Let's add two equations. Oxidative properties of nitric acid

Oxidation-reduction reactions, or ORR for short, are one of the fundamentals of the subject of chemistry, as they describe the interaction of individual chemical elements together. As the name of these reactions suggests, they involve at least two different chemicals one of which acts as an oxidizing agent, and the other as a reducing agent. Obviously, it is very important to be able to distinguish and identify them in various chemical reactions.

1. Oxidation is the process of losing electrons from the outer electron layer of a chemical element. In its turn oxidizing agent there will be an atom, molecule or ion that accepts electrons and thereby lowers its oxidation state, which is are being restored . After a chemical reaction of interaction with another substance, the oxidizing agent always acquires a positive charge.
2. Recovery called the process of adding electrons to an external electron layer chemical element. Restorer there will be an atom, molecule or ion that donates its electrons and thereby increases its oxidation state, that is oxidize . After a chemical reaction of interaction with another substance, a reducing agent always acquires a positive charge.
3. Simply put, an oxidizing agent is a substance that “takes” electrons, and a reducing agent is a substance that gives them to the oxidizing agent. It is possible to determine who in a redox reaction plays the role of an oxidizing agent, who is a reducing agent, and in what cases the oxidizing agent becomes a reducing agent and vice versa, knowing typical behavior in chemical reactions of individual elements.
4. Typical reducing agents are metals and hydrogen: Fe, K, Ca, Cu, Mg, Na, Zn, H). The less ionized they are, the greater their reducing properties. For example, partially oxidized iron, which has given up one electron and has a charge of +1, will be able to give up one electron less compared to “pure” iron. Also, reducing agents can be compounds of chemical elements in the lowest oxidation state, in which all free orbitals are filled and which can only donate electrons, for example, ammonia NH 3, hydrogen sulfide H 2 S, hydrogen bromide HBr, hydrogen iodide HI, hydrogen chloride HCl.
5. Typical oxidizing agents are many nonmetals (F, Cl, I, O, Br). Also, metals with a high oxidation state (Fe +3, Sn +4, Mn +4), as well as some compounds of elements with a high oxidation state: potassium permanganate KMnO4, sulfuric acid H 2 SO 4, nitric acid HNO 3, copper oxide CuO, iron chloride FeCl 3.
6. Chemical compounds in incomplete or intermediate oxidation states, for example monobasic nitric acid HNO 2, hydrogen peroxide H 2 O 2, sulfurous acid H 2 SO 3 can exhibit both oxidizing and reducing properties, depending on the redox properties of the second reagent involved in the interaction.

Ka follows from this example One sodium atom gives up its electron to one oxygen atom. Therefore, sodium is a reducing agent and oxygen is an oxidizing agent. In this case, sodium will be completely oxidized, since it will give up the maximum possible quantity electrons, but the oxygen atom will not be completely reduced, since it will be able to accept another electron from another oxygen atom.

Test

Task 10. Gas weighing 1.105 g at 27 0 C and P = 101.3 kPa. Occupies a volume of 0.8 liters. What is its relative molecular mass?

Given:

m(gas)=1.105 g=kg

t=27 0 C, T=300K

Р=101.3 kPa=101.3 ·10 3 Pa

V=0.8 l=0.8 l ·10 -3 m 3

Find: Mr(gas)-?

Solution.

According to the Clayperon-Mendeleev equation PV=n RT, where n is the number of moles of gas; P – gas pressure (for example, in atm), V – gas volume (in liters); T – gas temperature (K); R – gas constant (8.34 J/mol K).

Connection between thermodynamic temperature T(Kelvin scale) and temperature t according to the International Practical Scale (Celsius scale): T = (t+273), then T = 300K. The chemical amount of a gas is equal to the ratio of the mass of the gas to its molar mass: n=m/M, substituting this expression into the Clayperon-Mendeleev equation and expressing molar mass, we have:

M= = =34g/mol

Then the relative molecular mass of the gas is 34.

Answer: The relative molecular mass of the gas is 34.

Task 35. What volume of air is needed for complete combustion 25 kg methyl ethyl ether CH 3 OS 2 N 5, if t = -4 0 C, P = 1.2 × 10 5 Pa?

Solution.

Let's write the reaction equation: CH 3 OS 2 H 5 + 4.5O2 = 3CO 2 + 4H 2 O

Let's find the chemical amount of ether:

n(CH 3 OS 2 H 5) = m((CH 3 OS 2 H 5)/M(CH 3 OS 2 H 5) = 25000/60 = 4166.67 mol. Using the reaction, we will find the chemical amount of oxygen required for combustion this amount of ether:

When 1 mole of ether is burned, 4.5 moles of oxygen are consumed,

then during combustion 4166.67 mol of ether – x mol of oxygen.

Hence x=1875 mol. Let's find the volume of oxygen: V(O 2) = Vm n(O 2), where Vm is the molar volume, equal to 22.4 l/mol at normal conditions, that is, V(O 2) = 42000 l.

Considering that volume fraction oxygen in the air is 21%, then

V(air) = V(O2)/0.21 = 42000/0.21 = 200000 l

At t = -4 0 C, P = 1.2 × 10 5 Pa, this volume of air will be equal according to the formula of the combined gas law:

(P 1 V 1)/T 1 = (P 2 V 2)/T 2, hence

V 2 = (P 1 V 1 T 2)/(T 1 P 2) = (101.3 10 3 200000 269) / (273 1.2 × 10 5) = 166360 l or 166.36 m 3

269 ​​and 273 are the temperatures in Kelvin, corresponding to -4 0 C and 0 0 C, respectively.

Answer: 166.36 m 3

Task 85. What oxidation state can hydrogen exhibit in its compounds? Give examples of reactions in which hydrogen gas plays the role of an oxidizing agent and in which the role of a reducing agent. Describe the fire hazard of hydrogen. Oxidation states of the element hydrogen and examples of corresponding compounds.

Answer: Hydrogen is an element of the first period, the first A group, the electronic formula of which is 1s 1. It can take the following oxidation states: +1 (H 2 O, H 2 S. NH 3, etc.), 0 (H 2), -1 (metal hydrides: NaH, CaH 2).

Reactions involving compounds in which hydrogen exhibits an oxidation state of +1 are, for example, redox reactions involving water, in which hydrogen exhibits oxidizing properties.

2H +1 2 O + 2Li = 2LiOH + H 0 2

2H +1 + 2e = H 0 2 | oxidizer

Li 0 -1е= Li + |2 reducing agent

2H 2 O + 2Na = 2NaOH + H 2

Or the reactions of acids with metals that are present in the ECHR before hydrogen.

2H 2 S + 2K = K 2 S + H 2 V

Hydrogen is a reducing agent:

H 2 0 +Ca 0 =Ca +2 H -1 2

Ca 0 -2e=Ca 0 reducing agent

H 2 0 +2e= 2H -1 oxidizing agent

IN last decades Various possibilities for using hydrogen as an energy carrier are often discussed.

Many circumstances speak in favor of hydrogen as a universal energy carrier:

1. To produce hydrogen, water can be used, the reserves of which today seem to be significant.

2. Hydrogen combustion products are much more environmentally friendly than those of gasoline and diesel fuel.

3. Hydrogen can be used in existing engines with minor design modifications.

4. Hydrogen has a high specific heat combustion; good flammability of the hydrogen-air mixture over a wide temperature range; high anti-knock resistance, allowing operation at a compression ratio of up to 14; high speed and combustion completeness.

Practical use hydrogen encounters a number of significant difficulties, primarily due to the increased explosion hazard of the working fluid. Safety problems in hydrogen technology are associated with the combustion of hydrogen, its cryogenic state, corrosion resistance and a decrease in the strength properties of materials when low temperatures, high fluidity and penetrating ability. All this requires careful compliance with safety requirements when working with hydrogen. According to numerous reference data, the explosive properties of a hydrogen mixture with air are characterized by the following data: ignition area 4.12-75% of the volume, minimum ignition energy - 0.02 mJ, auto-ignition temperature - 783 K, normal speed flame spread - 2.7 m/s, critical diameter - 0.6-10-3 m, minimum explosive oxygen content - 5% volume.

