Reactions with active metals. active metals. Interaction of metals with aqueous solutions of salts
What information can be obtained from a series of voltages?
A number of metal stresses 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 some metal in the NRN. Let's discuss this issue in more detail.
The interaction of metals with acids
Metals that are in the series of voltages to the left of hydrogen react with acids - non-oxidizing agents. Metals located in the ERN to the right of H interact only with acids - oxidizing agents (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 \u003d ZnCl 2 + H 2
Zn + H 2 SO 4 \u003d 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), however, 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 draw attention to important point: when metals interact with oxidizing acids, not hydrogen 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 \u003d 2NaOH + H 2
2K + 2H 2 O = 2KOH + H 2
Ca + 2H 2 O \u003d Ca (OH) 2 + H 2
Metals located in the range of voltages 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 the removal of the oxide film from the metal surface. Iron does not react with water at room temperature, but interacts with water vapor. Cobalt, nickel, tin, 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 circumstances.
Interaction of metals with aqueous solutions of salts
We will talk about the following types of reactions:
metal (*) + metal salt (**) = metal (**) + metal salt (*)
I would like to emphasize that the asterisks in this case do not indicate the degree of oxidation, not the valence of the metal, but simply allow us to distinguish between metal No. 1 and metal No. 2.
For such a reaction to occur, three conditions must be met simultaneously:
- the salts involved in the process must be soluble in water (this is easy to check using the solubility table);
- metal (*) must be in a series of voltages to the left of metal (**);
- metal (*) should not react with water (which is also easily checked by ERN).
Example 4. Let's look at a few reactions:
Zn + CuSO 4 \u003d ZnSO 4 + Cu
K + Ni(NO 3) 2 ≠
The first reaction is easy to implement, all of the above conditions are met: copper sulfate is soluble in water, zinc is in the ERN 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 NRN). Finally, the fourth process will NOT result in nickel precipitation as potassium reacts with water; the resulting potassium hydroxide can react with a salt solution, but this is a completely different process.
The process of thermal decomposition 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 be different. The composition is determined by the position of the metal in the series of stresses.
Nitrates of metals located in the NER to the left of magnesium, when heated, form the corresponding nitrite and oxygen:
2KNO 3 \u003d 2KNO 2 + O 2
During the thermal decomposition of metal nitrates, located in a series of voltages from Mg to Cu inclusive, metal oxide, NO 2 and oxygen are formed:
2Cu(NO 3) 2 \u003d 2CuO + 4NO 2 + O 2
Finally, during the decomposition of nitrates of the least active metals (located in the NER to the right of copper), metal, nitrogen dioxide and oxygen are formed.
Metals that react easily are called active metals. These include alkali, alkaline earth metals and aluminium.
Position in the periodic table
The metallic properties of the elements weaken from left to right in Mendeleev's periodic table. Therefore, elements of groups I and II are considered the most active.
Rice. 1. Active metals in the periodic table.
All metals are reducing agents and easily part with electrons at the external energy level. Active metals have only one or two valence electrons. In this case, the metallic properties are enhanced from top to bottom with an increase in the number of energy levels, because. the farther an electron is from the nucleus of an atom, the easier it is for it to separate.
Alkali metals are considered the most active:
- lithium;
- sodium;
- potassium;
- rubidium;
- cesium;
- francium.
The alkaline earth metals are:
- beryllium;
- magnesium;
- calcium;
- strontium;
- barium;
- radium.
You can find out the degree of activity of a metal by the electrochemical series of metal voltages. The more to the left of hydrogen an element is located, the more active it is. The metals to the right of hydrogen are inactive and can only interact with concentrated acids.
Rice. 2. Electrochemical series of voltages of metals.
The list of active metals in chemistry also includes aluminum, located in group III and to the left of hydrogen. However, aluminum is located on the border of active and medium active metals and does not react with certain substances under normal conditions.
Properties
Active metals are soft (can be cut with a knife), light, and have a low melting point.
