1. What was observed in Oersted's experiment?
a) Interaction of two parallel conductors with current.
b) Interaction of two magnetic needles.
c) Rotation of the magnetic needle near the conductor when current is passed through it.
d) Emergence electric current in a coil when a magnet is placed in it.
2. How do two parallel conductors interact with each other if currents flow through them in the same direction?
a) are attracted. b) Repel. c) The force of interaction is zero. d) There is no correct answer.
3. When a direct electric current is passed through a conductor, a magnetic field arises around it. It is detected by the location of steel filings on a sheet of paper or by the turn of a magnetic needle located near the conductor. How can this magnetic field be moved in space?
a) Transfer of steel filings. b) By moving the magnet. c) The transfer of a conductor with current. d) The magnetic field cannot be moved.
4. How will the magnetic needles placed at points A and B inside the coil be located when the key K is opened?
a) Equally north pole to the right in the figure.
b) The same north pole to the left in the figure.
c) Arrows with north poles facing each other.
d) Arrows with south poles facing each other.
5. Why is the device of AC motors simpler than DC? Why are DC motors used in vehicles?
6. Determine the poles of the electromagnet.
7. Depict the magnetic field of the currents and determine the direction of the lines of force magnetic field.
8. Determine the direction of the force acting on a current-carrying conductor placed in a magnetic field.
9. You have three items - "device": wooden block, two non-attractive steel nails, and a permanent magnet.
Three "black boxes" contain respectively: a magnet, two nails and a wooden block. What instruments and in what sequence is it better to use to find out what is in each of the boxes?
10. A DC motor consumes a current of 2 A from a source with a voltage of 24 V. What is the mechanical power of the motor if the resistance of its winding is 3 ohms? What is its K.P.D.?
Determine the direction of the current in the conductor, the cross section of which and the magnetic field are shown in Figure 1.3. What direction does the current have in the conductor, the direction of the magnetic field lines of which is indicated by arrows (Fig. 3)?
5. In the direction of the magnetic lines of force shown in Figure 5, determine the direction of the circular current in the ring.
Electromagnetic waves arise: A. When electric charges move at a constant speed. B. With accelerated movement of electricalB. Around stationary charges.
G. Around a fixed conductor through which a direct electric current passes.
D. Around a fixed charged metal plate
1. Electric current is called... A). the movement of electrons. B). orderly movement of charged particles. B). orderly movement of electrons. 2.To create an electric current in a conductor, it is necessary... A). create an electric field in it. B). create electric charges in it. B). to separate electric charges in it. 3. What particles create an electric current in metals? A). Free electrons. B). positive ions. B). negative ions. ^ 4. What action of current is used in galvanometers? A. Thermal. B. Chemical. B. Magnetic. 5. The current strength in the circuit of the electric stove is 1.4 A. What electric charge passes through the cross section of its spiral in 20 minutes? A). 3200 cl. B). 1680 class B). 500 cl. ^ 6. In which diagram (Fig. 1) is the ammeter connected to the circuit correctly? BUT). 1. B). 2. B). 3. 7. When an electric charge equal to 6 C passes through the conductor, work of 660 J is performed. What is the voltage at the ends of this conductor? BUT). 110 V. B). 220 V. V). 330V. ^ 8. In which diagram (Fig. 2) is the voltmeter connected to the circuit correctly? BUT). 1. B). 2. 9. Two skeins copper wire of the same section have a length of 50 and 150 m, respectively. Which of them has a greater resistance and how many times? BUT). The first is 3 times. B). The second is 3 times. ^ 10. What is the strength of the current passing through a nickel wire 25 cm long and 0.1 mm2 in cross section if the voltage at its ends is 6 V? BUT). 2 A. B). 10 A. B). 6 A
1. In what units is the electric current strength measured? A. Ohm; B. J; W.W; G. A.2. What actions are always manifested when an electric current passes through any medium?
BUT. Thermal; B. Magnetic; AT. Chemical; G. Light.
4. Determine the voltage under which the light bulb is, if when moving a charge of 10 C, work is done 2200 J.
5. Determine the resistance of section AB in the circuit shown in the figure.
6. Calculate the resistance of a nichrome wire with a length of 150 m and a cross-sectional area of 0.2 mm2.
7. A copper conductor with a cross section of 3.5 mm2 and a length of 14.2 m carries a current of 2.25 A. Determine the voltage at the ends of this conductor.
8. How many electrons pass through the cross section of the conductor in 35 s at a current strength of 16 A in it?
9. Determine the mass of an iron wire with a cross-sectional area of 2 mm2, taken to make a resistor with a resistance of 6 ohms.
Electroactivated water solutions - catholytes and anolytes can be used in agriculture, to increase plant productivity, in animal husbandry, medicine, for water disinfection and for domestic purposes. Electrochemical water treatment includes several electro chemical processes associated with transfer in constant electric field electrons, ions and other particles (electrolysis, electrophoresis, electroflotation, electrocoagulation), the main of which is water electrolysis. This article introduces the reader to the main processes underlying the electrolysis of water.
Introduction
The phenomenon of electrochemical activation of water (EAW) is a combination of electrochemical and electrophysical effects on water in a double electric layer (DEL) of electrodes (anode and cathode) during non-equilibrium charge transfer through the DEL by electrons and under conditions of intensive dispersion in the liquid of the resulting gaseous products of electrochemical reactions. There are four main processes in the ECA process:
- electrolytic decomposition of water (electrolysis) due to redox reactions on the electrodes, caused by an external constant electric field;
- electrophoresis - the movement in an electric field of positively charged particles and ions to the cathode, and negatively charged particles and ions to the anode;
- electroflotation - the formation of gas floccules and aggregates consisting of finely dispersed gas bubbles (hydrogen at the cathode and oxygen at the anode) and coarsely dispersed water impurities;
- electrocoagulation - the formation of colloidal aggregates of particles of the precipitated dispersed phase due to the process of anodic dissolution of the metal and the formation of metal cations Al 3+ , Fe 2+ , Fe 3+ under the influence of a constant electric field.
As a result of water treatment with direct electric current, at potentials equal to or exceeding the water decomposition potential (1.25 V), water passes into a metastable state, characterized by abnormal values of electron activity and other physico-chemical parameters (pH, Eh, ORP, electrical conductivity). The passage of a constant electric current through the volume of water is accompanied by electrochemical processes, as a result of which redox reactions occur, leading to the destruction (destruction) of water pollution, coagulation of colloids, flocculation of coarse impurities and their subsequent flotation.
The phenomenon of electrochemical activation of water is a combination of electrochemical and electrophysical effects on water in a double electric layer of electrodes during nonequilibrium charge transfer.
Electrochemical treatment is used for lightening and bleaching natural waters, their softening, purification from heavy metals (Cu, Co, Cd, Pb, Hg), chlorine, fluorine and their derivatives, for the treatment of wastewater containing petroleum products, organic and organochlorine compounds, dyes, surfactants, phenol. The advantages of electrochemical water purification is that it allows you to adjust the values pH pH and redox potential E h , on which the possibility of various chemical processes in water depends; increases the enzymatic activity of activated sludge in aeration tanks; reduces resistivity and improves the conditions for coagulation and sedimentation of organic sediments.
In 1985, EXHAV was officially recognized as a new class of physical and chemical phenomena. Order of the Government of the Russian Federation of January 15, 1998 No. VCh-P1201044 gave recommendations to ministries and departments to use this technology in medicine, agriculture, and industry.
water electrolysis
The main stage of electrochemical water treatment is water electrolysis. When a constant electric current is passed through water, the entry of electrons into the water at the cathode, as well as the removal of electrons from the water at the anode, is accompanied by a series of redox reactions on the surfaces of the cathode and anode. As a result, new substances are formed, the system of intermolecular interactions changes, the composition of water, including the structure of water. Typical installation for electrochemical water treatment consists of a water treatment unit 1, an electrolyzer 2, a water treatment unit after electrochemical treatment 3 (Fig. 1).
In some electrochemical water treatment plants, preliminary mechanical water purification is provided, which reduces the risk of clogging of the electrolytic cell with coarse impurities with a large hydraulic resistance. Block for mechanical cleaning water is necessary if, as a result of electrochemical treatment, water is saturated with coarse impurities, for example, flakes of metal hydroxides (Al (OH) 3, Fe (OH) 3, Mg (OH) 2) after electrocoagulation. The main element of the installation is an electrolyzer, consisting of one or more electrolysis cells (Fig. 2).
The electrolysis cell is formed by two electrodes - a positively charged anode and a negatively charged cathode, connected to different poles of a direct current source. The interelectrode space is filled with water, which is an electrolyte capable of conducting electric current. As a result of the operation of the device, there is a transfer of electric charges through a layer of water - electrophoresis, that is, the migration of polar particles, charge carriers - ions, to electrodes of the opposite sign.
When a constant electric current is passed through water, the entry of electrons into the water at the cathode, as well as the removal of electrons from the water at the anode, is accompanied by a series of redox reactions on the surfaces of the cathode and anode.
In this case, negatively charged anions move to the anode, and positively charged cations move to the cathode. At the electrodes, charged ions lose their charge, depolarize, turning into decay products. In addition to charged ions, polar particles of various dispersity participate in electrophoresis, including coarse particles (emulsified particles, gas bubbles, etc.), but charged ions with the highest mobility play the main role in the transfer of electrochemical charges. Polar particles include polar particles from among water impurities and water molecules, which is explained by their special structure.
