Electrolytic production of calcium and its alloys. Physico-chemical properties of the electrolyte

Electrolysis is a redox reaction that occurs on electrodes if a constant electric current is passed through the melt or electrolyte solution.

The cathode is a reducing agent that donates electrons to cations.

The anode is an oxidizer that accepts electrons from anions.

Activity series of cations:	Na + , Mg 2+ , Al 3+ , Zn 2+ , Ni 2+ , Sn 2+ , Pb 2+ , H+ , Cu 2+ , Ag + _____________________________→ Strengthening the oxidizing power
Anion activity series:	I - , Br - , Cl - , OH - , NO 3 - , CO 3 2- , SO 4 2- ←__________________________________ Increasing recovery ability

Processes occurring on electrodes during the electrolysis of melts

(do not depend on the material of the electrodes and the nature of the ions).

1. Anions are discharged at the anode ( A m - ; oh-

A m - - m ē → A °; 4 OH - - 4ē → O 2 + 2 H 2 O (oxidation processes).

2. Cations are discharged at the cathode ( Me n + , H + ), turning into neutral atoms or molecules:

Me n + + n ē → Me ° ; 2 H + + 2ē → H 2 0 (recovery processes).

Processes occurring on the electrodes during the electrolysis of solutions

CATHODE (-) Do not depend on the cathode material; depend on the position of the metal in a series of stresses	ANOD (+) Depend on the anode material and the nature of the anions.
	The anode is insoluble (inert), i.e. made from coal, graphite, platinum, gold.	The anode is soluble (active), i.e. made fromCu, Ag, Zn, Ni, Feand other metals (exceptPt, Au)
1. First of all, metal cations are restored, standing in a series of voltages afterH 2 : Me n+ +nē → Me°	1. First of all, anions of oxygen-free acids are oxidized (exceptF - ): A m- - mē → A°	Anions are not oxidized. Anode metal atoms are oxidized: Me° - nē → Me n+ Cations Me n + go into solution. The mass of the anode is reduced.
2. Metal cations of medium activity, standing betweenAl And H 2 , are restored simultaneously with water: Me n+ + nē →Me° 2H 2 O + 2ē → H 2 + 2OH -	2. Anions of oxo acids (SO 4 2- , CO 3 2- ,..) And F - do not oxidize, molecules are oxidizedH 2 O : 2H 2 O - 4ē → O 2 + 4H +
3.Cations of active metals fromLi before Al (inclusive) are not restored, but molecules are restoredH 2 O : 2 H 2 O + 2ē → H 2 + 2OH -	3. During the electrolysis of alkali solutions, ions are oxidizedoh- : 4OH - - 4ē → O 2 +2H 2 O
4. During the electrolysis of acid solutions, cations are reduced H+: 2H + + 2ē → H 2 0

ELECTROLYSIS OF MELTS

Exercise 1. Make a diagram of the electrolysis of sodium bromide melt. (Algorithm 1.)

Sequencing	Taking Actions
	NaBr → Na + + Br -
	K - (cathode): Na +, A + (anode): Br -
	K + : Na + + 1ē → Na 0 (recovery), A +: 2 Br - - 2ē → Br 2 0 (oxidation).
	2NaBr \u003d 2Na +Br 2

Task 2. Make a diagram of the electrolysis of sodium hydroxide melt. (Algorithm 2.)

Sequencing	Taking Actions
	NaOH → Na + + OH -
2. Show the movement of ions to the corresponding electrodes	K - (cathode): Na +, A + (anode): OH -.
3. Draw up schemes of oxidation and reduction processes	K - : Na + + 1ē → Na 0 (recovery), A +: 4 OH - - 4ē → 2 H 2 O + O 2 (oxidation).
4. Make an equation for the electrolysis of an alkali melt	4NaOH \u003d 4Na + 2H 2 O + O 2

Task 3.Make a diagram of the electrolysis of a melt of sodium sulfate. (Algorithm 3.)

Sequencing	Taking Actions
1. Compose the salt dissociation equation	Na 2 SO 4 → 2Na + + SO 4 2-
2. Show the movement of ions to the corresponding electrodes	K - (cathode): Na + A + (anode): SO 4 2-
	K -: Na + + 1ē → Na 0, A +: 2SO 4 2- - 4ē → 2SO 3 + O 2
4. Make an equation for the electrolysis of molten salt	2Na 2 SO 4 \u003d 4Na + 2SO 3 + O 2

SOLUTION ELECTROLYSIS

Exercise 1.Draw up a scheme for the electrolysis of an aqueous solution of sodium chloride using inert electrodes. (Algorithm 1.)

Sequencing	Taking Actions
1. Compose the salt dissociation equation	NaCl → Na + + Cl -
	Sodium ions in the solution are not restored, so water is being restored. Chlorine ions are oxidized.
3. Draw up diagrams of the processes of reduction and oxidation	K -: 2H 2 O + 2ē → H 2 + 2OH - A +: 2Cl - - 2ē → Cl 2
	2NaCl + 2H 2 O \u003d H 2 + Cl 2 + 2NaOH

Task 2.Draw a scheme for the electrolysis of an aqueous solution of copper sulfate ( II ) using inert electrodes. (Algorithm 2.)

Sequencing	Taking Actions
1. Compose the salt dissociation equation	CuSO 4 → Cu 2+ + SO 4 2-
2. Select the ions that will be discharged at the electrodes	Copper ions are reduced at the cathode. At the anode in an aqueous solution, sulfate ions are not oxidized, so water is oxidized.
3. Draw up diagrams of the processes of reduction and oxidation	K - : Cu 2+ + 2ē → Cu 0 A + : 2H 2 O - 4ē → O 2 +4H +
4. Make an equation for the electrolysis of an aqueous salt solution	2CuSO 4 + 2H 2 O \u003d 2Cu + O 2 + 2H 2 SO 4

Task 3.Draw up a scheme for the electrolysis of an aqueous solution of an aqueous solution of sodium hydroxide using inert electrodes. (Algorithm 3.)

Sequencing	Taking Actions
1. Make an equation for the dissociation of alkali	NaOH → Na + + OH -
2. Select the ions that will be discharged at the electrodes	Sodium ions cannot be reduced, so water is reduced at the cathode. Hydroxide ions are oxidized at the anode.
3. Draw up diagrams of the processes of reduction and oxidation	K -: 2 H 2 O + 2ē → H 2 + 2 OH - A +: 4 OH - - 4ē → 2 H 2 O + O 2
4. Make an equation for the electrolysis of an aqueous solution of alkali	2 H 2 O \u003d 2 H 2 + O 2 , i.e. electrolysis of an aqueous solution of alkali is reduced to the electrolysis of water.

Remember.In the electrolysis of oxygen-containing acids (H 2 SO 4 etc.), bases (NaOH, Ca (OH) 2 etc.) , salts of active metals and oxygen-containing acids(K 2 SO 4 etc.) electrolysis of water occurs on the electrodes: 2 H 2 O \u003d 2 H 2 + O 2

Task 4.Draw up a scheme for the electrolysis of an aqueous solution of silver nitrate using an anode made of silver, i.e. the anode is soluble. (Algorithm 4.)

Sequencing	Taking Actions
1. Compose the salt dissociation equation	AgNO 3 → Ag + + NO 3 -
2. Select the ions that will be discharged at the electrodes	Silver ions are reduced at the cathode, and the silver anode is dissolved.
3. Draw up diagrams of the processes of reduction and oxidation	K-: Ag + + 1ē→ Ag 0 ; A+: Ag 0 - 1ē→ Ag +
4. Make an equation for the electrolysis of an aqueous salt solution	Ag + + Ag 0 = Ag 0 + Ag + electrolysis is reduced to the transfer of silver from the anode to the cathode.

^ CHAPTER 9. ELECTROCHEMICAL PRODUCTION

9.1 Theoretical foundations of industrial electrolysis

The technology of electrochemical production considers processes in which the main reactions take place in an environment of direct conversion of electrical energy into chemical energy, without intermediate conversion of energy into heat.

For this, special technological methods and equipment have been created, based on theoretical electrochemistry and differing from methods in other areas of chemical technology. In electrolysis, the desired reactions can be carried out, as a rule, with a high degree of selectivity, which makes it possible to obtain a product with relatively small impurities. The degree of useful use of electricity in electrolysis is relatively large.

Technological processes that can be carried out by electrochemical methods can in most cases also be carried out by other chemical methods.

The choice of technology should be made on the basis of a comparative techno-economic analysis, which takes into account the economics of production, the resources of the necessary raw materials, the complexity of hardware design and other issues.

The techno-economic advantages of electrochemical methods are determined by the fact that they can be used to obtain sufficiently pure products in relatively simple technological schemes. The disadvantages are associated with the need to spend an expensive type of energy (DC energy) and incur costs for the creation of sources for its production.

Electrochemical methods have found application for the production of hydrogen and oxygen, chlorine, sodium and potassium hydroxide, oxygen compounds of chlorine, for the electrosynthesis of inorganic substances, and also for the synthesis of organic substances.

An electrochemical method is used to produce hydroelectrometallurgical metals such as copper, nickel, zinc, cobalt, cadmium, manganese, chromium, iron, silver, gold, etc., as well as metal powders. Using the electrolysis of molten media, it is possible to obtain aluminum, magnesium, alkali and alkaline earth metals (sodium, calcium), beryllium, rare and rare earth metals, as well as elemental fluorine.

In electroplating, electrochemical methods are used for copper plating, nickel plating, chromium plating and other coatings, in mechanical engineering - for anodic mechanical processing of products (drilling, cutting, electropolishing, precise complex-shaped processing, etc.).

On the basis of electrochemical processes, chemical current sources, such as batteries and galvanic cells, have been created.

Electrochemical reactions take place in apparatuses called electrolyzers. In them, through electrolytes (solutions or melts - conductors of the second kind), a direct current passes from the anode to the cathode. Oxidation reactions take place at the anode, and reduction reactions take place at the cathode.

According to Faraday's laws, the amount of substance released on the electrodes is proportional to the amount of electricity passed. Several reactions can proceed in parallel on each electrode. The share of the total amount of electricity passed, spent on a given reaction, is its current output.

practically important current output for the main reaction, which characterizes the perfection of the process. The rate of a reaction in electrochemistry is current density - the amount of electricity that has passed per unit of time through the unit surface of the electrode at its border with the electrolyte.

In practice, the current density is determined by dividing its strength by the geometric area of ​​the electrode. Distinguish estimated And true current density, which is determined not by the geometric, but by the actual surface of the electrode. The latter depends on the porosity and topography of the surface (the presence of bulges and depressions) and practically cannot be determined. Obviously, the more developed the electrolyte surface, the lower the true current density and the more it differs from the calculated one.

In industrial plants, electrochemical reactions are mainly carried out by reactions that require the consumption of electrical energy. These costs are characterized by a potential jump that occurs at the electrode-electrolyte interface. If the electrode reaction takes place under reversible equilibrium conditions (with a current strength approaching zero), then the potential jump between the electrode and the electrolyte is called equilibrium potential. The equilibrium potential gives the value of the potential jump needed to start the reaction.

important concept "standard potential". This is the equilibrium potential, defined for the case when the activity of each active substance is equal to one. Standard potentials are given in reference tables. Taking into account real conditions and using the Nernst formula, equilibrium potentials can be calculated from them.

The sum of the equilibrium potentials at the anode and cathode is called decomposition stress. It is at this voltage, at a current close to zero, and in the presence of conditions for the reversibility of electrode reactions, that the electrolysis process begins.

In practice, a non-zero current passes through the electrodes, and therefore the electrode processes proceed under nonequilibrium conditions. The potential jump at the electrode-electrolyte interface in these cases is greater than the equilibrium one and is called electrode potential. It is impossible to experimentally measure the potential difference between the electrode and the electrolyte. Instead, the potential difference between the data and the standard (eg standard hydrogen electrode) electrodes is measured. This difference is taken as the potential of the electrode. The sign rule for potentials follows from this way of defining them.

The difference between the electrode potential and its equilibrium potential is called overvoltage. It is the greater, the higher the true current density. Overvoltage on the electrode also occurs when the reactions taking place on it are irreversible. The overvoltage is proportional to the energy that must be expended in order to carry out the electrode reaction at a certain rate.

The electrode reaction goes through the following stages:

1) supply of reacting substances from the electrolyte to the electrodes and removal of reaction products from them;

2) movement of electrons between electrodes and ions;

3) secondary reactions on the electrodes (for example, the formation of hydrogen and oxygen molecules from atoms).

In order for the process to proceed at a given speed on the electrodes, a certain amount of electrical energy is required for each stage. At the first stage, it is proportional to the concentration component of the overvoltage, or concentration polarization.