Main Chemical properties metals are presented in the table.
|
Reaction |
The equation |
Exception |
|
Alkali metals ignite spontaneously in air, interacting with oxygen |
K + O 2 → KO 2 |
Lithium reacts with oxygen only at high temperatures. |
|
Alkaline earth metals and aluminum form oxide films in air, and spontaneously ignite when heated. |
2Ca + O 2 → 2CaO |
|
|
React with simple substances to form salts |
Ca + Br 2 → CaBr 2; |
Aluminum does not react with hydrogen |
|
React violently with water, forming alkalis and hydrogen |
|
The reaction with lithium proceeds slowly. Aluminum reacts with water only after the removal of the oxide film. |
|
React with acids to form salts |
Ca + 2HCl → CaCl 2 + H 2; 2K + 2HMnO 4 → 2KMnO 4 + H 2 |
|
|
React with salt solutions, first reacting with water and then with salt |
2Na + CuCl 2 + 2H 2 O: 2Na + 2H 2 O → 2NaOH + H 2; |
Active metals easily react, therefore, in nature they are found only in mixtures - minerals, rocks.
Rice. 3. Minerals and pure metals.
What have we learned?
Active metals include elements of groups I and II - alkali and alkaline earth metals, as well as aluminum. Their activity is due to the structure of the atom - a few electrons are easily separated from the external energy level. These are soft light metals that quickly react with simple and complex substances, forming oxides, hydroxides, salts. Aluminum is closer to hydrogen and its reaction with substances requires additional conditions - high temperatures, destruction of the oxide film.
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If, from the whole series of standard electrode potentials, we single out only those electrode processes that correspond to the general equation
then we get a series of stresses of metals. In addition to metals, hydrogen is always included in this series, which makes it possible to see which metals are capable of displacing hydrogen from aqueous solutions of acids.
Table 19
A number of stresses for the most important metals are given in Table. 19. The position of a metal in a series of voltages characterizes its ability to redox interactions in aqueous solutions under standard conditions. Metal ions are oxidizing agents, and metals in the form of simple substances are reducing agents. At the same time, the further the metal is located in the series of voltages, the stronger the oxidizing agent in an aqueous solution are its ions, and vice versa, the closer the metal is to the beginning of the series, the stronger the reducing properties are exhibited by a simple substance - metal.
Electrode Process Potential
in a neutral medium it is B (see page 273). Active metals at the beginning of the series, having a potential much more negative than -0.41 V, displace hydrogen from water. Magnesium displaces hydrogen only from hot water. Metals located between magnesium and cadmium usually do not displace hydrogen from water. On the surface of these metals, oxide films are formed that have a protective effect.
Metals located between magnesium and hydrogen displace hydrogen from acid solutions. At the same time, protective films are also formed on the surface of some metals, which inhibit the reaction. So, the oxide film on aluminum makes this metal resistant not only in water, but also in solutions of certain acids. Lead does not dissolve in sulfuric acid at its concentration below , since the salt formed during the interaction of lead with sulfuric acid is insoluble and creates a protective film on the metal surface. The phenomenon of deep inhibition of metal oxidation, due to the presence of protective oxide or salt films on its surface, is called passivity, and the state of the metal in this case is called the passive state.
Metals are able to displace each other from salt solutions. The direction of the reaction is determined in this case by their mutual position in the series of voltages. Considering specific cases of such reactions, it should be remembered that active metals displace hydrogen not only from water, but also from any aqueous solution. Therefore, the mutual displacement of metals from solutions of their salts practically occurs only in the case of metals located in the row after magnesium.
The displacement of metals from their compounds by other metals was first studied in detail by Beketov. As a result of his work, he arranged the metals according to their chemical activity in a displacement series, which is the prototype of a series of metal stresses.
The mutual position of some metals in the series of voltages and in the periodic system at first glance does not correspond to each other. For example, according to the position in the periodic system, the reactivity of potassium must be greater than sodium, and sodium must be greater than lithium. In the series of voltages, lithium is the most active, and potassium occupies a middle position between lithium and sodium. Zinc and copper, according to their position in the periodic system, should have approximately equal chemical activity, but in the series of voltages, zinc is located much earlier than copper. The reason for this kind of inconsistency is as follows.
When comparing metals occupying a particular position in the periodic system, the measure of their chemical activity - reducing ability - is taken as the value of the ionization energy of free atoms. Indeed, during the transition, for example, from top to bottom along the main subgroup of group I of the periodic system, the ionization energy of atoms decreases, which is associated with an increase in their radii (i.e., with a large distance of external electrons from the nucleus) and with increasing screening of the positive charge of the nucleus by intermediate electron layers (see § 31). Therefore, potassium atoms exhibit greater chemical activity - they have stronger reducing properties - than sodium atoms, and sodium atoms are more active than lithium atoms.
When comparing metals in a series of voltages, the measure of chemical activity is taken as the work of converting a metal in a solid state into hydrated ions in an aqueous solution. This work can be represented as the sum of three terms: the energy of atomization - the transformation of a metal crystal into isolated atoms, the ionization energy of free metal atoms and the hydration energy of the formed ions. The atomization energy characterizes the strength of the crystal lattice of a given metal. The ionization energy of atoms - the detachment of valence electrons from them - is directly determined by the position of the metal in the periodic system. The energy released during hydration depends on the electronic structure of the ion, its charge and radius.
Lithium and potassium ions, having the same charge but different radii, will create unequal electric fields around them. The field generated near small lithium ions will be stronger than the field near large potassium ions. From this it is clear that lithium ions will hydrate with the release of more energy than potassium nones.
Thus, in the course of the transformation under consideration, energy is spent on atomization and ionization, and energy is released during hydration. The lower the total energy consumption, the easier the whole process will be and the closer to the beginning of the series of voltages the given metal will be located. But of the three terms of the total energy balance, only one - the ionization energy - is directly determined by the position of the metal in the periodic system. Consequently, there is no reason to expect that the mutual position of certain metals in a series of voltages will always correspond to their position in the periodic system. So, for lithium, the total energy consumption is less than for potassium, in accordance with which lithium is in the series of voltages before potassium.
For copper and zinc, the expenditure of energy for the ionization of free atoms and its gain during hydration of the ions are close. But metallic copper forms a stronger crystal lattice than zinc, which can be seen from a comparison of the melting points of these metals: zinc melts at , and copper only at . Therefore, the energy spent on the atomization of these metals is significantly different, as a result of which the total energy costs for the entire process in the case of copper are much greater than in the case of zinc, which explains the relative position of these metals in the voltage series.
When passing from water to non-aqueous solvents, the mutual position of metals in a series of voltages can change. The reason for this lies in the fact that the energy of solvation of ions of various metals varies in different ways when passing from one solvent to another.
In particular, the copper ion is very vigorously solvated in some organic solvents; this leads to the fact that in such solvents copper is located in a series of voltages up to hydrogen and displaces it from acid solutions.
Thus, in contrast to the periodic system of elements, a series of stresses in metals is not a reflection of the general Regularity, on the basis of which it is possible to give a versatile Characteristic of the chemical properties of metals. A series of voltages Characterizes only the redox ability Electrochemical system"metal - metal ion" under strictly defined conditions: the values \u200b\u200bgiven in it refer to an aqueous solution, temperature and a unit concentration (activity) of metal ions.
Grosse E., Weissmantel X.
Chemistry for the Curious. Fundamentals of chemistry and entertaining experiments.
Chapter 3 (continued)A SMALL COURSE OF 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 on a few simple experiments to study some of the laws of electrochemistry of aqueous solutions, galvanic cells, and also 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 metal objects, which causes great damage to the national economy, in most cases is an electrochemical process.
VOLTAGE RANGE OF METALS
The fundamental link for understanding electrochemical processes is the voltage series of metals. Metals can be arranged in a row that starts with reactive and ends with the least reactive 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 how, 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 of a row, then a negative voltage will appear on the material preceding in 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 a few simple experiments, for which we need a current source and electrical measuring instruments. Let's 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 to a shine with a thin emery cloth.) After a short time, the iron will be covered with a reddish layer of released copper. The more active iron displaces the copper from the solution, with the iron dissolving as ions and the copper liberated as a metal. The process continues as long as the solution is in contact with the iron. As soon as the copper covers the entire surface of the iron, it will practically stop. In this case, a rather porous copper layer is formed, so that protective coatings cannot be obtained without the use of current.
In the following experiments, we will lower small strips of zinc and lead tin into the copper sulfate solution. After 15 minutes, take them out, rinse and examine under a microscope. We can see beautiful, ice-like patterns that are red in reflected light and consist of liberated copper. Here, too, more active metals transferred copper from the ionic to the metallic state.
In turn, copper can displace metals that are lower in the series of voltages, that is, less active. On a thin strip of sheet copper or on a flattened copper wire (having previously cleaned the surface to a shine), we apply a few drops of a solution of silver nitrate. With the naked eye, it will be possible to notice the formed blackish coating, which under a microscope in reflected light looks like thin needles and plant patterns (the so-called dendrites).
