• Domestic science and medicine in the 19th - early 20th centuries Chemistry of the 19th century in medicine
• What are peptides? Types and types of peptides. All about peptides and anti-aging cosmetics with peptides Additional peptides are not formed
• Toxic effect on humans of hazardous chemicals
• Radioautography Method of autoradiography in cytology
• Identification of bacteria by antigenic structure
• Organic matter in wastewater What are organic compounds in water
• Hydrogen index (pH factor)
• What is the pH of water and why is it important to know it?
• Redox potential Redox systems
• Reversible redox system Reversible redox systems

Purpose: Study of redox reactions

Literature

Redox reactions are called chemical reactions, accompanied by a change in the oxidation state of the atoms of the elements. The oxidation state is the conditional charge of an atom in a molecule. It is calculated from the assumption that all bonds between atoms are ionic. Oxidation is the process of donating electrons, and reduction is the process of accepting electrons. Oxidation and reduction are interrelated. An oxidizing agent is a substance whose atoms accept electrons, while it is reduced. A reducing agent is a substance whose atoms donate electrons, while it is oxidized.

All redox reactions are classified as follows:

1. Intermolecular reactions. These are reactions in which the oxidizing agent and the reducing agent are different substances.

where Mn+4 is an oxidizing agent, Cl-1 is a reducing agent.

2. Reactions of intramolecular oxidation. These are reactions that occur with a change in the oxidation states of atoms of various elements of the same substance.

where Mn+7 is an oxidizing agent and O-2 is a reducing agent.

3. Reactions of disproportionation. In these reactions, both the oxidizing agent and the reducing agent are an element that is in an intermediate oxidation state in the composition of the same substance.

where Cl20 is an oxidizing agent and a reducing agent.

The possibility of a particular substance to exhibit oxidizing, reducing or dual properties can be judged by the degree of oxidation of the elements that perform these functions.

Elements in their highest oxidation state exhibit only oxidizing properties, while in their lowest oxidation state they exhibit only reducing properties. Elements with an intermediate oxidation state can exhibit both oxidizing and reducing properties. The main oxidizing and reducing agents are listed below.

Oxidizers	Reaction schemes
Simple substances: Halogens G2	(NG and their salts)
Oxygen O2	(H2O, oxides and their derivatives)
Nonmetals (S,P,N2)	(ENn and their salts)
Complex substances: HCl, H2SO4 (diluted), etc. (except HNO3).
	(SO2, H2SO3 and its salts) (H2S and its salts)
	(N2O3, HNO2 and its salts) (NH3, NH4+ and corresponding salts)
Oxidizers	Reaction schemes
Metal cations of the highest oxidation state (Fe+3, Sn+4, etc.)
Restorers	Reaction schemes
Simple substances: metals (Me)
Nonmetals (S,C,P,N2,…)
H2S and its salts, Na2S2O3
NG and their salts
Metal cations in lower oxidation states (Fe+2, Sn+2, Sb+3, Cr+3 etc.)
HNO2 and its salts.	(HNO3 or its salts).
H2SO3 and its salts.	(H2SO4 or its salts).

test questions

1. List all types of chemical reactions?

2. What reactions are called redox reactions?

3. What is the difference between redox. Reactions from other kinds of reactions?

4. What are the types of redox reactions?

5. What oxidizing and reducing agents do you know?

Lecture #12. ELECTROLYSIS OF SOLUTIONS. FARADAY'S LAWS. CORROSION OF METALS

Purpose: To give students knowledge about the electrolysis of solutions, Faraday's law and the types of corrosion and methods of its protection.

Literature

1. Akhmetova N.S. General and inorganic chemistry. Ed. "Chemistry", M. 1981

2. Glinka N.L. General chemistry. Ed. "Chemistry", Leningrad, 1987

3. Nekrasov V.B. Fundamentals of General Chemistry. Ed. "Chemistry", M. 1971

4. Karapetyants M.Kh., Drakin S.I. General and inorganic chemistry. Ed. "Chemistry", M.1983

5. Korzhukov N.G. Inorganic chemistry. Moscow "MISIS", 2001

6. Saveliev G.G., Smolova L.M. General Chemistry Ed. TPU. Tomsk 2003

7. Kurnakova N.S. Modern problems of general and inorganic chemistry. M. "Chemistry" M., 2004

Electrolysis is a redox process that takes place under the influence of electric current in electrolyte solutions or melts.

Here, the redox process is forced, due to the conversion of electrical energy into chemical energy.

When an electric current passes through an electrolyte solution or melt, the positive ions of the solution tend to the negative pole, and the negative ions tend to the positive pole. In the electrodes, the ions are discharged turning into neutral atoms.

As the electron current originates through the electrolyte solution or melt, the electrons pass from the anode to the cathode. The appearance of an excess of electrons at the cathode and a lack of them at the anode causes an ordered movement of ions in a solution or melt. Excess electrons of the cathode are transferred to a positively charged electrolyte solution, turning them into neutral atoms; negatively charged electrolyte ions at the anode, giving their electrons to the anode, are discharged. Thus, the reduction process occurs at the cathode and the oxidation process occurs at the anode.

The electrons from the anode go to the external circuit. Depending on the nature of the anode, the source of these electrons is the anode itself or anions from a solution or melt, in which case the anode is insoluble. Graphite, Comma, Au can be taken as an insoluble anode.

Electrolysis of aqueous solutions of electrolytes with insoluble electrodes.

During the electrolysis of aqueous solutions of electrolytes, not only the ions of the electrolyte are exposed to the action of the current, but the H and OH ions of the water formed during dissociation.

Therefore, two ions can be discharged at the cathode, the positive ion of the electrolyte and the H ion. Which of the ions is discharged is determined by the position of the metal in the series of voltages, as well as by the concentration of ions in the solution.

1. At the cathode, metal ions standing in the voltage series up to Al inclusive in an aqueous solution cannot be discharged; instead, hydrogen ions are discharged from water, i.e. electrons from the cathode are taken by the water itself, this is due to the fact that the difference between the electrodes is potentially very large.

Lithium, Barium, K, Na, Ca, Mg, Al, MS, Zinc, Cherry Red, Fe, Cd, Co, Nickel, Sn, H2, Cu, Ag, Hg, Comma, Au.

2. During electrolysis, a solution of metal salts consisting of Al to H2 in the voltage series at the cathode discharges the ions of these metals and partially discharges the H ions of water. From where it can be seen that metal ions more active than hydrogen are being reduced. This is due to the fact that in aqueous solutions, the electrolyte cations and the H ion of water are under the same conditions in relation to their concentration.

3. During the electrolysis of solutions of salts of metals that are in the voltage series after H2, only the ions of these metals are discharged at the cathode.

At the anode, ions of residues without oxygen acids are first of all discharged, since they easily lose their charge than the OH ion of water, and the ions of oxygen acid residues are not able to discharge at the anode, and instead of them, OH ions of water are oxidized.

Electrolysis of aqueous solutions of salts with soluble electrodes.

In this case of electrolysis, the regularities that are different in relation to the cathodic process with an insoluble anode remain valid.

Features of the anode process is that the source

the electron is the electrode from which the anode is made, i.e. the anode dissolves and goes into solution in the form of the Me+n ion.

For example: let's analyze the electrolysis of an aqueous solution of CuSO4 with a copper anode.

CuSO4 = Сu ++ + SO4-2

In this case, Cu is transferred from the anode to the cathode.

K / Cu ++ + OH- \u003d Copper (O) 2 secondary process

Soluble anode electrolysis is widely used to coat certain metals with others.

For example: when nickel-plating an object, the Nickel electrode serves as the anode, and the coated object serves as the cathode, a nickel salt solution is taken as an electrolyte.

NiSO4 with Nickel anode and Fe (coating material) cathode.

H2O+NiSO4 = Nickel ++ + SO4--

Coating one metal with another using electrolysis called electroplating. The same method is used to obtain pure copper from blister copper.

CuO + C = copper + CO

The anode is made from blister copper. First of all, Zinc, Sn passes into the solution from the anode.

Electrolysis of melts with an insoluble electrode.

