Preparing graduates for the exam. "The topic of electrolysis in the exam"
Electrolysis of melts and solutions (salts, alkalis)
If electrodes are lowered into a solution or molten electrolyte and a constant electric current is passed, then the ions will move in a directional way: cations to the cathode (negatively charged electrode), anions to the anode (positively charged electrode).
At the cathode, cations receive electrons and are reduced, at the anode, anions donate electrons and oxidize. This process is called electrolysis.
Electrolysis is a redox process that occurs on electrodes when an electric current passes through a melt or electrolyte solution.
Molten salt electrolysis
Consider the process of electrolysis of 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 to the cathode and receive electrons from it:
$ Na ^ (+) + ē → (Na) ↖ (0) $ (recovery).
Anions $ Cl ^ (-) $ move to the anode and donate electrons:
$ 2Cl ^ (-) - 2ē → (Cl_2) ↖ (0) $ (oxidation).
The total equation of the processes:
$ 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 $
Metallic sodium is formed at the cathode, and chlorine gas at the anode.
The main thing that you must remember: in the process of electrolysis due to electrical energy, a chemical reaction is carried out, which cannot spontaneously go.
Electrolysis of aqueous solutions of electrolytes
A more complicated case is electrolysis of electrolyte solutions.
In the salt solution, in addition to metal ions and acid residue, water molecules are present. Therefore, when considering the 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, there are the following rules:
1. Cathode process depends not on the material from which the cathode is made, but on the position of the metal (electrolyte cation) in the electrochemical series of voltages, while if:
1.1. The electrolyte cation is located in the series of voltages at the beginning of the series along $ Al $ inclusively, then the process of water reduction occurs at the cathode (hydrogen is released $ Н_2 $). Metal cations are not reduced, they remain in solution.
1.2. The electrolyte cation is in a series of voltages between aluminum and hydrogen, then both metal ions and water molecules are reduced at the cathode.
1.3. The electrolyte cation is in the series of voltages after hydrogen, then metal cations are reduced at the cathode.
1.4. The solution contains cations of different metals, then the metal cation, which is to the right in the series of voltages, is first reduced.
Cathodic processes
2. Anode processdepends on the material of the anode and on the nature of the anion.
Anodic processes
2.1. If a the anode dissolves (iron, zinc, copper, silver and all metals that are oxidized during electrolysis), then the anode metal is oxidized, despite the nature of the anion.
2.2. If a the anode does not dissolve (it is called inert - graphite, gold, platinum), then:
a) during the electrolysis of salt solutions anoxic acids (except fluorides) an anion is oxidized at the anode;
b) during the electrolysis of salt solutions oxygenated acids and fluorides the process of water oxidation is going on at the anode ($ О_2 $ is released). Anions are not oxidized, they remain in solution;
c) anions according to their ability to oxidize are arranged in the following order:
Let's try to apply these rules in specific situations.
Let us 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).
The solution is undergoing electrolytic dissociation:
Summary equation:
$ 2H_2O + 2Cl ^ (-) \u003d 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 \u003d Na ^ (+) + Cl ^ (-) $.
If the anode is soluble, then the anode metal will oxidize:
$ Cu ^ (0) -2ē \u003d Cu ^ (2 +) $.
The $ Cu ^ (2 +) $ cations in the series of voltages are after ($ H ^ (+) $); therefore, 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 (II) sulfate on insoluble anode:
$ Cu ^ (2 +) + 2ē \u003d Cu ^ (0) | 2 $
$ 2H_2O-4ē \u003d O_2 + 4H ^ (+) | 1 $
Total ionic equation:
$ 2Cu ^ (2 +) + 2H_2O \u003d 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 \u003d H_2 + 2OH ^ (-) | 2 $
$ 4OH ^ (-) - 4ē \u003d O_2 + 2H_2O | 1 $
Total ionic 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 that only water electrolysis takes place. We obtain a similar result 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:
- For the production of metals (aluminum, magnesium, sodium, cadmium are obtained only by electrolysis).
- For the production of hydrogen, halogens, alkalis.