To isolate zinc without current, it is necessary to use a more active metal. Excluding metals that violently interact with water, we find magnesium in the series of stresses above zinc. We place a few drops of zinc sulfate solution on a piece of magnesium tape or on a thin chip of an electron. We obtain a solution of zinc sulfate by dissolving a piece of zinc in dilute sulfuric acid. Simultaneously with zinc sulfate, add a few drops of denatured alcohol. On magnesium, after a short period of time, we notice, especially under a microscope, zinc that has separated out in the form of thin crystals.
In general, any member of the voltage series can be forced out of solution, where it is in the form of an ion, and transferred 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 stand out 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 this case is due to a thin oxide film on the aluminum surface. In such solutions, aluminum is called passive.
LET'S LOOK BEYOND THE SCENE
In order to formulate the patterns of the ongoing processes, we can restrict ourselves to considering cations, and exclude anions, since they themselves do not participate in the reaction. (However, the type of anions affects the rate of deposition.) If, for simplicity, we assume that both the liberated and dissolved metals give 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 = Сu.
So, the process consists in 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 get:
Fe = Fe 2+ + 2 e --
Сu 2+ + 2 e--=Cu
Now consider the case when the metal is immersed in water or in a salt solution, with the cation of which the exchange is impossible due to its position in the series of voltages. 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 is charged negatively with respect 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, then they will both be charged, but the less active one is somewhat weaker, due to the fact that its atoms are less prone to splitting off electrons.
Connect both metals with a conductor. Due to the potential difference, the 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 takes place in which the more active metal goes into solution, and the cations from the solution are released on the more noble metal. Let us now illustrate with a few experiments the above somewhat abstract reasoning (which, moreover, is a gross simplification).
First, fill a beaker with a capacity of 250 ml to the middle with a 10% sulfuric acid solution and immerse not too small pieces of zinc and copper into it. We solder or rivet a 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. Both metals form an electrical double layer. The potential difference between the electrodes is easiest to detect with a voltmeter. Immediately after turning on the device in the circuit, the arrow will indicate approximately 1 V, but then the voltage will quickly drop. If you connect a small light bulb to the element that consumes a voltage of 1 V, then it will light up - at first quite strongly, and then the glow will become weak.
By the polarity of the terminals of the device, we can conclude that the copper electrode is a positive pole. This can be proved even without a device by considering the electrochemistry of the process. Prepare a saturated solution of common salt in a small beaker or in a test tube, add about 0.5 ml alcohol solution indicator of phenolphthalein and immerse both electrodes closed with a wire into the solution. Near the negative pole, a slight reddish coloration will be observed, which is caused by the formation of sodium hydroxide at the cathode.
In other experiments, one can place various pairs of metals in the cell and determine the resulting voltage. For example, magnesium and silver will give a particularly large potential difference due to the significant distance between them in a series of voltages, while zinc and iron, on the contrary, will give a very small one, less than a tenth of a volt. Using aluminum, we will not get practically any current due to passivation.
All these elements, or, as electrochemists say, circuits, have the disadvantage that when a current is taken, the voltage drops very quickly on them. Therefore, electrochemists always measure the true value of the voltage in a de-energized state using the voltage compensation method, that is, by comparing it with the voltage of another current source.
Let us consider the processes in the copper-zinc element in more detail. At the cathode, zinc goes into solution according to the following equation:
Zn = Zn2+ + 2 e --
Sulfuric acid hydrogen ions are discharged on 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-- \u003d H 2
After a short period of time, copper will be covered with a thin layer of hydrogen bubbles. In this case, the copper electrode will turn into a hydrogen electrode, 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 that, the voltage will increase again, since potassium dichromate will oxidize hydrogen to water. Potassium dichromate acts in this case as a depolarizer.
In practice, galvanic circuits are used, the electrodes of which are not polarized, or circuits, the polarization of which can be eliminated by adding depolarizers.
As an example of a non-polarizable element, consider the Daniell 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 beaker 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 of copper sheet.
This element can be made from a glass jar, a commercially available clay cell (in a pinch, use flower pot by closing the hole in the bottom) and two suitable electrodes.
During the operation of the element, zinc dissolves with the formation of zinc sulfate, and copper ions are released on the copper electrode. But at the same time, the copper electrode is not polarized and the element gives a voltage of about 1 V. Actually, theoretically, the voltage at the terminals is 1.10 V, but when taking the current, we measure a slightly lower value, due to the electrical resistance of the cell.
If we do not remove the current from the cell, we must remove the zinc electrode from the sulfuric acid solution, because otherwise it will dissolve to form hydrogen.
