OVR direction criterion

The course of redox reactions depends on the nature of the interacting substances and on the reaction conditions. 1. From the concentration of the reagent

Zn+h2so4/1(razb)=znso4/2+h2 (OxH/+)

Zn+2h2so4/6(konc)=Znso4/2+s+2h2o (OxS/6)

2. Reaction temperatures

Cl2+2koh=kcl/-1+kclo/-1+h2o (cold)

3Cl2+6koh=5kcl/-1+kclo3/+5+3h20 (hot)
3. Presence of a catalyst.
4nh3/-3+3o2=2n2+6h2o/-2 (bez)
4nh3/-3+5o2=4no/2+6h2o/-2 (c)
4. Influence of the nature of the environment - redox reactions occur in different environments.
2kmno4/7+5na2so3/4+3h2so4=2mnso4/2+5na2so4/6+k2so4+3h20 (kisl)
2kmno4/7+3na2so3/4+h2o=2mno2/4+3na2so4/6+2koh (neu)
2kmno4/7+na2so3/4+2koh=2k2mno4/6+na2so4/6+h2o (shel)

Redox reactions always proceed spontaneously in the direction of converting a strong oxidizing agent into a weak conjugate reducing agent or a strong reducing agent into a weak conjugating oxidizing agent. The direction of a redox reaction can be predicted by analyzing the values of the redox potentials of conjugated pairs.

Irreversible DC sources. Examples

Chemical current sources in which direct conversion of chemical energy into electrical energy occurs include galvanic cells. Galvanic cells are primary chemical sources of current in which chemical reactions are irreversible. In its simplest form, the element consists of two electrodes made of different metals immersed in an electrolyte solution. In this case, at one of the electrodes (cathode), a reaction of dissolution of the electrode material - or oxidation - occurs, in which the electrode loses electrons, giving them to the external electrical circuit. At the other electrode (anode), a reduction reaction occurs - neutralization of ions of the material surrounding the electrode, due to electrons coming from the cathode through an external circuit. The potential difference (electromotive force) for various cells ranges from 0.85 to 6 V. In the most widely used galvanic cells (for powering radios, flashlights, etc.), the positive electrode is a carbon rod and a mass of activated carbon or a mixture of manganese dioxide with graphite, and the negative one - zinc plating 3 in the form of a cup or cup. A solution of ammonia is most often used as an electrolyte. The biscuit element is given a flat shape, convenient for connecting to a battery. A special electrically conductive layer is applied to the outer side of the zinc electrode, which does not allow the electrolyte to pass through. The assembled element is covered with a thin vinyl chloride film. For example, the Krona battery has such a device. The film coating insulates the individual elements from the sides, prevents the leakage of electrolyte, but easily allows gases formed inside the element to pass through. The activated carbon mass protrudes slightly from the biscuit for convenient contact with another biscuit. These batteries utilize the active material better and more completely than cup batteries. Dry galvanic cells are supplied to the consumer in finished form; Water-filled containers must be filled with clean water before use. The voltage of a galvanic cell is always less than the emf developed by it, firstly, due to the voltage drop inside the element across its internal resistance, and secondly, due to the phenomenon of polarization of the electrodes as a result of electrochemical reactions occurring on the surface of the electrodes under the influence current passing in the circuit. For example, the evolution of hydrogen at the cathode and oxygen at the anode is accompanied by the appearance of polarization potentials, which are directed towards the electrode potentials and reduce them. To reduce the influence of polarization on the operation of the element, depolarizers are used - substances that accept hydrogen or oxygen, react with them and thus help reduce the polarization potential. Galvanic cells with a carbon electrode use manganese dioxide as a depolarizer. In order to obtain a voltage in an electrical circuit that exceeds the voltage of one element, the elements are connected into a battery, including them in series, that is, the positive pole of each previous element is connected to the negative pole of the next one (Fig. 3.3, a). The total electromotive force of the battery in this case is equal to the sum of the electromotive forces of the individual elements:

E ob = E 1 + E 2 + E 3 + ... + E n.

When the elements are the same and their emf. equal, e.m.f. battery consisting of n elements,

E ob = n* E el.

Galvanic cells must be protected from short circuits and it is not recommended to test them for sparks. Their voltage should be measured under load. When there is no load, the voltmeter will show emf, which does not characterize the degree of battery utilization.

Let's consider the processes that will be observed if a metal plate (electrode) is immersed in water. Since all substances are to some extent soluble, in such a system the process of transition of metal cations into solution with their subsequent hydration will begin to occur. The electrons released in this case will remain on the electrode, giving it a negative charge. A negatively charged electrode will attract metal cations from the solution, resulting in equilibrium being established in the system:

M M n+ + ne - ,

in which the electrode will have a negative charge, and the adjacent solution layer will have a positive charge. The above equation describes a half-reaction for which the oxidized form is the Mn+ cations and the reduced form is the metal atoms M.

Rice. 26. Mechanisms of occurrence of potential difference at the interface

electrode - solution.

If a salt is introduced into the system under consideration, which eliminates Mn+ cations during dissociation, the equilibrium will shift towards the reverse reaction. At a sufficiently high Mn+ concentration, it becomes possible for metal ions to be deposited on the electrode, which will acquire a positive charge, while the solution layer adjacent to the electrode surface, containing an excess of anions, will be negatively charged. The sign of the electrode charge will ultimately be determined by the chemical activity of the metal, which promotes the appearance of a negative charge, and the concentration of the metal cation in the solution, an increase in which contributes to the appearance of a positive charge. However, in any case, an electric double layer is formed in such a system and a potential jump occurs at the electrode-solution interface (Fig. 26). The potential jump at the electrode-solution interface is called the electrode potential.

In the example we considered, the metal of the electrode underwent chemical changes. This condition is not necessary for the occurrence of electrode potential. If any inert electrode (graphite or platinum) is immersed in a solution containing the oxidized and reduced forms (RP and VF) of some half-reaction, then a potential jump will also occur at the electrode-solution interface. The appearance of the electrode potential in this case will be determined by the occurrence of the half-reaction:

OF + ne - VF

Since the exchange of electrons occurs through the surface of the electrode, which in this case plays the role of a mediator, a shift in equilibrium towards the forward reaction will contribute to the appearance of a positive charge on the electrode, and a negative charge towards the reverse reaction. The electrode will not change chemically; it will only serve as a source or receiver of electrons. Thus, any redox reaction can be characterized by a certain value redox potential – the potential difference that occurs on the surface of an inert electrode immersed in a solution containing the oxidized and reduced form of the substance.

The value of the electrode potential depends on the nature and concentration of the oxidized and reduced forms, as well as on temperature. This dependence is expressed by the Nernst equation:

where R is the universal gas constant, T is the absolute temperature, n is the number of electrons corresponding to the transition of the oxidized form to the reduced form, F is the Faraday number (96485 C mol -1), C ox and C red are the concentrations of the oxidized and reduced form, x and y are the coefficients in the half-reaction equation, E˚ is the electrode potential referred to standard conditions (p = 101.326 kPa, T = 298 K, C ox = C red = 1 mol/l). The E˚ values are called standard electrode potentials.

At a temperature of 298 K, the Nernst equation is easily transformed to a simpler form:

It is impossible to measure the absolute values of the electrode potentials, however, it is possible to determine the relative values of the electrode potentials by comparing the measured potential with another, taken as a standard. The standard potential of a hydrogen electrode is used as such a reference potential. The hydrogen electrode is a platinum plate coated with a layer of porous platinum (platinum black) and immersed in a solution of sulfuric acid with a hydrogen cation activity of 1 mol/l at a temperature of 298 K. The platinum plate is saturated with hydrogen under a pressure of 101.326 kPa (Fig. 27). Hydrogen absorbed by platinum is a more active component than platinum, and the electrode behaves as if it were made of hydrogen. As a result, an electrode potential arises in the system due to the half-reaction

Н 2 ¾ 2Н ¾® 2Н + + 2е -

This potential is conventionally assumed to be zero. If the oxidized form of a particular half-reaction is a more active oxidizing agent than the hydrogen cation, the value of the electrode potential of this half-reaction will be a positive value, otherwise it will be negative. The values of standard electrode potentials are given in reference tables.

Rice. 27. Scheme of the structure of a hydrogen electrode.

The Nernst equation allows you to calculate the values of electrode potentials under various conditions. For example, it is required to determine the electrode potential of the half-reaction:

MnO 4 - + 8H + + 5e - = Mn 2+ + 4H 2 O,

if the temperature is 320 K, and the concentrations of MnO 4 -, Mn 2+ and H + are equal to 0.800, respectively; 0.0050 and 2.00 mol/l. The E˚ value for this half-reaction is 1.51 V. Accordingly

Direction of redox reactions. Since the electrode potential is related to the change in Gibbs free energy by the relation:

electrode potentials can be used to determine the direction of redox processes.

