Task book on general and inorganic chemistry

2.2. Redox reactions

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Theoretical part

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

a) write down the formulas of the reactants 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

MnII¼ = MnIV

b) compose the equations of 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 MnII- 2 e - = MnIV

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

Cl V + 6 e - = Cl - I 1

MnII- 2 e - = Mn IV 3

d) put down (according to the factors found) stoichiometric coefficients in 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 course of the reaction (if there are two such elements, then it is enough to equalize the number of atoms of one of them, and check the second one). Get the equation of the chemical reaction:

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

Example 3. Fit Coefficients in Redox Equation

Fe 2 O 3 + CO ® Fe + CO2

Solution

Fe 2 O 3 + 3 CO \u003d 2 Fe + 3 CO 2

Fe III + 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 Fit Coefficients in Redox Equation

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

Solution

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

FeII- e - = Fe III

- 11 e - 4

2S - I - 10 e - = 2SIV

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 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 Find the coefficients in the equation of the redox reaction

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

Solution

2 (NH 4) 2 CrO 4 \u003d 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 dismutations (disproportionation, autoxidation- self-healing), in which atoms of the same element in the reagent are oxidized and reduced, additional factors are put down first on the right side of the equation, and then the coefficient for the reagent is found.

Example 6. Fit Coefficients in Dismutation Reaction Equation

H2O2 ® H 2 O + O 2

Solution

2 H 2 O 2 \u003d 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 the atoms of the same element of different reagents, as a result of their oxidation and reduction, receive the same oxidation state, additional factors are put down first on the left side of the equation.

Example 7 Select the coefficients in the commutation reaction equation:

H 2 S + SO 2 \u003d S + H 2 O

Solution

2 H 2 S + SO 2 \u003d 3 S + 2H 2 O

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 of selection of coefficients using the 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 - acid reaction medium, H 2 S - 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 - ), environments ( H+- more precisely, the oxonium cation H3O+ ) and reducing agent ( H2S):

Cr2O72 - + H + + H 2 S

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); these data are written on the next two lines, the electron-ion equations of the reduction and oxidation half-reactions are compiled, and additional factors are selected for the half-reaction equations:

half reaction reduction of Cr 2 O 7 2 - + 14 H + + 6 e - \u003d 2 Cr 3+ + 7 H 2 O 1

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

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

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

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

K 2 Cr 2 O 7 + 4H 2 SO 4 + 3H 2 S \u003d Cr 2 (SO 4) 3 + 7H 2 O + 3S ( m) + K 2 SO 4

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

oxidizedand restored forms of oxidizing and reducing agent often differ in oxygen content (compare Cr2O72 - and Cr3+ ). Therefore, when compiling half-reaction equations using the electron-ion balance method, they include H + / H 2 O pairs (for an acidic environment) and OH - / H 2 O (for an alkaline environment). If during the transition 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 a 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):

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

alkaline environment [ O 2 - ] + H 2 O \u003d 2 OH -

Lack of oxide ions in their original form (more often- reduced) in comparison with the final form is compensated by the addition of water molecules (in an acid medium) or hydroxide ions (in an alkaline medium):

acidic environment H 2 O \u003d [ 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 redox reaction equation:

® MnSO 4 + H 2 O + Na 2 SO 4 + ¼

Solution

2 KMnO 4 + 3 H 2 SO 4 + 5 Na 2 SO 3 \u003d

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 redox reaction equation:

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

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

SO 3 2 - + 2OH - - 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 reduction half-reaction equation is:

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

and if in a weakly alkaline medium, then

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

Often, a weakly acidic and weakly alkaline medium is conditionally called neutral, while only water molecules are introduced into the half-reaction equations on the left. In this case, when compiling the equation, one should (after selecting additional factors) write an additional equation that reflects the formation of water from H + and OH ions - .

Example 10. Select the coefficients in the equation for the reaction taking place in a neutral medium:

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

Solution

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

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

MNO 4 - + 2 H 2 O + 3 e - = MnO 2( m) + 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 simply draining aqueous solutions of potassium permanganate and sodium sulfite, then it proceeds in a conditionally neutral (and in fact, in a slightly alkaline) environment due to the formation of potassium hydroxide. If the solution of potassium permanganate is slightly acidified, then the reaction will proceed in a weakly acidic (conditionally neutral) medium.

