HNO3, HNO3 salts

only oxidizing agents

Compounds with intermediate oxidation states of nitrogen can act as oxidizing agents, being reduced to lower oxidation states, or as reducing agents, being oxidized to higher oxidation states

Figure 1 – Change in the degree of nitrogen oxidation

Electronic balance method equalization of redox reactions consists in fulfilling the following rule: the number of electrons donated by all particles of reducing agents is always equal to the number of electrons attached by all particles of oxidizing agents in a given reaction.

Example 2. Let us illustrate the electronic balance method using the example of the oxidation of iron with oxygen:
.

Fe 0 – 3ē = Fe +3 – oxidation process;

O 2 + 4ē = 2O –2 – reduction process.

In the reducing agent system (half-reaction of the oxidation process), the iron atom gives up 3 electrons (Appendix A).

In the oxidizing system (half-reaction of the reduction process), each oxygen atom accepts 2 electrons - a total of 4 electrons.

The least common multiple of the two numbers 3 and 4 is 12. Hence, iron gives up 12 electrons and oxygen accepts 12 electrons:

Coefficients 4 and 3, written to the left of the half-reactions during the summation of systems, are multiplied by all components of the half-reactions. The overall equation shows how many molecules or ions should appear in the equation. An equation is correct when the number of atoms of each element on both sides of the equation is the same.

Half-reaction method used to equalize reactions occurring in electrolyte solutions. In such cases, not only the oxidizing agent and the reducing agent, but also particles of the medium take part in the reactions: water molecules (H 2 O), H + and OH – ions. It is more correct for such reactions to use electron-ion systems (half-reactions). When composing half-reactions in aqueous solutions, H 2 O molecules and H + or OH – ions are introduced, if necessary, taking into account the reaction environment. Weak electrolytes, sparingly soluble (Appendix B) and gaseous compounds in ionic systems are written in molecular form (Appendix C).

Let us consider as examples the interaction of potassium sulfate and potassium permanganate in an acidic and alkaline environment.

Example 3. Reaction between potassium sulfate and potassium permanganate in an acidic environment:

Let us determine the change in the oxidation state of the elements and indicate them in the equation. Highest degree oxidation of manganese (+7) in KMnO 4 indicates that KMnO 4 is an oxidizing agent. Sulfur in the K 2 SO 3 compound has an oxidation state (+4) - it is a reduced form relative to sulfur (+6) in the K 2 SO 4 compound. Thus, K 2 SO 3 is a reducing agent. Real ions containing elements that change the oxidation state and their initial half-reactions take the following form:

The goal of further actions is to put equal signs in these half-reactions instead of arrows reflecting the possible direction of the reaction. This can be done when the types of elements, the number of their atoms and the total charges of all particles coincide in the left and right sides of each half-reaction. To achieve this, additional ions or molecules of the medium are used. Usually these are H + ions, OH – and water molecules. Half-reaction
the number of manganese atoms is the same, but the number of oxygen atoms is not equal, so we introduce four water molecules into the right side of the half-reaction: . Carrying out similar actions (equalizing oxygen) in the system
, we get
. Hydrogen atoms appeared in both half-reactions. Their number is equalized by the corresponding addition in the other part of the equations of an equivalent number of hydrogen ions.

Now all the elements included in the half-reaction equations have been equalized. It remains to equalize the charges of the particles. On the right side of the first half-reaction, the sum of all charges is +2, while on the left the charge is +7. Equality of charges is achieved by adding five negative charges in the form of electrons (+5 ē) to the left side of the equation. Similarly, in the equation of the second half-reaction, it is necessary to subtract 2 ē from the left. Now we can put equal signs in the equations of both half-reactions:

-recovery process;

– oxidation process.

In the example under consideration, the ratio of the number of electrons accepted during the reduction process to the number of electrons released during oxidation is equal to 5 ׃ 2. To obtain the total reaction equation, it is necessary to take into account this ratio by summing up the equations of the reduction and oxidation processes - multiply the reduction equation by 2, and oxidation equation – by 5.