- For the purification of metals - refining (purification of copper, nickel, lead is carried out by an electrochemical method).
- To protect metals from corrosion (chromium, nickel, copper, silver, gold) - electroplating.
- To obtain metal copies, records - electrotype.
The electrode on which the reduction takes place is called the cathode.
The electrode on which the oxidation takes place is the anode.
Let us consider the processes occurring during the electrolysis of molten salts of anoxic acids: HCl, HBr, HI, H 2 S (with the exception of hydrofluoric or hydrofluoric acid - HF).
In the melt, such a salt consists of metal cations and acid residue anions.
For example, NaCl \u003d Na + + Cl -
At the cathode: Na + + ē \u003d Na metallic sodium is formed (in general, the metal that is part of the salt)
At the anode: 2Cl - - 2ē \u003d Cl 2 gaseous chlorine is formed (in the general case - halogen, which is part of the acid residue - except for fluorine - or sulfur)
Let's 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 concentration of the electrolyte (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 \u200b\u200bof the standard electrode potential. It is enough for students to know that in the series of electrochemical intensity of metals (series of metal activity) the value of the standard electrode potential of the Me + n / Me pair:
- increases from left to right
- metals in the row before hydrogen have a negative value of this value
- hydrogen, when reduced by reaction 2H + + 2ē \u003d H 2, (i.e. from acids) has zero standard electrode potential
- metals in the row after hydrogen have a positive value of this value
! hydrogen upon reduction by reaction:
2H 2 O + 2ē \u003d 2OH - + H 2, (i.e. from water in a neutral medium) 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 that are formed when the anode dissolves:
Me - nē \u003d Me + n
The formed 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 determined:
1.The electrolyte cation is located in the electrochemical series of voltages of metals up to and including aluminum, the process of water recovery is in progress:
2H 2 O + 2ē \u003d 2OH - + H 2
Metal cations remain in solution, in the cathode space
2. The cation of the electrolyte 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 is taking place. Since the concentration is not indicated in the task, both possible processes are recorded:
2H 2 O + 2ē \u003d 2OH - + H 2
Me + n + nē \u003d Me
3. electrolyte cation is hydrogen ions, i.e. the electrolyte is acid. Hydrogen ions are reduced:
2H + + 2ē \u003d H 2
4. the electrolyte cation is located after hydrogen, metal cations are reduced.
Me + n + nē \u003d Me
The process at the anode depends on the material of the anode and the nature of the anion.
1. If the anode dissolves (for example, iron, zinc, copper, silver), then the anode metal is oxidized.
Me - nē \u003d Me + n
2. If the anode is inert, i.e. does not dissolve (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
S 2 - - 2ē \u003d S
b) During the electrolysis of alkali solutions, the process of oxidation of the OH - hydroxo group takes place:
4OH - - 4ē \u003d 2H 2 O + O 2
c) In 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, the process of water oxidation takes place.
2H 2 O - 4ē \u003d 4H + + O 2
d) In the electrolysis of acetates (salts of acetic or ethanic acid), the acetate ion is oxidized to ethane and carbon monoxide (IV) - carbon dioxide.
2СН 3 СОО - - 2ē \u003d C 2 H 6 + 2CO 2
Examples of tasks.
1. Establish a correspondence between the salt formula and the product formed on the inert anode during the electrolysis of its aqueous solution.
SALT FORMULA
A) NiSO 4
B) NaClO 4
B) LiCl
D) RbBr
PRODUCT ON ANODA
1) S 2) SO 2 3) Cl 2 4) O 2 5) H 2 6) Br 2
Decision:
Since the task specifies an inert anode, we consider only the changes that occur with acid residues formed during the dissociation of salts:
SO 4 2 - acid residue of oxygen-containing acid. The process of water oxidation is in progress, oxygen is released. Answer 4
ClO 4 - acid residue of oxygen-containing acid. The process of water oxidation is in progress, oxygen is released. Answer 4.
Cl - acid residue of anoxic acid. The process of oxidation of the acid residue itself is in progress. Chlorine is released. Answer 3.
Br - acid residue of anoxic acid. The process of oxidation of the acid residue itself is in progress. Bromine is released. Answer 6.
