Hydrochloric acid electrolysis equation. Electrolysis
Solution electrolysis
and molten salts (2 hours)
Classes of the elective course "Electrochemistry"
Goals of the first lesson:
First lesson plan
1. Repetition of the studied methods for obtaining metals.
2. Explanation of new material.
3. Solving problems from the textbook by G.E. Rudzitis, F.G. Feldman "Chemistry-9" (M .: Education, 2002), p. 120, no. 1, 2.
4. Checking the assimilation of knowledge on test tasks.
5. Report on the application of electrolysis.
Goals of the first lesson: to teach how to write schemes for the electrolysis of solutions and molten salts and apply the knowledge gained to solve calculation problems; continue the formation of skills in working with a textbook, test materials; discuss the application of electrolysis in the national economy.
PROGRESS OF THE FIRST LESSON
Repetition of learned methods obtaining metals on the example of obtaining copper from copper(II) oxide.
Recording the equations of the corresponding reactions:
Another way to obtain metals from solutions and melts of their salts is electrochemical, or electrolysis.
Electrolysis is a redox process that occurs on electrodes when an electric current is passed through a melt or electrolyte solution..
Electrolysis of sodium chloride melt:
NaCl Na + + Cl – ;
cathode (–) (Na +): Na + + e= Na 0 ,
anode (–) (Cl –): Cl – – e\u003d Cl 0, 2Cl 0 \u003d Cl 2;
2NaCl \u003d 2Na + Cl 2.
Electrolysis of sodium chloride solution:
NaCl Na + + Cl – ,
H 2 O H + + OH -;
cathode (–) (Na +; H +): H + + e= H 0 , 2H 0 = H 2
(2H 2 O + 2 e\u003d H 2 + 2OH -),
anode (+) (Cl - ; OH -): Cl - - e\u003d Cl 0, 2Cl 0 \u003d Cl 2;
2NaCl + 2H 2 O \u003d 2NaOH + Cl 2 + H 2.
Electrolysis of copper(II) nitrate solution:
Cu(NO 3) 2 Cu 2+ +
H 2 O H + + OH -;
cathode (–) (Cu 2+; H +): Cu 2+ + 2 e= Cu 0 ,
anode (+) (OH -): OH - - e=OH0,
4H 0 \u003d O 2 + 2H 2 O;
2Cu(NO 3) 2 + 2H 2 O \u003d 2Cu + O 2 + 4HNO 3.
These three examples show why it is more profitable to carry out electrolysis than to carry out other methods of obtaining metals: metals, hydroxides, acids, gases are obtained.
We wrote the electrolysis schemes, and now we will try to write the electrolysis equations right away, without referring to the schemes, but only using the ion activity scale:
Examples of electrolysis equations:
2HgSO 4 + 2H 2 O \u003d 2Hg + O 2 + 2H 2 SO 4;
Na 2 SO 4 + 2H 2 O \u003d Na 2 SO 4 + 2H 2 + O 2;
2LiCl + 2H 2 O \u003d 2LiOH + H 2 + Cl 2.
Problem solving from the textbook by G.E. Rudzitis and F.G. Feldman (9th grade, p. 120, No. 1, 2).
Task 1. During the electrolysis of a solution of copper (II) chloride, the mass of the cathode increased by 8 g. What gas was released, what is its mass?
Solution
CuCl 2 + H 2 O \u003d Cu + Cl 2 + H 2 O,
(Cu) \u003d 8/64 \u003d 0.125 mol,
(Cu) \u003d (Сl 2) \u003d 0.125 mol,
m(Cl 2) \u003d 0.125 71 \u003d 8.875 g.
Answer. The gas is chlorine with a mass of 8.875 g.
Task 2. During the electrolysis of an aqueous solution of silver nitrate, 5.6 liters of gas were released. How many grams of metal deposited on the cathode?
