Hydrogen is released during electrolysis of an aqueous solution. Example: Electrolysis of an aqueous solution of magnesium chloride on inert electrodes

U min

Rice. 3.1. Polarization curves during electrolysis.

Example 3.1. Consider electrolysis melt potassium chloride salts KCl on insoluble Pt electrodes. Write the equations electrode processes. Calculate the minimum potential difference U min electrolysis.

Solution. 1) Let's write down the ionic composition of the electrolyte:

KCl→ K + +Cl -

2) and 3) standard potentials of electrode processes:

TO - :
IN

A+:
IN.

4)Electrode processes:

K - : K + + → K

A + : 2Cl - → Cl 2 + 2 .

This electrolysis can be used to produce lithium and chlorine.

U min =E = E 0 Cl - / Cl 2 - E 0 K + /K =1.36V – (-2.925V) = 4.285V

Polarization curves:

Example 3.2. Determine the minimum potential difference U min , which must be supplied to the Pt electrodes to carry out the electrolysis of an aqueous solution of KOH, pH = 12. Write the equations for the electrolysis processes. Calculate the volumes of gases (reduced to normal conditions) that are formed on the electrodes in 10 hours at a current of 5A.

Solution. 1) In order to determine the ionic composition of the electrolyte, we write the dissociation equations of the electrolyte solution:

KOH → K + + OH - ;H 2 O H + +OH -

2) Distribution of ions over the electrodes:

A () (OH -), K () (K +, H +)

3) let us determine the equilibrium potentials of possible electrode processes:

TO - :
B,
IN,

A+:
IN.

4) Since E 0 K + /K is significantly more negative
, then only the reduction process of H + ions will occur at the cathode, and the oxidation process of OH - ions at the anode:

K - : 2H 2 O + 2e → H 2 + 2OH - ,

A + : 2OH - - 2e → 1/2O 2 + H 2 O.

Minimum potential difference for electrolysis of a given solution (back EMF):

The volume of gases released on the electrodes will be calculated using Faraday’s law (normal conditions):

l,

l.

Electrolysis of an aqueous solution of potassium hydroxide is widely used for the electrochemical production of hydrogen.

Example 3.3. Consider the electrolysis of an aqueous solution of CuCl 2 on graphite (insoluble) electrodes. Write the electrode processes, show the course of the polarization curves. Calculate the mass of copper formed at the cathode if, during the same time, 5.6 mlCl 2 and 5.6 mlO 2 were released at the anode.

Solution.

CuCl 2 → Cu 2+ + 2Cl -

H2O H + + OH - .

The salt CuCl 2 is formed by a weak base Cu(OH) 2 and a strong acid HCl, therefore, when it is dissolved in water, a hydrolysis process will occur with the formation of an excess of H + ions, the electrolyte solution will have a weakly acidic reaction of the medium (assuming pH = 5).

Let's determine the potentials of possible processes on the anode and cathode and write down the equations of electrode processes:

TO - :
B,
B,

because
more positive than
, then only the process of reduction of copper ions Cu 2+ from the electrolyte solution will take place at the cathode.

A+:
IN,
IN,

because
more negative than
, then first of all the process of oxidation of OH - ions will take place at the anode. However, due to polarization at high current densities, the potentials of the processes of oxygen and chlorine evolution are quite close, so the process of oxidation of Cl – ions from the electrolyte solution will also occur at the anode. Thus, the following processes occur on the electrodes:

K - : Cu 2+ + 2 → Cu

A + : 2H 2 O → O 2 + 4H + + 4

2Cl - → Cl 2 + 2 .

Electrolysis of this solution can be carried out to apply a copper coating to the product, as well as to produce oxygen and chlorine gases.

Rice. 3.2. Polarization curves of the electrolysis process of an aqueous solution of copper chloride on insoluble electrodes.

Let us determine the mass of copper formed at the cathode, for which we first calculate the volumes of moles of gas equivalents at normal conditions. and the mass of a mole equivalent of copper:

l/mol,
l/mol,
g/mol.

Using Faraday's law, we determine the amount of electricity required to release given volumes of oxygen and chlorine at the anode (no):

Kl,

Cl.

The total amount of electricity passing through the anode is:

Cl.

The same amount of electricity at the cathode ( Q K = Q A) will go through only one process of copper formation. Using Faraday's law, we determine the mass of released copper:

g = 48.3 mg

Let us determine the current output ( B j ) for all electrolysis processes:

%, (since there is one process at the cathode);

% ;
%.

Example 3.4. Consider the electrolysis of an aqueous solution of CuCl 2 on copper electrodes. Write equations for electrode processes and show the course of polarization curves. How does the course of the polarization curves in this version differ from the version considered in example 3.3?

Solution. The ionic composition of the electrolyte solution is the same as in example 3.3. Therefore, at the cathode, as in the case of electrolysis on insoluble electrodes, only the process of reduction of copper ions will occur.

Potentials of possible processes at the anode:

IN,
V, (see example 3.3), potential of the anode material
B. Since the equilibrium potential of copper oxidation is much more negative than the equilibrium potentials of oxygen and chlorine release, then at the anode, first of all, the process will begin oxidation of the copper electrode. If during electrolysis at the cathode and anode the equilibrium potentials of the systems are not achieved
And
(small polarizations  E K,  E A and current density i), then the electrode processes will be as follows:

K - : Cu 2+ + 2 → Cu

A+ : Cu → Cu 2+ + 2 .

At high cell voltages U, can be achieved
,
And
, then gas evolution will begin and the equations from example 3.3 will be added to the indicated equations of electrode processes.

Due to the dissolution of the copper anode under the influence of current, the supply of Cu 2+ ions in the electrolyte solution will be replenished, and the process of formation of a copper coating on the cathode will proceed more intensively than in the case of using inert electrodes (Ex. 3.3.).

Rice. 3.3. Polarization curves of the electrolysis process of an aqueous solution of copper chloride on copper electrodes.

Example 3.5. Consider the electrolysis of an aqueous solution of a mixture of salts Pb(NO 3) 2 and Sn(NO 3) 2 on graphite (insoluble) electrodes. Write the equations of electrode processes. Calculate the current efficiency of substances if 30 g of Sn, 52 g of Pb and 2.8 l of H 2 are simultaneously formed at the cathode (normal conditions).

Solution. Let's determine the ionic composition of the electrolyte solution and estimate the hydrogen index of the medium. Let us write down the equations for the dissociation of salt and water molecules:

Pb(NO 3) 2 → Pb 2+ + 2NO 3 -

Sn(NO 3) 2 → Sn 2+ + 2NO 3 -

H2O H + +OH - .

Salts Sn(NO 3) 2 and Pb(NO 3) 2 are formed weak grounds and strong acid, therefore, when they are dissolved in water, a hydrolysis process will occur with the formation of an excess of H + ions, the electrolyte solution will have a slightly acidic reaction of the medium (let’s assume pH ≈ 5).

Let us determine the equilibrium potentials of possible processes at the anode and cathode:

TO - :
B,
B,

IN.

because
,
And
have close value, then at the cathode the processes of reduction of Pb 2+, Sn 2+ and H + ions from the electrolyte solution will proceed in parallel. At the anode, NO 3 - ions, as complex oxygen-containing ions, will not be oxidized, and in this electrolyte solution at the insoluble anode only the oxidation process of OH - ions will occur.

Thus, the following processes occur on the electrodes:

K - : Pb 2+ + 2e → Pb

Sn 2+ + 2e → Sn

A + :H 2 O→O 2 + 4H + + 4 .

Let us write down the masses and volume (under normal conditions) of moles of equivalents of substances formed at the cathode:

g/mol,
g/mol,
l/mol (n.s.).

Using Faraday's law, we determine the amount of electricity required to be generated at the cathode given quantity substances (n.s.):

Kl,

Kl,

Cl.

The total amount of electricity passing through the cathode:

Let us determine the current output ( B j) for all electrolysis processes:

o / o , (since one process occurs at the anode);

100 % =
100% = 40,2%;

100%= 39,9%;
100% = 19,9%.

Rice. 3.4. Polarization curves of the electrolysis process of an aqueous solution of a mixture of salts Pb(NO 3) 2 and Sn(NO 3) 2 on graphite (insoluble) electrodes.

Example 3.6. Consider the process of refining nickel containing impurities of zinc and copper in an aqueous solution of H 2 SO 4. What processes will take place at the anode and cathode? How long does it take to carry out refining at a current of 500 A to separate 5 kg of nickel with a current efficiency of 98%?

Solution. Refining is the purification of metal from impurities using electrolysis. The base metal and impurities, the potential of which is more negative than the base metal, dissolve at the anode. Impurities having more positive potential, do not dissolve and fall out of the anode in the form of sludge. At the cathode, the metal with the most positive potential is released first.

The anode is a purified Ni metal with Zn and Cu impurities. Ionic composition of the electrolyte solution: H + , SO 4 2- , OH - . Let us write down the equilibrium potentials of possible electrode processes at pH = 2:

B,
B,
B,

IN,
B.

Because



, then the first process at the anode during refining will be the oxidation of zinc impurities, then the oxidation of the base metal (nickel), copper impurities do not dissolve, but precipitate (sludge) in the form of metal particles at the end of the process.

Because

, and the concentration of nickel ions is higher than the concentration of zinc ions, then pure nickel is deposited on the cathode. However, at the beginning of the process, when there are no Ni 2+ ions in the electrolyte solution, the process of hydrogen evolution occurs at the cathode.

Let us write down the equations of electrode processes:

A + :Zn→Zn 2+ + 2e

K - : 2H + + 2e→H 2

Ni 2+ + 2e→Ni.

We calculate the time required for refining according to Faraday’s law (
g/mol) :

or τ = 9.27 hours.

