Task book on general and inorganic chemistry
2.2. Redox reactions
See tasks >>>Theoretical part
Redox reactions include chemical reactions that are accompanied by a change 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 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
Mn II¼ = Mn IV
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 Mn II- 2 e - = Mn IV
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 of electrons donated in the oxidation half-reaction:
Cl V + 6 e - = Cl - I 1
Mn II- 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
Decision
Fe 2 O 3 + 3 CO \u003d 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 Fit Coefficients in Redox Equation
Fe(S ) 2 + O 2 = Fe 2 O 3 + SO 2
Decision
4 Fe(S ) 2 + 11 O 2 = 2 Fe 2 O 3 + 8 SO 2
FeII- e - = FeIII
- 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
Decision
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 the 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
Decision
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
Decision
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 the 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 the elements in the left and right parts of the equation (usually it is enough to check 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
Decision
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
Decision
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 equations of half-reactions 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
Decision
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 of 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
Decision
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 oxidized and reduced forms is determined, therefore, by the chemical properties of a given element in various degrees of oxidation, 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.
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 a mass fraction of 20%, hydrogen peroxide is a rather 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
Of great practical importance is the dismutation of nitrogen dioxide during its reaction 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 heating is carried out to a sufficiently high temperature, or the corresponding oxide is unstable (Ag 2 O, HgO), then a 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 happens 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
2KNO 3 \u003d 2KNO 2 + O 2 |
Zn(NO 3) 2 ZnO + NO 2 + 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 switching reactions, since the oxidation state of nitrogen atoms is equalized here: NH 4 NO 3 \u003d N 2 O + 2H 2 O (about 200 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) 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. To compile the equations of the OVR commutation, both electron-ionic and electronic balances are used, 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 made of a material that does not react (graphite, platinum, etc.) are usually used, 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 is of great importance that 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 Having written down the equations of reactions occurring on the electrodes in this way, 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:
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.
4K+ + 2CO 3 2 4K + 2CO 2 + O 2 3. Electrolysis of water (H 2 O). The electrodes are inert.
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: |
Task number 1
Si + HNO 3 + HF → H 2 SiF 6 + NO + ...
N +5 + 3e → N +2 │4 reduction reaction
Si 0 - 4e → Si +4 │3 oxidation reaction
N +5 (HNO 3) - oxidizing agent, Si - reducing agent
3Si + 4HNO 3 + 18HF → 3H 2 SiF 6 + 4NO + 8H 2 O
Task number 2
Using the electron balance method, write the equation for the reaction:
B+ HNO 3 + HF → HBF 4 + NO 2 + …
Determine the oxidizing agent and reducing agent.
N +5 + 1e → N +4 │3 reduction reaction
B 0 -3e → B +3 │1 oxidation reaction
N +5 (HNO 3) - oxidizing agent, B 0 - reducing agent
B+ 3HNO 3 + 4HF → HBF 4 + 3NO 2 + 3H 2 O
Task number 3
Using the electron balance method, write the equation for the reaction:
K 2 Cr 2 O 7 + HCl → Cl 2 + KCl + … + …
Determine the oxidizing agent and reducing agent.
2Cl -1 -2e → Cl 2 0 │3 oxidation reaction
Cr +6 (K 2 Cr 2 O 7) - oxidizing agent, Cl -1 (HCl) - reducing agent
K 2 Cr 2 O 7 + 14HCl → 3Cl 2 + 2KCl + 2CrCl 3 + 7H 2 O
Task number 4
Using the electron balance method, write the equation for the reaction:
Cr 2 (SO 4) 3 + ... + NaOH → Na 2 CrO 4 + NaBr + ... + H 2 O
Determine the oxidizing agent and reducing agent.