To ensure minimal danger when handling hydrogen, it is necessary to comply with following conditions:

1. Wide familiarization of personnel with the characteristics of hydrogen as a chemical product.

2. Constantly improving the reliability of means and methods for ensuring safety when performing various technological operations with hydrogen.

3. Creation of reliable means of indicating hydrogen leaks.

It is absolutely unacceptable for air (oxygen) to enter containers and pipelines filled with liquid hydrogen. The air freezes and settles on the walls above the level of hydrogen liquid or sinks to the bottom of the container. Breaking crystals of oxygen or solid air can be a source of ignition or explosion. for this reason, nitrogen, which is used to blow through lines and containers before filling them with hydrogen, should contain no more than 0.5-1% oxygen.

Spilled liquid hydrogen is dangerous because... it quickly evaporates, forming flammable and explosive mixtures.

The hydrogen flame is almost invisible in daylight. In this regard, it is necessary to use sensors to detect it. The most common optical sensors detect ultraviolet and infrared radiation. Blowing paints are also successfully used for this purpose. These paints char and swell at relatively low temperatures (about 470K) and release corrosive gases.

Safety measures when handling liquid hydrogen should exclude the possibility of uncontrolled leakage, as well as ensure rapid evacuation of leaked gas.

For structures located in open areas and liquid hydrogen storage facilities, the following measures can be recommended:

1. In the area of ​​work with liquid hydrogen, it is necessary to have a water shower, a fire hose or a special tank of water to wash off the liquid product from splashed areas of the process equipment.

2. Tanks and tanks for storing liquid products should be periodically cleaned of solid deposits (oxygen, nitrogen, etc.) at intervals of 1-2 years by defrosting them.

3. A thorough check of process equipment for leaks is necessary. A sign of hydrogen leakage from storage is the formation of frost on equipment parts.

4. Protective walls should not be constructed near storage tanks. For good circulation of gases, tanks should be installed in such a way that they are open to access of air with possible more sides

5. Zone possible danger around the tank in accordance with the safety instructions must be marked.

In addition, during long-term storage of organophosphorus toxic substances inside a sealed cavity, along with other decay products, hydrogen fluoride is released in noticeable quantities. When it interacts with the iron of the product body, intensive formation of hydrogen occurs - an extremely chemically active substance. Diatomic molecule Hydrogen forms compounds with all elements (except noble gases), dissolves well in metals and penetrates them relatively easily. Hydrogen directly combines with fluorine (even at a temperature of - 252°C).

Taking these features of molecular hydrogen into account suggests that in the body of a chemical munition or a sealed container with a toxic substance, a process of hydrogen accumulation occurs to a certain pressure, after which this element begins to diffuse through the metal body of the container. At a certain pressure, the process stabilizes and can subsequently change only by changing the amount of hydrogen fluoride released or the outside air temperature. Hydrogen absorbed by a metal causes the metal to lose its ductility and strength. This effect is known as hydrogen embrittlement. It causes cracks to appear as a result of the accumulation of hydrogen on various defects crystal structure metal

Hydrogen released from containers and ammunition inside concrete storage facilities will accumulate near the ceiling and can, in addition, be a source of fire and explosion hazards, since when mixed with oxygen in the air it forms an emergency explosive gas.

Similar problems arise when storing radioactive waste. When water enters the storage facility, it decomposes under the influence of ionizing radiation. Radiolysis of water creates hydrogen, which, at a concentration of more than 4 percent by volume, can form an “explosive” mixture. The concentration of hydrogen in the storage facility, due to the convective nature of its dissipation, is proportional to the outside air temperature, which leads to the need for forced ventilation of radioactive waste storage facilities in hot weather.

Task 60. Which orbitals of the atom are filled with electrons first: 3d or 4s, 5s or 4p? Why? Compose electronic formula element with serial number 21.

Answer. It should be taken into account that the electron occupies the energy sublevel at which it has the lowest energy - the smaller sum n + ℓ (Klechkovsky’s rule). Filling sequence energy levels and sublevels are as follows:

1s→2s→ 2р→ 3s→ 3р→ 4s→ 3d→ 4р→ 5s→ 4d→ 5р→ 6s→ 5d 1 →4f→ 5d→ 6р→ 7s →6d 1 →5f→ 6d→ 7r.

In our case

D 4s 5s 4р

Meaning n 3 4 5 4

Meaning l 2 0 0 1

Sum ( n +l ) 5 4 5 5

Filling sequence (based on Klechkovsky’s rules):

1 – 4s then 3d; 1-4 r then – 5s. 4р is filled in first, despite an equal amount (n +l ), since n=4, and 5s n=5, and with identical values of this amount, the sublevel with lower value main quantum number n.

Cu +2 +2e Cu 0 |3 oxidizing agent

2N -3 -6е N 2 0 |1 reducing agent

3CuO +2 NH 3 = 3Cu + N 2 + 3H 2 O

Task 135. When 1 liter of methanol vapor CH 3 OH is burned, 32.3 kJ of heat is released. Calculate the enthalpy of formation of methanol. The conditions are standard.

V(CH 3 OH) = 1l

DНр = -32.3 kJ

Find: DH 0 (CH 3 OH) -?

Solution. Let's find the heat of combustion of 1 mol (22.4 l) of methanol. During the combustion of 1 liter, 32.3 kJ was released, then with 22.4 mol of methanol combustion - x kJ, x = 723.52 kJ/mol, that is, DH 0 hor (CH 3 OH) = - 723.52 kJ/mol.

Let's write the reaction equation: CH 3 OH + 1.5 O 2 = CO 2 + 2H 2 O

To calculate the enthalpy of methanol formation we use the corollary

from Hess’s law: ΔH (H.R.) = ΣΔH 0 (cont.) - ΣΔH 0 (out.).

We use the enthalpy of combustion of methanol that we found and the enthalpies of formation of all participants in the process (except methanol) given in the appendix.

According to the 1st corollary of Hess’s law, the thermal effect of this reaction DН 0 р-i can be written as follows:

DH 0 r-i = DH 0 (CO 2) + 2DH 0 (H 2 O) - DH 0 (CH 3 OH). (1)

DH 0 (CO 2), DH 0 (H 2 O), DH 0 (CH 3 OH) – enthalpies of formation of substances. According to the conditions of the problem, the enthalpy of methanol formation must be calculated. According to the 2nd corollary of Hess’s law, the thermal effect of the same reaction is equal to the enthalpy of combustion of ethyl acetate.

DH 0 r-i = DH 0 hor (CH 3 OH). (2)

We found the value of DH 0 mountains (CH 3 OH). Combining equations (1) and (2) we can write:

DH 0 hor (CH 3 OH) = DH 0 (CO 2) + 2DH 0 (H 2 O) - DH 0 (CH 3 OH).

Then the enthalpy of formation of ethyl acetate DH 0 (CH 3 OH) can be calculated as follows:

DH 0 (CH 3 OH) = DH 0 (CO 2) + 2DH 0 (H 2 O) -DH 0 hor (CH 3 OH) = (–393.5) + 2×(–241.8) – (- 723.52) = - 153.57 kJ/mol.

The obtained value means that when 1 mole of methanol is formed, 153.57 kJ of heat is released ( DH<0 ).

Consider the diagrams of reaction equations below. What is their significant difference? Did the oxidation states of the elements change in these reactions?

In the first equation, the oxidation states of the elements did not change, but in the second they changed - for copper and iron.

The second reaction is a redox reaction.

Reactions that result in changes in the oxidation states of the elements that make up the reactants and reaction products are called oxidation-reduction reactions (ORR).

COMPILATION OF EQUATIONS FOR REDOX REACTIONS.

There are two methods for composing redox reactions - the method electronic balance and the half-reaction method. Here we will look at the electronic balance method.
In this method, the oxidation states of atoms in the starting substances and in the reaction products are compared, and we are guided by the rule: the number of electrons donated by the reducing agent must be equal to the number of electrons gained by the oxidizing agent.
To create an equation, you need to know the formulas of the reactants and reaction products. Let's look at this method with an example.