Equilibrium potentials are calculated based on the average activity (concentration) of the reactants in solution. At the electrodes, they either work out or work out, so their activity there differs from the average.

The equilibrium reaction potential calculated from the value of the true activity of substances at the electrodes differs from the potential calculated from the average activity. The difference between them is the concentration polarization. It is proportional to the work of concentration or dilution of solutions from the average value of activity to the true one, which is created at the electrodes, and the greater, the higher the reaction rate on the electrodes.

The appearance of the second component of the overvoltage depends on the other two stages of the electrode process - chemical surge or polarization. From the energy side, it is explained as follows. It is known from chemical kinetics that only active molecules with energies above a certain level (activation energy) enter into reactions.

By increasing the potential jump at the electrode-electrolyte interface against the equilibrium one, it is possible, as it were, to lower the energy barrier and thereby increase the fraction of active particles without changing the temperature. In this part, there is an analogy between catalytic and electrochemical processes. The additional potential jump against the equilibrium (chemical polarization) is proportional to the work required to activate the required number of ions or molecules so that the reaction can proceed at a given rate. The higher the rate of reaction, the higher the chemical polarization.

The physical picture on the electrodes, which explains the occurrence of chemical polarization, is considered in the double layer theory and the related theory of delayed discharge. These theories show that the magnitude of chemical polarization depends on the structure of the double layer, which is largely determined by the composition of the solution and the ability of the electrode material to adsorb certain components of the solution. Thus, by selecting the composition of the solution and the electrode material, it is possible to control the chemical polarization.

The actual voltage that must be applied to the terminals of the cell is electrolysis voltage - in order to carry out the reaction at a given rate, the decomposition voltage is greater by the magnitude of the overvoltages on the electrodes and by the sum of the losses in the conductors of the first and second kind.

The consumption of electrical energy per unit of product is directly proportional to the product of the electrolysis voltage and the amount of electricity required to produce the product (taking into account the current output). Of the total electricity consumption, only a part passes into chemical energy. This part is proportional to the voltage, which is called Thompson tension. It differs from the decomposition voltage for the following reason: the electrical energy required for the course of the process on the electrodes (at constant temperature and under equilibrium reversible conditions), proportional to the decomposition voltage, does not fully correspond to the change in the internal energy of the system. It may happen (depending on the properties of the substances involved in the reaction) that part of it in the course of the reaction will turn into heat, which passes into the internal energy of the system.

The difference between the electrolysis voltage and the Thompson voltage is proportional to the excess heat generated during electrolysis. This is the heating voltage, or the thermal component of the voltage.

The share of the total cost of electricity, transferred as a result of the reaction into the internal energy of the target product, is called energy output.

The main technological indicators of electrolysis include: current output, energy utilization factor (energy output) and energy consumption coefficient.

The current output (Wt,%) is calculated by the formula:

V t \u003d (m f / m t) ∙ 100%, (9.1)

Where: m f - the amount of substance actually obtained by electrolysis, kg; m t - the amount of a substance that should have been released according to Faraday's law, kg.

M t = k ∙ I∙ τ, (9.2)

Where: I - current strength, A; τ is the electrolysis time, h; k is the electrochemical equivalent of the released substance.

K = M/(F∙z), (9.3)

Where: F is the Faraday constant equal to 96,500 C or 26.8 Ah; z is the charge of the ion released on the electrode.

The energy utilization factor (V e, %) is calculated by the formula:

V e \u003d (Wt / W f) ∙ 100%, (9.4)

Where: W T - theoretical energy consumption, kWh; W f - actual energy consumption, kW∙h.

W t/p = V t/p ∙ J ∙ τ / m t/p (9.5)

Where: V t - theoretical decomposition voltage, V; V p - actual voltage on the electrodes, V.

The theoretical electricity consumption (kWh/t) can also be calculated using the equation:

W t = 10 3 ∙ V t / k (9.6)

The electrolysis process begins if the voltage applied to the electrolyzer (V p) exceeds by an infinitesimal value (∆V) the theoretical decomposition voltage (V t), i.e. condition will be provided:

V p \u003d V t + ∆V (9.7)

The theoretical decomposition voltage on the electrodes of the bath is determined by the equation:

V t \u003d E k - E a (9.8)

Where: E k is the actual potential of the ion discharge at the cathode, V; E a - the actual potential of the discharge of ions at the anode, V.

The actual discharge potentials of ions differ from their equilibrium discharge potentials by the magnitude of the overvoltage, respectively, cathode E k per and anodic E a per, which increase the equilibrium potentials:

E k \u003d E k p + E k ln and E a \u003d E a p + E a ln (9.9)

Where: E k p and E a p are the equilibrium potentials of the discharge of the cation and anion.

The equilibrium potentials of the discharge of ions are equal in magnitude and opposite in sign to the equilibrium electrode potentials: E k p \u003d - φ k and E a p \u003d - φ a, which can be calculated using the Nernst formula:

φ k/a = φ 0 k/a ± R∙ T ∙Iga k/a /z ∙ F, (9.10)

Where: φ 0 k / a - standard electrode potential, V; R - universal gas constant, J/mol∙K; T is temperature, K; a k / a - ion activity in solution (melt), mol/l; F is Faraday's constant equal to 96500 Cul.; z is the charge of the electrolyte ion.

Explanation of the condition presented in 9.7 gives the "stress balance" equation:

V p \u003d V t + J ∙∑R \u003d E k - E a + J (R e + R d + R tp) (9.11)

Where: J - current strength, A; R is the total resistance of the electrolysis process, Ohm; R e - electrolyte resistance, Ohm; R d - resistance of the diaphragm of the cell, Ohm; R tp - resistance of current-carrying paths, Ohm.

^ CONTROL QUESTIONS FOR TOPIC 9.1

9-1 . What processes are called electrochemical processes and how do they differ from electrothermal processes? Give examples of both.

9-2. What are the advantages of electrochemical methods of obtaining substances over chemical ones?

9-3. Name the areas of application of electrochemical methods.

9-4 . What is the condition of the electrolysis process? What is overvoltage and how does it affect the ion discharge sequence in electrolysis?

9-5 . List the quantitative characteristics of industrial electrolysis and give them a definition.

^ PROBLEM FOR TOPIC 9-1

9-1. How much hydrochloric acid theoretically can be obtained from electrolytic chlorine and hydrogen per day if the current supplied to the electrolyzer is 1500 A. The mass fraction of hydrochloric acid in the solution is 37.23% (density 1.19 g / ml). Express your answer in kilograms and liters.

9-2. From a diaphragm-type chlorine electrolyzer with a load of 40 kA per day, liquor with a volume of 10.6 m 3 containing 130 kg/m 3 of sodium hydroxide was obtained. Determine the alkali output by current.

9-3. How many baths should be in a copper refining plant with a capacity of 182.5 kt/y of cathode copper if the baths are operated with a load of 12 kA and the copper current efficiency is 96%? Bath utilization factor 0.96.

9-4. Determine the masses of gaseous chlorine and 50% sodium hydroxide solution produced by electrolysis of an aqueous solution of sodium chloride per day, if the current through the cell is 150 kA and the current efficiency is 0.95.

9-5. Determine the theoretical electricity consumption for the production of 3 tons of 85% sodium hydroxide and 3 tons of chlorine gas, if the theoretical electrolysis voltage is 2.2 V.

9-6. During the electrolysis of a melt of 24 g of a certain substance, 33.6 liters of hydrogen (n.o.) were released at the anode. Determine the substance that was taken for electrolysis and the volume of a 20% hydrochloric acid solution (density 1.1 g/ml) required for the reaction.

9-7. When a current of 1 A was passed through the melt of some binary inorganic compound for 8 hours, 2.068 g of metal was obtained. What compound underwent electrolysis if the ratio of components in it is 1: 0.145 wt.%?

9-8. When a current of 0.8 A was passed through 80 ml of a solution containing a mixture of AgNO 3 and Cu(NO 3) 2 for 117 min, a mixture of metals with a total mass of 3.0 g was released on the cathode. Write the electrolysis equations for each salt and determine the molar concentrations of salts in the initial solution, if it is known that gases have evolved at the anode, and after the completion of electrolysis, the solution does not contain metal ions.

9-9. During the electrolysis of a solution of chromium(III) nitrate, 31.2 g of chromium was released on the cathode, which was dissolved in hydrochloric acid. The solution was left in air, and then a 25% sodium hydroxide solution (density 1.28 g/mL) was gradually added to it. The precipitate that formed at first then completely dissolved. How many ml of sodium hydroxide solution did it take to dissolve the precipitate?

9-10. Two samples of a binary compound of some metal were investigated. The first sample, weighing 16 g, was melted and subjected to electrolysis, which gave 26.312 liters of hydrogen, measured at 720 mm. Hg and 31 o C. The second sample weighing 37.23 g, when exposed to water, gave 9.308 g of hydrogen. Set the formula for the unknown compound and write the equations for the ongoing processes.

9-11. A current of 2 A was passed through a solution of an organic acid salt for 5 hours. As a result of electrolysis, 12.195 g of metal was released at the cathode, and carbon monoxide (IV) and hydrogen were released at the anode. Determine which salt was subjected to electrolysis.

9-12. During the electrolysis of an aqueous solution of sodium chloride with a mercury cathode, an amalgam was obtained, which was treated with water. For titration of the resulting solution, 7.46 ml of a 0.5 M solution of sulfuric acid was used. Determine the strength of the current passed through the solution if the electrolysis time is 1 hour.

9-13. An aqueous solution of an unknown metal nitrate was subjected to electrolysis. In this case, 3.78 g of metal and 196 ml of oxygen (n.c.) were released on platinum electrodes. Determine which metal nitrate is subjected to electrolysis.

9-14. An aqueous solution of copper nitrate was subjected to electrolysis using inert (carbon) electrodes. The electrodes were weighed after completion of electrolysis and one hour after its completion. Will these masses be the same? Justify the answer.

9-15. Determine the energy efficiency during the electrolysis of alumina in cryolite, if the theoretical electrolysis voltage is 1.12 V, the practical voltage is 4.6 V, the metal current output is 0.8.

9-16. Calculate the degree of sodium chloride conversion in the electrolyzer, the catholyte of which contains sodium hydroxide 120 g/l and sodium chloride 190 g/l.

9-17 . Calculate the current efficiency for an electrolytic cell at a current of 14,000 A if there were 4,000 liters of electrolytic liquor containing 120 g/l of sodium hydroxide in 24 hours.

^ 9-18. For the conditions of the problem 10-17 Calculate the energy utilization factor if the practical sodium chloride decomposition voltage is 3.6V and the current output is 96 %.

9-20. At the factory of medical instruments, the surface of most products is covered with a nickel layer 5.0·10 -5 m thick from an electrolyte based on NiSO 4 . Determine the duration of electrolysis to obtain a coating of the required thickness on tweezers, the surface of which is 4.3·10 -3 m 2 , if the density of nickel metal is 8.9 t/m 3 and the current efficiency is 9 6%. Current during electrolysis 1.9 A.

^ 9.2. Electrolysis of an aqueous solution of sodium chloride

Electrolysis of an aqueous solution of sodium chloride is used in industry to produce chlorine, hydrogen and sodium hydroxide.

Currently, two methods of electrolysis are used in industry - diaphragm and mercury. The main process in both methods is the electrolysis of a saturated sodium chloride solution. In both methods, the anodic processes are similar; their main product is chlorine gas. The cathodic processes are different.

At diaphragm method a steel cathode is used, to which a solution of sodium chloride is supplied. Part of the sodium chloride is converted to sodium hydroxide and hydrogen is released. Sodium chloride is separated from sodium hydroxide by evaporating the solution. At the same time, due to a decrease in solubility, it precipitates. Commercial product - sodium hydroxide solution with a concentration of 42-50% (wt.) contains 2-4% (wt.) sodium chloride.

IN mercury electrolysis mercury cathode. Sodium ions, discharging on it, form a sodium amalgam. In a separate apparatus - a decomposer - sodium amalgam is decomposed by water, forming hydrogen and sodium hydroxide solution. In the decomposer, a solution of sodium hydroxide with a commercial concentration of 42-50% (wt) without sodium chloride impurities can be immediately obtained.

Salt solution (brine) is purified before electrolysis. The brine is purified from calcium and magnesium salts. Purification is carried out by precipitation of impurities with strictly metered precipitation reagents: soda suspension and lime milk.

The precipitation of impurities occurs according to the reactions:

Mg 2+ + Ca (OH) 2 \u003d Ca 2+ + Mg (OH) 2 ↓

Ca 2 + + Na 2 CO 3 \u003d 2 Na + + CaCO 3 ↓

In addition to chemical purification, the brine is freed from mechanical impurities by settling and filtering.