4. ELECTROCHEMICAL CORROSION OF METALS

Corrosion is the spontaneous destruction of metal materials under the influence of environmental components. As a result of corrosion, a total redox reaction occurs between the metal and the oxidizing agent:

nM+mOx→M n Red m , (4.1).

In this case, oxidation of the metal and destruction of metal structures occurs.

For a metal, it is the oxidized state that is thermodynamically more stable. Therefore, the corrosion process is always spontaneous, i.e. the change in Gibbs energy during corrosion is negative
.

Thermodynamic calculation
allows only to determine the possibility of the corrosion process occurring, but does not give a real idea of ​​​​the corrosion rate. For example, for a process:

4Al + 3O 2 + 6H 2 O = 4Al(OH) 3,
kJ< 0.

It would seem that aluminum should corrode intensively under the influence of oxygen dissolved in water. However, aluminum is widely used as a structural material. The reason is the properties of corrosion products. Corrosion is a heterogeneous process occurring at the M – Ox interface. Corrosion products can form oxide, hydroxide, salt and other films on the metal surface that have protective properties that make it difficult for the metal to contact the oxidizing agent and inhibit the further corrosion process. As a result of this reaction, a dense protective film is formed on the surface of aluminum, causing passivation of the metal and protecting it from corrosion, therefore aluminum is stable in the atmosphere.

According to the mechanism of flow, a distinction is made between chemical (in a medium that does not conduct electric current, for example, in dry gas, in aggressive organic liquids) and electrochemical (in a medium with ionic conductivity, for example, in aqueous solutions of salts, acids, bases, etc.) sea ​​water, in the atmosphere, in the soil) corrosion. The most common is electrochemical corrosion.

In electrochemical corrosion, the destruction of the metal occurs as a result of its anodic oxidation. The metal surface is energetically inhomogeneous. In areas with a more negative potential value, the process of metal oxidation occurs. Such areas play the role of anodes of corrosive galvanic cells and are oxidized:

A – : M → M n + +n .

In areas of the metal that have a more positive potential value, processes of reduction of oxidizing agents present in the environment take place:

K+:Ox+n →Red.

The Gibbs energy of the electrochemical corrosion process is directly related to the EMF of the corrosive galvanic cell:

, (4.2).

The EMF of a corrosive galvanic cell is equal to the difference in the equilibrium potentials of the metal and the oxidizing agent:

, (4.3).

Therefore, electrochemical corrosion is possible if
or
. To establish the possibility of oxidation of a given metal under the influence of a possible oxidizing agent, it is necessary to compare the potentials of the metal and the oxidizing agent in a given environment. The equilibrium potential of the anodic reaction of metal oxidation and the equilibrium potentials of reduction of oxidizing agents (H +, O 2) are calculated using the Nernst equation. For evaluation calculations, standard electrode potentials of metals can be used
.

The most common oxidizing agents in electrochemical corrosion are air oxygen O 2 dissolved in the electrolyte and hydrogen ions H + . In this regard, the following may be observed:

Corrosion with oxygen depolarization, corrosion with oxygen absorption, if
, (for example, corrosion of Cu, Ag in a neutral environment in air), dissolved O 2 acts as an oxidizing agent:

O 2 + 2H 2 O+ 4 → 4OH - , (pH7);

O2 + 4H + + 4 → 2H 2 O, (pH< 7);

Corrosion with hydrogen depolarization, corrosion with hydrogen evolution, if
, (for example, corrosion of Fe, Cd in acid), in this case H + acts as an oxidizing agent:

2H + + 2 → H 2 , (pH< 7);

2H2O+2 → H 2 + 2OH – (pH  7);

Corrosion with mixed depolarization, if
,
, (for example, corrosion of Mg in a neutral environment in air), simultaneously dissolved O 2 and H + act as an oxidizing agent.

If
,
, then under these conditions the process of electrochemical corrosion of the metal will not occur (for example, Pt, Au do not corrode in a neutral environment in air).

The main characteristics of electrochemical corrosion are the stationary corrosion potential E core, installed on the surface of the metal, during which conjugate reactions of ionization M and reduction Ox and corrosion current occur I core, or corrosion current density i core, reflecting the rate of the corrosion process in electrical units. The corrosion rate can be expressed in terms of metal loss per unit time, in terms of the magnitude of the current or the corrosion current density, calculated according to Faraday's law.

The rate of the electrochemical corrosion process is determined according to the laws of electrochemical kinetics. The rate of electrochemical corrosion is generally limited by the rate of the slowest stage of the process. For most metals, the cathodic reaction is the limiting reaction.

If corrosion with hydrogen depolarization occurs, then the rate of the corrosion process is determined by the rate of cathodic hydrogen evolution. The slowest stage of this process, which determines the rate of the entire corrosion process as a whole, is the reaction of the reduction of hydrogen ions to atomic hydrogen adsorbed by the surface H ads:

N + + → N ads

The rate of this process depends on the nature of the cathode sites on the surface of which it occurs. The presence of cathode impurities Hg, Pb, Cd, Zn in the metal composition slows down the rate of hydrogen evolution and the rate of corrosion in general.

If electrochemical corrosion with oxygen depolarization occurs, then the rate of the process is determined by the rate of cathodic reduction of oxygen. The limiting stage of this cathodic process is the process of diffusion of oxygen molecules through the diffusion layer. Changing the composition of cathode impurities in a metal alloy has little effect on the rate of cathodic reduction of oxygen. The rate of cathodic reduction of oxygen is determined by the limiting current density i etc:

i etc = 4F . D O2 . c O2 . δ -1 , (4.4)

Where D O2- oxygen diffusion coefficient;

c O2- oxygen concentration in solution;

δ - thickness of the diffusion layer.

Stirring the corrosive environment significantly increases the rate of corrosion with oxygen absorption.

In some cases, the rate of electrochemical corrosion is limited by the anodic oxidation reaction of the metal. This is typical for metals that can be passivated (Cr, Al, Ti). Passivation is caused by the formation on the metal surface of a dense, poorly soluble protective film of corrosion products, which inhibits the anodic process and the rate of electrochemical corrosion in general.

Corrosion protection methods include:

– alloying (usually with components that increase the passivation of the metal (Cr, Ni, Al, Mn, Mo, Cu);

– protective metal (anodic and cathodic) and non-metallic coatings;

– electrochemical protection: a) cathodic protection – connecting the protected product to the negative pole of an external current source, in which case it becomes a cathode and does not oxidize, b) connecting a protector to the protected product – metal with a more negative potential value, c) anodic protection – connecting the protected metal to the positive pole of an external current source and transferring it to a passive state, applicable to metals capable of passivation (Cr, Al, Ti, Zr, etc.);

– change in the properties of the corrosive environment (removal of dissolved oxygen, increase pH, adding corrosion inhibitors).

Let's consider a method of protecting metal from electrochemical corrosion using a metal coating. Based on the nature of the behavior of metal coatings during corrosion, they can be divided into cathodic and anodic. Cathodic coatings include coatings whose potentials in a given environment have a more positive value than the potential of the base metal. When the coating is damaged, a corrosion element appears in which the base metal serves as an anode and dissolves, and the coating metal acts as a cathode on which the oxidizing agent is reduced. Anodic coatings have a more negative potential than the potential of the base metal. In this case, the base metal serves as the cathode of the corrosion element, so it does not corrode when the coating is damaged.

Example 4.1. There is a junction of tin (Sn) with silver (Ag). Determine the possibility of corrosion when operating the product in an alkaline environment (pH = 9) in contact with oxygen. Write equations for possible corrosion processes.

Solution. Using Table 1 of the appendix, we determine the standard electrode potentials of metals:

B,
B,

because

, then in the resulting galvanic pair the anode will be tin and the cathode will be silver.

Because the
<
<
, then only corrosion of tin under the influence of oxygen will be possible:

A – : Sn → Sn 2+ + 2e.

K + : O 2 + 2H 2 O + 4e → 4OH – .

Example 4.2. During the corrosion of an iron product with the absorption of oxygen, 0.125 gFe(OH) 2 was formed in 3 minutes.

Solution. Calculate the volume of oxygen consumed for the corrosion of iron, the strength of the corrosion current and the mass of the metal destroyed by corrosion.

Let's calculate the number of mole equivalents of the formed Fe(OH) 2:

mol-eq.

Let's calculate the number of mole equivalents of the formed Fe(OH) 2:

Since all substances interact in equivalent quantities, 2.8 was destroyed. 10 -3 mole equivalents of Fe and the same number of mole equivalents of O 2 were consumed.

Then the volume of oxygen (no.) consumed for the corrosion of iron:

Using Faraday's law, we calculate the strength of the corrosion current:

A.

Let us determine the mass of corroded iron: Example 4.3 . Suggest an anodic coating to protect iron products from electrochemical corrosion in an oxygen-containing environment at pH = 7, R

Solution. gas =1. Write the equations of corrosion processes when the integrity of the coating is damaged.
Metals with a more negative potential value (for example, Zn, Cr, Al, etc.) can be used as an anodic coating for Fe. For example, choose chrome, standard potential
IN.

B, more negative than

because
<
, then if the integrity of the chrome coating on an iron product is damaged, chromium will play the role of an anode.

Because
<
,
, then in this environment corrosion of chromium with oxygen and hydrogen depolarization is thermodynamically possible:

A - : Cr → Cr 3+ + 3e

K + : O 2 + 2H 2 O + 4e → 4OH -

2H 2 O + 2e → H 2 + 2OH -

In a neutral environment, chromium has high corrosion resistance due to its tendency to passivation. Chromium corrosion products (Cr 2 O 3 , Cr(OH) 3, etc.) form dense, sparingly soluble oxide-salt films on the metal surface, which have protective properties that make it difficult for the metal to contact the oxidizing agent and inhibit the further corrosion process. Therefore, although corrosion is thermodynamically possible, an iron product coated with chrome is not actually destroyed by corrosion.

Example 4.4. Propose a cathodic coating to protect iron products from electrochemical corrosion in an oxygen-containing environment at pH = 8 and . Suggest an anodic coating to protect iron products from electrochemical corrosion in an oxygen-containing environment at pH = 7, gas =1. Write the equations of processes in a corrosive galvanic cell when the integrity of the coating is damaged.

Solution. Metals with a more positive potential value (for example, Ni, Cu, Ag, etc.) can be used as a cathode coating for Fe.
For example, let's choose copper, standard potential
.

Because
<
B, more positive than

B, more negative than

Because
<
<
, then if the integrity of the copper coating on an iron product is damaged, the role of the anode will be played by the iron.

, then in this environment corrosion of iron with oxygen depolarization is thermodynamically possible and corrosion with hydrogen evolution is impossible. Equations of corrosion processes:

A - : Fe → Fe 2+ + 2e

K + : O 2 + 2H 2 O + 4e → 4OH --

The iron product will be destroyed. Example 4.5.

Write the equations for electrochemical corrosion of the Sn-Zn pair at pH = 5 and 298 K. How much and what metal was corroded if 56 ml of oxygen was absorbed during the corrosion process and 22.4 ml of hydrogen was released? Determine what the corrosion current is if the corrosion duration is 20 minutes. Solution

B,
B,

because
<
. Standard metal potentials:

B, more negative than

Because
<
,
then in a given galvanic pair the anode will be zinc and the cathode will be tin.

, then in this environment electrochemical corrosion of zinc with oxygen and hydrogen depolarization is thermodynamically possible:

A - :Zn→Zn 2+ + 2e

K + :O 2 + 2H 2 O+ 4e→ 4OH -

l/mol,
2H 2 O+ 2e→H 2 + 2OH - .

l/mol):

Let's calculate the number of mole equivalents of the formed Fe(OH) 2:

mol-eq,
Thus, the cathode has undergone a change of 1.2. 10 -2 mol equivalents of the substance. According to the law of equivalents, the same amount of substance will dissolve at the anode:
. Mass of corroded zinc (taking into account the mass of a mole of zinc equivalent

The magnitude of the corrosion current is determined by Faraday's law:

Using Faraday's law, we calculate the strength of the corrosion current:

Example 4.6. Select a protector to protect steel structure (Fe) in an acidic environment (pH=4) in air. Write equations for corrosion processes. Calculate how the mass of the tread will change if, over a period of time, during the corrosion process, 112 ml of oxygen are absorbed and 112 ml of hydrogen are released.

Solution. In sacrificial protection, a metal or alloy with a more negative potential value than the potential of the protected product is connected to a metal product directly or through a metal conductor. For iron (
C) magnesium can be used as an anode protector (
B), zinc (
B), aluminum (
IN). Upon contact with the oxidizing agent, the protector metal dissolves, but the protected product is not destroyed. For example, let's choose magnesium. Because
<
, then paired with iron, magnesium will be the anode.

According to the Nernst equation, the equilibrium potentials of the probable oxidizing agents (H + and O 2) are equal:

Because
<
,
, then in this environment electrochemical corrosion of the magnesium protector with oxygen and hydrogen depolarization is thermodynamically possible:

A - : Мg → Мg 2+ + 2e

K + : O 2 + 4H + + 4e → 2H 2 O

2H + + 2e→H 2 .

In accordance with the assignment, we determine the number of mole equivalents of absorbed oxygen and released hydrogen (we consider the conditions to be normal,
l/mol,
2H 2 O+ 2e→H 2 + 2OH - .

l/mol):

Let's calculate the number of mole equivalents of the formed Fe(OH) 2:

Thus, 3 has undergone a change at the cathode. 10 -2 mol equivalents of oxidizing agent. According to the law of equivalents, the same amount of protective material dissolved on the anode:
mol-eq. Mass of dissolved protector (taking into account the molar mass of magnesium equivalent
g/mol) is equal to.

Hydrogen is a valuable raw material that has wide and varied applications. A large amount of hydrogen is used as a raw material for a number of important processes in the chemical industry: the synthesis of ammonia and benzene. In metallurgy, hydrogen is used for the selective reduction of non-ferrous metals from ammonia solutions and for the reduction of ores. Hydrogen is used to create the necessary atmosphere in furnaces, for cutting and welding metals, etc.

Industrial methods for producing hydrogen are divided into physical, chemical and electrochemical.

Physical methods are based on the fractional separation of hydrogen from a hydrogen-containing gas mixture by changing the physical state of the mixture (for example, the method of deep cooling of coke oven gas with the conversion of all components except hydrogen). Chemical methods are based on the thermal decomposition of hydrocarbons or on processes of conversion of carbons and carbon monoxide, for example:

CH 4 +N 2 ABOUT→CO+ 3N 2 ;CO+N 2 ABOUT→CO 2 +N 2 

and on the recovery of water vapor, for example:

4N 2 ABOUT+ 3Fe →Fe 3 ABOUT 4 + 4N 2 

Electrochemical method hydrogen production is based on the electrolytic decomposition of water. Electrochemical methods account for approximately 3% of the world's hydrogen production, but current estimates suggest that the share of electrolytic hydrogen produced by electrochemical methods will increase due to declining reserves of natural gases and oil. In recent years, the prospect of using hydrogen as a fuel has been widely discussed, the combustion of which in fuel cells produces virtually no environmentally harmful substances.

It is not advisable to subject pure water to electrolysis due to its low specific conductivity (4∙10 -6 S/m for distilled water and 1∙10 -1 S/m for tap water). Electrolysis of water is carried out with the addition of acid, alkali, or salt to increase the electrical conductivity of the electrolyte and reduce energy consumption. The electrical conductivity of sulfuric acid solutions is higher than that of alkali solutions; however, alkaline solutions are used in industry because in them, conventional construction materials are sustainable.

The main electrode processes during electrolysis are the release of hydrogen at the cathode and oxygen at the anode according to the total reaction 2 N 2 ABOUT→ 2N 2 +ABOUT 2. Oxygen is a by-product during electrolysis; this product has no independent significance, since it is more economical to obtain oxygen from air. Basics of reaction in an alkaline environment:

on cathode 2 N 2 ABOUT+ 2→N 2 + 2HE- (7.a)

on anode 4 HE - → 4N 2 ABOUT+ 2ABOUT 2 + 4; (7.b)

in an acidic environment:

on cathode 2 N + +→N 2 (7.v)

on anode 2 N 2 ABOUT→ 4Н + + ABOUT 2 (7.g)

There are no cations in an alkaline electrolyte that could be discharged at the cathode and lead to the occurrence of electrode reactions other than the formation of hydrogen gas.

The only side reaction at significantly more positive potentials than the reaction of cathodic formation of hydrogen is the reaction of electroreduction of dissolved oxygen

ABOUT 2 + 2H 2 O + 4 → 4OH - (7.d)

However, its rate is limited by the low solubility of oxygen in alkaline solutions, especially at high temperatures. Only a fraction of a percent of the current is wasted. Therefore, electrolysis baths all operate with very high cathode current outputs (about 97-98%, taking into account current leakage).

The process of oxygen evolution at the anode is accompanied by oxidation of the anode material with the formation of surface oxides such as MeO. Therefore, during long-term electrolysis, the discharge of anions occurs not on the metal, but on the oxidized surface. Over time, the overvoltage of oxygen evolution increases slightly until it reaches a constant value after a long period of time. Therefore, the value of the anode potential in an industrial, long-running bath is more positive than that determined in laboratory conditions.

Electrode materials. There is a requirement for materials for electrodes: the overvoltage of hydrogen and oxygen evolution on them should be as low as possible. The choice of electrode materials is dictated by the need to reduce wasteful energy consumption for electrode polarization.

As can be seen in Fig. 7.1 the best cathode material is platinized platinum, however, due to the high cost and instability of the sponge layer, platinum cannot be used as an electrode material.

Fig. 7.1 Polarization curves of hydrogen evolution on some metals from solutionNaOH (16 wt%): at 25 WITH; - - - at 80 WITH.

Metals of the iron group are stable in alkaline solutions, have low overvoltage and are suitable as materials for cathodes. The overvoltage on iron and cobalt is several tens of millivolts less than on nickel. Other metals ( Ti, Pb) are characterized by higher values ​​of hydrogen overvoltage on them and are not used in practice.

In Fig. Figure 7.2 shows the anodic polarization curves of oxygen evolution from an alkaline solution, from which it follows that on metals of the iron group the overvoltage of oxygen evolution is also small. Consequently, this group of metals is quite suitable as materials not only for cathodes, but also for anodes.

For the manufacture of cathodes, ordinary steel is used. The cathode is sometimes activated by depositing nickel containing sulfur or platinum group metals onto its surface.

Fig. 7.2 Polarization curves of oxygen release on some metals from solutionNaOH (16 wt%) (Ni(cottage 1. During the electrolysis of a copper sulfate solution, 48 g of copper was released at the cathode.) – sulfur-containing nickel coating): at 25°C; - - - at 80°C

Carbon steel is used as anodes in the electrolysis of aqueous alkaline solutions, onto which a nickel coating 100 microns thick is electrochemically applied. Such an anode retains sufficient corrosion resistance in alkaline solutions even in the presence of 1.5∙10 3 pores per 1 m 2. The low wear of such an anode, even with a higher porosity of the galvanic coating, is explained by clogging of the pores with corrosion products of the steel base. Nickel plating of the anode transfers it to a passive state and makes it insoluble in the potential region at which oxygen is released.

In all industrial baths, diaphragms made of asbestos fabric are used to separate gases. The role of the diaphragm is to prevent the mixing of gases. The mechanical strength of asbestos fabric is enhanced by incorporating nickel wires into the yarn.

Composition of the solution. The choice of composition and concentration of the electrolyte, as well as the design of the bath and its operating modes are dictated by the intended purpose of the electrolyzer and are determined by the need to minimize the unproductive consumption of electricity due to ohmic losses in the electrolyte and contacts. Solutions of potassium hydroxide and sodium hydroxide solutions are used as an electrolyte for the electrolysis of water. 2 – 3 g/l is introduced into the electrolyte TO 2 WITHr 2 ABOUT 7 to suppress steel corrosion. Water with a specific electrical conductivity of no higher than 10 -3 Sm∙m -1 and containing no more than 10 mg/l of chlorides and up to 3 mg/l of iron is considered suitable for the electrolyzer. However, it is recommended to use purer water to power electrolyzers; conductivity not higher than 10 -4 S∙m -1, iron content not higher than 1 mg/l, chlorides 2 mg/l and dry residue 3 mg/l.

During the electrolysis process, impurities accumulate in the electrolyte - carbonates, chlorides, sulfates, silicates, as well as iron, formed as a result of the destruction of parts of the electrolyzer and diaphragm. Foreign anions that accumulate in the solution do not participate in electrochemical reactions, with the exception of chlorine ions, which can cause depassivation of the anode.

The iron ions present in the electrolyte are discharged at the cathode to form an iron sponge. The thickness of the sponge increases during the electrolysis process, its layer reaches the diaphragm, causing its metallization, as a result of which hydrogen begins to be released on the anode side. It has been established that when potassium or sodium chromate is introduced into a solution, the electrodeposition of iron on the cathode decreases. A film of products of incomplete reduction of chromates forms on the cathode, making the electroreduction of iron compounds difficult.

The main impurity in hydrogen is oxygen, and in oxygen it is hydrogen. The alkali content in hydrogen is usually up to 20 mg/m3, in oxygen up to 100 mg/m3. Electrolytic hydrogen must have the following composition: no less than 99.7% (vol.) hydrogen, no more than 0.3% (vol.) oxygen. Electrolytic oxygen must contain no more than 0.7% (vol.) hydrogen.

Gas purification from alkaline fog is carried out in packed filters filled with glass wool. Hydrogen is purified from oxygen impurities in contact devices on nickel-aluminum and nickel-chromium catalysts at 100 - 130 °C. The purification of oxygen from hydrogen occurs on catalysts: platinized asbestos, platinum deposited on aluminum oxide. After cooling, the gases are dried with sorbents (silica gel, amomogel).

To select the optimal alkali concentration, it is necessary to know the dependence of the specific electrical conductivity of solutions on concentration at different temperatures. The curves expressing this dependence pass through a maximum at all temperatures (Fig. 7.3).

Fig.7.3. Dependence of specific electrical conductivity of solutionsNaOH (a) and KOH (b) on concentration at different temperatures

Maximum electrical conductivity of solutions CON more than NaOH, but the cost of sodium hydroxide is less. Therefore, how CON, so Nand he can equally be used in electrolysis baths. In industrial baths, a 21% solution has the maximum electrical conductivity. Nand he and accordingly 32% CON. In practice, a 16–20% solution is used Nand he and 25 - 30% solution CON.

Temperature. Electrolysis of aqueous solutions of alkalis is carried out at elevated temperatures in order to reduce the overvoltage of gas evolution and the resistivity of the electrolyte. In Fig. Figure 7.4 shows the relationship between temperature, resistivity and optimal concentration of alkali solutions.

Rice. 7.4 Dependence of the optimal concentration (1, 2) and resistivity (3, 4) of KOH solutions (1, 3) andNaOH (2, 4) on temperature.

Electrolysis process indicators behind ( U p) The voltage at the electrolyzer terminals consists of the water decomposition voltage E pr – oxygen equilibrium potential differences and hydrogen
(E pr = -
), hydrogen and oxygen overvoltage
And on these electrodes and the amounts of ohmic losses Σ IR Ohm, the main of which is the voltage drop in the electrolyte IR El + IR gas, in diaphragm IR aperture and in electrodes and contacts IR Ohm, V:

U p = E pr +
+ +I(R email + R gas + R cont.), (7.1)

Where I– operating current in the electrolyzer, A.

The hydrogen (oxygen) overvoltage at a given operating current density can be determined using the Tafel equation η =HOH+b lg i, using the reference values ​​of the coefficients HOH And b, corresponding to the electrode material and electrolyte solution.

The voltage drop in the electrolyte is calculated from the area of ​​the electrodes S, the distances between them l and specific electrical conductivity of the electrolyte  , the value of which can be taken from the reference book (taking into account the temperature of the electrolyte during the experiment).

The increase in voltage drop due to gas evolution is approximately 10% of the voltage drop in the electrolyte. The voltage drop in the electrodes and contacts is assumed to be 0.1 - 0.5V.

Approximate voltage balance of a filterpress bipolar bath having an interelectrode distance of 3.6 cm and operating at a pressure of up to 100 atm. with a current density of about 1000 A/dm 2, is given in Table 7.1.