The central oxygen atom, which is part of the water molecule, which has a greater electronegativity than hydrogen atoms, draws electrons towards itself, giving the molecule asymmetry. As a result, the electron density is redistributed: the water molecule is polarized, taking on the properties of an electric dipole having a dipole moment of 1.85 D (Debye), with positive and negative charges at the poles (Fig. 3).
The products of electrode reactions are neutralized aqueous impurities, gaseous hydrogen and oxygen formed during the electrolytic destruction of water molecules, metal cations (Al 3+, Fe 2+ , Fe 3+) in the case of using metal anodes made of aluminum and steel, molecular chlorine, etc. In this case, gaseous hydrogen is generated at the cathode, and oxygen is generated at the anode. The composition of water contains a certain amount of hydronium ion H 3 O +, which depolarizes on the cathode surface with the formation of atomic hydrogen H:
H 3 O + + e - → H + H 2 O.
In an alkaline environment, H 3 O + is absent, but water molecules are destroyed, accompanied by the formation of atomic hydrogen H - and hydroxide OH -:
H 2 O + e - → H + OH -.
Reactive hydrogen atoms are adsorbed on the surfaces of cathodes and, after recombination, form molecular hydrogen H 2 , which is released from water in gaseous form:
H + H → H 2.
At the same time, atomic oxygen is released at the anodes. In an acidic environment, this process is accompanied by the destruction of water molecules:
2H 2 O - 4e - → O 2 + 4H +.
In an alkaline environment, OH hydroxide ions always serve as a source of oxygen formation, moving under the action of electrophoresis on the electrodes, from the cathode to the anode:
4 OH - → O 2 + 2 H 2 O + 4 e -.
The normal redox potentials of these reactions are +1.23 and +0.403 V, respectively, but the process proceeds under conditions of some
overvoltage. The electrolysis cell can be considered as a generator of the above products, some of which, entering into chemical interaction with each other and with water pollution in the interelectrode space, provide additional chemical water purification (electroflotation, electrocoagulation). These secondary processes do not occur on the surface of the electrodes, but in the volume of water. Therefore, unlike electrode processes, they are designated as volumetric. They are initiated by an increase in water temperature during electrolysis and an increase in pH during cathodic destruction of water molecules.
Distinguish between cathodic and anodic oxidation. During the cathodic oxidation of the molecule organic matter, being sorbed on cathodes, accept free electrons, are restored, transforming into compounds that are not contaminants. In some cases, the recovery process takes place in one stage:
R + H + + e - → RH, where R is an organic compound; RH is the hydrated form of the compound and is not a contaminant.
In other cases, cathodic reduction takes place in two stages: at the first stage (I), the organic molecule is converted into an anion, at the second (II), the anion is hydrated by interacting with a water proton:
R + e - → R - , (I) R - + H + → RH. (II)
Distinguish between cathodic and anodic oxidation. During cathodic oxidation, molecules of organic substances, being sorbed on cathodes, accept free electrons and are reduced.
Cathodes made of materials that require a high overvoltage (lead, cadmium) allow, at a high cost of electricity, to destroy organic molecules and generate reactive free radicals - particles that have free unpaired electrons (Cl*, O*, OH*) in the outer orbits of atoms or molecules , NO*2, etc.). The latter circumstance gives free radicals the property of reactivity, that is, the ability to enter into chemical reactions with aqueous impurities and destroy them.
RH → R + H + + e - .
Anodic oxidation of organic compounds often leads to the formation of free radicals, the further transformations of which are determined by their reactivity. The processes of anodic oxidation are multistage and proceed with the formation of intermediate products. Anodic oxidation reduces the chemical stability of organic compounds and facilitates their subsequent degradation during bulk processes.
In volumetric oxidative processes, a special role is played by the products of water electrolysis - oxygen (O 2), hydrogen peroxide (H 2 O 2) and oxygen-containing chlorine compounds (HClO). In the process of electrolysis, an extremely reactive compound is formed - H 2 O 2, the formation of molecules of which occurs due to hydroxyl radicals (OH *), which are the products of discharge of hydroxyl ions (OH-) at the anode:
2OH - → 2OH* → H 2 O 2 + 2e -, where OH* is a hydroxyl radical.
The reactions of interaction of organic substances with oxidizing agents proceed for a certain period of time, the duration of which depends on the value of the redox potential of the element and the concentration of the reacting substances. As the concentration of the pollutant decreases and the concentration of the pollutant decreases, the oxidation process decreases.
The rate of the oxidation process during electrochemical treatment depends on the temperature of the treated water and pH. In the process of oxidation of organic compounds, intermediate products are formed that differ from the initial one both in resistance to further transformations and in toxicity indicators.
The sources of active chlorine and its oxygen-containing compounds generated in the electrolyzer are chlorides in the treated water and sodium chloride (NaCl), which is introduced into the treated water before electrolysis. As a result of anodic oxidation of Cl– anions, gaseous chlorine Cl 2 is generated. Depending on the pH of the water, it either hydrolyzes to form hypochlorous acid HOCl or forms hypochlorite ions ClO - . The equilibrium of the reaction depends on the pH value.