Basic principles of the theory of redox reactions

1. Oxidation called process of losing electrons by an atom, molecule, or ion.

For example :

Al – 3e - = Al 3+

Fe 2+ - e - = Fe 3+

H 2 – 2e - = 2H +

2Cl - - 2e - = Cl 2

During oxidation, the oxidation state increases.

2. Recovery called process of gaining electrons by an atom, molecule, or ion.

For example:

S + 2е - = S 2-

WITH l 2 + 2е- = 2Сl -

Fe 3+ + e - = Fe 2+

During reduction, the oxidation state decreases.

3. Atoms, molecules or ions that donate electrons are called restorers . During the reactionthey oxidize.

Atoms, molecules or ions that gain electrons are called oxidizing agents . During the reactionthey are recovering.

Since atoms, molecules and ions are part of certain substances, these substances are called accordingly restorers or oxidizing agents.

4. Redox reactions represent the unity of two opposing processes - oxidation and reduction.

The number of electrons given up by the reducing agent is equal to the number of electrons gained by the oxidizing agent.

EXERCISES

Simulator No. 1 Oxidation-reduction reactions

Simulator No. 2 Electronic balance method

Simulator No. 3 Test “Oxidation-reduction reactions”

ASSIGNMENT TASKS

No. 1. Determine the oxidation state of atoms of chemical elements using the formulas of their compounds: H 2 S, O 2, NH 3, HNO 3, Fe, K 2 Cr 2 O 7

No. 2. Determine what happens to the oxidation state of sulfur during the following transitions:

A) H 2 S → SO 2 → SO 3

B ) SO 2 → H 2 SO 3 → Na 2 SO 3

What conclusion can be drawn after completing the second genetic chain?

What groups can chemical reactions be classified into based on changes in the oxidation state of atoms of chemical elements?

No. 3. Arrange the coefficients in CHR using the electronic balance method, indicate the processes of oxidation (reduction), oxidizing agent (reducing agent); write the reactions in complete and ionic form:

A) Zn + HCl = H 2 + ZnCl 2

B) Fe + CuSO 4 = FeSO 4 + Cu

No. 4. Given diagrams of reaction equations:
СuS + HNO 3 (diluted ) = Cu(NO 3) 2 + S + NO + H 2 O

K + H 2 O = KOH + H 2
Arrange the coefficients in the reactions using the electronic balance method.
Indicate the substance - an oxidizing agent and a substance - a reducing agent.

