Classification of chemical reactions in inorganic and organic chemistry. Redox reactions and reactions that occur without changing the oxidation state of atoms Reactions without changing the oxidation state of atoms

Oxidation-reduction reactions (ORR) – reactions that occur with a change in the oxidation state of the atoms that make up the reacting substances as a result of the transfer of electrons from one atom to another.

Oxidation state – the formal charge of an atom in a molecule, calculated on the assumption that the molecule consists only of ions.

The most electronegative elements in a compound have negative oxidation states, and the atoms of elements with lower electronegativity have positive oxidation states.

Oxidation state is a formal concept; in some cases, the oxidation state does not coincide with the valence.

For example: N 2 H 4 (hydrazine)

nitrogen oxidation degree – -2; nitrogen valency – 3.

Calculation of oxidation state

To calculate the oxidation state of an element, the following provisions should be taken into account:

1. The oxidation states of atoms in simple substances are equal to zero (Na 0; H 2 0).

2. The algebraic sum of the oxidation states of all atoms that make up a molecule is always equal to zero, and in a complex ion this sum is equal to the charge of the ion.

3. The atoms have a constant oxidation state: alkali metals (+1), alkaline earth metals (+2), hydrogen (+1) (except for hydrides NaH, CaH 2, etc., where the oxidation state of hydrogen is -1), oxygen (-2 ) (except for F 2 -1 O +2 and peroxides containing the –O–O– group, in which the oxidation state of oxygen is -1).

4. For elements, the positive oxidation state cannot exceed a value equal to the group number of the periodic system.

V 2 +5 O 5 -2; Na 2 +1 B 4 +3 O 7 -2; K +1 Cl +7 O 4 -2 ; N -3 H 3 +1 ; K 2 +1 H +1 P +5 O 4 -2 ; Na 2 +1 Cr 2 +6 O 7 -2

Reactions with and without changes in oxidation state

There are two types of chemical reactions:

A Reactions in which the oxidation state of elements does not change:

Addition reactions: SO 2 + Na 2 O Na 2 SO 3

Decomposition reactions: Cu(OH) 2  CuO + H 2 O

Exchange reactions: AgNO 3 + KCl AgCl + KNO 3

NaOH + HNO 3 NaNO 3 + H 2 O

B Reactions in which there is a change in the oxidation states of the atoms of the elements that make up the reacting compounds:

2Mg 0 + O 2 0 2Mg +2 O -2

2KCl +5 O 3 -2 – t  2KCl -1 + 3O 2 0

2KI -1 + Cl 2 0 2KCl -1 + I 2 0

Mn +4 O 2 + 4HCl -1 Mn +2 Cl 2 + Cl 2 0 + 2H 2 O

Such reactions are called redox reactions .

Oxidation, reduction

In redox reactions, electrons are transferred from one atom, molecule, or ion to another. The process of losing electrons is oxidation. During oxidation, the oxidation state increases:

H 2 0 − 2ē 2H +

S -2 − 2ē S 0

Al 0 − 3ē Al +3

Fe +2 − ē Fe +3

2Br - − 2ē Br 2 0

The process of adding electrons is reduction. During reduction, the oxidation state decreases.

Mn +4 + 2ē Mn +2

Сr +6 +3ē Cr +3

Cl 2 0 +2ē 2Cl -

O 2 0 + 4ē 2O -2

Atoms or ions that gain electrons in a given reaction are oxidizing agents, and those that donate electrons are reducing agents.

Redox properties of a substance and the oxidation state of its constituent atoms

Compounds containing atoms of elements with the maximum oxidation state can only be oxidizing agents due to these atoms, because they have already given up all their valence electrons and are only able to accept electrons. The maximum oxidation state of an element's atom is equal to the number of the group in the periodic table to which the element belongs. Compounds containing atoms of elements with a minimum oxidation state can only serve as reducing agents, since they are only capable of donating electrons, because the outer energy level of such atoms is completed by eight electrons. The minimum oxidation state of metal atoms is 0, for non-metals - (n–8) (where n is the number of the group in the periodic table). Compounds containing atoms of elements with intermediate oxidation states can be both oxidizing and reducing agents, depending on the partner with which they interact and the reaction conditions.

REDOX REACTIONS

Oxidation state

The oxidation state is the nominal charge of an atom in a molecule, calculated under the assumption that the molecule consists of ions and is generally electrically neutral.

The most electronegative elements in a compound have negative oxidation states, and the atoms of elements with less electronegativity have positive oxidation states.

Oxidation state is a formal concept; in some cases, the oxidation state does not coincide with the valency.

For example:

N2H4 (hydrazine)

nitrogen oxidation degree – -2; nitrogen valency – 3.

Calculation of oxidation state

To calculate the oxidation state of an element, the following provisions should be taken into account:

1. The oxidation states of atoms in simple substances are equal to zero (Na 0; H2 0).

2. The algebraic sum of the oxidation states of all atoms that make up a molecule is always equal to zero, and in a complex ion this sum is equal to the charge of the ion.

3. The atoms have a constant oxidation state: alkali metals (+1), alkaline earth metals (+2), hydrogen (+1) (except for hydrides NaH, CaH2, etc., where the oxidation state of hydrogen is -1), oxygen (-2) (except for F 2 -1 O +2 and peroxides containing the –O–O– group, in which the oxidation state of oxygen is -1).

4. For elements, the positive oxidation state cannot exceed a value equal to the group number of the periodic system.

Examples:

V 2 +5 O 5 -2 ;Na 2 +1 B 4 +3 O 7 -2 ;K +1 Cl +7 O 4 -2 ;N -3 H 3 +1 ;K2 +1 H +1 P +5 O 4 -2 ;Na 2 +1 Cr 2 +6 O 7 -2

Reactions without and with changes in oxidation state

There are two types of chemical reactions:

AReactions in which the oxidation state of elements does not change:

Addition reactions

SO 2 +Na 2 O → Na 2 SO 3

Decomposition reactions

Cu(OH) 2 → CuO + H 2 O

Exchange reactions

AgNO 3 + KCl → AgCl + KNO 3

NaOH + HNO 3 → NaNO 3 + H 2 O

BReactions in which there is a change in the oxidation states of the atoms of the elements that make up the reacting compounds:

2Mg 0 + O 2 0 → 2Mg +2 O -2

2KCl +5 O 3 -2 →2KCl -1 + 3O 2 0

2KI -1 + Cl 2 0 → 2KCl -1 + I 2 0

Mn +4 O 2 + 4HCl -1 ® Mn +2 Cl 2 + Cl +1 2 0 + 2H 2 O

Such reactions are called redox reactions

Redox reactions are reactions in which the oxidation states of atoms change. Redox reactions are very common. All combustion reactions are redox.
The redox reaction consists of two processes that cannot occur separately from each other. The process of increasing the oxidation state is called oxidation. Simultaneously with oxidation, reduction occurs, that is, the process of decreasing the oxidation state.

Oxidation, reduction

Accordingly, in redox reactions there are two main participants: an oxidizing agent and a reducing agent. The process of losing electrons is oxidation. During oxidation, the oxidation state increases. During the reaction, the oxidizing agent lowers its oxidation state and is reduced. Here it is necessary to distinguish between an oxidizing chemical element and an oxidizing substance.

N +5 - oxidizer; HN +5 O3 and NaN +5 O 3 - oxidizing agents.
If we say that nitric acid and its salts are strong oxidizing agents, then by this we mean that the oxidizing agent is nitrogen atoms with an oxidation state of +5, and not the entire substance as a whole.
The second obligatory participant in the redox reaction is called the reducing agent. The process of adding electrons is reduction. During reduction, the oxidation state decreases.

The reducing agent increases its oxidation number by being oxidized during the reaction. Just as in the case of an oxidizing agent, one should distinguish between a reducing substance and a reducing chemical element. When carrying out the reaction of reducing an aldehyde to alcohol, we cannot take just hydrogen with an oxidation state of -1, but take some kind of hydride, preferably lithium aluminum hydride.

N -1 - reducing agent; NaH -1 and LiAlH -1 4 - reducing agents.
In redox reactions, a complete transfer of electrons from a reducing agent to an oxidizing agent is extremely rare, since there are few compounds with ionic bonds. But when arranging the coefficients, we proceed from the assumption that such a transition does occur. This makes it possible to correctly determine the main coefficients in front of the formulas of the oxidizing agent and reducing agent.
5H 2 SO 3 + 2KMnO 4 = 2H 2 SO 4 + 2MnSO 4 + K 2 SO 4 + 3H 2 O
S +4 – 2e → S +6 5 - reducing agent, oxidation
Mn +7 + 5e → Mn +2 2 - oxidizer, reduction

Atoms or ions that gain electrons in a given reaction are oxidizing agents, and those that donate electrons are reducing agents.