Example 11. Select the coefficients in the equation for the reaction taking place in a weakly acidic environment:

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

Solution

2KMnO 4 + H 2 SO 4 + 3Na 2 SO 3 \u003d 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. So, it is known from chemical practice (and this needs to be remembered) that the permanganate ion in an acidic medium forms a manganese cation ( II ) (pair MNO 4 - + H + / Mn 2+ + H 2 O ), in a weakly alkaline medium- 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 the oxidized and reduced forms is determined, therefore, chemical properties of this element in various oxidation states, i.e. unequal stability of specific forms in various media of an aqueous solution. All redox pairs used in this section are given in problems 2.15 and 2.16.

With an increase in the degree of oxidation an oxidation process takes place, and the substance itself is a reducing agent. When the oxidation state decreases, the reduction process proceeds, and the substance itself is an oxidizing agent.

The described method of OVR equalization is called the “oxidation state balance method”.

Stated in most chemistry textbooks and widely used in practice electronic balance method for equalization, OVR can be used with the caveat that the oxidation state is not equal to the charge.

2. Method of half-reactions.

In those cases, when the reaction proceeds in an aqueous solution (melt), when drawing up equations, they proceed not from a change in the oxidation state of the atoms that make up the reactants, but from a change in the charges of real particles, that is, they take into account the form of existence of substances in a solution (simple or complex ion, atom or a molecule of an undissolved or weakly dissociating substance in water).

In this case when compiling ionic equations of redox reactions, one should adhere to the same form of notation that is adopted for ionic equations of an exchange nature, namely: poorly soluble, poorly dissociated and gaseous compounds should be written in molecular form, and ions that do not change their state should be excluded from the equation . In this case, the processes of oxidation and reduction are recorded as separate half-reactions. Having equalized them according to the number of atoms of each type, the half-reactions are added, multiplying each by a coefficient that equalizes the change in the charge of the oxidizing agent and reducing agent.

The half-reaction method more accurately reflects the true changes in substances in the process of redox reactions and facilitates the formulation of equations for these processes in ionic-molecular form.

Because the from the same reagents different products can be obtained depending on the nature of the medium (acidic, alkaline, neutral), for such reactions in the ionic scheme, in addition to particles that perform the functions of an oxidizing agent and a reducing agent, a particle characterizing the reaction of the medium (that is, an H + ion or an OH ion -, or an H 2 O molecule).

Example 5 Using the half-reaction method, arrange the coefficients in the reaction:

KMnO 4 + KNO 2 + H 2 SO 4 ® MnSO 4 + KNO 3 + K 2 SO 4 + H 2 O.

Solution. We write the reaction in ionic form, given that all substances, except water, dissociate into ions:

MnO 4 - + NO 2 - + 2H + ® Mn 2+ + NO 3 - + H 2 O

(K + and SO 4 2 - remain unchanged, therefore they are not indicated in the ionic scheme). It can be seen from the ionic diagram that the oxidizing agent permanganate ion(MnO 4 -) is converted into Mn 2+ -ion and four oxygen atoms are released.

In an acidic environment each oxygen atom released by the oxidizing agent binds to 2H + to form a water molecule.

this implies: MnO 4 - + 8H + + 5® Mn 2+ + 4H 2 O .

We find the difference in the charges of products and reagents: Dq = +2-7 = -5 (the "-" sign indicates that the reduction process is taking place and 5 is attached to the reagents). For the second process, the conversion of NO 2 - to NO 3 - , the missing oxygen comes from the water to the reducing agent, and as a result, an excess of H + ions is formed, while the reagents lose 2 :

NO 2 - + H 2 O - 2® NO 3 - + 2H + .

Thus we get:

2 | MnO 4 - + 8H + + 5® Mn 2+ + 4H 2 O (reduction),

5 | NO 2 - + H 2 O - 2® NO 3 - + 2H + (oxidation).

Multiplying the terms of the first equation by 2, and the second - by 5 and adding them, we get the ion-molecular equation for this reaction:

2MnO 4 - + 16H + + 5NO 2 - + 5H 2 O \u003d 2Mn 2+ + 8H 2 O + 5NO 3 - + 10H +.

Having canceled identical particles on the left and right sides of the equation, we finally obtain the ion-molecular equation:

2MnO 4 - + 5NO 2 - + 6H + = 2Mn 2+ + 5NO 3 - + 3H 2 O.

