Tasks on the topic of electrolysis with the solution. Preparation of graduates to the exam. "Theme of electrolysis in the exam." Electrolysis of molten salts

What is electrolysis? For a simpler understanding of the answer to this question, let's imagine any source of direct current. Each DC source can always find a positive and negative pole:

Connect two chemically resistant electrically conductive plates that call electrodes. The plate attached to the positive pole is called an anode, and to a negative cathode:

Sodium chloride is an electrolyte, during its melting, dissociation on sodium cations and chloride ions:

NaCl \u003d Na + + Cl -

It is obvious that the charged negative chlorine anions will go to a positively charged electrode - anode, and positively charged Na + cations will go to a negatively charged electrode - cathode. As a result, Na + cations and CL anions are discharged, that is, they will become neutral atoms. The discharge occurs by the purchase of electrons in the case of Na + ions and the loss of electrons in the case of CL ions. That is, the process proceeds on the cathode:

Na + + 1e - \u003d na 0,

And on the anode:

Cl - - 1e - \u003d Cl

Since each chlorine atom has an unprofitable electron, a single existence of them is unprofitable and chlorine atoms are combined into a molecule of two chlorine atoms:

CL ∙ + ∙ CL \u003d CL 2

Thus, the total, the process flowing on the anode is more correctly recorded:

2CL - - 2E - \u003d CL 2

That is, we have:

Cathode: Na + + 1e - \u003d Na 0

Anode: 2CL - - 2E - \u003d Cl 2

Let's submit an electronic balance:

Na + + 1e - \u003d Na 0 | ∙ 2

2CL - - 2E - \u003d Cl 2 | ∙ 1<

Moving the left and right parts of both equations sereactionWe will get:

2NA + + 2E - + 2Cl - - 2E - \u003d 2NA 0 + CL 2

We will reduce the two electrons in the same way how this is done in the algebra we obtain the ionic electrolysis equation:

2nacl (g.) \u003d\u003e 2NA + CL 2

The case considered above is the most simple case from a theoretical point of view, since the sodium chloride melt from positively charged ions was only sodium ions, and from negative - only chlorine anions.

In other words, none of the Na + cations, nor at the anions CL - there was no "competitors" for the cathode and anode.

And what will happen, for example, if instead of the melt of sodium chloride, skip through its aqueous solution? Sodium chloride dissociation is observed in this case, but it becomes impossible for the formation of metal sodium in an aqueous solution. After all, we know that sodium is a representative of alkali metals - an extremely active metal that reacts with water is very violently. If sodium is not able to recover in such conditions, what then will be restored at the cathode?

Let's remember the structure of the water molecule. She is a dipole, that is, she has a negative and positive poles:

It is due to this property, it is capable of "ripping" both the surface of the cathode and the surface of the anode:

In this case, processes may 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 the solution of any electrolyte, we will see that the cations and anions formed during electrolyte dissociation compete with water molecules for recovery on the cathode and oxidation on the anode.

So what processes will happen on the cathode and on the anode? Discharge of ions formed during electrolyte dissociation or oxidation / restoration of water molecules? Or maybe all the indicated processes will happen at the same time?

Depending on the type of electrolyte, a variety of situations are possible during electrolysis of its aqueous solution. For example, cations of alkaline, alkaline earth metals, aluminum and magnesium are simply not able to recover in an aqueous medium, since it should be obtained, alkaline, alkaline earth metals, aluminum or magnesium, alkaline earth metals, aluminum or magnesium, ie. Water reacting metals.

In this case, only the restoration of water molecules on the cathode is possible.

Remember what the process will flow at the cathode at the electrolysis of the solution of any electrolyte can be followed by following the following principles:

1) If the electrolyte consists of a metal cation, which in the free state in normal conditions reacts with water, the cathode is processed:

2H 2 O + 2E - \u003d 2OH - + H 2

This applies to the metals at the beginning of a number of ALC activity inclusive.