1 . C + HNO 3 = CO 2 + NO + H 2 O

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

3. V 2 O 5 + Ca = CaO + V

4. Mn 2 O 3 + Si = SiO 2 + Mn

5. TiCl 4 + Mg = MgCl 2 + Ti

6. P 2 O 5 + C = P + CO

7. KClO 3 + S = KCl + SO 2

8. H 2 S + HNO 3 = S + NO 2 + H 2 O

9. KNO 2 + KClO 3 = KCl + KNO 3

10. NaI + NaIO 3 + H 2 SO 4 = I 2 + Na 2 SO 4 + H 2 O

11. Na 2 S 2 O 3 + Br 2 + NaOH = Na Br + Na 2 SO 4 + H 2 O

12. Mn(NO 3) 2 + NaBiO 3 + HNO 3 = HMnO 4 + Bi(NO 3) 3 + NaNO 3 + H 2 O

13. Cr 2 O 3 + Br 2 + NaOH = Na 2 CrO 4 + NaBr + H 2 O

14. HCl + KMnO 4 = MnCl 2 + Cl 2 + KCl + H 2 O

15. KBr + KMnO 4 + H 2 SO 4 = Br 2 + MnSO 4 + K 2 SO 4 + H 2 O

16. Cu + H 2 SO 4 = CuSO 4 + SO 2 + H 2 O

17. Mg + H 2 SO 4 = MgSO 4 + H 2 S + H 2 O

18. K + H 2 SO 4 = K 2 SO 4 + S + H 2 O

19. Ag + HNO 3 = AgNO 3 + NO 2 + H 2 O

20. Cu + HNO 3 = Cu(NO 3) 2 + NO + H 2 O

21. Ca + HNO 3 = Ca(NO 3) 2 + N 2 O + H 2 O

22. Zn + HNO 3 = Zn(NO 3) 2 + N 2 + H 2 O

23. Mg + HNO 3 = Mg(NO 3) 2 + NH 4 NO 3 + H 2 O

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

25. K 2 S + KMnO 4 + H 2 SO 4 = S + MnSO 4 + K 2 SO 4 + H 2 O

26. Zn + K 2 Cr 2 O 7 + H 2 SO 4 = ZnSO 4 + Cr 2 (SO 4) 3 + K 2 SO 4 + H 2 O

27. SnSO 4 + KMnO 4 + H 2 SO 4 = Sn(SO 4) 2 + MnSO 4 + K 2 SO 4 + H 2 O

28. NaI + KMnO 4 + KOH = I 2 + K 2 MnO 4 + NaOH

29. S + KClO 3 + H 2 O = Cl 2 + K 2 SO 4 + H 2 SO 4

30. Na 2 SO 3 + KIO 3 + H 2 SO 4 = I 2 + Na 2 SO 4 + K 2 SO 4 + H 2 O

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

32. Cu(NO 3) 2 = CuO + NO 2 + O 2

33. NH 4 NO 3 = N 2 O + H 2 O

34. KNO 3 = KNO 2 + O 2

35. KClO 3 = KCl + O 2

36. KClO = KCl + O 2

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

38. K 2 MnO 4 + CO 2 = KMnO 4 + MnO 2 + K 2 CO 3

39. KClO 3 = KClO 4 + KCl

40. Cl 2 + KOH = KCl + KClO 3 + H 2 O

41. KClO = KCl + KClO 3

42. S + KOH = K 2 S + K 2 SO 3 + H 2 O

43. Na 2 SO 3 = Na 2 S + Na 2 SO 4

44. H 2 C 2 O 4 + KMnO 4 = CO 2 + K 2 CO 3 + MnO 2 + H 2 O

45. CH 3 OH+K 2 Cr 2 O 7 +H 2 SO 4 = HCOOH+Cr 2 (SO 4) 3 +K 2 SO 4 +H 2 O

46. ​​C 12 H 22 O 11 + K 2 Cr 2 O 7 + H 2 SO 4 = CO 2 + Cr 2 (SO 4) 3 + K 2 SO 4 + H 2 O

47. CH 2 O + KMnO 4 + H 2 SO 4 = HCOOH + MnSO 4 + K 2 SO 4 + H 2 O

48. Mn 3 O 4 + Al = Al 2 O 3 + Mn

49. Fe 3 O 4 + H 2 = FeO + H 2 O

50. NaN 3 = Na + N 2

51. Na 2 S 4 O 6 +KMnO 4 +HNO 3 =Na 2 SO 4 +H 2 SO 4 +Mn(NO 3) 2 +KNO 3 +H 2 O

52. Mn 3 O 4 + KClO 3 + K 2 CO 3 = K 2 MnO 4 + KCl + CO 2

53. As 2 S 3 + HNO 3 = H 3 AsO 4 + SO 2 + NO 2 + H 2 O

54. KMnO 4 = K 2 MnO 4 + MnO 2 + O 2

55. Cu 2 S + O 2 + CaCO 3 = CuO + CaSO 3 + CO 2

56. FeCl 2 + KMnO 4 + HCl = FeCl 3 + Cl 2 + MnCl 2 + KCl + H 2 O

57. Pb(NO 3) 2 = PbO + NO 2 + O 2

58. KNO 2 + KI + H 2 SO 4 = I 2 + NO + K 2 SO 4 + H 2 O

59. KMnO 4 + NO + H 2 SO 4 = MnSO 4 + NO 2 + K 2 SO 4 + H 2 O

60. CuO + NH 3 = Cu + N 2 + H 2 O

61. Cl 2 + Br 2 + KOH = KCl + KBrO 3 + H 2 O

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

63. Ti 2 (SO 4) 3 + KClO 3 + H 2 O = TiOSO 4 + KCl + H 2 SO 4

64. Fe(NO 3) 2 + MnO 2 + HNO 3 = Fe(NO 3) 3 + Mn(NO 3) 2 + H 2 O

65. KCNS+K 2 Cr 2 O 7 +H 2 SO 4 =Cr 2 (SO 4) 3 +SO 2 +CO 2 +NO 2 +K 2 SO 4 +H 2 O

66. CuFeS 2 + HNO 3 = Cu(NO 3) 2 + Fe(NO 3) 3 + H 2 SO 4 + NO + H 2 O

67. H 2 O 2 + HI = I 2 + H 2 O

68. H 2 O 2 + HIO 3 = I 2 + O 2 + H 2 O

69. H 2 O 2 + KMnO 4 + HNO 3 = Mn(NO 3) 2 + O 2 + KNO 3 + H 2 O

70. H 2 O 2 + CrCl 3 + KOH = K 2 CrO 4 + KCl + H 2 O

71. H 2 O 2 + H 2 S = H 2 SO 4 + H 2 O

Task options

101. Reactions No. 1, 26, 51,

102. Reactions No. 2, 27, 52,

103. Reactions No. 3, 28, 53,

104. Reactions No. 4, 29, 54,

105. Reactions No. 5, 30, 55,

106. Reactions No. 6, 31, 56,

107. Reactions No. 7, 32, 57,

108. Reactions No. 8, 33, 58,

109. Reactions No. 9, 34, 59,

110. Reactions No. 10, 35, 60,

111. Reactions No. 11, 36, 61,

112. Reactions No. 12, 37, 62,

113. Reactions No. 13, 38, 63,

114. Reactions No. 14, 39, 64,

115. Reactions No. 15, 40, 65,

116. Reactions No. 16, 41, 66,

117. Reactions No. 17, 42, 67,

118. Reactions No. 18, 43, 68,

119. Reactions No. 19, 44, 69,

120. Reactions No. 20, 45, 70,

121. Reactions No. 21, 46, 71,

122. Reactions No. 22, 47, 62,

123. Reactions No. 23, 48, 64,

124. Reactions No. 24, 49, 66,

125. Reactions No. 25, 50, 38.

6. THERMOCHEMICAL CALCULATIONS

(problems no. 126 – 150).

Literature:

You can use the data to complete tasks table 1 applications.

Task options

126. What is the enthalpy of formation of pentane? C 5 H 12, if the combustion of 24 g of pentane released 1176.7 kJ of heat?

127. How much heat is released when 92 g of ethyl alcohol is burned C 2 H 5 OH?

128. Calculate the thermal effect of the formation of 156 g of benzene C 6 H 6, if its enthalpy of combustion DN mountains = - 3267.5 kJ/mol?

129. When burning 1 liter of acetylene C 2 H 2 58.2 kJ of heat is released. Calculate the enthalpy of combustion of acetylene.

130. Calculate the thermal effect of the formation of 20 g of toluene C 7 H 8, if its enthalpy of combustion DN mountains = - 3912.3 kJ/mol?

131. What is the enthalpy of formation of hexane? C 6 H 14, if the combustion of 43 g of hexane released 2097.4 kJ of heat?

132. How much heat is released when 11 g of ethyl acetate is burned CH 3 SOOS 2 H 5?

133. Calculate the thermal effect of the formation of 1 mole of cyclopentane C 5 H 10, if its enthalpy of combustion DN mountains = - 3290 kJ/mol?

134. When burning 267 g of anthracene C 14 N 10 10601.2 kJ of heat is released. Calculate the enthalpy of formation of anthracene.

135. When burning 1 liter of methanol vapor CH 3 OH 32.3 kJ of heat was released. Calculate the enthalpy of formation of methanol. The conditions are standard.

136. Calculate the thermal effect of the formation of 1 m 3 of propane C 3 H 8

137. What is the enthalpy of formation of pentane? C 5 H 12, if the combustion of 12 g of pentane released 588.35 kJ of heat?

138. How much heat will be released when 84 g of ethylene is burned C 2 H 4?

139. Calculate the thermal effect of the formation of 156 g of ethane C 2 H 6

140. How much heat will be released when 10 liters of methane are burned CH 4? Conditions are normal.

141. When burning 10 liters of butane C 4 H 10 1191 kJ of heat was released. Calculate the enthalpy of formation of butane. The conditions are standard.

142. Calculate the thermal effect of the formation of 100 liters of propane C 3 H 8, if its enthalpy of combustion DN mountains = - 2223.2 kJ/mol?

143. When burning 1 liter of butane C 4 H 10 119.1 kJ of heat was released. Calculate the enthalpy of formation of butane. Conditions are normal.

144. Calculate the thermal effect of the formation of 200 liters of propanol C 3 H 7 OH, if its enthalpy of combustion DN mountains = - 2010.4 kJ/mol?

145. How much heat will be released when burning 1 kg of carbon monoxide (II) CO?

146. Calculate the thermal effect of the formation of 15 g of ethane C 2 H 6, if its enthalpy of combustion DN mountains = - 1560 kJ/mol?

147. Upon combustion of 30.8 g of biphenyl C 12 H 10 124.98 kJ of heat is released. Calculate the enthalpy of formation of biphenyl.

148. How much heat is released when 1 m 3 of acetylene is burned C 2 H 2? Conditions are normal.

149. When burning 184 g of ethyl alcohol C 2 H 5 OH 4482.7 kJ of heat is released. Calculate the enthalpy of formation of ethyl alcohol.

150. How much heat will be released when burning 100 kg of coal WITH?

7. CALCULATIONS USING THERMODYNAMIC

STATUS FUNCTIONS

(problems no. 151 – 175).

Literature:

1. Glinka N.L. "General chemistry". - L.: Chemistry, 1986, ch. 6, pp.158-162, 182-191.

2. Kozhevnikova N.Yu., Korobeynikova E.G., Kutuev R.Kh., Malinin V.R., Reshetov A.P. "General chemistry". Tutorial. - L.: SPbVPTSh, 1991, topic 6, p.40-53.

3. Glinka N.L. "Problems and exercises in general chemistry." - L.: Chemistry, 1987, chapter 5, p. 73-88, Appendix No. 5.

4. Korobeynikova E.G., Kozhevnikova N.Yu. "Collection of problems and exercises in general chemistry" part 2. Energy of chemical reactions. Chemical kinetics. - L.: SPbVPTSh, 1991, p.2-14.

Task options

151. Prove by calculations the possibility of oxygen oxidation with fluorine at 10 0 C:

0.5O 2 + F 2 = ОF 2.

152. Determine the possibility of the following reaction occurring in the gas phase at 30 0 C: H 2 + C 2 H 2 = C 2 H 4.

153. Prove by calculations that under standard conditions and at 200 0 C

reaction 0.5N 2 + O 2 = NO 2 impossible.

154. Calculate at what temperature the acetylene trimerization reaction will begin:

3C 2 H 2 (g) = C 6 H 6 (g).

155. At what temperature is the following reaction possible:

H 2 S + 0.5O 2 = SO 2 + H 2 O (g)?

156. Establish the possibility (or impossibility) of the spontaneous occurrence of the following reaction at a temperature of 100 0 C:

C 2 H 4 = H 2 + C 2 H 2 .

C 2 H 4 = H 2 + C 2 H 2 ?

158. At what temperature will the oxidation of iron begin according to the reaction:

Fe + 0.5O 2 = FeO?

159. Calculate at what temperature the following reaction can occur:

2CH 4 = C 2 H 2 + 3H 2.

160. Determine at what temperature the reaction of calcium oxide reduction with coal begins:

CaO + 3C = CaC 2 + CO.

161. Determine the fire hazard of contact between carbon disulfide and oxygen at –20 0 C if the reaction proceeds according to the equation:

CS 2 + 3O 2 = CO 2 + 2SO 2.

162. Is it possible for the following reaction to occur at 200 0 C:

CO + 0.5O 2 = CO 2?

163. Calculate at what temperature the combustion reaction of hydrogen in oxygen will go in the opposite direction.

164. At what temperature does the reaction occur?

Al + 0.75O 2 =0.5Al 2 O 3 impossible?

CO 2 + H 2 = CH 4 + H 2 O (l).

167. At what temperature is the following reaction possible:

C 2 H 4 + H 2 O (l) = C 2 H 5 OH?

168. At what temperature will the decomposition of water into hydrogen and oxygen begin?

4HCl + O 2 = 2H 2 O (g) + 2Cl 2?

170. Determine the possibility of the following reaction occurring at 40 0 ​​C:

2C + 0.5O 2 + 3H 2 = C 2 H 5 OH.

171. At what temperature will equilibrium occur in the system:

CO + 2H 2 = CH 3 OH?

172. Determine the possibility of a spontaneous reaction at 50 0 C:

C + CO 2 = 2CO.

173. Is it possible for a spontaneous reaction to occur at 400 0 C?

H2 + Cl2 = 2HCl.

175. Is it possible for the following reaction to occur at a temperature of 100 0 C:

CH 4 = C + 2H 2?

8. CHEMICAL KINETICS

(problems no. 176 – 200).

Literature:

Task options

176. What is the temperature coefficient of the reaction rate if, with an increase in temperature by 30 0, the reaction rate increases by 15.6 times?

177. How many times should the concentration of hydrogen in the system be increased?

N2 + 3H2 = 2NH3 so that the reaction rate increases 100 times?

178. How many times should the concentration of carbon monoxide in the system be increased? 2CO = CO 2 + C so that the rate of the forward reaction increases 4 times?

179. How many times should the pressure be increased so that the rate of formation NO by reaction 2NO + O 2 = 2NO 2 increased 1000 times?

180. Write the equation for the reaction rate of coal combustion ( WITH) in oxygen and determine how many times the reaction rate will increase:

a) with an increase in oxygen concentration by 3 times;

b) when replacing oxygen with air.

181. By how many degrees should the temperature of the system be increased so that the rate of the reaction occurring in it increases 30 times, if the temperature coefficient of the reaction rate is 2.5?

182. How many times will the rate of forward and reverse reactions in the system change?

2SO2 + O2 = 2SO3, if the volume of the gas mixture increases 4 times?

183. The temperature coefficient of the reaction rate is 2. How will the reaction rate change when the temperature increases by 40 0?

184. The oxidation of sulfur and its dioxide proceeds according to the equations:

A) S (cr) + O 2 = SO 2 b) 2SO2 + O2 = 2SO3.

How will the rate of these reactions change if the volume of each system is reduced by 4 times?

185. Calculate how many times the rate of a reaction occurring in the gas phase will change, N2 + 3H2 = 2NH3, If

a) reduce the system pressure by 2 times;

b) increase the hydrogen concentration by 3 times?

186. How will the rate of a reaction occurring in the gas phase change when the temperature decreases by 30 0, if the temperature coefficient of the reaction rate is 2.

187. How many times will the rate of the direct reaction change?

CO + Cl 2 = COCl 2, If

concentration CO increase from 0.03 to 0.12 mol/l, and the concentration Cl2 decrease from 0.06 to 0.02 mol/l?

189. The temperature coefficient of the reaction rate is 3. How will the rate of the reaction occurring in the gas phase change when the temperature increases from 140 to 170 0?

190. The reaction proceeds according to the equation CO (g) + S (tv) = COS (tv)

a) reduce concentration CO 5 times;

b) reduce the volume of the system by 3 times?

191. How many times will the rate of the reaction occurring in the gas phase change when the temperature increases from 150 to 180 0? The temperature coefficient of the reaction rate is 2.

192. The reaction proceeds according to the equation: NH 3 + CO 2 +H 2 O = NH 4 HCO 3. How will the rate of the forward reaction change if

a) increase the volume of the system by 3 times;

b) reduce the concentration of ammonia and water vapor by 2 times?

193. How will the rate of a reaction occurring in the gas phase change when the temperature decreases by 40 0 ​​if the temperature coefficient of the reaction rate is 3?

194. How will the rate of the forward reaction change?

2CH 4 + O 2 + 2H 2 O (g) = 2CO 2 + 6H 2, If

a) reduce the concentration of methane and oxygen by 3 times;

b) reduce the volume of the system by 2 times?

195. The temperature coefficient of the reaction rate is 2. How will the reaction rate change when the temperature increases by 30 0?

196. The reaction follows the equation 2CH 4 + O 2 = 4H 2 + 2CO.

How will the rate of the reverse reaction change if

a) reduce the volume of the system by 4 times;

b) increase the hydrogen concentration by 2 times?

197. How many times will the rate of the reaction occurring in the gas phase change when the temperature increases from 30 to 70 0 C, if the temperature coefficient of the reaction rate is 2?

198. The reaction follows the equation: Cl 2 O (g) + H 2 O (g) = 2HClO (l).

The concentration of the starting substances is = 0.35 mol/l and

=1.3 mol/l. How will the rate of the forward reaction change if the concentrations of the substances are changed to 0.4 mol/L and 0.9 mol/L, respectively?

199. Calculate how many times the reaction rate occurring in the gas phase will change if the temperature is lowered from 130 to 90 0 C. The temperature coefficient of the reaction rate is 2.

200. How will the rate of the reverse reaction occurring in the gas phase change according to the equation: 2N 2 O 5 = 4NO 2 + O 2, If

a) reduce concentration NO 2 2 times;

b) reduce the pressure in the system by 3 times?

CHEMICAL EQUILIBRIUM

(tasks no. 201 – 225).

Literature:

1. Glinka N.L. "General chemistry". - L.: Chemistry, 1986, ch. 6, pp.163-181.

2. Kozhevnikova N.Yu., Korobeynikova E.G., Kutuev R.Kh., Malinin V.R., Reshetov A.P. "General chemistry". Tutorial. - L.: LVPTSH, 1991, topic 7, p.54-65.

3. Glinka N.L. "Problems and exercises in general chemistry." - L.: Chemistry, 1987, chapter 5, p. 89-105.

4. Korobeynikova E.G., Kozhevnikova N.Yu. "Collection of problems and exercises in general chemistry" part 2. Energy of chemical reactions. Chemical kinetics. - L.: LVPTSH, 1991, pp. 15-31.

Task options

201. How will the reaction rate change:

2 NO (g) + O 2 (g) « 2NO 2 (g)

if we increase the volume of the reaction vessel four times?

202. In what direction will the equilibrium of the system shift:

H 2 (g) + 2 S (tv) « 2 H 2 S (g) Q = 21.0 kJ,

b) increase the concentration of hydrogen?

203. In what direction will the equilibrium shift in the systems:

A) CO (g) + Cl 2 (g) « COCl 2 (g) ,

b) H 2(g) + I 2(g) « 2HI (g),

if at a constant temperature the pressure is reduced by increasing the volume of the gas mixture?

204. In what direction will the equilibrium shift in the system occur:

2 CO (g) « CO 2 (g) + C (tv) Q= 171 kJ,

if a) lower the system temperature,

b) reduce the pressure in the system?

205. How many times will the reaction rate change?

2 A + B « A 2 V,

if the concentration of the substance A increase by 2 times, and the concentration of the substance IN reduce by 2 times?

206. In what direction will the equilibrium shift in the system occur:

2 SO 3(g) « 2 SO 2(g) + O 2(g) Q= - 192 kJ,

if a) increase the temperature of the system,

b) reduce concentration SO 2?

207. How many times should the concentration of the substance be increased? AT 2 in system

2 A 2(g) + B 2(g) « 2 A 2 V (g), so that when the concentration of a substance decreases A 2 4 times the rate of the forward reaction did not change?

208. In what direction will the equilibrium shift in the system occur:

COCl 2(g) « CO (g) + Cl 2(g),

if a) increase the pressure in the system,

b) increase concentration COCl 2?

209. How will the reaction rate change:

2 NO (g) + O 2 (g) « 2 NO 2 (g) ,

if a) increase the pressure in the system by 3 times,

b) reduce the volume of the system by 3 times,

c) increase concentration NO 3 times?

210. In what direction will the equilibrium shift in the system occur:

CO (g) + Cl 2 (g) « COCl 2 (g) ,

if a) increase the volume of the system,

b) increase concentration CO?

211. In what direction will the equilibrium shift in the system occur:

2 N 2 O 5 (g) « 4 NO 2 (g) + 5 O 2 (g),

if a) increase the concentration O 2,

b) expand the system?