Metals standing in the voltage range up to Al inclusive are obtained by electrolysis of melts of their salts, because. the strongest reducing agent is electric current.

For example: electrolysis of NaC1 melt.

NaС1 Na + + Сl-

Receiving Na.

Na can be obtained from NaCl and NaOH. Tm NaCl = 805o C, Tm NaOH = 400o C

According to Tm, it is advantageous to use NaOH, but it is an expensive raw material than NaCl.

NaOH \u003d Na + + OH-

		2OH--2e \u003d 2H2O + O: O + O \u003d O2

Laws of electrolysis

The quantitative aspects of electrolysis were first studied by the English physicist M. Faraday, who established the following laws.

1. Faraday's law.

The weight of the substance released during electrolysis is proportional to the amount of electricity flowing through the solution and is completely independent of other factors.

2. Faraday's Law

When passing equal amounts of electricity from various chemical compounds, equivalent amounts of substances are released on the electrodes.

To isolate one gram equivalent of any substance, it is necessary to spend 96,500 pendants of electricity.

Faraday's law can also be expressed by the following equation:

m is the mass of the released substance, E is the equivalent of the substance, F is the Faraday number, Q is the amount of electricity.

Q = JJ-current, A.

Electrolysis duration, sec.

The following experiment is a clear illustration of Faraday's II law. Electric current flowing through solutions of HCl, AgNO3, CuSO4, FePO4, SnC14. The solutions are preliminarily placed in devices in which, at the end of the experiment, it is possible to determine the amount of released substances.

After a while, when the electrodes have a sufficient amount of electrolysis products, the current flow is stopped and measurements are taken. It turns out that during the time during which 1 g of H2 is released from the solution of HC1, those 1 g of the latter, the indicated amounts of metals are released from the remaining solutions. Comparisons of the amount of substances released at the cathode with atomic weights show that substances are released in an amount equal to their equivalents. Measurements of the amount of substances released at the anode lead to the same result. In 1 and 5, 35.5 g of osmor are released, in 2, 3, 4, 8 g of oxygen are released.

For example: how much copper will be released if through an aqueous solution

CuSO4 to pass a current of 2A for 2 hours.

2 hour = 7200sec

E= (Ав) / В: CuSO4 Cu+2 + SO4--

m = (31.8 * 2 * 7200) / 96500 = 4.74 g.

Polarization during electrolysis.

Oxidation and reduction processes occurring under the action of an electric current can cause significant changes in the electrodes. If electrolysis of water is carried out, a CuCl solution with an undissolved electrode.

Сu Сl2 = Сu ++ + 2 Closed

Chlorine is adsorbed on the electrode surface Comma and a closed layer is formed. Thus, the CuCl2 solution will not directly come into contact with the plate, but Ca and Closed.

If we now remove the current source and connect the ends of the electrodes through an external circuit through a galvanometer, then the galvanometer will show the presence of an electric current in the circuit - an electrochemical polarization current, its direction will be the opposite of that given by the current source. The EMF of the resulting galvanic cell is equal to the potential difference of the electrodes.

Copper / CuCl2 / C12 (comma)

c12 / closed = + 1.36 Copper ++ / copper = 0.34

Based on normal electrode potentials,

then EMF = c12 / closed-Copper ++ / copper = 1.02

and it is the polarization current that prevents electrolysis. In order for the electrolysis to continue with the desired intensity, the voltage of the current source must be applied to the electrodes somewhat higher than the EMF of the polarization current.

The smallest potential difference required for continuous electrolysis is called the decomposition potential.

The electrolyte decomposition potential is always greater than the polarization EMF.

The difference between the decomposition potential and the polarization EMF is called overvoltage.

Overvoltage depends on the following factors:

1. from the material from which the electrodes are made;

2. on the state of the surface of the electrodes;

3. from the state of aggregation of substances released on the electrodes;

4. on the current density and on the temperature of the solution.

Batteries

The introduction of electrode polarization is used in practice in devices that serve to store chemical energy, which is easily converted into electrical energy at the right time. Such devices are called batteries.

Batteries differ in the chemical nature of the electrodes and electrolyte, as well as in their design. Practical applications are mainly acid and alkaline batteries.

Acid (lead) batteries.

The lead battery consists of lattice lead plates filled with PbO lead oxide paste and immersed in a 25 - 30% H2SO4 solution. As a result of the interaction of PbO with an H2SO4 solution, a layer of hardly soluble PbSO4 is formed on the surface of the Pb plate.

PbO + H2SO4 \u003d PbSO4 + H2O

To charge the battery, i.e. to accumulate chemical energy in it, it is necessary to connect one of its lead plates to the negative, and the other to the positive pole of the current source. The reactions occurring in this case can be expressed by the negative pole of the cathode.

To PbSO4 + 2е = Pb + SO4--

+A PbSO4 - 2e + 2 H2O \u003d PbO2 + SO4 - + 4H +

As can be seen from the equation at the negative pole, the ions, attaching two electrons each, turn into a metal one. At the positive pole, the oxidative process leads to the conversion of PbO2.

If we add these reactions, then the general expression of the process will take

2 PbSO4 + H2O \u003d PbO2 + SO4 - + 4H +

When the battery is charged, water reacts and acid is formed.

The batteries are charged until the electrolysis of water begins with the vigorous release of hydrogen at the cathode and oxygen at the anode.

So, when the battery is charged, the electrodes become chemically different and a potential difference appears between them.

The electrical circuit characterizing the resulting galvanic cell has the form.

Pb / H2SO4 / PbO2 (Pb) +

If you connect a plate of a charged battery with a conductor, then electrons will move from the plate covered with lead to the plate covered with PbO2, i.e. an electric current appears, the battery works as a galvanic cell. The following reactions take place on its electrodes.

Pb - 2e + SO4-2 = Pb S04

PbO2 + 2е = 4Н+ = SO4-2- = PbSO4+ 2Н2О

During discharge, H2SO4 is consumed and the concentration of H2SO4 in the solution decreases. A decrease in acid concentration serves as an indicator of the degree of discharge of the battery.

The EMF of a lead battery is slightly more than 2 V.

Alkaline batteries.

Of the alkaline batteries, Fe - Ni, Cd - Ni, Ag - Zn batteries have found the greatest practical application. In a charged Fe - Ni battery, the active mass of the negative electrode is powdered iron, compressed with a small amount of mercury oxide, the active mass of the positive electrode is Ni (OH) 3 with a small admixture of graphite. the electrolyte is 23% KOH.

During discharging, the following processes occur

A (-) Fe - 2e \u003d Fe

K(+) Ni (OH) 3 + e \u003d Ni (OH) 2

reactions occurring during charging have the opposite phenomenon and the general equation of charge and discharge has the general form

Fe + 2 Ni(OH)3 Fe(OH)2 + 2 Ni(OH)2

The emf of such a battery is about 1.2 V.

Silver-zinc battery

The diagram of this battery is as follows

(+) Ag2O/KOH/ Zn(-)

Ag - Zn batteries are significantly superior to the acid and alkaline batteries discussed above in terms of specific energy and power density.

These batteries are characterized by a very small self-discharge and the ability to use them in a wide temperature range - from 30 to 70 degrees Celsius.

Use them in a wide temperature range from - 30 to + 70 s.

In it, the negative electrode is a pressed mixture of ZnO with Zn powder, and the positive electrode is a frame of Ag wire pressed with Ag2O. The electrolyte solution is 39% KOH 1ml ZnO solution.

Ag + ZnO + Zn (OH) 2 2 Zn + H2O + 2 Ag2O

When charging

electrode (+) 2Ag + 2 OH - 2e = Ag2O + H20

(-) ZnO + 2e = Zn

zinc oxide turns into zinc sponge.

ZnO + KOH + H2O = K

K + 2e \u003d Zn + KOH + 2OH

CORROSION OF METALS.

Most metals, coming into contact with the environment, are subjected to destruction from the surface. The reason for this is the chemical interaction of metals with gases in the air, with water and substances dissolved in it. In this case, as a result of oxidative processes, substances are formed that have properties that differ sharply from those of the original metal.

Any process of chemical destruction of metals under the influence of the environment is called corrosion.

There are several forms of manifestation of corrosion. The most common are uniform, local and intergranular.

Of these, intergranular corrosion is the most dangerous, it spreads between crystallites, and can imperceptibly lead to structural damage to a great depth.

According to the mechanism of chemical processes, two types of corrosion are distinguished: chemical and electrochemical.

1. Chemical corrosion is the destruction of metal without the occurrence of electric current in the system (with direct contact of the metal with an oxidizing agent).

Chemical corrosion is divided into:

a) gas corrosion is caused by dry gases. H:

O2, SO2, C12, F2, Br2, CO2, etc.

It is observed mainly in the high-temperature processing of metals, in internal combustion engines, etc.

b) liquid chemical corrosion - proceeds under the action of organic liquids without the participation of water: oil derivative, gasoline, cresol, benzene, toluene, etc.

c) electrochemical corrosion is the destruction of the metal in the electrolyte environment with the appearance of an electric current inside the system.

Electrochemical corrosion is divided into:

1. Atmospheric.

2. Soil.

3. Corrosion due to stray currents.

Just like a galvanic cell, galvanic corrosion requires two different electrodes and an electrolyte solution. From this it can be argued that pure metals theoretically should not be subjected to electrochemical corrosion at all. If, for example, we consider the corrosion of iron with the inclusion of copper in humid air (Fe + Cu), this forms a galvanic cell

chemistry reaction catalysis solution

A - Fe / H2O / Cu + K

Fe is the anode, Cu is the cathode, and corrodes as a result.

These Fe2+ electrons on

Cu surfaces (cathode) reduce air oxygen

O2 + 2H2O + 4e \u003d 4OH

Fe2+ ​​+ OH- = Fe(OH)2,

Iron in moist air quickly turns into 3-ion iron.

4Fe(OH)2 + O2 + 2H2O= 4Fe(OH)3

From this example, it can be seen that during the formation of a galvanic cell, the more active metal is corroded.

The area of ​​the surface from which the ions pass into solution, i.e. where the metal corrodes, is called the anode, the area where electrolyte cations are discharged is called the cathode.

The nature of cathodic processes during corrosion is determined by the substances present in the solution. In a strongly acidic environment, ionic hydrogens are reduced:

2 H+ + 2 C = H2.

In atmospheric corrosion, the pH of the medium is close to neutral, and therefore oxygen dissolved in water is reduced at the cathode.

O2 + 2H2O + 4e \u003d 4OH

Immerse the plastic of pure zinc in a dilute acid solution, then the evolution of hydrogen is, indeed, almost not observed. The absence of a reaction can be explained by the fact that zinc ions, which begin to go into solution, create a layer of positively charged hydrogenated ions near the surface of the plate.

This layer is a barrier that prevents hydrogen ions from coming close to the zinc plate and receiving electrons from it, and the dissolution of zinc stops. If you touch the zinc surface with some less active metal (Cu) as a result of the formation of a galvanic cell

A-Zn / K-TA / Cu+K

vigorous evolution of hydrogen begins on the surface of a less active metal

These electrons passing to Cu eliminate y. Cu surface protective barrier of its ions, and hydrogen ions are freely restored

Metals in the stress row to the left are easily corroded. Pure metals, also Au, Ag, Pt, do not corrode. And the following metals: Mg, Al, Cu, Cr, Ni, during corrosion form a dense protective oxide film, which prevents further corrosion.

Soil corrosion - This type of corrosion is a complex form of corrosion of metals in soil. The chemical and physical properties of soils play a role here. Corrosion in this case depends on the following factors

1. Humidity and soil environment.

2. From the electrical and air permeability of the soil.

3. From the electrode potential of the metal in contact with

soil, etc.

Corrosion due to stray currents.

An important role in the processes of underground corrosion is played by stray currents (currents of foreign sources)

In zone K near the rail, oxygen dissolved in soil moisture is reduced. As a result, an excess of OH- ions is created.

The presence of these ions shifts the equilibrium at the surface of the underground metal, the wire. The binding of ions to ions leads to the appearance of an increased concentration of excess electrons in a given place in the tube. These electrons begin to fit along the tube. At the same time, an oxidation process takes place on the rail in zone A. Rail metals are destroyed. Metal ions pass into soil moisture. This is facilitated by OH- ions formed at the surface of the pipe in zone A under the influence of electrons transferred here from zone K. Thus, an underground pipe corrodes in zone K, and a rail in zone A.

Methods for protecting metals from corrosion.

Based on the fact that electrochemical corrosion is the most common, various protection methods primarily take into account this type of corrosion.

Methods of protecting metals from corrosion are diverse, we will focus only on the main ones.

1. Insulation of metal from a corrosive environment.

This method consists in isolating the protected metal from moisture, because. in the absence of it, a galvanic cell does not occur, and therefore there will be no corrosion.

Insulating coatings can be very diverse: coatings of metals with non-metallic substances, i.e. oils, varnishes, paints.

2. Coating of metals with metals. There are two types of metal coatings, cathodic and anodic. An example of an anode coating is the Fe c Zn coating. In this case, the metal-protecting Zn is more active than the protected Fe.

If the integrity of the coating is violated with the access of moisture, a galvanic cell A-Zn / H2O + O2 / Fe appears, in which the anode Zn is destroyed, and the cathode - iron remains until the entire protective layer is destroyed

Zn-2e=Zn

Zn + 2 + 2OH- \u003d Zn (OH) 2

However, protection is otherwise called tread protection, i.e. the protector is the anode. This method of protection is used, for example, to protect against corrosion of the turbine blades of the underwater parts of the ship, in most cases Zn is used as protectors.

cathodic protection. A coating of a less active metal is called cathodic. In this case, if the integrity of the coating is violated, the protected metal is intensively corroded.

test questions

1. What process occurs at the cathode and at the anode during electrolysis?

2. What kind of electrodes do you know?

3. Name the types of batteries.

4. What is chemical corrosion? Types of corrosion?

5. How should corrosion be dealt with?

No matter how you formulate the title of the article, it will still be correct. Chemistry and energy are tied together in the design of a battery.

Lead-acid batteries can operate for several years in charge-discharge modes. They quickly recharge and quickly release the stored energy. The secret of these metamorphoses lies in chemistry, because it is she who helps to convert electricity, but how?

The “mystery” of energy conversion in a battery is provided by a set of reagents, among which there is an oxidizing agent and a reducing agent interacting through an electrolyte. The reducing agent (spongy lead Pb) has a negative charge. During a chemical reaction, it oxidizes, and its electrons travel to the oxidizing agent, which has a positive charge. The oxidizing agent (lead dioxide PbO2) is reduced, and the result is an electric current.

The electrolyte is a liquid that is a poor conductor of current, but a good conductor for ions. This is an aqueous solution of sulfuric acid (H2SO4). In a chemical reaction, a process takes place that is known to everyone from school - electrolytic dissociation.

During the reaction, - positively charged ions (H+) are sent to the positive electrode, and negatively charged ions (SO42-) to the negative. When the battery is discharged, then from the reducing agent (spongy lead), through the electrolyte to the positive electrode, ions with a positive charge of Pb2 + are sent.

Quadrivalent lead ions (Pb4+) are converted into divalent ions (Pb4+). However, this is not all chemical reactions. When ions of acidic residues with a negative charge (SO42-) combine with positively charged lead ions (Pb2+), lead sulfate (PbSO4) is formed on both electrodes. But this is already bad for the battery. Sulfation shortens battery life and can build up over time, leading to battery failure. A side effect of the chemical reactions in conventional lead-acid batteries are gases.

What happens when the battery is recharged?

The electrons are sent to an electrode with a negative charge, where they perform their function - they neutralize lead ions (Pb2+). The chemical reactions that take place in batteries can be described by the following formula:

The density of the electrolyte, and its level in the battery, depends on whether the battery is charged or discharged. Changes in electrolyte density can be described by the following formula:

Where the battery discharge rate, which is measured as a percentage, is Cp. The density of the electrolyte when fully charged is Rz. Electrolyte density at full discharge - Pр.