According to the ionic equation, we compose a molecular equation:

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

In alkaline and neutral environments you can be guided by the following rules: in an alkaline and neutral environment, each oxygen atom released by the oxidizing agent combines with one water molecule, forming two hydroxide ions (2OH -), and each missing one goes to the reducing agent from 2 OH - ions with the formation of one molecule water in an alkaline environment, and in a neutral one it comes from water with the release of 2 H + ions.

If a involved in redox reactions hydrogen peroxide(H 2 O 2), it is necessary to take into account the role of H 2 O 2 in a particular reaction. In H 2 O 2, oxygen is in an intermediate oxidation state (-1), therefore, hydrogen peroxide in redox reactions exhibits redox duality. In cases where H 2 O 2 is oxidizing agent, the half-reactions have the following form:

H 2 O 2 + 2H + + 2? ® 2H 2 O (acid medium);

H 2 O 2 +2? ® 2OH - (neutral and alkaline environments).

If hydrogen peroxide is reducing agent:

H 2 O 2 - 2? ® O 2 + 2H + (acid medium);

H 2 O 2 + 2OH - - 2? ® O 2 + 2H 2 O (alkaline and neutral).

Example 6 Equalize the reaction: KI + H 2 O 2 + H 2 SO 4 ® I 2 + K 2 SO 4 + H 2 O.

Solution. We write the reaction in ionic form:

I - + H 2 O 2 + 2H + ® I 2 + SO 4 2 - + H 2 O.

We compose half-reactions, given that H 2 O 2 in this reaction is an oxidizing agent and the reaction proceeds in an acidic environment:

1 2I - - 2= I 2 ,

1 H 2 O 2 + 2H + + 2® 2H 2 O.

Final equation: 2KI + H 2 O 2 + H 2 SO 4 ® I 2 + K 2 SO 4 + 2H 2 O.

There are four types of redox reactions:

1 . Intermolecular redox reactions, in which the oxidation states of the atoms of the elements that make up different substances change. The reactions discussed in examples 2-6 are of this type.

2 . Intramolecular redox reactions in which the oxidation state is changed by atoms of different elements of the same substance. According to this mechanism, reactions of thermal decomposition of compounds proceed. For example, in the reaction

Pb(NO 3) 2 ® PbO + NO 2 + O 2

changes the oxidation state of nitrogen (N +5 ® N +4) and oxygen atom (O - 2 ® O 2 0) located inside the Pb(NO 3) 2 molecule.

3. Self-oxidation-self-healing reactions(disproportionation, dismutation). In this case, the oxidation state of the same element both increases and decreases. Disproportionation reactions are characteristic of compounds or elements of substances corresponding to one of the intermediate oxidation states of the element.

Example 7 Using all the above methods, equalize the reaction:

Solution.

a) The method of balance of oxidation states.

Let us determine the oxidation states of the elements involved in the redox process before and after the reaction:

K 2 MnO 4 + H 2 O ® KMnO 4 + MnO 2 + KOH.

It follows from a comparison of oxidation states that manganese simultaneously participates in the oxidation process, increasing the oxidation state from +6 to +7, and in the reduction process, lowering the oxidation state from +6 to +4.2 Mn +6 ® Mn +7 ; Dw = 7-6 = +1 (oxidation process, reducing agent),

1 Mn +6 ® Mn +4 ; Dw = 4-6 = -2 (reduction process, oxidizing agent).

Since in this reaction the same substance (K 2 MnO 4) acts as an oxidizing and reducing agent, the coefficients in front of it are summed up. We write down the equation:

3K 2 MnO 4 + 2H 2 O = 2KMnO 4 + MnO 2 + 4KOH.

b) Method of half-reactions.

The reaction takes place in a neutral environment. Compiling ionic scheme reactions, taking into account that H 2 O is a weak electrolyte, and MnO 2 is an oxide that is poorly soluble in water:

MnO 4 2 - + H 2 O ® MnO 4 - + ¯MnO 2 + OH - .

We write down the half-reactions:

2 MnO 4 2 - - ? ® MnO 4 - (oxidation),

1 MnO 4 2 - + 2H 2 O + 2? ® MnO 2 + 4OH - (recovery).

We multiply by the coefficients and add both half-reactions, we get the total ionic equation:

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

Molecular equation: 3K 2 MnO 4 + 2H 2 O = 2KMnO 4 + MnO 2 + 4KOH.

In this case, K 2 MnO 4 is both an oxidizing agent and a reducing agent.

