The electrochemical series of metal voltages is complete. Active metals
Sections: Chemistry, Competition "Presentation for the lesson"
Class: 11
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Goals and objectives:
- Educational: Consideration of the chemical activity of metals based on their position in periodic table DI. Mendeleev and in electrochemical series metal stress.
- Developmental: To promote the development of auditory memory, the ability to compare information, think logically and explain ongoing chemical reactions.
- Educational: Forming a skill independent work, the ability to reasonably express one’s opinion and listen to classmates, we cultivate in the children a sense of patriotism and pride in their compatriots.
Equipment: PC with media projector, individual laboratories with a set of chemical reagents, models crystal lattices metals
Lesson type: using technology for the development of critical thinking.
During the classes
I. Challenge stage.
Updating knowledge on the topic, awakening cognitive activity.
Bluff game: “Do you believe that...” (Slide 3)
- Metals occupy the upper left corner in the PSHE.
- In crystals, metal atoms are connected by metallic bonds.
- The valence electrons of metals are tightly bound to the nucleus.
- Metals in the main subgroups (A) usually have 2 electrons in their outer level.
- In the group from top to bottom there is an increase in the reducing properties of metals.
- To assess the reactivity of a metal in solutions of acids and salts, it is enough to look at the electrochemical voltage series of metals.
- To assess the reactivity of a metal in solutions of acids and salts, just look at the periodic table of D.I. Mendeleev
Question for the class? What does the entry mean? Me 0 – ne —> Me +n(Slide 4)
Answer: Me0 is a reducing agent, which means it interacts with oxidizing agents. The following can act as oxidizing agents:
- Simple substances (+O 2, Cl 2, S...)
- Complex substances(H 2 O, acids, salt solutions...)
II. Understanding new information.
As a methodological technique, it is proposed to draw up a reference diagram.
Question for the class? What factors determine the reducing properties of metals? (Slide 5)
Answer: From the position in the periodic table of D.I. Mendeleev or from the position in the electrochemical series of voltage of metals.
The teacher introduces the concepts: chemical activity and electrochemical activity.
Before starting the explanation, the children are asked to compare the activity of atoms TO And Li position in the periodic table D.I. Mendeleev and the activity of simple substances formed by these elements according to their position in the electrochemical voltage series of metals. (Slide 6)
A contradiction arises:In accordance with the position of alkali metals in PSCE and according to the patterns of changes in the properties of elements in the subgroup, the activity of potassium is greater than that of lithium. By position in the voltage series, lithium is the most active.
New material. The teacher explains the difference between chemical and electrochemical activity and explains that the electrochemical series of voltages reflects the ability of a metal to transform into a hydrated ion, where the measure of metal activity is energy, which consists of three terms (atomization energy, ionization energy and hydration energy). We write down the material in a notebook. (Slides 7-10)
Let's write it down together in a notebook. conclusion: The smaller the radius of the ion, the greater the electric field around it is created, the more energy is released during hydration, hence the stronger reducing properties of this metal in reactions.
Historical reference: student's speech about Beketov's creation of a displacement series of metals. (Slide 11)
The action of the electrochemical voltage series of metals is limited only by the reactions of metals with solutions of electrolytes (acids, salts).
Memo:
- The reducing properties of metals decrease during reactions in aqueous solutions under standard conditions (250°C, 1 atm);
- The metal to the left displaces the metal to the right from their salts in solution;
- Metals standing before hydrogen displace it from acids in solution (except: HNO3);
- Me (to Al) + H 2 O -> alkali + H 2
Other Me (up to H 2) + H 2 O -> oxide + H 2 (harsh conditions)
Me (after H 2) + H 2 O -> do not react
(Slide 12)
Reminders are handed out to the guys.
Practical work:“Interaction of metals with salt solutions” (Slide 13)
Make the transition:
- CuSO 4 —> FeSO 4
- CuSO 4 —> ZnSO 4
Demonstration of experience of interaction between copper and mercury(II) nitrate solution.
III. Reflection, reflection.
We repeat: in which case do we use the periodic table, and in which case is a series of metal voltages needed? (Slides 14-15).
Back to initial questions lesson. We display questions 6 and 7 on the screen. We analyze which statement is incorrect. There is a key on the screen (checking task 1). (Slide 16).
Let's summarize the lesson:
- What new did you learn?
- In what case is it possible to use the electrochemical voltage series of metals?
Homework: (Slide 17)
- Repeat the concept of “POTENTIAL” from the physics course;
- Complete the reaction equation, write the electron balance equations: Сu + Hg(NO 3) 2 →
- Metals are given ( Fe, Mg, Pb, Cu)– propose experiments confirming the location of these metals in the electrochemical voltage series.
We evaluate the results for the bluff game, work at the board, oral answers, communication, and practical work.
Used Books:
- O.S. Gabrielyan, G.G. Lysova, A.G. Vvedenskaya “Handbook for teachers. Chemistry 11th grade, part II” Bustard Publishing House.
- N.L. Glinka "General Chemistry".
Grosse E., Weissmantel H.
Chemistry for the curious. Basics of chemistry and entertaining experiments.
Chapter 3 (continued)SMALL COURSE IN ELECTROCHEMISTRY OF METALS
We have already become acquainted with the electrolysis of solutions of alkali metal chlorides and the production of metals using melts. Now let’s try to study some of the laws of the electrochemistry of aqueous solutions using several simple experiments, galvanic cells, as well as get acquainted with the production of protective galvanic coatings.Electrochemical methods are used in modern analytical chemistry, serve to determine the most important quantities of theoretical chemistry.