Br 2 0 + 2e → 2Br -1 │3 reduction reaction
2Cr +3 - 6e → 2Cr +6 │1 oxidation reaction
Br 2 - oxidizing agent, Cr +3 (Cr 2 (SO 4) 3) - reducing agent
Cr 2 (SO 4) 3 + 3Br 2 + 16NaOH → 2Na 2 CrO 4 + 6NaBr + 3Na 2 SO 4 + 8H 2 O
Task number 5
Using the electron balance method, write the equation for the reaction:
K 2 Cr 2 O 7 + ... + H 2 SO 4 → l 2 + Cr 2 (SO 4) 3 + ... + H 2 O
Determine the oxidizing agent and reducing agent.
2Cr +6 + 6e → 2Cr +3 │1 reduction reaction
2I -1 -2e → l 2 0 │3 oxidation reaction
Cr +6 (K 2 Cr 2 O 7) - oxidizing agent, l -1 (Hl) - reducing agent
K 2 Cr 2 O 7 + 6HI + 4H 2 SO 4 → 3l 2 + Cr 2 (SO 4) 3 + K 2 SO 4 + 7H 2 O
Task number 6
Using the electron balance method, write the equation for the reaction:
H 2 S + HMnO 4 → S + MnO 2 + ...
Determine the oxidizing agent and reducing agent.
3H 2 S + 2HMnO 4 → 3S + 2MnO 2 + 4H 2 O
Task number 7
Using the electron balance method, write the equation for the reaction:
H 2 S + HClO 3 → S + HCl + ...
Determine the oxidizing agent and reducing agent.
S -2 -2e → S 0 │3 oxidation reaction
Mn +7 (HMnO 4) - oxidizing agent, S -2 (H 2 S) - reducing agent
3H 2 S + HClO 3 → 3S + HCl + 3H 2 O
Task number 8
Using the electron balance method, write the equation for the reaction:
NO + HClO 4 + ... → HNO 3 + HCl
Determine the oxidizing agent and reducing agent.
Cl +7 + 8e → Cl -1 │3 reduction reaction
N +2 -3e → N +5 │8 oxidation reaction
Cl +7 (HClO 4) - oxidizing agent, N +2 (NO) - reducing agent
8NO + 3HClO 4 + 4H 2 O → 8HNO 3 + 3HCl
Task number 9
Using the electron balance method, write the equation for the reaction:
KMnO 4 + H 2 S + H 2 SO 4 → MnSO 4 + S + ... + ...
Determine the oxidizing agent and reducing agent.
S -2 -2e → S 0 │5 oxidation reaction
Mn +7 (KMnO 4) - oxidizing agent, S -2 (H 2 S) - reducing agent
2KMnO 4 + 5H 2 S + 3H 2 SO 4 → 2MnSO 4 + 5S + K 2 SO 4 + 8H 2 O
Task number 10
Using the electron balance method, write the equation for the reaction:
KMnO 4 + KBr + H 2 SO 4 → MnSO 4 + Br 2 + ... + ...
Determine the oxidizing agent and reducing agent.
Mn +7 + 5e → Mn +2 │2 reduction reaction
2Br -1 -2e → Br 2 0 │5 oxidation reaction
Mn +7 (KMnO 4) - oxidizing agent, Br -1 (KBr) - reducing agent
2KMnO 4 + 10KBr + 8H 2 SO 4 → 2MnSO 4 + 5Br 2 + 6K 2 SO 4 + 8H 2 O
Task number 11
Using the electron balance method, write the equation for the reaction:
PH 3 + HClO 3 → HCl + ...
Determine the oxidizing agent and reducing agent.
Cl +5 + 6e → Cl -1 │4 reduction reaction
Cl +5 (HClO 3) - oxidizing agent, P -3 (H 3 PO 4) - reducing agent
3PH 3 + 4HClO 3 → 4HCl + 3H 3 PO 4
Task number 12
Using the electron balance method, write the equation for the reaction:
PH 3 + HMnO 4 → MnO 2 + … + …
Determine the oxidizing agent and reducing agent.