Let the redox reaction correspond to the half-reaction:

X(1) + n 1 e - = Y(1); ΔG° 1 = -n 1 FE° 1,

X(2) + n 2 e - = Y(2); ΔG° 2 = -n 2 FE° 2

It is quite obvious that one of these half-reactions must proceed from left to right (reduction process), and the other - from right to left (oxidation process). The change in the Gibbs energy for the reaction under consideration will be determined by the difference in the electrode potentials of the half-reactions

ΔG° = aΔG° 2 - bΔG° 1 = -nF(E° 2 - E° 1);

where a and b are factors that equalize the number of electrons donated and added during the reaction (n = an 1 = bn 2). For the reaction to proceed spontaneously, the value of ΔG must be negative, and this will occur when E 2 > E 1 . Thus, in the process of ORR, the one for which the electrode potential is greater is reduced from two oxidized forms, and the one for which the electrode potential is lower is oxidized from two reduced forms.

Example. Determine the direction of the reaction under standard conditions:

MnO 4 - + 5Fe 2+ + 8H + = Mn 2+ + 5Fe 3+ + 4H 2 O

Let's write down the equations for the transition of two oxidized forms into reduced ones and use the lookup tables to find the corresponding values of the electrode potentials:

Fe 3+ + 1e - = Fe 2+ │5 E° 1 = 0.77 B

MnO 4 - + 8H + + 5e - = Mn 2+ + 4H 2 O │1 E° 2 = 1.51 V

Since E° 2 > E° 1, the second half-reaction will proceed from left to right, and the first half-reaction will proceed from right to left. Thus, the process will proceed in the direction of a direct reaction.

Galvanic cell

Redox reactions, as already indicated, are accompanied by the transfer of electrons from the reducing agent to the oxidizing agent. If you separate the processes of oxidation and reduction in space, you can get a directed flow of electrons, i.e. electricity. Devices in which the chemical energy of a redox reaction is converted into the energy of an electric current are called chemical current sources or galvanic cells.

In the simplest case, a galvanic cell consists of two half-elements - vessels filled with solutions of the corresponding salts, into which metal electrodes are immersed. The half-cells are connected by a U-shaped tube (siphon) filled with an electrolyte solution, or a semi-permeable membrane, which allows ions to pass from one half-cell to another. If the electrodes are not connected by an external conductor, then the half-cells are in a state of equilibrium provided by a certain charge on the electrodes. If the circuit is closed, the equilibrium is disrupted, since electrons will begin to move from the electrode with a lower electrode potential to the electrode with a higher electrode potential. As a result, an oxidation-reduction reaction will begin to occur in the system, with a reduction process occurring at the electrode with a higher potential value, and an oxidation process at the electrode with a lower potential value. The electrode at which the reduction reaction occurs is called the cathode; the electrode on which the oxidation reaction occurs is the anode.

Rice. 28. Scheme of the structure of a copper-zinc galvanic cell.

As an example, consider a Daniel-Jacobi element, which consists of copper and zinc electrodes immersed in solutions of sulfates of these metals (Fig. 28). In this element, the oxidized forms are the cations Zn 2+ and Cu 2+, the reduced forms are zinc and copper. The half-reaction equations for the system have the form:

Zn 2+ + 2e - = Zn 0 ; E° 1 = -0.76 V

Cu 2+ + 2e - = Cu 0 ; E° 2 = 0.34 V

Since E° 2 > E° 1, the second half-reaction will proceed from left to right, and the first - from right to left, i.e. the reaction will occur in the system:

Zn + Cu 2+ = Zn 2+ + Cu

The process will continue until the zinc electrode dissolves or all copper ions are reduced. In the case of a copper-zinc cell, the cathode is a copper electrode (on which Cu 2+ ions are reduced to metallic copper), and the anode is a zinc electrode (on which zinc atoms are oxidized to Zn 2+ ions). The electromotive force of the element is equal to the difference between the electrode potentials of the cathode and anode:

ΔE = E cathode - E anode

Under standard conditions ΔE = 0.34 - (-076) = 1.10 (V).

To record the circuit of galvanic cells, use the form below:

Anode │ Anodic solution ││ Cathode solution │ Cathode

For anodic and cathodic solutions, indicate the concentrations of the corresponding ions at the moment the galvanic cell begins operation. Thus, the Daniel-Jacobi element with concentrations of CuSO 4 and ZnSO 4 equal to 0.01 mol/l corresponds to the following scheme:

Zn │ Zn 2+ (0.01 M) ││ Cu 2+ (0.01 M)│ Cu

By measuring the EMF of galvanic cells, the standard electrode potentials of certain half-reactions are determined. Let, for example, it is necessary to establish E˚ half-reaction:

Fe 3+ + 1e - = Fe 2+

To do this, it is enough to assemble a galvanic cell:

Pt│H 2 (g) (101.3 kPa), H + (1M)││Fe 3+ (1M), Fe 2+ (1M) │Pt

and measure its EMF, the latter is 0.77 V. Hence:

E°(Fe +3 /Fe +2) = DE + E°(H + /H) = 0.77 V + 0 = +0.77 V

Electrolysis

By passing an electric current through a solution or melt of an electrolyte, it is possible to carry out redox reactions that do not occur spontaneously. The process of separate oxidation and reduction at the electrodes, carried out by the flow of electric current from an external source, is called electrolysis.