By multiplying the coefficients by all terms of the half-reaction equations and summing together only their right-hand and only their left-hand sides, we obtain the final reaction equation in ionic-molecular form:

Reducing similar terms by subtracting the same number of H + ions and H 2 O molecules, we obtain:

The total ionic equation is written correctly, there is a correspondence between the medium and the molecular one. We transfer the obtained coefficients to the molecular equation:

Example 4. Reactions between potassium sulfate and potassium permanganate in an alkaline environment:

We determine the oxidation states of elements that change the oxidation state (Mn +7 → Mn +6, S +4 → S +6). Real ions, which include these elements (
,
). Processes (half-reactions) of oxidation and reduction:

2
– recovery process

1 – oxidation process

Summary equation:

In the total ionic equation there is a correspondence of the medium. We transfer the coefficients into the molecular equation:

Oxidation-reduction reactions are divided into the following types:

intermolecular oxidation-reduction;

self-oxidation-self-healing (disproportionation);

intramolecular oxidation - reduction.

Intermolecular oxidation-reduction reactions - these are reactions when the oxidizing agent is in one molecule and the reducing agent is in another.

Example 5. When iron hydroxide oxidizes in a humid environment, the following reaction occurs:

4Fe(OH) 2 + OH – – 1ē = Fe(OH) 3 – oxidation process;

1 O 2 + 2H 2 O + 4ē = 4OH – – reduction process.

In order to ensure that electron-ion systems are written correctly, it is necessary to check: the left and right parts of the half-reactions must contain the same number of element atoms and charge. Then, by equalizing the number of electrons accepted and donated, we summarize the half-reactions:

4Fe(OH) 2 + 4OH – + O 2 +2H 2 O = 4Fe(OH) 3 + 4OH –

4Fe(OH) 2 + O 2 +2H 2 O = 4Fe(OH) 3

Autoxidation-self-healing reactions (disproportionation reactions) - these are reactions during which part of the total amount of an element is oxidized, and the other part is reduced, typical for elements with an intermediate oxidation state.

Example 6. When chlorine reacts with water, a mixture of hydrochloric and hypochlorous (HClO) acids is obtained:

Here chlorine undergoes both oxidation and reduction:

1Cl 2 + 2H 2 O – 2ē = 2HClO +2H + – oxidation process;

1 Cl 2 + 2ē = 2Cl – – reduction process.

2Cl 2 + 2H 2 O = 2HClO + 2HCl

Example 7 . Disproportionation of nitrous acid:

In this case, oxidation and reduction undergoes containing HNO 2:

Summary equation:

HNO 2 + 2HNO 2 + H 2 O + 2H + = NO + 3H + + 2NO + 2H 2 O

3HNO2 = HNO3 + 2NO + H2O

Intramolecular oxidation-reduction reactions is a process when one component one molecule serves as an oxidizing agent, and the other serves as a reducing agent. Examples of intramolecular oxidation-reduction include many thermal dissociation processes.

Example 8. Thermal dissociation of NH 4 NO 2:

Here the ion is NH is oxidized, and the NO ion is reduced to free nitrogen:

12NH – 6 ē = N 2 + 8H +

1 2NO + 8Н + + 6 ē = N 2 + 4H 2 O

2NH +2NO + 8H + = N 2 + 8H + + N 2 + 4H 2 O

2NH 4 NO 2 = 2N 2 + 4H 2 O

Example 9 . Decomposition reaction of ammonium dichromate:

12NH – 6 ē = N 2 + 8H +

1 Cr 2 O + 8Н + + 6 ē = Cr 2 O 3 + 4H 2 O

2NH + Cr 2 O + 8H + = N 2 + 8H + + Cr 2 O 3 + 4H 2 O

(NH 4) 2 Cr 2 O 7 = N 2 + Cr 2 O 3 + 4H 2 O

Redox reactions involving more than two elements that change the oxidation state.

Example 10. An example is the reaction of iron sulfide with nitric acid, where during the reaction three elements (Fe, S, N) change the oxidation state:

FeS 2 + HNO 3
Fe 2 (SO 4) 3 + NO + ...

