When passing electric current through metals (conductors of the 1st kind), chemical reactions do not occur, and the metals remain unchanged. If an electric current passes through a melt or an electrolyte solution (conductors of the 2nd kind), various chemical reactions (electrolysis) occur at the electrolyte-metal conductor (electrode) interface and new compounds are formed.

Electrolysis is a set of processes that occur when an electric current passes through an electrochemical system consisting of two electrodes and a melt or electrolyte solution.

During electrolysis, cations move to the negative electrode (cathode), and anions move to the positive electrode (anode). In this case, however, the cations and anions of the electrolyte are not always discharged, accepting or donating electrons. Often, a solvent-electrolyte, such as water, takes part in electrolysis reactions.

The fundamental difference between reactions in a galvanic cell and an electrolyzer lies only in their direction and spontaneity. In a closed circuit of a galvanic cell, an electrochemical reaction proceeds spontaneously, and in an electrolyzer, only under the influence of an electric current from an external source.

Pay attention to the name of the electrodes: in a galvanic cell, the negative electrode is the anode, and the positive electrode is the cathode; in the cell, on the contrary, the negative electrode is the cathode, and the positive one is the anode.

It should be remembered that the terms "negative" and "positive" always refer to the poles of the current source, this is how they designate the electrodes of the cell. The common thing in these processes is that both in the galvanic cell and in the electrolyzer, an excess of electrons is created on the negative electrode, and their deficiency is created on the positive one. At the cathode, ions or molecules are reduced under the action of electrons; at the anode, particles are oxidized, giving their electrons to the electrode.

In the cell, cations (M n+) move to the cathode (-), and anions (A n–) - to the anode (+).

Decomposition voltage electrolyte during electrolysis is called the minimum voltage (external EMF) that must be applied to the electrodes. For example, for a solution of zinc chloride at standard conditions:

Zn 2+ + 2 ē = Zn φ° = - 0.76 B,

Cl2 + 2 ē \u003d 2Cl - φ ° \u003d + 1.36 V,

and the decomposition voltage is equal (in absolute value) to the sum of the standard electrode potentials of both electrodes: 0.76 + 1.36 = 2.12 V, i.e. the decomposition voltage cannot be lower than the EMF of the corresponding galvanic cell.

Decomposition voltage is made up of the potentials of two electrodes - the potentials of the discharge of ions.

Vacuum potential cation is sometimes called deposition potential metal. This is the minimum potential that must be applied to the electrode in order for the cation to lose charge and metal deposition to occur. For some ions (Cu 2+ , Ag + , Cd 2+) the deposition potential is close to the electrode potential, while for other ions (Fe 2 + , Co 2 + , Ni 2 +) the deposition potentials significantly exceed the electrode potentials of metals - for electrolysis it is necessary certain overvoltage.

Distinguish electrolysis of solutions and melt electrolysis. The electrolysis of solutions is divided into electrolysis with inert electrodes and soluble anode electrolysis. Metallic (Pt, Au) and non-metallic (graphite) electrodes can be inert. Anodes from Cr, Ni, Cd, Zn, Ag, Cu, etc. are used as soluble anodes.

Some metals practically do not dissolve due to high anodic polarization, such as Ni and Fe in an alkaline solution, Pb in H 2 SO 4 .

Electrolysis of solutions with inert electrodes. During the electrolysis of aqueous solutions of electrolytes, instead of metal, not metal, but hydrogen is released at the cathode. In acidic environments, hydrogen is formed by the reaction:

2H++2 ē =H2.

In neutral and alkaline media, hydrogen is formed by a reaction involving water molecules:

2H2O+2 ē = H 2 + OH -.

Cations such as Na + or K + do not discharge at all in an aqueous solution, but hydrogen is released.

Cations can be grouped according to their ability to discharge, ranging from non-discharging to easily discharging. At the same time, the products of electrolysis also change. For some cations, the simultaneous formation of metal and hydrogen is possible.

Below are the cations in order of decreasing difficulty of their discharge and the products of electrolysis:

Cations Products of electrolysis

Li + , K + , Na + , Mg 2+ , Al 3+ , H + (redirect) H 2

Mn 2+ , Zn 2+ , Cr 3 + , Fe 2 + , H + (pH 7) M + H 2

Co 2+, Ni 2+, Sr 2+, Pb 2+, H + (pH 0) M + H 2

Cu 2+ , Ag + , Au 3 + M

The different position of hydrogen in this series is explained by the following reasons. The position of hydrogen between lead and copper corresponds to the numerical values ​​of the standard electrode potentials at FROM M n+ = FROM H + = 1 mol/l, i.e. at pH=0. The position of hydrogen between iron and cobalt corresponds to the electrode potential of hydrogen in water at pH=7 (φº H 2 / H + = –0.414 V). Under these conditions, all metals can be precipitated from solutions, the value of φ ° which are greater than –0.414 V. However, in practice, in addition to cobalt, nickel, tin and lead, it is also possible to precipitate zinc, chromium and iron from aqueous solutions. This is due to the fact that the precipitation at the cathode hydrogen gas hampered by hydrogen overvoltage.

Thus, in the series of cations from Li + to A1 3+ metal is not formed, and hydrogen is released during electrolysis due to the reduction of water. In the series of cations from Mn 2+ to Pb 2+ during electrolysis, metal and hydrogen are formed simultaneously, and, finally, in a series Cu 2+ - Au 3+ only metal is formed.

Consequently, the further to the left (closer to the beginning) a metal is in a series of standard electrode potentials (a series of voltages), the more difficult it is to isolate this metal by electrolysis of an aqueous solution.

If a gradually increasing voltage is applied to a solution containing several cations, then electrolysis begins when the deposition potential of the cation with the highest electrode potential (most positive) is reached. During the electrolysis of a solution containing zinc ions (φ °= -0.76 V) and copper (φ ° = +0.34 V), copper is first released on the cathode, and only after almost all Cu 2+ ions are discharged, zinc will begin to be released. In this way, if the solution simultaneously contains different cations, then during electrolysis they can be isolated sequentially in accordance with the values ​​of their electrode potentials. In this case, it is assumed that the overvoltage of metal release for them is approximately the same (and small).

Concerning anion discharge potentials, then here the picture is much more complicated due to the ability of water to participate in the electrolysis process. In general, we can say that anions with the lowest potential (least positive) are first discharged at the anode. If the solution contains Cl - (φº = 1.36 V), Br - (φº = 1.09 V) and I - (φº = 0.54 V) ions, then iodine will be formed first, then bromine and, finally, chlorine. Fluoride ions in an aqueous solution cannot be discharged at all (φ ° = 2.87 V).

Most oxygen-containing anions (except for the acetate ion) do not discharge in an aqueous solution, instead of them in acidic and neutral solutions water decomposes:

2H 2 O - 4 ē \u003d O 2 + 4H +,

and in alkaline solutions - the discharge of hydroxide ions:

2OH - - 2 ē \u003d 1/2 O 2 + H 2 O.

According to their ability to be discharged during the electrolysis of aqueous solutions, anions are arranged in the following row from anions of oxygen-containing acids, such as SO 4 2–, NO 3, that do not discharge in an aqueous solution, to easily discharged ones:

Anions Electrolysis products

SO 4 2–, NO 3 - etc., OH - O 2

Cl -, Br -, I - Cl 2 (ClO -, ClO 3 -), Br 2, I 2 (+ O 2)

S 2– S, SO 2 (+ O 2)

Thus, it is possible to formulate the following main rules for the electrolysis of aqueous solutions of electrolytes with insoluble electrodes:

1. Of the anions of electrolytes, anions of oxygen-free acids (Cl -, Br -, S 2-, etc.) are first discharged at the anode.

2. Anions of oxygen-containing acids (SO 4 2–, NO 3 –, CO 3 2–, etc.) do not discharge in the presence of water, instead of them, water is oxidized according to the reaction:

2H2O-4 ē \u003d O 2 + 4H +.

3. active metals located in a series of voltages up to Al (inclusive) at the cathode are not restored, water is restored instead:

2H2O+2 ē \u003d H 2 + 2OH -.

4. Metals located in the series of voltages after aluminum, but before hydrogen, are reduced at the cathode along with water molecules:

K: 1) Zn 2+ + 2 ē = Zn

2) 2H 2 O + 2 ē \u003d H 2 + 2OH -.

5. Metals with a positive value of the electrode potential are reduced at the cathode in the first place:

Cu 2+ + 2 ē = Cu

For example, during the electrolysis of sulfuric acid (graphite electrodes), the following processes occur:

at the cathode 2H + + 2 ē = H 2

at the anode 2H 2 O - 4 ē \u003d O 2 + H +.

Summary Equation:

2H 2 O \u003d 2H 2 + O 2,

those. during the electrolysis of a solution of sulfuric acid, hydrogen and oxygen are released due to the decomposition of water molecules. Electrolysis products: hydrogen and oxygen.

Electrolysis of copper sulfate solution:

at the cathode Cu 2 + + 2 ē = Cu,

at the anode 2H 2 O - 4 ē \u003d O 2 + 4H +

Summary Equation:

2Cu 2+ + 2H 2 O \u003d 2Cu + O 2 + 4H +

2CuSO 4 + 2H 2 O \u003d 2Cu + O 2 + 2H 2 SO 4.

Electrolysis products: copper, oxygen, sulfuric acid.

The possibility of anion discharge depends on its concentration. Thus, the electrolysis products of concentrated and dilute NaCl solutions are chlorine and oxygen, respectively.

The electrolysis of a dilute solution of sodium chloride takes place without the discharge of Cl - ions (and, accordingly, Na + ions), i.e. water is decomposed. As the salt concentration at the anode increases, chlorine is released along with oxygen, and chlorine is formed in concentrated solutions (with an admixture of oxygen):

at the cathode 2H2O+2 ē \u003d H 2 + 2OH -

at the anode 2Cl - - 2 ē = Cl2 .

Summary Equation:

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

2NaCl + 2H 2 O \u003d H 2 + Cl 2 + 2NaOH.

Electrolysis products: hydrogen, chlorine and sodium hydroxide.

In the case of the release of chlorine during the electrolysis of chloride solutions, the main process of chlorine formation is superimposed by the reactions of interaction of chlorine with water (hydrolysis) and subsequent transformations of the resulting substances. The hydrolysis of chlorine proceeds with the formation of weak hypochlorous acid and chloride ions (hydrochloric acid):

Cl 2 + H 2 O \u003d H + + Cl - + HC1O.

Hypochlorous acid with alkali formed during electrolysis (more precisely, Na + + OH -) gives sodium hypochlorite NaClO as a product. In an alkaline environment, the overall reaction equation has the form:

Cl 2 + 2NaOH \u003d NaCl + NaClO + H 2 O.

At elevated temperatures (boiling water), the hydrolysis of chlorine proceeds with the formation of the chlorate ion. Possible reaction equations:

3Cl 2 + 3H 2 O \u003d ClO 3 - + 5 Cl - + 6H +,

3HClO \u003d ClO 3 - + 2Cl - + 3H +,

3СlО - = СlO 3 - + 2Сl -.

In an alkaline medium, the overall equation has the form

3Cl 2 + 6NaOH \u003d NaClO 3 + 5NaCl + 3H 2 O.

Diaphragm electrolysis. During the electrolysis of a dilute solution of sodium chloride, Na + ions move to the cathode, but hydrogen is released:

2H 2 O+2 ē \u003d H 2 + OH -

and the sodium hydroxide solution is concentrated.

Chloride ions move to the anode, but due to their low concentration, it is not chlorine that is mainly formed, but oxygen:

2H2O-4 ē \u003d O 2 + 4H +

and the solution is concentrated of hydrochloric acid.

If the electrolysis is carried out in a beaker or other similar vessel, the alkali and acid solutions are mixed and the electrolysis is reduced to the formation of hydrogen and oxygen due to the decomposition of water. If, on the other hand, the anode and cathode spaces are separated by a partition (diaphragm) that allows current-carrying ions to pass through but prevents the near-electrode solutions from mixing, then acid and alkali solutions can be obtained as electrolysis products.

During the electrolysis of a sodium chloride solution, hydroxide ions formed at the cathode according to the reaction:

2H2O+2 ē \u003d H 2 + 2OH -

they immediately begin to participate in the transfer of electricity and, together with C1 ions, move to the anode, where both ions are discharged and a mixture of oxygen and chlorine is formed. Therefore, the output of chlorine falls. If the anode is made of coal (graphite), then it is oxidized by oxygen and carbon oxides CO and CO 2 are formed, polluting chlorine. Further, chlorine formed at the anode interacts with hydroxide ions:

C1 2 + OH - \u003d H + + Cl - + OSl -.

The formation of hypochlorite ions is also an undesirable process (if obtaining a sodium hypochlorite solution is not the goal). All these undesirable consequences can be avoided by using a diaphragm separating the cathode and anode spaces and retaining OH - ions, but passing Cl - ions. Finally, the diaphragm prevents the diffusion of gases and makes it possible to obtain purer hydrogen.

If the solution contains several anions, it is more difficult to predict the sequence of their discharge at the anode than for cations, but, generally speaking, the rule is observed that the anion with the lowest potential value (or the highest negative value of the electrode potential of the reaction occurring at the anode).

Electrolysis of solutions with a soluble anode. Electrolysis with a soluble anode is possible when the metal gives off electrons more easily than Cl -, OH - ions or water molecules. For example, on a copper anode in a solution of copper chloride or copper sulfate, chlorine or oxygen is not released, but there is a transition to the solution of Cu 2+ ions. At the same time, the same ions are discharged at the cathode and metallic copper is deposited. Thus, electrolysis with a soluble anode is reduced to the transfer of copper from the anode to the cathode.

The reaction at the anode is in most cases complicated by numerous side and often undesirable processes. For example, the resulting ions can form oxides, hydroxides and their films:

M 2+ + 2OH - \u003d MO + H 2 O.

Oxygen evolution is also possible at the anode:

2H2O-4 ē \u003d O 2 + 4H +,

which can participate in a wide variety of reactions of the electrolytic system.

In the formation of gaseous products, especially oxygen, in most cases the decomposition potentials do not correspond to the electrode potentials due to high overvoltage values . Overvoltage is the difference between the real decomposition voltage and theoretically calculated from the electrode potentials of the EMF of the corresponding reaction. The nature of the liberated substance (for chlorine, bromine and iodine, the overvoltage is very insignificant) and the material of the electrode have a particularly strong influence on the magnitude of the overvoltage. Below are the data on the overvoltage during the evolution of hydrogen and oxygen at various cathodes and anodes.

Electrode Overvoltage, V

Hydrogen Oxygen

Pt blackened 0.00 0.2–0.3

Pt bright 0.1 0.4–0.5

Fe 0.1–0.2 0.2–0.3

Ni 0.1–0.2 0.1–0.3

Сu 0.2 0.2–0.3

Pb 0.4–0.6 0.2–0.3

The overvoltage also depends on the shape of the electrodes, the state of their surface, the current density, the temperature of the solution, the intensity of the mixing of the solution, and other factors.

The overvoltage of hydrogen on iron is ~ 0.1 V, and oxygen on the same material is ~ 0.3 V. Therefore, the overvoltage during electrolysis on iron electrodes will be 0.1 + 0.3 = 0.4 V. The sum of this value and theoretically calculated will be the minimum value of the discharge voltage of the corresponding electrolyte.

The attitude to overvoltage is ambivalent. On the one hand, overvoltage leads to increased power consumption, on the other hand, due to overvoltage, many metals can be deposited from aqueous solutions, which, according to the values ​​of their standard electrode potentials, should not be deposited. These are Fe, Pb, Sn, Ni, Co, Zn , cr. It is due to the overvoltage, as well as the effect of the concentration of the solution on the electrode potential, that electrolytic chromium plating and nickel plating of iron products are possible, and even sodium can be obtained from an aqueous solution on a mercury electrode.

The rarefaction in an aqueous solution of Cl - ions, and not OH - in solutions with a high electrolyte concentration is also explained by oxygen overvoltage. However, this overvoltage is not enough for the discharge of F ions and the release of free fluorine.

The magnitude of the overvoltage is influenced by many other kinetic factors - the rate of particle transfer to the electrodes and the removal of electrolysis products, the rate of the process of destruction of hydrate and other shells of discharged ions, the rate of joining of atoms into diatomic gas molecules, etc.

Solving chemical problems
aware of Faraday's law
high school

Author's development

Among the great variety of various chemical problems, as the practice of teaching at school shows, the greatest difficulties are caused by problems for the solution of which, in addition to solid chemical knowledge, it is required to have a good command of the material of the physics course. And although not every secondary school pays attention to solving at least the simplest problems using the knowledge of two courses - chemistry and physics, problems of this type are sometimes found in entrance exams in universities where chemistry is a major discipline. And therefore, without analyzing problems of this type in the classroom, a teacher can unintentionally deprive his student of the chance to enter a university in a chemical specialty.
This author's development contains over twenty tasks, one way or another related to the topic "Electrolysis". To solve problems of this type, it is necessary not only to have a good knowledge of the topic "Electrolysis" of the school chemistry course, but also to know Faraday's law, which is studied in the school physics course.
Perhaps this selection of tasks will not be of interest to absolutely all students in the class or is available to everyone. Nevertheless, tasks of this type are recommended to be analyzed with a group of interested students in a circle or optional class. It can be noted with certainty that tasks of this type are complicated and at least not typical for a school chemistry course (we are talking about an average general education school), and therefore tasks of this type can be safely included in the variants of the school or district chemical olympiad for 10th or 11th graders.
Availability detailed solution for each task makes development a valuable tool, especially for beginning teachers. Having analyzed several tasks with students in an optional lesson or a circle lesson, a creatively working teacher will certainly set several tasks of the same type at home and use this development in the process of checking homework, which will significantly save valuable teacher time.

Theoretical information on the problem

chemical reactions, flowing under the action of an electric current on electrodes placed in a solution or melt of an electrolyte, is called electrolysis. Consider an example.

In a glass at a temperature of about 700 ° C there is a melt of sodium chloride NaCl, electrodes are immersed in it. Before passing an electric current through the melt, Na + and Cl - ions move randomly, however, when an electric current is applied, the movement of these particles becomes ordered: Na + ions rush to the negatively charged electrode, and Cl - ions - to the positively charged electrode.

And he A charged atom or group of atoms that has a charge.

Cation is a positively charged ion.

Anion is a negatively charged ion.

Cathode- a negatively charged electrode (positively charged ions - cations) move towards it.

Anode- a positively charged electrode (negatively charged ions - anions) move towards it.

Electrolysis of sodium chloride melt on platinum electrodes

Total reaction:

Electrolysis of an aqueous solution of sodium chloride on carbon electrodes

Total reaction:

or in molecular form:

Electrolysis of an aqueous solution of copper(II) chloride on carbon electrodes

Total reaction:

In the electrochemical series of activity of metals, copper is located to the right of hydrogen, therefore copper will be reduced at the cathode, and chlorine will be oxidized at the anode.

Electrolysis of an aqueous solution of sodium sulfate on platinum electrodes

Total reaction:

Similarly, the electrolysis of an aqueous solution of potassium nitrate occurs (platinum electrodes).

Electrolysis of an aqueous solution of zinc sulfate on graphite electrodes

Total reaction:

Electrolysis of an aqueous solution of iron(III) nitrate on platinum electrodes

Total reaction:

Electrolysis of an aqueous solution of silver nitrate on platinum electrodes

Total reaction:

Electrolysis of an aqueous solution of aluminum sulfate on platinum electrodes

Total reaction:

Electrolysis of an aqueous solution of copper sulfate on copper electrodes - electrochemical refining

The concentration of CuSO 4 in the solution remains constant, the process is reduced to the transfer of the anode material to the cathode. This is the essence of the process of electrochemical refining (obtaining pure metal).

When drawing up schemes for the electrolysis of a particular salt, it must be remembered that:

– metal cations having a higher standard electrode potential (SEP) than that of hydrogen (from copper to gold inclusive) are almost completely reduced at the cathode during electrolysis;

– metal cations with small SEP values ​​(from lithium to aluminum inclusive) are not reduced at the cathode, but instead water molecules are reduced to hydrogen;

– metal cations, whose SEC values ​​are less than those of hydrogen, but greater than those of aluminum (from aluminum to hydrogen), are reduced simultaneously with water during electrolysis at the cathode;

- if the aqueous solution contains a mixture of cations of various metals, for example, Ag +, Cu 2+, Fe 2+, then silver will be the first to be reduced in this mixture, then copper, and the last iron;

- on an insoluble anode during electrolysis, anions or water molecules are oxidized, and anions S 2–, I –, Br – , Cl – are easily oxidized;

– if the solution contains anions of oxygen-containing acids , , , , then water molecules are oxidized to oxygen at the anode;

- if the anode is soluble, then during electrolysis it itself undergoes oxidation, i.e. it sends electrons to the external circuit: when electrons are released, the balance between the electrode and the solution is shifted and the anode dissolves.

If from the whole series of electrode processes we single out only those that correspond to the general equation

M z+ + ze=M,

then we get metal stress range. Hydrogen is also always placed in this row, which makes it possible to see which metals are able to displace hydrogen from aqueous solutions of acids, and which are not (table).

Table

A range of stress metals

The equation electrode process	Standard electrode potential at 25 °С, V	The equation electrode process	Standard electrode potential at 25 °C, V
Li + + 1 e= Li0	–3,045	Co2+ + 2 e= Co0	–0,277
Rb + + 1 e= Rb0	–2,925	Ni 2+ + 2 e= Ni0	–0,250
K++1 e= K0	–2,925	Sn 2+ + 2 e= Sn0	–0,136
Cs + + 1 e= Cs 0	–2,923	Pb 2+ + 2 e= Pb 0	–0,126
Ca 2+ + 2 e= Ca0	–2,866	Fe 3+ + 3 e= Fe0	–0,036
Na + + 1 e= Na 0	–2,714	2H++2 e=H2	0
Mg 2+ + 2 e=Mg0	–2,363	Bi 3+ + 3 e= Bi0	0,215
Al 3+ + 3 e=Al0	–1,662	Cu 2+ + 2 e= Cu 0	0,337
Ti 2+ + 2 e= Ti0	–1,628	Cu + +1 e= Cu 0	0,521
Mn 2+ + 2 e=Mn0	–1,180	Hg 2 2+ + 2 e= 2Hg0	0,788
Cr 2+ + 2 e=Cr0	–0,913	Ag + + 1 e= Ag0	0,799
Zn 2+ + 2 e= Zn0	–0,763	Hg 2+ + 2 e= Hg0	0,854
Cr 3+ + 3 e=Cr0	–0,744	Pt 2+ + 2 e= Pt0	1,2
Fe 2+ + 2 e= Fe0	–0,440	Au 3+ + 3 e= Au 0	1,498
CD 2+ + 2 e= CD 0	–0,403	Au++1 e= Au 0	1,691

In a simpler form, a series of metal stresses can be represented as follows:

To solve most electrolysis problems, knowledge of Faraday's law is required, the formula expression of which is given below:

m = M I t/(z F),

where m is the mass of the substance released on the electrode, F- Faraday number, equal to 96 485 A s / mol, or 26.8 A h / mol, M – molar mass an element that is reduced in the process of electrolysis, t– the time of the electrolysis process (in seconds), I- current strength (in amperes), z is the number of electrons involved in the process.

Task Conditions

1. What mass of nickel will be released during the electrolysis of a nickel nitrate solution for 1 hour at a current of 20 A?

2. At what current strength is it necessary to carry out the process of electrolysis of a solution of silver nitrate in order to obtain 0.005 kg of pure metal within 10 hours?

3. What mass of copper will be released during the electrolysis of a copper (II) chloride melt for 2 hours at a current of 50 A?

4. How long does it take to electrolyze an aqueous solution of zinc sulfate at a current of 120 A in order to obtain 3.5 g of zinc?

5. What mass of iron will be released during the electrolysis of an iron(III) sulfate solution at a current of 200 A for 2 hours?

6. At what current strength is it necessary to carry out the process of electrolysis of a solution of copper (II) nitrate in order to obtain 200 g of pure metal within 15 hours?

7. During what time is it necessary to carry out the process of electrolysis of a melt of iron (II) chloride at a current of 30 A in order to obtain 20 g of pure iron?

8. At what current strength is it necessary to carry out the process of electrolysis of a solution of mercury (II) nitrate in order to obtain 0.5 kg of pure metal within 1.5 hours?

9. At what current strength is it necessary to carry out the process of electrolysis of a sodium chloride melt in order to obtain 100 g of pure metal within 1.5 hours?

10. The potassium chloride melt was subjected to electrolysis for 2 hours at a current of 5 A. The resulting metal reacted with water weighing 2 kg. What concentration of alkali solution was obtained in this case?

11. How many grams of a 30% hydrochloric acid solution will be required for complete interaction with iron obtained by electrolysis of a solution of iron (III) sulfate for 0.5 h at current strength
10 A?

12. In the process of electrolysis of a melt of aluminum chloride, carried out for 245 min at a current of 15 A, pure aluminum was obtained. How many grams of iron can be obtained by the aluminothermic method when a given mass of aluminum interacts with iron(III) oxide?

13. How many milliliters of a 12% solution of KOH with a density of 1.111 g / ml will be required to react with aluminum (with the formation of potassium tetrahydroxyaluminate) obtained by electrolysis of an aluminum sulfate solution for 300 minutes at a current of 25 A?

14. How many milliliters of a 20% sulfuric acid solution with a density of 1.139 g / ml will be required to interact with zinc obtained by electrolysis of a zinc sulfate solution for 100 minutes at a current of 55 A?

15. What volume of nitric oxide (IV) (n.o.) will be obtained by the interaction of an excess of hot concentrated nitric acid with chromium obtained by electrolysis of a solution of chromium(III) sulfate for 100 min at a current of 75 A?

16. What volume of nitric oxide (II) (n.o.) will be obtained when an excess of nitric acid solution reacts with copper obtained by electrolysis of a copper(II) chloride melt for 50 minutes at a current strength of 10.5 A?

17. During what time is it necessary to carry out the electrolysis of a melt of iron (II) chloride at a current of 30 A in order to obtain the iron necessary for complete interaction with 100 g of a 30% hydrochloric acid solution?

18. How long does it take to electrolyze a nickel nitrate solution at a current of 15 A in order to obtain the nickel necessary for complete interaction with 200 g of a 35% sulfuric acid solution when heated?

19. The sodium chloride melt was electrolyzed at a current of 20 A for 30 minutes, and the potassium chloride melt was electrolyzed for 80 minutes at a current of 18 A. Both metals were dissolved in 1 kg of water. Find the concentration of alkalis in the resulting solution.

20. Magnesium obtained by electrolysis of a magnesium chloride melt for 200 min at current strength
10 A, dissolved in 1.5 l of a 25% sulfuric acid solution with a density of 1.178 g / ml. Find the concentration of magnesium sulfate in the resulting solution.

21. Zinc obtained by electrolysis of a solution of zinc sulfate for 100 min at current strength

17 A, was dissolved in 1 l of a 10% sulfuric acid solution with a density of 1.066 g/ml. Find the concentration of zinc sulfate in the resulting solution.

22. Iron obtained by electrolysis of a melt of iron(III) chloride for 70 min at a current of 11 A was powdered and immersed in 300 g of an 18% copper(II) sulfate solution. Find the mass of copper precipitated.

23. Magnesium obtained by electrolysis of a magnesium chloride melt for 90 minutes at current strength
17 A, were immersed in an excess of hydrochloric acid. Find the volume and amount of hydrogen released (n.o.s.).

24. A solution of aluminum sulfate was subjected to electrolysis for 1 hour at a current of 20 A. How many grams of a 15% hydrochloric acid solution would be required for complete interaction with the resulting aluminum?

25. How many liters of oxygen and air (N.O.) will be required for the complete combustion of magnesium obtained by electrolysis of a magnesium chloride melt for 35 minutes at a current of 22 A?

See the following numbers for answers and solutions

Commercial grades of commercially pure iron (Armco type) obtained by the pyrometallurgical method have a purity of 99.75-99.85%. Further removal of mainly non-metallic impurities (C, O, S, P, N) contained in this iron is possible by special remelting in high vacuum or annealing in an atmosphere of dry hydrogen. However, even after such treatment, the content of impurities reaches 2000-1500 parts per million parts of iron, and the main impurities are C, P, S, Mn and O.
iron more high degree purity is obtained by electrolytic and chemical methods, but it also requires additional complex purification.
By electrolytic methods, iron is obtained from moderately concentrated or concentrated solutions of iron chloride or sulfate, respectively, at low current densities and room temperatures or high densities and temperatures of the order of 100°.
According to one of the methods, iron was precipitated from a solution of the following composition, g/l: 45-60 Fe2+ (in the form of FeCl2), 5-10 BaCl2 and 15 NaHCOs. Plates made of Armco iron or Ural roofing iron were used as anodes, and pure aluminum was used as cathodes. The electrolysis was carried out at room temperature and a current density of 0.1 A/dm2. A precipitate with a coarse crystalline structure was obtained, containing about 0.01% C, traces of phosphorus and not containing sulfur.
The purity of the electrolytic iron depends on the purity of the electrolyte and the purity of the metal of the anodes. During precipitation, impurities nobler than iron, such as tin, zinc, copper, can be removed. He amenable to removal nickel, cobalt, manganese. The total content of impurities in electrolytic iron is approximately the same as in commercially pure iron. It usually contains a significant amount of oxygen (up to 0.1-0.2%), as well as sulfur (0.015-0.05%), if the precipitation was carried out from sulfate baths.
Removal of oxygen from electrolytic iron is carried out by reduction processes: treatment of liquid or solid metal with hydrogen or deoxidation of the melt with carbon in a vacuum. Annealing in a stream of dry hydrogen at 900-1400°C reduces the oxygen content to 0.003%.
To obtain high-purity iron on a semi-industrial scale, a hydrogen reduction method is used in a vacuum melting plant. Electrolytic iron is first subjected to desulfurization with manganese additive in a crucible of lime and fluorspar in an atmosphere of carbon monoxide (sulfur content decreased from 0.01 to 0.004%), then the melt is reduced with hydrogen by blowing or blowing in a crucible of aluminum oxide. In this case, it was possible to reduce the oxygen content to 0.004-0.001%. Metal desulfurization can also be carried out in high vacuum, using additions to the melt of such metals (tin, antimony, bismuth), which form volatile sulfides. By deoxidizing the melt with carbon in high-vacuum furnaces, it is possible to obtain iron with an oxygen and carbon content of up to 0.002% each.
Obtaining iron with a lower oxygen content by deoxidation in high vacuum is hampered by the interaction of the metal with the crucible material, which is accompanied by the transition of oxygen into the metal. best material The crucibles providing the minimum oxygen transfer are ZrO2 and ThO2.
High-purity iron is also obtained by the carbonyl method from pentacarbonyl Fe (CO) 5 by its decomposition at 200-300 °. Carbonyl iron does not usually contain impurities associated with iron - sulfur, phosphorus, copper, manganese, nickel, cobalt, chromium, molybdenum, zinc, silicon. Specific impurities in it are carbon and oxygen. The presence of oxygen is due to secondary reactions between the resulting carbon dioxide and iron. The carbon content reaches 1%; it can be reduced to 0.03% by adding a small amount of ammonia to the iron carbonyl vapor or by treating iron powder in hydrogen. The removal of carbon and oxygen is achieved by the same vacuum melting methods that are used for electrolytic iron.
The purest iron can be obtained chemically, but this method is very complicated and makes it possible to obtain metal in small quantities. AT chemical methods recrystallization, precipitation reactions or extraction of impurities by precipitation are used to purify iron salts from Co, Ni, Cu, Cr, Mn impurities.
One of chemical methods, which allows to obtain iron of a very high degree of purity (less than 30-60 parts per million of impurities), includes the following successive stages:
1) extraction of the FeCl3 complex with ether from a 6-n HCl solution with regeneration of the aqueous solution and subsequent extraction of the ether;
2) reduction of FeCls to FeCl2 with high purity iron;
3) additional purification of FeCl2 from copper by treatment with a sulfur reagent and then with ether;
4) electrolytic deposition of metal from FeCl2 solution;
5) annealing of metal grains in hydrogen (to remove oxygen and carbon);
6) obtaining compact iron by powder metallurgy (pressing into bars and sintering in hydrogen)
The last stage can be carried out by crucibleless zone melting, which eliminates the disadvantage of vacuum processing - the transfer of oxygen from the crucible to the metal.

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Making iron (read cast iron and steel) by electrolysis rather than conventional smelting could prevent the emission of a billion tons carbon dioxide into the atmosphere every year. So says Donald Sadoway of the Massachusetts Institute of Technology (MIT), who has developed and tested a "green" way to produce iron by electrolysis of its oxides.

If the process, demonstrated in a laboratory setting, could be scaled up, it could eliminate the need for conventional smelting, which releases almost a ton of carbon dioxide into the atmosphere for every ton of steel produced.

In conventional technology, iron ore is combined with coke. The coke reacts with iron to produce CO2 and carbon monoxide, and leaving an alloy of iron and carbon - cast iron, which can then be melted into steel.

In the Sadoway method, iron ore is mixed with a solvent - silicon dioxide and quicklime - at a temperature of 1600 degrees Celsius - and an electric current is passed through this mixture.

The negatively charged oxygen ions migrate to the positively charged anode, from where the oxygen escapes. The positively charged iron ions migrate to the negatively charged cathode, where they are reduced to iron, which collects at the base of the cell and is pumped out.

A similar process is used in the production of aluminum (and requires a decent amount of electricity), the oxide of which is so stable that it cannot actually be reduced with carbon in a blast furnace, in which, for example, pig iron is produced. And it is clear that the steel industry never had any reason to switch to the electrolysis of iron ore, since it is easily reduced by carbon.

But if governments different countries begin to impose heavy taxes on greenhouse gas emissions - carbon dioxide, in particular, then a new method of producing pig iron could become more attractive. True, from laboratory installations of this kind to industrial installations, as scientists estimate, it will take 10-15 years.

The author of the work says that the biggest obstacle is to find a practical material for the anode. In experiments, he used an anode made of graphite. But, unfortunately, carbon reacts with oxygen, releasing just as much carbon dioxide into the air as normal iron smelting.

Ideal platinum anodes, for example, are too expensive for large scale production. But there may be a way out - in the selection of some resistant metal alloys that form an oxide film on their outer surface, but still conduct electricity. Conductive ceramics can also be used.

Another problem is that new process uses a lot of electricity - about 2 thousand kilowatt-hours per ton of iron produced. So, the economic and even ecological sense in a new method of iron production will appear only on the condition that this electricity will be generated in some ecological, and at the same time cheap, way, without carbon dioxide emissions. This is acknowledged by the author of the method himself.