Solution
4AgNO 3 + 2H 2 O \u003d 4Ag + O 2 + 4HNO 3,
(O 2) \u003d 5.6 / 22.4 \u003d 0.25 mol,
(Ag) \u003d 4 (O 2) \u003d 4 25 \u003d 1 mol,
m(Ag) \u003d 1 107 \u003d 107 g.
Answer. 107 g of silver.
Testing
Option 1
1. During the electrolysis of a potassium hydroxide solution at the cathode, the following is released:
a) hydrogen; b) oxygen; c) potassium.
2. During the electrolysis of a solution of copper(II) sulfate in solution, the following is formed:
a) copper(II) hydroxide;
b) sulfuric acid;
3. During the electrolysis of a solution of barium chloride at the anode, the following is released:
a) hydrogen; b) chlorine; c) oxygen.
4. During the electrolysis of an aluminum chloride melt, the following is released at the cathode:
a) aluminum; b) chlorine;
c) electrolysis is impossible.
5. The electrolysis of a solution of silver nitrate proceeds according to the following scheme:
a) AgNO 3 + H 2 O Ag + H 2 + HNO 3;
b) AgNO 3 + H 2 O Ag + O 2 + HNO 3;
c) AgNO 3 + H 2 O AgNO 3 + H 2 + O 2.
Option 2
1. During the electrolysis of a sodium hydroxide solution at the anode, the following is released:
a) sodium; b) oxygen; c) hydrogen.
2. During the electrolysis of a solution of sodium sulfide in solution, the following is formed:
a) hydrosulphuric acid;
b) sodium hydroxide;
3. During the electrolysis of a mercury(II) chloride melt, the following is released at the cathode:
a) mercury; b) chlorine; c) electrolysis is impossible.
4.
5. The electrolysis of a solution of mercury(II) nitrate proceeds according to the following scheme:
a) Hg (NO 3) 2 + H 2 O Hg + H 2 + HNO 3;
b) Hg (NO 3) 2 + H 2 O Hg + O 2 + HNO 3;
c) Hg (NO 3) 2 + H 2 O Hg (NO 3) 2 + H 2 + O 2.
Option 3
1. During the electrolysis of a solution of copper (II) nitrate, the following is released at the cathode:
a) copper; b) oxygen; c) hydrogen.
2. During the electrolysis of a solution of lithium bromide in solution, the following is formed:
b) hydrobromic acid;
c) lithium hydroxide.
3. During the electrolysis of a silver chloride melt, the following is released at the cathode:
a) silver; b) chlorine; c) electrolysis is impossible.
4. During the electrolysis of an aluminum chloride solution, aluminum is released into:
a) cathode; b) anode; c) remains in solution.
5. The electrolysis of a solution of barium bromide proceeds according to the following scheme:
a) BaBr 2 + H 2 O Br 2 + H 2 + Ba (OH) 2;
b) BaBr 2 + H 2 O Br 2 + Ba + H 2 O;
c) BaBr 2 + H 2 O Br 2 + O 2 + Ba (OH) 2.
Option 4
1. During the electrolysis of a barium hydroxide solution at the anode, the following is released:
a) hydrogen; b) oxygen; c) barium.
2. During the electrolysis of a solution of potassium iodide in solution, the following is formed:
a) hydroiodic acid;
b) water; c) potassium hydroxide.
3. During the electrolysis of a melt of lead (II) chloride, the following is released on the cathode:
a) lead; b) chlorine; c) electrolysis is impossible.
4. During the electrolysis of a silver nitrate solution at the cathode, the following is released:
a) silver; b) hydrogen; c) oxygen.
5. The electrolysis of sodium sulfide solution proceeds according to the following scheme:
a) Na 2 S + H 2 O S + H 2 + NaOH;
b) Na 2 S + H 2 O H 2 + O 2 + Na 2 S;
c) Na 2 S + H 2 O H 2 + Na 2 S + NaOH.