4. Intramolecular oxidation-reduction reactions, in which the oxidation states of atoms of the same element are aligned (that is, the reverse of those previously considered), are processes counterdisproportionation(switching), for example

NH 4 NO 2 ® N 2 + 2H 2 O.

1 2N - 3 - 6? ® N 2 0 (oxidation process, reducing agent),

1 2N +3 + 6?® N 2 0 (reduction process, oxidizing agent).

The most difficult are redox reactions in which atoms or ions of not one, but two or more elements are simultaneously oxidized or reduced.

Example 8 Equalize the reaction using the above methods:

3 -2 +5 +5 +6 +2

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

Ministry of Education and Science of the Russian Federation

Federal State Budgetary Educational Institution of Higher Professional Education

"Siberian State Industrial University"

Department of General and Analytical Chemistry

Redox reactions

Guidelines for performing laboratory and practical exercises

in the disciplines "Chemistry", "Inorganic Chemistry",

"General and Inorganic Chemistry"

Novokuznetsk

UDC 544.3(07)

Reviewer

Candidate of Chemical Sciences, Associate Professor,

head Department of Physical Chemistry and TMP SibSIU

A.I. Poshevneva

O-504 Redox reactions: method. decree. / Sib. state industry un-t; comp. : P.G. Permyakov, R.M. Belkina, S.V. Zentsov. - Novokuznetsk: Ed. center SibGIU 2012. - 41 p.

Theoretical information, examples of solving problems on the topic "Oxidation-reduction reactions" in the disciplines "Chemistry", "Inorganic chemistry", "General and inorganic chemistry" are given. Laboratory works and questions for self-control developed by the team of authors, control and test tasks for performing control and independent work are presented.

Designed for first-year students of all areas of training.

Foreword

Methodical instructions in chemistry are compiled according to the program for technical areas of higher educational institutions, are designed to organize independent work on the topic "Oxidation-reduction reactions" on educational material during classroom and non-classroom time.

Independent work in the study of the topic "Oxidation-reduction reactions" consists of several elements: the study of theoretical material, the implementation of control and test tasks according to this methodological instruction and individual consultations with the teacher.

As a result of independent work, it is necessary to master the basic terms, definitions, concepts and master the technique of chemical calculations. The implementation of control and test tasks should be started only after a deep study of the theoretical material and a thorough analysis of the examples of typical tasks given in the theoretical section.

The authors hope that the guidelines will allow students not only to successfully master the proposed material on the topic "Oxidation-reduction reactions", but will also become useful for them in educational process when mastering the disciplines "Chemistry", "Inorganic chemistry".

Redox reactions Terms, definitions, concepts

Redox reactions- these are reactions accompanied by the transition of electrons from one atom or ion to another, in other words, these are reactions that result in changes in the oxidation states of elements.

Oxidation state is the charge of an atom of an element in a compound, calculated from the conditional assumption that all bonds in the molecule are ionic.

The degree of oxidation is usually indicated by an Arabic numeral above the symbol of the element with a plus or minus sign in front of the number. For example, if the bond in the HCl molecule is ionic, then hydrogen and chlorine ions with charges (+1) and (–1), therefore
.

Using the above rules, we calculate the oxidation states of chromium in K 2 Cr 2 O 7, chlorine in NaClO, sulfur in H 2 SO 4, nitrogen in NH 4 NO 2:

2(+1) + 2 x + 7(–2) = 0, x = +6;

+1 + x + (–2) = 0, x = +1;

2(+1) + x + 4(–2) = 0, x = +6;

x+4(+1)=+1, y+2(-2)=-1,

x = -3, y = +3.

Oxidation and reduction. Oxidation is the release of electrons, as a result of which the oxidation state of an element increases. Reduction is the addition of electrons, as a result of which the oxidation state of an element decreases.

Oxidation and reduction processes are closely related, since a chemical system can only donate electrons when another system attaches them ( redox system). The electron-accepting system ( oxidizer) is itself reduced (turns into the corresponding reducing agent), and the electron donating system ( reducing agent), oxidizes itself (turns into the corresponding oxidizing agent).

Example 1 Consider the reaction:

The number of electrons donated by the atoms of the reducing agent (potassium) is equal to the number of electrons attached by the molecules of the oxidizing agent (chlorine). Therefore, one chlorine molecule can oxidize two potassium atoms. Equalizing the number of received and given electrons, we obtain:

To typical oxidizers include:

Elementary substances - Cl 2, Br 2, F 2, I 2, O, O 2.