Diaphragm production (Fig. 9.1) includes the following stages:

1) preparation and purification of brine. At this stage, solid table salt is dissolved and the brine is purified from calcium and magnesium ions. The prepared brine is sent for electrolysis;

2) electrolysis;

3) evaporation of electrolytic liquors. At this stage, weak solutions of sodium hydroxide and sodium chloride, obtained by electrolysis, are evaporated to a commercial concentration of sodium hydroxide. The resulting salt is separated from the solution, dissolved in water and transferred to the brine preparation stage, where this brine is added to the brine prepared from fresh salt;

4) removal of sulfates. This stage receives sodium chloride, obtained at the last stage of evaporation of electrolytic liquors and containing an increased amount of sulfates. Sodium sulfate is isolated from salt in the form of commercial products. The purified salt solution is transferred to the brine preparation stage;

5) cooling and drying of chlorine;

6) cooling and drying of hydrogen.

The reactions taking place in a diaphragm cell depend on the materials and designs of the cells, brine concentration, pH of the medium, current density, temperature, and content of oxygen-containing ions.

Rice. 9.1. Structural diagram of the diaphragm method:

1- preparation and purification of brine; 2 - electrolysis; 3 - evaporation of electrolytic liquors; 4 - withdrawal of sulfides: 5 - cooling, drying and compression of chlorine; 6 - cooling, drying and compression of hydrogen.

In industrial electrolyzers, the anode is made of graphite, the cathode is made of iron.

On an iron cathode, the main process is the evolution of hydrogen:

2 H + + 2ē \u003d H 2

2 H 2 O + 2ē \u003d H 2 + 2 OH -

The discharge of sodium ions is impossible, since the equilibrium potential of the discharge of a sodium ion on an iron cathode in a neutral saturated solution of sodium chloride is much higher (-2.71 V) than that of hydrogen (-0.415 V).

The main reaction on the graphite anode:

2 Cl - + 2ē \u003d C1 2

In addition to this reaction, side reactions occur at the anode:

2OH - - 2e\u003d 0.5 O 2 + H 2 O H 2 O - 2ē \u003d 0.5 O 2 + 2 H +

The equilibrium electrode potential of the discharge of hydroxide ions in a neutral saturated solution of sodium chloride is +0.82 V, and of chloride ions +1.32 V. Therefore, oxygen should be released first at the anode with a small overvoltage.

As is known from theoretical electrochemistry, parallel electrode reactions take place with such partial current densities that give the same electrode potential. Therefore, we can write:

φ a \u003d φ (C1 2) + ψ (C1 2) \u003d φ (O 2) + ψ (O 2) (9.12)

Where: φ a - anode potential, V; φ (С1 2 ), φ (О 2 ) - equilibrium potentials of chlorine and oxygen release are determined by the Nernst formula and depend on the concentration (activity) of chlorine or hydroxide ions, as well as on temperature; ψ (С1 2 ), ψ (О 2) - overvoltage of chlorine and oxygen; the magnitude of the overvoltage increases with increasing current density.

The overvoltage for chlorine evolution decreases with increasing temperature to a greater extent than for oxygen. With an increase in the current density, the process at the anode also shifts towards the release of chlorine. As can be seen from fig. 9.2, with increasing current density, the potential for chlorine evolution increases to a lesser extent than that of oxygen. Hypochlorite ions can be discharged at the anode. As a result, oxygen is released:

3 ClO - + 3 H 2 O - 6ē \u003d ClO 3 - + 1.5 O 2 + 2 Cl - + 3 H 2

The presence of hypochlorite ions is caused by the partial hydrolysis of chlorine.

In diaphragm electrolysis, oxygen is always released along with chlorine. The normal level of oxygen evolution is determined by the established technological regime (anode material, current density, temperature, brine composition, etc.). The most important condition in this case is the normal acidity of the anolyte (solution located in the anode space).

Rice. 9.2. Anode polarization curves on graphite at 250°C in 22.6% (wt.) sodium chloride solution:

1- release of chlorine; 2 - release of oxygen.

The presence of hypochlorite and hypochlorate ions in the electrolyte can cause side reactions to occur at the cathode:

ClO 3 - + 3 H 2 \u003d 3 H 2 O + Cl - ClO - + H 2 \u003d H 2 O + Cl -

Increasing the alkalinity of the anolyte increases the intensity of oxygen evolution at the anode. Therefore, the electrolysis process in diaphragm electrolyzers is built in such a way as to minimize the electrolytic transfer of the hydroxide ion to the anode. This can be achieved through the use filter diaphragm.

The filtering diaphragm is made in the form of a porous partition separating the cathode and anode spaces. It prevents mixing of electrolysis products. Anolyte flow continuously passes through it from the anode space to the cathode.

The permeability of the diaphragm and the degree of conversion (approximately) control the concentration of sodium hydroxide in the catholyte (electrolyte located in the cathode space). In practice, in modern industrial electrolyzers, the limiting value of the degree of conversion corresponds to the concentration of sodium hydroxide in the catholyte of 140–150 g/l. When the alkali concentration exceeds its value, the course of electrolysis deviates from the norm.

The data showing the dependence of the current efficiency on the alkali concentration are shown in fig. 9.3. There is a decrease in current efficiency when working with a catholyte having an alkali concentration above 150 g/l.

Rice. 9.3. Dependence of the current output on the concentration of sodium hydroxide in the catholyte

Elevated temperatures of electrolysis and condensation of sodium chloride in the electrolyte reduce the solubility of chlorine, which reduces the likelihood of side reactions, and therefore increases the flow rate. In addition, an increase in temperature increases the electrical conductivity of the electrolyte, thereby reducing the voltage across the bath. Thus, the power consumption is reduced, so the electrolysis of sodium chloride solutions is carried out at temperatures of 70 - 80 °C.

Industrial electrolyzers with a filtering baffle are widely used in industry.

The diagram of a modern diaphragm electrolyzer is shown in fig. 9.4. Cell housing ^ 7 divided into two cavities: anode 4 and cathodic 5 space. Graphite anodes are placed in the anode space. The anode and cathode spaces are separated by a diaphragm, the basis for which is the cathode. 3 . The diaphragm covers the cathode from the side facing the anode. Brine is fed into the anode space - a saturated solution of table salt.

The anolyte level is above the upper limit of the diaphragm. Chlorine formed at the anode is collected in the gas space above the anolyte level. From here, chlorine is discharged to the collector. The anolyte, due to the difference in liquid levels in the anode and cathode spaces, flows through the diaphragm.

Hydrogen is reduced at the cathode, and the anolyte flowing to the cathode changes its composition and becomes enriched in hydroxide ions. The catholyte contains common salt, sodium hydroxide and a small amount of sodium chlorate. It is removed from the cathode space through the drain tube 9 ; its device allows you to adjust the level of the solution in the cell. Hydrogen is collected in the gas space above the level of the catholyte and then directed to the collector.

TO
the atolyte that came out of the electrolytic cell, otherwise called electrolytic liquor, contains sodium hydroxide 110-120 g/l and sodium chloride 170-180 g/l.

Rice. 9.4. Scheme of a diaphragm electrolyzer:

1- anode; 2 - diaphragm; 3 - cathode; 4 - anode space; 5 - cathode space; 6 - dropper; 7- housing of the electrolytic cell; 8 - cover; 9 - drain tube for catholyte

The ratio of the concentration of sodium hydroxide to the concentration of sodium chloride in the catholyte is determined by an important indicator of the technological regime - degree of conversion (X) sodium chloride during electrolysis. This is the ratio of the number of moles of sodium chloride converted to sodium hydroxide to the number of moles of sodium chloride received for electrolysis.

The degree of conversion is calculated by the formula:

X \u003d 1.46 C NaOH / (9.13)

The processes occurring in the electrolyzers and their technical indicators largely depend on the functioning of the diaphragm. In order for the diaphragm to perform its functions, it must meet the following requirements:

Be dense and strong enough to ensure complete separation of gas products and exclude displacement of anolyte and catholyte;

Have low electrical resistance to avoid loss of voltage across the diaphragm;

Have a sufficiently low hydraulic resistance;

Be chemically resistant to acids and alkalis so that the diaphragm works for a long time;

Have similar properties and uniformity in all areas.

The best material for the diaphragm is chrysolite asbestos.

The main requirement for the anode material is the greatest overvoltage to the release of oxygen than chlorine.

No material has yet been found that is absolutely resistant to the processes of joint electrochemical evolution of chlorine and oxygen. In practice, the aim is to ensure that the materials used are destroyed at a relatively low rate.

Preference is given to materials with low electrical resistivity, since the lower the resistance, the lower the voltage drop in the anode and the more uniform the current density distribution over it. In practice, platinum, graphite and magnetite can be used. The best in all respects (except cost) is platinum. In industry, anodes are made exclusively from artificial graphite.

^ Mercury method of electrolysis of an aqueous solution of sodium chloride contains the same stages as the diaphragm, with the exception of evaporation (Fig. 9.5). The preparation and purification of brine in this production have features and, according to the technological scheme, differ from the corresponding stage of diaphragm production. This is due to the special requirements for the anolyte returned for electrolysis.

Anolyte after electrolysis contains 260-270 g/l of sodium chloride, about 0.6 g/l of dissolved chlorine, about 5 mg/l of calcium, magnesium, heavy metals, graphite dust impurities.

Rice. 9.5. Block diagram of the electrolysis of an aqueous solution of sodium chloride with a mercury cathode:

1- preparation and cleaning of the brine; 2- electrolysis: 3 - dechlorination and purification of the anolyte; 4 - cooling, drying and compression of chlorine; 5 - cooling, drying and compression of hydrogen.

To remove chlorine from the anolyte, acidification, evacuation, air blowing, and destruction of chlorine residues by reducing agents are successively used. The anolyte is acidified with hydrochloric acid. Vacuuming is carried out at a pressure of 400-450 mm Hg.

The reactions occurring at the anode in electrolyzers with a mercury cathode are similar to those considered for diaphragm electrolyzers.

The cathode process in electrolyzers with a mercury cathode is fundamentally different from that in a diaphragm electrolyzer, in which hydrogen ions are discharged on a steel cathode.

In mercury electrolyzers, hydrogen evolution at the cathode is a side and harmful process. Its development is hampered by the fact that hydrogen is released at a mercury cathode or sodium amalgam cathode with a large overvoltage.

The characteristic polarization curve of this process is shown in Fig. . 9.6. The figure shows that intense hydrogen evolution occurs at cathode potentials that are more negative than -1.9 V. However, at a lower negative potential, another electrode reaction occurs on the mercury cathode - the formation of sodium mercury amalgam, for which the main part of the current is spent.

At the moment of release, metallic sodium reacts with mercury, forming an intermetallic compound NaHg n (sodium amalgam dissolved in mercury). In this case, the work required to reduce the sodium ion is reduced by the amount of energy released during the formation of the amalgam. Sodium amalgam formation potential φ k = -1.80 V.

The change in the release potential of substances in the electrochemical process due to the occurrence of a secondary reaction on the electrode is called depolarization. Due to depolarization, sodium can be released on the mercury cathode in the form of an amalgam according to the reaction:

Na + + n Hg + ē = NaHgn

This process takes place almost without overvoltage.

The main side reaction at the cathode:

2
H + + 2e - = H 2

Rice. 9.6. Polarization curve

Hydrogen release on mercury

Other side processes also take place at the cathode. Sodium amalgam reacts with chlorine dissolved in anolyte according to the equations:

NaHg n + Cl 2 = Na + + CI - + nHg Hg + Cl 2 = Hg 2+ + 2 Cl -

Under the action of water, the amalgam decomposes, releasing alkali:

NaHg n + 2 H 2 O \u003d H 2 + Na + + 2 OH - + n hg

The reaction in the decomposer consists of two coupled reactions:

2 H 2 O + e - \u003d H 2 + 2 OH - NaHg n - e - \u003d Na + + n Hg

The electrolytic process in an electrolyzer with a mercury cathode takes place in two stages. In the first stage, by electrolyzing an aqueous solution of sodium chloride, chlorine and strong sodium amalgam are obtained. The amalgam obtained after electrolysis contains 0.3-0.5% sodium. In the second stage, the amalgam is treated with purified water. Part of the amalgam decomposes to form sodium hydroxide and hydrogen. Weak amalgam mercury pump is re-served for electrolysis.

Carrying out the process in two stages makes it possible to obtain a solution of sodium hydroxide with very small impurities of sodium chloride in electrolyzers with a mercury cathode.

The scheme of the electrolytic cell with a mercury cathode is shown in fig. 9.7. It consists of three main parts: an electrolytic bath 9 , decomposer 12 and mercury pump 10.

Rice. 9.7. Scheme of an electrolytic cell with a mercury cathode:

1 - amalgam; 2 - outlet pocket of the cell; 3 - cell cover; 4 - anode; 5 - anode conductor and its seal; 6 - anolyte; 7 - space for collecting chlorine; 8 - input pocket of the cell; 9 - electrolytic bath; 10 - mercury pump; 11 - nozzle decomposer; 12 - decomposer; 13- sodium hydroxide solution.

In an electrolytic bath 9 saturated sodium chloride solution and a weak amalgam are continuously supplied. Chlorine is removed from the electrolyzer together with water vapor and strong amalgam. Separately, the solution of sodium chloride depleted as a result of electrolysis with chlorine dissolved in it is removed from the amalgam.

into the decomposer 12 strong amalgam and purified water are continuously supplied. Hydrogen with water vapor, a solution of sodium hydroxide in water and a weak amalgam are removed.

Mercury electrolyzers are designed to operate with high current density (5000-10,000 A/m2). As the density increases, the flow rate improves. In addition, the calculated cathode surface decreases (at the same current load), and therefore the required amount of mercury decreases.

Currently, horizontal electrolyzers are common. They represent an inclined chute of rectangular section, along the bottom of which amalgam flows by gravity. The gutter is covered with a lid 3 , on which graphite flat anode plates are fixed 4 . The distance between the electrodes is 3-5 mm. The plates are placed so closely that the working surface area of ​​the anodes approaches the surface area of ​​the cathode. Each anode plate has a current lead through the cell cover. There is a seal in the place where the current supply passes through the cover 5 preventing the release of chlorine into the atmosphere.

Graphite is destroyed during electrolysis. As a result, the interelectrode distance increases and the electrolysis voltage increases. Therefore, in modern designs, the anodes are equipped with a device that allows you to adjust the interelectrode distance.

For this, two different types of devices are used. The first type is designed for lowering each anode separately, the second - for lowering the whole group of anodes simultaneously.

Anolyte moves in the same direction over the amalgam layer in the electrolyzer ^ 6 .

A gas space is formed above the anolyte layer 7 . It collects released chlorine. Chlorine and anolyte are discharged from the electrolyzer either jointly or separately.

The second stage of the electrochemical process takes place in the decomposer. Horizontal decomposers have the form of a steel, hermetically sealed chute installed with a slope. Graphite plates are laid at the bottom of the decomposer. 12 . The flow of amalgam moves by gravity along its bottom. A sodium hydroxide solution moves countercurrently to the amalgam and is discharged together with hydrogen at the end of the decomposer.

^ CONTROL QUESTIONS FOR TOPIC 9.2

9-1. What are the industrial methods for the electrolysis of an aqueous solution of sodium chloride?

9-2. Name the main stages of the diaphragm electrolysis method.

9-3 . What reaction takes place at the cathode in diaphragm electrolysis? What side reactions can occur on the cathode in the diaphragm electrolysis method?

9-4 . What basic substance is released at the anode during diaphragm electrolysis? What by-product is released at the anode during diaphragm electrolysis?

9-5. What are the features of the electrolysis of an aqueous solution of sodium chloride with a mercury cathode? What is the role of a diaphragm in a diaphragm cell?

^ TASKS FOR TOPIC 9.2

9-1. The liquor flowing out of the diaphragm chlorine cell contains 130 kg/m 3 of lye. The bath operates with a load of 25 kA, current efficiency for CI 2 and NaOH 96%, and for hydrogen 98%. Calculate: a) the daily performance of the bath for chlorine and hydrogen (by mass and volume) and for alkali; b) the volume of liquor flowing out of the bath in 1 hour. Conditions are normal.

9-2. How many hours should the BGK-17-25 electrolyzer work to produce chlorine with a volume of 800 m 3, if the current efficiency is 96%, the current strength is 30 kA? Conditions are normal.

9-3. Calculate the theoretical value (V) of the decomposition voltage during the electrolysis of an aqueous solution of sodium chloride. Anolyte concentration 270 kg/m 3 , catholyte 120 kg/m 3 .

9-4. Calculate the energy utilization factor for an electrolyser equipped with an iron cathode, where the theoretical decomposition voltage is 2.16 V, and the practical voltage is 3.55 V when electrolyzing an aqueous solution of sodium chloride. Current output 93%.

9-5. Determine the current output for the BGK-17-50 cell, where during the day at a current of 40 kA, 9821 m 3 of electrolytic liquor containing 140 kg/m 3 of sodium hydroxide was obtained.

9-6. In the diaphragm method for producing caustic soda, the electrolysis process is completed when the mass fraction of caustic soda in the solution reaches 10%. Calculate what mass fraction of sodium chloride underwent electrolysis if the initial brine concentration was 310 kg/m 3 and the density was 1.197 t/m 3 .

9-7. Determine the conversion rate for a catholyte containing 120 kg/m 3 sodium hydroxide if the initial sodium chloride content was 293 kg/m 3 . Ignore production losses.

9-8. Determine the additional power consumption for obtaining hydrogen with a mass of 1 t, caused by the overvoltage of gas evolution h = 0.2 V.

9-9. Calculate the energy consumption for obtaining chlorine with a mass of 1 ton in the BGK-17-50 electrolytic cell, if the current at the terminals is 25 kA, the voltage is 3.6 V, the current output is 96%.

9-10. Determine the current output for a Hooker electrolytic cell producing 225 m 3 /h of catholyte containing 135 kg/m 3 of sodium hydroxide. The cell operates with a load of 40 kA.

9-11. Determine the weekly need of the enterprise for railway tanks with a carrying capacity of 50 tons for the transportation of liquid chlorine, if the enterprise has 3 series of electrolyzers BGK-17-50, 68 pieces in each series. The load of the cell is 50 kA, the current output is 96%.

9-12. Calculate the theoretical power consumption for producing caustic soda with a mass of 1 ton and chlorine with a mass of 1 ton in a diaphragm electrolyzer if the theoretical decomposition voltage of a sodium chloride solution is 2.2 V.

9-13. Calculate the energy consumption for the production of 1 ton of sodium hydroxide in an electrolyzer with a mercury cathode type "Solve" V-200, if the voltage on the electrodes is 4.56 V, the current output is 96%, the current strength is 190 kA.

9-14. The electrolysis shop has 66 baths with mercury cathodes. From a direct current source, they are supplied with a voltage of 250 V at a current strength of 30 kA. Determine the productivity of such a workshop per day for lye with a concentration of caustic soda of 140 kg / m 3 and chlorine at a current output of 96%; voltage on each bath and energy consumption per 1 ton of chlorine and 1 ton of caustic soda (separately).

9-15. On the bottom of the mercury electrolyzer, having a length of 10 m, a width of 1.5 m, mercury flows in a layer of 5 mm. At the entrance to the electrolyzer, the mass fraction of sodium in mercury is 0.01%, and at the exit it is 0.2%. Current output 95%. The cathode current density is 5000 A/m 2 . Determine the mass of a 40% sodium hydroxide solution, which can be obtained from 1 m 2 of a mercury cathode, and the linear flow rate of mercury. Ignore the change in mercury density during amalgam formation.

9-16. Determine the energy output for the mercury cell R-101, if here: the anode potential is -1.42 V; cathode potential 1.84 V; bath voltage 3.55 V; current output 93.7%.

9-17. Calculate the volumetric circulation rate of mercury in the chlorine electrolyzer if the mass fraction of sodium in the incoming mercury is 0.015%, and in the outgoing mercury it is 0.21%. The sodium current output is 97%, the cell load is 25 kA.

9-18. In the horizontal decomposer, which receives 23 tons of sodium amalgam per hour, hydrogen was released in a volume of 56 m 3 . Determine the mass fraction of sodium in the amalgam (at n.a.).

9-19 . The design annual capacity of one of the enterprises for the production of hydrochloric acid is 80 thousand tons of a product with a mass fraction of hydrogen chloride of 34%. Will this enterprise provide with chlorine and hydrogen a workshop with 84 R-30 baths, operating according to the enterprise's schedule? Current output 96%, load of one cell 30 kA. The acid yield is 95% of theoretical.

9-20. Diaphragm chlorine electrolyzer has the following performance indicators: current output of chlorine 95%; hydrogen current efficiency 99%; load 20 kA. What mass of hydrochloric acid with a mass fraction of hydrogen chloride of 35% can be obtained from the total chlorine produced in 30 days of operation of the electrolyzer? What volume of hydrogen in m 3 should be produced by the electrolyser to obtain this mass of acid, if the volume fraction of hydrogen is 5% more against stoichiometry?

ELECTROLYSIS

One of the ways to obtain metals is electrolysis. Active metals occur in nature only in the form of chemical compounds. How to isolate from these compounds in the free state?

Solutions and melts of electrolytes conduct electric current. However, when current is passed through an electrolyte solution, chemical reactions can occur. Consider what will happen if two metal plates are placed in an electrolyte solution or melt, each of which is connected to one of the poles of the current source. These plates are called electrodes. Electric current is a moving stream of electrons. As a result of the fact that the electrons in the circuit move from one electrode to another, an excess of electrons appears on one of the electrodes. The electrons have a negative charge, so this electrode becomes negatively charged. It is called the cathode. On the other electrode, a lack of electrons is created, and it is positively charged. This electrode is called the anode. An electrolyte in a solution or melt dissociates into positively charged ions - cations and negatively charged ions - anions. Cations are attracted to a negatively charged electrode - the cathode. Anions are attracted to a positively charged electrode - the anode. On the surface of the electrodes, interaction between ions and electrons can occur.

Electrolysis refers to the processes that occur when an electric current is passed through solutions or melts of electrolytes.

The processes occurring during the electrolysis of solutions and melts of electrolytes are quite different. Let's consider both of these cases in detail.

Melt electrolysis

As an example, consider the electrolysis of a sodium chloride melt. In the melt, sodium chloride dissociates into ions Na+
and Cl - : NaCl = Na + + Cl -

Sodium cations move to the surface of a negatively charged electrode - the cathode. There is an excess of electrons on the cathode surface. Therefore, there is a transfer of electrons from the electrode surface to sodium ions. At the same time, ions Na+ are converted into sodium atoms, that is, cations are reduced Na+ . Process equation:

Na + + e - = Na

Chloride ions Cl - move to the surface of a positively charged electrode - the anode. A lack of electrons is created on the anode surface and electrons are transferred from anions Cl- to the surface of the electrode. At the same time, negatively charged ions Cl- are converted into chlorine atoms, which immediately combine to form chlorine molecules C l2 :

2C l - -2e - \u003d Cl 2

Chloride ions lose electrons, that is, they are oxidized.

Let us write together the equations of the processes occurring at the cathode and anode

Na + + e - = Na

2 C l - -2 e - \u003d Cl 2

One electron is involved in the process of reduction of sodium cations, and 2 electrons are involved in the process of oxidation of chlorine ions. However, the law of conservation of electric charge must be observed, that is, the total charge of all particles in the solution must be constant. Therefore, the number of electrons involved in the reduction of sodium cations must be equal to the number of electrons involved in the oxidation of chloride ions. Therefore, we multiply the first equation by 2:

Na + + e - \u003d Na 2

2C l - -2e - \u003d Cl 2 1

We add both equations together and get the general equation for the reaction.

2 Na + + 2C l - \u003d 2 Na + Cl 2 (ionic reaction equation), or

2 NaCl \u003d 2 Na + Cl 2 (molecular reaction equation)

So, in the considered example, we see that electrolysis is a redox reaction. At the cathode, the reduction of positively charged ions - cations, at the anode - the oxidation of negatively charged ions - anions. To remember which process happens where, you can use the "T rule":

cathode - cation - reduction.

Example 2Electrolysis of sodium hydroxide melt.

Sodium hydroxide in solution dissociates into cations and hydroxide ions.