General answer: 4436
2. Establish a correspondence between the salt formula and the product formed at 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 ANODA
1) hydrogen 2) aluminum 3) mercury 4) copper 5) oxygen 6) sodium
Decision:
Since the task specifies the cathode, we consider only the changes that occur with the 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 recovery will take place. Hydrogen is evolved. Answer 1.
Hg 2+ in accordance with the position of mercury (after hydrogen), the process of reduction of mercury ions will take place. Mercury is formed. Answer 3.
Cu 2+ in accordance with the position of copper (after hydrogen), the process of reduction of copper ions will take place. Answer 4.
Na + in accordance with the position of sodium (from the beginning of the row to aluminum inclusive), the process of water recovery will take place. Answer 1.
General answer: 1341
Topic 6. "Electrolysis of solutions and molten salts"
1. Electrolysis is a redox process that occurs on the electrodes when an electric current is passed through a solution or an electrolyte melt.
2. The cathode is a negatively charged electrode. The reduction of metal and hydrogen cations (in acids) or water molecules occurs.
3. The anode is a positively charged electrode. Anions of the acid residue and the hydroxo group (in alkalis) are oxidized.
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. Distinguish between 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 the stress range:
Position of metal in a series of stresses
Recovery at the cathode
from Li to Al
Water molecules are reduced: 2H2O + 2e- → H20 + 2OH-
from Mn to Pb
Both water molecules and metal cations are reduced:
2H2O + 2e- → H20 + 2OH-
Men + + ne- → Me0
from Cu to Au
Metal cations are reduced: Men + + ne- → Me0
ANODE PROCESSES
Acidic residue
Asm-
Anode
Soluble
(iron, zinc, copper, silver)
Insoluble
(graphite, gold, platinum)
Oxygen free
Anode metal oxidation
М0 - n- \u003d Mn +
anode solution
Anion oxidation (except F-)
Acm- - me- \u003d Ac0
Oxygenated
Fluoride - ion (F-)
In acidic and neutral environments:
2 H2O - 4e- → О20 + 4H +
In an alkaline environment:
4OH- - 4- \u003d O20 + 2H2O
Examples of electrolysis processes of melts with inert electrodes
The electrolyte melt contains only its ions; therefore, electrolyte cations are reduced at the cathode, and anions are oxidized at the anode.
1. Consider the electrolysis of potassium chloride melt.
Thermal dissociation КСl → K + + Cl-
K (-) K + + 1e- → K0
А (+) 2Сl- - 2e- → Cl02
Cumulative equation:
2KSl → 2K0 + Cl20
2. Consider the electrolysis of a calcium chloride melt.
Thermal dissociation CaCl2 → Ca2 + + 2Cl-
К (-) Ca2 + + 2e- → Ca0
А (+) 2Сl- - 2e- → Cl02
Cumulative equation:
