Electrolysis is a process by which electrical energy is converted into chemical energy. This process occurs on the electrodes under the influence of direct current. What are the products of electrolysis of melts and solutions, and what is included in the concept of “electrolysis”.

Electrolysis of molten salts

Electrolysis is a redox reaction that occurs on electrodes when a direct electric current is passed through a solution or melt of an electrolyte.

Rice. 1. The concept of electrolysis.

The chaotic movement of ions under the influence of current becomes ordered. Anions move to the positive electrode (anode) and oxidize there, giving up electrons. Cations move to the negative pole (cathode) and are reduced there, accepting electrons.

Electrodes can be inert (metallic from platinum or gold or non-metallic from carbon or graphite) or active. The anode in this case dissolves during the electrolysis process (soluble anode). It is made from metals such as chromium, nickel, zinc, silver, copper, etc.

During the electrolysis of molten salts, alkalis, and oxides, metal cations are discharged at the cathode to form simple substances. Melt electrolysis is an industrial method for producing metals such as sodium, potassium, calcium (electrolysis of molten salts) and aluminum (electrolysis of molten aluminum oxide Al 2 O 3 in Na 3 AlF 6 cryolite, used to facilitate the transfer of the oxide into the melt). For example, the electrolysis scheme for molten sodium chloride NaCl goes like this:

NaCl Na + + Cl -

Cathode(-) (Na+): Na++ e= Na 0

Anode(-) (Cl -): Cl - - e= Cl 0, 2Cl 0 = Cl 2

Summary process:

2Na+ +2Cl- = electrolysis 2Na + 2Cl 2

2NaCl = electrolysis 2Na + Cl 2

Simultaneously with the production of the alkali metal sodium, chlorine is obtained by electrolysis of the salt.

Electrolysis of salt solutions

If salt solutions are subjected to electrolysis, then, along with the ions formed during the dissociation of the salt, water can also be oxidized or reduced at the electrodes.

There is a certain sequence of discharge of ions at the electrodes in aqueous solutions.

1. The higher the standard electrode potential of the metal, the easier it is to recover. In other words, the more to the right a metal is in the electrochemical voltage series, the easier its ions will be reduced at the cathode. During the electrolysis of solutions of metal salts from lithium to aluminum inclusive, water molecules are always reduced at the cathode:

2H 2 O+2e=H 2 +2OH-

If solutions of metal salts are subjected to electrolysis, starting with copper and to the right of copper, only metal cations are reduced at the cathode. During the electrolysis of metal salts from manganese MN to lead Pb, both metal cations and, in some cases, water can be reduced.

2. Anions of acidic residues (except F-) are oxidized at the anode. If salts of oxygen-containing acids undergo electrolysis, then the anions of the acidic residues remain in solution, and the water is oxidized:

2H 2 O-4e=O 2 +4H+

3. If the anode is soluble, then oxidation and dissolution of the anode itself occurs:

Example: electrolysis of an aqueous solution of sodium sulfate Na 2 SO 4:

Solving chemical problems
aware of Faraday's law
high school

Author's development

Among the great variety of different chemical problems, as teaching practice at school shows, the greatest difficulties are caused by problems whose solution, in addition to solid chemical knowledge, requires a good command of the physics course material. And although not every high school pays attention to solving even the simplest problems using the knowledge of two courses - chemistry and physics, problems of this type are sometimes found on entrance exams in universities where chemistry is a major discipline. Therefore, without examining problems of this type in class, a teacher may unintentionally deprive his student of the chance to enter a university to major in chemistry.
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 problems will not be of interest to absolutely all students in the class or accessible to everyone. Nevertheless, it is recommended that tasks of this type be discussed with a group of interested students in a circle or elective lesson. It can be noted with confidence that problems of this type are complicated and, at least, are not typical for a school chemistry course (we are talking about a secondary school), and therefore problems of this type can be safely included in the versions of the school or district chemistry Olympiad for the 10th or 11th grade.
Having a detailed solution for each problem makes the development a valuable tool, especially for beginning teachers. Having gone through several problems with students during an elective lesson or a club lesson, a creative teacher will certainly assign several similar problems at home and use this development in the process of checking homework, which will significantly save invaluable teacher time.