Cathode (-)<-- Na + + OH - à Анод (+)

On the cathode surface, sodium cations are reduced, and sodium atoms are formed:

cathode (-) Na + +e à Na

Hydroxide ions are oxidized on the anode surface, while oxygen is released and water molecules are formed:

cathode (-) Na + + e à Na

anode (+)4 OH - - 4 e à 2 H 2 O + O 2

The number of electrons involved in the reduction reaction of sodium cations and in the oxidation reaction of hydroxide ions should be the same. So let's multiply the first equation by 4:

cathode (-) Na + + e à Na 4

anode (+)4 OH - – 4 e à 2 H 2 O + O 2 1

Putting both equations together, we get the equation for the electrolysis reaction:

4 NaOH à 4 Na + 2 H 2 O + O 2

Example 3Consider the electrolysis of the melt Al2O3

Using this reaction, aluminum is obtained from bauxite, a natural compound that contains a lot of aluminum oxide. The melting point of aluminum oxide is very high (more than 2000º C), so special additives are added to it, lowering the melting point to 800-900º C. In the melt, aluminum oxide dissociates into ions Al 3+ and O 2-. H cations are reduced at the cathode Al 3+ , turning into aluminum atoms:

Al +3 e a Al

Anions are oxidized at the anode O 2- turning into oxygen atoms. Oxygen atoms immediately combine into O 2 molecules:

2 O 2- – 4 e à O 2

The number of electrons involved in the reduction of aluminum cations and the oxidation of oxygen ions must be equal, so we multiply the first equation by 4, and the second by 3:

Al 3+ +3 e à Al 0 4

2 O 2- – 4 e à O 2 3

Let's add both equations and get

4 Al 3+ + 6 O 2- a 4 Al 0 +3 O 2 0 (ionic reaction equation)

2 Al 2 O 3 à 4 Al + 3 O 2

Solution electrolysis

In the case of passing an electric current through an aqueous electrolyte solution, the matter is complicated by the presence of water molecules in the solution, which can also interact with electrons. Recall that in a water molecule, hydrogen and oxygen atoms are connected by a polar covalent bond. The electronegativity of oxygen is greater than the electronegativity of hydrogen, so the shared electron pairs are shifted towards the oxygen atom. A partial negative charge arises on the oxygen atom, it is denoted δ-, and on hydrogen atoms it has a partial positive charge, it is denoted δ+.

δ+

H-O δ-

│

H δ+

Due to this shift of charges, the water molecule has positive and negative "poles". Therefore, water molecules can be attracted by a positively charged pole to a negatively charged electrode - the cathode, and by a negative pole - to a positively charged electrode - anode. At the cathode, water molecules can be reduced, and hydrogen is released:

Oxidation of water molecules can occur at the anode with the release of oxygen:

2 H 2 O - 4e - \u003d 4H + + O 2

Therefore, either electrolyte cations or water molecules can be reduced at the cathode. These two processes seem to compete with each other. What process actually takes place at the cathode depends on the nature of the metal. Whether metal cations or water molecules will be reduced at the cathode depends on the position of the metal in series of metal stresses .

Li K Na Ca Mg Al ¦¦ Zn Fe Ni Sn Pb (H 2) ¦¦ Cu Hg Ag Au

If the metal is in the voltage series to the right of hydrogen, metal cations are reduced at the cathode and free metal is released. If the metal is in the voltage series to the left of aluminum, water molecules are reduced at the cathode and hydrogen is released. Finally, in the case of metal cations from zinc to lead, either metal evolution or hydrogen evolution can occur, and sometimes both hydrogen and metal are evolved simultaneously. In general, this is a rather complicated case, much depends on the reaction conditions: the concentration of the solution, the current strength, and others.

One of two processes can also occur at the anode - either the oxidation of electrolyte anions, or the oxidation of water molecules. Which process actually takes place depends on the nature of the anion. During the electrolysis of salts of anoxic acids or the acids themselves, anions are oxidized at the anode. The only exception is the fluoride ion F- . In the case of oxygen-containing acids, water molecules are oxidized at the anode and oxygen is released.

Example 1Let's look at the electrolysis of an aqueous solution of sodium chloride.

In an aqueous solution of sodium chloride there will be sodium cations Na + , chlorine anions Cl - and water molecules.

2 NaCl a 2 Na + + 2 Cl -

2Н 2 О а 2 H + + 2 OH -

cathode (-) 2 Na + ; 2 H + ; 2Н + + 2е а Н 0 2

anode (+) 2 Cl - ; 2OH-; 2 Cl - – 2e a 2 Cl 0

2NaCl + 2H 2 O à H 2 + Cl 2 + 2NaOH

Chemical activity anions hardly decreases.

Example 2What if the salt contains SO 4 2- ? Consider the electrolysis of a nickel sulfate solution ( II ). nickel sulfate ( II ) dissociates into ions Ni 2+ and SO 4 2-:

NiSO 4 à Ni 2+ + SO 4 2-

H 2 O à H + + OH -

Nickel cations are between metal ions Al 3+ and Pb 2+ , occupying a middle position in the voltage series, the recovery process at the cathode occurs according to both schemes:

2 H 2 O + 2e - \u003d H 2 + 2OH -

Anions of oxygen-containing acids are not oxidized at the anode ( anion activity series ), water molecules are oxidized:

anode e à O 2 + 4H +

Let us write together the equations of the processes occurring at the cathode and anode:

cathode (-) Ni 2+ ; H + ; Ni 2+ + 2е а Ni 0

2 H 2 O + 2e - \u003d H 2 + 2OH -

anode (+) SO 4 2- ; OH -; 2H 2 O - 4 e à O 2 + 4H +

4 electrons are involved in the reduction processes, and 4 electrons are also involved in the oxidation process. Putting these equations together, we get the general reaction equation:

Ni 2+ +2 H 2 O + 2 H 2 O à Ni 0 + H 2 + 2OH - + O 2 + 4 H +

On the right side of the equation, there are simultaneously H + ions and oh- , which combine to form water molecules:

H + + OH - à H 2 O

Therefore, on the right side of the equation, instead of 4 H + ions and 2 ions oh- we write 2 water molecules and 2 H + ions:

Ni 2+ +2 H 2 O + 2 H 2 O à Ni 0 + H 2 +2 H 2 O + O 2 + 2 H +

Let's reduce two water molecules on both sides of the equation:

Ni 2+ +2 H 2 O à Ni 0 + H 2 + O 2 + 2 H +

This is a short ionic equation. To get the full ionic equation, you need to add to both parts of the sulfate ion SO 4 2- , formed during the dissociation of nickel sulfate ( II ) and not participating in the reaction:

Ni 2+ + SO 4 2- + 2H 2 O à Ni 0 + H 2 + O 2 + 2H + + SO 4 2-

Thus, during the electrolysis of a solution of nickel sulfate ( II ) hydrogen and nickel are released at the cathode, and oxygen is released at the anode.

NiSO 4 + 2H 2 O à Ni + H 2 + H 2 SO 4 + O 2

Example 3 Write the equations of the processes occurring during the electrolysis of an aqueous solution of sodium sulfate with an inert anode.

Standard electrode potential of the system Na + + e = Na 0 is much more negative than the potential of the water electrode in a neutral aqueous medium (-0.41 V). Therefore, electrochemical reduction of water will occur on the cathode, accompanied by hydrogen evolution

2Н 2 О а 2 H + + 2 OH -

and Na ions + coming to the cathode will accumulate in the adjacent part of the solution (cathode space).

At the anode, electrochemical oxidation of water will occur, leading to the release of oxygen.

2 H 2 O - 4e à O 2 + 4 H +

because corresponding to this system standard electrode potential (1.23 V) is significantly lower than the standard electrode potential (2.01 V) that characterizes the system

2 SO 4 2- + 2 e \u003d S 2 O 8 2-.

Ions SO 4 2- moving towards the anode during electrolysis will accumulate in the anode space.

Multiplying the equation of the cathode process by two, and adding it with the equation of the anode process, we obtain the total equation of the electrolysis process:

6 H 2 O \u003d 2 H 2 + 4 OH - + O 2 + 4 H +

Taking into account that ions are simultaneously accumulated in the cathode space and ions in the anode space, the overall process equation can be written in the following form:

6H 2 O + 2Na 2 SO 4 \u003d 2H 2 + 4Na + + 4OH - + O 2 + 4H + + 2SO 4 2-

Thus, simultaneously with the release of hydrogen and oxygen, sodium hydroxide (in the cathode space) and sulfuric acid (in the anode space) are formed.

Example 4Electrolysis of copper sulfate solution ( II) CuSO4.

Cathode (-)<-- Cu 2+ + SO 4 2- à анод (+)

cathode (-) Cu 2+ + 2e à Cu 0 2

anode (+) 2H 2 O - 4 e à O 2 + 4H + 1

H + ions remain in the solution and SO 4 2- , since sulfuric acid accumulates.

2CuSO 4 + 2H 2 O à 2Cu + 2H 2 SO 4 + O 2

Example 5 Electrolysis of copper chloride solution ( II) CuCl 2 .

Cathode (-)<-- Cu 2+ + 2Cl - à анод (+)

cathode (-) Cu 2+ + 2e à Cu 0

anode (+) 2Cl - – 2e à Cl 0 2

Both equations involve two electrons.

Cu 2+ + 2e à Cu 0 1

2Cl - -– 2e à Cl 2 1

Cu 2+ + 2 Cl - à Cu 0 + Cl 2 (ionic equation)

CuCl 2 à Cu + Cl 2 (molecular equation)

Example 6 Electrolysis of silver nitrate solution AgNO3.

Cathode (-)<-- Ag + + NO 3 - à Анод (+)

cathode (-) Ag + + e à Ag 0

anode (+) 2H 2 O - 4 e à O 2 + 4H +

Ag + + e à Ag 0 4

2H 2 O - 4 e à O 2 + 4H + 1

4 Ag + + 2 H 2 O à 4 Ag 0 + 4 H + + O 2 (ionic equation)

4 Ag + + 2 H 2 Oà 4 Ag 0 + 4 H + + O 2 + 4 NO 3 - (full ionic equation)

4 AgNO 3 + 2 H 2 Oà 4 Ag 0 + 4 HNO 3 + O 2 (molecular equation)

Example 7 Electrolysis of hydrochloric acid solutionHCl.

Cathode (-)<-- H + + Cl - à anode (+)

cathode (-) 2H + + 2 eà H 2

anode (+) 2Cl - – 2 eà Cl 2

2 H + + 2 Cl - à H 2 + Cl 2 (ionic equation)

2 HClà H 2 + Cl 2 (molecular equation)

Example 8 Electrolysis of sulfuric acid solutionH 2 SO 4 .

Cathode (-) <-- 2H + + SO 4 2- à anode (+)

cathode (-)2H+ + 2eà H2

anode(+) 2H 2 O - 4eà O2+4H+

2H+ + 2eà H 2 2

2H2O-4eà O 2 + 4H+1

4H+ + 2H2Oà 2H 2 + 4H+ + O 2

2H2Oà 2H2+O2

Example 9. Electrolysis of potassium hydroxide solutionKOH.

Cathode (-)<-- K + + Oh - à anode (+)

Potassium cations will not be reduced at the cathode, since potassium is in the voltage series of metals to the left of aluminum, instead water molecules will be reduced:

2H2O + 2eà H 2 + 2OH - 4OH - -4eà 2H 2 O +O 2

cathode(-)2H2O+2eà H 2 + 2OH - 2

anode(+) 4OH - - 4eà 2H 2 O + O 2 1

4H 2 O + 4OH -à 2H 2 + 4OH - + 2H 2 O + O 2

2 H 2 Oà 2 H 2 + O 2

Example 10 Electrolysis of potassium nitrate solutionKNO 3 .

Cathode (-) <-- K + + NO 3 - à anode (+)

2H2O + 2eà H 2 + 2OH - 2H 2 O - 4eà O2+4H+

cathode(-)2H2O+2eà H 2 + 2OH-2

anode(+) 2H 2 O - 4eà O 2 + 4H+1

4H2O + 2H2Oà 2H2+4OH-+4H++ O2

2H2Oà 2H2+O2

When an electric current is passed through solutions of oxygen-containing acids, alkalis and salts of oxygen-containing acids with metals that are in the voltage series of metals, to the left of aluminum, water electrolysis practically occurs. In this case, hydrogen is released at the cathode, and oxygen at the anode.

Conclusions. When determining the products of electrolysis of aqueous solutions of electrolytes, in the simplest cases, one can be guided by the following considerations:

1. Metal ions with a small algebraic value of the standard potential - fromLi + beforeAl 3+ inclusive - have a very weak tendency to reattach electrons, yielding in this respect to ionsH + (cm. Cation activity series). In the electrolysis of aqueous solutions of compounds containing these cations, the function of an oxidizing agent on the cathode is performed by ionsH + , while restoring according to the scheme:

2 H 2 O+ 2 eà H 2 + 2OH -

2. Metal cations with positive values ​​of standard potentials (Cu 2+ , Ag + , hg 2+ etc.) have a greater tendency to attach electrons than ions. During the electrolysis of aqueous solutions of their salts, these cations emit the function of an oxidizing agent on the cathode, while being reduced to a metal according to the scheme, for example:

Cu 2+ +2 eà Cu 0

3. During the electrolysis of aqueous solutions of metal saltsZn, Fe, CD, Niand others, occupying a middle position between the listed groups in the voltage series, the reduction process at the cathode occurs according to both schemes. The mass of the released metal does not correspond in these cases to the amount of electric current flowing, part of which is spent on the formation of hydrogen.

4. In aqueous solutions of electrolytes, monatomic anions (Cl - , Br - , J - ), oxygen-containing anions (NO 3 - , SO 4 2- , PO 4 3- and others), as well as hydroxyl ions of water. Of these, halide ions have the stronger reducing properties, with the exception ofF. ionsOhoccupy an intermediate position between them and polyatomic anions. Therefore, during the electrolysis of aqueous solutionsHCl, HBr, HJor their salts on the anode, halide ions are oxidized according to the scheme:

2 X - -2 eà X 2 0

During the electrolysis of aqueous solutions of sulfates, nitrates, phosphates, etc. the function of the reducing agent is performed by ions, while being oxidized according to the scheme:

4 HOH – 4 eà 2 H 2 O + O 2 + 4 H +

.

Tasks.

W but dacha 1. During the electrolysis of a solution of copper sulfate, 48 g of copper was released at the cathode. Find the volume of gas released at the anode and the mass of sulfuric acid formed in the solution.

Copper sulfate in solution dissociates neither ionsC 2+ andS0 4 2 ".

CuS0 4 \u003d Cu 2+ + S0 4 2 "

Let us write down the equations of the processes occurring at the cathode and anode. Cu cations are reduced at the cathode, electrolysis of water occurs at the anode:

Cu 2+ + 2e- \u003d Cu12

2H 2 0-4e- = 4H + + 0 2 |1

General electrolysis equation:

2Cu2+ + 2H2O = 2Cu + 4H+ + O2 (short ionic equation)

Add to both sides of the equation 2 sulfate ions each, which are formed during the dissociation of copper sulfate, we get the complete ionic equation:

2Cu2+ + 2S042" + 2H20 = 2Cu + 4H+ + 2SO4 2" + O2

2CuSO4 + 2H2O = 2Cu + 2H2SO4 + O2

The gas released at the anode is oxygen. Sulfuric acid is formed in the solution.

The molar mass of copper is 64 g / mol, we calculate the amount of copper substance:

According to the reaction equation, when 2 mol of copper is released from the anode, 1 mol of oxygen is released. 0.75 mol of copper was released at the cathode, let x mol of oxygen be released at the anode. Let's make a proportion:

2/1=0.75/x, x=0.75*1/2=0.375mol

0.375 mol of oxygen was released at the anode,

v(O2) = 0.375 mol.

Calculate the volume of released oxygen:

V(O2) \u003d v (O2) "VM \u003d 0.375 mol" 22.4 l / mol \u003d 8.4 l

According to the reaction equation, when 2 mol of copper is released at the cathode, 2 mol of sulfuric acid is formed in the solution, which means that if 0.75 mol of copper is released at the cathode, then 0.75 mol of sulfuric acid is formed in the solution, v (H2SO4) = 0.75 mol . Calculate the molar mass of sulfuric acid:

M(H2SO4) = 2-1+32+16-4 = 98 g/mol.

Calculate the mass of sulfuric acid:

m (H2S04) \u003d v (H2S04> M (H2S04) \u003d \u003d 0.75 mol \u003d 98 g / mol \u003d 73.5 g.

Answer: 8.4 liters of oxygen were released at the anode; 73.5 g of sulfuric acid was formed in the solution

Task 2. Find the volume of gases released at the cathode and anode during the electrolysis of an aqueous solution containing 111.75 g of potassium chloride. What substance is formed in solution? Find its mass.

Potassium chloride in solution dissociates into K+ and Cl ions:

2KS1 \u003d K + + Cl

Potassium ions are not reduced at the cathode; instead, water molecules are reduced. Chloride ions are oxidized at the anode and chlorine is released:

2H2O + 2e "= H2 + 20H-|1

2SG-2e "= C12|1

General electrolysis equation:

2CHl + 2H2O \u003d H2 + 2OH "+ C12 (short ionic equation) The solution also contains K + ions formed during the dissociation of potassium chloride and not participating in the reaction:

2K+ + 2Cl + 2H20 = H2 + 2K+ + 2OH" + C12

Let's rewrite the equation in molecular form:

2KS1 + 2H2O = H2 + C12 + 2KOH

Hydrogen is released at the cathode, chlorine is released at the anode, and potassium hydroxide is formed in solution.

The solution contained 111.75 g of potassium chloride.

Calculate the molar mass of potassium chloride:

M(KC1) = 39+35.5 = 74.5 g/mol

Calculate the amount of potassium chloride substance:

According to the reaction equation, electrolysis of 2 mol of potassium chloride releases 1 mol of chlorine. Let the electrolysis of 1.5 mol of potassium chloride release x mol of chlorine. Let's make a proportion:

2/1=1.5/x, x=1.5 /2=0.75 mol

0.75 mol of chlorine will be released, v (C! 2) \u003d 0.75 mol. According to the reaction equation, when 1 mol of chlorine is released at the anode, 1 mol of hydrogen is released at the cathode. Therefore, if 0.75 mol of chlorine is released at the anode, then 0.75 mol of hydrogen is released at the cathode, v(H2) = 0.75 mol.

Let us calculate the volume of chlorine released at the anode:

V (C12) \u003d v (Cl2) -VM \u003d 0.75 mol \u003d 22.4 l / mol \u003d 16.8 l.

The volume of hydrogen is equal to the volume of chlorine:

Y (H2) \u003d Y (C12) \u003d 16.8 l.

According to the reaction equation, during the electrolysis of 2 mol of potassium chloride, 2 mol of potassium hydroxide is formed, which means that during the electrolysis of 0.75 mol of potassium chloride, 0.75 mol of potassium hydroxide is formed. Calculate the molar mass of potassium hydroxide:

M (KOH) \u003d 39 + 16 + 1 - 56 g / mol.

Calculate the mass of potassium hydroxide:

m(KOH) \u003d v (KOH> M (KOH) \u003d 0.75 mol-56 g / mol \u003d 42 g.

Answer: 16.8 liters of hydrogen were released at the cathode, 16.8 liters of chlorine were released at the anode, and 42 g of potassium hydroxide formed in the solution.

Task 3. During the electrolysis of a solution of 19 g of divalent metal chloride at the anode, 8.96 liters of chlorine were released. Determine which metal chloride was subjected to electrolysis. Calculate the volume of hydrogen released at the cathode.

We denote the unknown metal M, the formula of its chloride is MC12. At the anode, chloride ions are oxidized and chlorine is released. The condition says that hydrogen is released at the cathode, therefore, water molecules are reduced:

2H20 + 2e- = H2 + 2OH|1

2Cl -2e "= C12! 1

General electrolysis equation:

2Cl + 2H2O \u003d H2 + 2OH "+ C12 (short ionic equation)

The solution also contains M2+ ions, which do not change during the reaction. We write the full ionic reaction equation:

2SG + M2+ + 2H2O = H2 + M2+ + 2OH- + C12

Let's rewrite the reaction equation in molecular form:

MS12 + 2H2O - H2 + M(OH)2 + C12

Find the amount of chlorine released at the anode:

According to the reaction equation, during the electrolysis of 1 mol of chloride of an unknown metal, 1 mol of chlorine is released. If 0.4 mol of chlorine was released, then 0.4 mol of metal chloride was subjected to electrolysis. Calculate the molar mass of metal chloride:

The molar mass of chloride of an unknown metal is 95 g/mol. There are 35.5"2 = 71 g/mol per two chlorine atoms. Therefore, the molar mass of the metal is 95-71 = 24 g/mol. Magnesium corresponds to this molar mass.

According to the reaction equation, for 1 mole of chlorine released at the anode, there is 1 mole of hydrogen released at the cathode. In our case, 0.4 mol of chlorine was released at the anode, which means that 0.4 mol of hydrogen was released at the cathode. Calculate the volume of hydrogen:

V (H2) \u003d v (H2> VM \u003d 0.4 mol \u003d 22.4 l / mol \u003d 8.96 l.

Answer: subjected to electrolysis solution of magnesium chloride; 8.96 liters of hydrogen were released at the cathode.

*Problem 4. During the electrolysis of 200 g of a solution of potassium sulfate with a concentration of 15%, 14.56 liters of oxygen were released at the anode. Calculate the concentration of the solution at the end of the electrolysis.

In a solution of potassium sulfate, water molecules react both at the cathode and at the anode:

2H20 + 2e "= H2 + 20H-|2

2H2O - 4e "= 4H+ + O2! 1

Let's put both equations together:

6H2O \u003d 2H2 + 4OH "+ 4H + + O2, or

6H2O \u003d 2H2 + 4H2O + O2, or

2H2O = 2H2 + 02

In fact, during the electrolysis of a solution of potassium sulfate, the electrolysis of water occurs.

The concentration of a solute in a solution is determined by the formula:

C=m(solute) 100% / m(solution)

To find the concentration of the potassium sulfate solution at the end of the electrolysis, it is necessary to know the mass of potassium sulfate and the mass of the solution. The mass of potassium sulfate does not change during the reaction. Calculate the mass of potassium sulfate in the initial solution. Let us denote the concentration of the initial solution as C

m(K2S04) = C2 (K2S04) m(solution) = 0.15 200 g = 30 g.

The mass of the solution changes during electrolysis, as part of the water is converted into hydrogen and oxygen. Calculate the amount of released oxygen:

(O 2) \u003d V (O2) / Vm \u003d 14.56 l / 22.4 l / mol \u003d 0.65 mol

According to the reaction equation, 1 mole of oxygen is formed from 2 moles of water. Let 0.65 mol of oxygen be released during the decomposition of x mol of water. Let's make a proportion:

1.3 mol of water decomposed, v(H2O) = 1.3 mol.

Calculate the molar mass of water:

M(H2O) \u003d 1-2 + 16 \u003d 18 g / mol.

Calculate the mass of decomposed water:

m(H2O) \u003d v (H2O> M (H2O) \u003d 1.3 mol * 18 g / mol \u003d 23.4 g.

The mass of the potassium sulfate solution decreased by 23.4 g and became equal to 200-23.4 = 176.6 g. Let us now calculate the concentration of the potassium sulfate solution at the end of the electrolysis:

С2 (K2 SO4)=m(K2 SO4) 100% / m(solution)=30g 100% / 176.6g=17%

Answer: the concentration of the solution at the end of the electrolysis is 17%.

* 3 problem 5. 188.3 g of a mixture of sodium and potassium chlorides were dissolved in water and an electric current was passed through the resulting solution. During electrolysis, 33.6 liters of hydrogen were released at the cathode. Calculate the composition of the mixture in percent by weight.

After dissolving a mixture of potassium and sodium chlorides in water, the solution contains K+, Na+ and Cl- ions. Neither potassium ions nor sodium ions are reduced at the cathode, water molecules are reduced. Chloride ions are oxidized at the anode and chlorine is released:

Let's rewrite the equations in molecular form:

2KS1 + 2H20 = H2 + C12 + 2KOH

2NaCl + 2H2O = H2 + C12 + 2NaOH

Let us denote the amount of potassium chloride substance contained in the mixture, x mol, and the amount of sodium chloride substance, y mol. According to the reaction equation, during the electrolysis of 2 mol of sodium or potassium chloride, 1 mol of hydrogen is released. Therefore, during electrolysis x mol of potassium chloride, x / 2 or 0.5x mol of hydrogen is formed, and during electrolysis, y mol of sodium chloride is 0.5y mol of hydrogen. Find the amount of hydrogen substance released during the electrolysis of the mixture:

Let's make the equation: 0.5x + 0.5y \u003d 1.5

Calculate the molar masses of potassium and sodium chlorides:

M(KC1) = 39+35.5 = 74.5 g/mol

M (NaCl) \u003d 23 + 35.5 \u003d 58.5 g / mol

Mass x mole of potassium chloride is:

m (KCl) \u003d v (KCl) -M (KCl) \u003d x mol-74.5 g / mol \u003d 74.5 x g.

The mass of a mole of sodium chloride is:

m (KCl) \u003d v (KCl) -M (KCl) \u003d y mol-74.5 g / mol \u003d 58.5 u g.

The mass of the mixture is 188.3 g, we make the second equation:

74.5x + 58.5y = 188.3

So, we solve a system of two equations with two unknowns:

0.5(x + y)= 1.5

74.5x + 58.5y = 188.3g

From the first equation, we express x:

x + y \u003d 1.5 / 0.5 \u003d 3,

x = 3-y

Substituting this value of x into the second equation, we get:

74.5-(3-y) + 58.5y = 188.3

223.5-74.5y + 58.5y = 188.3

-16y = -35.2

y \u003d 2.2 100% / 188.3g \u003d 31.65%

Calculate the mass fraction of sodium chloride:

w(NaCl) = 100% - w(KCl) = 68.35%

Answer: the mixture contains 31.65% potassium chloride and 68.35% sodium chloride.

Physico-chemical properties of the electrolyte

The melting point of calcium chloride is 774°. In some cases, potassium chloride (melting point 768°) and sometimes sodium chloride (melting point 800°) are added to the electrolyte.
The melting diagram of the CaCl2-KCl system was studied by O. Menge. The compound CaCl2 KCl is formed in the system and there are two eutectics, at 75% (mol.) CaCl2 with a melting point of 634° and at 25% (mol.) CaCl2 with a melting point of 587°.
The CaCl2-NaCl system gives a eutectic at 53% (mol.) CaCl2 with a melting point of about 494°.
The state diagram of the CaCl2-KCl-NaCl system was studied by K. Scholich. In it, at 508 °, a eutectic of composition is formed - 52% CaCl2, 41% NaCl, 7% KCL
The electrolyte recommended by Ruff and Plateau contains 85.8% CaCl2 and 14.2% CaF2 and melts at 660° The density of calcium chloride, according to Arndt, is expressed by the equation: d = 2.03-0.00040 (t° - 850°) .
According to V.P. Borzakovskii, the density of CaCl2 at 800° is 2.049; at 900° 2.001, at 1000° 1.953 Additions of potassium chloride or sodium chloride lower the density of the melt. However, even with significant additions of alkali metal chlorides, the difference in the densities of the melt and metallic calcium is still sufficient for the metal to easily float to the surface of the electrolyte
The value of viscosity and surface tension of calcium chloride at the boundary with the gas phase, according to V.P. Borzakovsky, are given below

Additives of potassium chloride and sodium chloride to calcium chloride reduce the viscosity of the melt and increase the surface tension at the interface with the gas phase
The electrical conductivity of calcium chloride is, according to Borzakovsky: at 800° 2.02 ohm-1/cm3, at 900° 2.33 ohm-1/cm3; a value close to these data was obtained by Sandonini. Additives up to 25% (mol.) potassium chloride, or up to 55% (mol.) sodium chloride lower the electrical conductivity; further increase in additives increases the electrical conductivity of the melt
The vapor pressure of calcium chloride is much higher than that of KCl, NaCl, MgCl2. The boiling point of calcium chloride is approximately 1900°. The total vapor pressure in a mixture of calcium chloride with the indicated chloride salts was studied by V.A. Ilyichev and K.D. Muzhzhavlev.
Calcium chloride decomposition voltage (v) measured by Combi and Devato using emf. polarization in the temperature range 700-1000°, is expressed by the formula

E \u003d 3.38 - 1.4 * 10v-3 (t ° -700 °)

Below is a comparison of the decomposition stresses of several chloride salts at a temperature of 800°.

In practice, with a current output of 60-85%, the back emf in the bath is 2.8-3.2 V. Drossbach points out that the reverse e. d.s. emf answers. cells

Ca / CaCl / CaCl2 / Cl2.

The decomposition voltage of salts decreases with increasing temperature Ho, since the temperature coefficients of change in the decomposition voltage for different salts are different, the sequence of extraction of a particular metal from a mixture of salts can change with temperature. At temperatures of electrolysis of calcium chloride, a discharge of magnesium and sodium ions is possible. Therefore, the calcium bath electrolyte must be free from impurities of these salts.

Electrolysis with touch cathode

Basics of the theory

During the electrolysis of molten calcium chloride, the calcium released at the cathode, as in the production of magnesium or sodium, is much lighter than the electrolyte and therefore floats to the surface of the bath. However, it is not possible to obtain liquid calcium in the same way as magnesium is obtained. Magnesium slightly dissolves in the electrolyte and is protected by an electrolyte film held on the metal surface. Magnesium floating on the electrolyte surface is periodically scooped out. Calcium is much more active than magnesium and is not protected by an electrolyte film. Its solubility in the electrolyte is high, according to Lorentz's research, 13% of the metal dissolves in calcium chloride. When it dissolves, CaCl subchloride is formed, which, reacting with chlorine, turns into CaCl2. Under the action of oxygen and air moisture, subchlorides form a suspension of calcium oxide in the melt. If the molten calcium is allowed to remain in contact with the electrolyte, then, due to the circulation of the latter, the calcium will be carried away to the anode chlorine region and eventually all will turn into calcium chloride. In addition to dissolving in the electrolyte, calcium, being on the surface of the bath, actively reacts with the gases surrounding it.
When calcium is released below its melting point, a spongy dendritic metal is formed, permeated with salt, with a large oxidation surface. The melting of such metal is very difficult. Therefore, metallic calcium with an acceptable current output can only be obtained by the method of Rathenau and Suter - electrolysis with a touch cathode / The essence of the method lies in the fact that the cathode initially touches the molten electrolyte. At the point of contact, a liquid drop of metal that wets the cathode is formed, which, when the cathode is slowly and evenly raised, is removed from the melt with it and solidifies. In this case, the solidifying drop is covered with a solid electrolyte film that protects the metal from oxidation and nitriding. By continuously and carefully lifting the cathode, the calcium is drawn into the rods.
The conditions for electrolysis with a touch cathode on an electrolyte of calcium chloride and calcium fluoride were further studied and improved by Goodwin, who developed an apparatus for laboratory experiments, by Freri, who drew attention to practical methods in electrolysis, by Brace, who built a 200 A bath, and others.
In Russia, this method was studied and improved on baths with a current strength of 100 to 600 A (Z.V. Vasiliev, V.P. Mashovets, B.V. Popov and A.Yu. Taits, V.M. Guskov and M.T. Kovalenko , A.Yu. Taits and M.I. Pavlov, Yu.V. Baimakov).
One of the conditions for achieving a satisfactory current efficiency is the use of a high current density at the cathode. This is necessary so that the amount of metal released per unit time significantly exceeds its dissolution. Depending on the working surface of the cathode, the power of the cell and other factors, the cathode current density is selected in the range of 50–250 A/cm2. For the normal course of the process, it is important to ensure accurate control of the rise of the cathode. Too fast rise of the cathode causes separation of the liquid metal drop and its dissolution in the electrolyte. With a slow rise, calcium overheats and melts away from the rod. Metal separation can also be caused by overheating of the electrolyte. The dissolution of calcium in the electrolyte with the formation of subchloride and calcium oxide causes thickening of the electrolyte and the formation of foam, which disrupts the normal operation of the bath. During the cold course of the bath, the metal on the cathode grows in the form of dendrites.
The current density at the anode is chosen as low as possible (of the order of 0.7-1.5 A/cm2) in order to avoid the anode effect. The anode effect sets in when the current density on graphite reaches 8 A/cm2, and on a carbon anode 5.6 A/cm2. The temperature of the calcium chloride electrolyte without additives is maintained at 800-810°, while with the addition of other salts it decreases. Around the cathode, due to the high current concentration, there is a rim of an overheated electrolyte having a temperature of 820-850 °. In view of the need to maintain the electrolyte temperature close to the melting point of calcium (851°), additives to lower the melting point of the electrolyte are not essential, but their role is positive in reducing the solubility of calcium in the electrolyte.
The electrolyte used should be as dehydrated as possible and not contain harmful impurities. The moisture contained in the electrolyte decomposes with the release of hydrogen at the cathode, which, combining with calcium, forms calcium hydride, which is accompanied by an increase in temperature at the cathode. In addition, moisture contributes to the formation of foam in the electrolyte. All this disrupts the normal course of electrolysis. Another harmful electrolyte impurity is silica, which, even in small amounts, causes calcium to dissolve in the electrolyte. As a result, subchloride is formed and the electrolyte thickens, which makes it difficult to separate calcium at the cathode. Magnesium and sodium impurities are undesirable, since they are released during electrolysis and alloy with calcium, reducing the melting point of the cathode metal and making it difficult to draw.

The practice of electrolysis

The industrial production of calcium by electrolysis with a touch cathode was started even before the First World War in Germany (Biterfeld) and France (Jarry). Montel and Hardy indicate that the power consumption ranged from 30,000-50,000 kWh per 1 g of metal, depending on the size of the cell, its design features and the duration of the electrolysis campaign. The consumption of calcium chloride was 4.5 kg per 1 kg of metal.

The working chamber of the German bath (Fig. 2) has an octagonal shape with a diameter of 400 mm and a height of 350 mm. It is lined with coal blocks that serve as an anode. The space between the blocks and the casing of the bath is lined and covered with thermal insulation. An iron cathode with a diameter of 60 mm is fixed above the working chamber of the bath, which moves in the vertical direction and in the horizontal direction to regulate the voltage on the bath. Air cooling is connected to the cathode, and the air, together with the anode gases, is removed through a channel arranged in the wall of the bath. Bath capacity 40 l for 90 kg of melt. Electrolyte composition, %: 35.46 Ca, 63 Cl, 0.35 CaO, 0.03 SiO2 (max.), 0.04 Fe2O3+Al2O3 (max.). In addition, 1-1.5 kg of potassium chloride is added to the bath, and sometimes a small addition of fluoride salt is also given. Electrolyte temperature 800-820°C, cathodic current density 50-100 A/cm2, anode 1-1.5 A/cm2, bath current 900-2000 A, voltage 20-25 V. The current output fluctuates greatly at different times of the year and depending on the humidity of the air, and averages 35-40%. However, the bath gives from 6 to 15 kg of calcium per day. About 70 kWh of electricity and 8 kg of salt are consumed per 1 kg of calcium. Analysis of impurities in the cathode metal, % (wt.): 0.01-0.08 Mg 0.01-0.05 Si, 0.1-0.3 Fe + Al, 0.05-0.07 Mn, 0.008 -0.03 N, 0.7-1.6 Cl.
According to Bagley's description, in the USA (Michigan) in 1939 a pilot plant was built of three baths for a current of 2000 A, which was soon doubled (Fig. 3). The cathode control was automated, while the operations of periodic addition of electrolyte and removal of calcium rods were performed manually. Subsequently, new series of baths for 4000 A, then for 5000 A and finally for 10000 A were delivered.

The resulting calcium rods have a diameter of 175 to 350 mm and a length of up to 600 mm. The rod is covered with an electrolyte crust on the outside. The inner metal part of the rod is quite compact.
Nevertheless, it should be noted that, despite the existing technical achievements, electrolysis with a touch cathode has serious drawbacks: low current efficiency, high power consumption, low extraction of calcium from raw materials, the need to use an electrolyte that is completely free from impurities of H2O, SiO2, etc. compounds, the difficulty of designing a bath of greater power, etc. All this forced in the last decade, when the demand for calcium has greatly increased, to look for fundamentally different methods of obtaining. The search was unsuccessful.

Liquid cathode electrolysis and production of calcium alloys

Basics of the theory

Obtaining calcium on a liquid metal cathode eliminates the main difficulties encountered in the isolation of pure liquid metal. The fusion of calcium with the cathode metal located at the bottom of the bath under the electrolyte prevents its dissolution in the electrolyte and recombination with chlorine and makes it impossible for calcium to be oxidized by surrounding gases. This results in a high current output. The possibility of close proximity between the electrodes, the absence of a high cathode current density required for electrolysis with a touch cathode, and depolarization during the release of calcium on a liquid cathode can significantly reduce the voltage on the bath. Achieving high performance depends on the choice of cathode, cathode current density, temperature and other process conditions. The cathode metal must be alloyed with calcium, and the value of the cathode current density must correspond to the rate of diffusion of calcium into the alloy. Therefore, agitation of the cathode alloy is useful. Of great importance is the nature of the state diagram of calcium and the cathode metal. So, for example, during the electrolysis of calcium chloride with a liquid lead cathode, it is not possible to obtain rich alloys with a good current efficiency due to the fact that during the formation of the alloy, the melting temperature, according to the state diagram (Fig. 4), increases sharply, reaching at 28% Ca 1106°.

V.M. Guskov and V.F. Fedorov obtained a good current efficiency (89.3%) by stirring the Pb-Ca alloy and saturating it with calcium to 4.4%; the electrolysis temperature was 800-810°. With an increase in the calcium content in the alloy and with an increase in temperature, the current efficiency decreases sharply.
Before the amount of calcium in the alloy reaches 1-2%, the cathode current density can only be increased up to 2 A/cm2. With a further increase in the amount of calcium in the alloy, the current density should be reduced. A similar pattern was established by A.F. Alabyshev.
Due to the different nature of the Ca-Al state diagram, A. Yu. Taits and A.V. Golynskaya electrolysis of calcium chloride with a liquid aluminum cathode produced alloys containing 62% Ca at a temperature of 840–880°C and a cathode current density of 1.5 A/cm2. To prevent the calcium-rich alloy from floating up, 15% potassium chloride was added to the bath, which reduced the density of the electrolyte from 2.03 to 1.84.
According to the state diagram of Zn-Ca (Fig. 5), the electrolytic precipitation of calcium on a zinc cathode with bringing the Ca content to 90% in the alloy is possible at temperatures not exceeding 720°. However, it is difficult to obtain very rich alloys on a zinc cathode due to the floating and suspension of the alloy particles.