The standard temperature at which measurements are made is + 25°C, The density of the electrolyte in accordance with the temperature is + 25°C, g / cm3 - P25.
During a chemical reaction, positive electrodes use 1.6 times more acid than negative electrodes. When the battery is discharged, the volume of electrolyte increases, and when it is charged, on the contrary, it decreases.
In this way, with the help of chemical reactions, the battery receives and then gives off electrical energy.

A ready-to-use lead-acid battery consists of latticed lead plates, some of which are filled with lead dioxide and others with lead metal spongy. The plates are immersed in a solution at this concentration, the electrical conductivity of the sulfuric acid solution is maximum.

During battery operation - when it is discharged - a redox reaction occurs in it, during which metallic lead is oxidized

and lead dioxide is reduced:

The electrons donated by metal lead atoms during oxidation are accepted by lead atoms during reduction; electrons are transferred from one electrode to another via an external circuit.

Thus, metallic lead serves as an anode in a lead battery and is negatively charged, and serves as a cathode and is positively charged.

In the internal circuit (in solution), when the battery is operating, ion transport occurs. Ions move towards the anode and ions move towards the cathode. The direction of this movement is determined by the electric field resulting from the occurrence of electrode processes: anions are consumed at the anode, and cations are consumed at the cathode. As a result, the solution remains electrically neutral.

If we add the equations corresponding to the oxidation of lead and reduction, then we get the total equation of the reaction that occurs in the lead battery during its operation (discharge):

E. d. s. of a charged lead battery is approximately 2 V. As the battery charges, its cathode and anode materials (Pb) are consumed. Sulfuric acid is also consumed. In this case, the voltage at the battery terminals drops. When it becomes less than the value allowed by the operating conditions, the battery is charged again.

For charging (or charging), the battery is connected to an external current source (plus to plus and minus to minus). In this case, the current flows through the battery in the opposite direction to that in which it passed when the battery was discharged. As a result, the electrochemical processes on the electrodes “reverse”.

The lead electrode is now undergoing a reduction process

i.e. this electrode becomes the cathode.

The lead battery electrolyte is a sulfuric acid solution containing a relatively small amount of ions. The concentration of hydrogen ions in this solution is much greater than the concentration of lead ions. In addition, lead in the series of voltages is before hydrogen. However, when a battery is charged, it is lead that is reduced at the cathode, not hydrogen. This is because the hydrogen evolution overvoltage on lead is especially high (see Table 20 on page 295).

Electronic representations of oxidation and reduction. Chemical reactions can proceed without change or with a change in the oxidation state of elements, for example:

If in the first example (neutralization reaction) none of the elements changes the oxidation state, then in the second - the oxidation state of zinc changes from +2 to 0 and the oxidation state of carbon changes from 0 to +2.

Reactions that occur with a change in the oxidation state of elements are called redox reactions.

The change in oxidation states was apparently due to the transfer of two electrons from carbon to zinc, which can be expressed electronic equations of half-reactions of oxidation and reduction, which, when added together, give redox reaction equation:

reducing agent oxidation;

oxidizer recovery;

Element, giving electrons is called reducing agent during the course of the reaction oxidizes his degree oxidation increases.

Element, receiving electrons is called oxidizing agent during the course of the reaction recovering, his degree oxidation is reduced.

The concepts of oxidizing agent and reducing agent also apply to simple and complex substances containing the corresponding elements. In the given example, the reducing agent is a simple substance: carbon C, the oxidizing agent is a complex substance: zinc oxide ZnO.

In the general case, the redox reaction can be reversible, as a result, the reducing agent turns into an oxidizing agent, and the oxidizing agent into a reducing agent:

reducing agent - ne ↔ oxidizer

oxidizer + ne ↔ reducer

Thus, the redox reaction is an inseparable unity of two half-reactions - oxidation and reduction, and the number of electrons donated by the reducing agent and accepted by the oxidizing agent are equal.

Redox properties of simple and complex substances. Simple substances - metals, having a small electronegativity, relatively easily lose electrons, showing exclusively restorative properties. They are most pronounced in alkali metals. For simple substances - non-metals with high electronegativity are characteristic oxidizing properties. Fluorine is an absolute oxidizing agent, and oxygen also has pronounced oxidizing properties (except for the reaction with fluorine, where oxygen plays the role of a reducing agent). However, non-metals with a relatively low electronegativity, such as carbon, hydrogen, along with oxidizing properties, can also exhibit reducing properties, donating electrons to stronger oxidizing agents.

Complex substances can be oxidizing or reducing agents, depending on the degree of oxidation of the elements in their composition.

If the oxidation state of an element in a given compound is high, it has the ability to lower it by accepting electrons. The substance in this case will be an oxidizing agent. The most important oxidizing agents are: nitric acid H NO h and its salts - nitrates, nitrogen tetroxide N 2 O 4 , salts of perchloric acid HC1O 4 - perchlorates, potassium permanganate KMnO 4, etc.

If the compound contains an element with a low degree of oxidation, it can increase it by donating electrons. A complex substance containing such an element will exhibit reducing properties. Ammonia is the most important reducing agent. N H 3, hydrazine N 2 H 4 and its organic derivatives, hydrocarbons, alcohols, amines and other substances.

Obviously, if a compound contains an element with an intermediate oxidation state, it can lower it by accepting electrons or raise it by donating electrons. The concepts of an oxidizing agent and a reducing agent in this case become relative: the substance, depending on the properties of the reaction partner, will exhibit either oxidizing or reducing properties. An example is hydrogen peroxide H 2 O 2 , the oxidation state of oxygen in which is -1. Its value can decrease to -2 by adding one electron or increase to 0 by giving it away. Therefore, when interacting with energetic reducing agents, hydrogen peroxide behaves as an oxidizing agent, and when reacting with energetic oxidizing agents, it acts as a reducing agent.

Drawing up equations of redox reactions.

Redox reactions are often expressed by complex equations. To select the coefficients in them, two methods are used: the method of electronic equations and the method of electron-ion equations.

Method of electronic equations based on the concept of oxidation state. It is universal and applicable to all types of redox reactions. The method includes the following operations:

1. Write down the reaction scheme indicating the degree of oxidation of the elements, for example:

2. Determine the elements that have changed the degree of oxidation. In this reaction, the oxidation state was changed by carbon and nitrogen, while for hydrogen and oxygen, the oxidation state remained unchanged.

3. Compose electronic equations for the half-reactions of oxidation and reduction in compliance with equality of masses and charges:

The number of electrons donated by the reducing agent and accepted by the oxidizing agent must be equal, so the first equation should be multiplied by three, and the second by four. The indicated multipliers are the coefficients for the reducing agent C, the oxidizing agent HNO3 and products of their transformations CO and NO:

3C + 4HNO 3 ® 3CO 2 + 4NO + H 2 O,

4. The coefficients for the remaining substances, consisting of elements with a constant oxidation state, are found from the balance of the corresponding atoms on the left and right sides of the equation. In the considered reaction, such a substance is water, before the formula of which you need to put a factor of two. The final equation will be written as:

3C + 4HNO 3 ® 3CO 2 + 4NO + 2H 2 O

Method of electron-ion equations used in the preparation of equations for reactions occurring in electrolyte solutions. In this case, the oxidation state is not determined, and the processes of oxidation and reduction are recorded for real ions and molecules in solution.

In order to maintain the mass balance, particles of the medium in which the reaction proceeds are used. In any aqueous solutions, these are water molecules, in acidic solutions, additional H + ions, and in alkaline solutions, OH - ions.