Compounds in which the elements exhibit the highest oxidation state (determined by the group number) -

Cation H + and metal ions in their highest oxidation state - Sn 4+, Cu 2+, Fe 3+, etc.

To typical reducing agents include:

Redox duality.Compounds of the highest oxidation state inherent in a given element can only act as oxidizing agents in redox reactions, the oxidation state of the element can only decrease in this case. Compounds of the lowest oxidation state they can be, on the contrary, only reducing agents; here the oxidation state of the element can only increase. If the element is in an intermediate oxidation state, then its atoms can, depending on the conditions, accept electrons, acting as an oxidizing agent, or donate electrons, acting as a reducing agent.

So, for example, the degree of nitrogen oxidation in compounds varies from (-3) to (+5) (Figure 1):

NH 3 , NH 4 OH only

reducing agents

HNO 3 , HNO 3 salts

only oxidizers

Compounds with intermediate nitrogen oxidation states 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 oxidation of nitrogen

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 electron balance method using the example of iron oxidation with oxygen:
.

Fe 0 - 3ē \u003d Fe +3 - the oxidation process;

O 2 + 4ē \u003d 2O -2 - the recovery process.

In the reducing agent system (half-reaction of the oxidation process), the iron atom donates 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 donates 12 electrons, and oxygen accepts 12 electrons:

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

Half reaction method used to equalize reactions occurring in electrolyte solutions. In such cases, not only an oxidizing agent and a reducing agent, but also particles of the medium take part in the reactions: water molecules (H 2 O), H + and OH - - ions. More correct for such reactions is the use of electron-ionic systems (half-reactions). When compiling half-reactions in aqueous solutions, H 2 O molecules and H + or OH ions are introduced, if necessary, taking into account the reaction medium. Weak electrolytes, poorly 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:

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

The purpose 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 parts of each half-reaction. To achieve this, additional ions or molecules of the medium are used. Usually these are H + , OH - ions and water molecules. In half reaction
the number of manganese atoms is the same, but the number of oxygen atoms is not equal, therefore, we introduce four water molecules into the right side of the half-reaction: . Having carried 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 to the equivalent number of hydrogen ions.

Now all the elements included in the half-reaction equations are 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 side the charge is +7. The equality of charges is carried out 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 you can put equal signs in the equations of both half-reactions:

– recovery process;

is the process of oxidation.

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

Multiplying the coefficients by all terms of the half-reaction equations and summing only their right and only their left parts, we obtain the final reaction equation in the ion-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 the 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 to the molecular equation:

Redox 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 is oxidized 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 \u003d 4OH - - reduction process.

In order to make sure that the recording of electron-ionic systems is correct, it is necessary to check: the left and right parts of the half-reactions must contain the same number of elemental atoms and charge. Then, equalizing the number of received and given electrons, 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

Self-oxidation-self-healing reactions (disproportionation reactions) - these are reactions during which part of the total amount of the element is oxidized, and the other part is reduced, which is 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 \u003d 2HClO + 2H + - oxidation process;

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

2Cl 2 + 2H 2 O \u003d 2HClO + 2HCl

Example 7 . Disproportionation of nitrous acid:

In this case, oxidation and reduction undergoes as part of HNO 2:

Summary Equation:

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

3HNO 2 \u003d HNO 3 + 2NO + H 2 O

Intramolecular oxidation-reduction reactions - This is a process when one component of the molecule serves as an oxidizing agent, and the other as a reducing agent. Many thermal dissociation processes can be examples of intramolecular redox.

Example 8 Thermal dissociation of NH 4 NO 2:

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

12NH - 6 ē \u003d N 2 + 8H +

1 2NO + 8Н + + 6 ē \u003d N 2 + 4H 2 O

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

2NH 4 NO 2 \u003d 2N 2 + 4H 2 O

Example 9 . The decomposition reaction of ammonium dichromate:

12NH - 6 ē \u003d N 2 + 8H +

1 Cr 2 O + 8Н + + 6 ē \u003d 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 \u003d N 2 + Cr 2 O 3 + 4H 2 O

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

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

FeS2 + HNO3
Fe 2 (SO 4) 3 + NO + ...

The equation is not written down to the end and the use of electron-ionic 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 restored to NO:

S–1 → ()

We write down the oxidation half-reaction of FeS 2:

FeS2 → Fe3+ +

The presence of two Fe 3+ ions in Fe 2 (SO 4) 3 suggests doubling the number of iron atoms with further recording of 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 + 16Н 2 O → 2Fe 3+ + 4
.

