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

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

The anode is an oxidizer that accepts electrons from anions.

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

Processes occurring on electrodes during the electrolysis of melts

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

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

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

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

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

Processes occurring on the electrodes during the electrolysis of solutions

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

ELECTROLYSIS OF MELTS

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

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

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

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

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

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

SOLUTION ELECTROLYSIS

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

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

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

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

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

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

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

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

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

Solution electrolysis
and molten salts (2 hours)

Classes of the elective course "Electrochemistry"

Goals of the first lesson:

First lesson plan

1. Repetition of the studied methods for obtaining metals.

2. Explanation of new material.

3. Solving problems from the textbook by G.E. Rudzitis, F.G. Feldman "Chemistry-9" (M .: Education, 2002), p. 120, no. 1, 2.

4. Checking the assimilation of knowledge on test tasks.

5. Report on the application of electrolysis.

Goals of the first lesson: to teach how to write schemes for the electrolysis of solutions and molten salts and apply the knowledge gained to solve calculation problems; continue the formation of skills in working with a textbook, test materials; discuss the application of electrolysis in the national economy.

PROGRESS OF THE FIRST LESSON

Repetition of learned methods obtaining metals on the example of obtaining copper from copper(II) oxide.

Recording the equations of the corresponding reactions:

Another way to obtain metals from solutions and melts of their salts is electrochemical, or electrolysis.

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

Electrolysis of sodium chloride melt:

NaCl Na + + Cl – ;

cathode (–) (Na +): Na + + e= Na 0 ,

anode (–) (Cl –): Cl – – e\u003d Cl 0, 2Cl 0 \u003d Cl 2;

2NaCl \u003d 2Na + Cl 2.

Electrolysis of sodium chloride solution:

NaCl Na + + Cl – ,

H 2 O H + + OH -;

cathode (–) (Na +; H +): H + + e= H 0 , 2H 0 = H 2

(2H 2 O + 2 e\u003d H 2 + 2OH -),

anode (+) (Cl - ; OH -): Cl - - e\u003d Cl 0, 2Cl 0 \u003d Cl 2;

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

Electrolysis of copper(II) nitrate solution:

Cu(NO 3) 2 Cu 2+ +

H 2 O H + + OH -;

cathode (–) (Cu 2+; H +): Cu 2+ + 2 e= Cu 0 ,

anode (+) (OH -): OH - - e=OH0,

4H 0 \u003d O 2 + 2H 2 O;

2Cu(NO 3) 2 + 2H 2 O \u003d 2Cu + O 2 + 4HNO 3.

These three examples show why it is more profitable to carry out electrolysis than to carry out other methods of obtaining metals: metals, hydroxides, acids, gases are obtained.

We wrote the electrolysis schemes, and now we will try to write the electrolysis equations right away, without referring to the schemes, but only using the ion activity scale:

Examples of electrolysis equations:

2HgSO 4 + 2H 2 O \u003d 2Hg + O 2 + 2H 2 SO 4;

Na 2 SO 4 + 2H 2 O \u003d Na 2 SO 4 + 2H 2 + O 2;

2LiCl + 2H 2 O \u003d 2LiOH + H 2 + Cl 2.

Problem solving from the textbook by G.E. Rudzitis and F.G. Feldman (9th grade, p. 120, No. 1, 2).

Task 1. During the electrolysis of a solution of copper (II) chloride, the mass of the cathode increased by 8 g. What gas was released, what is its mass?

Solution

CuCl 2 + H 2 O \u003d Cu + Cl 2 + H 2 O,

(Cu) \u003d 8/64 \u003d 0.125 mol,

(Cu) \u003d (Сl 2) \u003d 0.125 mol,

m(Cl 2) \u003d 0.125 71 \u003d 8.875 g.

Answer. The gas is chlorine with a mass of 8.875 g.

Task 2. During the electrolysis of an aqueous solution of silver nitrate, 5.6 liters of gas were released. How many grams of metal deposited on the cathode?

Solution

4AgNO 3 + 2H 2 O \u003d 4Ag + O 2 + 4HNO 3,

(O 2) \u003d 5.6 / 22.4 \u003d 0.25 mol,

(Ag) \u003d 4 (O 2) \u003d 4 25 \u003d 1 mol,

m(Ag) \u003d 1 107 \u003d 107 g.

Answer. 107 g of silver.

Testing

Option 1

1. During the electrolysis of a potassium hydroxide solution at the cathode, the following is released:

a) hydrogen; b) oxygen; c) potassium.

2. During the electrolysis of a solution of copper(II) sulfate in solution, the following is formed:

a) copper(II) hydroxide;

b) sulfuric acid;

3. During the electrolysis of a solution of barium chloride at the anode, the following is released:

a) hydrogen; b) chlorine; c) oxygen.

4. During the electrolysis of an aluminum chloride melt, the following is released at the cathode:

a) aluminum; b) chlorine;

c) electrolysis is impossible.

5. The electrolysis of a solution of silver nitrate proceeds according to the following scheme:

a) AgNO 3 + H 2 O Ag + H 2 + HNO 3;

b) AgNO 3 + H 2 O Ag + O 2 + HNO 3;

c) AgNO 3 + H 2 O AgNO 3 + H 2 + O 2.

Option 2

1. During the electrolysis of a sodium hydroxide solution at the anode, the following is released:

a) sodium; b) oxygen; c) hydrogen.

2. During the electrolysis of a solution of sodium sulfide in solution, the following is formed:

a) hydrosulphuric acid;

b) sodium hydroxide;

3. During the electrolysis of a mercury(II) chloride melt, the following is released at the cathode:

a) mercury; b) chlorine; c) electrolysis is impossible.

4.

5. The electrolysis of a solution of mercury(II) nitrate proceeds according to the following scheme:

a) Hg (NO 3) 2 + H 2 O Hg + H 2 + HNO 3;

b) Hg (NO 3) 2 + H 2 O Hg + O 2 + HNO 3;

c) Hg (NO 3) 2 + H 2 O Hg (NO 3) 2 + H 2 + O 2.

Option 3

1. During the electrolysis of a solution of copper (II) nitrate, the following is released at the cathode:

a) copper; b) oxygen; c) hydrogen.

2. During the electrolysis of a solution of lithium bromide in solution, the following is formed:

b) hydrobromic acid;

c) lithium hydroxide.

3. During the electrolysis of a silver chloride melt, the following is released at the cathode:

a) silver; b) chlorine; c) electrolysis is impossible.

4. During the electrolysis of an aluminum chloride solution, aluminum is released into:

a) cathode; b) anode; c) remains in solution.

5. The electrolysis of a solution of barium bromide proceeds according to the following scheme:

a) BaBr 2 + H 2 O Br 2 + H 2 + Ba (OH) 2;

b) BaBr 2 + H 2 O Br 2 + Ba + H 2 O;

c) BaBr 2 + H 2 O Br 2 + O 2 + Ba (OH) 2.

Option 4

1. During the electrolysis of a barium hydroxide solution at the anode, the following is released:

a) hydrogen; b) oxygen; c) barium.

2. During the electrolysis of a solution of potassium iodide in solution, the following is formed:

a) hydroiodic acid;

b) water; c) potassium hydroxide.

3. During the electrolysis of a melt of lead (II) chloride, the following is released at the cathode:

a) lead; b) chlorine; c) electrolysis is impossible.

4. During the electrolysis of a silver nitrate solution at the cathode, the following is released:

a) silver; b) hydrogen; c) oxygen.

5. The electrolysis of sodium sulfide solution proceeds according to the following scheme:

a) Na 2 S + H 2 O S + H 2 + NaOH;

b) Na 2 S + H 2 O H 2 + O 2 + Na 2 S;

c) Na 2 S + H 2 O H 2 + Na 2 S + NaOH.

Answers

Option	Question 1	Question 2	Question 3	Question 4	Question 5
1	A	b	b	A	b
2	b	b	A	A	b
3	A	V	A	V	A
4	b	V	A	A	A

The use of electrolysis in the national economy

1. To protect metal products from corrosion, a thin layer of another metal is applied to their surface: chromium, silver, gold, nickel, etc. Sometimes, in order not to waste expensive metals, a multi-layer coating is produced. For example, the exterior parts of a car are first covered with a thin layer of copper, a thin layer of nickel is applied to the copper, and a layer of chromium is applied to it.

When applying coatings to metal by electrolysis, they are obtained even in thickness and durable. In this way, you can cover products of any shape. This branch of applied electrochemistry is called electroplating.

2. In addition to corrosion protection, galvanic coatings give a beautiful decorative look to products.

3. Another branch of electrochemistry, close in principle to electroplating, is called electroplating. This is the process of obtaining exact copies of various items. To do this, the object is covered with wax and a matrix is ​​obtained. All recesses of the copied object on the matrix will be bulges. The surface of the wax matrix is ​​coated with a thin layer of graphite, making it electrically conductive.

The resulting graphite electrode is immersed in a bath of copper sulfate solution. The anode is copper. During electrolysis, the copper anode dissolves, and copper is deposited on the graphite cathode. Thus, an exact copper copy is obtained.

With the help of electroforming, clichés for printing, gramophone records are made, various objects are metallized. Galvanoplasty was discovered by the Russian scientist B.S. Jacobi (1838).

Making record dies involves applying a thin layer of silver to a plastic record to make it electrically conductive. Then an electrolytic nickel coating is applied to the plate.

What should be done to make a plate in an electrolytic bath - anode or cathode?

(About the e t. Cathode.)

4. Electrolysis is used to obtain many metals: alkali, alkaline earth, aluminum, lanthanides, etc.

5. To clean some metals from impurities, the metal with impurities is connected to the anode. The metal is dissolved during the electrolysis process and precipitated on the metal cathode, while the impurity remains in solution.

6. Electrolysis is widely used to obtain complex substances (alkalis, oxygen-containing acids), halogens.

Practical work
(second lesson)

Lesson goals. Conduct water electrolysis, show electroplating in practice, consolidate the knowledge gained in the first lesson.

Equipment.On student tables: a flat battery, two wires with terminals, two graphite electrodes, a beaker, test tubes, a tripod with two legs, 3% sodium sulfate solution, a spirit lamp, matches, a torch.

On the teacher's desk: the same + a solution of copper sulfate, a brass key, a copper tube (a piece of copper).

Student briefing

1. Attach the wires with terminals to the electrodes.

2. Place the electrodes in a glass so that they do not touch.

3. Pour the electrolyte solution (sodium sulfate) into the beaker.

4. Pour water into the test tubes and, putting them upside down in a glass with electrolyte, put them on the graphite electrodes one by one, fixing the upper edge of the test tube in the foot of the tripod.

5. After the device is mounted, attach the ends of the wires to the battery.

6. Observe the evolution of gas bubbles: less of them are released at the anode than at the cathode. After almost all the water in one test tube is displaced by the released gas, and in the other - by half, disconnect the wires from the battery.

7. Light the spirit lamp, carefully remove the test tube, where the water is almost completely displaced, and bring it to the spirit lamp - a characteristic pop of gas will be heard.

8. Light a torch. Remove the second test tube, check with a smoldering splint of gas.

Assignments for students

1. Sketch the device.

2. Write an equation for the electrolysis of water and explain why it was necessary to carry out electrolysis in a solution of sodium sulfate.

3. Write reaction equations that reflect the release of gases on the electrodes.

Teacher demonstration experiment
(can be performed by the best students in the class
with appropriate equipment)

1. Connect the wire terminals to the copper tube and brass key.

2. Lower the tube and key into a beaker with copper(II) sulfate solution.

3. Connect the second ends of the wires to the battery: "minus" of the battery to the copper tube, "plus" to the key!

4. Observe the release of copper on the surface of the key.

5. After performing the experiment, first disconnect the terminals from the battery, then remove the key from the solution.

6. Disassemble the electrolysis circuit with a soluble electrode:

CuSO 4 \u003d Cu 2+ +

anode (+): Сu 0 - 2 e\u003d Cu 2+,

cathode (–): Cu 2+ + 2 e= Сu 0 .

The overall equation for electrolysis with a soluble anode cannot be written.

The electrolysis was carried out in a solution of copper(II) sulfate, because:

a) an electrolyte solution is needed in order for an electric current to flow, tk. water is a weak electrolyte;

b) no by-products of the reactions will be released, but only copper at the cathode.

7. To consolidate the past, write a scheme for the electrolysis of zinc chloride with carbon electrodes:

ZnCl 2 \u003d Zn 2+ + 2Cl -,

cathode (–): Zn 2+ + 2 e= Zn 0 ,

2H2O+2 e\u003d H 2 + 2OH -,

anode (+): 2Cl – – 2 e=Cl2.

The overall reaction equation in this case cannot be written, because it is not known what part of the total amount of electricity goes to the reduction of water, and what part - to the reduction of zinc ions.

Scheme of the demonstration experiment

Homework

1. Write an equation for the electrolysis of a solution containing a mixture of copper(II) nitrate and silver nitrate with inert electrodes.

2. Write the equation for the electrolysis of sodium hydroxide solution.

3. To clean a copper coin, it must be hung on a copper wire connected to the negative pole of the battery, and lowered into a 2.5% NaOH solution, where the graphite electrode connected to the positive pole of the battery should also be immersed. Explain how a coin becomes clean. ( Answer. Hydrogen ions are being reduced at the cathode:

2H + + 2 e\u003d H 2.

Hydrogen reacts with copper oxide on the surface of the coin:

CuO + H 2 \u003d Cu + H 2 O.

This method is better than powder cleaning, because. the coin is not erased.)

The electrode where the reduction takes place is called the cathode.

The electrode at which oxidation occurs is the anode.

Consider the processes occurring during the electrolysis of molten salts of oxygen-free acids: HCl, HBr, HI, H 2 S (with the exception of hydrofluoric or hydrofluoric - HF).

In the melt, such a salt consists of metal cations and anions of the acid residue.

For example, NaCl = Na + + Cl -

On the cathode: Na + + ē = Na metallic sodium is formed (in the general case, a metal that is part of the salt)

On the anode: 2Cl - - 2ē \u003d Cl 2 gaseous chlorine is formed (in the general case, a halogen, which is part of the acid residue - except for fluorine - or sulfur)

Let us consider the processes occurring during the electrolysis of electrolyte solutions.

The processes occurring on the electrodes are determined by the value of the standard electrode potential and the electrolyte concentration (Nernst equation). The school course does not consider the dependence of the electrode potential on the electrolyte concentration and does not use the numerical values ​​of the standard electrode potential. It is enough for students to know that in the series of electrochemical intensity of metals (the activity series of metals), the value of the standard electrode potential of the Me + n / Me pair:

1. increases from left to right
2. metals in the row up to hydrogen have a negative value of this quantity
3. hydrogen, when reduced by the reaction 2H + + 2ē \u003d H 2, (i.e. from acids) has a value of zero standard electrode potential
4. metals in the row after hydrogen have a positive value of this quantity

! hydrogen during reduction according to the reaction:

2H 2 O + 2ē \u003d 2OH - + H 2 , (i.e. from water in a neutral environment) has a negative value of the standard electrode potential -0.41

The anode material can be soluble (iron, chromium, zinc, copper, silver and other metals) and insoluble - inert - (coal, graphite, gold, platinum), so the solution will contain ions formed when the anode is dissolved:

Me - nē = Me + n

The resulting metal ions will be present in the electrolyte solution and their electrochemical activity will also need to be taken into account.

Based on this, for the processes occurring at the cathode, the following rules can be defined:

1. The electrolyte cation is located in the electrochemical series of metal voltages up to and including aluminum, the process of water reduction is in progress:

2H 2 O + 2ē \u003d 2OH -+H2

Metal cations remain in solution, in the cathode space

2. The electrolyte cation is located between aluminum and hydrogen, depending on the concentration of the electrolyte, either the process of water reduction or the process of reduction of metal ions takes place. Since the concentration is not specified in the task, both possible processes are recorded:

2H 2 O + 2ē \u003d 2OH -+H2

Me + n + nē = Me

3. electrolyte cation - these are hydrogen ions, i.e. electrolyte is acid. Hydrogen ions are restored:

2H + + 2ē \u003d H 2

4. The electrolyte cation is located after hydrogen, metal cations are reduced.

Me + n + nē = Me

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

1. If the anode is dissolved (for example, iron, zinc, copper, silver), then the anode metal is oxidized.

Me - nē = Me + n

2. If the anode is inert, i.e. insoluble (graphite, gold, platinum):

a) During the electrolysis of solutions of salts of anoxic acids (except for fluorides), the anion is oxidized;

2Cl - - 2ē \u003d Cl 2

2Br - - 2ē \u003d Br 2

2I - - 2ē \u003d I 2

S2 - - 2ē = S

b) During the electrolysis of alkali solutions, the process of oxidation of the hydroxo group OH - :

4OH - - 4ē \u003d 2H 2 O + O 2

c) During the electrolysis of solutions of salts of oxygen-containing acids: HNO 3 , H 2 SO 4 , H 2 CO 3 , H 3 PO 4 , and fluorides, water is oxidized.

2H 2 O - 4ē \u003d 4H + + O 2

d) During the electrolysis of acetates (salts of acetic or ethanoic acid), the acetate ion is oxidized to ethane and carbon monoxide (IV) - carbon dioxide.

2SN 3 SOO - - 2ē \u003d C 2 H 6 + 2CO 2

Task examples.

1. Establish a correspondence between the salt formula and the product formed on an inert anode during the electrolysis of its aqueous solution.

SALT FORMULA

A) NiSO 4

B) NaClO 4

B) LiCl

D) RbBr

PRODUCT ON ANODE

1) S 2) SO 2 3) Cl 2 4) O 2 5) H 2 6) Br 2

Solution:

Since the task specifies an inert anode, we consider only the changes that occur with acidic residues formed during the dissociation of salts:

SO 4 2 - acid residue of an oxygen-containing acid. Water is oxidized and oxygen is released. Answer 4

ClO4 - acid residue of an oxygen-containing acid. Water is oxidized and oxygen is released. Answer 4.

Cl - acid residue of an oxygen-free acid. There is a process of oxidation of the acid residue itself. Chlorine is released. Answer 3.

Br - acid residue of an oxygen-free acid. There is a process of oxidation of the acid residue itself. Bromine is released. Answer 6.

General response: 4436

2. Establish a correspondence between the salt formula and the product formed on the cathode during the electrolysis of its aqueous solution.

SALT FORMULA

A) Al (NO 3) 3

B) Hg (NO 3) 2

B) Cu (NO 3) 2

D) NaNO 3

PRODUCT ON ANODE

1) hydrogen 2) aluminum 3) mercury 4) copper 5) oxygen 6) sodium

Solution:

Since the task specifies the cathode, we consider only the changes that occur with metal cations formed during the dissociation of salts:

Al 3+ in accordance with the position of aluminum in the electrochemical series of metal voltages (from the beginning of the series to aluminum inclusive), the process of water reduction will proceed. Hydrogen is released. Answer 1.

Hg2+ in accordance with the position of mercury (after hydrogen), the process of reduction of mercury ions will take place. Mercury is formed. Answer 3.

Cu2+ in accordance with the position of copper (after hydrogen), the process of reduction of copper ions will proceed. Answer 4.

Na+ in accordance with the position of sodium (from the beginning of the series to aluminum inclusive), the process of water reduction will proceed. Answer 1.

General answer: 1341