In electrolysis, the anode is the positive electrode on which the oxidation process occurs, and the cathode is the negative electrode on which the reduction process occurs. The names "anode" and "cathode" are thus not related to the charge of the electrode: in electrolysis the anode is positive and the cathode negative, and in the operation of a galvanic cell it is the other way around. In the electrolysis process, the anode is an oxidizing agent, and the cathode is a reducing agent. The electrolysis process is quantitatively described by the laws of M. Faraday (1833):

1. The mass of the substance released on the electrode is proportional to the amount of electricity passing through the solution or melt.

2. To release one mole equivalent of any substance at the electrode, the same amount of electricity is consumed.

In general, Faraday's laws are expressed by the following equation:

where m is the mass of the electrolysis product, I is the current strength, t is the passage time of the current, F is a constant equal to 96485 C. mol -1 (Faraday number), M e - equivalent mass of the substance.

As already indicated, both solutions and melts of electrolytes are subjected to electrolysis. Electrolysis of melts proceeds most simply. In this case, reduction of the cation occurs at the cathode, and oxidation of the electrolyte anion occurs at the anode. For example, the electrolysis of molten sodium chloride proceeds according to the equations:

Cathode process: Na + + 1e - = Na | 2

Electrolysis equation: 2NaCl = 2Na + Cl 2

Electrolysis of solutions is much more complicated, since in this case water molecules can undergo electrolysis. During electrolysis, water can be both oxidized and reduced according to the following half-reactions:

1. Water recovery (cathode process):

2H 2 O + 2e - = H 2 + 2OH -; E° = -0.83 V

2. Water oxidation (anodic process):

2H 2 O - 4e - = 4H + + O 2; E° = 1.23 V

Therefore, during the electrolysis of aqueous solutions, competition is observed between electrode processes with different values of electrode potentials. Ideally, a half-reaction with the highest electrode potential should occur at the cathode, and a half-reaction with the lowest electrode potential should occur at the anode. However, for real processes, the value of electrode potentials is not the only factor influencing the nature of the interaction.

In most cases, the choice between competing reactions in electrolysis can be made based on the following rules:

1. If the metal in the series of standard electrode potentials is to the right of hydrogen, then the metal is reduced at the cathode.

2. If the metal in the series of standard electrode potentials is to the left of aluminum (inclusive), hydrogen is released at the cathode due to the reduction of water.

3. If a metal occupies a place between aluminum and hydrogen in the series of standard electrode potentials, parallel reduction of the metal and hydrogen occurs at the cathode.

4. If the electrolyte contains anions of oxygen-containing acids, hydroxyl or fluoride anions, water is oxidized at the anode. In all other cases, the electrolyte anion is oxidized at the anode. This order of oxidation of reducing agents at the anode is explained by the fact that the half-reactions:

F 2 + 2e - = 2F -

a very high electrode potential responds (E° = 2.87 V), and it is almost never realized if another competing reaction is possible. As for oxygen-containing anions, the product of their oxidation is molecular oxygen, which corresponds to a high overvoltage (0.5 V on the platinum electrode). For this reason, during the electrolysis of aqueous solutions of chlorides, chlorine ions are oxidized at the anode, although the electrode half-reaction potential

2Cl - - 2e - = Cl 2 ; E° = 1.36 V

higher than the electrode potential of water oxidation (E° = 1.23 V).

The electrolysis process is also influenced by the electrode material. There are inert electrodes that do not change during electrolysis (graphite, platinum), and active electrodes that undergo chemical changes during electrolysis.

Let's look at some examples of electrolysis of solutions.

Example 1. Electrolysis of an aqueous solution of copper(II) sulfate with inert electrodes.

CuSO 4 = Cu 2+ + SO

Cathode process: Cu 2+ + 2e - = Cu | 2

Electrolysis equation: 2Cu 2+ + 2H 2 O = 2Cu + 4H + + O 2

or 2СuSO 4 + 2H 2 O = 2Cu + 2H 2 SO 4 + O 2

Example 2. Electrolysis of an aqueous solution of copper sulfate with a copper anode.

