Metal oxidation series. Active metals. Interaction of metals with water

08.02.2022 Kinds

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 using several simple experiments to study some of the laws of the electrochemistry of aqueous solutions and galvanic cells, and also get acquainted with the production of protective galvanic coatings.
Electrochemical methods are used in modern analytical chemistry and 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.

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 dip small strips of zinc and lead sheet 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 copper, iron, lead and zinc salts, these metals should be released on it. But this, however, does not happen. The reason for the failure does not lie in an error in the voltage series, but is based on a special inhibition of the reaction, which in this case is due to a thin oxide film on the surface of the 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 --

Cu 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 the somewhat abstract reasoning above (which, moreover, represents a gross simplification) with several experiments.
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, one can place various pairs of metals in a 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 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 a flashlight battery, to which we connect a light bulb. In parallel with the lamp, that is, directly to the terminals, we connect a voltmeter.
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

The deposition 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 sulfuric acid in small portions ( 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 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 run 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. 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 a given number of equivalents from a solution 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 (small aquarium or 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. It is better not to use objects with complex shapes. 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 10 g solution 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.
Using a variable resistance, we will maintain the current density equal to 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 the 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 when 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.

The potential difference “electrode substance – solution” precisely serves as a quantitative characteristic of the ability of a substance (both metals andnon-metals) go into solution in the form of ions, i.e. characterstability of the OB ability of the ion and its corresponding substance.

This potential difference is calledelectrode potential.

However, direct methods for measuring such potential differencesdoes not exist, so we agreed to define them in relation toso-called standard hydrogen electrode, potentialal which is conventionally taken to be zero (often also calledreference electrode). A standard hydrogen electrode consists offrom a platinum plate immersed in a solution of acid containingconcentration of H ions + 1 mol/l and washed by a stream of gaseoushydrogen under standard conditions.

The emergence of a potential on a standard hydrogen electrode can be imagined as follows. Hydrogen gas, adsorbed by platinum, goes into the atomic state:

H22H.

A state of dynamic equilibrium is realized between atomic hydrogen formed on the surface of the plate, hydrogen ions in the solution and platinum (electrons!):

HH + + e.

The overall process is expressed by the equation:

H 2 2H + + 2e.

Platinum does not take part in redox reactions And tive process, but is only a carrier of atomic hydrogen.

If a plate of a certain metal, immersed in a solution of its salt with a concentration of metal ions equal to 1 mol/l, is connected to a standard hydrogen electrode, a galvanic cell is obtained. Electromotive force of this element(EMF), measured at 25° C, characterizes the standard electrode potential of the metal, usually designated as E 0 .

In relation to the H 2 /2H + system, some substances will behave as oxidizing agents, others as reducing agents. Currently, standard potentials have been obtained for almost all metals and many non-metals, which characterize the relative ability of reducing agents or oxidizing agents to donate or capture electrons.

The potentials of the electrodes that act as reducing agents with respect to hydrogen have a “-” sign, and the “+” sign indicates the potentials of the electrodes that are oxidizing agents.

If we arrange metals in increasing order of their standard electrode potentials, then the so-called electrochemical voltage series of metals:

Li, Rb, K, Ba, Sr, Ca, N a, M g, A l, M n, Zn, C r, F e, C d, Co, N i, Sn, P b, H, Sb, B i, С u, Hg, А g, Р d, Р t, А u.

A number of stresses characterize Chemical properties metals

1. The more negative the electrode potential of a metal, the greater its reducing ability.

2. Each metal is capable of displacing (reducing) from salt solutions those metals that are in the series of metal stresses after it. The only exceptions are alkali and alkaline earth metals, which will not reduce ions of other metals from solutions of their salts. This is due to the fact that in these cases the reactions of interaction of metals with water occur at a higher rate.

3. All metals that have a negative standard electrode potential, i.e. those located in the voltage series of metals to the left of hydrogen are capable of displacing it from acid solutions.