Mn +7 + 3e → Mn +4 │8 reduction reaction
P -3 - 8e → P +5 │3 oxidation reaction
Mn +7 (HMnO 4) - oxidizing agent, P -3 (H 3 PO 4) - reducing agent
3PH 3 + 8HMnO 4 → 8MnO 2 + 3H 3 PO 4 + 4H 2 O
Task number 13
Using the electron balance method, write the equation for the reaction:
NO + KClO + … → KNO 3 + KCl + …
Determine the oxidizing agent and reducing agent.
Cl +1 + 2e → Cl -1 │3 reduction reaction
N +2 − 3e → N +5 │2 oxidation reaction
Cl +1 (KClO) - oxidizing agent, N +2 (NO) - reducing agent
2NO + 3KClO + 2KOH → 2KNO 3 + 3KCl + H 2 O
Task number 14
Using the electron balance method, write the equation for the reaction:
PH 3 + AgNO 3 + ... → Ag + ... + HNO 3
Determine the oxidizing agent and reducing agent.
Ag +1 + 1e → Ag 0 │8 reduction reaction
P -3 - 8e → P +5 │1 oxidation reaction
Ag +1 (AgNO 3) - oxidizing agent, P -3 (PH 3) - reducing agent
PH 3 + 8AgNO 3 + 4H 2 O → 8Ag + H 3 PO 4 + 8HNO 3
Task number 15
Using the electron balance method, write the equation for the reaction:
KNO 2 + ... + H 2 SO 4 → I 2 + NO + ... + ...
Determine the oxidizing agent and reducing agent.
N +3 + 1e → N +2 │ 2 reduction reaction
2I -1 - 2e → I 2 0 │ 1 oxidation reaction
N +3 (KNO 2) - oxidizing agent, I -1 (HI) - reducing agent
2KNO 2 + 2HI + H 2 SO 4 → I 2 + 2NO + K 2 SO 4 + 2H 2 O
Task number 16
Using the electron balance method, write the equation for the reaction:
Na 2 SO 3 + Cl 2 + ... → Na 2 SO 4 + ...
Determine the oxidizing agent and reducing agent.
Cl 2 0 + 2e → 2Cl -1 │1 reduction reaction
Cl 2 0 - oxidizing agent, S +4 (Na 2 SO 3) - reducing agent
Na 2 SO 3 + Cl 2 + H 2 O → Na 2 SO 4 + 2HCl
Task number 17
Using the electron balance method, write the equation for the reaction:
KMnO 4 + MnSO 4 + H 2 O → MnO 2 + ... + ...
Determine the oxidizing agent and reducing agent.
Mn +7 + 3e → Mn +4 │2 reduction reaction
Mn +2 − 2e → Mn +4 │3 oxidation reaction
Mn +7 (KMnO 4) - oxidizing agent, Mn +2 (MnSO 4) - reducing agent
2KMnO 4 + 3MnSO 4 + 2H 2 O → 5MnO 2 + K 2 SO 4 + 2H 2 SO 4
Task number 18
Using the electron balance method, write the equation for the reaction:
KNO 2 + ... + H 2 O → MnO 2 + ... + KOH
Determine the oxidizing agent and reducing agent.
Mn +7 + 3e → Mn +4 │2 reduction reaction
N +3 − 2e → N +5 │3 oxidation reaction
Mn +7 (KMnO 4) - oxidizing agent, N +3 (KNO 2) - reducing agent
3KNO 2 + 2KMnO 4 + H 2 O → 2MnO 2 + 3KNO 3 + 2KOH
Task #19
Using the electron balance method, write the equation for the reaction:
Cr 2 O 3 + ... + KOH → KNO 2 + K 2 CrO 4 + ...
Determine the oxidizing agent and reducing agent.
N +5 + 2e → N +3 │3 reduction reaction
2Cr +3 − 6e → 2Cr +6 │1 oxidation reaction
N +5 (KNO 3) - oxidizing agent, Cr +3 (Cr 2 O 3) - reducing agent
Cr 2 O 3 + 3KNO 3 + 4KOH → 3KNO 2 + 2K 2 CrO 4 + 2H 2 O
Task number 20
Using the electron balance method, write the equation for the reaction:
I 2 + K 2 SO 3 + ... → K 2 SO 4 + ... + H 2 O
Determine the oxidizing agent and reducing agent.
I 2 0 + 2e → 2I -1 │1 reduction reaction
S +4 - 2e → S +6 │1 oxidation reaction
I 2 - oxidizing agent, S +4 (K 2 SO 3) - reducing agent
I 2 + K 2 SO 3 + 2KOH → K 2 SO 4 + 2KI + H 2 O
Task number 21
Using the electron balance method, write the equation for the reaction:
KMnO 4 + NH 3 → MnO 2 +N 2 + ... + ...
Determine the oxidizing agent and reducing agent.
Mn +7 + 3e → Mn +4 │2 reduction reaction
2N -3 - 6e → N 2 0 │1 oxidation reaction
Mn +7 (KMnO 4) - oxidizing agent, N -3 (NH 3) - reducing agent
2KMnO 4 + 2NH 3 → 2MnO 2 + N 2 + 2KOH + 2H 2 O
Task #22
Using the electron balance method, write the equation for the reaction:
NO 2 + P 2 O 3 + ... → NO + K 2 HPO 4 + ...
Determine the oxidizing agent and reducing agent.
N +4 + 2e → N +2 │2 reduction reaction
2P +3 - 4e → 2P +5 │1 oxidation reaction
N +4 (NO 2) - oxidizing agent, P +3 (P 2 O 3) - reducing agent
2NO 2 + P 2 O 3 + 4KOH → 2NO + 2K 2 HPO 4 + H 2 O
Task #23
Using the electron balance method, write the equation for the reaction:
KI + H 2 SO 4 → I 2 + H 2 S + … + …
Determine the oxidizing agent and reducing agent.
S +6 + 8e → S -2 │1 reduction reaction
2I -1 - 2e → I 2 0 │4 oxidation reaction
S +6 (H 2 SO 4) - oxidizing agent, I -1 (KI) - reducing agent
8KI + 5H 2 SO 4 → 4I 2 + H 2 S + 4K 2 SO 4 + 4H 2 O
Task #24
Using the electron balance method, write the equation for the reaction:
FeSO 4 + ... + H 2 SO 4 → ... + MnSO 4 + K 2 SO 4 + H 2 O
Determine the oxidizing agent and reducing agent.
Mn +7 + 5e → Mn +2 │2 reduction reaction
2Fe +2 − 2e → 2Fe +3 │5 oxidation reaction
Mn +7 (KMnO 4) - oxidizing agent, Fe +2 (FeSO 4) - reducing agent
10FeSO 4 + 2KMnO 4 + 8H 2 SO 4 → 5Fe 2 (SO 4) 3 + 2MnSO 4 + K 2 SO 4 + 8H 2 O
Task #25
Using the electron balance method, write the equation for the reaction:
Na 2 SO 3 + ... + KOH → K 2 MnO 4 + ... + H 2 O
Determine the oxidizing agent and reducing agent.
Mn +7 + 1e → Mn +6 │2 reduction reaction
S +4 − 2e → S +6 │1 oxidation reaction
Mn +7 (KMnO 4) - oxidizing agent, S +4 (Na 2 SO 3) - reducing agent
Na 2 SO 3 + 2KMnO 4 + 2KOH → 2K 2 MnO 4 + Na 2 SO 4 + H 2 O
Task #26
Using the electron balance method, write the equation for the reaction:
H 2 O 2 + ... + H 2 SO 4 → O 2 + MnSO 4 + ... + ...
Determine the oxidizing agent and reducing agent.