Cathode process: Cu 2+ + 2e - = Cu

Anodic process: Cu 0 - 2e - = Cu 2+

Electrolysis boils down to the transfer of copper from the anode to the cathode.

Example 3. Electrolysis of an aqueous solution of sodium sulfate with inert electrodes.

Na 2 SO 4 = 2Na + + SO

Cathodic process: 2H 2 O + 2e - = H 2 + 2OH - | 2

Anodic process: 2H 2 O - 4e - = 4H + + O 2 | 1

Electrolysis equation: 2H 2 O = 2H 2 + O 2

Electrolysis comes down to the decomposition of water.

Example 4. Electrolysis of an aqueous solution of sodium chloride with inert electrodes.

NaCl = Na + + Cl -

Cathode process: 2H 2 O + 2e - = H 2 + 2OH - | 1

Anodic process: 2Cl - - 2e - = Cl 2 | 1

Electrolysis equation: 2Cl - + 2H 2 O = H 2 + Cl 2 + 2OH -

or 2NaCl + 2H 2 O = 2NaOH + H 2 + Cl 2

Electrolysis is widely used in industry to produce a number of active metals (aluminum, magnesium, alkali and alkaline earth metals), hydrogen, oxygen, chlorine, sodium hydroxide, hydrogen peroxide, potassium permanganate and a number of other practically important substances. Electrolysis is used to deposit durable metal films to protect metals from corrosion.

Colloidal solutions

= Our discussions =

Criteria for the occurrence of OVR. Standard conditions and standard potentials

Methodological development for teachers and students

In most cases, chemists (both beginners and quite experienced) have to answer the question: is it possible for a redox reaction to occur between a given reagent, and if so, what is the completeness of such a reaction? This article is devoted to solving this problem, which often causes difficulties.

It is assumed that reactions at the interfaces proceed quite quickly, and in the bulk of the solution - almost instantly. In inorganic chemistry, where reactions often involve ions, this assumption is almost always justified (certain examples of thermodynamically possible reactions that actually do not occur due to kinetic hindrances will be discussed below).

When studying OVR in general and inorganic chemistry, one of our tasks is to teach students the conscious use of simple qualitative criteria for the spontaneous course of OVR in one direction or another under standard conditions and completeness of their course under conditions actually used in chemical practice (without taking into account possible kinetic difficulties).

First of all, let’s ask ourselves this question: what does the oxidizing ability of the oxidizing agent involved in one or another half-reaction depend on, for example, MnO 4 – + 8H + + 5e = Mn 2+ + 4H 2 O? Although when writing the equations of half-reactions and ionic ORR equations in general, we traditionally use the equal sign rather than the reversibility sign, in reality all reactions are chemically reversible to one degree or another, and therefore, in a state of equilibrium, both forward and reverse reactions always occur at equal rates .
Increasing the concentration of the oxidizing agent MnO 4 increases, as is known from a school chemistry course, the rate of the direct reaction, i.e. leads to a shift of equilibrium to the right; in this case, the completeness of oxidation of the reducing agent increases. The use of Le Chatelier's principle naturally leads to the same conclusion.
Thus, the oxidizing ability of an oxidizing agent always increases with increasing its concentration.
This conclusion seems quite trivial and almost self-evident.
However, reasoning in exactly the same way, we can conclude that the oxidizing ability of the permanganate ion in an acidic environment will increase with increasing concentration of hydrogen ions and decrease with increasing concentration of the Mn 2+ cation (in particular, when it accumulates in the solution as the reaction proceeds) .
In general, the oxidizing ability of the oxidizer depends on the concentrations of all particles appearing in the half-reaction equation. At the same time, its increase, i.e. the process of reduction of the oxidizer is facilitated by an increase in the concentration of particles in the left part of the half-reaction; an increase in the concentration of particles in its right side, on the contrary, prevents this process.
Exactly the same conclusions can be drawn regarding the oxidation half-reactions of the reducing agent and redox reactions in general (if they are written in ionic form).

Redox potentials

A quantitative measure of the oxidizing ability of an oxidizer (and at the same time the reducing ability of its reduced form) is the electric potential of the electrode φ (electrode potential), on which the half-reaction of its reduction and the reverse half-oxidation half-reaction of the corresponding reduced form occur simultaneously and at equal rates.
This redox potential is measured relative to a standard hydrogen electrode and characterizes the pair “oxidized form - reduced form” (therefore, the expressions “oxidizing potential” and “reducing potential” are strictly speaking incorrect). The higher the potential of the pair, the more pronounced the oxidizing ability of the oxidizing agent and, accordingly, the weaker the reducing ability of the reducing agent.
And vice versa: the lower the potential (up to negative values), the more pronounced are the reducing properties of the reduced form and the weaker are the oxidizing properties of the oxidizing agent conjugated with it.
Electrode types, the design of a standard hydrogen electrode, and methods for measuring potentials are covered in detail in a physical chemistry course.