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

2FeS 2 + 16Н 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), you need to subtract 30 ē:

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

10 NO + 4H + + 3 ē \u003d NO + 2H 2 O - recovery.

2FeS 2 + 16Н 2 O + 10NO + 40Н + = 2Fe 3+ + 4
+ 32H + + 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 +) by subtraction and get:

2FeS 2 + 10HNO 3 \u003d Fe 2 (SO 4) 3 + 10NO + H 2 SO 4 + 4H 2 O

Energetics of redox reactions . The condition for the spontaneous occurrence of any process, including the 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 oxidizer in the elementary oxidation-reduction act;

F is the Faraday number;

ε is the 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 \u003d E ok - E in,

Under standard conditions:

ε ° \u003d E ° ok - E ° in.

So, if the condition for the spontaneous flow 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 ° v.

Example 11. Determine the possibility of a redox reaction:

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

Cu - 2ē \u003d Cu 2+ E ° in \u003d +0.34 V

2H + + 2ē \u003d H 2 E ° ok \u003d 0.0 V

It can be seen from the half-reactions that E ° 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 the 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 \u003d 2Fe 3+ E ° in \u003d +0.77 V

∆G ° \u003d -n F ε ° \u003d -n F (E ° ok - E ° c),

where n \u003d 10, since the reducing agent gives 10 ē, the oxidizing agent takes 10 ē in the elementary act of oxidation-reduction.

∆G ° \u003d -10 69500 ​​(1.52-0.77) \u003d -725000 J,

∆G° = -725 kJ.

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

∆G ° = –RTlnK s or n F ε = RTlnK s,

where R \u003d 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 constant values ​​​​in the equation, converting the natural logarithm to decimal:

K c \u003d 10 127.

The data obtained indicate that the considered reduction reaction of potassium permanganate 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. OVR hydrogen peroxide

In hydrogen peroxide H 2 O 2 molecules, oxygen atoms are in the –I oxidation state. This is an intermediate and not the most stable oxidation state of the atoms of this element, so 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 ones and also depend on the concentration.

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

Depending on the reaction of the medium and on whether the oxidizing or reducing agent is hydrogen peroxide 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 for redox half-reactions of H 2 O 2 in solutions

Environment reaction	H 2 O 2 oxidizer	H 2 O 2 reducing agent
Acid
Neutral	H 2 O 2 + 2e - \u003d 2OH	H 2 O 2 + 2H 2 O - 2e - \u003d O 2 + 2H 3 O
alkaline	HO 2 + H 2 O + 2e - \u003d 3OH

Let us consider examples of OVR 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.

1	H 2 O 2 + 2H 3 O + 2e - = 4H 2 O
1	2I – 2e – = I 2

H 2 O 2 + 2H 3 O + 2I \u003d 4H 2 O + I 2
H 2 O 2 + H 2 SO 4 + 2KI \u003d 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.

2	MnO 4 + 8H 3 O + 5e - \u003d Mn 2 + 12H 2 O
5	H 2 O 2 + 2H 2 O - 2e - \u003d O 2 + 2H 3 O

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.

3	6	HO 2 + H 2 O + 2e - \u003d 3OH
1	2	I + 6OH - 6e - \u003d IO 3 + 3H 2 O

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 \u003d 3H 2 O + NaIO 3 (in the presence of NaOH)

The same equation will be obtained if the formation of hydroperoxide ions is not taken into account immediately (at the stage of compiling the balance).

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. Recovering under these conditions, it forms Pb 2 ions. In an alkaline environment, when PbO 2 is reduced, ions are formed.

1	PbO 2 + 2H 2 O + 2e - = + OH
1	HO 2 + OH - 2e - \u003d O 2 + H 2 O

PbO 2 + H 2 O + HO 2 \u003d + 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 assignment condition, the added hydrogen peroxide solution was alkaline, then the molecular equation should be written as follows:

PbO 2 + H 2 O + KHO 2 \u003d K + O 2

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

PbO 2 + KOH + H 2 O 2 \u003d K + O 2

18.6. OVR dismutations and intramolecular OVR

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

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 is reduced to the –I oxidation state:

Let us use the electron-ion balance method to compose an equation for a similar reaction that occurs when chlorine is passed through a cold alkali solution, for example, KOH:

1	Cl 2 + 2e - \u003d 2Cl
1	Cl 2 + 4OH - 2e - \u003d 2ClO + 2H 2 O

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

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

Cl 2 + 2OH \u003d Cl + ClO + H 2 O
Cl 2 + 2KOH \u003d KCl + KClO + H 2 O

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

5		Cl 2 + 2e - \u003d 2Cl
1		Cl 2 + 12OH - 10e - \u003d 2ClO 3 + 6H 2 O

3Cl 2 + 6OH = 5Cl + ClO 3 + 3H 2 O
3Cl 2 + 6KOH \u003d 5KCl + KClO 3 + 3H 2 O

big practical value has the dismutation of nitrogen dioxide when it reacts with water ( a) and with alkali solutions ( b):

a)		NO 2 + 3H 2 O - e - \u003d NO 3 + 2H 3 O			NO 2 + 2OH - e - \u003d NO 3 + H 2 O
		NO 2 + H 2 O + e - \u003d HNO 2 + OH			NO 2 + e - \u003d NO 2
	2NO 2 + 2H 2 O \u003d NO 3 + H 3 O + HNO 2			2NO 2 + 2OH \u003d NO 3 + NO 2 + H 2 O
	2NO 2 + H 2 O \u003d HNO 3 + HNO 2			2NO 2 + 2NaOH \u003d NaNO 3 + NaNO 2 + H 2 O

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

4KClO 3 \u003d KCl + 3KClO 4

A characteristic and very effective example of intramolecular OVR 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 decomposes rapidly. In this case, chromium(VI) transforms into chromium(III), the most stable state of chromium, while nitrogen(–III) transforms into nitrogen(0), also the most stable state. Taking into account the number of atoms in the formula unit of the electronic balance equation:

	2Cr + VI + 6e – = 2Cr + III
	2N -III - 6e - \u003d N 2,

and the reaction equation itself:

(NH 4) 2 Cr 2 O 7 \u003d Cr 2 O 3 + N 2 + 4H 2 O.

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

1		Cl + VII + 8e – = Cl –I
2		2O -II - 4e - \u003d O 2

and hence the reaction equation

KClO 4 \u003d KCl + 2O 2

Similarly, potassium chlorate KClO 3 decomposes when heated, if the decomposition is carried out in the presence of a catalyst (MnO 2): 2KClO 3 \u003d 2KCl + 3O 2

In the absence of a catalyst, the dismutation reaction proceeds.
The group of intramolecular OVR also includes reactions of thermal decomposition of nitrates.
Usually, 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 weak heating, dehydration of nitrate occurs [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 ], if the water is more strongly bound [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 at a higher temperature decompose to oxides .

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

A somewhat simplified scheme of the thermal decomposition of nitrates is shown in fig. 5.

Examples of successive 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(NO 3) 2 . 6H 2 O Cu(NO 3) 2 CuO Cu 2 O Cu;

Bi(NO3)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 ongoing processes, when answering the question of what will happen when the corresponding anhydrous nitrate is "calcined" (that is, at a temperature of 400 - 500 o C), they are 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 a series of voltages - from magnesium to copper) decompose to oxides;
3) nitrates of the least active metals (to the right of copper in the voltage series) 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.

Consider typical examples of reactions related to these three types:

KNO 3 KNO 2 + O 2 2 N + V + 2e– = N + III 1 2O– II – 4e– = O 2 2KNO 3 \u003d 2KNO 2 + O 2

Zn(NO 3) 2 ZnO + NO 2 + O 2 4S N + V + e– = N + IV 1½ 2O– II – 4e– = O 2 2Zn(NO 3) 2 \u003d 2ZnO + 4NO 2 + O 2