Calcium deposition on the copper cathode proceeds well. According to the Cu-Ca state diagram (Fig. 6), the melting point of the alloy lies below 750° when it contains from 25 to 70% Ca, the alloy of this composition does not float, its density, even at a content of 60% Ca, is 4.4 at a density electrolyte 2.2. The electrolytic production of calcium-copper alloys is of exceptional interest for the production of pure calcium. The large difference in the vapor pressure of copper (boiling point 2600°) and calcium (boiling point 1490°) allows distillation to isolate pure calcium from the alloy.

The practice of electrolysis

In industry, electrolysis with lead, zinc and copper cathodes is used. The production of lead alloys with calcium and barium is organized in the USA at the plant of the United Ltd. Company. Each bath is an iron crucible placed in a brickwork, in which external heating is arranged. Approximately 2 tons of lead ingots are loaded into the bath. Lead is covered with a melt layer of pure calcium and barium chlorides 75-100 mm high. A graphite anode is immersed in the center of the bath, with a device for lowering and raising, by moving which the temperature of the bath is regulated. At the bottom, as well as along the walls of the bath, a ledge is formed, which prevents current losses, which are possible due to its flow from the anode to the walls of the bath, bypassing the liquid lead cathode. Calcium and barium released during electrolysis are absorbed by molten lead. It is noted that the efficiency of the process is reduced due to anode effects, metal dissolution and the formation of calcium and barium carbides. The electrolysis is carried out to obtain an alloy containing 2% alkaline earth metals (approximately three days of electrolysis). When the desired concentration is reached, the current is turned off and the alloy is released into the ladle, from which it is poured into the general mixer.
In the GDR, a calcium-zinc alloy was produced at the IGF plant.
The bath (Fig. 7) consists of a cast-iron box measuring 2250x700x540 mm, walled up in brickwork. The anode is six coal blocks with a section of 200X200 mm, suspended on a common shaft with a manual drive for lowering and raising. Zinc is poured onto the bottom of the box, and the alloy accumulates in the bath, from where, at a content of 60-65% Ca, it is periodically scooped out without stopping the bath. The released chlorine is sucked off from above through the cap. Each bath consumes a current of 10,000 A at a voltage of 25 V. The electrolyte is an alloy of calcium chloride with 18% potassium chloride. Electrolysis temperature 750°. The capacity of the bath is 4 kg of calcium in the alloy per hour, the plant produced 10 tons of alloy per month.
In recent years, the electrolysis of calcium chloride with a liquid calcium-copper cathode, followed by distillation of calcium from the alloy, has received wide industrial application.
The electrolytic cell for obtaining a calcium-copper alloy (Fig. 8) is a rectangular cast-iron bath. The width of the bath is 0.90 m and the length is 3 m. The bath is lined with refractory bricks from the outside and enclosed in a metal casing for mechanical strength.

The anode is a package of graphite bars, which are attached to a metal traverse. Current is supplied to the anode through flexible busbars attached to the traverse. The anode can be raised and lowered using a handwheel. Chlorine is pumped out through gas ducts located on the side of the bath. At the bottom of the bath, a copper-calcium alloy is poured, which serves as the cathode. The current strength on such an electrolyzer is 15000 a. Recently, electrolyzers for high current strength have been created. The voltage on the bath is 7-9 V. The daily capacity of the electrolytic cell for 15,000 a is approximately 300 kg of calcium in the alloy.
The technological regime is ensured by the following conditions. The electrolyte temperature is 675°-715°. The composition of the electrolyte is 80-85% calcium chloride and 15-20% potassium chloride. The electrolyte level in the bath is 20-25 cm, the level of the cathode alloy is 5-20 cm. The alloy is saturated with calcium up to 60-65% - The return alloy after distillation contains approximately 30% Ca. The distance between the electrodes is 3-5 cm. The temperature of the electrolyte is regulated by changing the interpolar distance.
Cathodic current density 0.4-0.5 a/cm2, anode 1.0-1.2 a/cm2. There are indications of the use of almost twice as high current densities.
The bath is fed with small portions of solid calcium chloride (20-30 kg each). Unlike electrolyzers with a touch cathode, this bath can be fed with partially dehydrated raw materials containing up to 10% moisture. Its final dehydration occurs on the surface of the bath.
The alloy is removed when the calcium content is not higher than 65%. With a richer alloy, there is a danger of its floating. The alloy is scooped out using a vacuum ladle to a level in the bath of ~5 cm. After draining the rich alloy, the poor alloy and calcium chloride are loaded into the bath
During the electrolysis of calcium chloride with a liquid calcium-copper cathode, the current efficiency is 70-75%. The specific energy consumption is 15,000 - 18,000 kW/h per 1 ton of calcium in the alloy, the consumption of calcium chloride is 3.5 g, and graphite anodes are 60-70 kw per 1 g of calcium in the alloy. Cast iron bathtubs work 10-14 months.

3. Receipt. Calcium is obtained by electrolysis of its molten chloride.

4. Physical properties. Calcium is a silver-white metal, very light (ρ \u003d 1.55 g / cm 3), like alkali metals, but incomparably harder than them and has a much higher melting point, equal to 851 0 C.

5. Chemical properties. Like alkali metals, calcium is a strong reducing agent, which can be schematically depicted as follows:

Calcium compounds color the flame brick red. Like the alkali metals, calcium metal is usually stored under a layer of kerosene.

6. Application. Due to its high chemical activity, metallic calcium is used to reduce some refractory metals (titanium, zirconium, etc.) from their oxides. Calcium is also used in the production of steel and iron, to purify the latter from oxygen, sulfur and phosphorus, to obtain some alloys, in particular, lead-calcium, necessary for the manufacture of bearings.

7. The most important calcium compounds obtained in industry.

Calcium oxide is produced in industry by roasting limestone:

CaCO 3 → CaO + CO 2

Calcium oxide is a refractory substance of white color (melts at a temperature of 2570 0 C), has chemical properties inherent in the main oxides of active metals (I, Table II, p. 88).

The reaction of calcium oxide with water proceeds with the release of a large amount of heat:

CaO + H 2 O ═ Ca (OH) 2 + Q

Calcium oxide is the main component of quicklime, and calcium hydroxide is the main constituent of slaked lime.

The reaction of calcium oxide with water is called lime slaking.

Calcium oxide is mainly used to produce slaked lime.

Calcium hydroxide Ca(OH) 2 is of great practical importance. It is used in the form of slaked lime, lime milk and lime water.

Hydrated lime is a thin loose powder, usually gray in color (a component of calcium hydroxide), slightly soluble in water (1.56 g dissolves in 1 liter of water at 20 0 C). A doughy mixture of slaked lime with cement, water and sand is used in construction. Gradually the mixture hardens:

Ca (OH) 2 + CO 2 → CaCO 3 ↓ + H 2 O

Lime milk is a suspension (suspension) similar to milk. It is formed by mixing excess slaked lime with water. Lime milk is used to produce bleach, in the production of sugar, to prepare mixtures necessary in the fight against plant diseases, and to whitewash tree trunks.

Lime water is a clear solution of calcium hydroxide obtained by filtering milk of lime. Use it in the laboratory to detect carbon monoxide (IV):

Ca(OH) 2 + CO 2 → CaCO 3 ↓ + H 2 O

With prolonged transmission of carbon monoxide (IV), the solution becomes transparent:

CaCO 3 + CO 2 + H 2 O → Ca(HCO 3) 2

If the resulting transparent solution of calcium bicarbonate is heated, then cloudiness occurs again:

Similar processes also take place in nature. If water contains dissolved carbon monoxide (IV) and acts on limestone, then some of the calcium carbonate is converted into soluble calcium bicarbonate. On the surface, the solution warms up, and calcium carbonate again falls out of it.

* Chlorine is of great practical importance. It is obtained by the reaction of slaked lime with chlorine:

2 Ca(OH) 2 + 2 Cl 2 → Ca(ClO) 2 + CaCl 2 + 2H 2 O

The active ingredient in bleach is calcium hypochlorite. Hypochlorites undergo hydrolysis. This releases hypochlorous acid. Hypochlorous acid can be displaced from its salt even by carbonic acid:

Ca(ClO) 2 + CO 2 + H 2 O → CaCO 3 ↓+ 2 HClO

2 HClO → 2 HCl + O 2

This property of bleach is widely used for bleaching, disinfection and degassing.

8. Gypsum. There are the following types of gypsum: natural - CaSO 4 ∙ 2H 2 O, burnt - (CaSO 4) 2 ∙ H 2 O, anhydrous - CaSO 4.

Burnt (semiaqueous) gypsum, or alabaster, (CaSO 4) 2 ∙ H 2 O is obtained by heating natural gypsum to 150–180 0 С:

2 → (CaSO 4) 2 ∙ H 2 O + 3H 2 O

If you mix alabaster powder with water, a semi-liquid plastic mass is formed, which quickly hardens. The hardening process is explained by the addition of water:

(CaSO 4) 2 ∙ H 2 O + 3H 2 O → 2

The property of burnt gypsum to harden is used in practice. So, for example, alabaster mixed with lime, sand and water is used as a plaster. Art objects are made from pure alabaster, and in medicine it is used to apply plaster bandages.

If natural gypsum CaSO 4 ∙ 2H 2 O is heated at a higher temperature, then all the water is released:

CaSO 4 ∙ 2H 2 O → CaSO 4 + 2H 2 O

The resulting anhydrous gypsum CaSO 4 is no longer able to attach water, and therefore it was called dead gypsum.

Hardness of water and ways to eliminate it.

Everyone knows that soap lathers well in rainwater (soft water), while in spring water it usually lathers poorly (hard water). Analysis of hard water shows that it contains significant amounts of soluble calcium and magnesium salts. These salts form insoluble compounds with soap. Such water is unsuitable for cooling internal combustion engines and powering steam boilers, since when hard water is heated, scale forms on the walls of cooling systems. Scale does not conduct heat well; therefore, overheating of motors, steam boilers is possible, in addition, their wear is accelerated.

What are the types of hardness?

Carbonate, or temporary, hardness is due to the presence of calcium and magnesium bicarbonates. It can be fixed in the following ways:

1) boiling:

Ca(HCO 3) 2 → CaCO 3 ↓ + H 2 O + CO 2

Mg(HCO 3) 2 → MgCO 3 ↓ + H 2 O + CO 2

2) the action of lime milk or soda:

Ca(OH) 2 + Ca(HCO 3) 2 → 2CaCO 3 ↓ + 2H 2 O

Ca(HCO 3) 2 + Na 2 CO 3 → CaCO 3 ↓ + 2NaHCO 3

Ca 2+ + 2 HCO 3 - + 2 Na + + CO 3 2- → CaCO 3 ↓ + 2 Na + + 2HCO 3 -

Ca 2+ + CO 3 2- → CaCO 3 ↓

Non-carbonate, or permanent, hardness is due to the presence of sulfates and chlorides of calcium and magnesium.

It is eliminated by the action of soda:

CaSO 4 + Na 2 CO 3 → CaCO 3 ↓ + Na 2 SO 4

MgSO 4 + Na 2 CO 3 → MgCO 3 ↓ + Na 2 SO 4

Mg 2+ + SO 4 2- + 2Na + + CO 3 2- → MgCO 3 ↓ + 2Na + + SO 4 2-

Mg 2+ + CO 3 2- → MgCO 3 ↓

Carbonate and non-carbonate hardness add up to the total hardness of water.

IV. Consolidation of knowledge (5 min.)

1. On the basis of the periodic system and the theory of the structure of atoms, explain what properties of magnesium and calcium are common. Write equations for the corresponding reactions.

2. What minerals contain calcium and how are they used?

3. Tell us how to distinguish one natural mineral from another.

V. Homework (3 min.)

Answer the questions and complete exercises 1-15, § 48,49, solve exercises 1-4, pp. 132-133.

This is exactly what the lesson plan looks like at school on the topic “Calcium and its compounds”.

Based on the foregoing, it is obvious that the school chemistry course needs to be filled with environmental content. The results of this work will be presented in the third chapter.

One-time) - 0.01%. 4 Contents Introduction .................................................. ................................................. ..........................4 Chapter 1. Interdisciplinary connections in the course of the school subject of chemistry on the example of carbon and its compounds .............. ................................................. .........5 1.1 Using interdisciplinary connections to form students ...

Activity. The search for methods and forms of teaching that contribute to the education of a creative personality has led to the emergence of some specific teaching methods, one of which is game methods. The implementation of game teaching methods in the study of chemistry in the conditions of compliance with didactic and psychological and pedagogical features, increases the level of students' preparation. The word "game" in Russian...

and hygienic requirements); compliance of educational and physical activity with the age capabilities of the child; necessary, sufficient and rationally organized motor mode. By health-saving educational technology (Petrov) he understands a system that creates the maximum possible conditions for the preservation, strengthening and development of the spiritual, emotional, intellectual, ...