Theoretical information on the problem

Chemical reactions occurring under the influence of electric current on electrodes placed in a solution or molten electrolyte are called electrolysis. Let's look at 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 an electric current is passed through the melt, the Na + and Cl – ions move chaotically, but when an electric current is applied, the movement of these particles becomes ordered: the Na + ions rush towards the negatively charged electrode, and the Cl – ions towards the positively charged electrode.

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

Cation– a positively charged ion.

Anion– 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 metal activities, 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:

Electrolysis of an aqueous solution of potassium nitrate occurs similarly (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 comes down to the transfer of anode material to the cathode. This is the essence of the electrochemical refining process (obtaining pure metal).

When drawing up electrolysis schemes for a particular salt, you need to remember that:

– metal cations that have a higher standard electrode potential (SEP) than 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 SEP 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 an aqueous solution contains a mixture of cations of various metals, for example Ag +, Cu 2+, Fe 2+, then in this mixture silver will be reduced first, then copper and iron last;

– on the insoluble anode during the electrolysis process, oxidation of anions or water molecules occurs, and the 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, that is, it sends electrons to the external circuit: when electrons are released, the equilibrium between the electrode and the solution shifts and the anode dissolves.

If from the entire series of electrode processes we select 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 allows you to see which metals are capable of displacing hydrogen from aqueous solutions of acids and which are not (table).

Table

Metal stress range

The equation electrode process	Standard electrode potential at 25 °C, V	The equation electrode process	Standard electrode potential at 25 °C, V
Li + + 1 e= Li 0	–3,045	Co 2+ + 2 e= Co 0	–0,277
Rb + + 1 e= Rb 0	–2,925	Ni 2+ + 2 e= Ni 0	–0,250
K + + 1 e= K 0	–2,925	Sn 2+ + 2 e= Sn 0	–0,136
Cs + + 1 e= Cs 0	–2,923	Pb 2+ + 2 e= Pb 0	–0,126
Ca 2+ + 2 e= Ca 0	–2,866	Fe 3+ + 3 e= Fe 0	–0,036
Na + + 1 e= Na 0	–2,714	2H + + 2 e=H2	0
Mg 2+ + 2 e= Mg 0	–2,363	Bi 3+ + 3 e= Bi 0	0,215
Al 3+ + 3 e= Al 0	–1,662	Cu 2+ + 2 e= Cu 0	0,337
Ti 2+ + 2 e= Ti 0	–1,628	Cu + +1 e= Cu 0	0,521
Mn 2+ + 2 e= Mn 0	–1,180	Hg 2 2+ + 2 e= 2Hg 0	0,788
Cr 2+ + 2 e= Cr 0	–0,913	Ag + + 1 e= Ag 0	0,799
Zn 2+ + 2 e= Zn 0	–0,763	Hg 2+ + 2 e= Hg 0	0,854
Cr 3+ + 3 e= Cr 0	–0,744	Pt 2+ + 2 e= Pt 0	1,2
Fe 2+ + 2 e= Fe 0	–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, the series of metal stresses can be represented as follows:

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

m = M I t/(z F),

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

Problem 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 silver nitrate solution 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 a solution of iron(III) sulfate 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. How long does it take to electrolyze 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 molten sodium chloride 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?

11. How many grams of a 30% hydrochloric acid solution will be required to completely react with iron obtained by electrolysis of a solution of iron(III) sulfate for 0.5 hours at current strength
10 Huh?

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

13. How many milliliters of a 12% KOH solution with a density of 1.111 g/ml will be required to react with aluminum (to form 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 react with zinc obtained by electrolysis of a solution of zinc sulfate for 100 minutes at a current of 55 A?

15. What volume of nitrogen(IV) oxide (n.o.) will be obtained by reacting 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 nitrogen(II) oxide (n.o.) will be obtained by the interaction of an excess nitric acid solution with copper obtained by electrolysis of a copper(II) chloride melt for 50 minutes at a current of 10.5 A?