The equation is not completely written and the use of electron-ion systems (half-reactions) will allow us to complete the equation. Considering the oxidation states of the elements involved in the reaction, we determine that in FeS 2 two elements (Fe, S) are oxidized, and the oxidizing agent is
(), which is reduced to NO:

S –1 → ()

We write the oxidation half-reaction of FeS 2:

FeS 2 → Fe 3+ +

The presence of two Fe 3+ ions in Fe 2 (SO 4) 3 suggests doubling the number of iron atoms when further writing the half-reaction:

2FeS 2 → 2Fe 3+ + 4

At the same time we equalize the number of sulfur and oxygen atoms, we get:

2FeS 2 + 16H 2 O → 2Fe 3+ + 4
.

32 hydrogen atoms, by introducing into left side equations consisting of 16 H 2 O molecules are equalized by adding the equivalent number of hydrogen ions (32 H +) to the right side of the equation:

2FeS 2 + 16H 2 O → 2Fe 3+ + 4
+ 32H +

The charge on the right side of the equation is +30. In order for the left side to have the same thing (+30), it is necessary to subtract 30 ē:

1 2FeS 2 + 16Н 2 O – 30 ē = 2Fe 3+ + 4
+ 32H + – oxidation;

10 NO + 4Н + + 3 ē = NO + 2H 2 O – reduction.

2FeS 2 +16Н 2 O+10NO +40H + = 2Fe 3+ + 4
+ 32Н + + 10NO + 20H 2 O

2FeS 2 +10НNO 3 + 30Н + = Fe 2 (SO 4) 3 + 10NO +
+ 32Н + + 4H 2 O

H 2 SO 4 +30H +

We reduce both sides of the equation by the same number of ions (30 H +) using the subtraction method and get:

2FeS 2 +10HNO 3 = Fe 2 (SO 4) 3 + 10NO + H 2 SO 4 + 4H 2 O

Energy of redox reactions . The condition for the spontaneous occurrence of any process, including a redox reaction, is the inequality ∆G< 0, где ∆G – энергия Гиббса и чем меньше ∆G, т.е. чем больше его отрицательное значение, тем более реакционноспособнее окислительно-восстановительная система. Для реакций окисления-восстановления:

∆G = –n·F·ε,

where n is the number of electrons transferred by the reducing agent to the oxidizing agent in the elementary act of oxidation-reduction;

F – Faraday number;

ε – electromotive force (EMF) of the redox reaction.

The electromotive force of a redox reaction is determined by the potential difference between the oxidizing agent and the reducing agent:

ε = E ok – E in,

Under standard conditions:

ε ° = E ° ok – E ° in.

So, if the condition for the spontaneous occurrence of the process is the inequality ∆G °< 0, то это возможно, когда n·F·ε ° >0. If n and F are positive numbers, then it is necessary that ε ° > 0, and this is possible when E ° ok > E ° in. It follows that the condition for the spontaneous occurrence of a redox reaction is the inequality E ° ok > E ° in.

Example 11. Determine the possibility of a redox reaction occurring:

Having determined the oxidation states of elements that change the oxidation state, we write down the half-reactions of the oxidizing agent and the reducing agent, indicating their potentials:

Сu – 2ē = Сu 2+ Е ° в = +0.34 V

2H + + 2ē = H 2 E ° ok = 0.0 V

From the half-reactions it is clear that E° is ok< Е ° в, это говорит о том, что рассматриваемый процесс термодинамически невозможен (∆G ° >0). This reaction is only possible in the opposite direction, for which ∆G °< 0.

Example 12. Calculate the Gibbs energy and equilibrium constant for the reduction of potassium permanganate with iron (II) sulfate.

Half-reactions of oxidizing agent and reducing agent:

2 E ° ok = +1.52V

5 2Fe 2+ – 2 ē = 2Fe 3+ E ° in = +0.77 V

∆G ° = –n·F·ε ° = –n·F(E ° ok – E ° in),

where n = 10, since the reducing agent gives up 10 ē, the oxidizing agent accepts 10 ē in the elementary act of oxidation-reduction.

∆G ° = –10·69500(1.52–0.77) = –725000 J,

∆G ° = –725 kJ.

Considering that the standard change in the Gibbs energy is related to its equilibrium constant (K c) by the relation:

∆G ° = –RTlnК s or n·F·ε = RTlnК s,

where R = 8.31 J mol –1 K –1,

F
96500 C mol –1, T = 298 K.

