Electrolysis of melts and solutions (salts, alkalis)

If the electrodes are lowered into the electrolyte solution or melt and a direct electric current is passed through, then the ions will move in a direction: cations to the cathode (negatively charged electrode), anions to the anode (positively charged electrode).

At the cathode, cations accept electrons and are reduced; at the anode, anions donate electrons and are oxidized. This process is called electrolysis.

Electrolysis is a redox process that occurs on the electrodes during the passage electric current through a melt or electrolyte solution.

Electrolysis of molten salts

Consider the process of electrolysis of a sodium chloride melt. The process of thermal dissociation takes place in the melt:

$NaCl→Na^(+)+Cl^(-).$

Under the action of an electric current, $Na^(+)$ cations move towards the cathode and receive electrons from it:

$Na^(+)+ē→(Na)↖(0)$ (restoration).

Anions $Cl^(-)$ move towards the anode and donate electrons:

$2Cl^(-)-2ē→(Cl_2)↖(0)$ (oxidation).

The total equation of processes:

$Na^(+)+ē→(Na)↖(0)|2$

$2Cl^(-)-2ē→(Cl_2)↖(0)|1$

$2Na^(+)+2Cl^(-)=2(Na)↖(0)+(Cl_2)↖(0)$

$2NaCl(→)↖(\text"electrolysis")2Na+Cl_2$

Sodium metal is formed at the cathode, and chlorine gas is formed at the anode.

The main thing to remember is that in the process of electrolysis, electrical energy is used to chemical reaction, which cannot go spontaneously.

Electrolysis of aqueous solutions of electrolytes

A more complicated case is the electrolysis of electrolyte solutions.

In a salt solution, in addition to metal ions and an acidic residue, there are water molecules. Therefore, when considering processes on electrodes, it is necessary to take into account their participation in electrolysis.

For determination of electrolysis products aqueous solutions electrolytes, there are the following rules:

1. Process at the cathode does not depend on the material from which the cathode is made, but on the position of the metal (electrolyte cation) in the electrochemical series of voltages, and if:

1.1. The electrolyte cation is located in the voltage series at the beginning of the series in $Al$ inclusive, then the process of water reduction is going on at the cathode (hydrogen $H_2$ is released). Metal cations are not reduced, they remain in solution.

1.2. The electrolyte cation is in a series of voltages between aluminum and hydrogen, then both metal ions and water molecules are reduced at the cathode.

1.3. The electrolyte cation is in a series of voltages after hydrogen, then metal cations are reduced at the cathode.

1.4. The solution contains cations of different metals, then the metal cation, which is to the right in the series of voltages, is first restored.

cathodic processes

2. process at the anode depends on the material of the anode and on the nature of the anion.

Anode processes

2.1. If anode dissolves(iron, zinc, copper, silver and all metals that are oxidized during electrolysis), then the anode metal is oxidized, regardless of the nature of the anion.

2.2. If the anode does not dissolve(it is called inert - graphite, gold, platinum), then:

a) during the electrolysis of salt solutions anoxic acids (except for fluorides) the anion is oxidized at the anode;

b) during the electrolysis of salt solutions oxygenated acids and fluorides the process of water oxidation is going on at the anode ($O_2$ is released). Anions are not oxidized, they remain in solution;

c) anions according to their ability to be oxidized are arranged in the following order:

Let's try to apply these rules in specific situations.

Consider the electrolysis of a sodium chloride solution if the anode is insoluble and if the anode is soluble.

1) Anode insoluble(for example, graphite).

In solution, the process of electrolytic dissociation takes place:

Summary Equation:

$2H_2O+2Cl^(-)=H_2+Cl_2+2OH^(-)$.

Taking into account the presence of $Na^(+)$ ions in the solution, we compose the molecular equation:

2) Anode soluble(for example, copper):

$NaCl=Na^(+)+Cl^(-)$.

If the anode is soluble, then the anode metal will oxidize:

$Cu^(0)-2ē=Cu^(2+)$.

The $Cu^(2+)$ cations come after ($H^(+)$) in the voltage series, so they will be reduced at the cathode.