A diagram of a simple cell, which does not require a porous partition, is shown in the figure. The zinc electrode is located in glass jar at the top, and copper - near the bottom. The entire cell is filled with a saturated sodium chloride solution. At the bottom of the jar we pour a handful of copper sulfate crystals. The resulting concentrated solution of copper sulfate will mix with the common salt solution very slowly. Therefore, during the operation of the cell, copper will be released on the copper electrode, and zinc in the form of sulfate or chloride will dissolve in the upper part of the cell.
Batteries now use almost exclusively dry cells, which are more convenient to use. Their ancestor is the Leclanchet 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 coal. To avoid polarization, the carbon rod is lowered into a linen bag with a mixture of coal powder and pyrolusite. The carbon powder increases the surface of the electrode, 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 rapidly, they " get tired"due to polarization. Only after some time does the oxidation of hydrogen occur with pyrolusite. Thus, the elements" rest", if you do not pass current for some time. Let's check this on a flashlight battery, to which we connect a light bulb. Parallel to the lamp, that is, directly to the terminals, we connect a voltmeter.
At first, the voltage will be about 4.5 V. (Most often, three cells are connected in series in such batteries, each with a theoretical voltage of 1.48 V.) After a while, the voltage will drop, the light bulb will weaken. By reading the voltmeter, we can judge how long the battery needs to rest.
A special place is occupied by regenerating elements, known as accumulators. Reversible reactions take place in them, and they can be recharged after the cell is discharged by connecting to an external DC source.
Currently, lead-acid batteries are the most common; in them, the electrolyte is dilute sulfuric acid, into which two lead plates are immersed. The positive electrode is coated with lead dioxide PbO 2 , the negative electrode is metallic lead. The voltage at the terminals is approximately 2.1 V. During discharge, lead sulfate is formed on both plates, which again turns into metallic lead and into lead peroxide during charging.
PLATED COATINGS
Precipitation of metals from aqueous solutions using electric current is the reverse process of electrolytic dissolution, which we met when considering galvanic cells. First of all, let us examine the precipitation of copper, which is used in a copper coulometer to measure the amount of electricity.Metal is deposited by current
Having bent the ends of two plates of thin sheet copper, we hang them on opposite walls of a beaker or, better, a small glass aquarium. We attach the wires to the plates with terminals.
Electrolyte cook according to next 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, in small portions, add sulfuric acid ( The heating! Liquid may splash!), then pour in alcohol and bring water to a volume of 1 liter.
We fill the coulometer with the prepared solution and include a variable resistance, an ammeter and a lead battery in the circuit. With the help of 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 cm 2, then the current strength 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 a current in the coulometer for about half an hour. Then we take out the cathode, dry it carefully with filter paper and weigh it accurately. We install an electrode in the cell, close the circuit with a rheostat and maintain a constant current, for example 1 A. After an hour, we open the circuit and weigh the dried cathode again. At a current of 1 A per hour of operation, its mass will increase by 1.18 g.
Therefore, an amount of electricity equal to 1 ampere-hour, when 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 passed 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 the Avogadro number N A:
e*N A \u003d 1.6021 * 10 -19 * 6.0225 * 10 23 \u003d 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 a given number of equivalents from a solution n e through the solution, an amount of electricity equal to F*n e A * s * mol -1. In other words,
I*t =F*n e Here I- current, t is the time it takes for the current to pass 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 by the equivalent number:
n e = n*Z Consequently:
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 you to calculate all the processes that occur during electrolysis:
I*t =F*m*Z / M
Using this formula, you can calculate the current:
I = F*m*Z/(t*M)\u003d 9.65 * 10 4 * 1.18 * 2 / (3600 * 63.54) A * s * g * mol / (s * mol * g) \u003d 0.996 A
If we introduce a ratio for electrical work W email
W email = U*I*t and W email / U = I*t
Then knowing the tension U, you can calculate:
W email = F*m*Z*U/M
You can also calculate how long it takes for the electrolytic release of a certain amount of a substance, or how much of a substance will be released in a certain time. During the experiment, the current density must be maintained within the specified limits. If it is less than 0.01 A / cm 2, 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 finish.
In addition, metals, especially copper and lead, are refined 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 elutriated chalk and sequentially degrease it with a dilute solution of caustic soda, water and alcohol. If the object is covered with rust, it is necessary to pickle it in advance in a 10-15% sulfuric acid solution.
We hang the cleaned product in an electrolytic bath ( small aquarium or beaker) where it will serve as the 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, a copper plate will serve as the anode. The surface of the anode should be approximately equal to the surface of the coated object. 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.