17. How long does it take to electrolyze a melt of iron(II) chloride at a current of 30 A to obtain the iron necessary for complete reaction with 100 g of a 30% hydrochloric acid solution?

18. How long does it take to electrolyze a solution of nickel nitrate at a current of 15 A to obtain the nickel required for complete reaction with 200 g of a 35% solution of sulfuric acid 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 magnesium chloride melt for 200 minutes at current strength
10 A, dissolved in 1.5 l of 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 minutes at current strength

17 A, dissolved in 1 liter of 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 minutes at a current of 11 A, was turned into powder and immersed in 300 g of an 18% solution of copper(II) sulfate. Find the mass of copper that precipitated.

23. Magnesium obtained by electrolysis of magnesium chloride melt for 90 minutes at current strength
17 A, was immersed in a solution of hydrochloric acid taken in excess. Find the volume and amount of hydrogen released (n.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% solution of hydrochloric acid will be required to completely react with the resulting aluminum?

25. How many liters of oxygen and air (n.o.) will be required to completely burn magnesium obtained by electrolysis of magnesium chloride melt for 35 minutes at a current of 22 A?

For answers and solutions, see the following issues

Electrolysis is a redox reaction that occurs on electrodes when a direct electric current is passed through a melt or electrolyte solution.

The cathode is a reducing agent and gives electrons to cations.

The anode is an oxidizing agent and 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 + _____________________________→ Increased oxidative capacity
Anion activity series:	I - , Br - , Cl - , OH - , NO 3 - , CO 3 2- , SO 4 2- ←__________________________________ Increased recovery ability

Processes occurring on electrodes during 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 electrodes during electrolysis of solutions

CATHODE (-) Does not depend on the cathode material; depend on the position of the metal in the stress series	ANODE (+) Depends 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 reduced that are in the series of stresses afterH 2 : Me n+ +nē → Me°	1.First of all, the anions of oxygen-free acids are oxidized (exceptF - ): A m- - mē → A°	Anions do not oxidize. The metal atoms of the anode are oxidized: Me° - nē → Me n+ Men + cations go into solution. The anode mass decreases.
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.Oxoacid anions (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 reduced, 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. Draw up a scheme for the electrolysis of molten sodium bromide. (Algorithm 1.)

Sequencing	Performing 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 = 2Na +Br 2

Task 2. Draw up a scheme for the electrolysis of molten sodium hydroxide. (Algorithm 2.)

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

Task 3.Draw up a scheme for the electrolysis of molten sodium sulfate. (Algorithm 3.)

Sequencing	Performing Actions
1. Create an equation for the dissociation of salt	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. Create an equation for the electrolysis of molten salt	2Na 2 SO 4 = 4Na + 2SO 3 + O 2

ELECTROLYSIS OF SOLUTIONS

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

Sequencing	Performing Actions
1. Create an equation for the dissociation of salt	NaCl → Na + + Cl -
	Sodium ions in the solution are not reduced, so water is reduced. 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 + 2H2O = H2 + Cl2 + 2NaOH

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

Sequencing	Performing Actions
1. Create an equation for the dissociation of salt	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. Create an equation for the electrolysis of an aqueous salt solution	2CuSO 4 +2H 2 O = 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	Performing Actions
1. Create 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.Draw up an equation for the electrolysis of an aqueous alkali solution	2 H 2 O = 2 H 2 + O 2 , i.e. Electrolysis of an aqueous alkali solution is reduced to the electrolysis of water.

Remember.During 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 = 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	Performing Actions
1. Create an equation for the dissociation of salt	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 dissolves.
3.Draw up diagrams of the processes of reduction and oxidation	K - : Ag + + 1ē→ Ag 0 ; A+: Ag 0 - 1ē→ Ag +
4. Create an equation for the electrolysis of an aqueous salt solution	Ag + + Ag 0 = Ag 0 + Ag + electrolysis boils down to the transfer of silver from the anode to the cathode.