The concentration of $NaCl$ in the solution does not change.

Consider the electrolysis of a solution of copper sulfate (II) on insoluble anode:

$Cu^(2+)+2ē=Cu^(0)|2$

$2H_2O-4ē=O_2+4H^(+)|1$

Total ionic equation:

$2Cu^(2+)+2H_2O=2Cu^(0)+O_2+4H^(+)$

The overall molecular equation, taking into account the presence of $SO_4^(2-)$ anions in solution:

Consider the electrolysis of a potassium hydroxide solution on insoluble anode:

$2H_2O+2ē=H_2+2OH^(-)|2$

$4OH^(-)-4ē=O_2+2H_2O|1$

Total ionic equation:

$4H_2O+4OH^(-)=2H_2+4OH^(-)+O_2+2H_2O$

Overall molecular equation:

$2H_2O(→)↖(\text"electrolysis")2H_2+O_2$

In this case, it turns out that only the electrolysis of water takes place. A similar result will be obtained in the case of electrolysis of solutions $H_2SO_4, NaNO_3, K_2SO_4$, etc.

Electrolysis of melts and solutions of substances is widely used in industry:

1. To obtain metals (aluminum, magnesium, sodium, cadmium are obtained only by electrolysis).
2. For the production of hydrogen, halogens, alkalis.
3. For the purification of metals - refining (purification of copper, nickel, lead is carried out by the electrochemical method).
4. To protect metals from corrosion (chromium, nickel, copper, silver, gold) — electroplating.
5. To obtain metal copies, records - electrotype.

Establish a correspondence between the salt formula and the product formed on an inert anode during the electrolysis of its aqueous solution: for each position indicated by a letter, select the corresponding position indicated by a number.

SALT FORMULA		PRODUCT ON ANODE

A	B	IN	G

Solution.

In the electrolysis of aqueous solutions of salts, alkalis and acids on an inert anode:

Water is discharged and oxygen is released if it is a salt of an oxygen-containing acid or a salt of hydrofluoric acid;

Hydroxide ions are discharged and oxygen is released if it is alkali;

The acid residue that is part of the salt is discharged, and the corresponding simple substance is released if it is a salt of an oxygen-free acid (except for).

The process of electrolysis of salts of carboxylic acids takes place in a special way.

Answer: 3534.

Answer: 3534

Source: Yandex: training work USE in chemistry. Option 1.

Establish a correspondence between the formula of a substance and the product formed on the cathode during the electrolysis of its aqueous solution: for each position indicated by a letter, select the corresponding position indicated by a number.

SUBSTANCE FORMULA		ELECTROLYSIS PRODUCT, PRODUCED AT THE CATHODE

Write down the numbers in response, arranging them in the order corresponding to the letters:

A	B	IN	G

Solution.

During the electrolysis of aqueous solutions of salts, the following is released at the cathode:

Hydrogen, if it is a salt of a metal that is in the series of metal stresses to the left of aluminum;

Metal, if it is a salt of a metal that is in the series of metal voltages to the right of hydrogen;

Metal and hydrogen, if it is a salt of a metal in the series of metal voltages between aluminum and hydrogen.

Answer: 3511.

Answer: 3511

Source: Yandex: Training USE work in chemistry. Option 2.

Establish a correspondence between the salt formula and the product formed on an inert anode during the electrolysis of its aqueous solution: for each position indicated by a letter, select the corresponding position indicated by a number.

SALT FORMULA		PRODUCT ON ANODE

Write down the numbers in response, arranging them in the order corresponding to the letters:

A	B	IN	G

Solution.

During the electrolysis of aqueous solutions of salts of oxygen-containing acids and fluorides, oxygen is oxidized from water, so oxygen is released at the anode. During the electrolysis of aqueous solutions of anoxic acids, the acid residue is oxidized.

Answer: 4436.

Answer: 4436

Establish a correspondence between the formula of a substance and the product that is formed on an inert anode as a result of the electrolysis of an aqueous solution of this substance: for each position indicated by a letter, select the corresponding position indicated by a number.

