Electrolysis (Greek elektron - amber + lysis - decomposition) - chemical reaction that occurs when a direct current is passed through the electrolyte. This is the decomposition of substances into their component parts under the influence of an electric current.
The process of electrolysis is the movement of cations (positively charged ions) to the cathode (negatively charged), and negatively charged ions (anions) to the anode (positively charged).
So, anions and cations rush to the anode and cathode, respectively. This is where the chemical reaction takes place. In order to successfully solve tasks on this topic and write reactions, it is necessary to separate the processes at the cathode and anode. This is how this article will be built.
Cathode
Cations are attracted to the cathode - positively charged ions: Na +, K +, Cu 2+, Fe 3+, Ag +, etc.
To establish what reaction takes place at the cathode, first of all, you need to determine the activity of the metal: its position in the electrochemical series of metal voltages.
If an active metal (Li, Na, K) appears on the cathode, then water molecules are restored instead of it, from which hydrogen is released. If the metal is of medium activity (Cr, Fe, Cd), both hydrogen and the metal itself are released at the cathode. Inactive metals are isolated at the cathode in pure form (Cu, Ag).
I note that aluminum is considered the boundary between active and medium activity metals in a series of voltages. During electrolysis on the cathode, metals up to aluminum (inclusive!) are not restored, instead of them, water molecules are restored - hydrogen is released.
In the event that hydrogen ions - H + are supplied to the cathode (for example, during the electrolysis of acids HCl, H 2 SO 4), hydrogen is reduced from acid molecules: 2H + - 2e = H 2
Anode
Anions are attracted to the anode - negatively charged ions: SO 4 2-, PO 4 3-, Cl -, Br -, I -, F -, S 2-, CH 3 COO -.
During the electrolysis of oxygen-containing anions: SO 4 2-, PO 4 3- - not anions are oxidized on the anode, but water molecules, from which oxygen is released.
Oxygen-free anions are oxidized and release the corresponding halogens. Sulfide ion in the oxidation of sulfur oxidation. An exception is fluorine - if it hits the anode, then a water molecule is discharged and oxygen is released. Fluorine is the most electronegative element, and therefore is an exception.
Anions of organic acids are oxidized in a special way: the radical adjacent to the carboxyl group is doubled, and the carboxyl group itself (COO) is converted into carbon dioxide - CO 2 .
Solution examples
In the process of training, you may come across metals that are omitted in the activity series. At the training stage, you can use an extended range of metal activity.
Now you will know exactly what is released on the cathode ;-)
So, let's practice. Let us find out what is formed on the cathode and anode during the electrolysis of solutions of AgCl, Cu(NO 3) 2 , AlBr 3 , NaF, FeI 2 , CH 3 COOLi.
Sometimes in tasks it is required to record the reaction of electrolysis. I inform you: if you understand what is formed at the cathode and what is at the anode, then writing the reaction is not difficult. Take, for example, the electrolysis of NaCl and write the reaction:
NaCl + H 2 O → H 2 + Cl 2 + NaOH
Sodium is an active metal, so hydrogen is released at the cathode. The anion does not contain oxygen, halogen - chlorine is released. We write the equation so we can't make the sodium evaporate without a trace :) Sodium reacts with water to form NaOH.
Let's write down the electrolysis reaction for CuSO 4:
CuSO 4 + H 2 O → Cu + O 2 + H 2 SO 4
Copper belongs to low-active metals, therefore, in its pure form, it is released at the cathode. The anion is oxygen-containing, so oxygen is released in the reaction. The sulfate ion does not disappear anywhere, it combines with the hydrogen of water and turns into sulfuric acid.
Melt electrolysis
Everything we have discussed up to this point has been about the electrolysis of solutions where the solvent is water.
Industrial chemistry faces an important task - to obtain metals (substances) in a pure form. Inactive metals (Ag, Cu) can be easily obtained by electrolysis of solutions.
But what about active metals: Na, K, Li? After all, during the electrolysis of their solutions, they are not released at the cathode in their pure form, instead of them, water molecules are reduced and hydrogen is released. This is where melts that do not contain water come in handy.
In anhydrous melts, reactions are written even more simply: substances break down into their constituent parts:
AlCl 3 → Al + Cl 2
LiBr → Li + Br2
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Establish a correspondence between the salt formula and the product formed on an inert anode during the electrolysis of its aqueous solution: for each position indicated by a letter, select the corresponding position indicated by a number.
| SALT FORMULA | PRODUCT ON ANODE | |
| A | B | AT | G |
Solution.
In the electrolysis of aqueous solutions of salts, alkalis and acids on an inert anode:
Water is discharged and oxygen is released if it is a salt of an oxygen-containing acid or a salt of hydrofluoric acid;
Hydroxide ions are discharged and oxygen is released if it is alkali;
The acid residue that is part of the salt is discharged, and the corresponding simple substance is released if it is a salt of an oxygen-free acid (except for).
The process of electrolysis of salts of carboxylic acids takes place in a special way.
Answer: 3534.
Answer: 3534
Source: Yandex: training work USE in chemistry. Option 1.
Establish a correspondence between the formula of a substance and the product formed on the cathode during the electrolysis of its aqueous solution: for each position indicated by a letter, select the corresponding position indicated by a number.
| SUBSTANCE FORMULA | ELECTROLYSIS PRODUCT, PRODUCED AT THE CATHODE |
|
Write down the numbers in response, arranging them in the order corresponding to the letters:
| A | B | AT | G |
Solution.
During the electrolysis of aqueous solutions of salts, the following is released at the cathode:
Hydrogen, if it is a salt of a metal that is in the series of metal stresses to the left of aluminum;
Metal, if it is a salt of a metal that is in the series of metal voltages to the right of hydrogen;
Metal and hydrogen, if it is a salt of a metal in the series of metal voltages between aluminum and hydrogen.
Answer: 3511.
Answer: 3511
Source: Yandex: Training USE work in chemistry. Option 2.
Establish a correspondence between the salt formula and the product formed on an inert anode during the electrolysis of its aqueous solution: for each position indicated by a letter, select the corresponding position indicated by a number.
| SALT FORMULA | PRODUCT ON ANODE | |
Write down the numbers in response, arranging them in the order corresponding to the letters:
| A | B | AT | G |
Solution.
During the electrolysis of aqueous solutions of salts of oxygen-containing acids and fluorides, oxygen is oxidized from water, so oxygen is released at the anode. During the electrolysis of aqueous solutions of anoxic acids, the acid residue is oxidized.
Answer: 4436.
Answer: 4436
Establish a correspondence between the formula of a substance and the product that is formed on an inert anode as a result of the electrolysis of an aqueous solution of this substance: for each position indicated by a letter, select the corresponding position indicated by a number.
| SUBSTANCE FORMULA | PRODUCT ON ANODE |
2) sulfur oxide (IV) 3) carbon monoxide (IV) 5) oxygen 6) nitric oxide (IV) |
Write down the numbers in response, arranging them in the order corresponding to the letters:
| A | B | AT | G |
Topic 6. "Electrolysis of solutions and salt melts"
1. Electrolysis is a redox process that occurs on electrodes when an electric current is passed through an electrolyte solution or melt.
2. Cathode - negatively charged electrode. There is a reduction of metal and hydrogen cations (in acids) or water molecules.
3. Anode - a positively charged electrode. Oxidation of the anions of the acid residue and the hydroxyl group (in alkalis) occurs.
4. During the electrolysis of a salt solution, water is present in the reaction mixture. Since water can exhibit both oxidizing and reducing properties, it is a "competitor" for both cathodic and anodic processes.
5. There are electrolysis with inert electrodes (graphite, carbon, platinum) and an active anode (soluble), as well as electrolysis of melts and electrolyte solutions.
CATHODE PROCESSES
If the metal is in a series of voltages:
The position of the metal in a series of stresses
Recovery at the cathode
from Li to Al
Water molecules are reduced: 2H2O + 2e- → H20+ 2OH-
Mn to Pb
Both water molecules and metal cations are restored:
2H2O + 2e- → H20+ 2OH-
Men+ + ne- → Me0
from Cu to Au
Metal cations are reduced: Men+ + ne- → Me0
ANODIC PROCESSES
acid residue
Asm-
Anode
Soluble
(iron, zinc, copper, silver)
Insoluble
(graphite, gold, platinum)
anoxic
Anode metal oxidation
М0 – ne- = Mn+
anode solution
Anion oxidation (except F-)
Acm- - me- = Ac0
Oxygen-containing
Fluoride - ion (F-)
In acidic and neutral environments:
2 H2O - 4e- → O20 + 4H+
In an alkaline environment:
4OH- - 4e- \u003d O20 + 2H2O
Examples of melt electrolysis processes with inert electrodes
In the electrolyte melt, only its ions are present, therefore, electrolyte cations are reduced at the cathode, and anions are oxidized at the anode.
1. Consider the electrolysis of a potassium chloride melt.
Thermal dissociation KCl → K+ + Cl-
K(-) K+ + 1e- → K0
A (+) 2Cl- - 2e- → Cl02
Summary equation:
2KCl → 2K0 + Cl20
2. Consider the electrolysis of a calcium chloride melt.
Thermal dissociation CaCl2 → Ca2+ + 2Cl-
K(-) Ca2+ + 2e- → Ca0
A (+) 2Cl- - 2e- → Cl02
Summary equation:
CaCl2 → Ca0 + Cl20
3. Consider the electrolysis of a melt of potassium hydroxide.
Thermal dissociation of KOH → K+ + OH-
K(-) K+ + 1e- → K0
A (+) 4OH- - 4e- → O20 + 2H2O
Summary equation:
4KOH → 4K0 + O20 + 2H2O
Examples of electrolysis processes of electrolyte solutions with inert electrodes
Unlike melts, in an electrolyte solution, in addition to its ions, there are water molecules. Therefore, when considering the processes on the electrodes, it is necessary to take into account their participation. The electrolysis of a salt solution formed by an active metal, standing in a series of voltages up to aluminum and an acidic residue of an oxygen-containing acid, is reduced to the electrolysis of water. 1. Consider the electrolysis of an aqueous solution of magnesium sulfate. MgSO4 is a salt that is formed by a metal standing in a series of stresses up to aluminum and an oxygen-containing acid residue. Dissociation equation: MgSO4 → Mg2+ + SO42- K (-) 2H2O + 2e- \u003d H20 + 2OH- A (+) 2H2O - 4e- \u003d O20 + 4H + Total equation: 6H2O \u003d 2H20 + 4OH- + O20 + 4H + 2H2O \u003d 2H20 + O20 2. Consider the electrolysis of an aqueous solution of copper (II) sulfate. СuSO4 is a salt formed by a low-active metal and an oxygen-containing acid residue. In this case, electrolysis produces metal, oxygen, and the corresponding acid is formed in the cathode-anode space. Dissociation equation: CuSO4 → Cu2+ + SO42- K (-) Cu2+ + 2e- = Cu0 A (+) 2H2O – 4e- = O20 + 4H+ Summary equation: 2Cu2+ + 2H2O = 2Cu0 + O20 + 4H+ 2CuSO4 + 2H2O = 2Cu0 + О20 + 2Н2SO4
3. Consider the electrolysis of an aqueous solution of calcium chloride. CaCl2 is a salt that is formed by an active metal and an oxygen-free acid residue. In this case, hydrogen, halogen are formed during electrolysis, and alkali is formed in the cathode-anode space. Dissociation equation: CaCl2 → Ca2+ + 2Cl- K (-) 2H2O + 2e- = H20 + 2OH- A (+) 2Cl- – 2e- = Cl20 Summary equation: 2H2O + 2Cl- = Cl20 + 2OH- CaCl2 + 2H2O = Ca (OH)2 + Cl20 + H20 4. Consider the electrolysis of an aqueous solution of copper (II) chloride. CuCl2 is a salt that is formed by a low-active metal and an acidic residue of an oxygen-free acid. In this case, a metal and a halogen are formed. Dissociation equation: CuCl2 → Cu2+ + 2Cl- K (-) Cu2+ + 2e- = Cu0 A (+) 2Cl- – 2e- = Cl20 Summary equation: Cu2+ + 2Cl- = Cu0 + Cl20 CuCl2 = Cu0 + Cl20 5. Consider the process electrolysis of sodium acetate solution. CH3COOHa is a salt formed by the active metal and the acidic residue of a carboxylic acid. Electrolysis produces hydrogen and alkali. Dissociation equation: CH3COONa → CH3COO - + Na+ K (-) 2H2O + 2e- = H20 + 2OH- A (+) 2CH3COO¯- 2e = C2H6 + 2CO2 Summary equation: 2H2O + 2CH3COO¯ = H20 + 2OH - + C2H6 + 2CO2 2Н2О + 2CH3COONa = 2NaОH + Н20 + C2H6 + 2CO2 6. Consider the process of electrolysis of nickel nitrate solution. Ni(NO3)2 is a salt, which is formed by a metal 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- K (-) Ni2+ +2e- = Ni0 2H2O + 2e- = H20 + 2OH- A (+) 2H2O – 4e- = O20 + 4H+ Overall equation: Ni2+ + 2H2O + 2H2O = Ni0 + H20 + 2OH- + O20 + 4H+ Ni(NO3)2 + 2H2O = Ni0 + 2HNO3 + H20 + O20 7. Consider the process of electrolysis of a sulfuric acid solution. Dissociation equation: H2SO4 → 2H+ + SO42- K (-) 2H+ + 2e- = H20 A (+) 2H2O – 4e- = O20 + 4H+ Overall equation: 2H2O + 4H+ = 2H20 + O20 + 4H+ 2H2O = 2H20 + O20
8. Consider the process of electrolysis of sodium hydroxide solution. In this case, only water electrolysis takes place. The electrolysis of solutions of H2SO4, NaNO3, K2SO4, etc. proceeds similarly. Dissociation equation: NaOH → Na+ + OH- K (-) 2H2O + 2e- = H20 + 2OH- A (+) 4OH- – 4e- = O20 + 2H2O Overall equation: 4H2O + 4OH- = 2H20 + 4OH- + O20 + 2H2O 2H2O = 2H20 + O20
Examples of electrolysis processes of electrolyte solutions with soluble electrodes
The soluble anode undergoes oxidation (dissolution) during electrolysis. 1. Consider the process of electrolysis of copper sulfate (II) with a copper anode. During the electrolysis of a solution of copper sulfate with a copper anode, the process is reduced to the release of copper at the cathode and the gradual dissolution of the anode, despite the nature of the anion. The amount of copper sulfate in solution remains unchanged. Dissociation equation: CuSO4 → Cu2+ + SO42- K (-) Cu2+ +2e- → Cu0 A (+) Cu0 - 2e- → Cu2+ transition of copper ions from anode to cathode
Examples of tasks on this topic in the USE options
AT 3. (Var.5)
Establish a correspondence between the formula of a substance and the products of electrolysis of its aqueous solution on inert electrodes.