Answers
| Option | Question 1 | Question 2 | Question 3 | Question 4 | Question 5 |
| 1 | a | b | b | a | b |
| 2 | b | b | a | a | b |
| 3 | a | in | a | in | a |
| 4 | b | in | a | a | a |
The use of electrolysis in the national economy
1. To protect metal products from corrosion, a thin layer of another metal is applied to their surface: chromium, silver, gold, nickel, etc. Sometimes, in order not to waste expensive metals, a multi-layer coating is produced. For example, the exterior parts of a car are first covered with a thin layer of copper, a thin layer of nickel is applied to the copper, and a layer of chromium is applied to it.
When applying coatings to metal by electrolysis, they are obtained even in thickness and durable. In this way, you can cover products of any shape. This branch of applied electrochemistry is called electroplating.
2. In addition to corrosion protection, galvanic coatings give a beautiful decorative look to products.
3. Another branch of electrochemistry, close in principle to electroplating, is called electroplating. This is the process of obtaining exact copies of various items. To do this, the object is covered with wax and a matrix is obtained. All recesses of the copied object on the matrix will be bulges. The surface of the wax matrix is coated with a thin layer of graphite, making it electrically conductive.
The resulting graphite electrode is immersed in a bath of copper sulfate solution. The anode is copper. During electrolysis, the copper anode dissolves, and copper is deposited on the graphite cathode. Thus, an exact copper copy is obtained.
With the help of electroforming, clichés for printing, gramophone records are made, various objects are metallized. Galvanoplasty was discovered by the Russian scientist B.S. Jacobi (1838).
Making record dies involves applying a thin layer of silver to a plastic record to make it electrically conductive. Then an electrolytic nickel coating is applied to the plate.
What should be done to make a plate in an electrolytic bath - anode or cathode?
(About the e t. Cathode.)
4. Electrolysis is used to obtain many metals: alkali, alkaline earth, aluminum, lanthanides, etc.
5. To clean some metals from impurities, the metal with impurities is connected to the anode. The metal is dissolved during the electrolysis process and precipitated on the metal cathode, while the impurity remains in solution.
6. Electrolysis is widely used to obtain complex substances (alkalis, oxygen-containing acids), halogens.
Practical work(second lesson)
Lesson goals. Conduct water electrolysis, show electroplating in practice, consolidate the knowledge gained in the first lesson.
Equipment.On student tables: a flat battery, two wires with terminals, two graphite electrodes, a beaker, test tubes, a tripod with two legs, 3% sodium sulfate solution, a spirit lamp, matches, a torch.
On the teacher's desk: the same + a solution of copper sulfate, a brass key, a copper tube (a piece of copper).
Student briefing
1. Attach the wires with terminals to the electrodes.
2. Place the electrodes in a glass so that they do not touch.
3. Pour the electrolyte solution (sodium sulfate) into the beaker.
4. Pour water into the test tubes and, putting them upside down in a glass with electrolyte, put them on the graphite electrodes one by one, fixing the upper edge of the test tube in the foot of the tripod.
5. After the device is mounted, attach the ends of the wires to the battery.
6. Observe the evolution of gas bubbles: less of them are released at the anode than at the cathode. After almost all the water in one test tube is displaced by the released gas, and in the other - by half, disconnect the wires from the battery.
7. Light the spirit lamp, carefully remove the test tube, where the water is almost completely displaced, and bring it to the spirit lamp - a characteristic pop of gas will be heard.
8. Light a torch. Remove the second test tube, check with a smoldering splint of gas.
Assignments for students
1. Sketch the device.
2. Write an equation for the electrolysis of water and explain why it was necessary to carry out electrolysis in a solution of sodium sulfate.