SUBSTANCE FORMULA	PRODUCT ON ANODE
	2) sulfur oxide (IV) 3) carbon monoxide (IV) 5) oxygen 6) nitric oxide (IV)

Write down the numbers in response, arranging them in the order corresponding to the letters:

A	B	IN	G

Topic 6. "Electrolysis of solutions and salt melts"
1. Electrolysis is a redox process that occurs on electrodes when an electric current is passed through an electrolyte solution or melt.
2. Cathode - negatively charged electrode. There is a reduction of metal and hydrogen cations (in acids) or water molecules.
3. Anode - a positively charged electrode. Oxidation of the anions of the acid residue and the hydroxyl group (in alkalis) occurs.
4. During the electrolysis of a salt solution, water is present in the reaction mixture. Since water can exhibit both oxidizing and reducing properties, it is a "competitor" for both cathodic and anodic processes.
5. There are electrolysis with inert electrodes (graphite, carbon, platinum) and an active anode (soluble), as well as electrolysis of melts and electrolyte solutions.
CATHODE PROCESSES
If the metal is in a series of voltages:
The position of the metal in a series of stresses
Recovery at the cathode
from Li to Al
Water molecules are reduced: 2H2O + 2e- → H20+ 2OH-
Mn to Pb
Both water molecules and metal cations are restored:
2H2O + 2e- → H20+ 2OH-
Men+ + ne- → Me0
from Cu to Au
Metal cations are reduced: Men+ + ne- → Me0
ANODIC PROCESSES
acid residue
Asm-
Anode
Soluble
(iron, zinc, copper, silver)
Insoluble
(graphite, gold, platinum)
anoxic
Anode metal oxidation
М0 – ne- = Mn+
anode solution
Anion oxidation (except F-)
Acm- - me- = Ac0
Oxygen-containing
Fluoride - ion (F-)
In acidic and neutral environments:
2 H2O - 4e- → O20 + 4H+
In an alkaline environment:
4OH- - 4e- \u003d O20 + 2H2O
Examples of melt electrolysis processes with inert electrodes
In the electrolyte melt, only its ions are present, therefore, electrolyte cations are reduced at the cathode, and anions are oxidized at the anode.
1. Consider the electrolysis of a potassium chloride melt.
Thermal dissociation KCl → K+ + Cl-
K(-) K+ + 1e- → K0
A (+) 2Cl- - 2e- → Cl02
Summary equation:
2KCl → 2K0 + Cl20
2. Consider the electrolysis of a calcium chloride melt.
Thermal dissociation CaCl2 → Ca2+ + 2Cl-
K(-) Ca2+ + 2e- → Ca0
A (+) 2Cl- - 2e- → Cl02
Summary equation:
CaCl2 → Ca0 + Cl20
3. Consider the electrolysis of a melt of potassium hydroxide.
Thermal dissociation of KOH → K+ + OH-
K(-) K+ + 1e- → K0
A (+) 4OH- - 4e- → O20 + 2H2O
Summary equation:
4KOH → 4K0 + O20 + 2H2O
Examples of electrolysis processes of electrolyte solutions with inert electrodes
Unlike melts, in an electrolyte solution, in addition to its ions, there are water molecules. Therefore, when considering the processes on the electrodes, it is necessary to take into account their participation. The electrolysis of a salt solution formed by an active metal, standing in a series of voltages up to aluminum and an acidic residue of an oxygen-containing acid, is reduced to the electrolysis of water. 