Redox properties of a substance and the oxidation state of its constituent atoms

Compounds containing atoms of elements with the maximum oxidation state can only be oxidizing agents due to these atoms, because they have already given up all their valence electrons and are only able to accept electrons. The maximum oxidation state of an element's atom is equal to the number of the group in the periodic table to which the element belongs. Compounds containing atoms of elements with a minimum oxidation state can only serve as reducing agents, since they are only capable of donating electrons, because the outer energy level of such atoms is completed by eight electrons. The minimum oxidation state for metal atoms is 0, for non-metals - (n–8) (where n is the number of the group in the periodic table). Compounds containing atoms of elements with intermediate oxidation states can be both oxidizing and reducing agents, depending on the partner with which they interact and the reaction conditions.

The most important reducing and oxidizing agents

Reducers:

Metals,

hydrogen,

coal.

Carbon (II) monoxide (CO).

Hydrogen sulfide (H 2 S);

sulfur oxide (IV) (SO 2);

sulfurous acid H 2 SO 3 and its salts.

Hydrohalic acids and their salts.

Metal cations in lower oxidation states: SnCl 2, FeCl 2, MnSO 4, Cr 2 (SO 4) 3.

Nitrous acid HNO 2;

ammonia NH 3;

hydrazine NH 2 NH 2 ;

nitric oxide (II) (NO).

Cathode during electrolysis.

Oxidizing agents

Halogens.

Potassium permanganate (KMnO 4);

potassium manganate (K 2 MnO 4);

manganese (IV) oxide (MnO 2).

Potassium dichromate (K 2 Cr 2 O 7);

potassium chromate (K 2 CrO 4).

Nitric acid (HNO 3).

Sulfuric acid (H 2 SO 4) conc.

Copper(II) oxide (CuO);

lead(IV) oxide (PbO 2);

silver oxide (Ag 2 O);

hydrogen peroxide (H 2 O 2).

Iron(III) chloride (FeCl 3).

Berthollet's salt (KClO 3).

Anode during electrolysis.

Each such half-reaction is characterized by a standard redox potential E 0 (dimension - volts, V). The greater E0, the stronger the oxidizing form as an oxidizing agent and the weaker the reduced form as a reducing agent, and vice versa.

The half-reaction is taken as the reference point for potentials: 2H + + 2ē ® H 2, for which E 0 =0

For half-reactions M n+ + nē ® M 0, E 0 is called the standard electrode potential. Based on the magnitude of this potential, metals are usually placed in a series of standard electrode potentials (a series of metal voltages):

Li, Rb, K, Ba, Sr, Ca, Na, Mg, Al, Mn, Zn, Cr, Fe, Cd,

Co, Ni, Sn, Pb, H, Sb, Bi, Cu, Hg, Ag, Pd, Pt, Au

A number of stresses characterize the chemical properties of metals:

1. The further to the left a metal is located in the voltage series, the stronger its reducing ability and the weaker the oxidizing ability of its ion in solution (i.e., the easier it gives up electrons (oxidizes) and the more difficult it is for its ions to reattach electrons).

2. Each metal is capable of displacing from salt solutions those metals that are in the stress series to the right of it, i.e. reduces ions of subsequent metals into electrically neutral atoms, donating electrons and turning into ions itself.

3. Only metals that are in the voltage series to the left of hydrogen (H) are capable of displacing it from acid solutions (for example, Zn, Fe, Pb, but not Cu, Hg, Ag).

Galvanic cells

Every two metals, being immersed in solutions of their salts, which communicate with each other through a siphon filled with an electrolyte, form a galvanic cell. Metal plates immersed in solutions are called electrodes of the element.

If you connect the outer ends of the electrodes (poles of the element) with a wire, then electrons begin to move from the metal, which has a lower potential, to the metal, which has a higher potential (for example, from Zn to Pb). The departure of electrons disrupts the equilibrium that exists between the metal and its ions in the solution and causes a new number of ions to pass into the solution - the metal gradually dissolves. At the same time, electrons passing to another metal discharge the ions in the solution at its surface - the metal is released from the solution. The electrode at which oxidation occurs is called the anode. The electrode at which reduction occurs is called the cathode. In a lead-zinc cell, the zinc electrode is the anode and the lead electrode is the cathode.

Thus, in a closed galvanic cell, interaction occurs between a metal and a salt solution of another metal, which are not in direct contact with each other. Atoms of the first metal, giving up electrons, turn into ions, and ions of the second metal, adding electrons, turn into atoms. The first metal displaces the second from the solution of its salt. For example, during the operation of a galvanic cell composed of zinc and lead, immersed respectively in solutions of Zn(NO 3) 2 and Pb(NO 3) 2, the following processes occur at the electrodes:

Zn – 2ē → Zn 2+

Pb 2+ + 2ē → Pb

Summing up both processes, we obtain the equation Zn + Pb 2+ → Pb + Zn 2+, expressing the reaction occurring in the element in ionic form. The molecular equation for the same reaction would be:

Zn + Pb(NO 3) 2 → Pb + Zn(NO 3) 2

The electromotive force of a galvanic cell is equal to the potential difference between its two electrodes. When determining it, the smaller one is always subtracted from the larger potential. For example, the electromotive force (emf) of the considered element is equal to:

E.m.f. =	-0,13		(-0,76)	0.63v
	E Pb		E Zn

It will have this value provided that the metals are immersed in solutions in which the ion concentration is 1 g-ion/l. At other concentrations of solutions, the values of the electrode potentials will be somewhat different. They can be calculated using the formula:

E = E 0 + (0.058/n) logC

where E is the desired metal potential (in volts)

E 0 - its normal potential

n - valence of metal ions

C - concentration of ions in solution (g-ion/l)

Example

Find the electromotive force of the element (emf) formed by a zinc electrode immersed in a 0.1 M solution of Zn(NO 3) 2 and a lead electrode immersed in a 2 M solution of Pb(NO 3) 2.

Solution

We calculate the potential of the zinc electrode:

E Zn = -0.76 + (0.058 / 2) log 0.1 = -0.76 + 0.029 (-1) = -0.79 v

We calculate the potential of the lead electrode:

E Pb = -0.13 + (0.058 / 2) log 2 = -0.13 + 0.029 0.3010 = -0.12 v

Find the electromotive force of the element:

E.m.f. = -0.12 – (-0.79) = 0.67 v

Electrolysis

Electrolysis The process of decomposition of a substance by electric current is called.

The essence of electrolysis is that when current is passed through an electrolyte solution (or molten electrolyte), positively charged ions move to the cathode, and negatively charged ions move to the anode. Having reached the electrodes, the ions are discharged, as a result of which the components of the dissolved electrolyte or hydrogen and oxygen from the water are released at the electrodes.

To convert different ions into neutral atoms or groups of atoms, different voltages of electric current are required. Some ions lose their charges more easily, others more difficult. The degree of ease with which metal ions discharge (gain electrons) is determined by the position of the metals in the voltage series. The further to the left a metal is in the voltage series, the greater its negative potential (or less positive potential), the more difficult, other things being equal, are its ions to discharge (Au 3+, Ag + ions are the easiest to discharge; the most difficult are Li +, Rb +, K +).

If there are ions of several metals in a solution at the same time, then the ions of the metal with a lower negative potential (or a higher positive potential) are discharged first. For example, metallic copper is first released from a solution containing Zn 2+ and Cu 2+ ions. But the magnitude of the metal’s potential also depends on the concentration of its ions in the solution; the ease of discharge of ions of each metal also changes depending on their concentration: an increase in concentration facilitates the discharge of ions, a decrease makes it more difficult. Therefore, during the electrolysis of a solution containing ions of several metals, it may happen that the release of a more active metal will occur earlier than the release of a less active one (if the concentration of ions of the first metal is significant and the concentration of the second is very small).

In aqueous solutions of salts, in addition to salt ions, there are always also water ions (H + and OH -). Of these, hydrogen ions will be discharged more easily than the ions of all metals preceding hydrogen in the voltage series. However, due to the insignificant concentration of hydrogen ions during the electrolysis of all salts, except for the salts of the most active metals, metal, not hydrogen, is released at the cathode. Only during the electrolysis of salts of sodium, calcium and other metals up to aluminum inclusive, hydrogen ions are discharged and hydrogen is released.