3. Write reaction equations that reflect the release of gases on the electrodes.
Teacher demonstration experiment
(can be performed by the best students in the class
with appropriate equipment)
1. Connect the wire terminals to the copper tube and brass key.
2. Lower the tube and key into a beaker with copper(II) sulfate solution.
3. Connect the second ends of the wires to the battery: "minus" of the battery to the copper tube, "plus" to the key!
4. Observe the release of copper on the surface of the key.
5. After performing the experiment, first disconnect the terminals from the battery, then remove the key from the solution.
6. Disassemble the electrolysis circuit with a soluble electrode:
CuSO 4 \u003d Cu 2+ +
anode (+): Сu 0 - 2 e\u003d Cu 2+,
cathode (–): Cu 2+ + 2 e= Сu 0 .
The overall equation for electrolysis with a soluble anode cannot be written.
The electrolysis was carried out in a solution of copper(II) sulfate, because:
a) an electrolyte solution is needed in order for an electric current to flow, tk. water is a weak electrolyte;
b) no by-products of the reactions will be released, but only copper at the cathode.
7. To consolidate the past, write a scheme for the electrolysis of zinc chloride with carbon electrodes:
ZnCl 2 \u003d Zn 2+ + 2Cl -,
cathode (–): Zn 2+ + 2 e= Zn 0 ,
2H2O+2 e\u003d H 2 + 2OH -,
anode (+): 2Cl – – 2 e=Cl2.
The overall reaction equation in this case cannot be written, because it is not known what part of the total amount of electricity goes to the reduction of water, and what part - to the reduction of zinc ions.
 |
Scheme of the demonstration experiment |
Homework
1. Write an equation for the electrolysis of a solution containing a mixture of copper(II) nitrate and silver nitrate with inert electrodes.
2. Write the equation for the electrolysis of sodium hydroxide solution.
3. To clean a copper coin, it must be hung on a copper wire connected to the negative pole of the battery, and lowered into a 2.5% NaOH solution, where the graphite electrode connected to the positive pole of the battery should also be immersed. Explain how a coin becomes clean. ( Answer. Hydrogen ions are being reduced at the cathode:
2H + + 2 e\u003d H 2.
Hydrogen reacts with copper oxide on the surface of the coin:
CuO + H 2 \u003d Cu + H 2 O.
This method is better than powder cleaning, because. the coin is not erased.)
When considering the electrolysis of aqueous solutions, it must be borne in mind that, in addition to electrolyte ions, in any aqueous solution there are also ions that are products of the dissociation of water H + and OH -.
In an electric field, hydrogen ions move towards the cathode, and OH ions move towards the anode. Thus, both electrolyte cations and hydrogen cations can be discharged at the cathode. Similarly, at the anode, both electrolyte anions and hydroxide ions can be discharged. In addition, water molecules can also undergo electrochemical oxidation or reduction.
Which electrochemical processes will take place at the electrodes during electrolysis will primarily depend on the relative values of the electrode potentials of the corresponding electrochemical systems. Of the several possible processes, the one with the minimum energy consumption will proceed. This means that the oxidized forms of electrochemical systems with the highest electrode potential will be reduced at the cathode, while the reduced forms of systems with the lowest electrode potential will be oxidized at the anode. In the general case, those atoms, molecules and ions, whose potentials are the lowest under given conditions, are more easily oxidized on the anode, and those ions, molecules, atoms whose potentials are the highest are more easily reduced on the cathode. Let us consider the cathodic processes occurring during the electrolysis of aqueous solutions of salts. Here it is necessary to take into account the magnitude of the electrode potential of the process of reduction of hydrogen ions, which depends on the concentration of hydrogen ions. We know the general equation of the electrode potential for the hydrogen electrode (section 2.3).