The sequence of actions is as follows:

1. Make up an ionic reaction scheme, writing strong electrolytes in the form of ions, gaseous, insoluble substances and weak electrolytes in the form of molecules:

C + H + + NO 3 - ® CO 2 + NO + H 2 O

2. Write down the electron-ion equations of the oxidation half-reactions and recovery.

In this reaction, carbon C acts as a reducing agent, which, when oxidized, turns into carbon dioxide CO 2. To maintain the mass balance, two H 2 O molecules are added to the left side of the equation, and four H - ions are added to the right side. The charge balance is maintained by subtracting four electrons from the left side of the equation:

C + 2H 2 O - 4e ® C O 2 + 4H +

The oxidant is an ion. NO 3 - , turning into NO , The mass balance is provided in this case by the addition of two molecules H2O to the right side of the equation and four H + ions to its left side. Since the total charge of the particles on the left side of the equation is plus three, and on the right side it is zero, three electrons must be added to the left side:

NO 3 - + 4Н + + 3е ® NO + H 2 O

3. Summarize the equations of half-reactions, having previously equalized the numbers of given and received electrons:

After the reduction of similar terms, one obtains ionic equation:

ZS + 4H + + 4 NO 3 - ® ZSO 2 + 4 NO + 2H 2 O

4. Combine ions into molecules and get the final molecular equation reactions:

3C + 4H NO 3 ® 3CO 2 + 4 NO + 2H 2 O

Comparing the considered methods for compiling equations of redox reactions, it should be noted that both of them lead to the same final result. However, the method of electron-ion equations is more informative; it operates not with hypothetical, but with real ions and molecules that exist in electrolyte solutions. It is especially useful in describing electrochemical processes.

5.2. Electrochemical systems.

Electrode potentials. Electrochemical processes are called the processes of mutual transformation of chemical and electrical energy. These transformations are carried out as a result of redox reactions occurring at the phase boundary between the electronic and ionic conductors. An electronic conductor in contact with an ionic conductor is called electrode.

Consider an electrode consisting of a plate of active metal - zinc, immersed in an aqueous solution of zinc sulfate, dissociating into ions:

ZnSO 4 ↔ Zn 2+ SO 4 2-

Positively charged zinc cations located on the surface of the plate, as a result of interaction with polar water molecules, break away from the plate and go into solution, the electrons remain in the metal. Oxidation takes place:

Zn 0 – 2e ® Zn 2+

At the same time, the reverse process also takes place: zinc cations from the solution are attracted by the surface of the metal and are part of its crystal lattice. Restoration in progress:

Zn 2+ +2е ® Zn 0

As the concentration of zinc cations in the solution increases, the rate of release of ions from the metal decreases, and the rate of their transition into the metal increases. When the rates of these processes become equal, a redox equilibrium between the metal and its ions will be established at the metal-electrolyte interface, which we agreed to write as a reversible reduction process:

When recording the electrochemical circuit of an electrode, its oxidized form is separated from the reduced one by a line: Zn +2 / Zn .

Since zinc is an active metal, the equilibrium of the process is shifted to the left, that is, more ions go into solution than come back. As a result, the zinc plate acquires a negative potential (Fig. 5.1 a).

The same processes occur when a plate of low-active copper metal is immersed in a solution of copper sulfate, which dissociates into ions:

However, in this case, the metal sends an insignificant number of cations into the solution, the process of precipitation of cations on the metal predominates, and the equilibrium is shifted to the right:

Copper electrode C u 2+ /Cu acquires a positive potential (Fig. 5.1.).

Figure 5.1. Scheme of the occurrence of electrode potential

a) active metal; b) inactive metal

The absolute value of the electrode potential cannot be measured, therefore, it is measured relative to the potential of the reference electrode, which is used as standard hydrogen electrode (Fig. 5.2). It is a platinum plate in an aqueous solution of sulfuric acid with a concentration of hydrogen ions CH + = 1 mol / l, washed by hydrogen with a pressure of 101.3 kPa at a temperature of 293K.

Platinum has the ability to adsorb hydrogen and at the boundary

Figure 5.2. Scheme of the hydrogen electrode

phase separation, an equilibrium is established between molecules and hydrogen ions:

2Н + + 2е ↔ Н 2

The corresponding electrode potential is conditionally taken as zero, E 0 2H + / H2 = 0.

The standard electrode potential of a metal is the potential difference between a given metal immersed in a solution of its salt with a concentration of metal ions C M n + \u003d 1 mol / l at a temperature of 293K and a standard hydrogen electrode.

Standard electrode potential is a measure of the redox activity of a system.

FROM an increase in the value of the standard electrode potential, the reducing activity of the system decreases, and the oxidative- is growing.

So, with an increase in the value of the standard electrode potential of metals, the reducing activity of their atoms decreases and the oxidizing activity of ions increases.

Comparison of the electrode potentials of the half-reactions makes it possible to draw a conclusion about the direction of the redox process.

Consider a heterogeneous redox reaction that occurs when a zinc plate is immersed in a solution of copper sulfate, which dissociates into ions (Fig. 5.3a):

CuSO 4 ↔ Cu 2+ + SO 4 2-

The electrode potentials of zinc and copper have the following values:

Zn 2+ + 2e ↔ Zn 0; E 0 \u003d - 0.76 B

Cu 2+ + 2 e ↔ Cu 0; E 0 \u003d +0.34 V

As can be seen, the standard electrode potential for the second system is higher than for the first. Consequently, upon contact, the second system will act as an oxidizing agent, the first - as a reducing agent. In other words, the second reaction will go from left to right, and the first - in the opposite direction, that is, zinc will donate electrons to copper ions, thus displacing copper from its salt solution (Fig. 5.3 a):

The electrode potential arises not only as a result of the exchange of ions between the metal and its salt solution. Any redox half-reaction is characterized by a certain value of the electrode potential, for example:

CO 2 + 4H + + 4e ↔ C + 2H 2 O; E° = +0.21 b,

NO 3 - + 4Н + + 3 e ↔ NO + 2 H 2; E° = +0.96 B

In this case, the oxidizing properties are more pronounced for the ion NO 3 - , so this ion will oxidize carbon, being reduced to nitric oxide NO (see 5.1).

The value of the electrode potential is not constant; it depends on a number of factors, in particular, on the ratio of the oxidized and reduced forms of the substance. This dependence is expressed Nernst equation, which at a standard temperature of 293K is written as:

(5.1),

where: E - electrode potential at given concentrations of oxidized C ok and reduced C restore forms of the substance, mol / l,

E ° - standard electrode potential,

n is the number of electrons transferred.

For metal electrodes in salt solutions, the reduced form is metal atoms, the concentration of which is a constant value C M = const . In this case, the Nernst equation takes the form:

(5.2)

where:

C m + n - concentration of metal ions, mol/l;

n is the charge of the ion.

Chemical sources of electric current. In the considered systems, the transition of electrons from the reducing agent to the oxidizing agent is carried out chaotically, as a result, chemical energy is converted into thermal energy.

It is possible, however, by spatially dividing the processes of oxidation and reduction, to obtain a directed movement of electrons - an electric current. A device in which the chemical energy of a redox reaction is converted into electrical energy is called a chemical current source or a galvanic cell.

Figure 5.3. Heterogeneous redox process:

a - spatially undivided; c - spatially separated

Figure 5.3 shows a diagram of a Daniel-Jacobi galvanic cell, consisting of zinc and copper electrodes placed in aqueous solutions of sulfates of these metals. The solutions are connected by an electrolytic bridge - a tube filled with an electrolyte solution, for example, potassium chloride. When the electrodes are shorted through the load, the zinc electrode undergoes an oxidation process with the release of ions Zn2+ solution; the released electrons pass through the external circuit to the copper electrode, where C ions are reduced u+2 coming from a solution of copper sulfate.

The electrode on which the oxidation process takes place is called the anode, the electrode on which the reduction process takes place is called the cathode. In a copper-zinc cell, the zinc electrode is the anode, and the copper electrode is the cathode. During the operation of the element, the zinc anode gradually dissolves, and copper is deposited on the copper cathode. Thus, the zinc electrode is active, its material is directly involved in the redox process. The copper electrode plays a passive role as an electron conductor; its material does not participate in the redox process.

The redox process occurring on the electrodes disturbs the ionic equilibrium in solutions - an excess of ions is formed at the copper electrode SO 4 2- , in zinc - their lack. As a result, the movement of ions occurs in the internal circuit SO 4 2- from copper sulfate solution to zinc sulfate solution.

The electrochemical scheme of this element can be written as:

where one vertical bar denotes the boundary between the electronic and ionic conductors, and two vertical bars denote the boundary between two ionic conductors.

The electromotive force (EMF) of an element at standard ion concentrations (1 mol/l) can be calculated as the difference between the standard cathode and anode potentials:

EMF = D E ° \u003d E 0 cat - E 0 an \u003d E 0 C u 2 +/ C u - E 0 Zn 2 +/ Zn \u003d +0.34 - (- 0.76) \u003d 1.1 B.