2) If the electrolyte consists of a metal cation that does not react in free form with water, but reacts with non-acidic acids, there are two processes as the restoration of metal cations and water molecules:

ME N + + NE \u003d ME 0

Such metals include metals that are between Al and N in a number of activity.

3) If the electrolyte consists of hydrogen cations (acid) or metal cations that do not react with non-acid acids - only electrolyte cations are restored:

2N + + 2e - \u003d H 2 - in the case of acid

ME N + + NE \u003d ME 0 - in case of salt

On the anode in the meantime, the situation is as follows:

1) If the electrolyte contains anions of oxygen-free acid residues (except f -), then the anode is the process of their oxidation, the water molecules are not oxidized. For example:

2SL - - 2E \u003d Cl 2

S 2- - 2E \u003d S o

Fluoride ions are not oxidized on the anode since the fluorine is not able to form in an aqueous solution (reacts with water)

2) If the electrolyte includes hydroxide ions (alkali), they are oxidized instead of water molecules:

4on - - 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 -) on the anode there is a process of oxidation of water molecules:

2H 2 O - 4E - \u003d O 2 + 4H +

4) in the case of the acid residue of carboxylic acid on the anode there is a process:

2RCOO - - 2E - \u003d R-R + 2CO 2

Let's practice to record electrolysis equations for different situations:

Example №1

Write the equations of the processes of the cathode and anode with the electrolysis of the melt of zinc chloride, as well as the general electrolysis equation.

Decision

When moloring zinc chloride occurs its dissociation:

ZnCl 2 \u003d Zn 2+ + 2CL -

Next, it should be paid to the fact that the electrolysis is the melt of zinc chloride, and not an aqueous solution. In other words, without options, only the restoration of zinc cations may occur on the cathode, and the oxidation of chloride ions on the anode. There are 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 number 2.

Write the equations of processes flowing on the cathode and anode with the electrolysis of the aqueous solution of zinc chloride, as well as the general electrolysis equation.

Since in this case, an aqueous solution is subjected to electrolysis, then in electrolysis, theoretically, water molecules can take part. Since zinc is located in a row of activity between Al and it, it means that the cathode will occur both the restoration of zinc cations and water molecules.

2H 2 O + 2E - \u003d 2OH - + H 2

Zn 2+ + 2e - \u003d zn 0

The chloride ion is an acid residue of HCl oxygenic acid, so in competition for oxidation on the anode chloride ions "won" in water molecules:

2CL - - 2E - \u003d CL 2

In this particular case, the total electrolysis equation cannot be written, since it is unknown by the ratio between hydrogen and zinc released on the cathode.

Example number 3.

Write the equations of the processes of the cathode and anode with the electrolysis of the aqueous solution of copper nitrate, as well as the general electrolysis equation.

Copper nitrate in solution is in a predissal state:

Cu (NO 3) 2 \u003d Cu 2+ + 2NO 3 -

Copper is in a row of activity to the right of hydrogen, that is, the cations of copper will be recovered on the cathode:

Cu 2+ + 2e - \u003d Cu 0

Nitrate-ion NO 3 - - oxygen-containing acid residue, it means that in oxidation on anode nitrate ions "lose" in competition of water molecules:

2H 2 O - 4E - \u003d O 2 + 4H +

In this way:

Cathode: Cu 2+ + 2e - \u003d Cu 0 | ∙ 2

2CU 2+ + 2H 2 O \u003d 2CU 0 + O 2 + 4H +

The equation obtained as a result of the addition is the electrolysis ion equation. To obtain a complete molecular electrolysis equation, add 4 ion nitrate to the left and right-hand side of the obtained ion equation as counterions. Then we get:

2CU (NO 3) 2 + 2H 2 O \u003d 2CU 0 + O 2 + 4HNO 3

Example number 4.

Write the equations of the processes occurring at the cathode and anode with the electrolysis of aqueous solution of potassium acetate, as well as the general electrolysis equation.