At the anode, either ions of acidic residues or hydroxyl ions of water can be discharged. If the ions of acidic residues do not contain oxygen (Cl -, S 2-, CN -, etc.), then it is usually these ions that are discharged, and not hydroxyl ones, which lose their charge much more difficult, and Cl 2, S, etc. are released at the anode .d. On the contrary, if a salt of an oxygen-containing acid or the acid itself undergoes electrolysis, then hydroxyl ions are discharged, and not ions of oxygen residues. The neutral OH groups formed during the discharge of hydroxyl ions immediately decompose according to the equation:

4OH → 2H2O + O2

As a result, oxygen is released at the anode.

Electrolysis of nickel chloride solution NiCl 2

The solution contains Ni 2+ and Cl - ions, as well as H + and OH - ions in negligible concentrations. When current is passed, Ni 2+ ions move to the cathode, and Cl - ions move to the anode. Taking two electrons from the cathode, Ni 2+ ions turn into neutral atoms that are released from the solution. The cathode is gradually coated with nickel.

Chlorine ions, reaching the anode, give up electrons to it and turn into chlorine atoms, which, when combined in pairs, form chlorine molecules. Chlorine is released at the anode.

Thus, at the cathode there is recovery process, at the anode – oxidation process.

Electrolysis of potassium iodide solution KI

Potassium iodide is in solution in the form of K + and I - ions. When current is passed, K + ions move to the cathode, I - ions move to the anode. But since potassium is much to the left of hydrogen in the voltage series, it is not the potassium ions that are discharged at the cathode, but the hydrogen ions of water. The hydrogen atoms formed in this case combine into H 2 molecules, and thus hydrogen is released at the cathode.

As the hydrogen ions are discharged, more and more water molecules dissociate, as a result of which hydroxyl ions (released from the water molecule) accumulate at the cathode, as well as K + ions, which continuously move to the cathode. A KOH solution is formed.

Iodine is released at the anode, since I - ions are discharged more easily than hydroxyl ions of water.

Electrolysis of potassium sulfate solution

The solution contains K + ions, SO 4 2- and H + and OH - ions from water. Since K + ions are more difficult to discharge than H + ions, and SO 4 2- ions than OH - ions, then when an electric current is passed, hydrogen ions will be discharged at the cathode, and hydroxyl groups at the anode, that is, in fact, electrolysis of water. At the same time, due to the discharge of hydrogen and hydroxyl ions of water and the continuous movement of K + ions to the cathode, and SO 4 2- ions to the anode, an alkali solution (KOH) is formed at the cathode, and a sulfuric acid solution is formed at the anode.

Electrolysis of copper sulfate solution with a copper anode

Electrolysis occurs in a special way when the anode is made of the same metal whose salt is in solution. In this case, no ions are discharged at the anode, but the anode itself gradually dissolves, sending ions into the solution and donating electrons to the current source.

The whole process comes down to the release of copper at the cathode and the gradual dissolution of the anode. The amount of CuSO 4 in the solution remains unchanged.

Laws of electrolysis (M. Faraday)

1. The weight amount of the substance released during electrolysis is proportional to the amount of electricity flowing through the solution and practically does not depend on other factors.

2. Equal amounts of electricity are released during electrolysis from various chemical compounds in equivalent amounts of substances.

3. To isolate one gram equivalent of any substance from an electrolyte solution, 96,500 coulombs of electricity must be passed through the solution.

m (x) = ((I t) / F) (M (x) / n)

where m (x) is the amount of reduced or oxidized substance (g);

I is the strength of the transmitted current (a);

t - electrolysis time (s);

M(x) - molar mass;

n is the number of electrons acquired or given up in redox reactions;

F - Faraday's constant (96500 cool/mol).

Based on this formula, you can make a number of calculations related to the electrolysis process, for example:

1. Calculate the amounts of substances released or decomposed by a certain amount of electricity;

2. Find the current strength by the amount of substance released and the time spent on its release;

3. Determine how long it will take to release a certain amount of a substance at a given current.

Example 1

How many grams of copper will be released at the cathode when a current of 5 amperes is passed through a solution of copper sulfate CuSO 4 for 10 minutes?

Solution

Let us determine the amount of electricity flowing through the solution:

Q = It,

where I is the current in amperes;

t – time in seconds.

Q = 5A 600 s = 3000 coulombs

The equivalent of copper (at. mass 63.54) is 63.54: 2 = 31.77. Therefore, 96500 coulombs release 31.77 g of copper. Required amount of copper:

m = (31.77 3000) / 96500 » 0.98 g

Example 2

How long does it take to pass a current of 10 amperes through an acid solution to obtain 5.6 liters of hydrogen (at normal conditions)?

Solution

We find the amount of electricity that must pass through the solution in order for 5.6 liters of hydrogen to be released from it. Since 1 g-eq. hydrogen occupies at n. u. volume is 11.2 l, then the required amount of electricity

Q = (96500 5.6) / 11.2 = 48250 coulombs

Let us determine the current passage time:

t = Q / I = 48250 / 10 = 4825 s = 1 h 20 min 25 s

Example 3

When passing a current through a solution of silver salt at the cathode, it was released in 10 minutes. 1 g silver. Determine the current strength.

Solution

1 g-eq. silver is equal to 107.9 g. To release 1 g of silver, 96500 must pass through the solution: 107.9 = 894 coulombs. Hence the current strength

I = 894 / (10 60)" 1.5A

Example 4

Find the equivalent of tin if at a current of 2.5 amperes from a solution of SnCl 2 in 30 minutes. 2.77 g of tin is released.

Solution

The amount of electricity passing through the solution in 30 minutes.

Q = 2.5 30 60 = 4500 coulombs

Since for the release of 1 g-eq. 96,500 coulombs are required, then the equivalent of tin.

E Sn = (2.77 96500) / 4500 = 59.4

Corrosion

Before finishing our discussion of electrochemistry, let us apply the knowledge we have acquired to the study of one very important problem - corrosion metals Corrosion is caused by oxidation-reduction reactions in which a metal, as a result of interaction with some substance in its environment, transforms into an undesirable compound.

One of the most widely known corrosion processes is the rusting of iron. From an economic point of view, this is a very important process. It is estimated that 20% of the iron produced annually in the United States is used to replace iron products that have become unusable due to rusting.

It is known that oxygen is involved in the rusting of iron; iron does not oxidize in water in the absence of oxygen. Water also takes part in the rusting process; iron does not corrode in oxygenated oil as long as there are no traces of water in it. Rusting is accelerated by a number of factors, such as the pH of the environment, the presence of salts in it, the contact of iron with metal, which is more difficult to oxidize than iron, as well as under the influence of mechanical stress.

Corrosion of iron is in principle an electrochemical process. Some areas of the surface of iron serve as an anode on which its oxidation occurs:

Fe(solid) → Fe 2+ (aq) + 2e - Eº oxide = 0.44 V

The electrons generated in this case move through the metal to other areas of the surface, which play the role of a cathode. Oxygen reduction occurs on them:

O 2 (g.) + 4H + (aq.) + 4e - → 2H 2 O (l.) Eº restore = 1.23 V

Note that H + ions participate in the process of O 2 reduction. If the H+ concentration decreases (i.e., as the pH increases), O2 reduction becomes more difficult. It has been noticed that iron in contact with a solution whose pH is above 9-10 does not corrode. During the corrosion process, Fe 2+ ions formed at the anode are oxidized to Fe 3+. Fe 3+ ions form hydrated iron (III) oxide, which is called rust:

4Fe 2+ (aq.) + O 2 (g.) + 4H 2 O (l.) +2 X H 2 O (l.) → 2Fe 2 O 3 . x H2O( tv.) + 8H + (aq.)

Since the role of the cathode is usually played by that part of the surface that is best provided with an influx of oxygen, rust most often appears in these areas. If you carefully examine a shovel that has been standing for some time in open, humid air with dirt stuck to the blade, you will notice that depressions have formed under the dirt on the surface of the metal, and rust has appeared everywhere where O2 could penetrate.

Increased corrosion in the presence of salts is often encountered by motorists in areas where roads are generously sprinkled with salt in winter to combat icy conditions. The effect of salts is explained by the fact that the ions they form create the electrolyte necessary for the formation of a closed electrical circuit.

The presence of anodic and cathodic sites on the surface of iron leads to the creation of two different chemical environments on it. They can arise due to the presence of impurities or defects in the crystal lattice (apparently caused by stresses within the metal). At locations where such impurities or defects exist, the microscopic environment of a particular iron atom may cause its oxidation state to slightly increase or decrease from normal positions in the crystal lattice. Therefore, such places can play the role of anodes or cathodes. Ultra-pure iron, in which the number of such defects is reduced to a minimum, is much less likely to corrode than ordinary iron.

