Task EGE Chemistry Electrolysis. Hydrogen, when restoring the reaction. Electrolysis of molten salts

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

Soloi formula		Product on anode

A.	B.	IN	G.

Decision.

With 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 oxygen-containing acid or hydrochloric acid salt;

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

The acid residue is discharged, which is part of the salt, and the corresponding simple substance is distinguished, if it is a salt of oxygenic acid (except).

According to a special process of electrolysis of carboxylic salts.

Answer: 3534.

Answer: 3534.

Source: Yandex: Training work of the exam in chemistry. Option 1.

Set the correspondence between the formula of the substance and the product forming on the cathode at the electrolysis of its aqueous solution: to each position indicated by the letter, select the corresponding position indicated by the number.

Formula of substances		Electrolysis product, Cathode

Write the numbers in response by placing them in order corresponding to the letters:

A.	B.	IN	G.

Decision.

With the electrolysis of aqueous solutions of salts on the cathode, it is allocated:

Hydrogen, if it is a metal salt, standing in a row of metals to the left of the aluminum;

Metal, if it is a metal salt, standing in a row of metal stresses to the right of hydrogen;

Metal and hydrogen, if it is a metal salt, standing in a row of metal stresses between aluminum and hydrogen.

Answer: 3511.

Answer: 3511.

Source: Yandex: Training work of the exam in chemistry. Option 2.

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

Soloi formula		Product on anode

Write the numbers in response by placing them in order corresponding to the letters:

A.	B.	IN	G.

Decision.

With electrolysis of aqueous solutions of salts of oxygen-containing acids and fluorides oxidized oxygen from water, so oxygen is released on the anode. In the electrolysis of aqueous solutions of oxygenic acids, the acid residue is oxidation.

Answer: 4436.

Answer: 4436.

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

Formula of substances	Product on anode
	2) sulfur oxide (IV) 3) carbon oxide (IV) 5) Oxygen 6) Nitrogen oxide (IV)

Write the numbers in response by placing them in order corresponding to the letters:

A.	B.	IN	G.

Electrolysis of melts and solutions (salts, alkalis)

If you lower the electrodes into the solution or the melt of the electrolyte and skip the constant electric current, then the ions will move towards the cathipes (negatively charged electrode), anions to the anode (positively charged electrode).

At the cathode, the cations take electrons and restore, the anions are given electrons on the anode and oxidize. This process is called electrolysis.

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

Electrolysis of molten salts

Consider the process of sodium chloride melt electrolysis. The thermal dissociation is underway in the melt:

$ NaCl → Na ^ (+) + Cl ^ (-). $

Under the action of the electric current of the cations $ na ^ (+) $ move to the cathode and the electrons are taken from it:

$ Na ^ (+) + ē → (na) ↖ (0) $ (recovery).

Anions $ CL ^ (-) $ move to the anode and give electrons:

$ 2cl ^ (-) - 2ē → (Cl_2) ↖ (0) $ (oxidation).

Total processes equation:

$ Na ^ (+) + ē → (Na) ↖ (0) | 2 $

$ 2cl ^ (-) - 2ē → (Cl_2) ↖ (0) | 1 $

$ 2NA ^ (+) + 2cl ^ (-) \u003d 2 (Na) ↖ (0) + (CL_2) ↖ (0) $

$ 2NACL (→) ↖ (\\ Text "Electrolysis") 2NA + CL_2 $

A metallic sodium is formed on the cathode, chlorine gaseous gas on the anode.

The main thing is that you must remember: in the process of electrolysis, a chemical reaction is carried out at the expense of electrical energy, which cannot contain spontaneously.

Electrolysis of aqueous solutions of electrolytes

A more difficult case - electrolysis of electrolyte solutions.

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

To determine the products of electrolysis of aqueous solutions of electrolytes, the following rules exist:

1. The process at the cathode It does not depend on the material from which the cathode is made, but on the position of the metal (electrolyte cation) in the electrochemical row of stresses, and if:

1.1. The electrolyte cation is located in a row of voltages at the beginning of a series of $ Al $ inclusive, then the process of water recovery is in the cathode (hydrogen is highlighted $ n_2 $). Metal cations are not restored, they remain in solution.

1.2. The electrolyte cation is located in a row of stresses between aluminum and hydrogen, then metal ions and water molecules are restored at the cathode.

1.3. The electrolyte cation is located in a row of stresses after hydrogen, the metal cations are restored on the cathode.

1.4. The solution contains the cations of different metals, then the metal cation stands in the row of voltages is restored.

Cathodic processes

2. Process on the anodedepends on the material of the anode and on the nature of the anion.

Anode processes

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

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

a) with electrolysis solutions of salts beatless acids (in addition to fluoride) Anion oxidation process is underway on the anode;

b) with electrolysis solutions of salts oxygen-containing acids and fluorides On the anode there is a water oxidation process ($ o_2 is allocated). Anions are not oxidized, they remain in solution;

c) anions by their ability to oxidize are in the following order:

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

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

1) anode insoluble (for example, graphite).

The solution is the process of electrolytic dissociation:

Total equation:

$ 2H_2O + 2Cl ^ (-) \u003d H_2 + CL_2 + 2OH ^ (-) $.

Considering the presence of ions of $ na ^ (+) $ in solution, we compile a molecular equation:

2) anode soluble (for example, copper):

$ NaCl \u003d Na ^ (+) + Cl ^ (-) $.

If the anode is soluble, then the metal of the anode will be oxidized:

$ Cu ^ (0) -2ē \u003d Cu ^ (2 +) $.

Cations $ Cu ^ (2 +) $ are in a row of stresses after ($ n ^ (+) $), they will be restored on the cathode.

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

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

$ Cu ^ (2 +) + 2ē \u003d Cu ^ (0) | 2 $

$ 2h_2o-4ē \u003d o_2 + 4h ^ (+) | $ 1

Total ion equation:

$ 2cu ^ (2 +) + 2H_2O \u003d 2CU ^ (0) + O_2 + 4H ^ (+) $

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

Consider the electrolysis of potassium hydroxide solution insoluble anode:

$ 2H_2O + 2ē \u003d H_2 + 2OH ^ (-) | $ 2

$ 4oh ^ (-) - 4ē \u003d o_2 + 2H_2O | 1 $

Total ion equation:

$ 4H_2O + 4OH ^ (-) \u003d 2H_2 + 4OH ^ (-) + O_2 + 2H_2O $

Total molecular equation:

$ 2H_2O (→) ↖ (\\ Text "Electrolysis") 2H_2 + O_2 $

In this case, it turns out, only the electrolysis of water is. We obtain a similar result and in the case of the electrolysis of the solutions of $ H_2SO_4, NANO_3, K_2SO_4 $, etc.

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

1. For metals (aluminum, magnesium, sodium, cadmium is obtained by electrolysis).
2. To obtain hydrogen, halogen, alkalis.
3. For the purification of metals - refining (cleaning copper, nickel, lead is carried out with an electrochemical method).
4. To protect metals from corrosion (chromium, nickel, copper, silver, gold) - galvanotegia.
5. For metal copies, plates - electrotype.

The electrode on which the restoration occurs is called a cathode.

The electrode on which oxidation occurs is an anode.

Consider the processes occurring in the electrolysis of melts of salts of oxygenic acids: HCl, HBr, Hi, H 2 S (with the exception of fluoride or fluid - HF).

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

For example, NaCl \u003d Na + + Cl -

At the cathode: Na + + ē \u003d na metal sodium is formed (in the general case - metal included in the salt)

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

Consider the processes occurring in electrolysis of electrolyte solutions.

The processes flowing on the electrodes are determined by the value of the standard electrode potential and the electrolyte concentration (the Nernst equation). The school course does not consider the dependence of the electrode potential from the electrolyte concentration and the numerical values \u200b\u200bof the values \u200b\u200bof the standard electrode potential are not used. It is enough for students to know that in a number of electrochemical tensions of metals (a number of metal activity) The value of the standard electrode potential of the pair ME + N / ME:

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

! hydrogen when reconstructed by reaction:

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

Anode material can be soluble (iron, chrome, zinc, copper, silver, etc. Metals) and insoluble - inert - (coal, graphite, gold, platinum), Therefore, ions formed when dissolving the anode will be present in the solution:

Me - nē \u003d me + n

Metal ions formed will be present in the electrolyte solution and their electrochemical activity will also need to be considered.

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

1. The electrolyte cation is located in an electrochemical row of stresses of metals to aluminum inclusive, water recovery is processed:

2H 2 O + 2ē \u003d 2OH - + H 2

Metal cations remain in solution, in cathode space

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

2H 2 O + 2ē \u003d 2OH - + H 2

Me + n + nē \u003d me

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

2N + + 2ē \u003d H 2

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

Me + n + nē \u003d me

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

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

Me - nē \u003d me + n

2. If annert anode, i.e. Not dissolving (graphite, gold, platinum):

a) with the electrolysis of the solutions of oxygenic acid salts (except for fluorides), the process of oxidation of anion is underway;

2cl - - 2ē \u003d Cl 2

2br. - - 2ē \u003d br 2

2i. - - 2ē \u003d i 2

S 2. - - 2ē \u003d s

b) with electrolysis of alkalis solutions, the process of oxidation of the hydroxochroup is:

4oh. - - 4ē \u003d 2H 2 O + O 2

c) with electrolysis of solutions of oxygen-containing acids: HNO 3, H 2 SO 4, H 2 CO 3, H 3 PO 4, and fluorides, the water oxidation process is underway.

2H 2 O - 4ē \u003d 4H + + O 2

d) under the electrolysis of acetates (acetate or ethanic acid salts) is oxidated with an acetate ion to ethane and carbon oxide (IV) - carbon dioxide.

2 SO 3 SOO - - 2ē \u003d C 2 H 6 + 2So 2

Examples of tasks.

1. Install the correspondence between the salt formula and the product forming on an inert anode with the electrolysis of its aqueous solution.

Soloi formula

A) niso 4

B) Naclo 4

C) licl

D) rbbr.

Product on anode

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

Decision:

Since the inert anode is specified in the task, we consider only changes occurring with acid residues formed during salts dissociation:

SO 4 2. - acid residue of oxygen-containing acid. There is a water oxidation process, oxygen is released. Answer 4.

CLO 4. - acid residue of oxygen-containing acid. There is a water oxidation process, oxygen is released. Answer 4.

Cl. - acid residue of oxygenic acid. There is a process of oxidation of the acidic residue itself. Chlorine is distinguished. Answer 3.

Br. - acid residue of oxygenic acid. There is a process of oxidation of the acidic residue itself. Allocated in bromine. Answer 6.

Total answer: 4436

2. Install the correspondence between the salt formula and the product forming on the cathode at the electrolysis of its aqueous solution.

Soloi formula

A) Al (NO 3) 3

B) HG (NO 3) 2

C) Cu (NO 3) 2

D) nano 3

Product on anode

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

Decision:

Since the cathode is specified in the task, we consider only changes occurring with metals cations formed during salts dissociation:

Al 3+ in accordance with the position of aluminum in the electrochemical row of metals voltages (from the beginning of the row to aluminum inclusive) will go the process of water recovery. Hydrogen is distinguished. Answer 1.

HG 2+ in accordance with the position of mercury (after hydrogen) there will be a process of recovering mercury ions. It is formed mercury. Answer 3.

Cu 2+ in accordance with the position of copper (after hydrogen) there will be a process of restoring copper ions. Answer 4.

Na +. in accordance with the position of sodium (from the beginning of a number to aluminum inclusive) will go the process of water recovery. Answer 1.

Total answer: 1341