We determine the equilibrium constant for this reaction by putting in the equation constants, converting the natural logarithm to decimal:

Kc = 10,127.

The data obtained indicate that the reduction reaction of potassium permanganate under consideration is reactive (∆G ° = – 725 kJ), the process proceeds from left to right and is practically irreversible (K c = 10,127).

18. Redox reactions (continued 1)

18.5. ORR of hydrogen peroxide

In molecules of hydrogen peroxide H 2 O 2, oxygen atoms are in the oxidation state –I. This is an intermediate and not the most stable oxidation state of the atoms of this element, therefore hydrogen peroxide exhibits both oxidizing and reducing properties.

The redox activity of this substance depends on the concentration. In commonly used solutions with mass fraction 20% hydrogen peroxide is a fairly strong oxidizing agent; in dilute solutions its oxidizing activity decreases. The reducing properties of hydrogen peroxide are less characteristic than the oxidizing properties and also depend on the concentration.

Hydrogen peroxide is a very weak acid (see Appendix 13), therefore, in strongly alkaline solutions its molecules transform into hydroperoxide ions.

Depending on the reaction of the medium and whether hydrogen peroxide is the oxidizing or reducing agent in this reaction, the products of the redox interaction will be different. The half-reaction equations for all these cases are given in Table 1.

Table 1

Equations of redox half-reactions of H 2 O 2 in solutions

Let's consider examples of ORR involving hydrogen peroxide.

Example 1. Write an equation for the reaction that occurs when a solution of potassium iodide is added to a solution of hydrogen peroxide acidified with sulfuric acid.

H 2 O 2 + 2H 3 O +2I = 4H 2 O + I 2
H 2 O 2 + H 2 SO 4 + 2KI = 2H 2 O + I 2 + K 2 SO 4

Example 2. Write an equation for the reaction between potassium permanganate and hydrogen peroxide in an aqueous solution acidified with sulfuric acid.

2MnO 4 + 6H 3 O+ + 5H 2 O 2 = 2Mn 2 + 14H 2 O + 5O 2
2KMnO 4 + 3H 2 SO 4 + 5H 2 O 2 = 2MnSO 4 + 8H 2 O + 5O 2 + K 2 SO 4

Example 3. Write an equation for the reaction of hydrogen peroxide with sodium iodide in solution in the presence of sodium hydroxide.

3HO 2 + I = 3OH + IO 3
3NaHO 2 + NaI = 3NaOH + NaIO 3

Without taking into account the neutralization reaction between sodium hydroxide and hydrogen peroxide, this equation is often written as follows:

3H 2 O 2 + NaI = 3H 2 O + NaIO 3 (in the presence of NaOH)

The same equation will be obtained if we do not immediately (at the stage of drawing up the balance) take into account the formation of hydroperoxide ions.

Example 4. Write an equation for the reaction that occurs when lead dioxide is added to a solution of hydrogen peroxide in the presence of potassium hydroxide.

Lead dioxide PbO 2 is a very strong oxidizing agent, especially in an acidic environment. Reducing under these conditions, it forms Pb 2 ions. In an alkaline environment, when PbO 2 is reduced, ions are formed.

PbO 2 + H 2 O + HO 2 = + O 2

Without taking into account the formation of hydroperoxide ions, the equation is written as follows:

PbO 2 + H 2 O 2 + OH = + O 2 + 2H 2 O

If, according to the conditions of the task, the added solution of hydrogen peroxide was alkaline, then the molecular equation should be written as follows:

PbO 2 + H 2 O + KHO 2 = K + O 2

If a neutral solution of hydrogen peroxide is added to a reaction mixture containing an alkali, then the molecular equation can be written without taking into account the formation of potassium hydroperoxide:

PbO 2 + KOH + H 2 O 2 = K + O 2

Among the redox reactions there are dismutation reactions (disproportionation, self-oxidation-self-reduction).