Iron is often coated with paint or some other metal such as tin, zinc or chromium to protect its surface from corrosion. The so-called “tinplate” is obtained by covering sheet iron with a thin layer of tin. Tin protects iron only as long as the protective layer remains intact. As soon as it is damaged, air and moisture begin to affect the iron; Tin even accelerates the corrosion of iron because it serves as a cathode in the electrochemical corrosion process. A comparison of the oxidation potentials of iron and tin shows that iron is oxidized more easily than tin:

Fe (solid) → Fe 2+ (aq.) + 2e - Eº oxide = 0.44 V

Sn (tv.) → Sn 2+ (aq.) + 2e - Eº oxide = 0.14 V

Therefore, iron serves as an anode in this case and is oxidized.

"Galvanized" (galvanized) iron is made by coating the iron with a thin layer of zinc. Zinc protects iron from corrosion even after the integrity of the coating is damaged. In this case, iron plays the role of a cathode during the corrosion process, because zinc oxidizes more easily than iron:

Zn (solid) → Zn 2+ (aq.) + 2e - Eº oxide = 0.76 V

Consequently, zinc acts as an anode and corrodes instead of iron. This kind of metal protection, in which it plays the role of a cathode in the process of electrochemical corrosion, is called cathodic protection. Pipes laid underground are often protected from corrosion by making them the cathode of an electrochemical cell. To do this, blocks of some active metal, most often magnesium, are buried in the ground along the pipeline and connected with wire to the pipes. In moist soil, the active metal acts as an anode, and the iron pipe receives cathodic protection.

Although our discussion has focused on iron, it is not the only metal susceptible to corrosion. At the same time, it may seem strange that an aluminum can, carelessly left in the open air, corrodes immeasurably more slowly than an iron one. Judging by the standard oxidation potentials of aluminum (Eº oxide = 1.66 V) and iron (Eº oxide = 0.44 V), then it should be expected that corrosion of aluminum should occur much faster. The slow corrosion of aluminum is explained by the fact that a thin, dense film of oxide forms on its surface, protecting the underlying metal from further corrosion. Magnesium, which has a high oxidation potential, is protected from corrosion due to the formation of the same oxide film. Unfortunately, the oxide film on the surface of iron has a too loose structure and is not capable of creating reliable protection. However, a good protective oxide film forms on the surface of iron-chromium alloys. Such alloys are called stainless steel.

Redox processes. Compilation of oxidation-reduction reactions (ORR). Method for taking into account changes in oxidation states of elements. Types of OVR. Ion-electronic method for preparing OVR. The concept of standard electrode potential. Using standard redox potentials to determine the fundamental possibility of the redox process.

Topic 4.2.1. Oxidation state

The oxidation number is a positive or negative number assigned to each atom in a compound and is equal to the charge of the atom, assuming all chemical bonds in the compound are ionic. Since compounds with a purely ionic chemical bond do not exist, the actual charges on the atoms never coincide with the oxidation states. However, the use of oxidation states allows us to solve a number of chemical problems.

The degree of oxidation of an element in compounds is determined by the number of valence electrons involved in the formation of a chemical bond of a given element. But usually, to determine the oxidation states of elements, they do not describe the electronic configuration of the valence electrons, but use a number of empirical rules:

1. The sum of the oxidation states of atoms in a particle is equal to its electric charge.

2. In simple substances (consisting of atoms of only one element), the oxidation state of the element is zero.

3. In binary compounds (consisting of atoms of two elements), a negative oxidation state is assigned to the atom with higher electronegativity. Typically, formulas of chemical compounds are written in such a way that the more electronegative atom appears second in the formula, although some formulas can be written differently:

Or (common notation), or .

4. In complex compounds, certain atoms are assigned constant oxidation states:

– fluorine always has an oxidation state of -1;

– metal elements usually have a positive oxidation state;

– hydrogen usually has an oxidation state of +1 (,), but in compounds with metals (hydrides) its oxidation state is -1: , ;

– oxygen is characterized by an oxidation state of -2, but with more electronegative fluorine – , and in peroxide compounds – , , , (sodium superoxide);

– the maximum positive oxidation state of an element usually coincides with the number of the group in which the element is located (Table 1).

Exceptions:

1) the maximum oxidation state is less than the group number: F, O, He, Ne, Ar, cobalt subgroup: Co(+2,+3); Rh, Ir (+3,+4,+6), nickel subgroup: Ni (+2, rarely +4); Pd, Pt (+2,+4, rarely +6);

2) the maximum oxidation state is higher than the group number: elements of the copper subgroup: Cu (+1, +2), Au (+1, +3).

– the lowest negative oxidation state of non-metal elements is defined as the group number minus 8 (Table 4.1).

Table 4.1. Oxidation states of some elements

Element	Group number	Maximum positive oxidation state	Lowest negative oxidation state
Na
Al
N			5 – 8 = -3
S			6 – 8 = -2
Cl			7 – 8 = -1

Difficulties often arise in determining oxidation states in complex compounds - salts, the formula of which contains several atoms for which different oxidation states are possible. In this case, one cannot do without knowledge of the genetic relationship between the main classes of inorganic compounds, namely, knowledge of the formulas of acids, the derivatives of which are certain salts.

For example: determine the oxidation state of the elements in the compound Cr2(SO 4 ) 3 . The student’s reasoning in this case can be constructed in the following way: Cr2(SO 4 ) 3 - this is an average salt of sulfuric acid, in which the oxidation states of elements are quite simple to arrange. IN Cr2(SO 4 ) 3 sulfur and oxygen have the same oxidation states, while the sulfate ion has a charge of 2-:. Taking it as easy to determine the oxidation state of chromium: . That is, this salt is chromium (III) sulfate: .

Topic 4.2.2. Redox processes

Oxidation-reduction reactions (ORR) are reactions that occur with a change in the oxidation state of elements. A change in oxidation states occurs due to the transfer of electrons from one particle to another.

The process of a particle losing electrons is called oxidation, and the particle itself is oxidized. The process of a particle gaining electrons is called reduction, and the particle itself is reduced. That is, redox reactions are the unity of two opposing processes.

An oxidizing agent is a reagent that contains an element that, during redox reaction, reduces its oxidation state due to the addition of electrons. A reducing agent is a reagent that contains an element that increases its oxidation state by losing electrons.

For example:

reducing agent:

oxidizer:

reducing agent:

oxidizer:

Many redox reactions are accompanied by a change in the color of the solution.

For example:


violet	green	brown	colorless

Many redox reactions are widely used in practice.

BASIC TYPES

REDOX REACTIONS

1) Intermolecular (outer-sphere electron transfer reactions) are reactions in which electron transfer occurs between different reagents, that is, the oxidizing agent and the reducing agent are part of different substances.

Ok sssssssssssssssss

2) Intramolecular (intrasphere electron transfer reactions) - in these reactions, atoms of different elements of the same substance are the oxidizing agent and the reducing agent.

3) Self-oxidation reactions - self-healing (disproportionation) - in these reactions the oxidation state of the same element both increases and decreases.

Topic 4.2.3. Typical oxidizing agents

1) Potassium tetraoxomanganate (VII) -

The oxidizing properties of the ion depend on the nature of the medium:

Acidic environment:

Neutral environment:

Alkaline environment:

2) Potassium dichromate –

Oxidizing properties also depend on the nature of the environment:

Acidic environment:

Neutral environment:

Alkaline environment:

3) Halogens.

4) Hydrogen in dilute acids.

5) Concentrated sulfuric acid

The products of sulfur reduction depend on the nature of the reducing agent:

Low-active metal:

Medium activity metal:

Active Metal:

6) Nitric acid

In nitric acid of any concentration, the oxidizing agent is not protons, but nitrogen, which has an oxidation state of +5. Therefore, hydrogen is never released in these reactions. Because nitrogen has a wide variety of oxidation states, it also has a wide variety of reduction products. The reduction products of nitric acid depend on its concentration and the activity of the reducing agent.

When concentrated nitric acid reacts with metals, nitric oxide (IV) is usually released, and with non-metals, nitric oxide (II) is usually released:

Interaction with metal:

Interaction with non-metal:

When dilute nitric acid reacts with metals, the products depend on the activity of the metal:

Low-active metal:

Active Metal:

- active metal and very dilute acid:

7) They are also used as oxidizing agents PbO2 , MnO2 .

Topic 4.2.4. Typical reducing agents

1). Halide ions.

In the series, the reducing properties increase:

2). and its salts:

3). Ammonia and ammonium cation salts:

4). Derivatives:

In aqueous solutions, the complexes easily transform into complexes:

5). All metals are capable, although to varying degrees, of exhibiting reducing properties.

6). Industry uses hydrogen, carbon (in the form of coal or coke) and CO .

Topic 4.2.5. Compounds capable of exhibiting both oxidizing and reducing properties

Some elements in intermediate oxidation states have redox duality, i.e. with oxidizing agents they can act as reducing agents, and with reducing agents they behave as oxidizing agents.

NaNO3; Na 2 SO 4; S; NH 2 OH; H2O2 . For example:

H2O2 - reducing agent:

H2O2 - oxidizer:

For example , H2O2 may undergo disproportionation reactions:

Topic 4.2.3. Composition of redox reactions

To compile the OVR, two methods are used:

1) electronic balance method:

This method is based on the use of oxidation states.

The oxidation state of manganese decreases by 5 units,

in this case, the oxidation state of chlorine increases by 1 unit, but taking into account the resulting reaction product - a simple substance containing 2 moles of chlorine atoms - by 2 units.

Let's write these arguments in the form of a balance and find the main coefficients using the concept of a common multiple for numbers showing increased ie and decrease in oxidation states:

Let us put the obtained coefficients into the equation. Let us take into account that it is not only an oxidizing agent, but also binds reaction products - manganese and potassium ions (the degree of oxidation in this case does not change), that is, the coefficient before will be greater than it follows from the balance.

We find the remaining coefficients by calculating the balance of atoms, then using the balance of atoms we find the final coefficient before and using the balance of atoms we find the number of moles of water.

To check the correctness of the selected coefficients, we calculate the balance of moles of oxygen atoms. According to the final equation, it can be seen that out of 16 moles of acid taken for the reaction, 10 moles are spent on reduction, and 6 moles on binding the manganese (II) and potassium ions formed as a result of the reaction.

2) ion-electronic method (half-reaction method):

The oxidizing agent is , which is part of the ion.

In the partial equation of the reduction reaction for the balance of atoms, hydrogen cations must be added to the left side in order to bind oxygen atoms into water,

and to balance the charges, add 5 moles of electrons to the same left side of the equation. We get:

A reducing agent is an ion that contains .

In the particular equation of the oxidation reaction to balance the atoms, hydrogen cations must be added to the right side to bind excess oxygen atoms into water, and to balance the charges, add 2 moles of electrons to the same right side of the equation. We get:

Thus we have two half-reactions:

To equalize, multiply the first half-reaction by 2, and the second by 5. Add the two half-reactions.

Full ionic equation:

Let's reduce the same terms:

After reduction, the coefficients of the full ionic equation can be transferred to the molecular equation.

Topic 4.2.4. The concept of standard electrode potential

The possibility of a redox reaction occurring is judged by the values of the electrode potentials of individual half-reactions.

If a metal plate is immersed in a solution containing ions of this metal, then a potential difference will arise at the metal-solution interface, which is usually called the electrode potential φ. Electrode potentials are determined experimentally. For standard conditions (solution concentration 1 mol/l, T = 298 K), these potentials are called standard, denoted φ 0. Standard electrode potentials are usually measured relative to a standard hydrogen electrode and are given in reference tables.

2Н + + 2ē = Н 2 φ 0 = 0.

The standard electrode potential is related to the Gibbs free energy. For reaction under standard conditions:

ΔG = - nFφ 0

Faraday's F constant (F=96500 C/mol), n is the number of transferred electrons.

The value of the electrode potential depends on the concentration of the reagents and temperature. This dependence is expressed by the Nernst equation:

where φ is the value of the electrode potential, depending on temperature and concentration.

NO 3 - + 2ē + H 2 O = NO 2 - + 2OH - , φ 0 = - 0.01V

Let's take into account that = = 1 mol/l, pH + pH = 14, pH = -log, log = -log - 14.

The electrode potential depends on the acidity of the pH environment. With acidification of the solution (with a decrease in pH), the oxidative function of NO 3 - will increase.

Topic 4.2.5. Direction of OVR flow

redox reactions

By the value of the standard electrode potential φ o one can judge the reducing properties of the system: the more negative the value of φ o, the stronger the reducing properties, and the half-reaction proceeds more easily from right to left.

For example, let's compare the systems:

Li + + e ─ = Li, φ 0 = -3.045 V; Restorative

Ba 2+ + 2e ─ = Ba, φ 0 = - 2.91B activity of metals

Mg 2+ + 2e ─ = Mg, φ 0 = -2.363 V; falls as it increases

Zn 2+ + 2e – = Zn, φ о = -0.763 V standard value

Fe 2+ + 2e ─ = Fe, φ 0 = -0.44 V; electrode potential φ O

Cd 2+ + 2e ─ = Cd, φ 0 = - 0.403 V;

Pd 2+ + 2e – = Pd, φ о = 0.987 V

Pt 2+ + 2e – = Pt, φ о = 1.188 V

Au 3+ + 3e ─ = Au, φ 0 = 1.50 V.

In the series of the above systems, the decreasing negative value of φ o corresponds to a decrease in the regenerative capacity of the systems. Lithium has the greatest reducing ability, that is, lithium is the most active of the metals presented; it loses its electrons most easily and goes into a positive oxidation state. The reduction activity of metals decreases in the series Li - Ba - Mg - Zn - Fe - Cd - Pd - Pt - Au.

Based on the magnitude of electrode potentials, N. N. Beketov arranged metals in the so-called electrochemical series of metals, in which the electrode potential of a hydrogen electrode is taken as the point of comparison

Li Na K Mn Zn Cr Fe Co Ni H Cu Ag Pd Hg Pt Au

Metal activity decreases

1) Metals in the voltage series up to hydrogen (active metals, for which φ 0 < 0), взаимодействуют с разбавленными кислотами с вытеснением водорода.

2) Each subsequent metal displaces the previous metals from its salt.

The greater the value of φ o, the stronger the oxidizing properties of the system, and the half-reaction proceeds more easily from left to right.

For example, let's compare the systems:

As can be seen from the values of standard electrode potentials, F 2 is the strongest oxidizing agent; in the series F 2 - Cl 2 - Br 2 - I 2, the oxidizing properties of simple halogen substances decrease.

By comparing the values of the standard electrode potentials of different systems, one can judge the direction of the redox reaction as a whole: a system with a more positive value of φ o is an oxidizing agent, and a system with a less positive value of the standard electrode potential is a reducing agent.

For example:

a) to obtain Br 2 by oxidation of Br ions, you can use Cl 2:

Cl 2 + 2e – = 2Cl – , φо = 1.359 V

Br 2 + 2e – = 2Br – , φо = 1.065 V

Total reaction: Cl 2 + 2Br – = Br 2 + 2Cl –

Complete reaction: Cl 2 + 2 KBr = Br 2 + 2 KCl;

b) and to obtain F 2 by oxidation of F ions, Cl 2 cannot be used:

F 2 + 2e – = 2F – , φ о = 2.870 V

Cl 2 + 2e – = 2Cl – , φо = 1.359 V

Total reaction: F 2 + 2 Cl – = Cl 2 + 2F –, that is, the reaction Cl 2 + 2 КF = cannot occur.

It is also possible to determine the direction of occurrence of more complex redox reactions.

For example, let's answer the question: is it possible to reduce MnO 4 – ions with Fe 3+ ions in an acidic environment? That is, does the reaction proceed:

MnO 4 – + H + + Fe 3+ = Mn 2+ + Fe 2+ + H 2 O ?

Basic coefficient

MnO 4 – + 8H + + 5e – = Mn 2+ + 4H 2 O, φ о 1 = 1.505 V, 1

Since φ o 1 > φ o 2, the first half-reaction proceeds in the forward direction, and the second, relative to the first, proceeds in the opposite direction. Then, by equalizing the number of electrons transferred in the oxidation and reduction reactions, we obtain the following total reaction:

In this reaction, the coefficients in front of all compounds are doubled compared to the coefficients obtained in the ionic equation, since the reaction products produced iron (III) sulfate, having the formula Fe 2 (SO 4) 3 and containing 2 moles of Fe (III) atoms.

Practice 4.2. Redox reactions

1. Compilation of redox reactions using a method based on changes in the oxidation state of elements in a compound.

EXAMPLE 1.

KMnO 4 + Na 2 SO 3 + H 2 SO 4 → MnSO 4 + ...