At pH = 4-5, all chlorine is in the form of hypochlorous acid (HClO), and at pH = 7, half of the chlorine is in the form of hypochlorite ion (OCl -) and half is in the form of hypochlorous acid (HClO) (Fig. 4). The mechanism of interaction of hypochlorite ion (ClO -) with the oxidized substance is described by the following equation:
ClO - + A = C + Cl, where A is the oxidizable substance; C is an oxidation product.
The electrochemical oxidation of organic compounds with hypochlorithione (ClO -) is accompanied by an increase in the redox potential Eh, which indicates the predominance of oxidative processes. The growth of Eh depends on the ratio of the concentration of active chlorine in the interelectrode space to the content of organic impurities in water. As cleaning and reducing the amount of pollution, this ratio increases, which leads to an increase in Eh, but then this indicator stabilizes.
The amount of the substance that reacted on the electrodes when passing a direct electric current according to Faraday's law is directly proportional to the current strength and processing time:
G = AI cur τ, (1)
where A is the electrochemical equivalent of the element, g/(A⋅h); I cur - current strength, A; τ is the processing time, h. The electrochemical equivalent of the element is determined by the formula:
A = M / 26.8z , (2)
where M is the atomic mass of the element, g; z is its valency. The values of the electrochemical equivalents of some elements are given in Table. one.
The actual amount of the substance generated during electrolysis is less than the theoretical one, calculated by formula (1), since part of the electricity is spent on heating water and electrodes. Therefore, the calculations take into account the current utilization factor η< 1, величина которого определяется экспериментально.
During electrode processes, there is an exchange of charged particles and ions between the electrode and the electrolyte - water. To do this, under steady-state equilibrium conditions, it is necessary to create electrical potential, minimum value which depends on the type of redox reaction and on the water temperature at 25 °C (Table 2).
The main parameters of water electrolysis include current strength and density, voltage within the electrode cell, as well as the speed and duration of water stay between the electrodes.
The voltages generated in the electrode cell must be sufficient for the occurrence of redox reactions on the electrodes. The voltage value depends on the ionic composition of water, the presence of impurities in water, for example, surfactants, current density (its strength per unit area of the electrode), electrode material, etc. Other things being equal, the task of choosing an electrode material is to ensure that for the passage of oxidative reductive reactions on the electrodes, the required voltage was kept to a minimum as this reduces costs electrical energy.
Some redox reactions are competing - they proceed simultaneously and mutually inhibit each other. Their flow can be controlled by changing the voltage in the electrolytic cell. So, the normal potential of the reaction of formation of molecular oxygen is +0.401 V or +1.23 V; with an increase in voltage to +1.36 V (normal potential for the reaction of molecular chlorine formation), only oxygen will be released at the anode, and with a further increase in the potential, both oxygen and chlorine will be released simultaneously, and chlorine will be released with insufficient intensity. At a voltage of about 4-5 V, the evolution of oxygen will practically cease, and the electrolytic cell will only generate chlorine.
Calculation of the main parameters of the water electrolysis process
The main parameters of water electrolysis include current strength and density, voltage within the electrode cell, as well as the speed and duration of water stay in the interelectrode space.
The current strength I cur is a value determined depending on the required performance for the generated product [A], is determined by the formula:
Icur=G/A tη , (3)
This formula is obtained by transforming formula (1) taking into account the current utilization factor η. The current density is its strength, related to the unit area of the electrode [A / m 2], for example, the anode, is determined from the following expression:
i en = I cur / F en, (4)
where F an is the anode area, m 2 . The current density has the most decisive influence on the electrolysis process: that is, with an increase in current density, electrode processes are intensified and the surface area of the electrodes decreases, but at the same time, the voltage in the electrolysis cell increases and, as a result, the entire energy intensity of the process. An increased increase in current density intensifies the release of electrolysis gases, leading to bubbling and dispersion of insoluble products of electrical water treatment.
With an increase in current density, the passivation of the electrodes also increases, which consists in blocking the incoming electrons by surface deposits of the anode and cathode, which increases the electrical resistance in the electrode cells and inhibits the redox reactions occurring on the electrodes.
Anodes are passivated as a result of the formation of thin oxide films on their surfaces, as a result of sorption of oxygen and other components on the anodes, which, in turn, sorb particles of aqueous impurities. On the cathodes, mainly carbonate deposits are formed, especially in the case of water treatment with increased hardness. For these reasons, the current density during the electrolysis of water should be set to the minimum under the conditions for the stable occurrence of the necessary redox reactions in the course of the technological process.
The residence time of water in the interelectrode space of the electrolyzer is limited by the time required to generate the right amount electrolysis products.