AgNO 3 Ag + NO 2 + O 2 18.7. Redox switching reactions These reactions can be both intermolecular and intramolecular. For example, intramolecular OVR occurring during the thermal decomposition of ammonium nitrate and nitrite belong to commutation reactions, since the degree of oxidation of nitrogen atoms is equalized here: NH 4 NO 3 \u003d N 2 O + 2H 2 O (about 200 o C) NH 4 NO 2 \u003d N 2 + 2H 2 O (60 - 70 o C) At a higher temperature (250 - 300 o C), ammonium nitrate decomposes to N 2 and NO, and at an even higher temperature (above 300 o C) to nitrogen and oxygen, in both cases water is formed. An example of an intermolecular switching reaction is the reaction that occurs when hot solutions of potassium nitrite and ammonium chloride are poured: NH 4 + NO 2 \u003d N 2 + 2H 2 O NH 4 Cl + KNO 2 \u003d KCl + N 2 + 2H 2 O If a similar reaction is carried out by heating a mixture of crystalline ammonium sulfate and calcium nitrate, then, depending on the conditions, the reaction can proceed in different ways: (NH 4) 2 SO 4 + Ca(NO 3) 2 = 2N 2 O + 4H 2 O + CaSO 4 (t< 250 o C) (NH 4) 2 SO 4 + Ca (NO 3) 2 \u003d 2N 2 + O 2 + 4H 2 O + CaSO 4 (t\u003e 250 o C) 7(NH 4) 2 SO 4 + 3Ca(NO 3) 2 \u003d 8N 2 + 18H 2 O + 3CaSO 4 + 4NH 4 HSO 4 (t\u003e 250 o C) The first and third of these reactions are commutation reactions, the second is a more complex reaction, including both the commutation of nitrogen atoms and the oxidation of oxygen atoms. Which of the reactions will proceed at a temperature above 250 o C depends on the ratio of the reagents. Switching reactions leading to the formation of chlorine occur when salts of oxygen-containing chlorine acids are treated with hydrochloric acid, for example: 6HCl + KClO 3 \u003d KCl + 3Cl 2 + 3H 2 O Also, by the switching reaction, sulfur is formed from gaseous hydrogen sulfide and sulfur dioxide: 2H 2 S + SO 2 \u003d 3S + 2H 2 O OVR commutations are quite numerous and varied - they even include some acid-base reactions, for example: NaH + H 2 O \u003d NaOH + H 2. Both the electron-ionic and electronic balances are used to compile the equations of the OVR commutation, depending on whether a given reaction occurs in a solution or not. 18.8. Electrolysis In studying Chapter IX, you became acquainted with the electrolysis of melts of various substances. Since mobile ions are also present in solutions, solutions of various electrolytes can also be subjected to electrolysis. Both in the electrolysis of melts and in the electrolysis of solutions, electrodes are usually used made of a material that does not react (graphite, platinum, etc.), but sometimes electrolysis is also carried out with a "soluble" anode. "Soluble" anode is used in those cases when it is necessary to obtain an electrochemical connection of the element from which the anode is made. During electrolysis, it has great importance the anode and cathode spaces are separated, or the electrolyte is mixed during the reaction - the reaction products in these cases may turn out to be different. Consider the most important cases of electrolysis. 1. Electrolysis of NaCl melt. The electrodes are inert (graphite), the anode and cathode spaces are separated. As you already know, in this case, reactions take place on the cathode and on the anode: K: Na + e - = Na A: 2Cl - 2e - \u003d Cl 2 Having thus written the equations of reactions occurring on the electrodes, we obtain half-reactions with which we can act in exactly the same way as in the case of using the electron-ion balance method: 2 Na + e - = Na 1 2Cl - 2e - \u003d Cl 2 Adding these half-reaction equations, we obtain the ionic electrolysis equation 2Na + 2Cl 2Na + Cl2 and then molecular 2NaCl 2Na + Cl 2 In this case, the cathode and anode spaces must be separated so that the reaction products do not react with each other. In industry, this reaction is used to produce metallic sodium. 2. Electrolysis of K 2 CO 3 melt. The electrodes are inert (platinum). The cathode and anode spaces are separated. 4 K + e - = K 1 2CO 3 2 - 4e - \u003d 2CO 2 + O 2 4K+ + 2CO 3 2 4K + 2CO 2 + O 2 2K 2 CO 3 4K + 2CO 2 + O 2 3. Electrolysis of water (H 2 O). The electrodes are inert. 2 2H 3 O + 2e - \u003d H 2 + 2H 2 O 1 4OH - 4e - \u003d O 2 + 2H 2 O 4H 3 O + 4OH 2H 2 + O 2 + 6H 2 O 2H 2 O 2H 2 + O 2 Water is a very weak electrolyte, it contains very few ions, so the electrolysis of pure water is extremely slow. 4. Electrolysis of CuCl 2 solution. Graphite electrodes. The system contains Cu 2 and H 3 O cations, as well as Cl and OH anions. Cu 2 ions are stronger oxidizing agents than H 3 O ions (see the series of voltages), therefore, copper ions will first of all be discharged at the cathode, and only when there are very few of them left, oxonium ions will be discharged. For anions, you can follow the following rule: