Electrolysis of melts and solutions (salts, alkalis)
If the electrodes are lowered into the electrolyte solution or melt and a direct electric current is passed through, then the ions will move in a direction: cations to the cathode (negatively charged electrode), anions to the anode (positively charged electrode).
At the cathode, cations accept electrons and are reduced; at the anode, anions donate electrons and are oxidized. 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.
Electrolysis of molten salts
Consider the process of electrolysis of a 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 towards the cathode and receive electrons from it:
$Na^(+)+ē→(Na)↖(0)$ (restoration).
Anions $Cl^(-)$ move towards the anode and donate electrons:
$2Cl^(-)-2ē→(Cl_2)↖(0)$ (oxidation).
The total equation of processes:
$Na^(+)+ē→(Na)↖(0)|2$
$2Cl^(-)-2ē→(Cl_2)↖(0)|1$
$2Na^(+)+2Cl^(-)=2(Na)↖(0)+(Cl_2)↖(0)$
$2NaCl(→)↖(\text"electrolysis")2Na+Cl_2$
Sodium metal is formed at the cathode, and chlorine gas is formed at the anode.
The main thing to remember is that in the process of electrolysis, a chemical reaction is carried out due to electrical energy, which cannot go on spontaneously.
Electrolysis of aqueous solutions of electrolytes
A more complicated case is the electrolysis of electrolyte solutions.
In a salt solution, in addition to metal ions and an acidic residue, there are water molecules. Therefore, when considering processes on 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. Process at the cathode does not depend on the material from which the cathode is made, but on the position of the metal (electrolyte cation) in the electrochemical series of voltages, and if:
1.1. The electrolyte cation is located in the voltage series at the beginning of the series in $Al$ inclusive, then the process of water reduction is going on at the cathode (hydrogen $H_2$ is released). 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 a 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 is first restored, standing in the series of voltages to the right.
cathodic processes
2. process at the anode depends on the material of the anode and on the nature of the anion.
Anode processes
2.1. If anode dissolves(iron, zinc, copper, silver and all metals that are oxidized during electrolysis), then the anode metal is oxidized, regardless of the nature of the anion.
2.2. If the anode does not dissolve(it is called inert - graphite, gold, platinum), then:
a) during the electrolysis of salt solutions anoxic acids (except for fluorides) the 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 ($O_2$ is released). Anions are not oxidized, they remain in solution;
c) anions according to their ability to be oxidized are arranged in the following order:
Let's try to apply these rules in specific situations.
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).
In solution, the process of electrolytic dissociation takes place:
Summary Equation:
$2H_2O+2Cl^(-)=H_2+Cl_2+2OH^(-)$.
Taking into account the presence of $Na^(+)$ ions in the solution, we compose the molecular equation:
2) Anode soluble(for example, copper):
$NaCl=Na^(+)+Cl^(-)$.
If the anode is soluble, then the anode metal will oxidize:
$Cu^(0)-2ē=Cu^(2+)$.
The $Cu^(2+)$ cations come after ($H^(+)$) in the voltage series, so 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 sulfate (II) on insoluble anode:
$Cu^(2+)+2ē=Cu^(0)|2$
$2H_2O-4ē=O_2+4H^(+)|1$
Total ionic equation:
$2Cu^(2+)+2H_2O=2Cu^(0)+O_2+4H^(+)$
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:
$2H_2O+2ē=H_2+2OH^(-)|2$
$4OH^(-)-4ē=O_2+2H_2O|1$
Total ionic equation:
$4H_2O+4OH^(-)=2H_2+4OH^(-)+O_2+2H_2O$
Overall molecular equation:
$2H_2O(→)↖(\text"electrolysis")2H_2+O_2$
In this case, it turns out that only the electrolysis of water takes place. A similar result will be obtained 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:
- To obtain 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 the electrochemical method).
- To protect metals from corrosion (chromium, nickel, copper, silver, gold) — electroplating.
- To obtain metal copies, records - electrotype.
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The USE results show that tasks on the topic “Electrolysis” remain difficult for graduates. In the school curriculum, an insufficient number of hours are allotted for the study of this topic. Therefore, when preparing students for the exam, it is necessary to study this issue in great detail. Knowledge of the basics of electrochemistry will help the graduate to successfully pass the exam and continue their education in a higher educational institution. To study the topic “Electrolysis” at a sufficient level, it is necessary to carry out preparatory work with graduates passing the exam: - consider the definitions of the basic concepts in the topic “Electrolysis”; - analyze the process of electrolysis melts and solutions of electrolytes; - fix the rules for the reduction of cations at the cathode and the oxidation of anions at the anode (the role of water molecules during the electrolysis of solutions); - the formation of skills to draw up equations for the electrolysis process (cathode and anode processes); - teach students to perform standard tasks of the basic level ( tasks), high and high level of complexity. Electrolysis- redox process occurring in solutions and melts of electrolytes with the passage of a direct electric current. In a solution or melt of an electrolyte, it dissociates into ions. When the electric current is turned on, the ions acquire a directed motion, and redox processes can occur on the surface of the electrodes. Anode- a positive electrode, oxidation processes are taking place on it.
The cathode is a negative electrode, recovery processes are taking place on it.
Melt electrolysis used to obtain active metals located in a series of voltages up to aluminum (inclusive).
Electrolysis of sodium chloride melt
K(-) Na + + 1e -> Na 0
A(+) 2Cl - - 2e -> Cl 2 0
2NaCl (electronic current) -> 2Na + Cl 2 (only for melt electrolysis).
Aluminum is obtained by electrolysis of a solution of aluminum oxide in molten cryolite (Na 3 AlF 6).
2Al 2 O 3 (electronic current) -> 4Al + 3O 2
K(-)Al 3+ +3e‾ ->Al
A(+)2O 2‾ -2e‾ ->O 2
Electrolysis of a melt of potassium hydroxide.
KOH->K + +OH‾
K(-) K + + 1e -> K 0
A (+) 4OH - - 4e -> O 2 0 + 2H 2 O
4KOH (electric current) -> 4K 0 + O 2 0 + 2H 2 O
The electrolysis of aqueous solutions is more difficult, since water molecules can be reduced or oxidized on the electrodes in this case.
Electrolysis of aqueous solutions of salts is more complicated due to the possible participation of water molecules at the cathode and at the anode in the electrode processes.
Rules of electrolysis in aqueous solutions.
On the cathode:
1. Cations located in a series of metal voltages from lithium to aluminum (inclusive), as well as cations NH 4 + are not restored, water molecules are restored instead:
2H 2 O + 2e->H 2 + 2OH -
2. Cations located in the series of voltages after aluminum to hydrogen can be reduced together with water molecules:
2H 2 O + 2e->H 2 + 2OH -
Zn2+ + 2e->Zn 0
3. Cations located in a series of voltages after hydrogen are completely restored: Ag + + 1e->Ag 0
4. Hydrogen ions are reduced in acid solutions: 2H + + 2e->H 2
On the anode:
1. Oxygen-containing anions and F-- do not oxidize, instead of them, water molecules are oxidized:
2H 2 O - 4e->O 2 + 4H +
2.Anions of sulfur, iodine, bromine, chlorine (in this sequence) are oxidized to simple substances:
2Cl - - 2e->Cl 2 0 S 2- - 2e->S0
3. Hydroxide ions are oxidized in alkali solutions:
4OH - - 4e->O 2 + 2H 2 O
4. Anions are oxidized in solutions of salts of carboxylic acids:
2 R - SOO - - 2e->R - R + 2CO 2
5. When using soluble anodes, the anode itself sends electrons to the external circuit due to the oxidation of the atoms of the metal from which the anode is made:
Cu 0 - 2e->Сu 2+
Examples of electrolysis processes in aqueous electrolyte solutions
Example 1 K 2 SO 4 -> 2K + + SO 4 2-
K(-)2H 2 O + 2e‾ -> H 2 + 2OH -
A(+)2H 2 O – 4e‾ -> O 2 + 4H +
The general equation of electrolysis: 2H 2 O (el. current) -> 2 H 2 + O 2
Example 2. NaCl ->Na + +Cl‾
K(-)2H 2 O + 2e‾ -> H 2 + 2OH -
A(+) 2Cl - - 2e -> Cl 2 0
2NaCl + 2H 2 O (el. current) -> H 2 + 2NaOH + Cl 2
Example 3. Cu SO 4 -> Cu 2+ + SO 4 2-
K(-) Cu 2+ + 2e‾ -> Cu
A(+)2H 2 O – 4e‾ -> O 2 + 4H +
General electrolysis equation: 2 Cu SO 4 + 2H 2 O (el. current) -> 2Cu + O 2 + 2H 2 SO 4
Example 4. CH 3 COONa->CH 3 COO‾ +Na +
K(-)2H 2 O + 2e‾ -> H 2 + 2OH -
A(+)2CH 3 COO‾– 2e‾ ->C 2 H 6 +2CO 2
General electrolysis equation:
CH 3 COONa + 2H 2 O (el.current) -> H 2 + 2NaHCO 3 + C 2 H 6
Tasks of the basic level of complexity
Test on the topic “Electrolysis of melts and solutions of salts. A series of stresses of metals”.