Mn +7 + 5e → Mn +2 │2 reduction reaction
2O -1 - 2e → O 2 0 │5 oxidation reaction
Mn +7 (KMnO 4) - oxidizing agent, O -1 (H 2 O 2) - reducing agent
5H 2 O 2 + 2KMnO 4 + 3H 2 SO 4 → 5O 2 + 2MnSO 4 + K 2 SO 4 + 8H 2 O
Task number 27
Using the electron balance method, write the equation for the reaction:
K 2 Cr 2 O 7 + H 2 S + H 2 SO 4 → Cr 2 (SO 4) 3 + K 2 SO 4 + ... + ...
Determine the oxidizing agent and reducing agent.
2Cr +6 + 6e → 2Cr +3 │1 reduction reaction
S -2 - 2e → S 0 │3 oxidation reaction
Cr +6 (K 2 Cr 2 O 7) - oxidizing agent, S -2 (H 2 S) - reducing agent
K 2 Cr 2 O 7 + 3H 2 S + 4H 2 SO 4 → Cr 2 (SO 4) 3 + K 2 SO 4 + 3S + 7H 2 O
Task #28
Using the electron balance method, write the equation for the reaction:
KMnO 4 + HCl → MnCl 2 + Cl 2 + ... + ...
Determine the oxidizing agent and reducing agent.
Mn +7 + 5e → Mn +2 │2 reduction reaction
2Cl -1 - 2e → Cl 2 0 │5 oxidation reaction
Mn +7 (KMnO 4) - oxidizing agent, Cl -1 (HCl) - reducing agent
2KMnO 4 + 16HCl → 2MnCl 2 + 5Cl 2 + 2KCl + 8H 2 O
Task #29
Using the electron balance method, write the equation for the reaction:
CrCl 2 + K 2 Cr 2 O 7 + ... → CrCl 3 + ... + H 2 O
Determine the oxidizing agent and reducing agent.
2Cr +6 + 6e → 2Cr +3 │1 reduction reaction
Cr +2 − 1e → Cr +3 │6 oxidation reaction
Cr +6 (K 2 Cr 2 O 7) - oxidizing agent, Cr +2 (CrCl 2) - reducing agent
6CrCl 2 + K 2 Cr 2 O 7 + 14HCl → 8CrCl 3 + 2KCl + 7H 2 O
Task number 30
Using the electron balance method, write the equation for the reaction:
K 2 CrO 4 + HCl → CrCl 3 + ... + ... + H 2 O
Determine the oxidizing agent and reducing agent.
Cr +6 + 3e → Cr +3 │2 reduction reaction
2Cl -1 - 2e → Cl 2 0 │3 oxidation reaction
Cr +6 (K 2 CrO 4) - oxidizing agent, Cl -1 (HCl) - reducing agent
2K 2 CrO 4 + 16HCl → 2CrCl 3 + 3Cl 2 + 4KCl + 8H 2 O
Task number 31
Using the electron balance method, write the equation for the reaction:
KI + ... + H 2 SO 4 → I 2 + MnSO 4 + ... + H 2 O
Determine the oxidizing agent and reducing agent.
Mn +7 + 5e → Mn +2 │2 reduction reaction
2l -1 − 2e → l 2 0 │5 oxidation reaction
Mn +7 (KMnO 4) - oxidizing agent, l -1 (Kl) - reducing agent
10KI + 2KMnO 4 + 8H 2 SO 4 → 5I 2 + 2MnSO 4 + 6K 2 SO 4 + 8H 2 O
Task #32
Using the electron balance method, write the equation for the reaction:
FeSO 4 + KClO 3 + KOH → K 2 FeO 4 + KCl + K 2 SO 4 + H 2 O
Determine the oxidizing agent and reducing agent.