Pay attention to the fact that the plane of the anode and the surface to be coated are parallel to each other. It is better not to use objects of complex shape. By varying the duration of electrolysis, it is possible to obtain a copper coating of different thicknesses.
Preliminary copper plating is often resorted to in order to apply a durable coating of another metal to this layer. This is especially often used in iron chromium plating, zinc casting nickel plating and in other cases. True, very toxic cyanide electrolytes are used for this purpose.
To prepare an electrolyte for nickel plating in 450 ml of water, dissolve 25 g of crystalline nickel sulfate, 10 g boric acid or 10 g sodium citrate. Sodium citrate can be prepared by neutralizing a solution of 10 g citric acid diluted sodium hydroxide solution or soda solution. Let the anode be a nickel plate of the largest possible area, and take the battery as a voltage source.
The value of the current density with the help of a variable resistance will be maintained equal to 0.005 A/cm 2 . For example, with an object surface of 20 cm 2, it is necessary to work at a current strength of 0.1 A. After half an hour of work, the object will already be nickel plated. 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 passivate and the subsequent nickel coating will not adhere well.
In order to achieve a mirror shine without mechanical polishing, we introduce a so-called brightening additive into the plating bath. Such additives are, for example, glue, gelatin, sugar. You can enter into a nickel bath, for example, a few grams of sugar and study its effect.
To prepare an electrolyte for iron chromium plating (after preliminary copper plating), let's dissolve 40 g of CrO 3 chromic anhydride (Caution! Poison!) and exactly 0.5 g of sulfuric acid (in no case more!) in 100 ml of water. The process proceeds at a current density of about 0.1 A / cm 2, and a lead plate is used as an anode, the area of \u200b\u200bwhich should be several less area chromed surface.
Nickel and chrome baths are best heated slightly (up to about 35 °C). Please note that electrolytes for chromium plating, especially with a long process and high current strength, emit vapors containing chromic acid, which are very harmful to health. Therefore, chromium plating should be carried out under draft or on outdoors e.g. on the balcony.
In chromium plating (and, to a lesser extent, in nickel plating), not all of the current is used for metal deposition. At the same time, hydrogen is released. On the basis of a series of voltages, it would be expected that the metals standing 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 in the case of 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. Due to this circumstance, a lead battery can function. When the battery is charged, instead of PbO 2, hydrogen should appear on the cathode, but, due to overvoltage, hydrogen evolution begins when the battery is almost fully charged.
Restorative properties- These are the main chemical properties characteristic of all metals. They manifest themselves in interaction with a wide variety of oxidizing agents, including oxidizing agents from environment. AT general view the interaction of a metal with oxidizing agents can be expressed by the scheme:
Me + Oxidizer" Me(+X),
Where (+X) is the positive oxidation state of Me.
Examples of metal oxidation.
Fe + O 2 → Fe (+3) 4Fe + 3O 2 \u003d 2 Fe 2 O 3
Ti + I 2 → Ti(+4) Ti + 2I 2 = TiI 4
Zn + H + → Zn(+2) Zn + 2H + = Zn 2+ + H 2
Activity series of metals
The reducing properties of metals differ from each other. Electrode potentials E are used as a quantitative characteristic of the reducing properties of metals.
The more active the metal, the more negative its standard electrode potential E o.
Metals arranged in a row as their oxidative activity decreases form a row of activity.
Activity series of metals
| Me | Li | K | Ca | Na | mg | Al | Mn | Zn | Cr | Fe | Ni | sn | Pb | H2 | Cu | Ag | Au |
| Mez+ | Li + | K+ | Ca2+ | Na+ | Mg2+ | Al 3+ | Mn2+ | Zn2+ | Cr3+ | Fe2+ | Ni2+ | sn 2+ | Pb 2+ | H+ | Cu2+ | Ag+ | Au 3+ |
| E o ,B | -3,0 | -2,9 | -2,87 | -2,71 | -2,36 | -1,66 | -1,18 | -0,76 | -0,74 | -0,44 | -0,25 | -0,14 | -0,13 | 0 | +0,34 | +0,80 | +1,50 |
The reduction of a metal from a solution of its salt with another metal with a higher reducing activity is called cementation.. Cementation is used in metallurgical technologies.
In particular, Cd is obtained by reducing it from a solution of its salt with zinc.