Electrolysis of molten salts

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

1. Electrolysis of molten copper (II) chloride.

Electrode processes can be expressed by half-reactions:

on the cathode K(-): Cu 2+ + 2e = Cu 0 - cathodic reduction

at the anode A(+): 2Cl – - 2e = Cl 2 - anodic oxidation

The overall reaction of the electrochemical decomposition of a substance is the sum of two electrode half-reactions, and for copper chloride it will be expressed by the equation:

Cu 2+ + 2 Cl – = Cu + Cl 2

During the electrolysis of alkalis and oxoacid salts, oxygen is released at the anode:

4OH – - 4e = 2H 2 O + O 2

2SO 4 2– - 4e = 2SO 3 + O 2

2. Electrolysis of potassium chloride melt:

Electrolysis of solutions

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 competing reactions on the electrodes are:
at the anode - oxidation of anions and hydroxide ions, anodic dissolution of the metal - the anode material;
at the cathode - reduction of 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 SO 4 2-, NO - 3, PO 4 3- anions, as well as alkali solutions, water is oxidized at the anode and oxygen is released;

A + 2H 2 O - 4e - = 4H + + O 2

b) during the oxidation of Cl - , Br - , I - anions, chlorine, bromine, and iodine are released, respectively;

A + Cl - +e - = Cl 0

2. The following products can form at the cathode:

a) during the electrolysis of salt solutions containing ions located in the voltage series to the left of Al 3+, water is reduced at the cathode and hydrogen is released;

K - 2H 2 O + 2e - = H 2 + 2OH -

b) if the metal ion is located in the voltage series to the right of hydrogen, then metal is released at the cathode.

K - Me n+ + ne - = Me 0

c) during the electrolysis of salt solutions containing ions located in the voltage range between Al + and H +, competing processes of both cation reduction and hydrogen evolution can occur at the cathode.

Example: Electrolysis of an aqueous solution of silver nitrate on inert electrodes

Dissociation of silver nitrate:

AgNO 3 = Ag + + NO 3 -

During the electrolysis of an aqueous solution of AgNO 3, reduction of Ag + ions occurs at the cathode, and oxidation of water molecules occurs at the anode:

Cathode: Аg + + e = А g

Anode: 2H 2 O - 4e = 4H + + O 2

Summary equation:_______________________________________________

4AgNO 3 + 2H 2 O = 4Ag + 4HNO 3 + O 2

Draw up schemes for the electrolysis of aqueous solutions: a) copper sulfate; b) magnesium chloride; c) potassium sulfate.

In all cases, electrolysis is carried out using carbon electrodes.

Example: Electrolysis of an aqueous solution of copper chloride on inert electrodes

Dissociation of copper chloride:

CuCl 2 ↔ Cu 2+ + 2Cl -

The solution contains Cu 2+ and 2Cl - ions, which, under the influence of an electric current, are directed to the corresponding electrodes:

Cathode - Cu 2+ + 2e = Cu 0

Anode + 2Cl - - 2e = Cl 2

_______________________________

CuCl 2 = Cu + Cl 2

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

If in the considered example of electrolysis of a CuCl 2 solution we take a copper plate as an anode, then copper is released at the cathode, and at the anode, where oxidation processes occur, instead of discharging Cl 0 ions and releasing chlorine, oxidation of the anode (copper) occurs.

In this case, the anode itself dissolves, and it goes into solution in the form of Cu 2+ ions.

Electrolysis of CuCl 2 with a soluble anode can be written as follows:

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.

Example: Electrolysis of an aqueous solution of magnesium chloride on inert electrodes

Dissociation of magnesium chloride in aqueous solution:

MgCl 2 ↔ Mg 2+ +2Сl -

Magnesium ions cannot be reduced in an aqueous solution (water is being reduced), chloride ions are oxidized.

Electrolysis scheme:

Example: Electrolysis of an aqueous solution of copper sulfate on inert electrodes

In solution, copper sulfate dissociates into ions:

CuSO 4 = Cu 2+ + SO 4 2-

Copper ions can be reduced at the cathode in an aqueous solution.

Sulfate ions in an aqueous solution do not oxidize, so water oxidation will occur at the anode.

Electrolysis scheme:

Electrolysis of an aqueous solution of a salt of an active metal and an oxygen-containing acid (K 2 SO 4) on inert electrodes

Example: Dissociation of potassium sulfate in aqueous solution:

K 2 SO 4 = 2K + + SO 4 2-

Potassium ions and sulfate ions cannot be discharged at the electrodes in an aqueous solution, therefore, reduction will occur at the cathode, and oxidation of water will occur at the anode.

Electrolysis scheme:

or, given that 4H + + 4OH - = 4H 2 O (carried out with stirring),

H2O2H2+O2

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.

Melt electrolysis of sodium hydroxide

Electrolysis of water is always carried out in the presence of an inert electrolyte (to increase the electrical conductivity of a very weak electrolyte - water):

Faraday's law

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 (F = 96,500 C/mol = 26.8 Ah/mol).

Hydrolysis of inorganic compounds

The interaction of salt ions with water, leading to the formation of weak electrolyte molecules, is called salt hydrolysis.

If we consider a salt as a product of neutralization of a base with an acid, then we can divide the salts into four groups, for each of which hydrolysis will proceed in its own way.

1. A salt formed by a strong base and a strong acid KBr, NaCl, NaNO 3) will not undergo hydrolysis, since in this case a weak electrolyte is not formed. The environment's reaction remains neutral.

2. In a salt formed by a weak base and a strong acid FeCl 2, NH 4 Cl, Al 2 (SO 4) 3, MgSO 4), the cation undergoes hydrolysis:

FeCl 2 + HOH → Fe(OH)Cl + HCl

Fe 2+ + 2Cl - + H + + OH - → FeOH + + 2Cl - + H +

As a result of hydrolysis, a weak electrolyte, H + ion and other ions are formed. solution pH< 7 (раствор приобретает кислую реакцию).

3. A salt formed by a strong base and a weak acid (KClO, K 2 SiO 3, Na 2 CO 3, CH 3 COONa) undergoes hydrolysis at the anion, resulting in the formation of a weak electrolyte, hydroxide ion and other ions.

K 2 SiO 3 + HOH → KHSiO 3 + KOH

2K + +SiO 3 2- + H + + OH - → HSiO 3 - + 2K + + OH -

The pH of such solutions is > 7 (the solution becomes alkaline).

4. A salt formed by a weak base and a weak acid (CH 3 COONH 4, (NH 4) 2 CO 3, Al 2 S 3) is hydrolyzed by both the cation and the anion. As a result, a slightly dissociating base and acid are formed. The pH of solutions of such salts depends on the relative strength of the acid and base.

Algorithm for writing reaction equations for the hydrolysis of a salt of a weak acid and a strong base

There are several options for hydrolysis of salts:

1. Hydrolysis of a salt of a weak acid and a strong base: (CH 3 COONa, KCN, Na 2 CO 3).

Example 1. Hydrolysis of sodium acetate.

or CH 3 COO – + Na + + H 2 O ↔ CH 3 COOH + Na + + OH –

CH 3 COO – + H 2 O ↔ CH 3 COOH + OH –

Since acetic acid dissociates weakly, the acetate ion binds the H + ion, and the dissociation equilibrium of water shifts to the right according to Le Chatelier's principle.

OH - ions accumulate in the solution (pH >7)

If the salt is formed by a polybasic acid, then hydrolysis occurs in stages.

For example, carbonate hydrolysis: Na 2 CO 3

Stage I: CO 3 2– + H 2 O ↔ HCO 3 – + OH –

Stage II: HCO 3 – + H 2 O ↔ H 2 CO 3 + OH –

Na 2 CO 3 + H 2 O = NaHCO 3 + NaOH

Usually only the process that occurs in the first stage is of practical importance, which, as a rule, is limited to when assessing the hydrolysis of salts.

The equilibrium of hydrolysis in the second stage is significantly shifted to the left compared to the equilibrium of the first stage, since a weaker electrolyte (HCO 3 -) is formed in the first stage than in the second (H 2 CO 3)

Example 2. Hydrolysis of rubidium orthophosphate.

1. Determine the type of hydrolysis:

Rb 3 PO 4 ↔ 3Rb + + P.O. 4 3–

Rubidium is an alkali metal, its hydroxide is a strong base, phosphoric acid, especially in its third stage of dissociation, which corresponds to the formation of phosphates, is a weak acid.