Cl +5 + 6e → Cl -1 │2 reduction reaction
Fe +2 − 4e → Fe +6 │3 oxidation reaction
3FeSO 4 + 2KClO 3 + 12KOH → 3K 2 FeO 4 + 2KCl + 3K 2 SO 4 + 6H 2 O
Task number 33
Using the electron balance method, write the equation for the reaction:
FeSO 4 + KClO 3 + ... → Fe 2 (SO 4) 3 + ... + H 2 O
Determine the oxidizing agent and reducing agent.
Cl +5 + 6e → Cl -1 │1 reduction reaction
2Fe +2 − 2e → 2Fe +3 │3 oxidation reaction
Cl +5 (KClO 3) - oxidizing agent, Fe +2 (FeSO 4) - reducing agent
6FeSO 4 + KClO 3 + 3H 2 SO 4 → 3Fe 2 (SO 4) 3 + KCl + 3H 2 O
Task number 34
Using the electron balance method, write an equation for the reaction.
Reactions, which are called redox (ORR), occur with a change in the oxidation states of the atoms that are part of the molecules of the reagents. These changes occur in connection with the transition of electrons from atoms of one element to another.
The processes occurring in nature and carried out by man, for the most part, represent OVR. Essential processes such as respiration, metabolism, photosynthesis(6CO2 + H2O = C6H12O6 + 6O2) - all this is OVR.
In industry, with the help of OVR, sulfuric, hydrochloric acids and much more are obtained.
The recovery of metals from ores - in fact, the basis of the entire metallurgical industry - is also a redox process. For example, the reaction for obtaining iron from hematite: 2Fe2O3 + 3C = 4Fe + 3CO2.
Oxidizing and reducing agents: characteristic
Atoms that give up electrons in the process of chemical transformation are called reducing agents, their oxidation state (CO) increases as a result. The atoms that accept electrons are called oxidizing agents, and their CO is reduced.
It is said that oxidizing agents are reduced by accepting electrons, and reducing agents are oxidized by donating electrons.
The most important representatives of oxidizing and reducing agents are presented in the following table:
| Typical oxidizers | Typical reducing agents |
| Simple substances consisting of elements with high electronegativity (non-metals): iodine, fluorine, chlorine, bromine, oxygen, ozone, sulfur, etc. | Simple substances consisting of atoms of elements with low electronegativity (metals or non-metals): hydrogen H2 , carbon C (graphite), zinc Zn, aluminum Al, calcium Ca, barium Ba, iron Fe, chromium Cr and so on. |
Molecules or ions containing metal or non-metal atoms with high oxidation states:
|
Molecules or ions containing metal or non-metal atoms with low oxidation states:
|
| Ionic compounds containing cations of some metals with high CO: Pb3+, Au3+, Ag+, Fe3+ and others. | Organic compounds: alcohols, acids, aldehydes, sugars. |
On the basis of the periodic law of chemical elements, it is most often possible to assume the redox abilities of atoms of a particular element. According to the reaction equation, it is also easy to understand which of the atoms are the oxidizing agent and reducing agent.
How to determine whether an atom is an oxidizing or reducing agent: it is enough to write down CO and understand which atoms increased it during the reaction (reducing agents), and which decreased it (oxidizing agents).
Substances with a dual nature
Atoms having intermediate COs are capable of both accepting and donating electrons, as a result of which substances containing such atoms in their composition will be able to act as both an oxidizing agent and a reducing agent.
An example would be hydrogen peroxide. The oxygen contained in its composition in CO -1 can both accept an electron and give it away.
When interacting with a reducing agent, peroxide exhibits oxidizing properties, and with an oxidizing agent, it exhibits reducing properties.
You can take a closer look at the following examples:
- reduction (peroxide acts as an oxidizing agent) when interacting with a reducing agent;
SO2 + H2O2 = H2SO4
O -1 + 1e \u003d O -2
- oxidation (peroxide is in this case a reducing agent) when interacting with an oxidizing agent.
2KMnO4 + 5H2O2 + 3H2SO4 = 2MnSO4 + 5O2 + K2SO4 + 8H2O
2O -1 -2e \u003d O2 0
OVR classification: examples
There are the following types of redox reactions:
- intermolecular oxidation-reduction (the oxidizing agent and the reducing agent are in the composition of different molecules);
- intramolecular oxidation-reduction (the oxidizing agent is part of the same molecule as the reducing agent);
- disproportionation (an atom of the same element is an oxidizing and reducing agent);
- reproportionation (oxidizing agent and reducing agent form one product as a result of the reaction).
Examples of chemical transformations related to various types of OVR:
- Intramolecular OVR are most often reactions of thermal decomposition of a substance:
2KCLO3 = 2KCl + 3O2
(NH4)2Cr2O7 = N2 + Cr2O3 + 4H2O
2NaNO3 = 2NaNO2 + O2
- Intermolecular OVR:
3Cu + 8HNO3 = 3Cu(NO3)2 + 2NO + 4H2O
2Al + Fe2O3 = Al2O3 + 2Fe
- Disproportionation reactions:
3Br2 + 6KOH = 5KBr + KBrO3 + 6H2O
3HNO2 = HNO3 + 2NO + H2O
2NO2 + H2O = HNO3 + HNO2
4KClO3 = KCl + 3KClO4
- Reproportionation reactions:
2H2S + SO2 = 3S + 2H2O
HOCl + HCl = H2O + Cl2
Current and non-current OVR
Redox reactions are also divided into current and currentless.
The first case is the production of electrical energy through a chemical reaction (such energy sources can be used in car engines, in radio engineering devices, control devices), or electrolysis, that is, a chemical reaction, on the contrary, occurs due to electricity (using electrolysis, you can get various substances, treat the surfaces of metals and products from them).
Examples currentless OVR we can name the processes of combustion, corrosion of metals, respiration and photosynthesis, etc.
OVR electronic balance method in chemistry
The equations of most chemical reactions are equalized by a simple selection stoichiometric coefficients. However, when selecting coefficients for the OVR, one may encounter a situation where the number of atoms of some elements cannot be equalized without violating the equality of the numbers of atoms of others. In the equations of such reactions, the coefficients are selected by the method of compiling an electronic balance.
The method is based on the fact that the sum of electrons accepted by the oxidizing agent and the number of electrons given away by the reducing agent is brought to equilibrium.
The method consists of several stages:
- The reaction equation is written.
- CO elements are determined.
- The elements that have changed their oxidation states as a result of the reaction are determined. The oxidation and reduction half-reactions are recorded separately.
- The factors for the half-reaction equations are chosen so as to equalize the electrons received in the reduction half-reaction and given away in the oxidation half-reaction.
- The selected coefficients are entered into the reaction equation.
- The remaining reaction coefficients are selected.
On a simple example aluminum interactions with oxygen, it is convenient to write the equation step by step:
- Equation: Al + O2 = Al2O3
- The CO of atoms in simple substances of aluminum and oxygen is 0.
Al 0 + O2 0 \u003d Al +3 2O -2 3
- Let's make half-reactions:
Al 0 -3e \u003d Al +3;
O2 0 +4e = 2O -2
- We select the coefficients, when multiplied by which, the number of received and the number of given electrons will be the same:
Al 0 -3e \u003d Al +3 coefficient 4;
O2 0 +4e = 2O -2 coefficient 3.
- We put down the coefficients in the reaction scheme:
4 Al+ 3 O2 = Al2O3
- It can be seen that to equalize the entire reaction, it is enough to put a coefficient in front of the reaction product:
4Al + 3O2 = 2 Al2O3
Examples of tasks for compiling an electronic balance
The following may occur equalization tasks OVR:
- Interaction of potassium permanganate with potassium chloride in an acidic environment with the release of gaseous chlorine.
Potassium permanganate KMnO4 (potassium permanganate, "potassium permanganate") is a strong oxidizing agent due to the fact that in KMnO4 the oxidation state of Mn is +7. With it, chlorine gas is often obtained in the laboratory by the following reaction:
KCl + KMnO4 + H2SO4 = Cl2 + MnSO4 + K2SO4 + H2O
K +1 Cl -1 + K +1 Mn +7 O4 -2 + H2 +1 S +6 O4 -2 = Cl2 0 + Mn +2 S +6 O4 -2 + K2 +1 S +6 O4 -2 + H2 +1 O -2
Electronic balance:
As can be seen after the arrangement of CO, chlorine atoms donate electrons, increasing their CO to 0, and manganese atoms accept electrons:
Mn +7 +5e = Mn +2 multiplier two;
2Cl -1 -2e = Cl2 0 multiplier five.
We put the coefficients in the equation in accordance with the selected factors:
10 K +1 Cl -1 + 2 K +1 Mn +7 O4 -2 + H2SO4 = 5 Cl2 0 + 2 Mn +2 S +6 O4 -2 + K2SO4 + H2O
Equalize the number of other elements:
10KCl + 2KMnO4 + 8 H2SO4 = 5Cl2 + 2MnSO4 + 6 K2SO4 + 8 H2O
- The interaction of copper (Cu) with concentrated nitric acid (HNO3) with the release of gaseous nitric oxide (NO2):
Cu + HNO3(conc.) = NO2 + Cu(NO3)2 + 2H2O
Cu 0 + H +1 N +5 O3 -2 = N +4 O2 + Cu +2 (N +5 O3 -2) 2 + H2 +1 O -2
Electronic balance:
As you can see, copper atoms increase their CO from zero to two, and nitrogen atoms decrease from +5 to +4
Cu 0 -2e \u003d Cu +2 factor one;
N +5 +1e = N +4 multiplier two.
We put the coefficients in the equation:
Cu 0 + 4 H +1 N +5 O3 -2 = 2 N +4 O2 + Cu +2 (N +5 O3 -2)2 + H2 +1 O -2
Cu+ 4 HNO3(conc.) = 2 NO2 + Cu(NO3)2 + 2 H2O
- Interaction of potassium dichromate with H2S in an acidic medium:
Let's write down the reaction scheme, arrange CO:
K2 +1 Cr2 +6 O7 -2 + H2 +1 S -2 + H2 +1 S +6 O4 -2 = S 0 + Cr2 +3 (S +6 O4 -2) 3 + K2 +1 S +6 O4 -2 + H2O
S -2 -2e \u003d S 0 coefficient 3;
2Cr +6 +6e = 2Cr +3 coefficient 1.
We substitute:
K2Cr2O7 + 3H2S + H2SO4 = 3S + Cr2(SO4)3 + K2SO4 + H2O
Equalize the rest of the elements:
К2Сr2О7 + 3Н2S + 4Н2SO4 = 3S + Cr2(SO4)3 + K2SO4 + 7Н2О
Influence of the reaction medium
The nature of the environment affects the course of certain OVR. The role of the reaction medium can be traced by the example of the interaction of potassium permanganate (KMnO4) and sodium sulfite (Na2SO3) at different pH values:
- Na2SO3 + KMnO4 = Na2SO4 + MnSO4 + K2SO4 (pH<7 кислая среда);
- Na2SO3 + KMnO4 = Na2SO4 + MnO2 + KOH (pH = 7 neutral medium);
- Na2SO3 + KMnO4 = Na2SO4 + K2MnO4 + H2O (pH>7 alkaline).
It can be seen that a change in the acidity of the medium leads to the formation of different products of the interaction of the same substances. When the acidity of the medium changes, they also occur for other reagents entering the OVR. Similarly to the examples shown above, reactions involving the dichromate ion Cr2O7 2- will take place with the formation of different reaction products in different media:
in an acidic environment, the product will be Cr 3+ ;
in alkaline - CrO2 -, CrO3 3+;
in neutral - Cr2O3.
