Many chemical processes occur with a change in the oxidative degrees of atoms that form reacting compounds. Writing the equations of the reactions of the redox type is often accompanied by difficulty in arranging the coefficients in front of each formula of substances. For these purposes, methods have been developed related to the electronic or electron-ion balance of the charge distribution. The article describes in detail the second way of making equations.
Half-reaction method, essence
It is also called the electron-ion balance of the distribution of coefficient factors. The 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 reactions of oxidizing and reducing type electrolytes, ions with a negative or positive charge are involved. The equations of the molecular-ionic type, which are based on the half-reaction method, clearly demonstrate the essence of any process.
To form the balance, a special designation of electrolytes of the strong link is used as ionic particles, and weak compounds, gases and precipitates in the form of undissociated molecules. As part of the scheme, it is necessary to indicate particles in which the degree of their oxidation changes. To determine the solvent medium, acid (H + ), alkaline (OH - ), and neutral (H 2 O) conditions are indicated in the balance.
What is it used for?
In OVR, the half-reaction method is aimed at writing ionic equations separately for oxidative and reduction processes. The final balance will be their summation.
Stages of execution
The half-reaction method has its own writing features. The algorithm includes the following stages:
- The first step is to write down the formulas of all reacting substances. For instance:
H 2 S + KMnO 4 + HCl
- Then it is necessary to establish the function, from a chemical point of view, of each component of the process. In this reaction, KMnO 4 acts as an oxidizing agent, H 2 S is a reducing agent, and HCl determines the acidic environment.
- The third step is to write from a new line the formulas of ionic reacting compounds with a strong electrolyte potential, the atoms of which exhibit a change in their oxidation states. In this interaction, MnO 4 - acts as an oxidizing agent, H 2 S is a reducing agent, and H + or the oxonium cation H 3 O + determines the acidic environment. Gaseous, solid or weak electrolytic compounds are expressed in whole molecular formulas.
Knowing the initial components, try to determine what the oxidizing and reducing reagent will be reduced and oxidized form, respectively. Sometimes the final substances are already set in conditions, which facilitates the work. The following equations indicate the transition of H 2 S (hydrogen sulfide) to S (sulfur), and the MnO 4 anion to the Mn 2+ cation.
To balance the atomic particles in the left and right sections, the hydrogen cation H + or molecular water is added to the acidic medium. OH - or H 2 O hydroxide ions are added to the alkaline solution.
MnO 4 - → Mn 2+
In a solution, an oxygen atom from manganate ions together with H + form water molecules. To equalize the number of elements, the equation is written as: 8H + + MnO 4 - → 4H 2 O + Mn 2+ .
Then carry out electrical balancing. To do this, consider the total amount of charges in the left section, it turns out +7, and then in the right side, +2 comes out. To balance the process, five negative particles are added to the starting materials: 8H + + MnO 4 - + 5e - → 4H 2 O + Mn 2+ . It turns out half-recovery.
Now, the oxidation process follows in terms of the number of atoms. To do this, add hydrogen cations to the right side: H 2 S → 2H + + S.
After the charges are equalized: H 2 S -2e - → 2H + + S. It is seen that two negative particles are taken from the starting compounds. The result is a half-reaction of the oxidative process.
Both equations are written in a column and the given and received charges are aligned. According to the rule of determining the smallest multiples, they select their factor for each half-reaction. The oxidizing and reducing equation is multiplied by it.
Now you can summarize the two balances by folding the left and right sides together and reducing the number of electronic 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 H + number can be reduced by 10: 6H + + 2MnO 4 - + 5H 2 S → 8H 2 O + 2Mn 2+ + 5S.
We check the correctness of the ion balance by calculating the number of oxygen atoms before and after the arrow, which is 8. It is also necessary to verify the charges of the final and initial parts of the balance: (+6) + (-2) = +4. If everything matches, then it is composed correctly.
The half-reaction method ends with a transition from ionic writing to the molecular equation. For each anionic and cationic particle on the left side of the balance, an ion of the opposite charge is selected. Then they are transferred to the right side, in the same amount. Now 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 + 2MnCl 2 + 5S + 2KCl.
The half-reaction method, the algorithm of which reduces to writing the molecular equation, can be used along with writing electronic-type balances.
Determination of oxidizing agents
Such a role belongs to ionic, atomic or molecular particles that accept negatively charged electrons. Oxidizing substances undergo reduction in reactions. They have an electronic flaw that can be easily filled. Such processes include redox half-reactions.
Not all substances have the ability to attach electrons. Strong oxidizing agents include:
- halogen representatives;
- acid such as nitric, selenium and sulfuric;
- potassium permanganate, dichromate, manganate, chromate;
- manganese and lead tetravalent oxides;
- silver and ionic gold;
- gaseous oxygen compounds;
- copper divalent and silver monovalent oxides;
- chlorine-containing salt components;
- royal vodka;
- hydrogen peroxide.
Determination of reducing agents
This role belongs to ionic, atomic or molecular particles that give off a negative charge. In reactions, reducing substances undergo an oxidizing effect upon the elimination of electrons.
Restorative properties have :
- representatives of many metals;
- tetravalent sulfur and hydrogen sulfide;
- halogen-containing acids;
- iron, chromium and manganese sulfates;
- tin divalent chloride;
- nitrogen-containing reagents such as nitrous acid, divalent oxide, ammonia and hydrazine;
- natural carbon and its bivalent oxide;
- hydrogen molecules;
- phosphorous acid.
The advantages of the electron-ion method
To write redox reactions, the half-reaction method is used more often than the balance of the electronic form.
This is due to the benefits. electron-ion method :
- At the time of writing the equations, real ions and compounds that exist in the composition of the solution are considered.
- You may not initially have information about the resulting substances, they are determined at the final stages.
- Oxidation data are not always needed.
- Thanks to the method, you can find out the number of electrons that participate in half-reactions, how the pH of the solution changes.
- Using abbreviated equations of the ionic type, 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 hydrogen ions obeys the basic algorithm. The half-reaction method in an acidic environment begins with the recording of the constituent parts of any process. Then they are expressed in the form of equations of the ionic form in compliance with the balance of the atomic and electronic charge. Separately, processes of an oxidizing and reducing nature are recorded.
To align atomic oxygen to the side of reactions with its excess, hydrogen cations are introduced. Amounts of H + should be enough to produce molecular water. In the direction of oxygen deficiency, H 2 O is attributed.
Then carry out the balance of hydrogen atoms and electrons.
Do the summation of the parts of the equations before and after the arrow with the arrangement of the coefficients.
Carry out the reduction of the same ions and molecules. To the already recorded reagents in the total equation, the missing anionic and cationic particles are added. Their number after and before the arrow should match.
The OVR equation (the half-reaction method) is considered to be fulfilled when writing the finished expression of the molecular form. Each component should have a specific factor.
Examples for acidic environment
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 an indication of 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 changes from +3 to +5, and the charge of chlorine +5 becomes -1. Both products do not form a precipitate.
Half reactions for alkaline environment
Carrying out calculations with an excess of hydroxide ions corresponds to the calculations for acidic solutions. The half-reaction method 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 atomic oxygen numbers. So, molecular water is added to the reaction side with its excess, and hydroxide anions are added to the opposite side.
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, which acts as a reducing agent, takes away O atoms from hydroxyl anions.
The half-reaction method ends with the remaining steps of the algorithm, which coincide with processes having an acidic excess. The end result is a molecular equation.
Examples for alkaline environment
When iodine is mixed with sodium hydroxide, sodium iodide and iodate, water molecules, are formed. To obtain the balance of the process using the half-reaction method. Examples for alkaline solutions have their own specifics 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 + 5I - + IO 3 -
6Na + → Na + + 5Na +
6NaOH + 3I 2 → 5NaI + NaIO 3 + 3H 2 O.
The result of the reaction is the disappearance of the violet staining of molecular iodine. The oxidation state of this element changes from 0 to -1 and +5 with the formation of sodium iodide and iodate.
Reactions in a neutral environment
Usually this is the name of the processes that occur during hydrolysis of salts with the formation of a slightly acidic (with a hydrogen index of 6 to 7) or slightly alkaline (with a pH of 7 to 8) solution.
The half-reaction method in a neutral medium is written in several variants.
In the first method, salt hydrolysis is not taken into account. The medium is taken as neutral, and to the left of the arrow is attributed molecular water. In this embodiment, one half reaction is taken as acidic, and the other as alkaline.
The second method is suitable for processes in which you can set the approximate value of the hydrogen index. Then the reactions for the ion-electron method are considered in an alkaline or acidic solution.
Neutral Example
When hydrogen sulfide is combined with sodium dichromate in water, a precipitate of sulfur, sodium and chromium trivalent hydroxides is obtained. This is a typical reaction for a neutral solution.
Na 2 Cr 2 O 7 + H 2 S + H2O → 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 the surplus in front of the arrow.
H 2 O + 3H 2 S + Cr 2 O 7 2- → 3S + 2Cr (OH) 3 + 2OH -
2Na + → 2Na +
Na 2 Cr 2 O 7 + 3H 2 S + H 2 O → 2NaOH + 3S + 2Cr (OH) 3
At the end of the reaction, a precipitate is formed from blue chromium hydroxide 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.