, of each component process. In this reaction, KMnO ₄ acts as an oxidizing agent, H ₂ S is a reducing agent, and HCl determines the acid medium.

- The third step is to write down a new line of formula ionic reacting compounds with a strong electrolyte potential, whose atoms are observed to change the degrees of their oxidation. In this interaction, MnO ₄ acts as an oxidizing agent, H ₂ S is a reducing reagent, and H ⁺ or the oxonium cation H ₃ O ⁺ determines the acidic medium. Gaseous, solid or weak electrolytic compounds are expressed by molecular formulas as whole.

Knowing the original components, try to determine which oxidizing and reducing agent will have the reduced and oxidized form, respectively. Sometimes final substances are already set in conditions, which facilitates the work. In the following equations, the transition of H ₂ S (hydrogen sulphide) to S (sulfur) and the MnO ₄ anion to the Mn ²⁺ cation is indicated.

To balance the atomic particles in the left and right sections, hydrogen cation H ⁺ or molecular water is added to the acid medium. To the alkaline solution, hydroxide ions OH ^- or H ₂ O are added.

MnO ₄ ^- → Mn ²⁺

In a solution, the oxygen atom from manganate ions, together with H ^{+, is} formed by water molecules. To equalize the number of elements, the equation is written as: 8H ⁺ + MnO ₄ ^- → 4H ₂ O + Mn ²⁺ .

Then electric balancing is carried out. To do this, consider the total amount of charges in the left area, it turns out +7, and then on the right side, it's +2. To balance the process, five negative particles are added to the starting materials: 8H ⁺ + MnO ₄ ^- + 5e ^- → 4H ₂ O + Mn ²⁺ . 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: H ₂ S → 2H ⁺ + S.

After equalization of charges: H ₂ S -2e ^- → 2H ⁺ + S. It is seen that two negative particles are taken from the initial compounds. It turns out a half reaction of the oxidation process.

Record both equations in a column and equalize the given and received charges. By the rule of determining the least multiple, one chooses a different factor 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 ⁺ + MnO ₄ ^- + 5e ^- → 4H ₂ O + Mn ²⁺ | 2

H ₂ S -2e ^- → 2H ⁺ + S | 5

16H ⁺ + 2MnO ₄ ^- + 5H ₂ S → 8H ₂ O + 2Mn ²⁺ + 10H ⁺ + 5S

In the obtained equation, the number H ⁺ can be reduced by 10: 6H ⁺ + 2MnO ₄ ^- + 5H ₂ S → 8H ₂ O + 2Mn ²⁺ + 5S.

We check the correctness of the composition of the ion balance 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 from the ion record 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 ⁺ + 2MnO ₄ ^- + 5H ₂ S → 8H ₂ O + 2Mn ²⁺ + 5S

6Cl ^- + 2K ⁺ → 6Cl ^- + 2K ⁺

H ₂ S + KMnO ₄ + 6HCl → 8H ₂ O + ₂ MnCl ₂ + 5S + 2KCl.

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

Determination of oxidants

This role belongs to ionic, atomic or molecular 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 oxidation-reduction half-reactions.

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

• Halogenated representatives;
• Acid type of 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 or molecular particles that give off a negative charge. In reactions, the reducing substances undergo oxidative action when electrons are split off.

Restorative properties are :

• 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.

This is due to the advantages Electron-ion method :

1. At the time of writing, the equations consider the real ions and compounds that exist in the solution.
2. You can initially not have information about the resulting substances, they are determined at the final stages.
3. Data on the oxidative degree are not always necessary.
4. 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.
5. 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

The calculation of the excess hydrogen ions 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 oxidation and reduction are separately recorded.

To equalize the atomic oxygen in the direction of the reactions with its excess, hydrogen cations are added. The quantity of H ⁺ should be sufficient to obtain molecular water. To the side of lack of oxygen is attributed H ₂ O.

Then balance the hydrogen atoms and electrons.

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

Reduce the same ions and molecules. To the already recorded reagents, the addition of 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 chloric acid 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.

NaNO ₂ + HClO ₃ → NaNO ₃ + HCl

ClO ₃ ^- + 6H ⁺ + 6e ^- → 3H ₂ O + Cl ^- | 1

NO ₂ ^- + H ₂ O - 2e ^- → NO ₃ ^- + 2H ⁺ | 3

ClO ₃ ^- + 6H ⁺ + 3H ₂ O + 3NO ₂ ^- → 3H ₂ O + Cl ^- + 3NO ₃ ^- + 6H ⁺

ClO ₃ ^- + 3NO ₂ ^- → Cl ^- + 3NO ₃ ^-

3Na ⁺ + H ⁺ → 3Na ⁺ + H ⁺

3NaNO ₂ + HClO ₃ → 3NaNO ₃ + HCl.

In this process, sodium nitrate is obtained from nitrite, 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 hydroxide ions corresponds to the 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 H ₂ O molecule shows the difference in the amount of oxygen after and before the arrow, and for OH ions it is 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 steps 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 hydroxide, sodium iodide and iodate, water molecules, are formed. 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 + I ₂ → NaI + NaIO ₃ + H ₂ O

I + e ^- → I ^- | 5

6OH ^- + I - 5e ^- → I ^- + 3H ₂ O + IO ₃ ^- | 1

I + 5I + 6OH ^- → 3H ₂ O + 5 I ^- + IO ₃ ^-

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 another - for alkaline reaction.

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.

Na ₂ Cr ₂ O ₇ + H ₂ S + H2 O → NaOH + S + Cr (OH) ₃

H ₂ S - 2e ^- → S + H ⁺ | 3

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

7H ₂ O + 3H ₂ S + Cr ₂ O ₇ ^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.

H ₂ O + 3H ₂ S + Cr ₂ O ₇ ^2- → 3S + 2 Cr (OH) ₃ + 2OH ^-

2Na ⁺ → 2Na ⁺

Na ₂ Cr ₂ O ₇ + 3H ₂ S + H ₂ O → 2NaOH + 3S + 2 Cr (OH) ₃

At the end of the reaction, a precipitate is formed from chromium hydroxide of blue color 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.