212. Derive the equation of the chemical equilibrium constant for the reaction: MgO (tv) + CO 2 (g) « MgCO 3 (tv) Q > 0.

In what ways can you shift chemical equilibrium this reaction to the left?

213. How will the rates of forward and reverse reactions change and in which direction will the equilibrium in the system shift A (g) + 2 B (g) « AB 2 (g), if you increase the pressure of all substances by 3 times?

214. In a state of equilibrium in the system:

N 2(g) + 3 H 2(g) « 2 NH 3(g) Q= 92.4 kJ

determine in which direction the equilibrium will shift

a) with increasing temperature,

b) when the volume of the reaction vessel decreases?

215. In what direction will the equilibrium shift in the system occur:

CO 2 (g) + H 2 O (g) « H 2 CO 3 (tv) + Q,

with a) expansion of the system,

b) with increasing concentration carbon dioxide?

216. How will the speed of forward and reverse reactions in the system change:

2 SO 2(g) + O 2(g) « SO 3(g) ,

if we reduce the volume of the reactor by 2 times? Will this affect the equilibrium in the system?

217. Indicate what changes in the concentrations of reacting substances can shift the reaction equilibrium to the right:

CO 2 (g) + C (tv) « 2 CO (g).

218. Under what conditions is the reaction equilibrium:

4 Fe (tv) + 3 O 2 (g) « 2 Fe 2 O 3 (tv),

will shift towards the decomposition of the oxide?

219. Reversible reaction occurs in the gas phase and in the equation of the forward reaction the sum of the stoichiometric coefficients is greater than in the equation of the reverse reaction. How will a change in pressure affect the equilibrium in the system? Explain.

220. What conditions will promote greater yield IN by reaction: 2 A (g) + B 2 (g) « 2B (g) , Q=100 kJ.

221. Methanol is obtained as a result of the reaction:

CO (g) + 2 H 2 (g) « CH 3 OH (l) Q= 127.8 kJ.

How will the equilibrium shift as the

a) temperature,

b) pressure?

222. How will the yield of chlorine in the system be affected by:

4 HCl (g) + O 2 (g) « 2 Cl 2 (g) + 2 H 2 O (l), Q= 202.4 kJ,

a) increase in temperature in the system,

b) reducing the total volume of the mixture,

c) decrease in oxygen concentration,

d) increasing the total volume of the reactor,

e) introduction of a catalyst?

223. In what direction will the equilibrium in the systems shift:

1) 2 CO (g) + O 2 (g) « 2 CO 2 (g) , Q= 566 kJ,

2) = - 180 kJ,

if a) lower the temperature,

b) increase blood pressure?

224. In what direction will the equilibrium in the systems shift:

1) 2 CO (g) + O 2 (g) « 2 CO 2 (g) , Q= 566 KJ,

2) N 2(g) + O 2(g) « 2 NO (g) , Q= - 180 kJ,

if a) increase the temperature,

b) lower the pressure?

225. How will the equilibrium be affected in the following reaction:

CaCO 3(tv) « CaO (tv) + CO 2(g) , Q= - 179 kJ,

a) increase in pressure,

b) increase in temperature?

Oxidizing agents are particles (atoms, molecules or ions) that accept electrons during a chemical reaction. In this case, the oxidation state of the oxidizing agent goes down. Oxidizing agents are being restored.

Restorers are particles (atoms, molecules or ions) that donate electrons during a chemical reaction. In this case, the oxidation state of the reducing agent rises. Reductants in this case oxidize.

Chemicals can be divided into typical oxidizing agents, typical reducing agents , and substances that may exhibit both oxidizing and reducing properties. Some substances exhibit virtually no redox activity.

TO typical oxidizing agents include:

• simple substances-non-metals with the strongest oxidizing properties (fluorine F 2, oxygen O 2, chlorine Cl 2);
• ionsmetals or non-metals With high positive (usually higher) oxidation states : acids (HN +5 O 3, HCl +7 O 4), salts (KN +5 O 3, KMn +7 O 4), oxides (S +6 O 3, Cr +6 O 3)
• compounds containing some metal cations having high oxidation states: Pb 4+, Fe 3+, Au 3+, etc.

Typical reducing agents - this is, as a rule:

• simple substances - metals (restorative abilities metals are determined by a number of electrochemical activities);
• complex substances that contain atoms or ions of nonmetals with a negative (usually lowest) oxidation state: binary hydrogen compounds(H 2 S, HBr), salts of oxygen-free acids (K 2 S, NaI);
• some compounds containing cations with minimal positive degree oxidation(Sn 2+, Fe 2+, Cr 2+), which, giving up electrons, can increase their oxidation state;
• compounds containing complex ions consisting of nonmetals with an intermediate positive oxidation state(S +4 O 3) 2–, (НР +3 O 3) 2–, in which elements can, by donating electrons, increase its positive oxidation state.

Most other substances may exhibit both oxidizing and reducing properties.

Typical oxidizing and reducing agents are given in the table.

In laboratory practice the most commonly used are the following oxidizing agents :

potassium permanganate (KMnO 4);

potassium dichromate (K 2 Cr 2 O 7);

nitric acid (HNO 3);

concentrated sulfuric acid (H 2 SO 4);

hydrogen peroxide (H 2 O 2);

oxides of manganese (IV) and lead (IV) (MnO 2, PbO 2);

molten potassium nitrate (KNO 3) and melts of some other nitrates.

TO restoration workers , which apply V laboratory practice relate:

• magnesium (Mg), aluminum (Al), zinc (Zn) and other active metals;
• hydrogen (H 2) and carbon (C);
• potassium iodide (KI);
• sodium sulfide (Na 2 S) and hydrogen sulfide (H 2 S);
• sodium sulfite (Na 2 SO 3);
• tin chloride (SnCl 2).

Classification of redox reactions

Redox reactions are usually divided into four types: intermolecular, intramolecular, disproportionation (auto-oxidation-self-reduction) reactions, and counter-disproportionation reactions.

Intermolecular reactions occur with a change in the oxidation state different elements from different reagents. In this case, various oxidation and reduction products .

2Al 0 + Fe +3 2 O 3 → Al +3 2 O 3 + 2Fe 0,

C 0 + 4HN +5 O 3 (conc) = C +4 O 2 + 4N +4 O 2 + 2H 2 O.

Intramolecular reactions - these are reactions in which different elements from one reagent go to different products, for example:

(N -3 H 4) 2 Cr +6 2 O 7 → N 2 0 + Cr +3 2 O 3 + 4 H 2 O,

2 NaN +5 O -2 3 → 2 NaN +3 O 2 + O 0 2 .

Disproportionation reactions (auto-oxidation-self-healing) are reactions in which the oxidizing agent and the reducing agent are the same element of the same reagent, which then turns into different products:

3Br 2 + 6 KOH → 5KBr + KBrO 3 + 3 H 2 O,

Reproportionation (comproportionation, counter-disproportionation ) are reactions in which the oxidizing agent and the reducing agent are the same element, Which one of different reagents goes into one product. The reaction is the opposite of disproportionation.

2H 2 S -2 + S +4 O 2 = 3S + 2H 2 O

Basic rules for composing redox reactions

Redox reactions are accompanied by oxidation and reduction processes:

Oxidation is the process of donating electrons by a reducing agent.

Recovery is the process of gaining electrons by an oxidizing agent.

Oxidizer is being restored, and the reducing agent oxidizes .

In redox reactions it is observed electronic balance: The number of electrons that the reducing agent gives up is equal to the number of electrons that the oxidizing agent gains.

If the balance sheet is drawn up incorrectly, you will not be able to create complex OVRs.

Several methods for composing redox reactions (ORR) are used: the electron balance method, the electron-ion balance method (half-reaction method) and others. Let's take a closer look .

electronic balance method

It is quite easy to “identify” ORR - just arrange the oxidation states in all compounds and determine that the atoms change the oxidation state:

We write out separately the atoms of elements that change the oxidation state, in the state BEFORE the reaction and AFTER the reaction.