KMn +7 O 4 – oxidizing agent: in an acidic environment Mn +7 → Mn +2, the oxidation state decreases by 5 units; Na 2 S +4 O 3 – reducing agent: S +4 → S +6, the oxidation state increases by 2 units. To put the coefficients in the reaction equation, we find the multiple for the numbers showing the increase and decrease in oxidation states:

For 2 moles of Mn(VII) atoms, 5 moles of S(IV) atoms are required:

2 Mn +7 + 5 S +4 = 2 Mn +2 + 5 S +6 – these are the main coefficients for the oxidizing agent and the reducing agent. Let's add the reaction products, substitute the main coefficients into the reaction equation, then calculate the balance of other elements: K, Na, S and H:

To check the correctness of the selected coefficients, we calculate the balance of moles of oxygen atoms. The sum of the coefficients in the equation of the redox reaction is 21.

EXAMPLE 2.

Add and balance the redox reaction:

KMnO 4 + Na 2 SO 3 + H 2 O → MnO 2 +…

KMn +7 O 4 – oxidizing agent: in a neutral environment Mn +7 → Mn +4, the oxidation state decreases by 3 units; Na 2 S +4 O 3 – reducing agent: S +4 → S +6, the oxidation state increases by 2 units. To put the coefficients in the reaction equation, we find the multiple for the numbers showing the increase and decrease in oxidation states:

For 2 moles of Mn(VII) atoms, 3 moles of S(IV) atoms are required:

2 Mn +7 + 3 S +4 = 2 Mn +4 + 3 S +6 – these are the main coefficients for the oxidizing agent and the reducing agent. Let's add the reaction products, substitute the main coefficients into the reaction equation, then calculate the balance of other elements: K, Na and H:

To check the correctness of the selected coefficients, we calculate the balance of moles of oxygen atoms. The sum of the coefficients in the equation of the redox reaction is 13.

EXAMPLE 3

Add and balance the redox reaction:

KMnO 4 + Na 2 SO 3 + KOH → K 2 MnO 4 +…

KMn +7 O 4 – oxidizing agent: in an alkaline environment Mn +7 → Mn +6, the oxidation state decreases by 1 unit; Na 2 S +4 O 3 – reducing agent: S +4 → S +6, the oxidation state increases by 2 units. To put the coefficients in the reaction equation, we find the multiple for the numbers showing the increase and decrease in oxidation states:

For 2 moles of Mn(VII) atoms, 1 mole of S(IV) atoms is required:

2 Mn +7 + S +4 = 2 Mn +6 + S +6 – these are the main coefficients for the oxidizing agent and the reducing agent. Let's add the reaction products, substitute the main coefficients into the reaction equation, then calculate the balance of other elements: K, Na and H:

To check the correctness of the selected coefficients, we calculate the balance of moles of oxygen atoms.

The sum of the coefficients in the equation of the redox reaction is 9.

EXAMPLE 4

Add and balance the redox reaction:

K 2 Cr 2 O 7 + Na 2 SO 3 + H 2 SO 4 → Cr 2 (SO 4) 3 + ...

K 2 Cr 2 +6 O 7 – oxidizing agent: 2Cr +6 → 2Cr +3, the oxidation state decreases by 6 units; Na 2 S +4 O 3 – reducing agent: S +4 → S +6, the oxidation state increases by 2 units. To put the coefficients in the reaction equation, we find the multiple for the numbers showing the increase and decrease in oxidation states:

For 2 moles of Cr(VI) atoms, 3 moles of S(IV) atoms are required:

2 Cr +6 + 3 S +4 = 2 Cr +3 + 3 S +6 – these are the main coefficients for the oxidizing agent and the reducing agent. Let's add the reaction products, substitute the main coefficients into the reaction equation, then calculate the balance of other elements: K, Na, S and H:

To check the correctness of the selected coefficients, we calculate the balance of moles of oxygen atoms. The sum of the coefficients in the equation of the redox reaction is 17.

EXAMPLE 5

The sum of the coefficients in the equation of the redox reaction

K 2 MnO 4 + FeSO 4 + H 2 SO 4 → MnSO 4 + ...

K 2 Mn +6 O 4 – oxidizing agent: in an acidic environment Mn +6 → Mn +2, the oxidation state decreases by 4 units; Fe +2 SO 4 – reducing agent: Fe +2 → Fe +3, the oxidation state increases by 1 unit. To put the coefficients in the reaction equation, we find the multiple for the numbers showing the increase and decrease in oxidation states:

For 1 mole of Mn(VII) atoms, 4 moles of Fe(II) atoms are required:

Mn +6 + 4 Fe +2 = Mn +2 + 4 Fe +3 – these are the main coefficients for the oxidizing agent and the reducing agent. Let's add the reaction products, substitute the main coefficients into the reaction equation, then calculate the balance of other elements: K, S and H:

To check the correctness of the selected coefficients, we calculate the balance of moles of oxygen atoms. The sum of the coefficients in the equation of the redox reaction is 17.

2. Compilation of redox reactions using the electronic balance method

EXAMPLE 6

If an acidic solution of potassium tetraoxomanganate (VII) is used as an oxidizing agent:

then the reducing agent can be the system:

Fe 3+ + e – = Fe 2+, φ o = 0.771 V

Co 3+ + e – = Co 2+, φ o = 1.808 V

By the value of the standard redox potential φ o one can judge the redox properties of the system. A system with a more positive value of φ o is an oxidizing agent, and a system with a less positive value of the standard redox potential φ o is a reducing agent. Therefore, for the system MnO 4 – + 8H + + 5e – = Mn 2+ + 4H 2 O, φ o = 1.505 V, the reducing agent can be the system Fe 3+ + e – = Fe 2+, φ o = 0.771 V.

EXAMPLE 7

Rh 3+ + 3e – = Rh, φ о = 0.8 V

Bi 3+ + 3e – = Bi, φ о = 0.317 V

Ni 2+ + 2e – = Ni, φ о = -0.250 V

2H + + 2e – = H 2, φ o = 0.0 V

Which metal can dissolve in hydrochloric acid?

By the value of the standard electrode potential φ o one can judge the redox properties of the system. A system with a more positive value of φ o is an oxidizing agent, and a system with a less positive value of the standard electrode potential is a reducing agent. In hydrochloric acid (HCl), H + cations are an oxidizing agent, accept electrons and are reduced to H 2 , for this reaction φ o = 0 V. Therefore, only that metal is dissolved in HCl that can be a reducing agent under these conditions, that is, for which φ O< 0, а именно никель:

Ni + 2 HCl = NiCl 2 + H 2

EXAMPLE 8

Based on the values of standard electrode potentials of half-reactions:

Zn 2+ + 2e – = Zn, φ о = -0.763 V

Cd 2+ + 2e – = Cd, φ о = -0.403 V

Which metal is the most active?

The more active the metal, the greater its reducing properties. The reducing properties of the system can be judged by the value of the standard redox potential φ o: the more negative the value of φ o, the stronger the reducing properties of the system, and the half-reaction proceeds more easily from right to left. Consequently, zinc has the greatest reducing ability, that is, zinc is the most active of the metals presented.

EXAMPLE 9

If an acidic solution of iron(III) chloride is used as an oxidizing agent:

then which system can be a reducing agent:

I 2 + 2e – = 2I – , φ о = 0.536 V

Br 2 + 2e – = 2Br – , φо = 1.065 V

Pb 4+ + 2e – = Pb 2+, φ o = 1.694 V?

By the value of the standard redox potential φ o one can judge the redox properties of the system. A system with a more positive value of φ o is an oxidizing agent, and a system with a less positive value of the standard redox potential is a reducing agent. Therefore, for the system Fe 3+ + e – = Fe 2+, φ o = 0.771 V, the reducing agent can be the system I 2 + 2e – = 2I –, φ o = 0.536 V.

Basic coefficient

Fe 3+ + e – = Fe 2+, φ o 1 = 0.771 V 2

I 2 + 2e – = 2I – , φ o 2 = 0.536 V 1

Since φ o 1 >

2 Fe 3+ + 2I – = 2 Fe 2+ + I 2

By adding ions of the opposite sign, we get the complete equation:

2 FeCl 3 + 2 KI = 2 FeCl 2 + 2 KCl + I 2

EXAMPLE 10

Is it possible to reduce MnO 4 – ions with Fe 3+ ions in an acidic environment?

Let's write the question in the form of a reaction equation:

MnO 4 – + H + + Fe 3+ = Mn 2+ + Fe 2+ + H 2 O.