The voltage in the electrode cell [V] is determined by the formula:
V i = i en ΔK g / χ R , (5)
where i an is the current density, A/m 2 ; D is the distance between the electrodes (width of the interelectrode channel), m; χ R is the electrical conductivity of water, 1/(Ohm⋅m); K g - coefficient of gas filling of the interelectrode space, usually taken K g \u003d 1.05-1.2.
Formula (5) does not take into account electrical resistance electrode due to their low values, but with passivation, these resistances turn out to be significant. The width of the interelectrode channel is assumed to be minimal (3-20 mm) according to the conditions of non-clogging by impurities.
The specific electrical conductivity of water χ R depends on a number of factors, among which the most significant are temperature, pH, ionic composition and ion concentration (Fig. 5). With an increase in temperature, the electrical conductivity χ R increases, and the voltage decreases (Fig. 6). The minimum value of electrical conductivity corresponds to pH = 7. In addition, during the electrolysis process, the temperature and pH of water increase. If pH > 7, then we can expect a decrease in the specific electrical conductivity of water χ R , and at pH values< 7 удельная электропроводность воды χ R , наоборот, возрастает (рис. 5).
The specific electrical conductivity of natural waters of medium mineralization is 0.001-0.005 1 / (Ohm⋅m), urban wastewater is 10-0.01 1 / (Ohm⋅m) . During electrolysis, the electrical conductivity should be in the range of 0.1-1.0 1 / (Ohm⋅m) . If the source water has insufficient electrical conductivity, the salinity should be increased (Fig. 7). Usually, sodium chloride (NaCl) is used for this, the doses of which are determined experimentally and are most often 500-1500 mg / l (8-25 meq / l). Sodium chloride is not only convenient in terms of application and safety (storage, solution preparation, etc.), but in the presence of NaCl, electrode passivation slows down. Dissociating in water, NaCl saturates the water with chlorine anions Cl - and sodium cations Na + . Chlorine ions Cl - have small size and penetrating through passivating deposits to the anode surface destroy these deposits. In the presence of other anions, especially sulfate ions (SO 2- 4 ), the depassivating effect of chloride ions (Cl -) is reduced. Stable operation of the cell is possible if the ions - Cl - make up at least 30% of the total number of anions. As a result of electrophoresis, sodium cations Na + move to the cathodes, on which hydroxide ions OH - are generated, and, interacting with the latter, form sodium hydroxide (NaOH), which dissolves carbonate deposits on the cathodes.
The power consumption [W] of the electrolytic cell is determined by the following relationship:
N consumption = η e I cur V e, (6)
where η e is the efficiency of the cell, it is usually taken η e = 0.7-0.8; I cur - current strength, A; V e - voltage on the electrolyzer, V.
The residence time of water in the interelectrode space of the cell is limited by the time required to generate the required amount of electrolysis products, as well as the duration of the corresponding volumetric reactions, and is determined experimentally.
The speed of water movement in the interelectrode space is set taking into account the conditions for the removal of electrolysis products and other impurities from the electrolyzer; in addition, turbulent mixing depends on the speed of water movement, which affects the course of volumetric reactions. Like the residence time of the water, the speed of the water is chosen based on experimental data.
To be continued.
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Electrolysis is a set of redox processes that occur when a direct electric current is passed through an electrolyte solution or melt with electrodes immersed in it.
The device in which electrolysis is carried out is called an electrolyzer.
The electrode on which the oxidation processes take place is called the anode. In the electrolyzer, it is positively charged (connected to the positive pole of an external DC source).
The electrode on which the reduction processes take place is called the cathode. In the electrolyzer, it is negatively charged (connected to the negative pole of an external DC source).
When a voltage is applied, cations (positively charged particles) move towards the cathode, anions (negatively charged particles) move towards the anode, and there they are discharged. At the anode, ions donate electrons, and they are oxidized. At the cathode, ions accept electrons, they are reduced.
Only cations and anions of the electrolyte are not always involved in electrode processes; they are competed with solvent molecules, in particular water, if an aqueous solution is electrolyzed.
In addition, the participation of water in electrochemical processes during electrolysis can lead to another result. The free radicals OH formed as intermediates ( due to the oxidation of hydroxide ions at the anode) and H ( due to the reduction of hydrogen ions at the cathode) have a high reactivity and strongly pronounced oxidizing and reducing properties, respectively. At the electrode surface, they are able to be involved in interaction with substances dissolved in water. In such cases, one speaks of oxidation in the anode space and reduction in the cathode space.
Features of the flow of electrochemical processes in aqueous solutions are due to the ability of water molecules to undergo both oxidation (at the anode) and reduction (at the cathode).
Anode (+) pH=0 pH=7 pH=14
2H 2 O - 4e \u003d 2O + 2H + 4OH - - 4e \u003d 4OH 4OH - - 4e \u003d 4OH
2O \u003d O 2 4OH \u003d O 2 + 2H 2 O 4OH \u003d O 2 + 2H 2 O
2H 2 O - 4e \u003d 2O + 2H +
Cathode (–) pH=0 pH=7 pH=14
2H + + 2e = 2H 2H 2 O + 2e = 2H + 2OH - 2H 2 O + 2e = 2H + 2OH -
2H \u003d H 2 or 2H \u003d H 2
Distinguish primary and secondary electrode processes. The primary ones are electrochemical in nature, the secondary ones are non-electrochemical. As a result of electrolysis, the corresponding reduction and oxidation products (primary processes) are released on the electrodes (cathode and anode), which, depending on the conditions, can react with the solvent, the electrode material, with each other (recombination of atoms), etc. (secondary processes). In some cases, it is impossible to unambiguously separate the primary and secondary processes. In the example above, the free radicals OH (at the anode) and H (at the cathode) were formed as a result of primary processes, and the oxidation of manganate ions and the reduction of nitric acid were secondary processes. Let's consider one more example.
In some cases, side reactions are superimposed on the course of the main processes during electrolysis: the interaction between the electrolysis products or the reactions of products with water. To prevent secondary reactions between electrolysis products, diaphragms (partitions between the anode and cathode) are used to prevent the diffusion of certain ions. For example, in the above example with the electrolysis of a solution of sodium chloride, to prevent interaction between chlorine and hydroxide ions, the cathode is surrounded by a diaphragm that prevents the diffusion of sodium and chlorine ions. As a result, alkali (NaOH) is concentrated in the cathode space. Therefore, in most cases, one should expect a slight difference in the composition of the products during the electrolysis of the same solution with and without a diaphragm.
E dec \u003d E A - E K
For each electrolyte there is a specific minimum value voltage (from an external current source), which must be applied to the electrodes for electrolysis to proceed. It is called the expansion voltage (E decom).
The decomposition voltage is the difference between the electrode potentials of the anode and cathode processes.
E dec \u003d E A - E K
At the cathode, first of all, the ions or molecules that are part of the redox system with the most positive potential are reduced (which are the reduced form in redox systems with the most positive potential).
1) If subjected to electrolysis melt, containing several different cations metals, then in this case the reduction sequence is determined by the electrode potentials of the metals under the given conditions ( in this melt!). In this case, first of all, cations of metals with great value electrode potential (from the end of the series of voltages for a given melt).
2) Recovery processes at the cathode in aqueous solutions:
Metal cations located in a series of voltages after hydrogen (with a standard electrode potential greater than that of hydrogen): Cu 2+, Hg 2 2+, Ag +, Hg 2+, Pt 2+ ... Pt 4+. During electrolysis, they are almost completely reduced at the cathode and are released in the form of a metal.
· metal cations located at the beginning of the row (with a standard electrode potential less than that of aluminum): Li + , Na + , K + ... Al 3+ . During electrolysis, they are not restored, water molecules are restored instead.
Metal cations located in a row after aluminum and before hydrogen (with a standard electrode potential greater than that of aluminum, but less than that of hydrogen): Mn 2+, Zn 2+, Cr 3+, Fe 2+ ... H During electrolysis, these cations are reduced at the cathode simultaneously with water molecules.
3) If a gradually increasing voltage is applied to a solution containing several cations, then electrolysis begins when the decomposition potential of the cation with the most positive potential is reached. So, during the electrolysis of a solution containing Cu 2+ ions (Е 0 Cu 2+ / Cu = 0.35 V) and Zn 2+ (Е 0 Zn 2+ / Zn = – 0.76 V), copper is first released on the cathode, and only after after almost all the copper ions are discharged, zinc will begin to be released.
It would seem that according to the values of electrode potentials, in aqueous solution it would be possible to precipitate only metals standing in the series of stresses after hydrogen. However, due to hydrogen overvoltage, many metals can be deposited from aqueous solutions, which, according to the values of their standard potentials, should not be deposited (for example, Zn). In addition, the nature of the medium (acidic, neutral, alkaline) affects the nature of the discharged metal. This is due to the fact that - as shown above, the electrode potential depends on the reaction of the medium.
The formation of an insoluble substance as a result chemical reaction- this is only one of the conditions for obtaining a colloidal solution. Another equally important condition is the inequality of the starting materials taken into the reaction. The consequence of this inequality is the limitation of the growth of the size of particles in colloidal solutions, which would lead to the formation of a coarsely dispersed system.
Let us consider the mechanism of formation of a colloidal particle using the example of the formation of a silver iodide sol, which is obtained by the interaction of dilute solutions of silver nitrate and potassium iodide.
AgNO 3 + KI \u003d AgI + KNO 3
Ag + + NO 3 ¯ + K + + I ¯ = AgI ↓ + NO 3 ¯ + K +
Insoluble neutral molecules of silver iodide form the core of a colloidal particle.
At first, these molecules combine in disorder, forming an amorphous, loose structure, which gradually turns into a highly ordered crystalline structure of the core. In the example we are considering, the core is a silver iodide crystal, consisting of a large number (m) of AgI molecules:
m - the core of the colloidal particle
An adsorption process takes place on the surface of the core. According to the Peskov-Fajans rule, ions that are part of the particle core are adsorbed on the surface of the nuclei of colloidal particles, i.e. silver ions (Ag +) or iodine ions (I -) are adsorbed. Of these two types of ions, those that are in excess are adsorbed.
So, if a colloidal solution is obtained in an excess of potassium iodide, then iodine ions will be adsorbed on particles (nuclei), which complete the crystal lattice of the nucleus, naturally and firmly entering its structure. In this case, an adsorption layer is formed, which gives the nucleus a negative charge:
Ions adsorbed on the surface of the nucleus, giving it an appropriate charge, are called potential-forming ions.
At the same time, oppositely charged ions are also in solution, they are called counterions. In our case, these are potassium ions (K +), which are electrostatically attracted to the charged nucleus (the charge value can reach I c). Part of the K + counterions is strongly bound by electric and adsorption forces and enters the adsorption layer. A core with a double adsorption layer of ions formed on it is called a granule.
(m . nI - . (n-x) K + ) x - (granule structure)
The remaining part of the counterions (let us denote them by the number "x K +") forms a diffuse layer of ions.
The core with adsorption and diffusion layers is called a micelle. :
(m . nI -. (n-x) K + ) x - . x K + (micelle structure)
When a constant electric current is passed through a colloidal solution, the granules and counterions will move towards the oppositely charged electrodes, respectively.
The presence of the same charge on the surface of sol particles is important. factor in its sustainability. The charge prevents sticking and enlargement of particles. In a stable disperse system, particles are kept in suspension, i.e. no precipitation of the colloidal substance occurs. This property of sols is called kineti chesky stability.
The structure of micelles of the silver iodide sol obtained in excess of AgNO 3 is shown in Fig. 3. 1a, in excess of KCI - 1b .
Fig.1.5. The structure of micelles of silver iodide sol obtained in excess:
a) silver nitrate; b) potassium chloride.
3.4.2 Electrochemical production
Electrolysis is a redox reaction that occurs when a constant electric current is passed through a melt or electrolyte solution.
The essence of electrolysis is as follows: when an electric current is passed through a melt or electrolyte solution, positive electrolyte ions (metal or hydrogen ions) are attracted to the cathode, and negative ions (acid residues or hydroxyl groups) are attracted to the anode. The electrons brought to the cathode from the current source are attached to the positive ions of the electrolyte, restoring them. At the same time, the negative ions of the electrolyte give up their electrons to the anode, from which they move to the current source. Losing their electrons, they are oxidized into neutral atoms or groups of atoms. Thus, the reduction process takes place at the cathode, and the oxidation process occurs at the anode.
A (+): nA n - - ne - → nA p -
K (-): nB n + + ne - → nB p +
Both processes form a single redox reaction. But unlike conventional redox reactions, electrons from the reducing agent to the oxidizing agent do not pass directly, but through an electric current. The cathode, which brings electrons, is a reducing agent, and the anode, which carries them away, is an oxidizing agent.
The main indicators of electrochemical production are the current output, the degree of energy use. Consumption coefficient for energy, voltage applied to the electrolyzer, etc. Most of the calculations are based on Faraday's law, according to which the mass of a substance released during electrolysis is proportional to the current strength I, the electrolysis time t and the electrochemical equivalent of this substance E E
The mass of a substance is calculated by the formula
where, I - current strength, F - Faraday's constant (96500 C)
(g-eq) (1.3.2)
Mr - relative molecular mass substances
n is the charge of the ion (absolute value) in the form of which the substance is in solution or in the melt (ie, the number of given or received electrons).
The current output is determined by the ratio of the mass of the substance released during electrolysis to the mass of the substance that should theoretically be released according to Faraday's law, and is expressed as a percentage:
(1.3.3)
The mass m theor is found by the formula
The energy output is determined by the equation
where, E theor and E pr are the theoretical and practical voltage of decomposition during electrolysis, respectively, V; η - energy output,%.
The energy output can also be calculated from the amount of energy expended:
(1.3.6)
where w theor and w pr - the amount of energy theoretically necessary and practically spent to obtain a unit of product.
(1.3.7)
where 1000 is the conversion factor W*h to kW*h;
1*10 -6 is the number used to convert grams to tons.
The theoretical power consumption is in relation
(1.3.8)
where φ decom is the decomposition voltage.
Examples of problem solving
1. What processes occur during the electrolysis of sodium hydroxide melt?
Caustic soda melt contains Na + and OH ions. Oxidized at the anode, OH ions in the next stage decompose with the formation of water and oxygen. The process can be depicted as follows:
K(-): 2Na + + 2e - = 2Na;
A (+): 2OH - 2e - \u003d H 2 O + O 2
Two oxygen atoms, connecting with each other, form an oxygen molecule O 2. So the overall equation
4NaOH \u003d 4Na + 2H 2 O + O 2
During the electrolysis of melts of salts of oxygen acids, the oxidized ions of acidic residues immediately decompose into oxygen and the corresponding oxides.
Electrolysis proceeds in a peculiar way in an aqueous solution. The fact is that water itself is an electrolyte, although it is very weak. Thus, an aqueous solution actually contains two electrolytes - a solvent and a solute, and, accordingly, two types of both positive and negative ions. Which of them will be discharged depends on a number of conditions. As a rule, you can follow the following. If the positive electrolyte ions are ions of very active metals, such as Na + or K -, then during electrolysis, it is not the ions of these metals that are discharged, but hydrogen ions from water with the release of free hydrogen and the release of hydroxide ions, which can be expressed by the following electron-ionic equation :
2H + OH + 2e - \u003d H 2 + 2OH
If the negative ions of the electrolyte are the acidic residues of oxygenic acids, then during electrolysis, it is not the acidic residues of these acids that are discharged, but OH ions from water with the release of oxygen, which can be expressed by the equation:
4H 2 O - 4e - \u003d 4H + + 4OH
4OH - 2H 2 O + O 2
Adding both equations, we get:
2H 2 O - 4e - \u003d 4H + + O 2
2. Determine the current output (in%) if 4200 l of electrolytic alkali with a NaOH concentration of 125 kg/m
According to equation (1.3.4), the mass of sodium hydroxide theoretically should have been
practically received
Therefore, the current output according to the formula (1.3.3) will be equal to
Answer: 94.6% current output.
3. Determine the actual power consumption (in kilowatt-hours) for obtaining chlorine with a mass of 1 ton and the energy output (in%), if the average voltage on the cell is 3.35V, the current output is 96%, and the electrochemical equivalent of chlorine is 1.323 g / A * h.
Using formula (1.3.7), we determine the actual energy consumption
If we take the current efficiency as 100%, then at a theoretical NaCl decomposition voltage of 2.17V, the theoretical energy consumption per 1 ton of chlorine will be
In this case, the energy efficiency
Answer: energy efficiency 62.2%; 2637 kWh
Tasks for independent solution
1. One of the methods of industrial production of calcium is the electrolysis of molten calcium chloride. What mass of metal will be obtained if it is known that as a result of electrolysis, chlorine with a volume of 896 l (n.o.) was released?
2. During the electrolysis of a solution of sodium chloride in electrolysis, which worked for 24 hours at a current of 30,000 A, 8.5 m 3 of electrolytic alkali with a NaOH concentration of 120 kg / m 3 were obtained. calculate the current efficiency (for alkali)
3. Determine the current strength required to produce 100% sodium hydroxide weighing 1720 kg per day in an electrolytic cell with an iron cat during its continuous operation, if the current efficiency is 96%
4. Calculate the mass of chlorine produced per year by a plant that has 5 series of 150 electrolyzers with iron cathodes for continuous operation for 350 days, a current of 34,000 A and a current output of 95%. Determine the power of the alternator of the power plant, which provides the needs of the plant for electrical energy at a bottom series voltage of 550 V, if the efficiency of the rectifier is 95%.
5. Calculate the theoretical and practical power consumption per 1 ton of 100% NaOH for the electrolysis of sodium chloride solution with a mercury cathode. The theoretical decomposition voltage is 3.168 V. Determine the energy efficiency if the practical decomposition voltage is 4.4 V and the current efficiency is 92.5%.
6. What substances and in what quantity are released on carbon electrodes if the composition of the solution is 0.1 mol HgCl 2 and 0.2 mol CuCl 2 and a current of 10 A is passed through it for 1 hour?
7. When an electric current passed through a dilute solution of sulfuric acid for 10 minutes, 100 ml of hydrogen were released at 18C and pressure
755 mmHg Art. Calculate the current strength.
8. In the electrolytic production of magnesium, molten magnesium chloride can serve as an electrolyte. Calculate the current output if 72.6 kg of magnesium were released in a bath operating at a current of 40,000 A for 5 hours.
9. Determine the amount of electricity required to release 1 m 3 of hydrogen and 0.5 m 3 of oxygen, obtained by electrolysis of water. The theoretical voltage of water is 1.23 V, and the actual voltage exceeds it by 1.5 - 2 times. Calculate the actual consumption of electrical energy.
10. During the electrolysis of a solution containing 2.895 g of a mixture of FeCl 2 and FeCl 3, 1.12 g of metal was released on the cathode. Calculate mass fraction each of the components of the initial mixture, if the electrolysis was carried out until the complete precipitation of iron.
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