Decision:

Potassium acetate in aqueous solution dissociates potassium and acetate ions:

CH 3 COPS \u003d CH 3 SOO - + K +

Potassium is an alkaline metal, i.e. Located in a number of electrochemical row of stresses at the very beginning. This means that its cations are not able to discharge at the cathode. Water molecules will be restored instead:

2H 2 O + 2E - \u003d 2OH - + H 2

As mentioned above, the acid residues of carboxylic acids "won" in competition for oxidation in water molecules on the anode:

2 SO 3 SO - - 2E - \u003d CH 3 -CH 3 + 2CO 2

Thus, by summing up the electronic balance and folding two equations of semi-resources on the cathode and the anode we get:

Cathode: 2H 2 O + 2E - \u003d 2OH - + H 2 | ∙ 1

Anode: 2CH 3 soo - - 2e - \u003d CH 3 -CH 3 + 2CO 2 | ∙ 1

2H 2 O + 2CH 3 soo - \u003d 2OH - + H 2 + CH 3 -CH 3 + 2CO 2

We obtained the complete electrolysis equation in ion form. By adding two potassium ions to the left and right-hand part of the equation and resolved with counterions, we obtain the complete electrolysis equation in the molecular form:

2H 2 O + 2CH 3 coo \u003d 2KOH + H 2 + CH 3 -CH 3 + 2CO 2

Example number 5.

Write the equations of processes occurring at the cathode and anode with the electrolysis of the aqueous solution of sulfuric acid, as well as the general electrolysis equation.

Sulfuric acid dissociates on hydrogen cations and sulfate ions:

H 2 SO 4 \u003d 2H + + SO 4 2-

The cathode will occur on the cation of hydrogen H +, and on the anode oxidation of water molecules, since sulfate ions are oxygen-containing acid residues:

Cathode: 2n + + 2e - \u003d H 2 | ∙ 2

Anode: 2H 2 O - 4E - \u003d O 2 + 4H + | ∙ 1

4N + + 2H 2 O \u003d 2H 2 + O 2 + 4H +

Reduced hydrogen ions in the left and right and left part of the equation, we obtain the equation of the electrolysis of the aqueous solution of sulfuric acid:

2H 2 O \u003d 2H 2 + O 2

As you can see, the electrolysis of the aqueous solution of sulfuric acid is reduced to the electrolysis of water.

Example number 6.

Write the equations of the processes occurring at the cathode and anode with the electrolysis of the aqueous solution of sodium hydroxide, as well as the overall electrolysis equation.

Sodium hydroxide dissociation:

NaOH \u003d Na + + Oh -

On the cathode only water molecules will be restored, since sodium is a high-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

We reduce the two water molecules to the left and right and 4 hydroxide ions and come to the fact that, as in the case of sulfuric acid, the electrolysis of the aqueous solution of sodium hydroxide is reduced to the electrolysis of water.

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The results of the EE show that the tasks on the topic "Electrolysis" for graduates remain complex. In the school program, an insufficient number of hours is given to the study of this topic. Therefore, when preparing schoolchildren to the exam, it is necessary to explore this issue in very detailed. Knowledge of the fundamentals of electrochemistry will help the graduate to successfully pass the exam and continue their studies in the higher educational institution. For the study of the topic "Electrolysis" at a sufficient level, it is necessary to conduct preparatory work with graduates who pass the EGE: - consider identifying the basic concepts in the topic "Electrolysis"; - analyzing the electrolysis process Melts and solutions of electrolytes; - consolidate the rules for the reduction of cations on the cathode and oxidation of the anions on the anode (the role of water molecules during the electrolysis of solutions); - the formation of the skills to make the equations of the electrolysis process (cathode and anode processes); - teach students to perform model baseline tasks ( Tasks), high and high level of complexity. Electrolysis - the redox process flowing in solutions and melts of electrolytes during the passage of direct electric current. In solution or melt electrolyte, its dissociation on ions occurs. When the electrical current is turned on, the directional movement and on the surface of the electrodes may occur redox processes. Anode - Positive electrode, it goes oxidation processes.

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

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

Electrolysis of melt sodium chloride

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

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

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

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

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

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

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

Electrolysis of potassium hydroxide melt.

KOH-\u003e K + + OH~

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

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

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

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

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

Electrolysis rules in aqueous solutions.

At the cathode:

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

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

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

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

Zn 2+ + 2e-> Zn 0.

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

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

On the anode:

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

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

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

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

3. Hydroxide ions are oxidized in alkalis solutions:

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

4. Anions are oxidized in solutions of carboxylic salts:

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

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

Cu 0 - 2e-> Cu 2+

Examples of electrolysis processes in aqueous solutions of electrolytes

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

General electrolysis equation:

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

Quests of the basic level of complexity

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

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

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

2. With the electrolysis of aqueous solution of potassium nitrate on the anode allocated: 1) O 2.2) NO 2 3) N 2 4) H 23. Hydrogen is formed under the electrolysis of the aqueous solution: 1) CACI 2. 2) CUSO 4 3) Hg (NO 3) 2 4) AGNO 34. The reaction is possible between: 1) AG and K 2 SO 4 (P-P) 2) Zn and KCI (P-P) 3) MG and SNCI 2(P-p) 4) AG and CUSO 4 (P-P) 5. With the electrolysis of the solution of sodium iodide at the cathode of the lacmus color in the solution: 1) Red 2 ) Blue 3) Purple 4) Yellow6. With the electrolysis of aqueous solution of potassium fluoride on the cathode allocated: 1) hydrogen2) fluoride fluorine 3) fluorine 4) oxygen

Tasks on the topic "Electrolysis"

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

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

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

Answer: 73% of salt decomposed.

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

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

The solution of the problem.We compile the electrolysis reaction equation: 2Cr 2 (SO 4) 3 + 6H 2 O → 4Cr + 3O 2 + 6H 2 SO 4M (CR 2 (SO 4) 3) \u003d 200 ∙ 0.1 \u003d 20gν (CR 2 (SO 4) 3) \u003d 20/392 \u003d 0.051molν (H 2 O) \u003d 0.051 ∙ 3 \u003d 0.153 molm (H 2 O) \u003d 0.153 ∙ 18 \u003d 2.76 g

Quests of the elevated level of complexity B3

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

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

5. Install the correspondence between the name of the substance and the electrolysis products of its aqueous solution.

Answers: 1 - 3411, 2 - 3653, 3 - 2353, 4 - 2246, 5 - 145. In the way, studying the topic of electrolysis, graduates are well absorbed by this section and show good results on the exam. The study of material is accompanied by a presentation on this topic.

Back forward

Attention! Preview slides is used exclusively for informational purposes and may not provide ideas about all presentation capabilities. If you are interested in this work, please download the full version.

The results of the EE show that the tasks on the topic "Electrolysis" for graduates remain complex. In the school program, an insufficient number of hours is given to the study of this topic. Therefore, when preparing schoolchildren to the exam, it is necessary to explore this issue in very detailed. Knowledge of the fundamentals of electrochemistry will help the graduate to successfully pass the exam and continue their studies in the higher educational institution. For the study of the topic "Electrolysis" at a sufficient level, it is necessary to conduct preparatory work with graduates who pass the EGE: - consider identifying the basic concepts in the topic "Electrolysis"; - analyzing the electrolysis process Melts and solutions of electrolytes; - consolidate the rules for the reduction of cations on the cathode and oxidation of the anions on the anode (the role of water molecules during the electrolysis of solutions); - the formation of the skills to make the equations of the electrolysis process (cathode and anode processes); - teach students to perform model baseline tasks ( Tasks), high and high level of complexity. Electrolysis - the redox process flowing in solutions and melts of electrolytes during the passage of direct electric current. In solution or melt electrolyte, its dissociation on ions occurs. When the electrical current is turned on, the directional movement and on the surface of the electrodes may occur redox processes. Anode - Positive electrode, it goes oxidation processes.

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

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

Electrolysis of melt sodium chloride

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

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

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

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

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

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

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

Electrolysis of potassium hydroxide melt.

KOH-\u003e K + + OH~

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

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

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

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

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

Electrolysis rules in aqueous solutions.

At the cathode:

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

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

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

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

Zn 2+ + 2e-> Zn 0.

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

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

On the anode:

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

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

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

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

3. Hydroxide ions are oxidized in alkalis solutions:

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

4. Anions are oxidized in solutions of carboxylic salts:

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

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

Cu 0 - 2e-> Cu 2+

Examples of electrolysis processes in aqueous solutions of electrolytes

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

General electrolysis equation:

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

Quests of the basic level of complexity

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

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

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

2. With the electrolysis of aqueous solution of potassium nitrate on the anode allocated: 1) O 2.2) NO 2 3) N 2 4) H 23. Hydrogen is formed under the electrolysis of the aqueous solution: 1) CACI 2. 2) CUSO 4 3) Hg (NO 3) 2 4) AGNO 34. The reaction is possible between: 1) AG and K 2 SO 4 (P-P) 2) Zn and KCI (P-P) 3) MG and SNCI 2(P-p) 4) AG and CUSO 4 (P-P) 5. With the electrolysis of the solution of sodium iodide at the cathode of the lacmus color in the solution: 1) Red 2 ) Blue 3) Purple 4) Yellow6. With the electrolysis of aqueous solution of potassium fluoride on the cathode allocated: 1) hydrogen2) fluoride fluorine 3) fluorine 4) oxygen

Tasks on the topic "Electrolysis"

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

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

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

Answer: 73% of salt decomposed.

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

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

The solution of the problem.We compile the electrolysis reaction equation: 2Cr 2 (SO 4) 3 + 6H 2 O → 4Cr + 3O 2 + 6H 2 SO 4M (CR 2 (SO 4) 3) \u003d 200 ∙ 0.1 \u003d 20gν (CR 2 (SO 4) 3) \u003d 20/392 \u003d 0.051molν (H 2 O) \u003d 0.051 ∙ 3 \u003d 0.153 molm (H 2 O) \u003d 0.153 ∙ 18 \u003d 2.76 g

Quests of the elevated level of complexity B3

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

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

5. Install the correspondence between the name of the substance and the electrolysis products of its aqueous solution.

Answers: 1 - 3411, 2 - 3653, 3 - 2353, 4 - 2246, 5 - 145. In the way, studying the topic of electrolysis, graduates are well absorbed by this section and show good results on the exam. The study of material is accompanied by a presentation on this topic.

Electrolysis (Greek Elektron - Amber + Lysis - decomposition) - Chemical reaction occurring during the passage of DC through the electrolyte. This decomposition of substances on their components under the action of electric current.

The electrolysis process is to move the cations (positively charged ions) to the cathode (charged negatively), and adversely charged ions (anions) to the anode (charged positively).

So, the anions and cations rushed according to the anode and cathode, respectively. Here is a chemical reaction. To successfully solve tasks on this topic and write reactions, it is necessary to separate the processes on the cathode and anode. This is how this article will be built.

Cathode

Cations are attracted to the cathody - positively charged ions: Na +, K +, Cu 2+, Fe 3+, Ag +, etc.

To establish which reaction is on the cathode, first of all, it is necessary to determine the activity of the metal: its position in the electrochemical row of metals voltages.

If an active metal (Li, Na, k) appeared on the cathode, then water molecules are restored instead of it, of which hydrogen is distinguished. If the metal of medium activity (CR, FE, CD) - hydrogen is allocated on the cathode, and the metal itself. Non-effective metals are highlighted on the cathode in pure form (CU, AG).

I note that aluminum is considered to be the boundary between metals of active and medium activity in a row of stresses. With the electrolysis on the cathode, the metals to aluminum (inclusive!) Are not restored, water molecules are restored instead - hydrogen is released.

If hydrogen ions are received on the cathode - H + (for example, with HCl, H 2 SO 4 acid electrolysis), hydrogen is restored from an acid molecules: 2H + - 2E \u003d 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.

With electrolysis of oxygen-containing anions: SO 4 2-, PO 4 3- - no anions are oxidized on the anode, and water molecules from which oxygen is distinguished.

Hexless anions are oxidized and excreted by the corresponding halogens. Sulfide ion when oxiding sulfur oxidation. Exception is fluorine - if it enters the anode, the water molecule is discharged and oxygen is released. Fluorine is the electronegative element, therefore is an exception.

Anions of organic acids are oxidized in a special way: a radical adjacent to the carboxyl group doubles, and the carboxyl group (COO) itself turns into carbon dioxide - CO 2.

Examples of solutions

In the process of training, you may come across metals that are missed in a row of activity. At the study stage, you can use an expanded number of metals activity.

Now you will know exactly what it stands out on the cathode ;-)

So, practice. We find out what is formed on the cathode and anode with the electrolysis of AGCL solutions, Cu (NO 3) 2, ALBR 3, NAF, FEI 2, CH 3 Cooli.

Sometimes in tasks you need to record the electrolysis response. I inform: if you understand that it is formed on the cathode, and what is on the anode, then it is not difficult to write a reaction. Take, for example, electrolysis NaCl and write the reaction:

NaCl + H 2 O → H 2 + Cl 2 + NaOH

Sodium is an active metal, so hydrogen is distinguished on the cathode. Anion does not contain oxygen, halogen is highlighted - chlorine. We write equation, so we cannot force sodium to evaporate without a trace :) Sodium reacts with water, NaOH is formed.

We write the reaction of electrolysis for CUSO 4:

CUSO 4 + H 2 O → Cu + O 2 + H 2 SO 4

Copper refers to low-active metals, so it is in pure form allocated on the cathode. Anion oxygen-containing, therefore oxygen is released in the reaction. Sulfate ion does not disappear anywhere, it is connected to water hydrogen and turns into a sullen acid.

Electrolysis of melts

All that we discussed up to this point concerned the electrolysis of solutions, where the solvent is water.

In front of industrial chemistry is an important task - to obtain metals (substances) in pure form. Low-effective metals (AG, CU) can be easily obtained by the method of electrolysis of solutions.

But what about active metals: na, k, li? After all, with electrolysis of their solutions, they do not stand out on the cathode in its pure form, water molecules are restored instead and hydrogen is distinguished. Here we will use melt, which do not contain water.

In anhydrous reaction melts, it is even easier: substances are disintegrated by components:

AlCl 3 → Al + Cl 2

LIBR → Li + Br 2

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

The electrode on which oxidation occurs is an anode.

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

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

For example, NaCl \u003d Na + + Cl -

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

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

Consider the processes occurring in electrolysis of electrolyte solutions.

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

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

! hydrogen when reconstructed by reaction:

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

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

Me - nē \u003d me + n

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

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

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

2H 2 O + 2ē \u003d 2OH - + H 2

Metal cations remain in solution, in cathode space

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

2H 2 O + 2ē \u003d 2OH - + H 2

Me + n + nē \u003d me

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

2N + + 2ē \u003d H 2

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

Me + n + nē \u003d me

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

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

Me - nē \u003d me + n

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

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

2cl - - 2ē \u003d Cl 2

2br. - - 2ē \u003d br 2

2i. - - 2ē \u003d i 2

S 2. - - 2ē \u003d s

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

4oh. - - 4ē \u003d 2H 2 O + O 2

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

2H 2 O - 4ē \u003d 4H + + O 2

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

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

Examples of tasks.

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

Soloi formula

A) niso 4

B) Naclo 4

C) licl

D) rbbr.

Product on anode

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

Decision:

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

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

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

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

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

Total answer: 4436

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

Soloi formula

A) Al (NO 3) 3

B) HG (NO 3) 2

C) Cu (NO 3) 2

D) nano 3

Product on anode

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

Decision:

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

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

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

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

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

Total answer: 1341