Nernst equation

The dependence of the redox potential corresponding to the half-reaction of reduction of the permanganate ion in an acidic medium (and, as already noted, at the same time the half-reaction of oxidation of the Mn 2+ cation to the permanganate ion in an acidic medium) on the factors listed above that determine it is quantitatively described by the Nernst equation φ(MnO 4 – , H + / Mn 2+) = φ o (MnO 4 – , H + / Mn 2+) + RT / 5F ln 8/. In the general case, the Nernst equation is usually written in the conditional form φ(Ox/Red) = φ o (Ox/Red) + RT/(nF) ln /, corresponding to the conventional notation of the reduction half-reaction of the oxidizing agent Ox + ne- = Red

Each of the concentrations under the sign of the natural logarithm in the Nernst equation is raised to the power corresponding to the stoichiometric coefficient of a given particle in the half-reaction equation, n– number of electrons accepted by the oxidizer, R– universal gas constant, T- temperature, F– Faraday number.

Measure the redox potential in the reaction vessel during the reaction, i.e. under non-equilibrium conditions, it is impossible, since when measuring the potential, electrons must be transferred from the reducing agent to the oxidizing agent not directly, but through the metal conductor connecting the electrodes. In this case, the electron transfer rate (current strength) must be kept very low due to the application of an external (compensating) potential difference. In other words, measuring electrode potentials is possible only under equilibrium conditions, when direct contact between the oxidizing agent and the reducing agent is excluded.
Therefore, square brackets in the Nernst equation denote, as usual, the equilibrium (under measurement conditions) concentrations of particles. Although the potentials of redox pairs during a reaction cannot be measured, they can be calculated by substituting the current ones into the Nernst equation, i.e. concentrations corresponding to a given point in time.
If the change in potential as the reaction proceeds is considered, then first these are the initial concentrations, then the time-dependent concentrations, and finally, after the termination of the reaction, the equilibrium ones.
As the reaction proceeds, the potential of the oxidizing agent calculated using the Nernst equation decreases, and the potential of the reducing agent corresponding to the second half-reaction, on the contrary, increases. When these potentials are equalized, the reaction stops and the system returns to a state of chemical equilibrium.

Standard redox potentials

The first term on the right side of the Nernst equation is the standard redox potential, i.e. potential measured or more often calculated under standard conditions.
Under standard conditions, the concentrations of all particles in a solution are, by definition, 1 mol/l, and the second term on the right side of the equation becomes zero.
Under non-standard conditions, when at least one of the concentrations is not equal to 1 mol/l, the potential determined by the Nernst equation differs from the standard one. Potential in non-standard conditions is often called real potential.
The term “electrochemical potential,” strictly speaking, is not recommended, since it is assigned to another quantity (the sum of the chemical potential of the ion and the product of its charge and the electric potential), which students will encounter in the course of physical chemistry. If one or more gases take part in the ORR, their standard states are states at a pressure of 1 atm = 101300 Pa. The temperature when determining standard states and standard potentials is not standardized and can be any, but tables of standard potentials in reference books are compiled for T=298 K (25 o C).

The student must distinguish standard states of substances from normal conditions that have essentially nothing in common with them ( R= 1 atm, T=273 K), to which, using the equation of state of ideal gases pV = nRT, it is customary to give volumes of gases measured under other conditions.

The table of standard potentials, compiled in descending order, clearly ranks oxidizing agents (i.e. oxidized forms of various redox couples) according to their strength. At the same time, reducing agents are also ranked by strength ( restored forms of pairs).

Criterion for the direction of a reaction under standard conditions.

If the reaction mixture contains both the starting substances and the reaction products formed by them during the reaction process, or, in other words, two oxidizing agents and two reducing agents, then the direction of the reaction is determined by which of the oxidizing agents under given conditions, in accordance with the Nernst equation, will be stronger.
It is especially easy to determine the direction of a reaction under standard conditions, when all the substances (particles) participating in it are in their standard states. Under these conditions, the oxidizing agent of that pair, which is characterized by a higher standard potential, obviously turns out to be stronger.
Although the direction of the reaction under standard conditions is unambiguously determined by this, we, without knowing it in advance, can write the reaction equation or Right(the reaction under standard conditions actually proceeds in the direction we have accepted, i.e. in the forward direction) or wrong(the reaction goes in the opposite direction to the direction we assumed).
Any entry for the OVR equation presupposes a certain choice of oxidizing agent on the left side of the equation. If under standard conditions this oxidizing agent is stronger, the reaction will proceed in direct direction, if not - in reverse.
We will call the standard potential of a redox pair, in which the oxidized form is the oxidizing agent we have chosen, oxidizer potentialφ o Ok, and the standard potential of the other pair, in which the reduced form is the reducing agent we have chosen - reducing agent potentialφ o Sun.