In the case of neutral solutions (pH=7), the value of the electrode potential of the hydrogen ion reduction process is
|
φ = –0,059 . 7 = -0.41 V. |
1) during the electrolysis of salt solutions containing metal cations, the electrode potential of which is much more positive than –0.41 V, metal will be reduced from a neutral solution of such an electrolyte at the cathode. Such metals are in a series of voltages near hydrogen (starting approximately from tin and after it);
2) during the electrolysis of salt solutions containing metal cations, the electrode potential of which is much more negative than -0.41 V, the metal will not be reduced at the cathode, but hydrogen will be released. Such metals include alkali, alkaline earth, magnesium, aluminum, up to approximately titanium;
3) during the electrolysis of salt solutions containing metal cations, the electrode potential of which is close to -0.41 V (metals of the middle part of the series - Zn, Cr, Fe, Cd, Ni), then, depending on the concentration of the salt solution and the electrolysis conditions ( current density, temperature, solution composition), both metal reduction and hydrogen evolution are possible; sometimes there is a joint release of metal and hydrogen.
The electrochemical release of hydrogen from acidic solutions occurs due to the discharge of hydrogen ions:
2H + 2ē → 2H 0
2H 0 = H 2 .
In the case of neutral or alkaline media, hydrogen evolution occurs as a result of the electrochemical reduction of water:
HOH + ē → H 0 + OH –
H 0 + H 0 = H 2 ,
|
then 2HON + 2ē → H 2 + 2OH – |
Thus, the nature of the cathodic process during the electrolysis of aqueous solutions is determined primarily by the position of the corresponding metal in the series of standard electrode potentials of metals.
If an aqueous solution containing cations of various metals is subjected to electrolysis, then their release at the cathode, as a rule, will proceed in order of decreasing the algebraic value of the electrode potential of the metal. For example, from a mixture of cations Ag +, Cu 2+ and Zn 2+ with sufficient voltage at the terminals of the electrolyzer, silver cations (φ 0 \u003d +0.8 V), then copper (φ 0 \u003d +0.34 V) and , finally, zinc (φ 0 \u003d -0.76 V).
The electrochemical separation of metals from a mixture of cations is used in engineering and in quantitative analysis. In general, the ability to discharge (accept electrons) for metal ions is determined by the position of metals in a series of standard electrode potentials. The more to the left the metal is in the series of voltages, the greater its negative potential or the less positive potential, the more difficult it is for its ions to discharge. So, from metal ions in a series of voltages, trivalent gold ions are discharged most easily (at the lowest voltages of the electric current), then silver ions, etc. The most difficult (at the highest voltage of the electric current) is the discharge of potassium ions. But the value of the potential of a metal, as is known, varies depending on the concentration of its ions in solution; in the same way, the ease of discharge of ions of each metal varies depending on their concentration: an increase in concentration facilitates the discharge of ions, a decrease makes it difficult. Therefore, during the electrolysis of a solution containing ions of several metals, it may be that the release of a more active metal will occur earlier than the release of a less active one (if the concentration of the first metal ion is significant, and the second is very low).
Let us consider the anodic processes occurring during the electrolysis of aqueous solutions of salts. The nature of the reactions occurring at the anode depends both on the presence of water molecules and on the substance from which the anode is made. It should be borne in mind that the anode material may oxidize during electrolysis. In this regard, a distinction is made between electrolysis with an inert (insoluble) anode and electrolysis with an active (soluble) anode. Insoluble anodes are made from coal, graphite, platinum, iridium; soluble anodes - from copper, silver, zinc, cadmium, nickel and other metals. On an insoluble anode during electrolysis, anions or water molecules are oxidized. During the electrolysis of aqueous solutions of oxygen-free acids HI, HBr, HCl, H 2 S and their salts (except for HF and fluorides), anions are discharged at the anode and the corresponding halogen is released. Note that the release of chlorine during the electrolysis of HCl and its salts contradicts the mutual position of the systems
2Cl – – 2ē →Cl 2 (φ 0 = +1.36 V)
2 H 2 O– 4ē →O 2 + 4 H + (φ 0 = +1.23 V)
in a series of standard electrode potentials. This anomaly is associated with a significant overvoltage of the second of these two electrode processes - the anode material has an inhibitory effect on the process of oxygen evolution.