Table 7.1

Stress balance of filter press bipolar bath

Electrode current densities. The value of electrode current densities during the electrolytic decomposition of water varies widely depending on the design of the electrolyzer. Advances in the creation of designs of electrolyzers with electrodes that provide rapid removal of gases and a reduction in gas filling, activation of the surface of the electrodes, an increase in the temperature of the electrolyte and a resulting reduction in gas overvoltage have made it possible to increase the current density in new designs of industrial electrolyzers to 2.5 - 3.7 kA/m 2 (0.25 – 0.37 A/cm2)

Electrolyzer designs. All modern designs of electrolyzers are of the filter-press type with bipolar inclusion of electrodes. The diagram of a filter-press bipolar electrolyzer for producing hydrogen and oxygen is shown in Fig. 7.5

Rice. 7.5. Bipolar filter-press electrolyzer: 1 – remote electrode; 2 – monopolar electrode – anode; 3 – bipolar electrode; 4 – monopolar electrode – cathode; 5 – diaphragm; 6 – tie plate; 7 – coupling bolt; 8 – diaphragm frame.

Filter-press bipolar electrolysers can have an equivalent load of up to 1200 - 1800 kA and consist of 160 - 170 individual cells.

The set of redox reactions that occur on electrodes in solutions or melts of electrolytes when an electric current is passed through them is called electrolysis.

At the cathode of the current source, the process of transferring electrons to cations from a solution or melt occurs, so the cathode is a “reducing agent”.

At the anode, electrons are given away by anions, so the anode is an “oxidizing agent.”

During electrolysis, competing processes can occur at both the anode and cathode.

When electrolysis is carried out using an inert (non-consumable) anode (for example, graphite or platinum), as a rule, two oxidative and two reduction processes are competing: at the anode - oxidation of anions and hydroxide ions, at the cathode - reduction of cations and hydrogen ions.

When electrolysis is carried out using an active (consumable) anode, the process becomes more complicated and the competing reactions on the electrodes are the following:

at the anode - oxidation of anions and hydroxide ions, anodic dissolution of the metal - the anode material; at the cathode - reduction of the salt cation and hydrogen ions,

reduction of metal cations obtained by dissolving the anode.

When choosing the most probable process at the anode and cathode, one should proceed from the position that the reaction that requires the least amount of energy will proceed. In addition, to select the most probable process at the anode and cathode during the electrolysis of salt solutions with an inert electrode, the following rules are used.

1. The following products can form at the anode: a) during the electrolysis of solutions containing anions, as well as alkali solutions, oxygen is released; b) during the oxidation of anions, chlorine, bromine, and iodine are released, respectively; c) during the oxidation of anions of organic acids, the process occurs:

2. During the electrolysis of salt solutions containing ions located in the voltage series to the left, hydrogen is released at the cathode; if the ion is located in the voltage series to the right of hydrogen, then metal is deposited at the cathode.

3. During the electrolysis of salt solutions containing ions located in a series of voltages between, competing processes of both cation reduction and hydrogen evolution can occur at the cathode.

Let us consider, as an example, the electrolysis of an aqueous solution of copper chloride on inert electrodes. The solution contains ions that, under the influence of an electric current, are directed to the corresponding electrodes:

Metallic copper is released at the cathode, and chlorine gas is released at the anode.

If in the considered example of electrolysis of a solution we take a copper plate as the anode, then copper is released at the cathode, and at the anode, where oxidation processes occur, instead of discharging ions and releasing chlorine, the anode (copper) oxidizes. In this case, the anode itself dissolves, and in the form of ions it goes into solution. Electrolysis with a soluble anode can be written as follows:

Thus, the electrolysis of salt solutions with a soluble anode is reduced to the oxidation of the anode material (its dissolution) and is accompanied by the transfer of metal from the anode to the cathode. This property is widely used in the refining (cleaning) of metals from contaminants.

To obtain highly active metals (sodium, aluminum, magnesium, calcium, etc.), which easily interact with water, electrolysis of molten salts or oxides is used:

If an electric current is passed through an aqueous solution of a salt of an active metal and an oxygen-containing acid, then neither the metal cations nor the ions of the acid residue are discharged. Hydrogen is released at the cathode, and oxygen is released at the anode, and electrolysis is reduced to the electrolytic decomposition of water.

Let us finally note that electrolysis of electrolyte solutions is energetically more favorable than melts, since electrolytes - salts and alkalis - melt at very high temperatures.

The dependence of the amount of substance formed under the influence of electric current on time, current strength and the nature of the electrolyte can be established on the basis of Faraday’s generalized law:

where m is the mass of the substance formed during electrolysis (g); E is the equivalent mass of the substance (g/mol); M is the molar mass of the substance (g/mol); n is the number of electrons given or received; I - current strength (A); t - process duration (s); F is Faraday's constant, characterizing the amount of electricity required to release 1 equivalent mass of a substance.