We will call the quantity Δφ o = φ o Ok – φ o Sun standard redox potential difference.
After introducing these notations reaction direction criterion in standard conditions you can give it a simple form:

If Δφ o > 0, the reaction under standard conditions proceeds in the forward direction; if Δφ o< 0, то в обратном.

Criterion for completeness of ORR (or criterion for chemical irreversibility of ORR)

The degree of completeness of the reaction occurring in the forward direction at Δφ o > 0 depends on the value of Δφ o. For the reaction to proceed almost completely or "to the end", i.e. until at least one of the initial particles (ions, molecules) is exhausted, or, in other words, so that it is chemically irreversible, it is necessary that the standard potential difference be large enough.
Note that any reaction, regardless of its chemical reversibility, is always thermodynamically irreversible, if it occurs in a test tube or other chemical reactor, i.e. outside of a reversible galvanic cell or other special device). The authors of a number of manuals consider the criterion for the completeness of ORR to be Δφо > 0.1 V. For many reactions this condition is correct, however, the completeness of ORR (more precisely, the degree of reaction) at a given value of Δφо depends on the stoichiometric coefficients in its ionic equation, as well as on initial concentrations of reagents.
Calculations using the Nernst equation to find ORR equilibrium constant, and the law of mass action show that reactions are obviously chemically irreversible at Δφ o > 0.4.
In this case the reaction is always, i.e. under any initial conditions (we are, of course, not talking about standard conditions now), passes in the forward direction to the end.
In a completely similar way, if Δφ o< – 0,4 В, реакция всегда протекает до конца, но в обратном направлении.
Change the direction and completeness of such reactions, i.e. it is impossible to control them, with all the desire, in contrast to chemically reversible reactions, for which< Δφ о < 0,4 В или –0,4 В < Δφ о < 0.

In the first case, under standard conditions, the reaction always proceeds in the forward direction. This means that in the absence of reaction products at the initial moment of time, the reaction will even more (i.e., also always) proceed in the forward direction, but not completely.
Promote a more complete reaction excess one or more reagents and withdrawal from the sphere of reaction in one way or another of its products.
It is often possible to achieve fairly complete completion of such reactions despite their chemical reversibility.
On the other hand, it is usually also possible to create conditions under which such a reaction will proceed in the opposite direction. To do this, it is necessary to create high concentrations of “reagents” (until now we considered them products reaction), start the reaction in the absence of its “products” (i.e. reagents, during a direct reaction) and try to maintain their concentration as low as possible during the reaction.

In the same way, in general terms, we can consider chemically reversible ORRs with Δφ o< 0. Вместо этого обсудим возможности управления конкретной химической реакцией Cu (т) + 2H 2 SO 4 = CuSO 4 + SO 2(г) + 2H 2 O или в ионном виде: Cu (т) + 4H + + SO 4 2- = Cu 2+ + SO 2(г) + 2H 2 O с Δφ о = – 0,179 В. В стандартных условиях, когда концентрации ионов H + , SO 4 2- , Cu 2+ в водном растворе равны 1 моль/л, а давление SO 2 составляет 1 атм, эта реакция протекает в обратном направлении, т.е. диоксид серы восстанавливает катион Cu 2+ до порошка металлической меди.
Let us note, firstly, that we are not talking about any concentrated sulfuric acid yet.
Secondly, it is impossible to create a solution with standard ion concentrations using only sulfuric acid and copper sulfate, and if we wanted to solve this problem (why?) we would have to use other combinations of substances, for example, NaHSO 4 + CuCl 2 or HCl + CuSO 4, neglecting the possible influence of chloride ions on the course of the reaction.
The reduction reaction of copper cations is facilitated by an increase in SO 2 pressure and the removal of H + and SO 4 2- ions from the reaction sphere (for example, when adding Ba(OH) 2, Ca(OH) 2, etc.).
In this case, a high – close to 100% – completeness of copper recovery can be achieved. On the other hand, increasing the concentration of sulfuric acid, removing sulfur dioxide and water from the reaction sphere, or binding the latter in some other way contributes to the occurrence of a direct reaction, and already during the first laboratory work, students can directly observe the interaction of copper with concentrated sulfuric acid with the release of sulfur dioxide.
Since the reaction occurs only at the interface, its speed is low. Such heterogeneous (more precisely, heterophasic) reactions always proceed better (in the sense of faster) when heated. The effect of temperature on standard potentials is small and is not usually considered. Thus, reversible ORRs can be carried out in both forward and reverse directions. For this reason they are sometimes called bilateral, and we must admit that this term, which, unfortunately, has not received widespread use, is more successful, especially since it eliminates the difficulties arising due to the consonance of concepts that have nothing in common with each other chemical And thermodynamic reversibility and irreversibility (it is difficult for a student to understand that any chemically reversible reaction under ordinary conditions proceeds thermodynamically irreversibly, but without this understanding the true meaning of many branches of chemical thermodynamics is almost inaccessible to him).
Let us also note that in both directions, reversible ORRs proceed spontaneously (or, in other words, thermodynamically irreversible), like any other chemical reactions.
Non-spontaneous OVR occurs only during electrolysis or charging of batteries. Therefore, perhaps it would be more correct not to mention in passing the spontaneous course of reactions in one direction or another, and instead take a deeper look at the very concepts of spontaneous and non-spontaneous processes.