1. Consider the electrolysis of an aqueous solution of magnesium sulfate. MgSO4 is a salt that is formed by a metal standing in a series of stresses up to aluminum and an oxygen-containing acid residue. Dissociation equation: MgSO4 → Mg2+ + SO42- K (-) 2H2O + 2e- \u003d H20 + 2OH- A ​​(+) 2H2O - 4e- \u003d O20 + 4H + Total equation: 6H2O \u003d 2H20 + 4OH- + O20 + 4H + 2H2O \u003d 2H20 + O20 2. Consider the electrolysis of an aqueous solution of copper (II) sulfate. СuSO4 is a salt formed by a low-active metal and an oxygen-containing acid residue. In this case, electrolysis produces metal, oxygen, and the corresponding acid is formed in the cathode-anode space. Dissociation equation: CuSO4 → Cu2+ + SO42- K (-) Cu2+ + 2e- = Cu0 A (+) 2H2O – 4e- = O20 + 4H+ Summary equation: 2Cu2+ + 2H2O = 2Cu0 + O20 + 4H+ 2CuSO4 + 2H2O = 2Cu0 + О20 + 2Н2SO4
3. Consider the electrolysis of an aqueous solution of calcium chloride. CaCl2 is a salt that is formed by an active metal and an oxygen-free acid residue. In this case, hydrogen, halogen are formed during electrolysis, and alkali is formed in the cathode-anode space. Dissociation equation: CaCl2 → Ca2+ + 2Cl- K (-) 2H2O + 2e- = H20 + 2OH- A ​​(+) 2Cl- – 2e- = Cl20 Summary equation: 2H2O + 2Cl- = Cl20 + 2OH- CaCl2 + 2H2O = Ca (OH)2 + Cl20 + H20 4. Consider the electrolysis of an aqueous solution of copper (II) chloride. CuCl2 is a salt that is formed by a low-active metal and an acidic residue of an oxygen-free acid. In this case, a metal and a halogen are formed. Dissociation equation: CuCl2 → Cu2+ + 2Cl- K (-) Cu2+ + 2e- = Cu0 A (+) 2Cl- – 2e- = Cl20 Summary equation: Cu2+ + 2Cl- = Cu0 + Cl20 CuCl2 = Cu0 + Cl20 5. Consider the process electrolysis of sodium acetate solution. CH3COOHa is a salt formed by the active metal and the acidic residue of a carboxylic acid. Electrolysis produces hydrogen and alkali. Dissociation equation: CH3COONa → CH3COO - + Na+ K (-) 2H2O + 2e- = H20 + 2OH- A ​​(+) 2CH3COO¯- 2e = C2H6 + 2CO2 Summary equation: 2H2O + 2CH3COO¯ = H20 + 2OH - + C2H6 + 2CO2 2Н2О + 2CH3COONa = 2NaОH + Н20 + C2H6 + 2CO2 6. Consider the process of electrolysis of nickel nitrate solution. Ni(NO3)2 is a salt, which is formed by a metal standing in a series of voltages from Mn to H2 and an oxygen-containing acid residue. In the process, we get metal, hydrogen, oxygen and acid. Dissociation equation: Ni(NO3)2 → Ni2+ + 2NO3- K (-) Ni2+ +2e- = Ni0 2H2O + 2e- = H20 + 2OH- A ​​(+) 2H2O – 4e- = O20 + 4H+ Overall equation: Ni2+ + 2H2O + 2H2O = Ni0 + H20 + 2OH- + O20 + 4H+ Ni(NO3)2 + 2H2O = Ni0 + 2HNO3 + H20 + O20 7. Consider the process of electrolysis of a sulfuric acid solution. Dissociation equation: H2SO4 → 2H+ + SO42- K (-) 2H+ + 2e- = H20 A (+) 2H2O – 4e- = O20 + 4H+ Overall equation: 2H2O + 4H+ = 2H20 + O20 + 4H+ 2H2O = 2H20 + O20
8. Consider the process of electrolysis of sodium hydroxide solution. In this case, only water electrolysis takes place. The electrolysis of solutions of H2SO4, NaNO3, K2SO4, etc. proceeds similarly. Dissociation equation: NaOH → Na+ + OH- K (-) 2H2O + 2e- = H20 + 2OH- A ​​(+) 4OH- – 4e- = O20 + 2H2O Overall equation: 4H2O + 4OH- = 2H20 + 4OH- + O20 + 2H2O 2H2O = 2H20 + O20
Examples of electrolysis processes of electrolyte solutions with soluble electrodes
The soluble anode undergoes oxidation (dissolution) during electrolysis. 1. Consider the process of electrolysis of copper sulfate (II) with a copper anode. During the electrolysis of a solution of copper sulfate with a copper anode, the process is reduced to the release of copper at the cathode and the gradual dissolution of the anode, despite the nature of the anion. The amount of copper sulfate in solution remains unchanged. Dissociation equation: CuSO4 → Cu2+ + SO42- K (-) Cu2+ +2e- → Cu0 A (+) Cu0 - 2e- → Cu2+ transition of copper ions from anode to cathode
Examples of tasks on this topic in the USE options
AT 3. (Var.5)
Establish a correspondence between the formula of a substance and the products of electrolysis of its aqueous solution on inert electrodes.
FORMULA OF THE SUBSTANCE ELECTROLYSIS PRODUCTS
A) Al2(SO4)3 1. metal hydroxide, acid
B) СsOH 2. metal, halogen
C) Hg(NO3)2 3. metal, oxygen
D) AuBr3 4. hydrogen, halogen 5. hydrogen, oxygen 6. metal, acid, oxygen Argument: 1. During the electrolysis of Al2(SO4)3 and CsOH on the cathode, water is reduced to hydrogen. We exclude options 1, 2, 3 and 6. 2. For Al2(SO4)3, water is oxidized to oxygen at the anode. We choose option 5. For CsOH, the hydroxide ion is oxidized to oxygen at the anode. We choose option 5. 3. During the electrolysis of Hg(NO3)2 and АuBr3 on the cathode, metal cations are reduced. 4. For Hg(NO3)2, water is oxidized at the anode. Nitrate ions in solution bind with hydrogen cations, forming in the anode space nitric acid. We choose option 6. 5. For АuBr3, the Br- anion is oxidized at the anode to Br2. We choose option 2.
A
B
IN
G
5
5
6
2
AT 3. (Var.1)
Establish a correspondence between the name of the substance and the method of obtaining it.
NAME OF THE SUBSTANCE PRODUCTION BY ELECTROLYSIS A) lithium 1) LiF solution B) fluorine 2) LiF melt C) silver 3) MgCl2 solution D) magnesium 4) AgNO3 solution 5) Ag2O melt 6) MgCl2 melt Argument: 1. Similar to the electrolysis of sodium chloride melt , the process of electrolysis of the lithium fluoride melt proceeds. For options A and B, choose answers 2. 2. Silver can be restored from a solution of its salt - silver nitrate. 3. Magnesium cannot be restored from a salt solution. We choose option 6 - a melt of magnesium chloride.
A
B
IN
G
2
2
4
6
AT 3. (Var.9)
Establish a correspondence between the salt formula and the equation of the process occurring on the cathode during the electrolysis of its aqueous solution.
SALT FORMULA EQUATION OF THE CATHODE PROCESS
A) Al(NO3)3 1) 2H2O – 4e- → O2 + 4H+
B) CuCl2 2) 2H2O + 2e- → H2 + 2OH-
C) SbCl3 3) Cu2+ + 1e- → Cu+
D) Cu(NO3)2 4) Sb3+ - 2 e- → Sb5+ 5) Sb3+ + 3e- → Sb0
6) Cu2+ + 2e- → Cu0
The course of reasoning: 1. Processes of reduction of metal or water cations take place on the cathode. Therefore, we immediately exclude options 1 and 4. 2. For Al(NO3)3: the process of water reduction is going on at the cathode. Select option 2. 3. For CuCl2: Cu2+ metal cations are reduced. Choose option 6. 4. For SbCl3: Sb3+ metal cations are reduced. Select option 5. 5. For Cu(NO3)2: Cu2+ metal cations are reduced. We choose option 6.
A
B
IN
G
2

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USE results show that tasks on the topic “Electrolysis” for graduates remain difficult. IN school curriculum not enough hours are devoted to the study of this topic. Therefore, when preparing students for the exam, it is necessary to study this issue in great detail. Knowledge of the basics of electrochemistry will help the graduate to successfully pass the exam and continue their studies at a higher educational institution. To study the topic “Electrolysis” at a sufficient level, it is necessary to conduct preparatory work with graduates passing the exam: - consider the definitions of the basic concepts in the topic "Electrolysis"; - analysis of the process of electrolysis of melts and electrolyte solutions; - consolidate the rules for the reduction of cations at the cathode and the oxidation of anions at the anode (the role of water molecules during the electrolysis of solutions); - formation skills to make equations of the electrolysis process (cathode and anode processes); - to teach students to perform typical tasks basic level(tasks), advanced and high level difficulties. Electrolysis- redox process occurring in solutions and melts of electrolytes with the passage of a direct electric current. In a solution or melt of an electrolyte, it dissociates into ions. When the electric current is turned on, the ions acquire a directed motion, and redox processes can occur on the surface of the electrodes. Anode- a positive electrode, oxidation processes are taking place on it.

The cathode is a negative electrode, recovery processes are taking place on it.

Melt electrolysis used to obtain active metals located in a series of voltages up to aluminum (inclusive).

Electrolysis of sodium chloride melt

K(-) Na + + 1e -> Na 0

A(+) 2Cl - - 2e -> Cl 2 0

2NaCl (electronic current) -> 2Na + Cl 2 (only for melt electrolysis).

Aluminum is obtained by electrolysis of a solution of aluminum oxide in molten cryolite (Na 3 AlF 6).

2Al 2 O 3 (electronic current) -> 4Al + 3O 2

K(-)Al 3+ +3e‾ ->Al

A(+)2O 2‾ -2e‾ ->O 2

Electrolysis of a melt of potassium hydroxide.

KOH->K + +OH‾

K(-) K + + 1e -> K 0

A (+) 4OH - - 4e -> O 2 0 + 2H 2 O

4KOH (electric current) -> 4K 0 + O 2 0 + 2H 2 O

The electrolysis of aqueous solutions is more difficult, since water molecules can be reduced or oxidized on the electrodes in this case.

Electrolysis of aqueous solutions of salts is more complicated due to the possible participation of water molecules at the cathode and at the anode in the electrode processes.

Rules of electrolysis in aqueous solutions.

On the cathode:

1. Cations located in a series of metal voltages from lithium to aluminum (inclusive), as well as cations NH 4 + are not restored, water molecules are restored instead:

2H 2 O + 2e->H 2 + 2OH -

2. Cations located in the series of voltages after aluminum to hydrogen can be reduced together with water molecules:

2H 2 O + 2e->H 2 + 2OH -

Zn2+ + 2e->Zn 0

3. Cations located in a series of voltages after hydrogen are completely restored: Ag + + 1e->Ag 0

4. Hydrogen ions are reduced in acid solutions: 2H + + 2e->H 2

On the anode:

1. Oxygen-containing anions and F-- do not oxidize, instead of them, water molecules are oxidized:

2H 2 O - 4e->O 2 + 4H +

2. Anions of sulfur, iodine, bromine, chlorine (in this sequence) are oxidized to simple substances:

2Cl - - 2e->Cl 2 0 S 2- - 2e->S0

3. Hydroxide ions are oxidized in alkali solutions:

4OH - - 4e->O 2 + 2H 2 O

4. Anions are oxidized in solutions of salts of carboxylic acids:

2 R - SOO - - 2e->R - R + 2CO 2

5. When using soluble anodes, the anode itself sends electrons to the external circuit due to the oxidation of the atoms of the metal from which the anode is made:

Cu 0 - 2e->Сu 2+

Examples of electrolysis processes in aqueous electrolyte solutions

Example 1 K 2 SO 4 -> 2K + + SO 4 2-

K(-)2H 2 O + 2e‾ -> H 2 + 2OH -

A(+)2H 2 O – 4e‾ -> O 2 + 4H +

The general equation of electrolysis: 2H 2 O (el. current) -> 2 H 2 + O 2

Example 2. NaCl ->Na + +Cl‾

K(-)2H 2 O + 2e‾ -> H 2 + 2OH -

A(+) 2Cl - - 2e -> Cl 2 0

2NaCl + 2H 2 O (el.current) -> H 2 + 2NaOH + Cl 2

Example 3. Cu SO 4 -> Cu 2+ + SO 4 2-

K(-) Cu 2+ + 2e‾ -> Cu

A(+)2H 2 O – 4e‾ -> O 2 + 4H +

General electrolysis equation: 2 Cu SO 4 + 2H 2 O (el. current) -> 2Cu + O 2 + 2H 2 SO 4