CaCl2 → Ca0 + Cl20
3. Consider the electrolysis of potassium hydroxide melt.
Thermal dissociation of KOH → K + + OH-
K (-) K + + 1e- → K0
А (+) 4ОН- - 4e- → О20 + 2Н2О
Cumulative equation:
4KON → 4K0 + O20 + 2H2O
Examples of electrolysis processes of electrolyte solutions with inert electrodes
In contrast to melts, the electrolyte solution contains, in addition to its ions, water molecules. Therefore, when considering the processes on the electrodes, it is necessary to take into account their participation. The electrolysis of a solution of a salt formed by an active metal in the 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 in the stress range up to aluminum and an oxygen-containing acid residue. Dissociation equation: MgSO4 → Mg2 + + SO42- K (-) 2H2O + 2e \u003d H20 + 2OH- A \u200b\u200b(+) 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- \u003d Cu0 A (+) 2H2O - 4- \u003d O20 + 4H + Total equation: 2Cu2 + + 2H2O \u003d 2Cu0 + O20 + 4H + 2CuSO4 + 2H2O \u003d 2Cu0 + O20 + 2H2SO4
3. Consider the electrolysis of an aqueous solution of calcium chloride. CaCl2 is a salt formed by an active metal and an anoxic acid residue. In this case, during electrolysis, hydrogen, halogen are formed, and alkali is formed in the cathode - anode space. Dissociation equation: CaCl2 → Ca2 + + 2Cl- К (-) 2Н2О + 2e \u003d Н20 + 2OH- А (+) 2Сl- - 2e \u003d Cl20 Total equation: 2Н2О + 2Cl- \u003d Cl20 + 2OH- CaCl2 + 2Н2О \u003d Ca (OH) 2 + Cl20 + H20 4. Consider the electrolysis of an aqueous solution of copper (II) chloride. CuCl2 is a salt formed by a low-activity metal and an acidic residue of anoxic acid. In this case, metal and halogen are formed. Dissociation equation: CuCl2 → Cu2 + + 2Cl- К (-) Cu2 + + 2e- \u003d Cu0 А (+) 2Сl- - 2- \u003d Cl20 Total equation: Cu2 + + 2Cl- \u003d Cu0 + Cl20 CuCl2 \u003d Cu0 + Cl20 5. Consider the process electrolysis of sodium acetate solution. CH3COONa is a salt formed by an active metal and an acidic residue of a carboxylic acid. Electrolysis produces hydrogen and alkali. Dissociation equation: СН3СООNa → СН3СОО - + Na + К (-) 2Н2О + 2- \u003d Н20 + 2ОН- А (+) 2CH3COO¯− 2e \u003d C2H6 + 2CO2 Total equation: 2Н2О + 2CH3COO¯ \u003d Н20 + 2ОН - + C2H6 + 2CO2 2Н2О + 2CH3COONa \u003d 2NaОH + Н20 + C2H6 + 2CO2 6. Consider the process of electrolysis of a solution of nickel nitrate. Ni (NO3) 2 is a salt that is formed by a metal in the range 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- К (-) Ni2 + + 2e- \u003d Ni0 2Н2О + 2e- \u003d Н20 + 2ОН- A (+) 2H2O - 4e- \u003d O20 + 4H + Total equation: Ni2 + + 2Н2О + 2H2O \u003d Ni0 + Н20 + 2ОН- + O20 + 4H + Ni (NO3) 2 + 2Н2О \u003d Ni0 + 2HNO3 + Н20 + O20 7. Consider the process of electrolysis of a sulfuric acid solution. Dissociation equation: H2SO4 → 2H + + SO42- K (-) 2H + + 2e- \u003d H20 A (+) 2H2O - 4e- \u003d O20 + 4H + Total equation: 2H2O + 4H + \u003d 2H20 + O20 + 4H + 2H2O \u003d 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 in a similar way. The equation of dissociation: NaOH → Na + + OH- К (-) 2H2O + 2e- \u003d H20 + 2OH- A \u200b\u200b(+) 4OH- - 4e- \u003d O20 + 2H2O Total equation: 4H2O + 4OH- \u003d 2H20 + 4OH- + O20 + 2H2O 2H2O \u003d 2H20 + O20
Examples of electrolysis processes of electrolyte solutions with soluble electrodes
During electrolysis, a soluble anode itself undergoes oxidation (dissolution). 1. Consider the process of electrolysis of copper (II) sulfate with a copper anode. In the electrolysis of a copper sulfate solution 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 the solution remains unchanged. Dissociation equation: CuSO4 → Cu2 + + SO42- K (-) Cu2 + + 2e- → Cu0 A (+) Cu0 - 2e- → Cu2 + transition of copper ions from the anode to the cathode
Examples of assignments on this topic in the versions of the exam
AT 3. (Option 5)
Establish a correspondence between the formula of a substance and the products of electrolysis of its aqueous solution on inert electrodes.
FORMULA OF SUBSTANCE ELECTROLYSIS PRODUCTS
A) Al2 (SO4) 3 1.metal hydroxide, acid
B) СsOH 2.metal, halogen
B) Hg (NO3) 2 3.metal, oxygen
D) AuBr3 4. hydrogen, halogen 5. hydrogen, oxygen 6. metal, acid, oxygen Line of reasoning: 1. During the electrolysis of Al2 (SO4) 3 and СsOH, water is reduced to hydrogen at the cathode. We exclude options 1, 2, 3 and 6. 2. For Al2 (SO4) 3, water is oxidized to oxygen at the anode. Choose option 5. For СsOH, the hydroxide ion is oxidized to oxygen at the anode. We choose option 5. 3. During the electrolysis of Hg (NO3) 2 and AuBr3, metal cations are reduced at the cathode. 4. For Hg (NO3) 2, water is oxidized at the anode. Nitrate ions in solution bind with hydrogen cations, forming nitric acid in the anode space. Choose option 6. 5. For АuBr3, the Br- anion is oxidized to Br2 at the anode. We choose option 2.
AND
B
IN
D
5
5
6
2
AT 3. (Option 1)
Establish a correspondence between the name of the substance and the method of obtaining it
NAME OF SUBSTANCE PREPARATION 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 Line of reasoning: 1. Similar to the electrolysis of sodium chloride melt , the process of electrolysis of molten lithium fluoride is taking place. 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 recovered from a salt solution. We choose option 6 - magnesium chloride melt.
AND
B
IN
D
2
2
4
6
AT 3. (Option 9)
Establish a correspondence between the salt formula and the equation of the process occurring at the cathode during the electrolysis of its aqueous solution.
SALT FORMULA CATHODE PROCESS EQUATION
A) Al (NO3) 3 1) 2H2O - 4e- → O2 + 4H +
B) CuCl2 2) 2H2O + 2e- → H2 + 2OH-
B) SbCl3 3) Cu2 + + 1e- → Cu +
D) Cu (NO3) 2 4) Sb3 + - 2 e- → Sb5 + 5) Sb3 + + 3e- → Sb0
6) Cu2 + + 2e- → Cu0
Line of reasoning: 1. At the cathode, the processes of reduction of metal cations or water take place. Therefore, we immediately exclude options 1 and 4. 2. For Al (NO3) 3: the process of water reduction is in progress at the cathode. We select option 2. 3. For CuCl2: metal cations Cu2 + are reduced. Choose option 6. 4. For SbСl3: metal cations Sb3 + are reduced. We choose option 5. 5. For Cu (NO3) 2: metal cations Cu2 + are reduced. We choose option 6.
AND
B
IN
D
2
Establish a correspondence between the salt formula and the product formed on the inert anode during the electrolysis of its aqueous solution: for each position marked with a letter, select the corresponding position marked with a number.
| SALT FORMULA | PRODUCT ON ANODA | |
| A | B | IN | D |
Decision.
When 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, which is part of the salt, is discharged, and the corresponding simple substance is released, if it is a salt of anoxic acid (except).
The process of electrolysis of carboxylic acid salts takes place in a special way.
Answer: 3534.
Answer: 3534
Source: Yandex: USE training work in chemistry. Option 1.
Establish a correspondence between the formula of the substance and the product formed at the cathode during the electrolysis of its aqueous solution: for each position marked with a letter, select the corresponding position marked with a number.
| FORMULA OF SUBSTANCE | ELECTROLYSIS PRODUCT, FORMED ON THE CATHODE |
|
Write down the numbers in the answer, arranging them in the order corresponding to the letters:
| A | B | IN | D |
Decision.
When electrolysis of aqueous solutions of salts at the cathode is released:
Hydrogen, if it is a salt of a metal standing in the series of metal stresses to the left of aluminum;
Metal, if it is a salt of a metal standing in the series of metal voltages to the right of hydrogen
Metal and hydrogen, if it is a salt of a metal standing in the series of metal voltages between aluminum and hydrogen.
Answer: 3511.
Answer: 3511
Source: Yandex: USE training work in chemistry. Option 2.
Establish a correspondence between the salt formula and the product formed on the inert anode during the electrolysis of its aqueous solution: for each position marked with a letter, select the corresponding position marked with a number.
| SALT FORMULA | PRODUCT ON ANODA | |
Write down the numbers in the answer, arranging them in the order corresponding to the letters:
| A | B | IN | D |
Decision.
During the electrolysis of aqueous solutions of salts of oxygen-containing acids and fluorides, oxygen is oxidized from water, therefore, 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 forms on the inert anode as a result of electrolysis of an aqueous solution of this substance: for each position marked with a letter, select the corresponding position marked with a number.
| FORMULA OF SUBSTANCE | PRODUCT ON ANODA |
2) sulfur (IV) oxide 3) carbon monoxide (IV) 5) oxygen 6) nitric oxide (IV) |
Write down the numbers in the answer, arranging them in the order corresponding to the letters:
| A | B | IN | D |
What is electrolysis? For a simpler understanding of the answer to this question, let's imagine any DC source. Each DC power supply always has a positive and negative pole:
Let's connect two chemically resistant electrically conductive plates to it, which we will call electrodes. The plate connected to the positive pole is called the anode, and to the negative cathode:
Sodium chloride is an electrolyte; when it melts, it dissociates into sodium cations and chloride ions:
NaCl \u003d Na + + Cl -
It is obvious that negatively charged chlorine anions will go to the positively charged electrode - the anode, and positively charged Na + cations will go to the negatively charged electrode - the cathode. As a result, both Na + cations and Cl - anions are discharged, that is, they become neutral atoms. Discharge occurs through the acquisition of electrons in the case of Na + ions and the loss of electrons in the case of Cl - ions. That is, the process occurs at the cathode:
Na + + 1e - \u003d Na 0,
And at the anode:
Cl - - 1e - \u003d Cl
Since each chlorine atom has an unpaired electron, their single existence is disadvantageous and the chlorine atoms combine into a molecule of two chlorine atoms:
Сl ∙ + ∙ Cl \u003d Cl 2
Thus, in total, the process taking place at the anode is more correctly written as follows:
2Cl - - 2e - \u003d Cl 2
That is, we have:
Cathode: Na + + 1e - \u003d Na 0
Anode: 2Cl - - 2e - \u003d Cl 2
Let's sum up the electronic balance:
Na + + 1e - \u003d Na 0 | ∙ 2
2Cl - - 2e - \u003d Cl 2 | ∙ 1<
Let us add the left and right sides of both equations half-reactions, we get:
2Na + + 2e - + 2Cl - - 2e - \u003d 2Na 0 + Cl 2
Let us cancel two electrons in the same way as it is done in algebra, we obtain the ionic equation of electrolysis:
2NaCl (l) \u003d\u003e 2Na + Cl 2
The case considered above is the simplest from the theoretical point of view, since in the sodium chloride melt, of the positively charged ions there were only sodium ions, and of the negative ones, only chlorine anions.
In other words, neither the Na + cations, nor the Cl - anions had "competitors" for the cathode and anode.
And what will happen, for example, if instead of the sodium chloride melt, a current is passed through its aqueous solution? Dissociation of sodium chloride is observed in this case as well, but the formation of metallic sodium in an aqueous solution becomes impossible. After all, we know that sodium - a representative of alkali metals - is an extremely active metal that reacts very violently with water. If sodium is not able to be reduced under such conditions, then what will be reduced at the cathode?
Let's remember the structure of the water molecule. It is a dipole, that is, it has a negative and a positive pole:
It is due to this property that it is able to "stick" to both the cathode surface and the anode surface:
In this case, processes can occur:
2H 2 O + 2e - \u003d 2OH - + H 2
2H 2 O - 4e - \u003d O 2 + 4H +
Thus, it turns out that if we consider a solution of any electrolyte, then we will see that the cations and anions formed during the dissociation of the electrolyte compete with water molecules for reduction at the cathode and oxidation at the anode.
So what processes will take place at the cathode and at the anode? Discharge of ions formed during electrolyte dissociation or oxidation / reduction of water molecules? Or, perhaps, all these processes will take place simultaneously?
Depending on the type of electrolyte during the electrolysis of its aqueous solution, a variety of situations are possible. For example, cations of alkali, alkaline earth metals, aluminum and magnesium are simply not able to be reduced in an aqueous medium, since when they are reduced, respectively alkali, alkaline earth metals, aluminum or magnesium should be obtained, i.e. metals that react with water.
In this case, only the reduction of water molecules at the cathode is possible.
It is possible to remember what process will take place at the cathode during the electrolysis of a solution of any electrolyte, following the following principles:
1) If the electrolyte consists of a metal cation, which in a free state under normal conditions reacts with water, a process is going on at the cathode:
2H 2 O + 2e - \u003d 2OH - + H 2
This applies to metals at the beginning of the Al activity range, inclusive.
2) If the electrolyte consists of a metal cation, which does not react with water in its free form, but reacts with acids with non-oxidizing agents, there are two processes at once, both the reduction of metal cations and water molecules:
Me n + + ne \u003d Me 0
These metals include metals located between Al and H in the activity series.
3) If the electrolyte consists of hydrogen cations (acid) or metal cations that do not react with acids with non-oxidizing agents, only electrolyte cations are reduced:
2Н + + 2е - \u003d Н 2 - in the case of acid
Me n + + ne \u003d Me 0 - in the case of salt
At the anode, meanwhile, the situation is as follows:
1) If the electrolyte contains anions of anoxic acid residues (except for F -), then the process of their oxidation takes place at the anode, water molecules are not oxidized. For example:
2Сl - - 2e \u003d Cl 2
S 2- - 2e \u003d S o
Fluoride ions are not oxidized at the anode since fluorine is not able to form in aqueous solution (reacts with water)
2) If the electrolyte contains hydroxide ions (alkalis), they are oxidized instead of water molecules:
4OH - - 4e - \u003d 2H 2 O + O 2
3) If the electrolyte contains an oxygen-containing acid residue (except for organic acid residues) or fluoride ion (F -), the process of oxidation of water molecules takes place at the anode:
2H 2 O - 4e - \u003d O 2 + 4H +
4) In the case of an acidic residue of a carboxylic acid on the anode, the process is:
2RCOO - - 2e - \u003d R-R + 2CO 2
Let's practice writing down the electrolysis equations for different situations:
Example # 1
Write down the equations of the processes occurring at the cathode and anode during the electrolysis of zinc chloride melt, as well as the general equation of electrolysis.
Decision
When zinc chloride melts, its dissociation occurs:
ZnCl 2 \u003d Zn 2+ + 2Cl -
Next, you should pay attention to the fact that it is the zinc chloride melt that undergoes electrolysis, and not the aqueous solution. In other words, without options, only the reduction of zinc cations can occur at the cathode, and oxidation of chloride ions at the anode. no water molecules:
Cathode: Zn 2+ + 2e - \u003d Zn 0 | ∙ 1
Anode: 2Cl - - 2e - \u003d Cl 2 | ∙ 1
ZnCl 2 \u003d Zn + Cl 2
Example No. 2
Write the equations of the processes occurring at the cathode and anode during electrolysis of an aqueous solution of zinc chloride, as well as the general equation of electrolysis.
Since in this case, an aqueous solution is subjected to electrolysis, water molecules can theoretically take part in electrolysis. Since zinc is located in the line of activity between Al and H, this means that both the reduction of zinc cations and water molecules will occur at the cathode.
2H 2 O + 2e - \u003d 2OH - + H 2
Zn 2+ + 2e - \u003d Zn 0
Chloride ion is an acidic residue of anoxic acid HCl, therefore, in competition for oxidation at the anode, chloride ions “win” over water molecules:
2Cl - - 2e - \u003d Cl 2
In this particular case, it is impossible to write down the overall equation of electrolysis, since the ratio between the hydrogen and zinc released at the cathode is unknown.
Example No. 3
Write the equations of the processes occurring at the cathode and anode during the electrolysis of an aqueous solution of copper nitrate, as well as the general equation of electrolysis.
Copper nitrate in solution is in a dissociated state:
Cu (NO 3) 2 \u003d Cu 2+ + 2NO 3 -
Copper is in the line of activity to the right of hydrogen, that is, copper cations will be reduced at the cathode:
Cu 2+ + 2e - \u003d Cu 0
Nitrate ion NO 3 - is an oxygen-containing acid residue, which means that in oxidation at the anode nitrate ions "lose" in competition with water molecules:
2H 2 O - 4e - \u003d O 2 + 4H +
Thus:
Cathode: Cu 2+ + 2e - \u003d Cu 0 | ∙ 2
2Cu 2+ + 2H 2 O \u003d 2Cu 0 + O 2 + 4H +
The resulting equation is the ionic equation of electrolysis. To get the complete molecular equation of electrolysis, you need to add 4 nitrate ions to the left and right sides of the resulting ionic equation as counterions. Then we get:
2Cu (NO 3) 2 + 2H 2 O \u003d 2Cu 0 + O 2 + 4HNO 3
Example No. 4
Write the equations of the processes occurring at the cathode and anode during the electrolysis of an aqueous solution of potassium acetate, as well as the general equation of electrolysis.
Decision:
Potassium acetate in aqueous solution dissociates into potassium cations and acetate ions:
CH 3 COOK \u003d CH 3 COO - + K +
Potassium is an alkali metal, i.e. is in the series of electrochemical voltages at the very beginning. This means that its cations are not able to discharge at the cathode. Instead, water molecules will be restored:
2H 2 O + 2e - \u003d 2OH - + H 2
As mentioned above, acidic residues of carboxylic acids “win” in the competition for oxidation from water molecules at the anode:
2СН 3 СОО - - 2e - \u003d CH 3 −CH 3 + 2CO 2
Thus, summing up the electronic balance and adding two equations of half-reactions at the cathode and anode, we obtain:
Cathode: 2H 2 O + 2e - \u003d 2OH - + H 2 | ∙ 1
Anode: 2СН 3 СОО - - 2e - \u003d CH 3 −CH 3 + 2CO 2 | ∙ 1
2H 2 O + 2СН 3 СОО - \u003d 2OH - + Н 2 + CH 3 −CH 3 + 2CO 2
We have obtained the complete electrolysis equation in ionic form. By adding two potassium ions to the left and right sides of the equation and adding them with counterions, we get the complete electrolysis equation in molecular form:
2H 2 O + 2СН 3 СООK \u003d 2KOH + Н 2 + CH 3 −CH 3 + 2CO 2
Example No. 5
Write the equations of the processes occurring at the cathode and anode during the electrolysis of an aqueous solution of sulfuric acid, as well as the general equation of electrolysis.
Sulfuric acid dissociates into hydrogen cations and sulfate ions:
H 2 SO 4 \u003d 2H + + SO 4 2-
The reduction of hydrogen cations H + will occur at the cathode, and oxidation of water molecules at the anode, since sulfate ions are oxygen-containing acid residues:
Cathode: 2H + + 2e - \u003d H 2 | ∙ 2
Anode: 2H 2 O - 4e - \u003d O 2 + 4H + | ∙ 1
4H + + 2H 2 O \u003d 2H 2 + O 2 + 4H +
Reducing the hydrogen ions in the left and right and left sides of the equation, we obtain the equation for electrolysis of an aqueous solution of sulfuric acid:
2H 2 O \u003d 2H 2 + O 2
As you can see, the electrolysis of an aqueous solution of sulfuric acid is reduced to the electrolysis of water.
Example No. 6
Write the equations of the processes occurring at the cathode and anode during the electrolysis of an aqueous solution of sodium hydroxide, as well as the general equation of electrolysis.
Dissociation of sodium hydroxide:
NaOH \u003d Na + + OH -
Only water molecules will be reduced on the cathode, since sodium is a highly active metal, only hydroxide ions on the anode:
Cathode: 2H 2 O + 2e - \u003d 2OH - + H 2 | ∙ 2
Anode: 4OH - - 4e - \u003d O 2 + 2H 2 O | ∙ \u200b\u200b1
4H 2 O + 4OH - \u003d 4OH - + 2H 2 + O 2 + 2H 2 O
Let us reduce two water molecules on the left and right and 4 hydroxide ions and come to the conclusion that, as in the case of sulfuric acid, the electrolysis of an aqueous solution of sodium hydroxide is reduced to the electrolysis of water.