Hydrolysis occurs at the anion.

PO 3- 4 + H–OH ↔ HPO 2- 4 + OH – .

Products are hydrophosphate and hydroxide ions, the medium is alkaline.

3. Compose the molecular equation:

Rb 3 PO 4 + H 2 O ↔ Rb 2 HPO 4 + RbOH.

We obtained an acid salt - rubidium hydrogen phosphate.

Algorithm for writing reaction equations for the hydrolysis of a salt of a strong acid and a weak base

2. Hydrolysis of a salt of a strong acid and a weak base: NH 4 NO 3, AlCl 3, Fe 2 (SO 4) 3.

Example 1. Hydrolysis of ammonium nitrate.

NH 4 + + NO 3 – + H 2 O ↔ NH 4 OH + NO 3 – + H +

NH 4 + + H 2 O ↔ NH 4 OH + H +

In the case of a multiply charged cation, hydrolysis proceeds stepwise, for example:

Stage I: Cu 2+ + HOH ↔ CuOH + + H +

Stage II: CuOH + + HOH ↔ Cu(OH) 2 + H +

CuCl 2 + H 2 O = CuOHCl + HCl

In this case, the concentration of hydrogen ions and the pH of the medium in the solution are also determined mainly by the first stage of hydrolysis.

Example 2. Hydrolysis of copper(II) sulfate

1. Determine the type of hydrolysis. At this stage it is necessary to write the salt dissociation equation:

CuSO 4 ↔ Cu 2+ + SO 2- 4.

A salt is formed by a cation of a weak base (we emphasize) and an anion of a strong acid. Hydrolysis of the cation occurs.

2. We write the ionic equation of hydrolysis and determine the medium:

Cu 2+ + H-OH ↔ CuOH + + H + .

A hydroxycopper(II) cation and a hydrogen ion are formed, the medium is acidic.

3. Make up a molecular equation.

It must be taken into account that the compilation of such an equation is a certain formal task. From positive and negative particles in solution, we make up neutral particles that exist only on paper. In this case, we can create the formula (CuOH) 2 SO 4, but to do this we must mentally multiply our ionic equation by two.

We get:

2CuSO 4 + 2H 2 O ↔ (CuOH) 2 SO 4 + H 2 SO 4.

Please note that the reaction product belongs to the group of basic salts. The names of the main salts, as well as the names of the intermediate salts, should be composed of the names of the anion and cation; in this case, we will call the salt “hydroxycopper(II) sulfate”.

Algorithm for writing reaction equations for the hydrolysis of a salt of a weak acid and a weak base

3. Hydrolysis of a salt of a weak acid and a weak base:

Example 1. Hydrolysis of ammonium acetate.

CH 3 COO – + NH 4 + + H 2 O ↔ CH 3 COOH + NH 4 OH

In this case, two slightly dissociated compounds are formed, and the pH of the solution depends on the relative strength of the acid and base.

If the hydrolysis products can be removed from the solution, for example, in the form of a precipitate or gaseous substance, then the hydrolysis proceeds to completion.

Example 2. Hydrolysis of aluminum sulfide.

Al 2 S 3 + 6H 2 O = 2Al(OH) 3 + 3H 2 S

2А l 3+ + 3 S 2- + 6Н 2 О = 2Аl(ОН) 3 (precipitate) + ЗН 2 S (gas)

Example 3 Hydrolysis of aluminum acetate

1. Determine the type of hydrolysis:

Al(CH 3 COO) 3 = Al 3+ + 3CH 3 COO – .

A salt is formed by a cation of a weak base and anions of a weak acid.

2. We write ionic hydrolysis equations and determine the medium:

Al 3+ + H–OH ↔ AlOH 2+ + H + ,

CH 3 COO – + H–OH ↔ CH 3 COOH + OH – .

Considering that aluminum hydroxide is a very weak base, we assume that hydrolysis at the cation will occur to a greater extent than at the anion. Consequently, there will be an excess of hydrogen ions in the solution, and the medium will be acidic.