If the ion concentrations differ from the standard ones, it is necessary to calculate the potentials of the electrodes at given concentrations using formula 5.2 and then take their difference.

In principle, any redox reaction can be used to generate electrical energy, provided that the processes of oxidation and reduction are spatially separated. The active materials of the electrodes can be not only metals, but also non-metals, as well as oxides, hydroxides and other complex substances. So, in the case of the oxidation of carbon with nitric acid (see 5.1), an electric current can be obtained if carbon and platinum electrodes are immersed in a solution of nitric acid and closed with a metal conductor. In this case, the active carbon anode is oxidized with the formation of carbon dioxide C O2 , on a passive platinum cathode, nitrate ions are being reduced NO 3 - to nitric oxide NO . The element scheme can be written as:

EMF \u003d D E ° \u003d E ° cat - E ° an \u003d E ° NO 3 - / NO - E ° CO 2 / C \u003d 0.96 - 0.21 \u003d 0.75 AT

D value E° is related to the standard Gibbs energy of the reaction ( D G °) ratio:

D G °=- nF D E ° (5.3)

where n is the number of electrons transferred during the reaction, F - Faraday number (96500 C).

Equation 5.3 establishes the relationship between chemical and electricalforms of energy. It allows for a known value G calculate E galvanic cell and vice versa, knowing E to calculate G.

Fuel elements. A fuel cell is a kind of galvanic cell in which electrical energy is obtained as a result of a redox process between fuel components - a fuel (reductant) and an oxidizer, continuously supplied to the electrodes from the outside. Hydrogen, carbon monoxide, methane, alcohols can serve as a fuel, and oxygen, air, hydrogen peroxide and other substances can serve as an oxidizing agent. Thus, in fuel cells, unlike heat engines, the chemical energy of the fuel is directly converted into electrical energy, so their efficiency is 1.5 - 2.0 higher than that of heat engines. In addition, they significantly less pollute the environment.

Figure 5.4. Hydrogen-oxygen fuel cell

At present, a hydrogen-oxygen fuel cell has found practical application (Fig. 5.4).

It contains two porous metal or carbon electrodes with catalyst additives. An aqueous 40 - 85% solution of potassium hydroxide is used as an electrolyte. Electrochemical scheme of the element:

Gaseous hydrogen and oxygen supplied to the electrodes diffuse through the electrodes in the direction of the electrolyte, are adsorbed on the surface of the pores, and are activated by the catalyst. This facilitates and accelerates the processes of hydrogen oxidation at the anode and oxygen reduction at the cathode:

H 2 + 2 OH - - 2е ® 2 H 2 O

1/2O 2 + H 2 O + 2e ® 2OH -

Overall reaction equation:

H 2 + 1 / 2O 2 ® H 2 O

The reaction product, vaporous water, is removed by a stream of hydrogen, which, after separation of water, is returned back to the system. Thus, "cold combustion" of hydrogen in oxygen is carried out with the release of energy in electrical form.

Batteries.The redox processes occurring during the operation of galvanic cells can be either irreversible or reversible. Accordingly, chemical current sources can be of single and multiple action. Reusable galvanic cells are called batteries. When the battery is operating in the current source mode, a redox process spontaneously proceeds, leading to the conversion of chemical energy into electrical energy (the Gibbs energy of the reaction is negative D G <0). Химический состав электродов при этом меняется, аккумулятор разряжается. Обратная реакция самопроизвольно не идет ( D G >0). However, a discharged battery can be charged from an external current source, the voltage of which exceeds the EMF of the cell. In this case, the reverse process of converting electrical energy into chemical energy takes place, and the electrode material is regenerated.

The redox process that occurs when a current is passed through an electrolyte is called electrolysis.

As a result of electrolysis, the battery can again work as a current source. The number of charge-discharge cycles can reach several hundred. In aviation, lead, silver-zinc and cadmium-nickel batteries are used.

Lead (acid) battery in the charged state corresponds to the electrochemical circuit:

In the current source mode, during discharge, lead is oxidized at the negative electrode, and lead dioxide is reduced at the positive electrode. When charging, reverse processes occur: on the negative electrode - the reduction of lead sulfate, on the positive - its oxidation:

In a charged lead battery, depending on its type, the concentration of sulfuric acid is 27 - 30% of the mass. During discharge, it decreases, since water is released into the electrolyte. The density of the electrolyte also decreases. This makes it possible, by measuring the density of the electrolyte, to control the degree of battery discharge.

Cadmium nickel battery made according to the scheme:

When discharging at the negative electrode CD oxidized, on the positive - Ni(OH ) 3 is partially restored. When charging, the reverse processes take place:

Nickel-cadmium batteries are stable in operation, require less maintenance and have a longer service life than lead-acid batteries.

Silver-zinc accumulator in a charged state corresponds to the scheme:

During its operation, reversible reactions occur: on the negative electrode - oxidation of zinc, on the positive - reduction of silver oxide:

The main advantage of the silver-zinc battery is its high energy capacity; per unit mass, it gives 4 to 6 times more energy than the considered types of batteries.

Corrosion of metals. Corrosion is the destruction of a metal as a result of its physical and chemical interaction with the environment. The most dangerous and most common type of metal corrosion is electrochemical corrosion, which occurs when metals come into contact with electrolyte solutions. Most often, it is a consequence of the work of short-circuited galvanic cells, which are formed, for example, upon contact of parts made of dissimilar metals.

The role of the electrolyte in this case can be played by a thin film of moisture formed during the adsorption of water vapor from the atmosphere on metal surfaces. So, when parts made of copper and iron come into contact in the presence of water, a galvanic cell is formed (Fig. 5.5):

Figure 5.5. Scheme of electrochemical corrosion

Iron, as a more active metal, serves as an anode and undergoes oxidation, while on the copper cathode, air oxygen is reduced with the participation of water:

To protect metals from corrosion, various coatings are used: metal, non-metallic, paint and varnish, polymer.

Sample self-assessment questions:

1. What reactions are called redox reactions?

2. What is meant by a reducing agent or an oxidizing agent?

3. What is the meaning of the method of electronic equations?

4. What is the meaning of the method of electron-ion equations?

5. What processes are called electrochemical?

6. What is a standard hydrogen electrode?

7. What is a measure of the redox activity of a system?

8. What dependence does the Nerst equation express?

9. What is a galvanic cell?

10. What is a cathode and an anode?

11. How do redox processes occur in batteries?

12. What is electrolysis?

13. What is electrochemical corrosion?

Tasks for topic number 5

Task number 5.1.

Using the electronic balance method, compose the equations of redox reactions. Specify the oxidizing agent and reducing agent:

1. NH 3 + O 2 NO + H 2 O

2. HClO 3 ClO 2 + HClO 4 + H 2 O

3. AgNO 3 Ag + NO 2 + O 2

4. NH 4 NO 2 + H 2 O

5. H 2 O 2 + PbS PbSO 4 + H 2 O

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

7. Ca 3 (PO 4) 2 + C + SiO 2 CaSiO 3 + P + CO

8. FeS + O 2 Fe 2 O 3 + SO 2

9. N 2 H 2 + O 2 N 2 + H 2 O

10. S + KOH K 2 SO 3 + K 2 S + H 2 O

Task number 5.2.

Compose the equations of redox reactions:

1) electronic balance method;

2) ion-electronic method.

Specify the oxidizing agent and reducing agent.

1. P + NO 3 H 3 PO 4 + NO 2 + H 2 O

2. Zn + HNO 3 Zn (NO 3) 2 NO 2 + H 2 O

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

4. KMnO 4 + KNO 2 + H 2 O KNO 3 + MnO 2 + KOH

5. FeSO 4 + H 2 O 2 + H 2 SO 4 Fe 2 (SO 4) 3 + H 2 O

6. CrCl 3 + H 2 O 2 + NaOH Na 2 CrO 4 + NaCl + H 2 O

7. CrO 3 + KNO 3 + KOH K 2 CrO 4 + KNO 2 + H 2 O

8. PH 3 + KMnO 4 + H 2 SO 4 H 3 PO 4 + K 2 SO 4 + MnSO 4 + H 2 O

9. Si + NaOH + H 2 O Na 2 SiO 3 + H 2

10. HCl + KMnO 4 Ci 2 + MnCl 2 + KCl + H 2 O

Task number 5.3.

Solution:

The electrode potential is calculated by the Nernst formula, which for metal and hydrogen electrodes is written as:

where E is the electrode potential,

n is the charge of the metal (hydrogen) ion.

Ferrous sulfate dissociation equation:

Fe 2 SO 4 2 Fe 3+ +3 SO 4 2-

shows that upon dissociation of 0.05 mol Fe 2 (SO 4) 3, 0.05 2 = 0.1 mol Fe 3+ ions are formed.

Therefore C(Fe 3+ )=0.1 mol/l, n=3.

From table 3 we have E 0 (Fe3+/Fe)=-0.04.

Task number 5.4.

Task number 5.5.

How much will the potential of the zinc electrode change if the zinc salt solution in which it is immersed is diluted 10 times.

Problem number 5.6.

The potential of the cadmium electrode in a solution of its salt is 0.52V. Calculate the concentration of Cd + ions in the solution.

Task number 5.7.

Task number 5.8.

Calculate the pH of the solution in which the potential of the hydrogen electrode is -100 mV.

Task number 5.9.

Reaction equation

Ion concentration, С mol/l

pH

MnO 4 - + 8H + + 5 e Mn 2+ + 4H 2 O

C(MnO4-)=C(Mn2+)=1

ClO 3 - + 6H + + 6e Cl - + 3H 2 O

C(ClO 3 -)=C(Cl-)=0.1

Cr 2 O 7 2- + 14H + + 6e 2Cr 3 ++ 7H 2 O

C(Cr 2 O 7 2-)=C(Cr 3+)=1

PbO 2+ 4H + +2e Pb 2 ++2H 2 O

C(Pb2+)=0.1

Solution 1:

The potential of the redox electrode E is calculated using the Nerst equation:

where E 0 standard electrode potential;

n is the number of electrons involved in the reaction;

C ok, C restore - products of concentrations of substances in oxidized and reduced forms, respectively.

In this system, Mn 4 ions are in the oxidized form - and H+ , in the reduced one - the Mn 2+ ion and the H 2 O molecule. 5 electrons take part in the reaction. Given that the concentration of water remains practically constant and enters the value of E 0 , we have:

According to table 3: E 0 (MnO 4 - /Mn 2+)=+1.51V.

Substituting the numerical values, we finally get:

Task number 5.10.

Write the equations of electrode processes occurring during the operation of a galvanic cell. Calculate the EMF of the element at given concentrations, C mol / l.

Element scheme

С, mol/l

Zn/Zn 2+ //Pb 2+ /Pb

C(Zn2+)=0.2, C(Pb2+)=0.04

Mn/Mn 2+ //Ni 2+ /Ni

C(Mn 2+)=0.1, C(Ni 2+)=0.01

Fe/Fe 2+ //Cu 2+ /Cu

C(Fe 2+)=1, C(Cu 2+)=0.5

H 2 /2H + //Ag + /Ag

C(H+)=0.01, C(Ag+)=0.1

Ni/Ni 2+ (C 1)//Ni 2+ (C 2)/Ni

C 1 (Ni 2+)=0.1, C 2 (Ni 2+)=0.01

Cu/Cu 2+ //Fe 3+ /Fe 2+

C(Cu 2+)=1, C(Fe 3+)=C(Fe 2+)=1

Solution 1:

Based on the data in Table 3, it can be concluded that the more active zinc metal will be the anode in this element, and the less active lead metal will be the cathode.

The electromotive force of a galvanic cell is equal to the difference between the electrode potentials of the oxidizer (cathode) and reductant (anode):

Using the Nerst formula, we have:

Task number 5.11.

Determine in which direction spontaneous flow is possible under standard conditions of this reaction. Calculate the value of the equilibrium constant of the reaction.

Reaction equation

2С l - + 2Fe 3+ 2Fe 2+ +Cl 2

H 2 O 2 + HClO H + Cl + O 2 + H 2 O

5H 2 O 2 +H + +2IO 3 I 2 +5O 2 +6H 2 O

Sn4+ +2I - Sn2+ +I2

Sn4+ +H2S Sn2+ +S+2H+

H 2 S + 4H 2 O 2 2H + + SO4 2- + 4H 2 O

Solution 1:

To determine the direction of the redox reaction, it is necessary to find the EMF of a galvanic cell formed from a double oxidizer and a reducing agent.

where E 0 ok, E 0 restore - the standard potentials of the oxidizing agent and reducing agent.

The reaction is possible for which, since in this case the change in the Gibbs energy is a negative value.

where n is the number of electrons participating in the reaction;

F - Faraday number, equal to 96480 C / mol.

In turn, the change in the Gibbs energy is related to the equilibrium constant by the relation:

Consequently,

where

, .

Standard electrode potentials are equal (see table 3):

Cl2 + 2e 2Cl - E 0 (Cl 2 / 2Cl -) \u003d 1.36 B

Fe 3+ + e Fe 2+ E 0 (Fe 3+ /Fe 2+ = 0.77 B

Since E 0 (C l 2 / 2C l)> E 0 (Fe 3+ / Fe 2+ ) chlorine will serve as an oxidizing agent, and the Fe 2+ ion as a reducing agent.

Electrode process equations:

Summary Equation:

Cl 2+ 2Fe 2+ 2 Cl - + Fe 3+

Thus, the reaction under consideration will proceed from right to left.

K=10 20

Task number 5.12.

Calculate the value of the potential of the redox electrode obtained by immersing a platinum wire in an aqueous solution containing simultaneously two salts A and B with concentrations of C A and C B, mol/l at a given pH value.

S A

M B

pH

Na 2 Cr 2 O 7

Cr2(SO4)3

0,2

4

2

NaClO 2

NaClO

0,1

0,3

9

3

KClO 4

NaClO3

0,2

0,3

3

4

Na2SO4

K2SO3

0,05

0,08

10

5

CrCl 3

CrCl2

0,2

0,8

1

6

NaNO 3

NaNO 2

0,01

0,09

9

7

Na 2 S 2 O 8

Na2SO4

0,1

0,2

6

8

KMnO 4

K2MnO 4

0,3

0,6

8

9

Fe 2 (SO 4) 3

FeSO4

1

3

2

10

Ce(SO4)2

Ce 2 (SO 4) 3

0,002

0,001

0,5

Solution 1:

A solution containing both the oxidized and reduced forms of the same element (in this case, chromium) is called a redox system. In general, the redox reaction equation for a redox electrode is:

Oh+red,

wherenis the number of electrons involved in the reaction, and Ox andRed- oxidized and reduced form of the element. To determine the value of the electrode potential of such a system, one should use the Nerst equation:

Formally, in this case, the oxidation state in the electrode process changes chromium

Cr 6+ + 3 eCr 3+ ,

that is, the oxidized form will beCr 6+ - containing form, but this does not mean that it is possible to write down the activity value using the logarithm in the Nerst equationCr 6+ . This is due to the fact that the particle itselfr 6+ does not exist in an aqueous solution, it is part of a more complex particleCr 2 O 7 2- , therefore the concept of activity, that is, the apparent concentration of non-existent particlesCr 6+ meaningless. Can determine activity (or concentration) of particlesCr 2 O 7 2- , but then the equation of the electrode process should also be written with the participation of particlesCr 2 O 7 2-

Cr 2 O 7 2- +…. Cr 3+ +…,

however, in this case, oxygen is present on the left side, but not on the right side, so it is necessary to add particles containing O 2- to the right side. There are no O 2- ions in an aqueous solution, however, oxygen with this degree of oxidation is part of either H 2 O molecules or OH - ions. Since by condition the environment is acidic (pH<7), концентрация ионов ОН - в этом растворе крайне мала, значит следует записывать электродный процесс на с участием этих ионов, а с участием молекул Н 2 О

Cr 2 O 7 2- + 14Н+ + 6е 2Сr 3+ + 7 H 2 O

Thus, in the electrode process, in addition to ionsCr 2 O 7 2- and Cr 3+ H + ions are also involved, so their concentration will also affect the magnitude of the electrode potential, that is

By condition, the concentration of K 2Cr 2 O 7 andCr 2 (SO 4 ) 3 are 0.1 and 0.2 mol/L, respectively. Since these salts are strong electrolytes, that is, they completely dissociate in solution, the concentration of ionsCr 2 O 7 2- andCr 3+ will be 0.1 and 0.4 mol/l. At pH = 2, the concentration of H + ions is C (H +) \u003d 10 -pH \u003d 10 -2, from here:

Key concepts:

· oxidation reaction;

· recovery reaction;

· oxidizer;

· reducing agent;

· redox reaction equation;

· electrochemical system;

· standard hydrogen electrode;

· standard electrode potential;

· chemical current source;

· cathode;

· anode;

· fuel cell;

· battery;

· electrolysis;

· corrosion.

Battery		specific energy,	Specific power,	Life time, number of cycles
Pb-acid
Fe-air
Zn-air
Zn-chloride
Na-sulfide
Li-sulfide

Lead battery

Lead-acid batteries are by far the most widely used. It serves as a current source for starters of internal combustion engines, for emergency lighting, radio and telephone equipment, is used in underwater vehicles and stations, and for other purposes.

A Pb-acid battery consists of a lead anode and a cathode in the form of a lead grid stuffed with lead (IV) oxide. The electrolyte is sulfuric acid. During the operation of EA, reactions occur on one electrode (anode), in which the oxidation state of lead changes from 0 to +2 (discharge) and from +2 to 0 (charge), and on the other electrode (cathode), the oxidation state of lead changes from +4 to +2 (discharge) and vice versa (charge).

On the anode:
On the cathode:
The total current-generating reaction is described by the equation:

The current drawn from a lead battery can be increased by designing the cathode as a series of plates interleaved with several anode plates (Figure 9.4). Each such EA gives a voltage of approximately 2 V. Batteries used in automobiles usually consist of six such batteries connected in series and giving a voltage of about 12 V.

Electrolysis.

In solutions and melts of electrolytes, there are ions (cations and anions) of opposite signs, which, like all particles of a liquid, are in chaotic motion. If in such an electrolyte melt, for example, a NaCl melt ( ) immerse the electrodes and pass a constant electric current, then the ions will move to the electrodes: cations

Na + +=Na 0 (cathode)

2Cl - - 2e=Cl 2 (anode)

This reaction is OVR at the anode, the oxidation process takes place, at the cathode, the reduction process.

Electrolysis is a redox process that occurs on electrodes when an electric current passes through an electrolyte solution or melt.

The essence of electrolysis is the implementation of chemical reactions due to electrical energy - reduction at the cathode and oxidation at the anode. In this case, the cathode donates electrons to cations, and the anode receives electrons from anions.

The process of electrolysis is visually depicted by a diagram that shows the dissociation of the electrolyte, the direction of movement of ions, the processes of their electrodes and the released substances. NaCl electrolysis scheme:

Cathode Anode

For electrolysis, the electrodes are immersed in an electrolyte solution or melt and connected to a current source. The device on which electrolysis is carried out is called an electrolyser or an electrolytic bath.

Electrolysis of aqueous solutions of electrolytes.

During the electrolysis of electrolyte solutions, water molecules can participate in the processes. For reduction, a potential equal to B must be applied to the cathode, and for the reduction of water molecules B.

Therefore, water cations will be reduced at the cathode:

cathode

and chloride ions will be oxidized at the anode:

The ions accumulate near the cathode and together with the ions form sodium hydroxide.

Cathodic and anodic processes

Metal cations with a standard potential greater than that of

hydrogen (from up to and including), during electrolysis, the density is restored at the cathode.

Metal cations having a small value of the standard

electrode potential (from up to and including), are not restored at the cathode, but water molecules are restored instead.

If the aqueous solution contains cations of various metals, then during electrolysis, they are released at the cathode in order of decreasing standard electrode potential of the corresponding metal.

first .

The nature of the reactions occurring at the anode depends on the presence of molecules and on the substance from which the anode is made. usually anodes are divided into soluble (Cu, Ag, Zn, Cd, Ni) and insoluble (coal, graphite, Pt,).

On a soluble anode, during electrolysis, anions are oxidized (if the acids are oxygen-free -) if the solution contains anions of oxygen-containing acids (), then these ions are not oxidized on the anode, but water molecules:

The soluble anode is oxidized during electrolysis, i.e. sends to the external circuit.

and the anode dissolves.

How does electrolysis proceed with insoluble (carbon) electrodes?

Example 2 with insoluble electrode.

Cathode Anode

e

if the cathode and anode spaces are not separated by a partition, then:

Example 4 Solution Electrolysis

Copper electrodes

Cathode (Cu) Anode: e

5) Electrolysis with electrodes

Faraday's Law

This is the quantitative law of electrolysis

m is the mass of the substance. that are released on the electrodes (g)

n is the number of electrons exchanged between the oxidizing agent and the reducing agent

I - current strength (A)

M is the molar mass of the substance that is released at the electrode

F- Faraday constant 96485

t- time (sec)

The reason for the occurrence and flow of electric current in a galvanic cell is the difference in electrode potentials.

Standard Recovery Potential - a quantitative measure of the ability of a substance (molecule or ion) to enter into redox reactions in an aqueous solution.

A redox reaction is possible if

where

- standard oxidant reduction potential.

Standard reductant recovery potential.

The equation Nernst:

where is the electrode potential of the metal, V;

Standard electrode potential of the metal, V;

Universal gas constant (8.31 J/mol;

Absolute temperature, K;

The number of electrons involved in the reaction;

Faraday's constant (96,500 C/mol).

The EMF of any galvanic cell can be calculated from the difference in standard electronic potentials E o. In this case, it should be borne in mind that the EMF is always a positive value. Therefore, it is necessary from the potential of the electrode, which has a large algebraic value, to calculate the potential, the algebraic value of which is less.

E \u003d E o si - E o zn \u003d (+ 0.34) - (-0.76) \u003d 1.10 V

E=E about ok - E about resurrection

E o ok-l - the potential of the electrode with a larger algebraic value.

E o vos-l - the potential of the electrode with a smaller algebraic value.

Some standard electrode potentials are given in Appendix 4.

The quantitative characteristic of electrolysis processes is determined Faraday's law :

The mass of the electrolyte that has undergone transformation during electrolysis, as well as the mass of substances formed on the electrodes, are directly proportional to the amount of electricity that has passed through the electrolyte solution or melt, and the equivalent masses of the corresponding substances.

Faraday's law is expressed by the following equation:

Where is the mass of the substance formed or subjected to transformation;

E is its equivalent mass, g eq;

I - current strength, A;

t is time, sec;

F is the Faraday number (96,500 C/mol), i.e. the amount of electricity required to carry out the electrochemical transformation of one equivalent of a substance.

Example 1: How many grams of copper will be released on the cathode during the electrolysis of a CuSO 4 solution for 1 hour at a current of 4 A.

Solution: The equivalent mass of copper in CuSO 4 is equal to =, substituting the values ​​E = 32, I = 4 A, t = 6060 = 3600 s into the Faraday equation, we get

= 4.77 g.

Example 2: Calculate the equivalent of the metal, knowing that during the electrolysis of a chloride solution of this metal, 3880 C of electricity was expended and 11.74 g of metal is released at the cathode.

Solution: From the Faraday equation we derive E \u003d, where m \u003d 11.742 g; F = 96500 C/mol; It = Q = 3880 C.

E = = 29,35

Example 3: How many grams of potassium hydroxide was formed at the cathode during the electrolysis of a solution of K 2 SO 4 if 11.2 liters of oxygen (n.o.) were released at the anode?

Solution: Oxygen equivalent volume (N.O.) 22.4/4 = 5.6 L. Therefore, 11.2 liters contain 2 equivalent masses of oxygen. The same number of equivalent masses of KOH formed at the cathode. Or 56 2 \u003d 112.7 (56 g / mol - molar and equivalent mass of KOH).

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