Zn + Cd 2+ = Cd + Zn 2+
3.3. 1. Interaction of metals with oxygen
Oxygen is a strong oxidizing agent. It can oxidize the vast majority of metals exceptAuandPt . Metals in air come into contact with oxygen, therefore, when studying the chemistry of metals, attention is always paid to the features of the interaction of a metal with oxygen.
Everyone knows that iron in humid air is covered with rust - hydrated iron oxide. But many metals in a compact state at a not too high temperature show resistance to oxidation, since they form thin protective films on their surface. These films of oxidation products do not allow the oxidizing agent to come into contact with the metal. The phenomenon of the formation of protective layers on the surface of the metal that prevent the oxidation of the metal is called metal passivation.
An increase in temperature promotes the oxidation of metals by oxygen. The activity of metals increases in the finely divided state. Most metals in powder form burn in oxygen.
s-metals
The greatest restorative activity is showns-metals. The metals Na, K, Rb Cs are capable of igniting in air, and they are stored in sealed vessels or under a layer of kerosene. Be and Mg are passivated at low temperatures in air. But when ignited, the Mg strip burns with a dazzling flame.
MetalsIIA-subgroups and Li, when interacting with oxygen, form oxides.
2Ca + O 2 \u003d 2CaO
4 Li + O 2 \u003d 2 Li 2 O
Alkali metals, other thanLi, when interacting with oxygen, they form not oxides, but peroxidesMe 2 O 2 and superoxidesMeO 2 .
2Na + O 2 \u003d Na 2 O 2
K + O 2 = KO 2
p-metals
Metals ownedp- to the block on air are passivated.
When burning in oxygen
- IIIA-subgroup metals form oxides of the type Me 2 O 3,
- Sn is oxidized to SNO 2 , and Pb - up to PbO
- Bi goes to Bi 2 O 3.
d-metals
Alld- period 4 metals are oxidized by oxygen. Sc, Mn, Fe are most easily oxidized. Particularly resistant to Ti, V, Cr corrosion.
When burned in oxygen of alld
When burned in oxygen of alld- elements of the 4th period, only scandium, titanium and vanadium form oxides in which Me is in the highest oxidation state, equal to group number. The remaining d-metals of the 4th period, when burned in oxygen, form oxides in which Me is in intermediate but stable oxidation states.
Types of oxides formed by d-metals of 4 periods during combustion in oxygen:
- Meo form Zn, Cu, Ni, Co. (at T>1000оС Cu forms Cu 2 O),
- Me 2 O 3, form Cr, Fe and Sc,
- MeO 2 - Mn and Ti
- V forms the highest oxide - V 2 O 5 .
When burned in oxygend-metals of 5 and 6 periods, as a rule, form higher oxides, the exceptions are the metals Ag, Pd, Rh, Ru.
Types of oxides formed by d-metals of 5 and 6 periods during combustion in oxygen:
- Me 2 O 3- form Y, La; Rh;
- MeO 2- Zr, Hf; Ir:
- Me 2 O 5- Nb, Ta;
- MeO 3- Mo, W
- Me 2 O 7- Tc, Re
- Meo 4 - Os
- MeO- Cd, Hg, Pd;
- Me 2 O- Ag;
The interaction of metals with acids
In acid solutions, the hydrogen cation is an oxidizing agent.. The H + cation can oxidize metals in the activity series to hydrogen, i.e. having negative electrode potentials.
Many metals, when oxidized, in acidic aqueous solutions, many turn into cationsMez + .
Anions of a number of acids are capable of exhibiting oxidizing properties that are stronger than H + . Such oxidizing agents include anions and the most common acids H 2 SO 4 andHNO 3 .
Anions NO 3 - exhibit oxidizing properties at any concentration in solution, but the reduction products depend on the concentration of the acid and the nature of the oxidized metal.
Anions SO 4 2- exhibit oxidizing properties only in concentrated H 2 SO 4 .
Oxidizer reduction products: H + , NO 3 - , SO 4 2 -
2H + + 2e - =H 2
SO 4
2-
from concentrated H 2 SO 4 SO 4
2-
+ 2e -
+ 4
H +
=
SO 2
+ 2
H 2
O
(possible also the formation of S, H 2 S)
NO 3 - from concentrated HNO 3 NO 3 - + e -
+2H+=
NO 2 + H 2 O
NO 3 - from diluted HNO 3 NO 3 - + 3e -
+4H+=NO + 2H 2 O
(It is also possible to form N 2 O, N 2, NH 4 +)
Examples of reactions of interaction of metals with acids
Zn + H 2 SO 4 (razb.) "ZnSO 4 + H 2
8Al + 15H 2 SO 4 (c.) "4Al 2 (SO 4) 3 + 3H 2 S + 12H 2 O
3Ni + 8HNO 3 (deb.) " 3Ni(NO 3) 2 + 2NO + 4H 2 O
Cu + 4HNO 3 (c.) "Cu (NO 3) 2 + 2NO 2 + 2H 2 O
Metal oxidation products in acidic solutions
Alkali metals form a cation of the Me + type, s-metals of the second group form cations Me 2+.
The p-block metals, when dissolved in acids, form the cations indicated in the table.
Metals Pb and Bi dissolve only in nitric acid.
| Me | Al | Ga | In | Tl | sn | Pb | Bi |
| Mez+ | Al 3+ | Ga3+ | In 3+ | Tl+ | sn 2+ | Pb 2+ | Bi 3+ |
| Eo,B | -1,68 | -0,55 | -0,34 | -0,34 | -0,14 | -0,13 | +0,317 |
All d-metals 4 periods except Cu , can be oxidized by ionsH+ in acid solutions.
Types of cations formed by d-metals 4 periods:
- Me 2+(form d-metals ranging from Mn to Cu)
- Me 3+ ( form Sc, Ti, V, Cr and Fe in nitric acid).
- Ti and V also form cations MeO 2+
In acidic solutions, H + can oxidize: Y, La, Cd.
In HNO 3 can dissolve: Cd, Hg, Ag. Hot HNO 3 dissolves Pd, Tc, Re.
In hot H 2 SO 4 dissolve: Ti, Zr, V, Nb, Tc, Re, Rh, Ag, Hg.
Metals: Ti, Zr, Hf, Nb, Ta, Mo, W are usually dissolved in a mixture of HNO 3 + HF.
In aqua regia (HNO 3 + HCl mixtures) Zr, Hf, Mo, Tc, Rh, Ir, Pt, Au and Os can be dissolved with difficulty). The reason for the dissolution of metals in aqua regia or in a mixture of HNO 3 + HF is the formation of complex compounds.
Example. The dissolution of gold in aqua regia becomes possible due to the formation of a complex -
Au + HNO 3 + 4HCl \u003d H + NO + 2H 2 O
Interaction of metals with water
The oxidizing properties of water are due H(+1).
2H 2 O + 2e -" H 2 + 2OH -
Since the concentration of H + in water is low, its oxidizing properties are low. Metals can dissolve in water E< - 0,413 B. Число металлов, удовлетворяющих этому условию, значительно больше, чем число металлов, реально растворяющихся в воде. Причиной этого является образование на поверхности большинства металлов плотного слоя оксида, нерастворимого в воде. Если оксиды и гидроксиды металла растворимы в воде, то этого препятствия нет, поэтому щелочные и щелочноземельные металлы энергично растворяются в воде. Alls- metals, other than Be and Mg easily soluble in water.
2 Na + 2 HOH = H 2 + 2 Oh -
Na reacts vigorously with water, releasing heat. Emitted H 2 may ignite.
2H 2 + O 2 \u003d 2H 2 O
Mg dissolves only in boiling water, Be is protected from oxidation by an inert insoluble oxide
p-block metals are less powerful reducing agents thans.
Among p-metals, the reducing activity is higher for metals of the IIIA subgroup, Sn and Pb are weak reducing agents, Bi has Eo > 0.
p-metals do not dissolve in water under normal conditions. When the protective oxide is dissolved from the surface in alkaline solutions, Al, Ga, and Sn are oxidized by water.
Among d-metals, they are oxidized by water when heated Sc and Mn, La, Y. Iron reacts with water vapor.
Interaction of metals with alkali solutions
In alkaline solutions, water acts as an oxidizing agent..
2H 2 O + 2e - \u003dH 2 + 2OH - Eo \u003d - 0.826 B (pH \u003d 14)
The oxidizing properties of water decrease with increasing pH, due to a decrease in the concentration of H +. Nonetheless, some metals that do not dissolve in water dissolve in alkali solutions, for example, Al, Zn and some others. The main reason for the dissolution of such metals in alkaline solutions is that the oxides and hydroxides of these metals are amphoteric, dissolve in alkali, eliminating the barrier between the oxidizing agent and the reducing agent.
Example. Dissolution of Al in NaOH solution.
2Al + 3H 2 O + 2NaOH + 3H 2 O \u003d 2Na + 3H 2