Many chemical processes take place with a change in the oxidation states of the atoms that form the reacting compounds. Writing equations for reactions of the redox type is often accompanied by difficulty in arranging the coefficients in front of each formula of substances. For these purposes, techniques have been developed related to the electronic or electron-ion balance of charge distribution. The article describes in detail the second way of writing equations.
Semi-reaction method, entity
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 pH values.
In the reactions of electrolytes of the oxidizing and reducing type, ions with a negative or positive charge participate. Molecular-ionic equationstypes, based on the method of semi-reactions, clearly prove the essence of any process.
To form a balance, a special designation of electrolytes of a strong link is used as ionic particles, and weak compounds, gases and precipitation in the form of undissociated molecules. As part of the scheme, it is necessary to indicate the particles in which the degree of their oxidation changes. To determine the solvent medium in the balance, acidic (H+), alkaline (OH-) and neutral (H2 O) conditions.
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.
Execution steps
The half-reaction method has its own peculiarities of writing. The algorithm includes the following stages:
- The first step is to write down the formulas of all reactants. For example:
H2S + KMnO4 + HCl
- Then you need to establish the function, from a chemical point of view, of each constituent process. In this reaction, KMnO4 acts as an oxidizing agent, H2S is a reducing agent, and HCl defines an acidic environment.
- The third step is to write down from a new line the formulas of ionic reacting compounds with a strong electrolyte potential, the atoms of which have a change in their oxidation states. In this interaction, MnO4- acts as an oxidizing agent, H2S isreducing reagent, and H+ or oxonium cation H3O+ determines the acid environment. Gaseous, solid or weak electrolytic compounds are expressed by whole molecular formulas.
Knowing the initial components, try to determine which oxidizing and reducing reagents will have reduced and oxidized forms, respectively. Sometimes the final substances are already set in the conditions, which makes the work easier. The following equations indicate the transition of H2S (hydrogen sulfide) to S (sulfur), and the anion MnO4-to Mn cation2+.
To balance the atomic particles in the left and right sections, hydrogen cation H+ or molecular water is added to the acid medium. Hydroxide ions OH- or H2O.
are added to the alkaline solution
MnO4-→ Mn2+
In solution, an oxygen atom from manganate ions together with H+ form water molecules. To equalize the number of elements, the equation is written as follows: 2O + Mn2+.
Then electrical balancing is carried out. To do this, consider the total amount of charges in the left section, it turns out +7, and then in the right side, it turns out +2. To balance the process, five negative particles are added to the starting substances: 8H+ + MnO4- + 5e - → 4H2O + Mn2+. This results in a reduction half-reaction.
Now the oxidation process follows to equalize the number of atoms. For this, on the right sideadd hydrogen cations: H2S → 2H+ + S.
After the charges are equalized: H2S -2e- → 2H+ + S. It can be seen that two negative particles are taken away from the starting compounds. It turns out the half-reaction of the oxidative process.
Write down both equations in a column and equalize the given and received charges. According to the rule for determining the smallest multiples, a multiplier is selected for each half-reaction. The oxidation and reduction equation is multiplied by it.
Now you can add the two balances by adding the left and right sides together and reducing the number of electron particles.
8H+ + MnO4- + 5e-→ 4H2O + Mn2+ |2
H2S -2e- → 2H+ + S |5
16H+ + 2MnO4- + 5H2 S → 8H2O + 2Mn2+ + 10H+ + 5S
In the resulting equation, you can reduce the number H+ by 10: 6H+ + 2MnO4 - + 5H2S → 8H2O + 2Mn2+ + 5S.
Checking the correctness of the ion balance by counting the number of oxygen atoms before and after the arrow, which is equal to 8. It is also necessary to check the charges of the final and initial parts of the balance: (+6) + (-2)=+4. If everything matches, then it is correct.
The half-reaction method ends with the transition from ionic notation to a molecular equation. For each anionic andcationic particle of the left side of the balance, an ion opposite in charge is selected. Then they are transferred to the right side, in the same amount. Now ions can be combined into whole molecules.
6H+ + 2MnO4- + 5H2 S → 8H2O + 2Mn2+ + 5S
6Cl- + 2K+ → 6Cl- + 2K +
H2S + KMnO4 + 6HCl → 8H2O + 2MnCl 2 + 5S + 2KCl.
It is possible to apply the method of half-reactions, the algorithm of which boils down to writing a molecular equation, along with writing electronic type balances.
Determination of oxidizing agents
This role belongs to ionic, atomic or molecular particles that accept negatively charged electrons. Substances that oxidize undergo reduction in reactions. They have an electronic deficiency that can be easily filled. Such processes include redox half-reactions.
Not all substances have the ability to accept electrons. Strong oxidizing agents include:
- halogen representatives;
- acid like nitric, selenic and sulfuric;
- potassium permanganate, dichromate, manganate, chromate;
- manganese and lead tetravalent oxides;
- silver and gold ionic;
- gaseous oxygen compounds;
- divalent copper and monovalent silver oxides;
- chlorine-containing s alt 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 action when electrons are eliminated.
Restorative properties have:
- representatives of many metals;
- sulfur tetravalent compounds and hydrogen sulfide;
- halogenated 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 divalent oxide;
- hydrogen molecules;
- phosphorous acid.
Advantages of the electron-ion method
To write redox reactions, the half-reaction method is used more often than the electronic form balance.
This is due to the advantages of the electron-ion method:
- When writing an equation, consider the real ions and compounds that exist in the solution.
- You may not initially have information about the resulting substances, they are determined at the final stages.
- Oxidation grade data is 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.
- Singularityprocesses and the structure of the resulting substances.
Half-reactions in acid solution
Carrying out calculations with an excess of hydrogen ions obeys the main algorithm. The method of half-reactions in an acid medium begins with the recording of the constituent parts of any process. Then they are expressed in the form of equations of the ionic form with the balance of atomic and electronic charge. Processes of an oxidizing and reducing nature are recorded separately.
To equalize atomic oxygen in the direction of reactions with its excess, hydrogen cations are introduced. The amount of H+ should be enough to obtain molecular water. In the direction of lack of oxygen, H2O.
Then carry out the balance of hydrogen atoms and electrons.
They sum up the parts of the equations before and after the arrow with the arrangement of the coefficients.
Reduce identical ions and molecules. The missing anionic and cationic particles are added to the already recorded reagents in the overall equation. Their number after and before the arrow must match.
The OVR equation (half-reaction method) is considered fulfilled when writing a ready-made expression of a molecular form. Each component must have a certain multiplier.
Examples for sour environments
The interaction of sodium nitrite with chloric acid leads to the production of sodium nitrate and hydrochloric acid. To arrange the coefficients, the method of semi-reactions is used, examples of writing equationsassociated with indicating an acidic environment.
NaNO2 + HClO3 → NaNO3 + HCl
ClO3- + 6H+ + 6e- → 3H2O + Cl- |1
NO2- + H2O – 2e- → NO3- +2H+ |3
ClO3- + 6H+ + 3H2 O + 3NO2- → 3H2O + Cl - + 3NO3- +6H+
ClO3- + 3NO2-→ Cl- + 3NO3-
3Na+ + H+ → 3Na+ + H +
3NaNO2 + HClO3 → 3NaNO3 + HCl.
In this process, sodium nitrate is formed from nitrite, and hydrochloric acid is formed from chloric acid. The oxidation state of nitrogen changes from +3 to +5, and the charge of chlorine +5 becomes -1. Both products do not precipitate.
Semi-reactions for alkaline medium
Carrying out calculations with an excess of hydroxide ions 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. So, molecular water is added to the side of the reaction with its excess, and hydroxide anions are added to the opposite side.
The coefficient in front of the H2O molecule shows the difference in the amount of oxygen after and before the arrow, and for OH- ions it is doubled. During the oxidationa reagent that acts as a reducing agent removes O atoms from hydroxyl anions.
The method of half-reactions ends with the remaining steps of the algorithm, which coincide with processes that have an acidic excess. The end result is a molecular equation.
Alkaline examples
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 specifics associated with the equalization of atomic oxygen.
NaOH + I2 →NaI + NaIO3 + H2O
I + e- → I- |5
6OH- + I - 5e- → I- + 3H 2O + IO3- |1
I + 5I + 6OH- → 3H2O + 5I- + IO 3-
6Na+ → Na+ + 5Na+
6NaOH + 3I2 →5NaI + NaIO3 + 3H2O.
The result of the reaction is the disappearance of the violet color of molecular iodine. There is a change in the oxidation state of this element 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 take place during the hydrolysis of s alts with the formation of a slightly acidic (with a pH 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 down in severaloptions.
The first method does not take into account s alt hydrolysis. The medium is taken as neutral, and molecular water is attributed to the left of the arrow. In this version, 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 pH value. Then the reactions for the ion-electron method are considered in an alkaline or acidic solution.
Neutral environment example
When hydrogen sulfide is combined with sodium dichromate in water, a precipitate of sulfur, sodium and trivalent chromium hydroxides is obtained. This is a typical reaction for a neutral solution.
Na2Cr2O7 + H2 S +H2O → NaOH + S + Cr(OH)3
H2S - 2e- → S + H+ |3
7H2O + Cr2O72- + 6e- → 8OH- + 2Cr(OH)3 |1
7H2O +3H2S + Cr2O 72- → 3H+ +3S + 2Cr(OH)3 +8OH-. Hydrogen cations and hydroxide anions combine to form 6 water molecules. They can be removed on the right and left sides, leaving the excess in front of the arrow.
H2O +3H2S + Cr2O 72- → 3S + 2Cr(OH)3 +2OH-
2Na+ → 2Na+
Na2Cr2O7 + 3H2 S +H2O → 2NaOH + 3S + 2Cr(OH)3
At the end of the reaction, a precipitate of blue chromium hydroxide and yellowsulfur in alkaline solution with sodium hydroxide. The oxidation state of the element S with -2 becomes 0, and the chromium charge with +6 becomes +3.