Finally, corrosion of metal objects, which causes great damage to the national economy, in most cases is an electrochemical process.
METALS STRESS SERIES
The fundamental link for understanding electrochemical processes is the voltage series of metals. Metals can be arranged in a series that begins with the chemically active and ends with the least active noble metals:Li, Rb, K, Ba, Sr, Ca, Mg, Al, Be, Mn, Zn, Cr, Ga, Fe, Cd, Tl, Co, Ni, Sn, Pb, H, Sb, Bi, As, Cu, Hg, Ag, Pd, Pt, Au.
This is, according to the latest ideas, a series of voltages for the most important metals and hydrogen. If electrodes of a galvanic cell are made from any two metals in a row, then a negative voltage will appear on the material preceding the row.
Voltage value ( electrochemical potential) depends on the position of the element in the voltage series and on the properties of the electrolyte.
We will establish the essence of the voltage series from several simple experiments, for which we will need a current source and electrical measuring instruments. Dissolve about 10 g of crystalline copper sulfate in 100 ml of water and immerse a steel needle or a piece of iron sheet into the solution. (We recommend that you first clean the iron until it shines with fine sandpaper.) After a short time, the iron will be covered with a reddish layer of released copper. More active iron displaces copper from solution, with iron dissolving as ions and copper being released as metal. The process continues as long as the solution is in contact with the iron. Once the copper covers the entire surface of the iron, it will practically stop. In this case, a rather porous layer of copper is formed, so protective coatings cannot be obtained without the use of current.
In the following experiments, we will lower small strips of zinc and lead sheet metal into a solution of copper sulfate. After 15 minutes, we take them out, wash them and examine them under a microscope. We can discern beautiful ice-like patterns, which in reflected light are red in color and consist of released copper. Here, too, more active metals converted copper from the ionic to the metallic state.
In turn, copper can displace metals that are lower in the voltage series, that is, less active. Apply a few drops of silver nitrate solution to a thin strip of sheet copper or flattened copper wire (having previously cleaned the surface to a shine). With the naked eye you can see the resulting blackish coating, which under a microscope in reflected light looks like thin needles and plant patterns (so-called dendrites).
To isolate zinc without current, it is necessary to use a more active metal. Excluding metals that react violently with water, we find magnesium in the voltage series above zinc. Place a few drops of zinc sulfate solution on a piece of magnesium tape or on thin electron shavings. We obtain a solution of zinc sulfate by dissolving a piece of zinc in dilute sulfuric acid. At the same time as zinc sulfate, add a few drops of denatured alcohol. On magnesium, after a short period of time, we will notice, especially under a microscope, zinc released in the form of thin crystals.
In general, any member of the voltage series can be displaced from solution, where it exists as an ion, and converted to the metallic state. However, when trying all sorts of combinations, we may be disappointed. It would seem that if a strip of aluminum is immersed in solutions of salts of copper, iron, lead and zinc, these metals should be released on it. But this, however, does not happen. The reason for the failure lies not in an error in the series of voltages, but is based on a special inhibition of the reaction, which in in this case due to a thin oxide film on the surface of aluminum. In such solutions, aluminum is called passive.
LET'S LOOK BEHIND THE SCENES
To formulate the laws of the ongoing processes, we can limit ourselves to considering cations and exclude anions, since they themselves do not participate in the reaction. (However, the rate of deposition is affected by the type of anions.) If, for simplicity, we assume that both the precipitated and dissolved metals produce doubly charged cations, then we can write:Me 1 + Me 2 2+ = Me 1 2+ + Me 2
Moreover, for the first experiment Me 1 = Fe, Me 2 = Cu.
So, the process consists of the exchange of charges (electrons) between atoms and ions of both metals. If we separately consider (as intermediate reactions) the dissolution of iron or the precipitation of copper, we obtain:
Fe = Fe 2+ + 2 e --
Сu 2+ + 2 e-- = Cu
Now consider the case when a metal is immersed in water or in a salt solution, with a cation of which exchange is impossible due to its position in the stress series. Despite this, the metal tends to go into solution in the form of an ion. In this case, the metal atom gives up two electrons (if the metal is divalent), the surface of the metal immersed in the solution becomes negatively charged relative to the solution, and a double electric layer is formed at the interface. This potential difference prevents further dissolution of the metal, so that the process soon stops.
If two different metals are immersed in a solution, they will both charge, but the less active one will be somewhat weaker, due to the fact that its atoms are less prone to losing electrons.
Let's connect both metals with a conductor. Due to the potential difference, a flow of electrons will flow from the more active metal to the less active one, which forms the positive pole of the element. A process occurs in which the more active metal goes into solution, and cations from the solution are released on the more noble metal. Let us now illustrate with several experiments the somewhat abstract reasoning given above (which, moreover, represents a gross simplification).
First, fill a 250 ml beaker to the middle with a 10% solution of sulfuric acid and immerse not too small pieces of zinc and copper in it. We solder or rivet copper wire to both electrodes, the ends of which should not touch the solution.
As long as the ends of the wire are not connected to each other, we will observe the dissolution of zinc, which is accompanied by the release of hydrogen. Zinc, as follows from the voltage series, is more active than hydrogen, so the metal can displace hydrogen from the ionic state. An electrical double layer is formed on both metals. The easiest way to detect the potential difference between the electrodes is with a voltmeter. Immediately after connecting the device to the circuit, the arrow will indicate approximately 1 V, but then the voltage will quickly drop. If you connect a small light bulb that consumes 1 V to the element, it will light up - at first quite strongly, and then the glow will become weak.
Based on the polarity of the device terminals, we can conclude that the copper electrode is the positive pole. This can be proven without a device by considering the electrochemistry of the process. Let's prepare a saturated solution of table salt in a small beaker or test tube, add about 0.5 ml of an alcohol solution of the phenolphthalein indicator and immerse both electrodes closed with wire into the solution. A faint reddish color will be observed near the negative pole, which is caused by the formation of sodium hydroxide at the cathode.
In other experiments it can be placed in a cell various pairs metals and determine the resulting stress. For example, magnesium and silver will give a particularly large potential difference due to the significant distance between them and a series of voltages, while zinc and iron, on the contrary, will give a very small one, less than a tenth of a volt. By using aluminum, we will not receive practically any current due to passivation.
All these elements, or, as electrochemists say, circuits, have the disadvantage that when measuring current, the voltage across them drops very quickly. Therefore, electrochemists always measure the true magnitude of the voltage in the de-energized state using the voltage compensation method, that is, comparing it with the voltage of another current source.
Let us consider the processes in the copper-zinc element in a little more detail. At the cathode, zinc goes into solution according to the following equation:
Zn = Zn 2+ + 2 e --
Hydrogen ions of sulfuric acid are discharged at the copper anode. They attach electrons coming through the wire from the zinc cathode and as a result, hydrogen bubbles are formed:
2H + + 2 e-- = N 2
After a short period of time, the copper will be covered with a thin layer of hydrogen bubbles. In this case, the copper electrode will turn into a hydrogen one, and the potential difference will decrease. This process is called electrode polarization. The polarization of the copper electrode can be eliminated by adding a little potassium dichromate solution to the cell after the voltage drop. After this, the voltage will increase again, as potassium dichromate will oxidize hydrogen to water. Potassium dichromate acts in this case as a depolarizer.
In practice, galvanic circuits are used whose electrodes are not polarized, or circuits whose polarization can be eliminated by adding depolarizers.
As an example of a non-polarizable element, consider the Daniel element, which was often used in the past as a current source. This is also a copper-zinc element, but both metals are immersed in different solutions. The zinc electrode is placed in a porous clay cell filled with dilute (about 20%) sulfuric acid. The clay cell is suspended in a large glass containing a concentrated solution of copper sulfate, and at the bottom there is a layer of copper sulfate crystals. The second electrode in this vessel is a cylinder made of copper sheet.
This element can be made from a glass jar, a commercially available clay cell (in extreme cases, we use a flower pot, closing the hole in the bottom) and two electrodes of suitable size.
During operation of the element, zinc dissolves to form zinc sulfate, and copper ions are released at the copper electrode. But at the same time, the copper electrode is not polarized and the element produces a voltage of about 1 V. Actually, theoretically, the voltage at the terminals is 1.10 V, but when collecting current we measure a slightly lower value due to the electrical resistance of the cell.
If we do not remove the current from the element, we need to remove the zinc electrode from the sulfuric acid solution, because otherwise it will dissolve to form hydrogen.
A diagram of a simple cell that does not require a porous partition is shown in the figure. The zinc electrode is located at the top of the glass jar, and the copper electrode is located near the bottom. The entire cell is filled with a saturated solution of table salt. Place a handful of copper sulfate crystals at the bottom of the jar. The resulting concentrated copper sulfate solution will mix with the table salt solution very slowly. Therefore, when the cell operates, copper will be released on the copper electrode, and zinc will dissolve in the form of sulfate or chloride in the upper part of the cell.
Nowadays, batteries use almost exclusively dry cells, which are more convenient to use. Their ancestor is the Leclanche element. The electrodes are a zinc cylinder and a carbon rod. The electrolyte is a paste that mainly consists of ammonium chloride. Zinc dissolves in the paste, and hydrogen is released on the coal. To avoid polarization, the carbon rod is dipped into a linen bag containing a mixture of coal powder and pyrolusite. The carbon powder increases the electrode surface, and the pyrolusite acts as a depolarizer, slowly oxidizing the hydrogen.
True, the depolarizing ability of pyrolusite is weaker than that of the previously mentioned potassium dichromate. Therefore, when current is received in dry cells, the voltage drops quickly, they " get tired"due to polarization. Only after some time does the oxidation of hydrogen occur with pyrolusite. Thus, the elements " resting", if you do not pass current for some time. Let's check this on the battery for flashlight, to which we connect the light bulb. We connect a voltmeter parallel to the lamp, that is, directly to the terminals.
At first, the voltage will be about 4.5 V. (Most often, such batteries have three cells connected in series, each with a theoretical voltage of 1.48 V.) After some time, the voltage will drop and the glow of the light bulb will weaken. Based on the voltmeter readings, we can judge how long the battery needs to rest.
A special place is occupied by regenerating elements known as batteries. They undergo reversible reactions and can be recharged after the cell has been discharged by connecting to an external DC source.
Currently, lead-acid batteries are the most common; The electrolyte in them is dilute sulfuric acid, into which two lead plates are immersed. The positive electrode is coated with lead dioxide PbO 2, the negative is metallic lead. The voltage at the terminals is approximately 2.1 V. When discharging, lead sulfate is formed on both plates, which again turns into metallic lead and lead peroxide when charging.
APPLICATION OF GALVANIC COATINGS
Precipitation of metals from aqueous solutions using electric current is the reverse process of electrolytic dissolution, which we became familiar with when considering galvanic cells. First of all, we will examine copper deposition, which is used in a copper coulometer to measure the amount of electricity.Metal is deposited by current
Having bent the ends of two thin sheet copper plates, we hang them on opposite walls of a beaker or, better yet, a small glass aquarium. We attach the wires to the plates with terminals.
Electrolyte Let's prepare according to the following recipe: 125 g of crystalline copper sulfate, 50 g of concentrated sulfuric acid and 50 g of alcohol (denatured alcohol), the rest is water up to 1 liter. To do this, first dissolve copper sulfate in 500 ml of water, then carefully add in small portions sulfuric acid (Heating! Liquid may splash!), then add alcohol and add water to a volume of 1 liter.
Fill the coulometer with the prepared solution and connect a variable resistance, an ammeter and a lead battery to the circuit. Using resistance, we adjust the current so that its density is 0.02-0.01 A/cm 2 of the electrode surface. If the copper plate has an area of 50 cm2, then the current should be in the range of 0.5-1 A.
After some time, light red metallic copper will begin to precipitate at the cathode (negative electrode), and copper will go into solution at the anode (positive electrode). To clean the copper plates, we will pass current through the coulometer for about half an hour. Then we take out the cathode, dry it carefully with filter paper and weigh it accurately. Let's install an electrode in the cell, close the circuit using a rheostat and maintain a constant current, for example 1 A. After an hour, open the circuit and weigh the dried cathode again. At a current of 1 A, its mass will increase by 1.18 g per hour of operation.
Therefore, an amount of electricity equal to 1 ampere hour passing through a solution can release 1.18 g of copper. Or in general: the amount of substance released is directly proportional to the amount of electricity passing through the solution.
To isolate 1 equivalent of an ion, it is necessary to pass through the solution an amount of electricity equal to the product of the electrode charge e and Avogadro's number N A:
e*N A = 1.6021 * 10 -19 * 6.0225 * 10 23 = 9.65 * 10 4 A * s * mol -1 This value is indicated by the symbol F and is named after the discoverer of the quantitative laws of electrolysis Faraday number(exact value F- 96,498 A*s*mol -1). Therefore, to isolate from solution given number equivalents n e an amount of electricity should be passed through the solution equal to F*n e A*s*mol -1 . In other words,
I*t =F*n uh Here I- current, t- time of passage of current through the solution. In chapter " Titration Basics"It has already been shown that the number of equivalents of a substance n e is equal to the product of the number of moles and the equivalent number:
n e = n*Z Hence:
I*t = F*n*Z
In this case Z- ion charge (for Ag + Z= 1, for Cu 2+ Z= 2, for Al 3+ Z= 3, etc.). If we express the number of moles as the ratio of mass to molar mass ( n = m/M), then we get a formula that allows us to calculate all the processes occurring during electrolysis:
I*t =F*m*Z/M
Using this formula you can calculate the current:
I = F*m*Z/(t*M)= 9.65*10 4 *1.18*2 / (3600*63.54) A*s*g*mol/(s*mol*g) = 0.996 A
If we introduce the relation for electrical work W el
W el = U*I*t And W email/ U = I*t
Then, knowing the tension U, you can calculate:
W el = F*m*Z*U/M
It is also possible to calculate how long it takes for a certain amount of a substance to be electrolytically released, or how much of a substance will be released in a certain time. During the experiment, the current density must be maintained within specified limits. If it is less than 0.01 A/cm2, then too little metal will be released, since copper(I) ions will be partially formed. If the current density is too high, the adhesion of the coating to the electrode will be weak and when the electrode is removed from the solution, it may crumble.
In practice, galvanic coatings on metals are used primarily to protect against corrosion and to obtain a mirror-like shine.
In addition, metals, especially copper and lead, are purified by anodic dissolution and subsequent separation at the cathode (electrolytic refining).
To plate iron with copper or nickel, you must first thoroughly clean the surface of the object. To do this, polish it with washed chalk and successively degrease it with a diluted solution of caustic soda, water and alcohol. If the item is covered with rust, you need to pickle it in advance in a 10-15% solution of sulfuric acid.
We hang the cleaned product in an electrolytic bath (a small aquarium or a beaker), where it will serve as a cathode.
The solution for applying copper plating contains 250 g of copper sulfate and 80-100 g of concentrated sulfuric acid in 1 liter of water (Caution!). In this case, the copper plate will serve as the anode. The surface of the anode should be approximately equal to the surface of the object being coated. Therefore, you must always ensure that the copper anode hangs in the bath at the same depth as the cathode.
The process will be carried out at a voltage of 3-4 V (two batteries) and a current density of 0.02-0.4 A/cm 2 . The temperature of the solution in the bath should be 18-25 °C.
Let us pay attention to the fact that the anode plane and the surface to be coated are parallel to each other. Items complex shape It's better not to use it. By varying the duration of electrolysis, it is possible to obtain copper coatings of different thicknesses.
Often they resort to preliminary copper plating in order to apply a durable coating of another metal to this layer. This is especially often used when chrome plating iron, nickel plating zinc casting and in other cases. True, very poisonous cyanide electrolytes are used for this purpose.
To prepare an electrolyte for nickel plating, dissolve 25 g of crystalline nickel sulfate, 10 g of boric acid or 10 g of sodium citrate in 450 ml of water. You can prepare sodium citrate yourself by neutralizing a solution of 10 g of citric acid with a dilute solution of sodium hydroxide or soda solution. Let the anode be a nickel plate, perhaps larger area, and take the battery as a voltage source.
Using a variable resistance, we will maintain the current density at 0.005 A/cm 2 . For example, with an object surface of 20 cm 2, you need to work at a current strength of 0.1 A. After half an hour of work, the object will already be nickel-plated. Let's take it out of the bath and wipe it with a cloth. However, it is better not to interrupt the nickel plating process, since then the nickel layer may become passivated and the subsequent nickel coating will not adhere well.
To achieve a mirror shine without mechanical polishing, we introduce a so-called shine-forming additive into the galvanic bath. Such additives include, for example, glue, gelatin, sugar. You can add, for example, a few grams of sugar to a nickel bath and study its effect.
To prepare an electrolyte for chrome plating of iron (after preliminary copper plating), dissolve 40 g of chromic anhydride CrO 3 (Caution! Poison!) and exactly 0.5 g of sulfuric acid (in no case more!) in 100 ml of water. The process occurs at a current density of about 0.1 A/cm 2, and a lead plate is used as an anode, the area of which should be slightly less than the area of the chrome-plated surface.
Nickel and chrome baths are best heated slightly (to about 35 ° C). Please note that electrolytes for chrome plating, especially during a long process and high strength current, emit fumes containing chromic acid, which are very harmful to health. Therefore, chrome plating should be carried out under traction or in the open air, for example on a balcony.
When chrome plating (and to a lesser extent, nickel plating), not all of the current is used for metal deposition. At the same time, hydrogen is released. Based on a number of voltages, it would be expected that metals in front of hydrogen should not be released from aqueous solutions at all, but on the contrary, less active hydrogen should be released. However, here, as with the anodic dissolution of metals, the cathodic evolution of hydrogen is often inhibited and is observed only at high voltage. This phenomenon is called hydrogen overvoltage, and it is especially large, for example, on lead. Thanks to this circumstance, a lead-acid battery can function. When charging a battery, instead of PbO 2, hydrogen should appear at the cathode, but, due to overvoltage, the evolution of hydrogen begins when the battery is almost fully charged.
All metals, depending on their redox activity, are combined into a series called the electrochemical metal voltage series (since the metals in it are arranged in order of increasing standard electrochemical potentials) or the metal activity series:
Li, K, Ba, Ca, Na, Mg, Al, Zn, Fe, Ni, Sn, Pb, H2, Cu, Hg, Ag, Pt, Au
The most chemically active metals are in the activity series up to hydrogen, and the more to the left the metal is located, the more active it is. Metals occupying the place after hydrogen in the activity series are considered inactive.
Aluminum
Aluminum is a silvery-white color. Basic physical properties aluminum – lightness, high thermal and electrical conductivity. In the free state, when exposed to air, aluminum is covered with a durable film of Al 2 O 3 oxide, which makes it resistant to the action of concentrated acids.
Aluminum belongs to the p-family metals. Electronic configuration of external energy level– 3s 2 3p 1 . In its compounds, aluminum exhibits an oxidation state of “+3”.
Aluminum is produced by electrolysis of the molten oxide of this element:
2Al 2 O 3 = 4Al + 3O 2
However, due to the low yield of the product, the method of producing aluminum by electrolysis of a mixture of Na 3 and Al 2 O 3 is more often used. The reaction occurs when heated to 960C and in the presence of catalysts - fluorides (AlF 3, CaF 2, etc.), while the release of aluminum occurs at the cathode, and oxygen is released at the anode.
Aluminum is able to interact with water after removing the oxide film from its surface (1), interact with simple substances (oxygen, halogens, nitrogen, sulfur, carbon) (2-6), acids (7) and bases (8):
2Al + 6H 2 O = 2Al(OH) 3 + 3H 2 (1)
2Al +3/2O 2 = Al 2 O 3 (2)
2Al + 3Cl 2 = 2AlCl 3 (3)
2Al + N 2 = 2AlN (4)
2Al +3S = Al 2 S 3 (5)
4Al + 3C = Al 4 C 3 (6)
2Al + 3H 2 SO 4 = Al 2 (SO 4) 3 + 3H 2 (7)
2Al +2NaOH +3H 2 O = 2Na + 3H 2 (8)
Calcium
In its free form, Ca is a silvery-white metal. When exposed to air, it instantly becomes covered with a yellowish film, which is the product of its interaction with components air. Calcium is a fairly hard metal and has a face-centered cubic crystal lattice.
The electronic configuration of the outer energy level is 4s 2. In its compounds, calcium exhibits an oxidation state of “+2”.
Calcium is obtained by electrolysis of molten salts, most often chlorides:
CaCl 2 = Ca + Cl 2
Calcium is capable of dissolving in water to form hydroxides, exhibiting strong basic properties (1), reacting with oxygen (2), forming oxides, interacting with non-metals (3-8), dissolving in acids (9):
Ca + H 2 O = Ca(OH) 2 + H 2 (1)
2Ca + O 2 = 2CaO (2)
Ca + Br 2 = CaBr 2 (3)
3Ca + N2 = Ca3N2 (4)
2Ca + 2C = Ca 2 C 2 (5)
2Ca + 2P = Ca 3 P 2 (7)
Ca + H 2 = CaH 2 (8)
Ca + 2HCl = CaCl 2 + H 2 (9)
Iron and its compounds
Iron is a gray metal. In its pure form it is quite soft, malleable and viscous. The electronic configuration of the outer energy level is 3d 6 4s 2. In its compounds, iron exhibits oxidation states “+2” and “+3”.
Metallic iron reacts with water vapor, forming mixed oxide (II, III) Fe 3 O 4:
3Fe + 4H 2 O (v) ↔ Fe 3 O 4 + 4H 2
In air, iron easily oxidizes, especially in the presence of moisture (rusts):
3Fe + 3O 2 + 6H 2 O = 4Fe(OH) 3
Like other metals, iron reacts with simple substances, for example, halogens (1), and dissolves in acids (2):
Fe + 2HCl = FeCl 2 + H 2 (2)
Iron forms a whole spectrum of compounds, since it exhibits several oxidation states: iron (II) hydroxide, iron (III) hydroxide, salts, oxides, etc. Thus, iron (II) hydroxide can be obtained by the action of alkali solutions on iron (II) salts without access to air:
FeSO 4 + 2NaOH = Fe(OH) 2 ↓ + Na 2 SO 4
Iron(II) hydroxide is soluble in acids and oxidizes to iron(III) hydroxide in the presence of oxygen.
Iron (II) salts exhibit reducing agent properties and are converted into iron (III) compounds.
Iron (III) oxide cannot be obtained by the combustion of iron in oxygen; to obtain it, it is necessary to burn iron sulfides or calcinate other iron salts:
4FeS 2 + 11O 2 = 2Fe 2 O 3 +8SO 2
2FeSO 4 = Fe 2 O 3 + SO 2 + 3H 2 O
Iron (III) compounds exhibit weak oxidizing properties and are capable of entering into redox reactions with strong reducing agents:
2FeCl 3 + H 2 S = Fe(OH) 3 ↓ + 3NaCl
Iron and steel production
Steels and cast irons are alloys of iron and carbon, with the carbon content in steel up to 2%, and in cast iron 2-4%. Steels and cast irons contain alloying additives: steels – Cr, V, Ni, and cast iron – Si.
Highlight Various types Steels, for example, are divided into structural, stainless, tool, heat-resistant and cryogenic steels according to their intended purpose. By chemical composition carbon (low-, medium- and high-carbon) and alloyed (low-, medium- and high-alloy) are distinguished. Depending on the structure, austenitic, ferritic, martensitic, pearlitic and bainitic steels are distinguished.
Steels have found application in many industries National economy, such as construction, chemical, petrochemical, security environment, transport energy and other industries.
Depending on the form of carbon content in cast iron - cementite or graphite, as well as their quantity, several types of cast iron are distinguished: white (light color of the fracture due to the presence of carbon in the form of cementite), gray (gray color of the fracture due to the presence of carbon in the form of graphite ), malleable and heat resistant. Cast irons are very brittle alloys.
The areas of application of cast iron are extensive - artistic decorations (fences, gates), cabinet parts, plumbing equipment, household items (frying pans) are made from cast iron, and it is used in the automotive industry.
Examples of problem solving
EXAMPLE 1
| Exercise | An alloy of magnesium and aluminum weighing 26.31 g was dissolved in hydrochloric acid. In this case, 31.024 liters of colorless gas were released. Determine the mass fractions of metals in the alloy. |
| Solution | Both metals are capable of reacting with hydrochloric acid, resulting in the release of hydrogen: Mg +2HCl = MgCl 2 + H 2 2Al +6HCl = 2AlCl3 + 3H2 Let's find the total number of moles of hydrogen released: v(H 2) =V(H 2)/V m v(H 2) = 31.024/22.4 = 1.385 mol Let the amount of substance Mg be x mol, and Al be y mol. Then, based on the reaction equations, we can write the expression for the total number of moles of hydrogen: x + 1.5y = 1.385 Let us express the mass of metals in the mixture: Then, the mass of the mixture will be expressed by the equation: 24x + 27y = 26.31 We received a system of equations: x + 1.5y = 1.385 24x + 27y = 26.31 Let's solve it: 33.24 -36y+27y = 26.31 v(Al) = 0.77 mol v(Mg) = 0.23 mol Then, the mass of metals in the mixture is: m(Mg) = 24×0.23 = 5.52 g m(Al) = 27×0.77 = 20.79 g Let's find the mass fractions of metals in the mixture: ώ =m(Me)/m sum ×100% ώ(Mg) = 5.52/26.31 ×100%= 20.98% ώ(Al) = 100 – 20.98 = 79.02% |
| Answer | Mass fractions metals in alloy: 20.98%, 79.02% |
What information can be obtained from a series of voltages?
A range of metal voltages are widely used in inorganic chemistry. In particular, the results of many reactions and even the possibility of their implementation depend on the position of a certain metal in the NER. Let's discuss this issue in more detail.
Interaction of metals with acids
Metals located in the voltage series to the left of hydrogen react with acids - non-oxidizing agents. Metals located in the NER to the right of H interact only with oxidizing acids (in particular, with HNO 3 and concentrated H 2 SO 4).
Example 1. Zinc is located in the NER to the left of hydrogen, therefore, it is able to react with almost all acids:
Zn + 2HCl = ZnCl 2 + H 2
Zn + H 2 SO 4 = ZnSO 4 + H 2
Example 2. Copper is located in the ERN to the right of H; this metal does not react with “ordinary” acids (HCl, H 3 PO 4, HBr, organic acids), but it interacts with oxidizing acids (nitric, concentrated sulfuric):
Cu + 4HNO 3 (conc.) = Cu(NO 3) 2 + 2NO 2 + 2H 2 O
Cu + 2H 2 SO 4 (conc.) = CuSO 4 + SO 2 + 2H 2 O
I would like to draw your attention to an important point: when metals interact with oxidizing acids, it is not hydrogen that is released, but some other compounds. You can read more about this!
Interaction of metals with water
Metals located in the voltage series to the left of Mg easily react with water already at room temperature with the release of hydrogen and the formation of an alkali solution.
Example 3. Sodium, potassium, calcium easily dissolve in water to form an alkali solution:
2Na + 2H 2 O = 2NaOH + H 2
2K + 2H 2 O = 2KOH + H 2
Ca + 2H 2 O = Ca(OH) 2 + H 2
Metals located in the voltage range from hydrogen to magnesium (inclusive) in some cases interact with water, but the reactions require specific conditions. For example, aluminum and magnesium begin to interact with H 2 O only after removing the oxide film from the metal surface. Iron does not react with water at room temperature, but does react with water vapor. Cobalt, nickel, tin, and lead practically do not interact with H 2 O, not only at room temperature, but also when heated.
The metals located on the right side of the ERN (silver, gold, platinum) do not react with water under any conditions.
Interaction of metals with aqueous solutions of salts
We will talk about reactions of the following type:
metal (*) + metal salt (**) = metal (**) + metal salt (*)
I would like to emphasize that the asterisks in this case do not indicate the oxidation state or the valency of the metal, but simply allow one to distinguish between metal No. 1 and metal No. 2.
To carry out such a reaction, three conditions must be met simultaneously:
- the salts involved in the process must be dissolved in water (this can be easily checked using the solubility table);
- the metal (*) must be in the stress series to the left of the metal (**);
- the metal (*) should not react with water (which is also easily verified by ESI).
Example 4. Let's look at a few reactions:
Zn + CuSO 4 = ZnSO 4 + Cu
K + Ni(NO 3) 2 ≠
The first reaction is easily feasible, all the above conditions are met: copper sulfate is soluble in water, zinc is in the NER to the left of copper, Zn does not react with water.
The second reaction is impossible because the first condition is not met (copper (II) sulfide is practically insoluble in water). The third reaction is not feasible, since lead is a less active metal than iron (located to the right in the ESR). Finally, the fourth process will NOT result in nickel precipitation because potassium reacts with water; the resulting potassium hydroxide can react with the salt solution, but this is a completely different process.
Thermal decomposition process of nitrates
Let me remind you that nitrates are salts of nitric acid. All nitrates decompose when heated, but the composition of the decomposition products may vary. The composition is determined by the position of the metal in the stress series.
Nitrates of metals located in the NER to the left of magnesium, when heated, form the corresponding nitrite and oxygen:
2KNO 3 = 2KNO 2 + O 2
During the thermal decomposition of metal nitrates located in the voltage range from Mg to Cu inclusive, metal oxide, NO 2 and oxygen are formed:
2Cu(NO 3) 2 = 2CuO + 4NO 2 + O 2
Finally, during the decomposition of nitrates of the least active metals (located in the ERN to the right of copper), metal, nitrogen dioxide and oxygen are formed.
Electrochemical activity series of metals (voltage range, range of standard electrode potentials) - sequence in which metals are arranged in order of increasing their standard electrochemical potentials φ 0, corresponding to the half-reaction of reduction of the metal cation Me n+: Me n+ + nē → Me
A number of voltages characterize the comparative activity of metals in redox reactions in aqueous solutions.
Story
The sequence of metals in the order of changes in their chemical activity in general outline was already known to alchemists. The processes of mutual displacement of metals from solutions and their surface deposition (for example, the displacement of silver and copper from solutions of their salts by iron) were considered as a manifestation of the transmutation of elements.
Later alchemists came close to understanding the chemical side of the mutual precipitation of metals from their solutions. Thus, Angelus Sala in his work “Anatomia Vitrioli” (1613) came to the conclusion that products chemical reactions consist of the same “components” that were contained in the original substances. Subsequently, Robert Boyle proposed a hypothesis about the reasons why one metal displaces another from solution based on corpuscular concepts.
In the era of the emergence of classical chemistry, the ability of elements to displace each other from compounds became an important aspect of understanding reactivity. J. Berzelius, based on the electrochemical theory of affinity, constructed a classification of elements, dividing them into “metalloids” (the term “non-metals” is now used) and “metals” and placing hydrogen between them.
The sequence of metals according to their ability to displace each other, long known to chemists, was especially thoroughly and comprehensively studied and supplemented by N. N. Beketov in the 1860s and subsequent years. Already in 1859, he made a report in Paris on the topic “Investigation of the phenomena of the displacement of some elements by others.” Beketov included in this work whole line generalizations about the relationship between the mutual displacement of elements and their atomic weight, linking these processes with “ initial chemical properties elements - what is called chemical affinity". Beketov's discovery of the displacement of metals from solutions of their salts by hydrogen under pressure and the study of the reducing activity of aluminum, magnesium and zinc at high temperatures(metallothermy) allowed him to put forward a hypothesis about the connection between the ability of some elements to displace others from compounds with their density: lighter simple substances capable of displacing heavier ones (therefore this series is often also called Beketov's displacement series, or simply Beketov series).
Without denying Beketov’s significant merits in the formation modern ideas about a number of activities of metals, it should be considered erroneous in the domestic popular and educational literature the idea of him as the sole creator of this series. Numerous experimental data obtained in late XIX centuries, refuted Beketov’s hypothesis. Thus, William Odling described many cases of “reversal of activity.” For example, copper displaces tin from a concentrated acidified solution of SnCl 2 and lead from an acidic solution of PbCl 2 ; it is also capable of dissolving in concentrated hydrochloric acid with the release of hydrogen. Copper, tin and lead are in the series to the right of cadmium, but can displace it from a boiling slightly acidified solution of CdCl 2.
The rapid development of theoretical and experimental physical chemistry pointed to another reason for the differences in the chemical activity of metals. With the development of modern concepts of electrochemistry (mainly in the works of Walter Nernst), it became clear that this sequence corresponds to the “series of voltages” - the arrangement of metals according to the value of standard electrode potentials. Thus, instead of a qualitative characteristic - the “propensity” of a metal and its ion to certain reactions - Nerst introduced an exact quantitative value characterizing the ability of each metal to go into solution in the form of ions, as well as to be reduced from ions to the metal on the electrode, and the corresponding series got the name range of standard electrode potentials.