Let us select suitable half-reactions from the reference table and present their standard electrode potentials:

Basic coefficient

MnO 4 – + 8H + + 5e – = Mn 2+ + 4H 2 O, φ o 1 = 1.505 V, 1

Fe 3+ + e – = Fe 2+, φ o 2 = 0.771 V 5

MnO 4 – + 8H + + 5 Fe 3+ = Mn 2+ + 5Fe 2+ + 4H 2 O

That is, it is possible to reduce MnO 4 – ions with Fe 3+ ions in an acidic environment. The complete reaction looks like:

In this reaction, the coefficients for all compounds are doubled compared to the coefficients obtained in the ionic equation, since the reaction products produced iron (III) sulfate, having the formula Fe 2 (SO 4) 3.

TASKS FOR INDEPENDENT SOLUTION

1. Determine the oxidation states of elements in compounds:

H 3 P.O. 4 , K 3 P.O. 4 , N 2 O 5 , N.H. 3 , Cl 2 , KCl, KClO 3 , Ca(ClO 4 ) 2 , N.H. 4 Cl, HNO 2 , Li, Li 3 N, Mg 3 N 2 , NF 3 , N 2 , N.H. 4 NO 3 , H 2 O, H 2 O 2 , KOH, KH, K 2O 2 , BaO, BaO 2 , OF 2 , F 2 , NF 3 , Na 2 S, FeS, FeS 2 , NaHS, Na 2 SO 4 , NaHSO 4 , SO 2 , SOCl 2 , SO 2 Cl 2 , MnO 2 , Mn(OH) 2 , KMnO 4 , K 2 MnO 4 , Cr, Cr(OH) 2 , Cr(OH) 3 , K 2 CrO 4 , K 2 Cr 2 O 7 , (N.H. 4 ) 2 Cr 2 O 7 , K 3 [ Al(OH) 6 ], Na 2 [ Zn(OH) 4 ], K 2 [ ZnCl 4 ], H 2 SO 3 , FeSO 3 , Fe 2 (SO 3 ) 3 , H 3 P.O. 4 , Cu 3 P.O. 4 , Cu 3 (P.O. 4 ) 2 , Na 2 SiO 3 , MnSiO 3 , PbSO 4 , Al 2 (SO 4 ) 3 , Fe 2 (SO 4 ) 3 , N.H. 4 Cl, (N.H. 4 ) 2 SO 4 , Cr 2 (SO 4 ) 3 , CrSO 4 , NiSO 4 , [ Zn(OH 2 ) 6 ] SO 4 , Fe(NO 3 ) 2 , Fe(NO 3 ) 3 , PbCO 3 , Bi 2 (CO 3 ) 3 , Ag 2 S, Hg 2 S, HgS, Fe 2 S 3 , FeS, SnSO 4 .

2. Indicate the oxidizing agent and the reducing agent, draw up diagrams of changes in oxidation states, add and place the coefficients in the reaction equation:

A. MnO 2 + HCl(conc) →

b. KMnO 4 +H 2 S + H 2 SO 4 →

V. FeCl 3 + SnCl 2 →

g. KMnO 4 + H 2 O 2 + H 2 SO 4 → O 2

d. Br 2 + KOH →

e. Zn + HNO 3 → NH 4 NO 3 +…

and. Cu + HNO 3 → NO 2 + …

h. K 2 MnO 4 + FeSO 4 + H 2 SO 4 →

And. K 2 Cr 2 O 7 + (NH 4) 2 S + H 2 O → Cr(OH) 3 + …+ NH 3 +…

j. H 2 S + Cl 2 →

l. K 2 Cr 2 O 7 +HCl → CrCl 3 + ...

m. FeCl 3 + H 2 S →

n. KMnO 4 + NaNO 2 + H 2 SO 4 →

O. Cl 2 + KOH →

a) Based on standard values of electrode potentials, arrange the metals in order of increasing reducing properties:

Ba 2+ + 2e ─ = Ba, φ 0 = -2.91 B;

Au 3+ + 3e ─ = Au, φ 0 = 1.50 V;

Fe 2+ + 2e ─ = Fe, φ 0 = -0.44 B.

What happens when an iron plate is immersed in a solution of AuCl 3

b) Based on standard values of electrode potentials of half-reactions

MnO 4 – + 8H + + 5e – = Mn 2+ + 4H 2 O, φ о = 1.505 V,

Pb 4+ + 2e – = Pb 2+, φ o = 1.694 V

give a reasonable answer to the question - is it possible to oxidize Mn 2+ ions using Pb 4+ ions? Give the total reaction, indicate the oxidizing agent and the reducing agent.

c) Based on standard values of electrode potentials of half-reactions, give a reasonable answer to the question - is it possible to oxidize Fe 2+ ions using Pb 4+ ions? Give the total reaction, indicate the oxidizing agent and the reducing agent.

d) Based on standard values of electrode potentials, arrange the metals in order of increasing reducing properties:

Mg 2+ + 2e ─ = Mg

Cd 2+ + 2e ─ = Cd

Сu 2+ + 2e ─ = Cu

What happens when a copper plate is immersed in a solution of cadmium chloride?

e) Based on standard values of electrode potentials of half-reactions

Ir 3+ + 3e – = Ir,

NO 3 - + 4H + + 3e – = NO + 2H 2 O,

Give a reasoned answer to the question: is iridium soluble in nitric acid? Give the total reaction, indicate the oxidizing agent and reducing agent

f) Based on standard values of electrode potentials, arrange the halogens in order of increasing their oxidizing properties:

Cl 2 + 2e ─ = 2Cl ─ φ 0 = 1.359 V;

Br 2 + 2e ─ = 2Br ─ φ 0 = 1.065 V;

I 2 + 2e ─ = 2I ─ φ 0 = 0.536 V;

F 2 + 2e ─ = 2F ─ φ 0 = 2.87 V.

Prove whether it is possible to use the oxidation reaction of Br ions ─ chlorine Cl 2 to produce bromine?

g) Based on standard values of electrode potentials of half-reactions

Fe 3+ + e – = Fe 2+, φ o = 0.771 V,

Br 2 + 2e – = 2Br – , φо = 1.065 V

give a reasonable answer to the question - is it possible to oxidize Fe 2+ ions using Br 2? Give the total reaction, indicate the oxidizing agent and the reducing agent.

h) Based on standard values of electrode potentials, arrange the metals in order of increasing reducing properties:

Zn 2+ + 2e – = Zn, φ о = - 0.763 V

Hg 2+ + 2e – = Hg, φо = 0.850 V

Cd 2+ + 2e – = Cd, φ o = - 0.403 V.

What happens when a cadmium plate is immersed in a solution of zinc chloride?

The occurrence of chemical reactions is generally determined by the exchange of particles between reacting substances. Often the exchange is accompanied by the transfer of electrons from one particle to another. Thus, when zinc replaces copper in a solution of copper (II) sulfate:

Zn(s) + CuSO 4 (p) = ZnSO 4 (p) + Cu(s)

electrons from zinc atoms go to copper ions:

Zn 0 = Zn 2+ + 2 e,

Cu 2+ + 2 e= Cu 0 ,

or in total: Zn 0 + Cu 2+ = Zn 2+ + Cu 0.

The process of a particle losing electrons is called oxidation , and the process of acquiring electrons is restoration . Oxidation and reduction occur simultaneously, therefore interactions accompanied by the transfer of electrons from one particle to another are called redox reactions (ORR).

For the convenience of describing OVR, the concept is used oxidation states - a value numerically equal to the formal charge that an element acquires, based on the assumption that all the electrons from each of its bonds have transferred to a more electronegative atom of the given compound. The occurrence of redox reaction is accompanied by a change in the oxidation states of the elements involved in the reaction of substances . When reduced, the oxidation state of an element decreases; when oxidized, it increases. . A substance that contains an element that reduces its oxidation state is called oxidizing agent ; a substance that contains an element that increases the oxidation state is called reducing agent .

The oxidation state of an element in a compound is determined in accordance with the following rules:

1) the oxidation state of an element in a simple substance is zero;

2) the algebraic sum of all oxidation states of atoms in a molecule is equal to zero;

3) the algebraic sum of all oxidation states of atoms in a complex ion, as well as the oxidation state of an element in a simple monatomic ion, is equal to the charge of the ion;

4) a negative oxidation state is exhibited in the compound by the atoms of the element having the highest electronegativity;

5) the maximum possible (positive) oxidation state of an element corresponds to the number of the group in which the element is located in the Periodic Table D.I. Mendeleev.

A number of elements in compounds exhibit a constant state of oxidation:

1) fluorine, which has the highest electronegativity among elements, has an oxidation state of –1 in all compounds;

2) hydrogen in compounds exhibits an oxidation state of +1, except for metal hydrides (–1);

3) metals of subgroup IA in all compounds have an oxidation state of +1;

4) metals of subgroup IIA, as well as zinc and cadmium in all compounds have an oxidation state of +2;

5) oxidation state of aluminum in compounds +3;

6) the oxidation state of oxygen in compounds is –2, with the exception of compounds in which oxygen is present in the form of molecular ions: O 2 +, O 2 -, O 2 2 -, O 3 -, as well as fluorides O x F 2.

The oxidation states of atoms of elements in a compound are written above the symbol of a given element, first indicating the sign of the oxidation state, and then its numerical value, for example, K +1 Mn +7 O 4 -2, in contrast to the charge of the ion, which is written on the right, indicating first the charge number and then sign: Fe 2+, SO 4 2–.

The redox properties of atoms of various elements manifest themselves depending on many factors, the most important of which are the electronic structure of the element, its oxidation state in the substance, and the nature of the properties of other participants in the reaction.

Compounds that contain atoms of elements in their maximum (positive) oxidation state, for example, K +1 Mn +7 O 4 -2, K 2 +1 Cr +6 2 O 7 -2, H + N +5 O 3 -2, Pb +4 O 2 -2, can only be reduced, acting as oxidizing agents.

Compounds containing elements in their minimum oxidation state, for example, N -3 H 3, H 2 S -2, HI -1, can only be oxidized and act as reducing agents.

Substances containing elements in intermediate oxidation states, for example H + N +3 O 2, H 2 O 2 -1, S 0, I 2 0, Cr +3 Cl 3, Mn +4 O 2 -2, have redox duality. Depending on the reaction partner, such substances can both accept and donate electrons. The composition of reduction and oxidation products also depends on many factors, including the environment in which the chemical reaction occurs, the concentration of reagents, and the activity of the partner in the redox process. To create an equation for a redox reaction, you need to know how the oxidation states of elements change and what other compounds the oxidizing agent and reducing agent transform into.

Classification of redox reactions. There are four types of redox reactions.

1. Intermolecular– reactions in which the oxidizing agent and the reducing agent are different substances: Zn 0 + Cu + 2 SO 4 = Zn + 2 SO 4 + Cu 0 .

2. During the thermal decomposition of complex compounds, which include an oxidizing agent and a reducing agent in the form of atoms of different elements, redox reactions occur, called intramolecular: (N -3 H 4) 2 Cr +6 2 O 7 = N 2 0 + Cr +3 2 O 3 + 4H 2 O.

3. Reactions disproportionation can occur if compounds containing elements in intermediate oxidation states are exposed to conditions where they are unstable (for example, at elevated temperatures). The oxidation state of this element both increases and decreases: 2H 2 O 2 -1 = O 0 2 + 2 H 2 O -2.

4. Reactions counter-proportionation- these are processes of interaction between an oxidizing agent and a reducing agent, which include the same element in different oxidation states. As a result, the oxidation product and reduction product is a substance with an intermediate oxidation state of the atoms of a given element:

Na 2 S +4 O 3 + 2Na 2 S -2 + 6HCl = 3S 0 + 6NaCl + 3H 2 O.

There are also mixed reactions. For example, the intramolecular counterproportionation reaction includes the decomposition reaction of ammonium nitrate: N -3 H 4 N +5 O 3 = N +1 2 O + 2H 2 O.

Drawing up equations of redox reactions. To compile equations for redox reactions, the electron balance method and the electron-ion half-reaction method are most often used.

Electronic balance method usually used to draw up equations for redox reactions occurring between gases, solids and in melts. The sequence of operations is as follows:

1. Write down the formulas of the reagents and reaction products in molecular form: FeCl 3 + H 2 S → FeCl 2 + S + HCl;

2. Determine the oxidation state of atoms that change it during the reaction: Fe 3+ Cl 3 + H 2 S -2 → Fe 2+ Cl 2 + S 0 + HCl;

3. Based on the change in oxidation states, the number of electrons given up by the reducing agent and the number of electrons accepted by the oxidizing agent are determined; make up an electronic balance taking into account the principle of equality of the number of given and received electrons:

Fe +3 +1 e= Fe +2 ½ ∙2

S -2 – 2 e= S 0 ½ ∙1

4. Electronic balance factors are written into the redox reaction equation as basic stoichiometric coefficients: 2FeCl 3 + H 2 S → 2FeCl 2 + S + HCl.

5. Select the stoichiometric coefficients of the remaining participants in the reaction: 2FeCl 3 + H 2 S = 2FeCl 2 + S + 2HCl.

Electron-ion half-reaction method used in drawing up equations for reactions occurring in an aqueous solution, as well as reactions involving substances in which it is difficult to determine the oxidation states of elements. According to this method, the following main stages of composing the reaction equation are distinguished:

1. Write down the general molecular diagram of the process, indicating the reducing agent, oxidizing agent and the medium in which the reaction occurs (acidic, neutral or alkaline). For example:

SO 2 + K 2 Cr 2 O 7 + H 2 SO 4 (diluted) → ...

2. Taking into account the dissociation of electrolytes in an aqueous solution, this scheme is presented in the form of molecular-ionic interaction. Ions whose oxidation states of atoms do not change are not indicated in the diagram, with the exception of H + and OH - ions:

SO 2 + Cr 2 O 7 2– + H + → ...

3. Determine the oxidation states of the reducing agent and oxidizing agent, as well as the products of their interaction:

4. Write down the material balance of the oxidation and reduction half-reactions:

5. Sum up the half-reactions, taking into account the principle of equality of electrons given and received:

SO 2 + 2H 2 O – 2 e= SO 4 2– + 4H + ½ ∙3

Cr 2 O 7 2– + 14H + + 6 e= 2Cr 3+ + 7H 2 O ½ ∙1

3SO 2 + 6H 2 O + Cr 2 O 7 2– + 14H + = 3SO 4 2– + 12H + + 2Cr 3+ + 7H 2 O

by abbreviating the particles of the same name, we obtain the general ionic-molecular equation:

3SO 2 + Cr 2 O 7 2– + 2H + = 3SO 4 2– + 2Cr 3+ + H 2 O.

6. Add ions that did not participate in the oxidation-reduction process, equalize their amounts on the left and right, and write down the molecular equation of the reaction:

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

When compiling a material balance for half-reactions of oxidation and reduction, when the number of oxygen atoms included in the particles of the oxidizing agent and reducing agent changes, it should be taken into account that in aqueous solutions the binding or addition of oxygen occurs with the participation of water molecules and ions of the medium.

During the oxidation process, for one oxygen atom that attaches to a reducing agent particle, in acidic and neutral environments, one molecule of water is consumed and two H + ions are formed; in an alkaline environment, two hydroxide ions OH - are consumed and one molecule of water is formed.

During the reduction process, to bind one oxygen atom of an oxidizing agent particle in an acidic environment, two H + ions are consumed and one water molecule is formed; in neutral and alkaline environments, one H 2 O molecule is consumed and two OH - ions are formed (Table 2).

table 2

Balance of oxygen atoms

in redox reactions

When drawing up equations, it should be taken into account that the oxidizing agent (or reducing agent) can be consumed not only in the main redox reaction, but also when binding the resulting reaction products, i.e. act as a medium and salt former. An example when the role of the medium is played by an oxidizing agent is the oxidation reaction of a metal in nitric acid:

3Cu + 2HNO3(oxidizer) + 6HNO3(medium) = 3Cu(NO3)2 + 2NO + 4H2O

or 3Cu + 8HNO 3(dil) = 3Cu(NO 3) 2 + 2NO + 4H 2 O.

An example when the reducing agent is the medium in which the reaction occurs is the oxidation of hydrochloric acid with potassium dichromate: 6HCl (reducing agent) + K 2 Cr 2 O 7 + 8HCl (medium) = 2CrCl 3 + 3Cl 2 + 2KCl + 7H 2 O

or 14HCl + K 2 Cr 2 O 7 = 2CrCl 3 + 3Cl 2 + 2KCl + 7H 2 O.

When calculating the quantitative, mass and volume ratios of participants in redox reactions, the basic stoichiometric laws of chemistry are used, and, in particular, the law of equivalents, taking into account that equivalence number The equivalence number of an oxidizing agent is equal to the number of electrons that one formula unit of the oxidizing agent accepts, and the equivalence number of a reducing agent is equal to the number of electrons that one formula unit of the reducing agent gives up.

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