Lesson method of electronic balance solution of training exercises. Drawing up equations of redox reactions by the method of electronic balance. Application of the electronic balance method step by step. Example "a"

8. Classification of chemical reactions. OVR. Electrolysis

8.3. Redox reactions: general provisions

redox reactions(OVR) are called reactions that occur with a change in the oxidation state of the atoms of the elements. As a result of these reactions, some atoms donate electrons, while others accept them.

A reducing agent is an atom, ion, molecule or FE that donates electrons, an oxidizing agent is an atom, ion, molecule or FE that accepts electrons:

The process of giving off electrons is called oxidation, and the process of accepting - restoration. In OVR, there must be a reducing agent and an oxidizing agent. There is no oxidation process without a reduction process and there is no reduction process without an oxidation process.

The reducing agent donates electrons and is oxidized, while the oxidizing agent accepts electrons and is reduced.

The reduction process is accompanied by a decrease in the degree of oxidation of atoms, and the oxidation process is accompanied by an increase in the degree of oxidation of atoms of elements. It is convenient to illustrate the above with a diagram (CO - oxidation state):

Specific examples of oxidation and reduction processes (electron balance schemes) are given in Table. 8.1.

Table 8.1

Examples of electronic balance schemes

Scheme of electronic balance	Process characteristic
Oxidation process
	The calcium atom donates electrons, increases the degree of oxidation, is a reducing agent
	Ion Cr +2 donates electrons, increases the degree of oxidation, is a reducing agent
	The chlorine molecule donates electrons, chlorine atoms increase the oxidation state from 0 to +1, chlorine is a reducing agent
Recovery process
	The carbon atom accepts electrons, lowers the oxidation state, is an oxidizing agent
	The oxygen molecule accepts electrons, the oxygen atoms lower their oxidation state from 0 to -2, the oxygen molecule is an oxidizing agent
	The ion accepts electrons, lowers the oxidation state, is an oxidizing agent

The most important reducing agents: simple substances metals; hydrogen; carbon in the form of coke; carbon monoxide(II); compounds containing atoms in the lowest oxidation state (metal hydrides,, sulfides, iodides, ammonia); the strongest reducing agent electricity on the cathode.

The most important oxidizers: simple substances - halogens, oxygen, ozone; concentrated sulfuric acid; Nitric acid; a number of salts (KClO 3 , KMnO 4 , K 2 Cr 2 O 7); hydrogen peroxide H 2 O 2 ; the strongest oxidizing agent is an electric current at the anode.

Over the period, the oxidizing properties of atoms and simple substances are enhanced: fluorine - the strongest oxidizing agent of all simple substances. In each period, halogens form simple substances with the most pronounced oxidizing properties.

In groups A, from top to bottom, the oxidizing properties of atoms and simple substances weaken, while the reducing properties increase.

For atoms of the same type, the reducing properties increase with an increase in their radius; for example, the reducing properties of the anion
I - are more pronounced than the anion Cl - .

For metals, the redox properties of simple substances and ions in an aqueous solution are determined by the position of the metal in the electrochemical series: from left to right (top to bottom), the reducing properties of simple metals weaken: the strongest reducing agent- lithium.

For metal ions in an aqueous solution, from left to right in the same row, respectively, the oxidizing properties are enhanced: most powerful oxidizing agent- Au 3 + ions.

To arrange the coefficients in the OVR, you can use a method based on the mapping of oxidation and reduction processes. This method is called electronic balance method.

The essence of the electronic balance method is as follows.

1. Draw up a reaction scheme and determine the elements that have changed the oxidation state.

2. Compose electronic equations for half-reactions of reduction and oxidation.

3. Since the number of electrons donated by the reducing agent must be equal to the number of electrons accepted by the oxidizing agent, additional factors are found using the least common multiple (LCM) method.

4. Additional multipliers are put down before the formulas of the corresponding substances (coefficient 1 is omitted).

5. Equalize the number of atoms of those elements that have not changed the degree of oxidation (first - hydrogen in water, and then - the number of oxygen atoms).

An example of compiling an equation for a redox reaction

electronic balance method.

We find that the carbon and sulfur atoms have changed their oxidation state. We compose the equations of half-reactions of reduction and oxidation:

For this case, the LCM is 4, and the additional factors are 1 (for carbon) and 2 (for sulfuric acid).

We put down the additional factors found in the left and right parts of the reaction scheme in front of the formulas of substances containing carbon and sulfur:

C + 2H 2 SO 4 → CO 2 + 2SO 2 + H 2 O

We equalize the number of hydrogen atoms by putting a factor of 2 in front of the water formula, and we make sure that the number of oxygen atoms in both parts of the equation is the same. Therefore, the OVR equation

C + 2H 2 SO 4 \u003d CO 2 + 2SO 2 + 2H 2 O

The question arises: in which part of the OVR scheme should the found additional factors be placed - on the left or on the right?

For simple reactions, this does not matter. However, it should be borne in mind: if additional factors are defined on the left side of the equation, then the coefficients are put down before the formulas of substances on the left side; if the calculations were carried out for the right side, then the coefficients are put on the right side of the equation. For example:

According to the number of Al atoms on the left side:

According to the number of Al atoms on the right side:

In the general case, if substances of a molecular structure participate in the reaction (O 2, Cl 2, Br 2, I 2, N 2), then when selecting the coefficients, they proceed precisely from the number of atoms in the molecule:

If N 2 O is formed in a reaction involving HNO 3, then it is also better to write the electron balance scheme for nitrogen based on two nitrogen atoms .

In some redox reactions, one of the substances can perform the function of both an oxidizing agent (reducing agent) and a salt former (i.e., participate in the formation of salt).

Such reactions are typical, in particular, for the interaction of metals with oxidizing acids (HNO 3, H 2 SO 4 (conc)), as well as oxidizing salts (KMnO 4 , K 2 Cr 2 O 7 , KClO 3 , Ca (OCl) 2) with hydrochloric acid (due to Cl anions - hydrochloric acid has reducing properties) and other acids, the anion of which is a reducing agent.

Let's make an equation for the reaction of copper with dilute nitric acid:

We see that part of the nitric acid molecules is spent on the oxidation of copper, while being reduced to nitric oxide (II), and part is used to bind the formed Cu 2+ ions to the salt Cu (NO 3) 2 (in the composition of the salt, the degree of oxidation of the nitrogen atom is the same , as in acid, i.e. does not change). In such reactions, an additional factor for the oxidizing element is always placed on the right side before the reduction product formula, in this case, before the NO formula, and not HNO 3 or Cu(NO 3) 2 .

Before the formula HNO 3 we put a coefficient of 8 (two HNO 3 molecules are spent on the oxidation of copper and six on the binding of three Cu 2+ ions into a salt), we equalize the numbers of H and O atoms and get

3Cu + 8HNO 3 \u003d 3Cu (NO 3) 2 + 2NO + 4H 2 O.

In other cases, an acid, such as hydrochloric acid, can simultaneously be both a reducing agent and participate in the formation of a salt:

Example 8.5. Calculate what mass of HNO 3 is spent on salt formation, when in the reaction, the equation of which

zinc enters with a mass of 1.4 g.

Solution. From the reaction equation, we see that out of 8 moles of nitric acid, only 2 moles went to the oxidation of 3 moles of zinc (there is a factor of 2 in front of the formula for the acid reduction product, NO). Salt formation consumed 6 mol of acid, which is easy to determine by multiplying the coefficient 3 in front of the salt formula Zn(HNO 3) 2 by the number of acid residues in one formula unit of the salt, i.e. on 2.

n (Zn) \u003d 1.4 / 65 \u003d 0.0215 (mol).

x = 0.043 mol;

m (HNO 3) \u003d n (HNO 3) M (HNO 3) \u003d 0.043 ⋅ 63 \u003d 2.71 (g)

Answer: 2.71 g.

In some OVR, the oxidation state is changed by the atoms of not two, but three elements.

Example 8.6. Arrange the coefficients in the OVR flowing according to the scheme FeS + O 2 → Fe 2 O 3 + SO 2 using the electron balance method.

Solution. We see that the oxidation state is changed by the atoms of three elements: Fe, S and O. In such cases, the numbers of electrons donated by atoms of different elements are summed up:

Having placed the stoichiometric coefficients, we obtain:

4FeS + 7O 2 \u003d 2Fe 2 O 3 + 4SO 2.

Consider examples of solving other types of examination tasks on this topic.

Example 8.7. Indicate the number of electrons passing from the reducing agent to the oxidizing agent during the complete decomposition of copper(II) nitrate, with a mass of 28.2 g.

Solution. We write down the reaction equation for the decomposition of salt and the scheme of the electronic balance of the OVR; M = 188 g/mol.

We see that 2 mol O 2 is formed during the decomposition of 4 mol salt. At the same time, 4 mol of electrons pass from the atoms of the reducing agent ( in this case, these are ions) to the oxidizing agent ( i.e. to ions): . Since the chemical amount of salt is n = 28.2/188 = = 0.15 (mol), we have:

2 moles of salt - 4 moles of electrons

0.15 mol - x

n (e) \u003d x \u003d 4 ⋅ 0.15 / 2 \u003d 0.3 (mol),

N (e) \u003d N A n (e) \u003d 6.02 ⋅ 10 23 ⋅ 0.3 \u003d 1.806 ⋅ 10 23 (electrons).

Answer: 1.806 ⋅ 10 23 .

Example 8.8. During the interaction of sulfuric acid with a chemical amount of 0.02 mol with magnesium, sulfur atoms added 7.224 ⋅ 10 22 electrons. Find the formula for the acid recovery product.

Solution. In the general case, the schemes for the processes of reduction of sulfur atoms in the composition of sulfuric acid can be as follows:

those. 1 mole of sulfur atoms can accept 2, 6 or 8 moles of electrons. Given that 1 mol of acid contains 1 mol of sulfur atoms, i.e. n (H 2 SO 4) = n (S), we have:

n (e) \u003d N (e) / N A \u003d (7.224 ⋅ 10 22) / (6.02 ⋅ 10 23) \u003d 0.12 (mol).

We calculate the number of electrons accepted by 1 mol of acid:

0.02 mole of acid accept 0.12 mole of electrons

1 mol - x

n (e) \u003d x \u003d 0.12 / 0.02 \u003d 6 (mol).

This result corresponds to the process of reducing sulfuric acid to sulfur:

Answer: sulfur.

Example 8.9. In the reaction of carbon with concentrated nitric acid, water and two salt-forming oxides are formed. Find the mass of carbon that reacted if the atoms of the oxidizing agent took 0.2 mol of electrons in this process.

Solution. The interaction of substances proceeds according to the reaction scheme

We compose the equations for the half-reactions of oxidation and reduction:

From the schemes of the electronic balance, we see that if the atoms of the oxidizing agent () accept 4 mol of electrons, then 1 mol (12 g) of carbon enters into the reaction. Compose and solve the proportion:

4 moles of electrons - 12 g of carbon

0.2 - x

x = 0.2 ⋅ 12 4 = 0.6 (d).

Answer: 0.6 g.

Classification of redox reactions

There are intermolecular and intramolecular redox reactions.

When intermolecular OVR the atoms of the oxidizing agent and the reducing agent are part of different substances and are atoms of different chemical elements.

When intramolecular OVR The oxidizing and reducing atoms are in the same substance. Intramolecular reactions are disproportionation, in which the oxidizing agent and reducing agent are atoms of the same chemical element in the composition of the same substance. Such reactions are possible for substances containing atoms with an intermediate oxidation state.

Example 8.10. Specify the OVR disproportionation scheme:

1) MnO 2 + HCl → MnCl 2 + Cl 2 + H 2 O

2) Zn + H 2 SO 4 → ZnSO 4 + H 2

3) KI + Cl 2 → KCl + I 2

4) Cl 2 + KOH → KCl + KClO + H 2 O

Solution . Reactions 1)–3) are intermolecular OVR:

The disproportionation reaction is reaction 4), since it contains a chlorine atom and an oxidizing agent and a reducing agent:

Answer: 4).

It is possible to qualitatively assess the redox properties of substances based on the analysis of the oxidation states of atoms in the composition of the substance:

1) if the atom responsible for the redox properties is in the highest degree of oxidation, then this atom can no longer donate electrons, but can only accept them. Therefore, in OVR, this substance will exhibit only oxidizing properties. Examples of such substances (in the formulas the oxidation state of the atom responsible for the redox properties is indicated):

2) if the atom responsible for the redox properties is in the lowest oxidation state, then this substance in the OVR will show only restorative properties(A given atom can no longer accept electrons, it can only give them away). Examples of such substances:,. Therefore, all halogen anions (with the exception of F - for the oxidation of which an electric current at the anode is used), the sulfide ion S 2-, the nitrogen atom in the ammonia molecule, and the hydride ion H - show only reducing properties in the OVR. Metals (Na, K, Fe) have only reducing properties;

3) if an atom of an element is in an intermediate oxidation state (the oxidation state is greater than the minimum, but less than the maximum), then the corresponding substance (ion) will, depending on the conditions, exhibit dual oxidation-restorative properties: stronger oxidizing agents will oxidize these substances (ions), and stronger reducing agents will reduce them. Examples of such substances: sulfur, since highest degree oxidation of the sulfur atom +6, and the lowest -2, sulfur oxide (IV), nitric oxide (III) (the highest oxidation state of the nitrogen atom is +5, and the lowest is -3), hydrogen peroxide (the highest oxidation state of the oxygen atom is +2, and the lowest −2). Dual redox properties are exhibited by metal ions in an intermediate oxidation state: Fe 2+, Mn +4, Cr +3, etc.

Example 8.11. A redox reaction cannot proceed, the scheme of which is:

1) Cl 2 + KOH → KCl + KClO 3 + H 2 O

2) S + NaOH → Na 2 S + Na 2 SO 3 + H 2 O

3) KClO → KClO 3 + KClO 4

4) KBr + Cl 2 → KCl + Br

Solution. The reaction, the scheme of which is indicated at number 3), cannot proceed, since it contains a reducing agent, but no oxidizing agent:

Answer: 3).

For some substances, redox duality is due to the presence in their composition of various atoms in both the lowest and highest oxidation states; for example, hydrochloric acid (HCl) due to the hydrogen atom (highest oxidation state, equal to +1) is an oxidizing agent, and due to the anion Cl - - a reducing agent (lowest oxidation state).

OVR is impossible between substances that exhibit only oxidizing (HNO 3 and H 2 SO 4, KMnO 4 and K 2 CrO 7) or only reducing properties (HCl and HBr, HI and H 2 S)

OVR are extremely common in nature (metabolism in living organisms, photosynthesis, respiration, decay, combustion), are widely used by humans for various purposes (obtaining metals from ores, acids, alkalis, ammonia and halogens, creating chemical current sources, obtaining heat and energy during combustion of various substances). Note that OVR often complicate our lives (spoilage of food, fruits and vegetables, corrosion of metals - all this is associated with the occurrence of various redox processes).

Ion-electronic method (half-reaction method)

When compiling the OVR equations flowing in aqueous solutions, it is preferable to select the coefficients using the half-reaction method.

The procedure for selecting coefficients using the half-reaction method:

1. Write down the reaction scheme in molecular and ionic-molecular forms and determine the ions and molecules that change the oxidation state.

2. Determine the environment in which the reaction proceeds (H + - acidic; OH - alkaline; H 2 O - neutral)

3. Make up an ion-molecular equation for each half-reaction and equalize the number of atoms of all elements.

1. The number of oxygen atoms is equalized using water molecules or OH - ions.
2. If the original ion or molecule contains more oxygen atoms than the reaction product, then

• an excess of oxygen atoms in an acidic environment binds with H + ions into water molecules
• in neutral and alkaline environment excess oxygen atoms are bound by water molecules into OH groups -
  

1. If the original ion or molecule contains fewer oxygen atoms than the reaction product, then

· the lack of oxygen atoms in acidic and neutral solutions is compensated by water molecules

· in alkaline solutions - due to OH - ions.

4. Compose electron-ion equations of half-reactions.

To do this, electrons are added (or subtracted) to the left side of each half-reaction in such a way that the total charge on the left and right sides of the equations becomes the same. We multiply the resulting equations by the smallest factors, for the balance of electrons.

5. Summarize the resulting electron-ion equations. Cancel like terms and get the ion-molecular OVR equation

6. According to the obtained ion-molecular equation, a molecular equation is made.

Example :

1 . Na 2 SO 3 + KMnO 4 + H 2 SO 4 → Na 2 SO 4 + MnSO 4 + K 2 SO 4 + H 2 O

2Na + +SO 3 2- +K + +MnO 4 - +2H + +SO 4 2- →2Na + +SO 4 2- +Mn 2+ +SO 4 2- +2K + +SO 4 2- +H 2 O

SO 3 2- → SO 4 2-

MNO 4 - → Mn 2+

2 . Acidic environment - H +

3 .

MnO 4 - + 8 H + → Mn 2+ + 4 H 2 O

SO 3 2- + H 2 O → SO 4 2- + 2 H +

4 .

MnO 4 - + 8 H + + 5ē → Mn 2+ + 4 H 2 O│ X2

SO 3 2- + H 2 O - 2ē → SO 4 2- + 2 H + │ X5

5 .

2MnO 4 - + 16 H + + 10ē →2Mn 2+ + 8 H 2 O

5SO 3 2- + 5H 2 O - 10ē → 5SO 4 2- + 10 H +

2MnO 4 - + 16 H + + 5SO 3 2- + 5H 2 O →2Mn 2+ + 8 H 2 O + 5SO 4 2- + 10 H +

2MnO 4 - + 6 H + + 5SO 3 2- →2Mn 2+ + 3 H 2 O + 5SO 4 2-

6 . 5Na 2 SO 3 + 2KMnO 4 + 3H 2 SO 4 → 5Na 2 SO 4 + 2MnSO 4 + K 2 SO 4 + 3H 2 O

Reminder!

Restorers
Name of the reducing agent (oxidizing agent)	Electronic equation	Ion-electronic equation	Oxidation product ( recovery)
Chromium(III) ion ) in an alkaline environment	Cr +3 - 3ē = Cr +6	Cr 3+ + 8OH - - 3ē \u003d CrO 4 2- + 4H 2 O	CrO 4 2-
Chromium(III) ion in acid medium	Cr +3 - 3ē = Cr +6	2Cr 3+ + 7H 2 O - 6ē \u003d Cr 2 O 7 2- + 14 H +	Cr 2 O 7 2-
hydrogen sulfide	S -2 - 2ē \u003d S 0	H 2 S - 2ē \u003d S + 2H +
sulfite ion	S +4 - 2ē = S +6	SO 3 2- + H 2 O - 2ē \u003d SO 4 2- + 2 H +	SO 4 2-
Oxidizers
Permanganate ion in an acidic environment	Mn +7 + 5ē = Mn +2	MnO 4 - + 8H + + 5ē \u003d Mn 2+ + 4H 2 O	Mn2+
Permanganate ion in a neutral environment	Mn +7 + 3ē = Mn +4	MnO 4 - + 2H 2 O + 3ē \u003d MnO 2 + 4OH -	MnO2
Permanganate ion in an alkaline environment	Mn +7 + ē = Mn +6	MnO 4 - + ē \u003d MnO 4 2-	MnO 4 2-
dichromate ion	2Cr +6 + 6ē = 2Cr +3	Cr 2 O 7 2- + 14H + + 6ē \u003d 2Cr 3+ + 7H 2 O	Cr+3
Hydrogen peroxide in an acidic environment	2O - + 2ē \u003d 2O -2	H 2 O 2 + 2H + + 2ē \u003d 2H 2 O	H2O
Hydrogen peroxide in neutral and alkaline media	2O - + 2ē \u003d 2O -2	H 2 O 2 + 2ē \u003d 2 OH -	oh-

The specificity of many OVRs is that, when compiling their equations, the selection of coefficients causes difficulty. To facilitate the selection of coefficients, it is most often used electron balance method and ion-electronic method (half-reaction method). Consider the application of each of these methods with examples.

Electronic balance method

It is based on next rule: the total number of electrons donated by reducing atoms must match the total number of electrons received by oxidizing atoms.

As an example of compiling an OVR, consider the process of interaction of sodium sulfite with potassium permanganate in an acidic environment.

1. First you need to draw up a reaction scheme: write down the substances at the beginning and end of the reaction, given that in an acidic environment MnO 4 - is reduced to Mn 2+ ():

1. Next, we determine which of the compounds are; find their oxidation state at the beginning and end of the reaction:

Na 2 S +4 O 3 + KMn +7 O 4 + H 2 SO 4 = Na 2 S +6 O 4 + Mn +2 SO 4 + K 2 SO 4 + H 2 O

From the above diagram, it is clear that during the reaction, the oxidation state of sulfur increases from +4 to +6, so S +4 donates 2 electrons and is reducing agent. The oxidation state of manganese decreased from +7 to +2, i.e. Mn +7 accepts 5 electrons and is oxidizing agent.

1. We compose electronic equations and find the coefficients for the oxidizing agent and reducing agent.

S +4 - 2e - \u003d S +6 ¦ 5

Mn +7 +5e - = Mn +2 ¦ 2

In order for the number of electrons donated by the reducing agent to be equal to the number of electrons accepted by the reducing agent, it is necessary:

• Put the number of electrons donated by the reducing agent as a factor in front of the oxidizing agent.
• Put the number of electrons accepted by the oxidizing agent as a factor in front of the reducing agent.

Thus, 5 electrons received by the oxidizing agent Mn +7, we put the coefficient in front of the reducing agent, and 2 electrons given away by the reducing agent S +4 as a coefficient in front of the oxidizing agent:

5Na 2 S +4 O 3 + 2KMn +7 O 4 + H 2 SO 4 = 5Na 2 S +6 O 4 + 2Mn +2 SO 4 + K 2 SO 4 + H 2 O

1. Next, you need to equalize the number of atoms of elements that do not change the oxidation state, in the following sequence: the number of metal atoms, acid residues, the number of molecules of the medium (acid or alkali). Lastly, the number of water molecules formed is counted.

So, in our case, the number of metal atoms in the right and left parts are the same.

By the number of acid residues on the right side of the equation, we find the coefficient for the acid.

As a result of the reaction, 8 acid residues SO 4 2- are formed, of which 5 are due to the transformation 5SO 3 2- → 5SO 4 2-, and 3 are due to sulfuric acid molecules 8SO 4 2- - 5SO 4 2- \u003d 3SO 4 2 - .

Thus, sulfuric acid must take 3 molecules:

5Na 2 SO 3 + 2KMnO 4 + 3H 2 SO 4 = 5Na 2 SO 4 + 2MnSO 4 + K 2 SO 4 + H 2 O

1. Similarly, we find the coefficient for water by the number of hydrogen ions, in a given amount of acid

6H + + 3O -2 = 3H 2 O

The final form of the equation is as follows:

A sign that the coefficients are placed correctly is an equal number of atoms of each of the elements in both parts of the equation.

Ion-electronic method (half-reaction method)

Oxidation-reduction reactions, as well as exchange reactions, in electrolyte solutions occur with the participation of ions. That is why the ionic-molecular equations of the OVR more clearly reflect the essence of the redox reactions. When writing ion-molecular equations, strong electrolytes are written as , and weak electrolytes, precipitates and gases are written as molecules (in undissociated form). In the ionic scheme indicate the particles undergoing a change in their oxidation states, as well as characterizing the environment, particles: H + - acidic environment,OH — — alkaline environment and H 2 O - neutral environment.

Consider an example of compiling a reaction equation between sodium sulfite and potassium permanganate in an acidic environment.

1. First you need to draw up a reaction scheme: write down the substances at the beginning and end of the reaction:

Na 2 SO 3 + KMnO 4 + H 2 SO 4 = Na 2 SO 4 + MnSO 4 + K 2 SO 4 + H 2 O

1. We write the equation in ionic form, reducing those ions that do not take part in the redox process:

SO 3 2- + MnO 4 - + 2H + = Mn 2+ + SO 4 2- + H 2 O

1. Next, we define the oxidizing agent and reducing agent and compose the half-reactions of the reduction and oxidation processes.

In the above reaction oxidizing agent - MnO 4- accepts 5 electrons recovering in an acidic environment to Mn 2+. In this case, oxygen is released, which is part of MnO 4 -, which, combining with H +, forms water:

MnO 4 - + 8H + + 5e - \u003d Mn 2+ + 4H 2 O

Reducing agent SO 3 2-- oxidized to SO 4 2-, giving 2 electrons. As you can see, the resulting SO 4 2- ion contains more oxygen than the original SO 3 2- . The lack of oxygen is replenished by water molecules and as a result, 2H + is released:

SO 3 2- + H 2 O - 2e - \u003d SO 4 2- + 2H +

1. We find the coefficient for the oxidizing agent and reducing agent, considering that the oxidizing agent adds as many electrons as the reducing agent gives up in the oxidation-reduction process:

MnO 4 - + 8H + + 5e - \u003d Mn 2+ + 4H 2 O ¦2 oxidizing agent, reduction process

SO 3 2- + H 2 O - 2e - \u003d SO 4 2- + 2H + ¦5 reducing agent, oxidation process

1. Then it is necessary to sum both half-reactions, preliminarily multiplying by the found coefficients, we obtain:

2MnO 4 - + 16H + + 5SO 3 2- + 5H 2 O \u003d 2Mn 2+ + 8H 2 O + 5SO 4 2- + 10H +

Reducing like terms, we find the ionic equation:

2MnO 4 - + 5SO 3 2- + 6H + = 2Mn 2+ + 5SO 4 2- + 3H 2 O

1. Let's write the molecular equation, which has the following form:

5Na 2 SO 3 + 2KMnO 4 + 3H 2 SO 4 = 5Na 2 SO 4 + 2MnSO 4 + K 2 SO 4 + 3H 2 O

Na 2 SO 3 + KMnO 4 + H 2 O \u003d Na 2 SO 4 + MnO 2 + KOH

AT ionic form the equation becomes:

SO 3 2- + MnO 4 - + H 2 O \u003d MnO 2 + SO 4 2- + OH -

Also, as in the previous example, the oxidizing agent is MnO 4 -, and the reducing agent is SO 3 2-.

In a neutral and slightly alkaline environment, MnO 4 - accepts 3 electrons and is reduced to MnO 2. SO 3 2- - is oxidized to SO 4 2-, giving 2 electrons.

Half reactions have the following form:

MnO 4 - + 2H 2 O + 3e - \u003d MnO 2 + 4OH - ¦2 oxidizing agent, reduction process

SO 3 2- + 2OH - - 2e - \u003d SO 4 2- + H 2 O ¦3 reducing agent, oxidation process

We write the ionic and molecular equations, taking into account the coefficients for the oxidizing agent and reducing agent:

3SO 3 2- + 2MnO 4 - + H 2 O \u003d 2 MnO 2 + 3SO 4 2- + 2OH -

3Na 2 SO 3 + 2KMnO 4 + H 2 O \u003d 2MnO 2 + 3Na 2 SO 4 + 2KOH

And one more example - drawing up an equation for the reaction between sodium sulfite and potassium permanganate in an alkaline environment.

Na 2 SO 3 + KMnO 4 + KOH \u003d Na 2 SO 4 + K 2 MnO 4 + H 2 O

AT ionic form the equation becomes:

SO 3 2- + MnO 4 - + OH - \u003d MnO 2 + SO 4 2- + H 2 O

In an alkaline environment oxidizing agent MnO 4 - accepts 1 electron and is reduced to MnO 4 2-. Reductant SO 3 2- - is oxidized to SO 4 2-, giving 2 electrons.

Half reactions have the following form:

MnO 4 - + e - \u003d MnO 2 ¦2 oxidizing agent, reduction process

SO 3 2- + 2OH - - 2e - \u003d SO 4 2- + H 2 O ¦1 reducing agent, oxidation process

Let's write down the ionic and molecular equations, taking into account the coefficients for the oxidizing agent and reducing agent:

SO 3 2- + 2MnO 4 - + 2OH - \u003d 2MnO 4 2- + SO 4 2- + H 2 O

Na 2 SO 3 + 2KMnO 4 + H 2 O \u003d 2K 2 MnO 4 + 3Na 2 SO 4 + 2KOH

It should be noted that not always in the presence of an oxidizing agent and a reducing agent, spontaneous OVR may occur. Therefore, for quantitative characteristics the strength of the oxidizing agent and reducing agent and to determine the direction of the reaction, use the values ​​of redox potentials.

Categories ,

Restorers	Oxidizers
metals, hydrogen, coal	halogens
carbon monoxide (II) CO	manganese (VII) oxide - Mn 2 O 7
hydrogen sulfide H 2 S	manganese (IV) oxide - MnO 2
sodium sulfide Na 2 S	potassium permanganate - KMnO 4
sulfur oxide (IV) - SO 2	potassium manganate - K 2 MnO 4
sulfurous acid - H 2 SO 3 and its salts	chromium oxide (VI) - CrO 3
sodium thiosulfate - Na 2 S 2 O 3	potassium chromate - K 2 CrO 4
hydroiodic acid - HI	potassium dichromate - K 2 Cr 2 O 7
hydrobromic acid - HBr hydrochloric acid - HCl tin (II) chloride - SnCl 2 iron (II) sulfate - FeSO 4 manganese (II) sulfate - MnSO 4 chromium sulfate (III) - Cr 2 (SO 4) 3 nitrous acid - HNO 2 ammonia NH 3 hydrazine N 2 H 4 nitric oxide (II) NO phosphorous acid - H 3 PO 3 orthoarsenic acid - H 3 AsO 3 potassium hexacyanoferrate (II) - K 4	nitric acid - HNO 3 oxygen - O 2 ozone - O 3 hydrogen peroxide - H 2 O 2 sulfuric acid - H 2 SO 4 (conc.) selenic acid - H 2 SeO 4 copper(II) oxide - CuO silver oxide (I) - Ag 2 O lead(IV) oxide - PbO 2 noble metal ions (Ag+, Au 3+, etc.) sodium bismuthate - NaBiO 3 ammonium persulfate - (NH 4) 2 S 2 O 8 potassium hexacyanoferrate (III) -K 3 iron (III) chloride - FeCl 3 hypochlorites, chlorates, perchlorates aqua regia mixture of concentrated nitric and hydrofluoric acids

9.3. Influence of the environment on redox reactions

The nature of the environment (acidic, neutral, alkaline) affects the OVR. In different environments, the interaction of the same substances can produce different products. We were convinced of this by the examples considered in section 9.1, where the oxidizing agent is permanganate - the MnO ion:

oxidized form restored form

acidic medium Mn 2+ b / c or slightly pink

pH  7 coloring solution

7 neutral environment +4

MnO pH  7 MnO 2 (brown precipitate)

alkaline environment (MnO 4) 2- (green color

pH  7 solution)

Permanganate ion exhibits oxidizing properties to a greater extent in an acidic environment (greater decrease in the degree of oxidation).

Usually, to create an acidic environment in a solution, sulfuric acid. Nitric and hydrochloric (hydrochloric) acids are rarely used: the first is itself an oxidizing agent, the second is capable of being oxidized. To create an alkaline environment, solutions of potassium or sodium hydroxide are used.

Let us consider examples of the influence of the medium on the course of a reaction involving hydrogen peroxide. Hydrogen peroxide, depending on the medium, is reduced according to the scheme:

acidic pH 7

H 2 O 2 + 2H + + 2e - = H 2 O

neutral environment

alkaline medium H 2 O 2 + 2e - \u003d 2OH -

Here H 2 O 2 acts as an oxidizing agent. For example:

2FeSO 4 + H 2 O 2 + H 2 SO 4 = Fe 2 (SO 4) 3 + 2 H 2 O

2 Fe 2+ - e - \u003d Fe 3+

1 H 2 O 2 + 2H + + 2e \u003d 2 H 2 O

2Fe 2+ + H 2 O 2 + 2H + = 2Fe 3+ + 2 H 2 O

However, with a very strong oxidizing agent, such as KMnO 4 , hydrogen peroxide interacts as a reducing agent:

H 2 O 2 - 2e - \u003d O 2 + 2H +

For example:

5 H 2 O 2 + 2 KMnO 4 + 3H 2 SO 4 = 5 O 2 + 2 MnSO 4 + K 2 SO 4 + 8H 2 O

5 H 2 O 2 - 2e - \u003d O 2 + 2H +

2 MnO - 4 + 8H + + 5e = Mn 2+ + 4H 2 O

5 H 2 O 2 + 2 MnO - 4 + 6H + = 5 O 2 + 2 Mn 2+ + 8H 2 O

Chromium in its compounds has stable s.d. (+6) and (+3). In the first case, chromium compounds (chromate, dichromate ions) exhibit the properties of oxidizing agents, in the second - reducing agents. Chromate and dichromate ions are strong oxidizing agents, they are reduced to Cr 3+ compounds:

oxidized form restored form

Drawing up equations of redox reactions

In order to write the OVR equation, it is necessary, first of all, to know which substances are formed as a result of the reaction. In the general case, this issue is solved experimentally. However, knowledge of the chemical properties of various oxidizing and reducing agents often makes it possible to fairly reliably (although not with a 100% guarantee) predict the composition of the interaction products.

If the reaction products are known, the stoichiometric coefficients in the reaction equation can be found by equalizing the number of electrons added by oxidant atoms and lost by reducing agent atoms. Two methods of selection of coefficients in the OVR equations are used - the method of electronic balance and the method of ion-electron balance. Let's consider these methods.

The method is based on the principle of conservation of electric charge during a chemical reaction, as a result of which substances react in such a ratio that ensures the equality of the number of electrons given up by all atoms of the reducing agent and attached by all atoms of the oxidizing agent. To select the coefficients, it is advisable to use the following algorithm:

1. Write down the OVR scheme (starting substances and reaction products).

2. Determine the elements whose oxidation state changes during the reaction.

3. Make diagrams of the processes of oxidation and reduction.

4. Find the factors equalizing the number of electrons attached by the oxidizing agent atoms and lost by the reducing agent atoms (balancing factors). To do this, find the least common multiple for electrons attached by one atom of the oxidizing agent and given away by one atom of the reducing agent; the balancing factors will be equal to the least common multiple divided by the number of electrons attached (for an oxidizing agent) and donated electrons (for a reducing agent).

5. Determine and enter into the equation the coefficients for substances containing elements whose oxidation state changes (reference coefficients) by dividing the balancing factors by the number of oxidizing or reducing agent atoms in the formula unit of the substance. If the quotient is not an integer, the balancing factors should be increased by the required number of times.

6. Find and arrange additional coefficients that equalize the number of atoms that have not changed their oxidation state (except for hydrogen and oxygen); at the same time, if the medium is acidic, first equalize the metal atoms, and then the anions of the acids, if the medium is alkaline or neutral, vice versa.

7. Equalize the number of hydrogen atoms, adding water to the right or left side of the equation if necessary.

8. Check whether the coefficients for oxygen are correctly selected.

Consider, as an example, the formulation of the equation for the interaction of potassium permanganate with iron (II) sulfate in a sulfuric acid medium according to the stages of the proposed algorithm:

1. KMnO 4 + FeSO 4 + H 2 SO 4 → MnSO 4 + Fe 2 (SO 4) 3 + K 2 SO 4

2. KMn +7 O 4 + Fe +2 SO 4 + H 2 S0 4 → Mn +2 SO 4 + Fe(SO 4) 3 + K 2 SO 4

3. Fe +2 - 1e - = Fe +3 (oxidation)

Mn +7 +5e - = Mn +2 (recovery)

4. Fe +2 - 1e - = Fe +3 │5 │ 10

Mn +7 + 5e - = Mn +2 │1 │2

5. Reference coefficients: with KMnO 4 - 2:1=2, with FeSO 4 - 10:1=10, with MnSO 4 - 2:1=2, with Fe 2 (SO 4) 3 - 10:2=5.

2KMnO 4 + 10FeSO 4 + H 2 SO 4 → 2MnSO 4 + 5Fe 2 (SO 4) 3 + K 2 SO 4

6. The medium is acidic, so first we equalize the potassium atoms, then the sulfate ions.

2KMnO 4 + 10FeSO 4 + 5H 2 SO 4 → 2MnSO 4 + 5Fe 2 (SO 4) 3 + K 2 SO 4

7. Since the left side of the equation contains 10 hydrogen atoms, we add 5 water molecules to the right side:

2KMnO 4 + 10FeSO 4 + 5H 2 SO 4 = 2MnSO 4 + 5Fe 2 (SO 4) 3 + K 2 SO 4 + 5H 2 O

8. The number of oxygen atoms (excluding oxygen included in sulfate ions) in the right and left parts of the equation is 8. The coefficients are chosen correctly.

When OVR occurs, there may be cases when the oxidizing agent or reducing agent is partially spent on binding the oxidation or reduction products without changing the oxidation state of the corresponding element. In this case, the coefficient for a substance with a dual function is equal to the sum of the reference and additional coefficients and is introduced into the equation after the additional coefficient is found. Thus, the reaction between zinc and very dilute nitric acid proceeds according to the equation

4Zn + 10HNO 3 \u003d 4Zn (NO 3) 2 + NH 4 NO 3 + 3H 2 O

Zn 0 - 2e - = Zn +2 │4

N +5 + 8e - = N -3 │1

As follows from the redox schemes, the oxidation of four zinc atoms requires one nitric acid molecule (the reference factor for HNO 3 is 1); however, the formation of four molecules of zinc nitrate and one molecule of ammonium nitrate requires nine more HNO 3 molecules that react without changing the nitrogen oxidation state (additional coefficient for HNO 3 - 9). Accordingly, the coefficient for nitric acid in the reaction equation will be equal to 10, and 3 water molecules should be entered into the right side of the equation.

If one of the substances simultaneously performs the function of both an oxidizing agent and a reducing agent (disproportionation reactions) or is a product of both oxidation and reduction (counter-disproportionation reactions), then the coefficient for this substance is equal to the sum of the reference coefficients for the oxidizing agent and reducing agent. For example, in the reaction equation for the disproportionation of sulfur in an alkaline medium, the coefficient for sulfur is three.

3S 0 + 6NaOH \u003d Na 2 S +4 O 3 + Na 2 S -2 + 3H 2 O

S - 4e - = S +4 │1

S + 2e - = S -2 │2

Sometimes, during the course of OVR, a change in the oxidation state of more than two elements is observed; in this case, the coefficients of the equation can be uniquely determined if all oxidizing agents or all reducing agents are part of one molecule. In this case, the calculation of donated or attached electrons is rationally carried out for the formula unit of a substance containing these oxidizing or reducing agents. As an example, consider the interaction of arsenic(III) sulfide with nitric acid according to the stages of the above algorithm.

1. As 2 S 3 + HNO 3 → H 3 AsO 4 + H 2 SO 4 + NO

2. AsS+ HN +5 O 3 → H 3 As +5 O 4 + H 2 S +6 O 4 + N +2 O

The reaction involves two reducing agents (As +3 and S -2) and one oxidizing agent (N +5).

3. N +5 + 3e - = N +2 │28

As 2 S 3 - 28e - \u003d 2As +5 + 3S +6 │ 3

4. Least common multiple - 84, balancing factors - 28 and 3.

5. 3As 2 S 3 + 28HNO 3 → 6H 3 AsO 4 + 9H 2 SO 4 + 28NO

6. There are no additional coefficients.

7. Water molecules should be entered on the left side of the equation:

3As 2 S 3 + 28HNO 3 + 4H 2 O \u003d 6H 3 AsO 4 + 9H 2 SO 4 + 28NO

8. The number of oxygen atoms in both the left and right sides of the equation is 88. The coefficients are chosen correctly.

If organic substances are involved in the OVR, then the oxidation states are not determined for them, since in this case each atom can have its own oxidation state value, and often not integer. When drawing up redox schemes for such reactions, the following rules should be followed:

1. the addition of an oxygen atom is identical to the loss of two electrons by a molecule;

2. the loss of an oxygen atom is identical to the addition of two electrons;

3. the addition of a hydrogen atom is identical to the addition of one electron;

4. The loss of a hydrogen atom is identical to the loss of one electron.

Below, as an example, the reaction equation for the oxidation of ethyl alcohol with potassium dichromate is given:

3C 2 H 5 OH + 2K 2 Cr 2 O 7 + 8H 2 SO 4 \u003d 3CH 3 COOH + 2Cr 2 (SO 4) 3 + 2K 2 SO 4 + 11H 2 O

C 2 H 5 OH + [O] - 2 [H] - 4e - \u003d 3CH 3 COOH │3

Cr +6 + 3e - = Cr +3 │4

The conversion of ethyl alcohol to acetic acid requires the addition of an oxygen atom and the loss of two hydrogen atoms, which corresponds to the loss of four electrons.

The electron balance method is a universal method applicable to any OVR occurring in the gas phase, condensed systems, and solutions. The disadvantage of the method is that this technique is formal and operates with particles that do not really exist (Mn +7, N +5, etc.).