During the electrolysis of aqueous solutions of salts containing anions SO 4 2-, SO 3 2-, NO 3 -, PO 4 3-, etc., as well as hydrogen fluoride and fluorides, electrochemical oxidation of water occurs. Depending on the pH of the solution, this process proceeds differently and can be written in different equations. In an alkaline medium, the equation has the form
4OH – – 4ē → 2H 2 O+O 2 , (pH > 7)
and in acidic or neutral media we have
HOH– 2ē →O 0 + 2 H + (pH ≤ 7)
2 O 0 = O 2 ,
|
then 2H 2 О – 4ē → 4Н + + 2O 2 . |
In the cases under consideration, the electrochemical oxidation of water is the most energetically favorable process. Oxidation of oxygen-containing anions occurs at very high potentials. For example, the standard oxidation potential of the SO 4 2- ion - 2ē → S 2 O 8 2- is 2.01 V, which significantly exceeds the standard water oxidation potential of 1.228 V.
2H 2 O - 4ē → O 2 + 4H + (φ 0 = 1.228 V).
The standard ion oxidation potential F - is even more important
2F – – 2ē →F 2 (φ 0 = 2 ,87 AT).
In general, during the electrolysis of aqueous solutions of salts, metal and hydrogen cations simultaneously approach the cathode of the electrolyzer, while each of them “claims” to be reduced due to electrons coming from the cathode. How will the reduction process actually proceed at the cathode? The answer can be obtained based on a number of stresses of metals. In this case, the smaller the algebraic value of the standard electrode potential of the metal, the weaker electron acceptors are their cations and the more difficult it is to reduce them on the cathode. In this regard, three groups of cations are distinguished according to their relation to electroreduction.
1. Cations characterized by high electron-withdrawing activity (Cu 2+, Hg 2+, Ag+, Au 3+, Pt 2+, Pt 4+). During the electrolysis of salts of these cations, almost complete reduction of metal cations occurs; current output 100% or close to it.
2. Cations characterized by average values of electron-withdrawing ability (Mn 2+, Zn 2+, Cr 3+, Fe 2+, Ni 2+, Sn 2+, Pb 2+). During electrolysis at the cathode, cations of both the metal and water molecules are simultaneously reduced, which leads to a decrease in the current efficiency of the metal.
3. Cations exhibiting a low electron-withdrawing ability (K +, Ca 2+, Mg 2+, Al 3+). In this case, the electron acceptors to the cathode are not cations of the group under consideration, but water molecules. In this case, the cations themselves remain unchanged in the aqueous solution, and the current efficiency approaches zero.
The ratio of various anions to electrooxidation at the anode
Anions of oxygen-free acids and their salts (Cl ¯, Br ¯, J ¯, S 2-, CN¯, etc.) hold their electrons weaker than a water molecule. Therefore, during the electrolysis of aqueous solutions of compounds containing these anions, the latter will play the role of electron donors, they will be oxidized and transfer their electrons to the external circuit of the electrolytic cell.
Anions of oxygen acids (NO 3 ¯, SO 4 2-, PO 4 3-, etc.) are able to hold their electrons more firmly than water molecules. In this case, water is oxidized at the anode, while the anions themselves remain unchanged.
In the case of a soluble anode, the number of oxidative processes increases to three:
1) electrochemical oxidation of water with the release of oxygen; 2) anion discharge (i.e., its oxidation); 3) electrochemical oxidation of the anode metal (anodic dissolution of the metal).
Of the possible processes, the one that is energetically most favorable will take place. If the anode metal is located in a series of standard potentials earlier than both other electrochemical systems, then anodic dissolution of the metal will be observed. Otherwise, there will be an evolution of oxygen or an anion discharge. No close sequence has been established for the discharge of anions. By decreasing the ability to donate electrons, the most common anions are arranged as follows: S 2-, J ¯, Br ¯, Cl ¯, OH¯, H 2 O, SO 4 2-, NO 3 ¯, CO 3 2-, PO 4 3- .
Let us consider several typical cases of electrolysis of aqueous solutions.
Electrolysis of a CuCl 2 solution with an insoluble anode
In a series of voltages, copper is located after hydrogen, so Cu 2+ will be discharged at the cathode and metallic copper will be released, and chloride ions will be oxidized to molecular chlorine Cl 2 at the anode.
|
Cathode (-) |
|
|
Cu 2+ + 2ē → Cu 0 |
|
|
2Cl – – 2ē → Cl 2 |
Cu 2+ + 2 Cl – → Cu 0 +Cl 2
CuCl 2 → Cu 0 +Cl 2
Metal current output (95-100%).
Electrolysis of NaNO 3 solution
Since sodium in the series of voltages is much earlier than hydrogen, water will be discharged at the cathode. At the anode, water will also be discharged.
|
Cathode (-) |
|
2 H 2 O+ 2ē →H 2 + 2 Oh – |
|
2H 2 O–4ē → 4H + +O 2 . |
Thus, hydrogen is released at the cathode and an alkaline environment is created, oxygen is released at the anode and an acidic environment is created near the anode. If the anode and cathode spaces are not separated from each other, then the solution in all its parts will remain electrically neutral.
|
Cathode (-) |
|
|
2 H 2 O+ 2ē →H 2 + 2 Oh – |
|
|
2H 2 O–4ē → 4H + +O 2 . |
6H 2 O → 2H 2 + 4OH – + 4H + +O 2
6H 2 O → 2H 2 +O 2 + 4H 2 O
2 H 2 O → 2 H 2 + O 2
The current output of the metal is zero.
Therefore, during the electrolysis of the NaNO 3 solution, the electrolysis of water will occur. The role of NaNO 3 salt is reduced to an increase in the electrical conductivity of the solution.
Electrolysis of FeSO 4 solution
Reactions at the cathode (-) (reduction):
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a) Fe 2+ + 2ē → Fe 0 |
simultaneous reactions |
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b) 2 H 2 O+ 2ē →H 2 + 2 Oh – . |
Reaction at the anode (+) (oxidation):
2H 2 O–4ē → 4H + +O 2 .
The current output of the metal is average.
Electrolysis of KJ solution with insoluble anode
|
Cathode (-) |
|
|
2 H 2 O+ 2ē →H 2 + 2 Oh – |
|
|
2J – – 2ē → J 2 |
2 H 2 O + 2J – → H 2 + 2 Oh – + J 2 .
The final reaction equation for the electrolysis of solution KJ:
2KJ+2H 2 O→H 2 + J 2 + 2KOH.
Electrolysis of a CuSO 4 solution with a copper (soluble) anode.
The standard potential of copper is +0.337 V, which is much higher than -0.41 V; therefore, during the electrolysis of a solution of CuSO 4 at the cathode, a discharge of Cu 2+ ions occurs and metallic copper is released. At the anode, the opposite process occurs - the oxidation of the metal, since the copper potential is much less than the oxidation potential of water (+1.228 V), and even more so - the oxidation potential of the SO 4 2- ion (+2.01 V). Consequently, in this case, electrolysis is reduced to the dissolution of the metal (copper) of the anode and its separation at the cathode.
Scheme of electrolysis of copper sulfate solution:
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Cathode (-) |
|
Cu 2+ + 2ē → Cu 0 |
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Cu 0 – 2ē → Cu 2+ . |
This process is used for electrical refining of metals (so-called electrolytic refining).
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 produced alkali, alkaline earth metals, aluminum or magnesium, respectively. 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 example:
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 on the right and 4 hydroxide ions and come to the conclusion that, as in the case of sulfuric acid, the electrolysis of an aqueous solution of sodium hydroxide is reduced to the electrolysis of water.