The oxidation state is changed by manganese and sulfur atoms:

S -2 -2e = S 0

Mn +7 + 1e = Mn +6

Manganese absorbs 1 electron, sulfur gives up 2 electrons. In this case, it is necessary to comply electronic balance. Therefore, it is necessary to double the number of manganese atoms, and leave the number of sulfur atoms unchanged. We indicate balance coefficients both before the reagents and before the products!

Scheme for compiling OVR equations using the electronic balance method:

Attention! There may be several oxidizing or reducing agents in a reaction. The balance must be drawn up so that Total number the electrons given and received were the same.

General patterns of redox reactions

The products of redox reactions often depend on conditions for the process. Let's consider main factors influencing the course of redox reactions.

The most obvious determining factor is reaction solution environment — . Typically (but not necessarily), the substance defining the medium is listed among the reagents. The following options are possible:

• oxidative activity is enhanced in a more acidic environment and the oxidizing agent is reduced more deeply(for example, potassium permanganate, KMnO 4, where Mn +7 in an acidic environment is reduced to Mn +2, and in an alkaline environment - to Mn +6);
• oxidative activity increases in a more alkaline environment, and the oxidizing agent is reduced deeper (for example, potassium nitrate KNO 3, where N +5, when interacting with a reducing agent in an alkaline environment, is reduced to N -3);
• or the oxidizing agent is practically not subject to changes in the environment.

The reaction environment makes it possible to determine the composition and form of existence of the remaining OVR products. The basic principle is that products are formed that do not interact with reagents!

Note! E If the solution medium is acidic, then bases cannot be present among the reaction products and basic oxides, because they react with acid. And, conversely, in an alkaline environment the formation of acid is excluded and acid oxide. This is one of the most common and most serious mistakes.

The direction of the flow of OVR is also affected by nature of the reacting substances. For example, when interacting nitric acid HNO 3 with reducing agents there is a pattern - the greater the activity of the reducing agent, the more nitrogen N +5 is reduced.

When increasing temperature Most ODD tends to be more intense and deeper.

IN heterogeneous reactions the composition of products is often influenced degree of grinding solid . For example, powdered zinc with nitric acid forms some products, while granulated zinc forms completely different ones. How more degree grinding the reagent, the greater its activity, usually.

Let's look at the most typical laboratory oxidizing agents.

Basic schemes of redox reactions

Permanganate recovery scheme

Permanganates contain a powerful oxidizing agent - manganese in oxidation state +7. Manganese salts +7 color the solution in violet color.

Permanganates, depending on the environment of the reaction solution, are reduced in different ways.

IN acidic environment recovery occurs more deeply, to Mn 2+. Manganese oxide in the +2 oxidation state exhibits basic properties, therefore in acidic environment salt is formed. Manganese salts +2 colorless. IN neutral solution manganese is reduced to oxidation state +4 , with education amphoteric oxide MnO2 — brown precipitate insoluble in acids and alkalis. IN alkaline environment, manganese is restored minimally - to the nearest oxidation states +6 . Manganese compounds +6 exhibit acid properties, in an alkaline environment they form salts - manganates. Manganates impart to the solution green color .

Let's consider the interaction of potassium permanganate KMnO 4 with potassium sulfide in acidic, neutral and alkaline media. In these reactions, the oxidation product of the sulfide ion is S0.

5 K 2 S + 2 KMnO 4 + 8 H 2 SO 4 = 5 S + 2 MnSO 4 + 6 K 2 SO 4 + 8 H 2 O,

3 K 2 S + 2 KMnO 4 + 4 H 2 O = 2 MnO 2 ↓ + 3 S↓ + 8 KOH,

A common mistake in this reaction is to indicate the interaction of sulfur and alkali in the reaction products. However, sulfur interacts with alkali under rather harsh conditions (elevated temperature), which does not correspond to the conditions of this reaction. Under normal conditions, it would be correct to indicate molecular sulfur and alkali separately, and not the products of their interaction.

K 2 S + 2 KMnO 4 –(KOH)= 2 K 2 MnO 4 + S↓

Difficulties also arise when composing this reaction. The point is that in in this case writing a molecule of the medium (KOH or another alkali) in the reagents is not required to equalize the reaction. The alkali takes part in the reaction and determines the product of the reduction of potassium permanganate, but the reagents and products are equalized without its participation. This seemingly paradox can be easily resolved if we remember that chemical reaction- this is just a conditional record that does not indicate each ongoing process, but is simply a display of the sum of all processes. How to determine this yourself? If you act according classic scheme- balance-balance coefficients-metal equalization, then you will see that metals are equalized by balance coefficients, and the presence of alkali on the left side of the reaction equation will be superfluous.

Permanganates oxidize:

• nonmetals with negative oxidation state to simple substances (with oxidation state 0), exceptions — phosphorus, arsenic - up to +5 ;
• nonmetals with intermediate oxidation state before highest degree oxidation;
• active metals stable positive degree of oxidation of the metal.

KMnO 4 + neMe (lowest d.o.) = neMe 0 + other products

KMnO 4 + neMe (intermediate d.o.) = neMe (higher d.o.) + other products

KMnO 4 + Me 0 = Me (stable s.o.) + other products

KMnO 4 + P -3 , As -3 = P +5 , As +5 + other products

Chromate/bichromate recovery scheme

A special feature of chromium with valency VI is that it forms 2 types of salts in aqueous solutions: chromates and dichromates, depending on the solution environment. Active metal chromates (for example, K 2 CrO 4) are salts that are stable in alkaline environment. Dichromates (bichromates) of active metals (for example, K 2 Cr 2 O 7) - salts, stable in an acidic environment .

Chromium(VI) compounds are reduced to chromium(III) compounds . Chromium compounds Cr +3 are amphoteric, and depending on the solution environment they exist in solution in various forms: in an acidic environment in the form salts(amphoteric compounds form salts when interacting with acids), insoluble in a neutral environment amphoteric hydroxide chromium (III) Cr(OH) 3 , and in an alkaline environment chromium (III) compounds form complex salt, For example, potassium hexahydroxochromate (III) K 3 .

Chromium VI compounds oxidize:

• nonmetals V negative degree oxidation to simple substances (with oxidation state 0), exceptions — phosphorus, arsenic – up to +5;
• nonmetals in intermediate oxidation state to the highest degree of oxidation;
• active metals from simple substances (oxidation stage 0) to compounds with stable positive degree of oxidation of the metal.

Chromate/bichromate + NeMe (negative d.o.) = NeMe 0 + other products

Chromate/bichromate + neMe (intermediate positive d.o.) = neMe (higher d.o.) + other products

Chromate/bichromate + Me 0 = Me (stable d.o.) + other products

Chromate/bichromate + P, As (negative d.o.) = P, As +5 + other products

Nitrate decomposition

Nitrate salts contain nitrogen in oxidation state +5 - strong oxidizer. Such nitrogen can oxidize oxygen (O -2). This occurs when nitrates are heated. In most cases, oxygen is oxidized to oxidation state 0, i.e. before molecular oxygen O2 .

Depending on the type of metal forming the salt, the thermal (temperature) decomposition of nitrates produces various products: If active metal(in the series of electrochemical activity there are to magnesium), then nitrogen is reduced to the oxidation state +3, and during decomposition nitrite salts and molecular oxygen are formed .

For example:

2NaNO 3 → 2NaNO 2 + O 2 .

Active metals occur in nature in the form of salts (KCl, NaCl).

If a metal is in the series of electrochemical activity to the right of magnesium and to the left of copper (including magnesium and copper) , then upon decomposition it is formed metal oxide in a stable oxidation state, nitric oxide (IV)(brown gas) and oxygen. Metal oxide also forms during decomposition lithium nitrate .

For example, decomposition zinc nitrate:

2Zn(NO 3) 2 → 2ZnО + 4NO 2 + O 2.

Metals average activity most often found in nature in the form of oxides (Fe 2 O 3, Al 2 O 3, etc.).

Ions metals, located in the series of electrochemical activity to the right of copper are strong oxidizing agents. At decomposition of nitrates they, like N +5, participate in the oxidation of oxygen and are reduced to simple substances, i.e. metal is formed and gases are released - nitric oxide (IV) and oxygen .