An example of a dismutation reaction known to you is the reaction of chlorine with water:

Cl 2 + H 2 O HCl + HClO

In this reaction, half of the chlorine(0) atoms are oxidized to the +I oxidation state, and the other half are reduced to the –I oxidation state:

Using the electron-ion balance method, let’s compose an equation for a similar reaction that occurs when chlorine is passed through a cold alkali solution, for example KOH:

2Cl 2 + 4OH = 2Cl + 2ClO + 2H 2 O

All coefficients in this equation have a common divisor, therefore:

Cl 2 + 2OH = Cl + ClO + H 2 O
Cl 2 + 2KOH = KCl + KClO + H 2 O

The dismutation of chlorine in a hot solution proceeds somewhat differently:

3Cl 2 + 6OH = 5Cl + ClO 3 + 3H 2 O
3Cl 2 + 6KOH = 5KCl + KClO 3 + 3H 2 O

Big practical significance has dismutation of nitrogen dioxide when it reacts with water ( A) and with alkali solutions ( b):

Dismutation reactions occur not only in solutions, but also when heated solids, for example, potassium chlorate:

4KClO 3 = KCl + 3KClO 4

A typical and very effective example of intramolecular ORR is the reaction of thermal decomposition of ammonium dichromate (NH 4) 2 Cr 2 O 7. In this substance, nitrogen atoms are in their lowest oxidation state (–III), and chromium atoms are in their highest (+VI). At room temperature this compound is quite stable, but when heated it intensively decomposes. In this case, chromium(VI) transforms into chromium(III) - the most stable state of chromium, and nitrogen(–III) - into nitrogen(0) - also the most stable state. Taking into account the number of atoms in the formula unit of the electron balance equation:

and the reaction equation itself:

(NH 4) 2 Cr 2 O 7 = Cr 2 O 3 + N 2 + 4H 2 O.

Another important example intramolecular ORR – thermal decomposition of potassium perchlorate KClO 4 . In this reaction, chlorine(VII), as always when it acts as an oxidizing agent, transforms into chlorine(–I), oxidizing oxygen(–II) to simple substance:

and therefore the reaction equation

KClO 4 = KCl + 2O 2

Potassium chlorate KClO 3 decomposes similarly when heated, if the decomposition is carried out in the presence of a catalyst (MnO 2): 2KClO 3 = 2KCl + 3O 2

In the absence of a catalyst, a dismutation reaction occurs.
The group of intramolecular redox reactions also includes reactions of thermal decomposition of nitrates.
Typically, the processes that occur when nitrates are heated are quite complex, especially in the case of crystalline hydrates. If water molecules are weakly retained in the crystalline hydrate, then with low heating the nitrate dehydrates [for example, LiNO 3. 3H 2 O and Ca(NO 3) 2 4H 2 O are dehydrated to LiNO 3 and Ca(NO 3) 2 ], but if water is bound more tightly [as, for example, in Mg(NO 3) 2. 6H 2 O and Bi(NO 3) 3. 5H 2 O], then a kind of “intramolecular hydrolysis” reaction occurs with the formation of basic salts - hydroxide nitrates, which upon further heating can turn into oxide nitrates (and (NO 3) 6), the latter decompose to oxides at a higher temperature .

When heated, anhydrous nitrates can decompose into nitrites (if they exist and are still stable at this temperature), and nitrites can decompose into oxides. If heating is carried out to a sufficiently high temperature, or the corresponding oxide is unstable (Ag 2 O, HgO), then the product of thermal decomposition can also be a metal (Cu, Cd, Ag, Hg).

A somewhat simplified diagram of the thermal decomposition of nitrates is shown in Fig. 5.

Examples of sequential transformations that occur when certain nitrates are heated (temperatures are given in degrees Celsius):

KNO 3 KNO 2 K 2 O;

Ca(NO3)2. 4H 2 O Ca(NO 3) 2 Ca(NO 2) 2 CaO;

Mg(NO3)2. 6H 2 O Mg(NO 3)(OH) MgO;

Cu(NO3)2. 6H 2 O Cu(NO 3) 2 CuO Cu 2 O Cu;

Bi(NO 3) 3 . 5H 2 O Bi(NO 3) 2 (OH) Bi(NO 3)(OH) 2 (NO 3) 6 Bi 2 O 3.