Theoretical basis
The values of electrochemical potentials are a function of many variables and therefore exhibit a complex dependence on the position of metals in the periodic table. Thus, the oxidation potential of cations increases with an increase in the atomization energy of the metal, with an increase in the total ionization potential of its atoms, and with a decrease in the hydration energy of its cations.
In the very general view It is clear that metals located at the beginning of the periods are characterized by low values of electrochemical potentials and occupy places on the left side of the voltage series. In this case, the alternation of alkali and alkaline earth metals reflects the phenomenon of diagonal similarity. Metals located closer to the middle of the periods are characterized large values potentials and occupy places in the right half of the row. A consistent increase in the electrochemical potential (from −3.395 V for the Eu 2+ /Eu [ ] to +1.691 V for the Au + /Au pair) reflects a decrease in the reducing activity of metals (the ability to donate electrons) and an increase in the oxidizing ability of their cations (the ability to gain electrons). Thus, the strongest reducing agent is metallic europium, and the strongest oxidizing agent is the gold cations Au+.
Hydrogen is traditionally included in the voltage series, since practical measurement of electrochemical potentials of metals is made using a standard hydrogen electrode.
Practical use of a range of voltages
A number of voltages are used in practice for comparative [relative] assessment of the chemical activity of metals in reactions with aqueous solutions salts and acids and for assessing cathodic and anodic processes during electrolysis:
- Metals to the left of hydrogen are stronger reducing agents than metals to the right: they displace the latter from salt solutions. For example, the interaction Zn + Cu 2+ → Zn 2+ + Cu is possible only in the forward direction.
- Metals in the row to the left of hydrogen displace hydrogen when interacting with aqueous solutions of non-oxidizing acids; the most active metals (up to and including aluminum) - and when interacting with water.
- Metals in the series to the right of hydrogen do not interact with aqueous solutions of non-oxidizing acids under normal conditions.
- During electrolysis, metals to the right of hydrogen are released at the cathode; the reduction of moderately active metals is accompanied by the release of hydrogen; The most active metals (up to aluminum) cannot be isolated from aqueous salt solutions under normal conditions.
Table of electrochemical potentials of metals
| Metal | Cation | φ 0, V | Reactivity | Electrolysis (at the cathode): |
|---|---|---|---|---|
| Li+ | -3,0401 | reacts with water | hydrogen is released | |
| Cs+ | -3,026 | |||
| Rb+ | -2,98 | |||
| K+ | -2,931 | |||
| Fr+ | -2,92 | |||
| Ra 2+ | -2,912 | |||
| Ba 2+ | -2,905 | |||
| Sr 2+ | -2,899 | |||
| Ca2+ | -2,868 | |||
| Eu 2+ | -2,812 | |||
| Na+ | -2,71 | |||
| Sm 2+ | -2,68 | |||
| Md 2+ | -2,40 | reacts with aqueous solutions of acids | ||
| La 3+ | -2,379 | |||
| Y 3+ | -2,372 | |||
| Mg 2+ | -2,372 | |||
| Ce 3+ | -2,336 | |||
| Pr 3+ | -2,353 | |||
| Nd 3+ | -2,323 | |||
| Er 3+ | -2,331 | |||
| Ho 3+ | -2,33 | |||
| Tm 3+ | -2,319 | |||
| Sm 3+ | -2,304 | |||
| PM 3+ | -2,30 | |||
| Fm 2+ | -2,30 | |||
| Dy 3+ | -2,295 | |||
| Lu 3+ | -2,28 | |||
| Tb 3+ | -2,28 | |||
| Gd 3+ | -2,279 | |||
| Es 2+ | -2,23 | |||
| Ac 3+ | -2,20 | |||
| Dy 2+ | -2,2 | |||
| PM 2+ | -2,2 | |||
| Cf 2+ | -2,12 | |||
| Sc 3+ | -2,077 | |||
| Am 3+ | -2,048 | |||
| Cm 3+ | -2,04 | |||
| Pu 3+ | -2,031 | |||
| Er 2+ | -2,0 | |||
| Pr 2+ | -2,0 | |||
| Eu 3+ | -1,991 | |||
| Lr 3+ | -1,96 | |||
| Cf 3+ | -1,94 | |||
| Es 3+ | -1,91 | |||
| Th 4+ | -1,899 | |||
| Fm 3+ | -1,89 | |||
| Np 3+ | -1,856 | |||
| Be 2+ | -1,847 | |||
| U 3+ | -1,798 | |||
| Al 3+ | -1,700 | |||
| MD 3+ | -1,65 | |||
| Ti 2+ | -1,63 | competing reactions: both the release of hydrogen and the release of pure metal | ||
| Hf 4+ | -1,55 | |||
| Zr 4+ | -1,53 | |||
| Pa 3+ | -1,34 | |||
| Ti 3+ | -1,208 | |||
| Yb 3+ | -1,205 | |||
| No 3+ | -1,20 | |||
| Ti 4+ | -1,19 | |||
| Mn 2+ | -1,185 | |||
| V 2+ | -1,175 | |||
| Nb 3+ | -1,1 | |||
| Nb 5+ | -0,96 | |||
| V 3+ | -0,87 | |||
| Cr 2+ | -0,852 | |||
| Zn 2+ | -0,763 | |||
| Cr 3+ | -0,74 | |||
| Ga 3+ | -0,560 | |||