1. Alkali is one of the products of electrolysis in an aqueous solution:
1) KCI 2) CuSO 4 3) FeCI 2 4) AgNO 3
2. During the electrolysis of an aqueous solution of potassium nitrate, the following is released at the anode: 1) About 2 2) NO 2 3) N 2 4) H 23. Hydrogen is formed during the electrolysis of an 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 (solution) 2) Zn and KCI (solution) 3) Mg and SnCI 2(solution) 4) Ag and CuSO 4 (solution) 5. During the electrolysis of a solution of sodium iodide at the cathode, the color of litmus in solution: 1) red 2 ) blue 3) purple 4) yellow6. During the electrolysis of an aqueous solution of potassium fluoride, the following is released at the cathode: 1) hydrogen 2) hydrogen fluoride 3) fluorine 4) oxygen
Tasks on the topic “Electrolysis”
1. The electrolysis of 400 g of a 20% common salt solution was stopped when 11.2 liters (n.o.) of gas were released at the cathode. The degree of decomposition of the original salt (in%) is:
1) 73 2) 54,8 3) 36,8 4) 18
The solution of the problem. We compose the electrolysis reaction equation: 2NaCl + 2H 2 O → H 2 + Cl 2 + 2NaOHm (NaCl) \u003d 400 ∙ 0.2 \u003d 80 g of salt was in solution. ν (H 2) \u003d 11.2 / 22.4 \u003d 0 .5 mol ν(NaCl)=0.5∙2=1 mol(NaCl)= 1∙58.5=58.5 g of salt was decomposed during electrolysis. Degree of salt decomposition 58.5/80=0.73 or 73%.Answer: 73% of the salt has decomposed.
2. Conducted electrolysis of 200 g of a 10% solution of chromium (III) sulfate until the salt is completely consumed (metal is released on the cathode). The mass (in grams) of water used is:
1) 0,92 2) 1,38 3) 2,76 4) 5,52
The solution of the problem. We compose 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 20g ν (Cr 2 (SO 4) 3) \u003d 20 / 392 \u003d 0.051 mol ν (H 2 O) \u003d 0.051 ∙ 3 \u003d 0.153 mol (H 2 O) \u003d 0.153 18 \u003d 2.76 gTasks of an increased level of complexity B3
1. Establish a correspondence between the salt formula and the equation of the process occurring at the anode during the electrolysis of its aqueous solution.
3. Establish a correspondence between the salt formula and the equation of the process occurring on the cathode during the electrolysis of its aqueous solution.
5. Establish a 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. Thus, studying the topic of electrolysis, graduates master this section well and show good results in the exam. The study of the material is accompanied by a presentation on this topic.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 instance, 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
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 | V | 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: USE training work 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 | V | 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: USE training 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 | V | 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 | V | G |
What is electrolysis? For a simpler understanding of the answer to this question, let's imagine any source of direct current. For every DC source, you can always find a positive and a negative pole:
Let us connect to it two chemically resistant electrically conductive plates, which we will call electrodes. The plate connected to the positive pole is called the anode, and to the negative pole is called the 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 the negatively charged chlorine anions will go to the positively charged electrode - the anode, and the positively charged Na + cations will go to the negatively charged electrode - the cathode. As a result of this, both Na + cations and Cl - anions will be discharged, that is, they will become neutral atoms. The 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 proceeds at the cathode:
Na + + 1e − = Na 0 ,
And on the anode:
Cl − − 1e − = Cl
Since each chlorine atom has an unpaired electron, their single existence is unfavorable and the chlorine atoms combine into a molecule of two chlorine atoms:
Сl∙ + ∙Cl \u003d Cl 2
Thus, in total, the process occurring at the anode is more correctly written as follows:
2Cl - - 2e - = Cl 2
That is, we have:
Cathode: Na + + 1e − = Na 0
Anode: 2Cl - - 2e - = Cl 2
Let's sum up the electronic balance:
Na + + 1e − = Na 0 |∙2
2Cl − − 2e − = Cl 2 |∙1<
Add the left and right sides of both equations half reactions, we get:
2Na + + 2e − + 2Cl − − 2e − = 2Na 0 + Cl 2
We reduce two electrons in the same way as it is done in algebra, we get the ionic equation of electrolysis:
2NaCl (l.) => 2Na + Cl 2
From a theoretical point of view, the case considered above is the simplest, since in the sodium chloride melt, among the positively charged ions, there were only sodium ions, and among the negative ones, only chlorine anions.
In other words, neither Na + cations nor Cl − anions had "competitors" for the cathode and anode.
And what will happen, for example, if instead of a melt of sodium chloride, a current is passed through its aqueous solution? Dissociation of sodium chloride is also observed in this case, 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 cannot 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 around” both the cathode surface and the anode surface:
The following processes may take place:
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, 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 the dissociation of the electrolyte or oxidation / reduction of water molecules? Or, perhaps, all of these processes will occur simultaneously?
Depending on the type of electrolyte, a variety of situations are possible during the electrolysis of its aqueous solution. For example, cations of alkali, alkaline earth metals, aluminum and magnesium are simply not able to be reduced in the aquatic environment, since their reduction should have resulted in respectively alkali, alkaline earth metals, aluminum or magnesium, 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 on 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, the following process takes place on the cathode:
2H 2 O + 2e - \u003d 2OH - + H 2
This applies to metals that are at the beginning of the Al activity series, inclusive.
2) If the electrolyte consists of a metal cation, which in its free form does not react with water, but reacts with non-oxidizing acids, two processes take place at once, both the reduction of metal cations and water molecules:
Me n+ + ne = Me 0
These metals include those 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 non-oxidizing acids, only electrolyte cations are restored:
2H + + 2e - \u003d H 2 - in the case of acid
Me n + + ne = Me 0 - in the case of salt
At the anode, meanwhile, the situation is as follows:
1) If the electrolyte contains anions of oxygen-free acid residues (except F -), then the process of their oxidation takes place at the anode, water molecules are not oxidized. For instance:
2Cl - - 2e \u003d Cl 2
S 2- − 2e = S o
Fluoride ions are not oxidized at the anode because fluorine is not able to form in an 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 a fluoride ion (F -) on the anode, the process of oxidizing water molecules takes place:
2H 2 O - 4e - \u003d O 2 + 4H +
4) In the case of an acidic residue of a carboxylic acid on the anode, the following process takes place:
2RCOO - - 2e - \u003d R-R + 2CO 2
Let's practice writing electrolysis equations for various situations:
Example #1
Write the equations for the processes occurring at the cathode and anode during the electrolysis of a zinc chloride melt, as well as the general electrolysis equation.
Solution
When zinc chloride is melted, it dissociates:
ZnCl 2 \u003d Zn 2+ + 2Cl -
Further, attention should be paid 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 the oxidation of chloride ions at the anode. no water molecules
Cathode: Zn 2+ + 2e − = Zn 0 |∙1
Anode: 2Cl − − 2e − = Cl 2 |∙1
ZnCl 2 \u003d Zn + Cl 2
Example #2
Write the equations for the processes occurring at the cathode and anode during the electrolysis of an aqueous solution of zinc chloride, as well as the general electrolysis equation.
Since in this case, an aqueous solution is subjected to electrolysis, then, theoretically, water molecules can take part in electrolysis. Since zinc is located in the activity series 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 − = Zn 0
The chloride ion is the acidic residue of the oxygen-free acid HCl, therefore, in the competition for oxidation at the anode, chloride ions “win” over water molecules:
2Cl - - 2e - = Cl 2
In this particular case, it is impossible to write the overall electrolysis equation, since the ratio between hydrogen and zinc released at the cathode is unknown.
Example #3
Write the equations for the processes occurring at the cathode and anode during the electrolysis of an aqueous solution of copper nitrate, as well as the general electrolysis equation.
Copper nitrate in solution is in a dissociated state:
Cu(NO 3) 2 \u003d Cu 2+ + 2NO 3 -
Copper is in the activity series to the right of hydrogen, that is, copper cations will be reduced at the cathode:
Cu 2+ + 2e − = 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 +
In this way:
Cathode: Cu 2+ + 2e − = Cu 0 |∙2
2Cu 2+ + 2H 2 O = 2Cu 0 + O 2 + 4H +
The equation obtained as a result of addition is the ionic equation of electrolysis. To get the complete molecular electrolysis equation, you need to add 4 nitrate ions to the left and right sides of the resulting ionic equation as counterions. Then we will get:
2Cu(NO 3) 2 + 2H 2 O = 2Cu 0 + O 2 + 4HNO 3
Example #4
Write the equations for the processes occurring at the cathode and anode during the electrolysis of an aqueous solution of potassium acetate, as well as the general electrolysis equation.
Solution:
Potassium acetate in an 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 electrochemical series of voltages at the very beginning. This means that its cations are not capable of being discharged at the cathode. Instead, water molecules will be restored:
2H 2 O + 2e - \u003d 2OH - + H 2
As mentioned above, the acid residues of carboxylic acids “win” in the competition for oxidation from water molecules at the anode:
2CH 3 COO - - 2e - \u003d CH 3 -CH 3 + 2CO 2
Thus, summing up the electronic balance and adding the two equations of half-reactions at the cathode and anode, we obtain:
Cathode: 2H 2 O + 2e − = 2OH − + H 2 |∙1
Anode: 2CH 3 COO - - 2e - \u003d CH 3 -CH 3 + 2CO 2 | ∙ 1
2H 2 O + 2CH 3 COO - \u003d 2OH - + H 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 + 2CH 3 COOK \u003d 2KOH + H 2 + CH 3 -CH 3 + 2CO 2
Example #5
Write the equations for the processes occurring at the cathode and anode during the electrolysis of an aqueous solution of sulfuric acid, as well as the general electrolysis equation.
Sulfuric acid dissociates into hydrogen cations and sulfate ions:
H 2 SO 4 \u003d 2H + + SO 4 2-
Hydrogen cations H + will be reduced at the cathode, and water molecules will be oxidized at the anode, since sulfate ions are oxygen-containing acid residues:
Cathode: 2Н + + 2e − = H 2 |∙2
Anode: 2H 2 O - 4e - = 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 the electrolysis of an aqueous solution of sulfuric acid:
2H 2 O \u003d 2H 2 + O 2
As can be seen, the electrolysis of an aqueous solution of sulfuric acid is reduced to the electrolysis of water.
Example #6
Write the equations for the processes occurring at the cathode and anode during the electrolysis of an aqueous solution of sodium hydroxide, as well as the general electrolysis equation.
Dissociation of sodium hydroxide:
NaOH = Na + + OH -
Only water molecules will be reduced at the cathode, since sodium is a highly active metal, and only hydroxide ions at the anode:
Cathode: 2H 2 O + 2e − = 2OH − + H 2 |∙2
Anode: 4OH − − 4e − = O 2 + 2H 2 O |∙1
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 we 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.