For example, decomposition silver nitrate:

2AgNO3 → 2Ag + 2NO2 + O2.

Inactive metals occur in nature as simple substances.

Some exceptions!

Decomposition ammonium nitrate :

The ammonium nitrate molecule contains both an oxidizing agent and a reducing agent: nitrogen in the -3 oxidation state exhibits only reducing properties, while nitrogen in the +5 oxidation state exhibits only oxidative properties.

When heated, ammonium nitrate decomposes. At temperatures up to 270 o C, it forms nitric oxide (I)(“laughing gas”) and water:

NH 4 NO 3 → N 2 O + 2H 2 O

This is an example of a reaction counter-disproportionation .

The resulting oxidation state of nitrogen is the arithmetic mean of the oxidation state of nitrogen atoms in the original molecule.

With more high temperature Nitric oxide (I) decomposes into simple substances - nitrogen And oxygen:

2NH 4 NO 3 → 2N 2 + O 2 + 4H 2 O

At decomposition ammonium nitrite NH4NO2 counter-disproportionation also occurs.

The resulting oxidation state of nitrogen is also equal to the average arithmetic powers oxidation of the initial nitrogen atoms - oxidizing agent N +3 and reducing agent N -3

NH 4 NO 2 → N 2 + 2H 2 O

Thermal decomposition manganese(II) nitrate accompanied by metal oxidation:

Mn(NO 3) 2 = MnO 2 + 2NO 2

Iron(II) nitrate at low temperatures it decomposes to iron (II) oxide; when heated, iron oxidizes to the oxidation state +3:

2Fe(NO 3) 2 → 2FeO + 4NO 2 + O 2 at 60°C
4Fe(NO 3) 2 → 2Fe 2 O 3 + 8NO 2 + O 2 at >60°C

Nickel(II) nitrate decomposes to nitrite when heated.

Oxidative properties of nitric acid

Nitric acid HNO 3 when interacting with metals is practically never produces hydrogen , unlike most mineral acids.

This is due to the fact that the acid contains a very strong oxidizing agent - nitrogen in the oxidation state +5. When interacting with reducing agents - metals, various nitrogen reduction products are formed.

Nitric acid + metal = metal salt + nitrogen reduction product + H 2 O

Nitric acid upon reduction can transform into nitrogen oxide (IV) NO 2 (N +4); nitric oxide (II) NO (N +2); nitric oxide (I) N 2 O (“laughing gas”); molecular nitrogen N 2; ammonium nitrate NH 4 NO 3. As a rule, a mixture of products is formed with a predominance of one of them. Nitrogen is reduced to oxidation states from +4 to −3. The depth of restoration depends primarily by nature of a reducing agent And on the concentration of nitric acid . The rule works: the lower the acid concentration and the higher the activity of the metal, the more electrons nitrogen receives, and the more reduced products are formed.

Some regularities will allow you to correctly determine the main product of the reduction of nitric acid by metals in the reaction:

• upon action very dilute nitric acid on metals is usually formed ammonium nitrate NH 4 NO 3;

For example, reaction of zinc with very dilute nitric acid:

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

• concentrated nitric acid in the cold passivates some metals - chromium Cr, aluminum Al and iron Fe . When the solution is heated or diluted, the reaction occurs;

metal passivation - this is the conversion of the metal surface into inactive state due to the formation on the metal surface thin layers inert compounds, in this case mainly metal oxides, which do not react with concentrated nitric acid

• Nitric acid does not react with metals of the platinum subgroup — gold Au, platinum Pt, and palladium Pd;

• when interacting concentrated acid with no active metals And medium activity metals nitrogen acid is reduced to nitric oxide (IV) NO 2 ;

For example, oxidation of copper with concentrated nitric acid:

Cu+ 4HNO 3 = Cu(NO 3) 2 + 2NO 2 + 2H 2 O

• when interacting concentrated nitric acid with active metals is formed Nitric oxide (I)N2O ;

For example, oxidation sodium concentrated nitric acid:

Na+ 10HNO 3 = 8NaNO 3 + N 2 O + 5H 2 O

• when interacting dilute nitric acid with inactive metals (in the activity series to the right of hydrogen) the acid is reduced to nitric oxide (II) NO ;
• when interacting dilute nitric acid with medium activity metals is formed either nitric oxide (II) NO, or nitric oxide N 2 O, or molecular nitrogen N 2 - depending on additional factors (metal activity, degree of metal grinding, degree of acid dilution, temperature).
• when interacting dilute nitric acid with active metals is formed molecular nitrogen N 2 .

For an approximate determination of the products of reduction of nitric acid when interacting with different metals I propose to use the pendulum principle. The main factors that shift the position of the pendulum are: acid concentration and metal activity. To simplify, we use 3 types of acid concentrations: concentrated (more than 30%), dilute (30% or less), very dilute (less than 5%). We divide metals according to their activity into active (before aluminum), medium activity (from aluminum to hydrogen) and inactive (after hydrogen). We arrange the reduction products of nitric acid in descending order of oxidation state:

NO2; NO; N2O; N 2; NH4NO3

The more active the metal, the more we move to the right. The higher the concentration or the lower the degree of dilution of the acid, the more we shift to the left.

For example , interact concentrated acid and the inactive metal copper Cu. Consequently, we shift to the extreme left position, nitrogen oxide (IV), copper nitrate and water are formed.

Reaction of metals with sulfuric acid

Dilute sulfuric acid interacts with metals like normal mineral acid. Those. interacts with metals that are located in the series of electrochemical voltages up to hydrogen. The oxidizing agent here is H + ions, which are reduced to molecular hydrogen H 2 . In this case, metals are oxidized, as a rule, to minimum degree of oxidation.

For example:

Fe + H 2 SO 4 (dil) = FeSO 4 + H 2

interacts with metals in the voltage range both before and after hydrogen.

H 2 SO 4 (conc) + metal = metal salt + sulfur reduction product (SO 2, S, H 2 S) + water

When concentrated sulfuric acid interacts with metals, a metal salt (in a stable oxidation state), water and a sulfur reduction product are formed - sulphur dioxide S +4 O 2, molecular sulfur S or hydrogen sulfide H 2 S -2, depending on the degree of concentration, activity of the metal, degree of its grinding, temperature, etc. When concentrated sulfuric acid reacts with metals, molecular hydrogen is not formed!

Basic principles of interaction of concentrated sulfuric acid with metals:

1. Concentrated sulfuric acid passivates aluminum, chrome, iron at room temperature or in the cold;

2. Concentrated sulfuric acid doesn't interact With gold, platinum and palladium ;

3. WITH inactive metals concentrated sulfuric acid restored to sulfur(IV) oxide.

For example, copper is oxidized by concentrated sulfuric acid:

Cu 0 + 2H 2 S +6 O 4 (conc) = Cu +2 SO 4 + S +4 O 2 + 2H 2 O

4. When interacting with active metals and zinc concentrated sulfuric acid formssulfur S or hydrogen sulfide H 2 S 2- (depending on temperature, degree of grinding and activity of the metal).

For example , interaction of concentrated sulfuric acid with zinc:

8Na 0 + 5H 2 S +6 O 4 (conc) → 4Na 2 + SO 4 + H 2 S — 2 + 4H 2 O

Hydrogen peroxide

Hydrogen peroxide H 2 O 2 contains oxygen in the oxidation state -1. Such oxygen can both increase and decrease the oxidation state. Thus, hydrogen peroxide exhibits both oxidizing and reducing properties.

When interacting with reducing agents, hydrogen peroxide exhibits the properties of an oxidizing agent and is reduced to an oxidation state of -2. Typically, the product of hydrogen peroxide reduction is water or hydroxide ion, depending on the reaction conditions. For example:

S +4 O 2 + H 2 O 2 -1 → H 2 S +6 O 4 -2

When interacting with oxidizing agents, peroxide is oxidized to molecular oxygen(oxidation state 0): O 2 . For example :

2KMn +7 O 4 + 5H 2 O 2 -1 + 3H 2 SO 4 → 5O 2 0 + 2Mn +2 SO 4 + K 2 SO 4 + 8H 2 O