FORMULA OF THE SUBSTANCE ELECTROLYSIS PRODUCTS
A) Al2(SO4)3 1. metal hydroxide, acid
B) СsOH 2. metal, halogen
C) Hg(NO3)2 3. metal, oxygen
D) AuBr3 4. hydrogen, halogen 5. hydrogen, oxygen 6. metal, acid, oxygen Argument: 1. During the electrolysis of Al2(SO4)3 and CsOH on the cathode, water is reduced to hydrogen. We exclude options 1, 2, 3 and 6. 2. For Al2(SO4)3, water is oxidized to oxygen at the anode. We choose option 5. For CsOH, the hydroxide ion is oxidized to oxygen at the anode. We choose option 5. 3. During the electrolysis of Hg(NO3)2 and АuBr3 on the cathode, metal cations are reduced. 4. For Hg(NO3)2, water is oxidized at the anode. Nitrate ions in solution bind with hydrogen cations, forming nitric acid in the anode space. We choose option 6. 5. For АuBr3, the Br- anion is oxidized at the anode to Br2. We choose option 2.
BUT
B
AT
G
5
5
6
2
AT 3. (Var.1)
Establish a correspondence between the name of the substance and the method of obtaining it.
NAME OF THE SUBSTANCE PRODUCTION BY ELECTROLYSIS A) lithium 1) LiF solution B) fluorine 2) LiF melt C) silver 3) MgCl2 solution D) magnesium 4) AgNO3 solution 5) Ag2O melt 6) MgCl2 melt Argument: 1. Similar to the electrolysis of sodium chloride melt , the process of electrolysis of the lithium fluoride melt proceeds. For options A and B, choose answers 2. 2. Silver can be restored from a solution of its salt - silver nitrate. 3. Magnesium cannot be restored from a salt solution. We choose option 6 - a melt of magnesium chloride.
BUT
B
AT
G
2
2
4
6
AT 3. (Var.9)
Establish a correspondence between the salt formula and the equation of the process occurring on the cathode during the electrolysis of its aqueous solution.
SALT FORMULA EQUATION OF THE CATHODE PROCESS
A) Al(NO3)3 1) 2H2O – 4e- → O2 + 4H+
B) CuCl2 2) 2H2O + 2e- → H2 + 2OH-
C) SbCl3 3) Cu2+ + 1e- → Cu+
D) Cu(NO3)2 4) Sb3+ - 2 e- → Sb5+ 5) Sb3+ + 3e- → Sb0
6) Cu2+ + 2e- → Cu0
The course of reasoning: 1. Processes of reduction of metal or water cations take place on the cathode. Therefore, we immediately exclude options 1 and 4. 2. For Al(NO3)3: the process of water reduction is going on at the cathode. Choose option 2. 3. For CuCl2: Cu2+ metal cations are reduced. Choose option 6. 4. For SbCl3: Sb3+ metal cations are reduced. Select option 5. 5. For Cu(NO3)2: Cu2+ metal cations are reduced. We choose option 6.
BUT
B
AT
G
2
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:
- increases from left to right
- metals in the row up to hydrogen have a negative value of this quantity
- hydrogen, when reduced by the reaction 2H + + 2ē \u003d H 2, (i.e. from acids) has a value of zero standard electrode potential
- 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