It should be noted that the presented series characterizes the behavior of metals and their salts only in aqueous solutions, since the potentials take into account the peculiarities of the interaction of a particular ion with solvent molecules. That is why the electrochemical series begins with lithium, while the more chemically active rubidium and potassium are located to the right of lithium. This is due to the exceptionally high energy of the hydration process of lithium ions compared to ions of other alkali metals.

The algebraic value of the standard redox potential characterizes the oxidative activity of the corresponding oxidized form. Therefore, a comparison of the values of standard redox potentials allows us to answer the question: does this or that redox reaction occur?

Thus, all half-reactions of oxidation of halide ions to free halogens

2 Cl – – 2 e = C l 2 E 0 = -1.36 V (1)

2 Br – -2e = V r 2 E 0 = -1.07 V (2)

2I – -2 e = I 2 E 0 = -0.54 V (3)

can be implemented under standard conditions when using lead oxide as an oxidizing agent ( IV ) (E 0 = 1.46 V) or potassium permanganate (E 0 = 1.52 V). When using potassium dichromate ( E 0 = 1.35 V) only reactions (2) and (3) can be carried out. Finally, the use of nitric acid as an oxidizing agent ( E 0 = 0.96 V) allows only a half-reaction involving iodide ions (3).

Thus, the quantitative criterion for assessing the possibility of a particular redox reaction occurring is positive value differences in standard redox potentials of oxidation and reduction half-reactions.

Metals that react easily are called active metals. These include alkali, alkaline earth metals and aluminum.

Position in the periodic table

The metallic properties of elements decrease from left to right in the 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 outer energy level. Active metals have only one or two valence electrons. In this case, metallic properties increase from top to bottom with increasing number of energy levels, because The further 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;
French

Alkaline earth metals include:

beryllium;
magnesium;
calcium;
strontium;
barium;
radium.

The degree of activity of a metal can be determined by the electrochemical series of metal voltages. The further to the left of hydrogen an element is located, the more active it is. Metals to the right of hydrogen are inactive and can only react with concentrated acids.

Rice. 2. Electrochemical series of metal voltages.

The list of active metals in chemistry also includes aluminum, located in group III and to the left of hydrogen. However, aluminum is on the border of active and intermediately active metals and does not react with some substances under normal conditions.

Properties

Active metals are soft (can be cut with a knife), light, and have a low melting point.

The main chemical properties of metals are presented in the table.

Reaction	The equation	Exception
Alkali metals spontaneously ignite in air when 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; - 2Al + 3S → Al 2 S 3	Aluminum does not react with hydrogen
React violently with water, forming alkalis and hydrogen	- Ca + 2H 2 O → Ca(OH) 2 + H 2	The reaction with lithium is slow. Aluminum reacts with water only after removing the oxide film
React with acids to form salts	Ca + 2HCl → CaCl 2 + H 2; 2K + 2HMnO 4 → 2KMnO 4 + H 2
Interact with salt solutions, first reacting with water and then with salt	2Na + CuCl 2 + 2H 2 O: 2Na + 2H 2 O → 2NaOH + H 2; - 2NaOH + CuCl 2 → Cu(OH) 2 ↓ + 2NaCl

Active metals easily react, so 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 determined by 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, and 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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Electrochemical activity series of metals(a series of voltages, a series of standard electrode potentials) - a sequence in which metals are arranged in increasing order of their standard electrochemical potentials φ 0, corresponding to the half-reaction of reduction of the metal cation Me n+: Me n+ + nē → Me

Practical use of the activity series of metals

A number of voltages are used in practice for a comparative assessment of the chemical activity of metals in reactions with aqueous solutions of salts and acids and for the assessment of 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.

Alkali metals are considered the most active:

lithium;
sodium;
potassium;
rubidium;
cesium;
French

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. The electronic configuration of the outer energy level is 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 products of its interaction with the components of the 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 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. Based on their chemical composition, they are divided into carbon (low-, medium- and high-carbon) and alloyed (low-, medium- and high-alloy). Depending on the structure, austenitic, ferritic, martensitic, pearlitic and bainitic steels are distinguished.

Steels have found application in many sectors of the 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: