Thematic tests for chemistry EG electrolysis. Preparation of graduates to the exam

Install the correspondence between the salt formula and the product forming on an inert anode with its electrolysis aquatic solution: To each position indicated by the letter, select the appropriate 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).

The electrolysis process of salts occurs on a special carboxylic acids.

Answer: 3534.

Source: Yandex: Training work EGE 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.

Source: Yandex: Training work of the EME 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.

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 electrolysis process due to electrical energy is carried out chemical reactionwhich can not go spontaneously.

Electrolysis of aqueous solutions of electrolytes

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 cations different metals, I first restore the metal cation standing in the row of stresses to the right.

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, make up 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 caseIt turns out, only the electrolysis of water is underway. 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:

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

Topic 6. "Electrolysis of solutions and salts melts"
1. Electrolysis - oxidative - the reduction process flowing on the electrodes when the electric current is passed through the solution or the electrolyte melt.
2. Cathode - negative electrode. The restoration of metal and hydrogen cations (in acids) or water molecules occurs.
3. Anode is a positive electrode. There is oxidation of an anions of the acid residue and the gyrosogroup (in alkali).
4. With the electrolysis of the salt solution in the reaction mixture there are water. Since water can also show oxidative and restorative properties, it is a "competitor" and for cathode and for anode processes.
5. There are electrolysis with inert electrodes (graphite, coal, platinum) and an active anode (soluble), as well as electrolysis of melts and solutions of electrolytes.
Cathodic processes
If the metal is located in a row of voltages:
Metal position in row of stresses
Restoration at the cathode
from li to al
Restore water molecules: 2H2O + 2E- → H20 + 2OH-
from Mn to pb
Water molecules and metal cations are restored:
2H2O + 2E- → H20 + 2OH-
MEN + + NE- → ME0
from Cu to Au
Metal cations are restored: MEN + + NE- → ME0
Anode processes
Acid residue
Asm-
Anode
Soluble
(iron, zinc, copper, silver)
Insoluble
(Graphite, Gold, Platinum)
Oxless
Oxidation of metal anode
M0 - NE- \u003d Mn +
anode solution
Anion oxidation (except F-)
ASM- - ME- \u003d AC0
Oxygen-containing
Fluoride - Ion (F-)
In acidic and neutral environments:
2 H2O - 4E- → O20 + 4H +
In an alkaline environment:
4one- - 4E- \u003d O20 + 2N2O
Examples of electrolysis processes of melts with inert electrodes
In the melt of the electrolyte there are only its ions, so electrolyte cations are restored on the cathode, and anions are oxidized on the anode.
1. Consider the electrolysis of the melt of potassium chloride.
Thermal dissociation of KSL → K + + CL-
K (-) K + + 1E- → K0
A (+) 2SL- - 2E- → CL02
Summary Equation:
2xl → 2K0 + CL20
2. Consider the electrolysis of calcium chloride melt.
Thermal dissociation of SASL2 → CA2 + + 2SL
K (-) CA2 + + 2E- → CA0
A (+) 2SL- - 2E- → CL02
Summary Equation:
CACL2 → CA0 + CL20
3. Consider the electrolysis of potassium hydroxide melt.
Thermal dissociation Kon → K + +
K (-) K + + 1E- → K0
A (+) 4on- - 4e- → O20 + 2N2O
Summary Equation:
4Cone → 4k0 + O20 + 2N2O
Examples of electrolysis processes of electrolyte solutions with inert electrodes
In contrast to the melts in the electrolyte solution, besides its ions, there are water molecules. Therefore, when considering processes on the electrodes, it is necessary to take into account their participation. The electrolysis of the salt solution formed by the active metal, standing in a row of stresses to aluminum and an acidic residue of oxygen-containing acid comes down to the electrolysis of water. 1. Consider the electrolysis of the aqueous solution of magnesium sulfate. MgSO4 is a salt that is formed by a metal standing in a row of stresses to aluminum and oxygen-containing acid residue. The dissociation equation: MgSO4 → Mg2 + + SO42- K (-) 2N2O + 2E- \u003d H20 + 2N- A (+) 2N2O - 4E- \u003d O20 + 4N + Total equation: 6N2O \u003d 2N20 + 4H- + O20 + 4N + 2N2O \u003d 2N20 + O20 2. Consider the electrolysis of the aqueous solution of copper sulfate (II). SusO4 - salt, which is formed by a low-active metal and oxygen-containing acid residue. In this case, the electrolysis obtains metal, oxygen, and the corresponding acid is formed in the cathode-anode space. The dissociation equation: Cuso4 → Cu2 + + SO42- to (-) Cu2 + + 2e- \u003d Cu0 A (+) 2N2O - 4E- \u003d O20 + 4N + Total equation: 2CU2 + 2N2O \u003d 2CU0 + O20 + 4N + 2CU0 + 2N2O \u003d 2CU0 + O20 + 2N2SO4
3. Consider the electrolysis of the hydrogen solution of calcium chloride. CACL2 is a salt, which is formed by an active metal and an oxygenic acid residue. In this case, hydrogen is formed under electrolysis, halogen, and alkali is formed in the cathode-anode space. The dissociation equation: CAcl2 → Ca2 + + 2Cl- to (-) 2N2O + 2E- \u003d H20 + 2O- A (+) 2SL- 2E- \u003d CL20 The total equation: 2N2O + 2CL- \u003d CL20 + 2O / CACL2 + 2N2O \u003d CA (OH) 2 + CL20 + H20 4. Consider the electrolysis of the aqueous solution of copper chloride (II). CuCl2 is a salt that is formed by a low-active metal and acid residue of oxygenic acid. In this case, metal and halogen are formed. Dissociation equation: CUCL2 → CU2 + 2CL- to (-) Cu2 + + 2e- \u003d Cu0 A (+) 2SL- 2E- \u003d CL20 Total equation: Cu2 + 2cl- \u003d Cu0 + CL20 CUCL2 \u003d CU0 + CL20 5. Consider the process Electrolysis solution of sodium acetate. CH3COONA - salt, which is formed by the active metal and acid residue of carboxylic acid. At electrolysis, hydrogen is obtained, alkali. The dissociation equation: CH3SOONA → CH3SOO - + Na + K (-) 2N2O + 2E- \u003d H20 + 2N- A (+) 2CH3COO2E \u003d C2H6 + 2CO2 Total equation: 2N2O + 2CH3COO¯ \u003d H20 + 2Ho - + C2H6 + 2CO2 2N2O + 2CH3Coona \u003d 2NAOH + H20 + C2H6 + 2CO2 6. Consider the electrolysis process of nickel nitrate solution. Ni (NO3) 2 - salt, which is formed by a metal standing in a row of voltages from Mn to H2 and an oxygen-containing acid residue. In the process we get metal, hydrogen, oxygen and acid. The dissociation equation: Ni (NO3) 2 → Ni2 + + 2NO3- to (-) ni2 + 2e- \u003d ni0 2N2O + 2E- \u003d H20 + 2O- A (+) 2H2O - 4E- \u003d O20 + 4H + COMMARY EQUATION: Ni2 + + 2N2O + 2H2O \u003d Ni0 + H20 + 2H + O20 + 4H + Ni (NO3) 2 + 2N2O \u003d Ni0 + 2HNO3 + H20 + O20 7. Consider the electrolysis process of sulfuric acid solution. The dissociation equation: H2SO4 → 2H + + SO42- K (-) 2N + + 2E- \u003d H20 A (+) 2H2O - 4E- \u003d O20 + 4H + Summarine Equation: 2N2O + 4N + \u003d 2N20 + O20 + 4H + 2H2O \u003d 2N20 + O20
8. Consider the electrolysis process of sodium hydroxide solution. In this case, only the electrolysis of water is. The electrolysis of H2SO4 solutions, NaNO3, K2SO4, etc. The dissociation equation: NaOH → Na + + 2e- \u003d H20 + 2O- A (+) 4OH- - 4E- \u003d O20 + 2H2O Summarine equation: 4H2O + 4OH- \u003d 2H20 + 4OH- + O20 + 2H2O 2H2O \u003d 2H20 + O20
Examples of electrolysis processes of electrolyte solutions with soluble electrodes
The soluble anode during electrolysis itself is oxidation (dissolution). 1. Consider the electrolysis process of copper (II) sulfate with a copper anode. With electrolysis of copper sulfate solution with a copper anode, the process is reduced to the highlight of copper on the cathode and the gradual dissolution of the anode, despite the nature of the anion. The amount of copper sulfate in the solution remains unchanged. DISSOCIATION EQUATION: CUSO4 → CU2 + + SO42- K (-) Cu2 + + 2e- → Cu0 A (+) Cu0 - 2e- → Cu2 + copper ion transition from anode to cathode
Examples of tasks on this topic in the EGE options
IN 3. (Var.5)
Install the correspondence between the formula of the substance and the electrolysis products of its aqueous solution on the inert electrodes.
Formula substance Electrolysis products
A) Al2 (SO4) 3 1. Metal hydroxide, acid
B) CSOH 2. Metal, halogen
C) Hg (NO3) 2 3. Metal, oxygen
D) AUBR3 4. Hydrogen, halogen 5. Hydrogen, oxygen 6. Metal, acid, oxygen stroke of reasoning: 1. With Al2 electrolysis (SO4) 3 and CSOH on the cathode, water is restored to hydrogen. We exclude variants 1, 2, 3 and 6. 2. For Al2 (SO4) 3, water is oxidized on the anode to oxygen. We choose option 5. For CSOH, an ion hydroxide is oxidized on the anode to oxygen. We choose option 5. 3. With the electrolysis Hg (NO3) 2 and AUBR3 on the cathode there is a restoration of metal cations. 4. For HG (NO3) 2 water is oxidized on the anode. Nitrate ions in solution are associated with hydrogen cations, forming in anodic space nitric acid. We choose option 6. 5. For AUBR3, an anion BR2 is oxidized on the anode. Select option 2.
BUT
B.
IN
G.
5
5
6
2
IN 3. (VAR.1)
Set the correspondence between the name of the substance and the method of obtaining it.
The name of the substance is obtained by electrolysis a) lithium 1) solution of LIF b) fluorine 2) melt LIF c) silver 3) solution MgCl2 g) Magnesium 4) AGNO3 solution 5) melt AG2O 6) molten MgCl2 stroke of reasoning: 1. Similar to the sodium chloride electrolysis The process of lithium fluoride melt electrolysis is proceeded. For options A and B, we choose the answers 2. 2. Silver it is possible to restore it from the solution of its salt - silver nitrate. 3. From the solution of Magnesium Salt cannot be restored. We choose the option 6 - the melt of magnesium chloride.
BUT
B.
IN
G.
2
2
4
6
IN 3. (Var.9)
Install the correspondence between the salt formula and the equation of the process flowing 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. The processes of recovery of metal cations or water flow on the cathode. Therefore, immediately exclude options 1 and 4. 2. For Al (NO3) 3: the process of water recovery is on the cathode. Select option 2. 3. For CUCL2: CU2 + metal cations are restored. Select option 6. 4. For SBCl3: SB3 + metal cations are restored. Select option 5. 5. For CU (NO3) 2: CU2 + metal cations are restored. Select option 6.
BUT
B.
IN
G.
2

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). IN school course The dependence of the electrode potential from the electrolyte concentration is not considered 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:

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

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EME results Show that tasks on the topic "Electrolysis" for graduates remain complex. IN school Program The study of this topic is given an insufficient number of hours. Therefore, when preparing schoolchildren to the exam, it is necessary to explore this issue in very detailed. Knowledge of the fundamentals of electrochemistry will help a graduate to successfully pass the exam and continue training in the higher educational institution. For the study of the topic "Electrolysis" at a sufficient level need to be held preparatory work With graduates of the EGE: - consider defining the basic concepts in the topic "Electrolysis"; - analysis of the process of electrolysis of melts and electrolyte solutions; - consolidate the rules for restoring cations on the cathode and oxidation of the anions on 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); - teach students to perform typical tasks basic level (tasks), elevated and high level difficulties. Electrolysis - the redox process flowing in solutions and melts of electrolytes during the passage of direct electric current. In solution or melt electrolyte, its dissociation on ions occurs. When the electrical current is turned on, the directional movement and on the surface of the electrodes may occur redox processes. Anode - Positive electrode, it goes oxidation processes.

The cathode is a negative electrode, there are processes of recovery on it.

Electrolysis of melts It is used to obtain active metals located in a row of stresses to aluminum (inclusive).

Electrolysis of melt sodium chloride

K (-) Na + + 1e -\u003e Na 0

A (+) 2CL - - 2E -\u003e Cl 2 0

2NACL (email) -\u003e 2NA + CL 2 (only with melt electrolysis).

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

2AL 2 O 3 (email) -\u003e 4Al + 3O 2

K (-) Al 3+ + 3E~ -\u003e Al

A (+) 2O 2~ -2E~ -\u003e O 2

Electrolysis of potassium hydroxide melt.

KOH-\u003e K + + OH~

K (-) k + + 1e -\u003e k 0

A (+) 4OH - - 4E -\u003e O 2 0 + 2N 2

4Koh (email) -\u003e 4K 0 + O 2 0 + 2N 2 o

Electrolysis of aqueous solutions is more complicated, since the water molecules can be restored on the electrodes in this case.

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

Electrolysis rules in aqueous solutions.

At the cathode:

1. Cations, located in a row of metals voltage from lithium to aluminum (inclusive), as well as cations NN 4 +. Do not restore, water molecules are restored instead:

2N 2 O + 2E-> H 2 + 2H -

2. Cations, located in a row of stresses after aluminum to hydrogen, can be recovered along with water molecules:

2N 2 O + 2E-> H 2 + 2H -

Zn 2+ + 2e-> Zn 0.

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

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

On the anode:

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

2N 2 O - 4E-> O 2 + 4N +

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

2SL - - 2E-> Cl 2 0 S 2- - 2e-> S 0.

3. Hydroxide ions are oxidized in alkalis solutions:

4on - - 4E-> O 2 + 2N 2 o

4. Anions are oxidized in solutions of carboxylic salts:

2 R - SOO - - 2E-> R - R + 2SO 2

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

Cu 0 - 2e-> Cu 2+

Examples of electrolysis processes in aqueous solutions of electrolytes

Example 1.K 2 SO 4 -\u003e 2K + + SO 4 2-

K (-) 2H 2 O + 2E~ -\u003e H 2 + 2OH -

A (+) 2H 2 O - 4E~ -\u003e O 2 + 4H +

General electrolysis equation: 2H 2 O (email) -\u003e 2 H 2 + O 2

Example 2. NaCl -\u003e Na + + CL~

K (-) 2H 2 O + 2E~ -\u003e H 2 + 2OH -

A (+) 2CL - - 2E -\u003e Cl 2 0

2NACL + 2H 2 O (email) -\u003e H 2 + 2NAOH + CL 2

Example 3. Cu SO 4 -\u003e Cu 2+ + SO 4 2-

K (-) Cu 2+ + 2E~ -\u003e Cu

A (+) 2H 2 O - 4E~ -\u003e O 2 + 4H +

General Equation of Electrolysis: 2 Cu SO 4 + 2H 2 O (Current) -\u003e 2CU + O 2 + 2H 2 SO 4

Example 4. CH 3 COONA-\u003e CH 3 COO~ + NA +

K (-) 2H 2 O + 2E~ -\u003e H 2 + 2OH -

A (+) 2CH 3 COO~- 2E~ -\u003e C 2 H 6 + 2CO 2

General electrolysis equation:

CH 3 COONA + 2H 2 O (email) -\u003e H 2 + 2NAHCO 3 + C 2 H 6

Quests of the basic level of complexity

Test on the topic "Electrolysis of melts and solutions of salts. A number of metal voltages. "

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

1) Kci. 2) CUSO 4 3) FECI 2 4) AGNO 3

2. With the electrolysis of aqueous solution of potassium nitrate on the anode allocated: 1) O 2.2) NO 2 3) N 2 4) H 23. Hydrogen is formed under the electrolysis of the 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 (P-P) 2) Zn and KCI (P-P) 3) MG and SNCI 2(P-p) 4) AG and CUSO 4 (P-P) 5. With the electrolysis of the solution of sodium iodide at the cathode of the lacmus color in the solution: 1) Red 2 ) Blue 3) Purple 4) Yellow6. With the electrolysis of aqueous solution of potassium fluoride on the cathode allocated: 1) hydrogen2) fluoride fluorine 3) fluorine 4) oxygen

Tasks on the topic "Electrolysis"

1. Electrolysis 400 g of a 20% solution crash salt It was stopped when 11.2 liters (N.U.) gas was separated on the cathode. The degree of decomposition of the source salt (in%) is:

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

The solution of the problem.We compile the equation of the electrolysis reaction: 2NACL + 2H 2 O → H 2 + CL 2 + 2NAOHM (NaCl) \u003d 400 ∙ 0.2 \u003d 80 g of salts were in solution.ν (H 2) \u003d 11.2 / 22.4 \u003d 0 , 5 mole ν (naCl) \u003d 0.5 ∙ 2 \u003d 1 molm (NaCl) \u003d 1 ∙ 58.5 \u003d 58.5 g of salts were decomposed during electrolysis. Salt decomposition of 58.5 / 80 \u003d 0.73 or 73%.

Answer: 73% of salt decomposed.

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

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

The solution of the problem.We compile 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 20gν (CR 2 (SO 4) 3) \u003d 20/392 \u003d 0.051molν (H 2 O) \u003d 0.051 ∙ 3 \u003d 0.153 molm (H 2 O) \u003d 0.153 ∙ 18 \u003d 2.76 g

Tasks increased level Q3 complexity

1. Install the correspondence between the salt formula and the equation of the process flowing on the anode during the electrolysis of its aqueous solution.

3. Install the correspondence between the salt formula and the equation of the process flowing on the cathode at the electrolysis of its aqueous solution.

5. Install the 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. In the way, studying the topic of electrolysis, graduates are well absorbed by this section and show good results on the exam. The study of material is accompanied by a presentation on this topic.