There is no point in trying to create a summary equation for the reaction here. Both reactions are reversible, have no connection with each other, and such summation is meaningless.

3. Let's make a molecular equation:

Al(CH 3 COO) 3 + H 2 O = AlOH(CH 3 COO) 2 + CH 3 COOH.

This is also a formal exercise for training in drawing up formulas of salts and their nomenclature. Let's call the resulting salt hydroxoaluminum acetate.

Algorithm for writing reaction equations for the hydrolysis of a salt of a strong acid and a strong base

4. Salts formed by a strong acid and a strong base do not undergo hydrolysis, because the only weakly dissociating compound is H 2 O.

The salt of a strong acid and a strong base does not undergo hydrolysis, and the solution is neutral.

Module 2. Basic chemistry processes and properties of substances

Laboratory work No. 7

Topic: Electrolysis of aqueous salt solutions

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

When a direct electric current is passed through an electrolyte solution or melt, cations move towards the cathode and anions move towards the anode. Redox processes occur at the electrodes; The cathode is a reducing agent, since it gives electrons to cations, and the anode is an oxidizing agent, since it accepts electrons from anions. The reactions occurring on the electrodes depend on the composition of the electrolyte, the nature of the solvent, the material of the electrodes, and the operating mode of the electrolyzer.

Chemistry of the electrolysis process of molten calcium chloride:

CaCl 2 ↔ Ca 2+ + 2Cl -

at the cathode Ca 2+ + 2e→ Ca°

at the anode 2Сl - - 2е→ 2С1° → С1 2

The electrolysis of a potassium sulfate solution on an insoluble anode looks schematically like this:

K 2 SO 4 ↔ 2K + + SO 4 2 -

H 2 O ↔ H + + OH -

at the cathode 2Н + + 2е→2Н°→ Н 2 2

at the anode 4OH - 4e → O 2 + 4H + 1

K 2 SO 4 + 4H 2 O 2H 2 + O 2 + 2K0H + H 2 SO 4

Goal of the work: familiarization with the electrolysis of salt solutions.

Instruments and equipment: electric current rectifier, electrolyzer, carbon electrodes, sandpaper, cups, washing machine.

Rice. 1. Device for conducting

electrolysis

1 - electrolyzer;

2 - electrodes;

3-conducting wires; DC source.

Reagents and solutions: 5% solutions of copper chloride CuC1 2, potassium iodide KI , potassium hydrogen sulfate KHSO 4, sodium sulfate Na 2 SO 4, copper sulfate CuSO 4, zinc sulfate ZnSO 4, 20% sodium hydroxide solution NaOH, copper and nickel plates, phenolphthalein solution, nitric acid (conc.) HNO 3, 1% starch solution , neutral litmus paper, 10% sulfuric acid solution H 2 SO 4.

Experiment 1. Electrolysis of copper chloride with insoluble electrodes

Fill the electrolyzer up to half the volume with a 5% copper chloride solution. Lower a graphite rod into both elbows of the electrolyzer, secure them loosely with pieces of rubber tube. Connect the ends of the electrodes with conductors to direct current sources. If there is a slight smell of chlorine, immediately disconnect the electrolyzer from the power source. What happens at the cathode? Write down equations for electrode reactions.

Experiment 2. Electrolysis of potassium iodide with insoluble electrodes

Fill the electrolyzer with 5% potassium iodide solution. add 2 drops of phenolphthalein to each knee. Paste V each electrolyser elbow graphite electrodes and connect them to a DC source.

In which elbow and why did the solution become colored? Add 1 drop of starch paste to each knee. Where and why is iodine released? Write down equations for electrode reactions. What was formed in the cathode space?

Experiment 3. Electrolysis of sodium sulfate with insoluble electrodes

Fill half the volume of the electrolyzer with a 5% sodium sulfate solution and add 2 drops of methyl orange or litmus to each elbow. Insert electrodes into both elbows and connect them to a DC source. Record your observations. Why did the electrolyte solutions turn different colors at different electrodes? Write down equations for electrode reactions. What gases are released at the electrodes and why? What is the essence of the electrolysis process of an aqueous solution of sodium sulfate