Example 4. CH 3 COONa->CH 3 COO‾ +Na +

K(-)2H 2 O + 2e‾ -> H 2 + 2OH -

A(+)2CH 3 COO‾– 2e‾ ->C 2 H 6 +2CO 2

General electrolysis equation:

CH 3 COONa + 2H 2 O (el.current) -> H 2 + 2NaHCO 3 + C 2 H 6

Tasks of the basic level of complexity

Test on the topic “Electrolysis of melts and solutions of salts. A series of stresses of metals”.

1. Alkali is one of the products of electrolysis in an aqueous solution:

1) KCI 2) CuSO 4 3) FeCI 2 4) AgNO 3

2. During the electrolysis of an aqueous solution of potassium nitrate, the following is released at the anode: 1) About 2 2) NO 2 3) N 2 4) H 23. Hydrogen is formed during the electrolysis of an aqueous solution: 1) CaCI 2 2) CuSO 4 3) Hg (NO 3) 2 4) AgNO 34. The reaction is possible between: 1) Ag and K 2 SO 4 (solution) 2) Zn and KCI (solution) 3) Mg and SnCI 2(solution) 4) Ag and CuSO 4 (solution) 5. During the electrolysis of a solution of sodium iodide at the cathode, the color of litmus in solution: 1) red 2 ) blue 3) purple 4) yellow6. During the electrolysis of an aqueous solution of potassium fluoride, the following is released at the cathode: 1) hydrogen 2) hydrogen fluoride 3) fluorine 4) oxygen

Tasks on the topic “Electrolysis”

1. The electrolysis of 400 g of a 20% common salt solution was stopped when 11.2 liters (n.o.) of gas were released at the cathode. The degree of decomposition of the original salt (in%) is:

1) 73 2) 54,8 3) 36,8 4) 18

The solution of the problem. We compose the electrolysis reaction equation: 2NaCl + 2H 2 O → H 2 + Cl 2 + 2NaOHm (NaCl) \u003d 400 ∙ 0.2 \u003d 80 g of salt was in solution. ν (H 2) \u003d 11.2 / 22.4 \u003d 0 .5 mol ν(NaCl)=0.5∙2=1 mol(NaCl)= 1∙58.5=58.5 g of salt was decomposed during electrolysis. Degree of salt decomposition 58.5/80=0.73 or 73%.

Answer: 73% of the salt has decomposed.

2. Conducted electrolysis of 200 g of a 10% solution of chromium (III) sulfate until the salt is completely consumed (metal is released on the cathode). The mass (in grams) of water used is:

1) 0,92 2) 1,38 3) 2,76 4) 5,52

The solution of the problem. We compose the electrolysis reaction equation: 2Cr 2 (SO 4) 3 + 6H 2 O → 4Cr + 3O 2 + 6H 2 SO 4m (Cr 2 (SO 4) 3) \u003d 200 ∙ 0.1 \u003d 20g ν (Cr 2 (SO 4) 3) \u003d 20 / 392 \u003d 0.051 mol ν (H 2 O) \u003d 0.051 ∙ 3 \u003d 0.153 molm (H 2 O) \u003d 0.153 18 \u003d 2.76 g

Tasks advanced level difficulty B3

1. Establish a correspondence between the salt formula and the equation of the process occurring at the anode during the electrolysis of its aqueous solution.

3. Establish a correspondence between the salt formula and the equation of the process occurring on the cathode during the electrolysis of its aqueous solution.

5. Establish a correspondence between the name of the substance and the electrolysis products of its aqueous solution.

Answers: 1 - 3411, 2 - 3653, 3 - 2353, 4 - 2246, 5 - 145. Thus, studying the topic of electrolysis, graduates master this section well and show good results in the exam. The study of the material is accompanied by a presentation on this topic.