Despite the complexity of the processes taking place, when answering the question of what happens when the corresponding anhydrous nitrate is “calcined” (that is, at a temperature of 400 – 500 o C), one is usually guided by the following extremely simplified rules:

1) nitrates of the most active metals (in the series of voltages - to the left of magnesium) decompose to nitrites;
2) nitrates of less active metals (in the range of voltages - from magnesium to copper) decompose to oxides;
3) nitrates of the least active metals (in the series of voltages - to the right of copper) decompose to metal.

When using these rules, it should be remembered that in such conditions
LiNO 3 decomposes to oxide,
Be(NO 3) 2 decomposes to oxide at a higher temperature,
from Ni(NO 3) 2, in addition to NiO, Ni(NO 2) 2 can also be obtained,
Mn(NO 3) 2 decomposes to Mn 2 O 3,
Fe(NO 3) 2 decomposes to Fe 2 O 3;
from Hg(NO 3) 2, in addition to mercury, its oxide can also be obtained.

Let's look at typical examples of reactions belonging to these three types:

Problem book on general and inorganic chemistry

2.2. Redox reactions

Theoretical part

Redox reactions include chemical reactions, which are accompanied by changes in the oxidation states of elements. In the equations of such reactions, the selection of coefficients is carried out by compiling electronic balance. The method for selecting odds using an electronic balance consists of the following steps:

a) write down the formulas of the reagents and products, and then find the elements that increase and decrease their oxidation states and write them out separately:

MnCO 3 + KClO 3 ® MnO2+ KCl + CO2

Cl V¼ = Cl - I

Mn II¼ = Mn IV

b) compose equations for half-reactions of reduction and oxidation, observing the laws of conservation of the number of atoms and charge in each half-reaction:

half-reaction recovery Cl V + 6 e - = Cl - I

half-reaction oxidation Mn II- 2 e - = Mn IV

c) additional factors are selected for the equation of half-reactions so that the law of conservation of charge is satisfied for the reaction as a whole, for which the number of accepted electrons in the reduction half-reactions is made equal to the number of electrons donated in the oxidation half-reaction:

Cl V + 6 e - = Cl - I 1

Mn II- 2 e - = Mn IV 3

d) insert (using the found factors) stoichiometric coefficients into the reaction scheme (coefficient 1 is omitted):

3 MnCO 3 + KClO 3 = 3 MnO 2 + KCl+CO2

d) equalize the number of atoms of those elements that do not change their oxidation state during the reaction (if there are two such elements, then it is enough to equalize the number of atoms of one of them, and check for the second). The equation for the chemical reaction is obtained:

3 MnCO 3 + KClO 3 = 3 MnO 2 + KCl+ 3 CO 2

Example 3. Select the coefficients in the equation of the redox reaction

Fe 2 O 3 + CO ® Fe + CO 2

Solution

Fe 2 O 3 + 3 CO = 2 Fe +3 CO 2

FeIII + 3 e - = Fe 0 2

C II - 2 e - = C IV 3

With the simultaneous oxidation (or reduction) of atoms of two elements of one substance, the calculation is carried out for one formula unit of this substance.

Example 4. Select the coefficients in the equation of the redox reaction

Fe(S ) 2 + O 2 = Fe 2 O 3 + SO 2

Solution

4Fe(S ) 2 + 11 O 2 = 2 Fe 2 O 3 + 8 SO 2

FeII- e - = FeIII

- 11 e - 4

2S - I - 10 e - = 2S IV

O 2 0 + 4 e - = 2O - II+4 e - 11

In examples 3 and 4, the functions of the oxidizing and reducing agent are divided between different substances, Fe 2 O 3 and O 2 - oxidizing agents, CO and Fe(S)2 - reducing agents; Such reactions are classified as intermolecular redox reactions.

When intramolecular oxidation-reduction, when in the same substance the atoms of one element are oxidized and the atoms of another element are reduced, the calculation is carried out per one formula unit of the substance.

Example 5. Select the coefficients in the oxidation-reduction reaction equation

(NH 4) 2 CrO 4 ® Cr 2 O 3 + N 2 + H 2 O + NH 3

Solution

2 (NH 4) 2 CrO 4 = Cr 2 O 3 + N 2 +5 H 2 O + 2 NH 3

Cr VI + 3 e - = Cr III 2

2N - III - 6 e - = N 2 0 1

For reactions dismutation (disproportionation, autoxidation- self-healing), in which atoms of the same element in the reagent are oxidized and reduced, additional factors are first added to the right side of the equation, and then the coefficient for the reagent is found.

Example 6. Select the coefficients in the dismutation reaction equation

H2O2 ® H2O+O2

Solution

2 H 2 O 2 = 2 H 2 O + O 2

O - I+ e - = O - II 2

2O - I - 2 e - = O 2 0 1

For the commutation reaction ( synproportionation), in which atoms of the same element of different reagents, as a result of their oxidation and reduction, receive the same oxidation state, additional factors are first added to the left side of the equation.

Example 7. Select the coefficients in the commutation reaction equation:

H 2 S + SO 2 = S + H 2 O

Solution

2H2S + SO2 = 3S + 2H2O

S - II - 2 e - = S 0 2

SIV+4 e - = S 0 1

To select coefficients in the equations of redox reactions occurring in an aqueous solution with the participation of ions, the method is used electron-ion balance. The method for selecting coefficients using electron-ion balance consists of the following steps:

a) write down the formulas of the reagents of this redox reaction

K 2 Cr 2 O 7 + H 2 SO 4 + H 2 S

and establish the chemical function of each of them (here K2Cr2O7 - oxidizing agent, H 2 SO 4 - acidic reaction medium, H2S - reducing agent);

b) write down (on the next line) the formulas of the reagents in ionic form, indicating only those ions (for strong electrolytes), molecules (for weak electrolytes and gases) and formula units (for solids) that will take part in the reaction as an oxidizing agent ( Cr2O72 - ), environment ( H+- more precisely, oxonium cation H3O+ ) and reducing agent ( H2S):

Cr2O72 - +H++H2S

c) determine the reduced formula of the oxidizing agent and the oxidized form of the reducing agent, which must be known or specified (for example, here the dichromate ion passes chromium cations ( III), and hydrogen sulfide - into sulfur); This data is written down on the next two lines, the electron-ion equations for the reduction and oxidation half-reactions are drawn up, and additional factors are selected for the half-reaction equations:

half-reaction reduction of Cr 2 O 7 2 - + 14 H + + 6 e - = 2 Cr 3+ + 7 H 2 O 1

half-reaction oxidation of H 2 S - 2 e - = S (t) + 2 H + 3

d) compose, by summing up the half-reaction equations, the ionic equation of a given reaction, i.e. supplement entry (b):

Cr2O72 - + 8 H + + 3 H 2 S = 2 Cr 3+ + 7 H 2 O + 3 S ( T )

d) based on the ionic equation, make up the molecular equation of this reaction, i.e. supplement entry (a), and the formulas of cations and anions that are missing in the ionic equation are grouped into the formulas of additional products ( K2SO4):

K 2 Cr 2 O 7 + 4H 2 SO 4 + 3H 2 S = Cr 2 (SO 4) 3 + 7H 2 O + 3S ( t ) + K 2 SO 4

f) check the selected coefficients by the number of atoms of the elements on the left and right sides of the equation (usually it is enough to only check the number of oxygen atoms).

OxidizedAnd restored The oxidizing and reducing forms often differ in oxygen content (compare Cr2O72 - and Cr 3+ ). Therefore, when compiling half-reaction equations using the electron-ion balance method, they include the pairs H + / H 2 O (for an acidic medium) and OH - / H 2 O (for alkaline environment). If, when moving from one form to another, the original form (usually - oxidized) loses its oxide ions (shown below in square brackets), then the latter, since they do not exist in free form, must be combined with hydrogen cations in an acidic environment, and in an alkaline environment - with water molecules, which leads to the formation of water molecules (in an acidic environment) and hydroxide ions (in an alkaline environment):

acidic environment[ O2 - ] + 2 H + = H 2 O

alkaline environment[ O 2 - ] + H 2 O = 2 OH -

Lack of oxide ions in their original form (usually- in reduced) compared to the final form is compensated by the addition of water molecules (in an acidic environment) or hydroxide ions (in an alkaline environment):

acidic environment H 2 O = [ O 2 - ] + 2 H +

alkaline environment2 OH - = [ O 2 - ] + H 2 O

Example 8. Select the coefficients using the electron-ion balance method in the equation of the redox reaction:

® MnSO 4 + H 2 O + Na 2 SO 4 + ¼

Solution

2 KMnO 4 + 3 H 2 SO 4 + 5 Na 2 SO 3 =

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

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

MnO4 - + 8H + + 5 e - = Mn 2+ + 4 H 2 O2

SO 3 2 - +H2O - 2 e - = SO 4 2 - + 2 H + 5

Example 9. Select the coefficients using the electron-ion balance method in the equation of the redox reaction:

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

Solution

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

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

MnO4 - + 1 e - = MnO 4 2 - 2

SO 3 2 - + 2 OH - - 2 e - = SO 4 2 - + H 2 O 1

If the permanganate ion is used as an oxidizing agent in a weakly acidic environment, then the equation for the reduction half-reaction is:

MnO4 - + 4 H + + 3 e - = MnO 2( t) + 2 H 2 O

and if in a slightly alkaline environment, then

MnO 4 - + 2 H 2 O + 3 e - = MnO 2( t) + 4 OH -

Often, a weakly acidic and slightly alkaline medium is conventionally called neutral, and only water molecules are introduced into the half-reaction equations on the left. In this case, when composing the equation, you should (after selecting additional factors) write down an additional equation reflecting the formation of water from H + and OH ions - .

Example 10. Select the coefficients in the equation of the reaction occurring in a neutral medium:

KMnO 4 + H 2 O + Na 2 SO 3 ® Mn ABOUT 2( t) + Na 2 SO 4 ¼

Solution

2 KMnO 4 + H 2 O + 3 Na 2 SO 3 = 2 MnO 2( t) + 3 Na 2 SO 4 + 2 KOH

MnO4 - + H 2 O + 3 SO 3 2 - = 2 MnO 2( t ) + 3 SO 4 2 - + 2 OH -

MnO 4 - + 2 H 2 O + 3 e - = MnO 2( t) + 4 OH -

SO 3 2 - +H2O - 2 e - = SO 4 2 - +2H+

8OH - + 6 H + = 6 H 2 O + 2 OH -

Thus, if the reaction from example 10 is carried out by simple merging aqueous solutions potassium permanganate and sodium sulfite, then it proceeds in a conditionally neutral (and in fact, slightly alkaline) environment due to the formation of potassium hydroxide. If the potassium permanganate solution is slightly acidified, the reaction will proceed in a weakly acidic (conditionally neutral) environment.

Example 11. Select the coefficients in the equation of the reaction occurring in a weakly acidic environment:

KMnO 4 + H 2 SO 4 + Na 2 SO 3 ® Mn ABOUT 2( t) + H 2 O + Na 2 SO 4 + ¼

Solution

2KMnO 4 + H 2 SO 4 + 3Na 2 SO 3 = 2Mn O 2( T ) + H 2 O + 3Na 2 SO 4 + K 2 SO 4

2 MnO 4 - + 2 H + + 3 SO 3 2 - = 2 MnO 2( t ) + H 2 O + 3 SO 4 2 -

MnO4 - + 4H + + 3 e - = Mn O 2( t ) + 2 H 2 O2

SO 3 2 - +H2O - 2 e - = SO 4 2 - + 2 H + 3

Forms of existence of oxidizing agents and reducing agents before and after the reaction, i.e. their oxidized and reduced forms are called redox couples. Thus, from chemical practice it is known (and this must be remembered) that the permanganate ion in an acidic environment forms a manganese cation ( II) (pair MnO 4 - +H+/ Mn 2+ + H 2 O ), in a slightly alkaline environment- manganese(IV) oxide (pair MnO 4 - +H+ ¤ Mn O 2(t) + H 2 O or MnO 4 - + H 2 O = Mn O 2(t) + OH - ). The composition of oxidized and reduced forms is determined, therefore, chemical properties of a given element in various oxidation states, i.e. unequal stability of specific forms in different environments of aqueous solution. All redox couples used in this section are given in problems 2.15 and 2.16.