Kinetic difficulties in the interaction of ions

As already noted, reactions involving ions occurring throughout the entire volume of a solution almost always proceed very quickly. However, there are exceptions. Thus, the oxidation reaction of the ammonium cation in an acidic environment with the iron(III) cation 6Fe 3+ + 2NH 4 + ≠ N 2 + 6Fe 2+ + 8H + is thermodynamically possible (Δφ o = 0.499 V), but in fact does not occur.

Reason kinetic difficulties here is the Coulomb repulsion of the oxidizing and reducing cations, which prevents them from approaching each other at a distance at which an electronic transition is possible. For a similar reason (but due to the Coulomb repulsion of anions), the oxidation of iodide ion by nitrate ion in an acidic medium does not occur, although for this reaction Δφ o = 0.420 V.
After adding zinc, neutral molecules of nitrous acid appear in the system, which are not prevented from oxidizing iodide ions.

PRACTICAL LESSONS ON THE TOPIC

"REDOX
REACTIONS AND ELECTROCHEMICAL
PROCESSES"

IN THE DISCIPLINE "CHEMISTRY"

Educational and methodological manual

CHEREPOVETS

Practical classes on the topic “Oxidation-reduction reactions and electrochemical processes” in the discipline “Chemistry”: Educational method. allowance. Cherepovets: State Educational Institution of Higher Professional Education ChSU, 2005. 45 p.

Considered at a meeting of the Department of Chemistry, minutes No. 11 of 06/09/2004.

Approved by the editorial and publishing commission of the Institute of Metallurgy and Chemistry of the State Educational Institution of Higher Professional Education of ChSU, protocol No. 6 of June 21, 2004.

Compiled by: O.A. Kalko – Ph.D. tech. Sciences, Associate Professor; N.V. Kunina

Reviewers: T.A. Okuneva, associate professor (GOU VPO ChSU);

G.V. Kozlova, Ph.D. chem. Sciences, Associate Professor (GOU VPO ChSU)

Scientific editor: G.V. Kozlova – Ph.D. chem. Sciences, Associate Professor

Vienna University, 2005

INTRODUCTION

The manual includes brief theoretical information, examples of problem solving and test options on the topic “Oxidation-reduction reactions and electrochemical processes” of the general chemistry course. The content of the educational manual complies with the state standard of the discipline “Chemistry” for chemical and engineering specialties.

REDOX REACTIONS

Reactions that occur with changes in the oxidation states of elements in substances are called redox.

Oxidation state element (CO) is the number of electrons displaced to a given atom from others (negative CO) or from a given atom to others (positive CO) in a chemical compound or ion.

1. The CO of an element in a simple substance is zero, for example: , .

2. The CO of an element in the form of a monatomic ion in an ionic compound is equal to the charge of the ion, for example: , , .

3. In compounds with covalent polar bonds, the atom with the highest electronegativity (EO) value has a negative CO, and the following COs are characteristic of some elements:

– for fluorine – “-1”;

– for oxygen – “-2”, with the exception of peroxides, where CO = -1, superoxides (CO = -1/2), ozonides (CO = -1/3) and ОF 2 (CO = +2);

For alkali and alkaline earth metals CO = +1 and +2, respectively.

4. The algebraic sum of the COs of all elements in a neutral molecule is equal to zero, and in an ion – the charge of the ion.

Most elements in substances exhibit variable CO. For example, let’s determine the CO of nitrogen in various substances:

Any redox reaction (ORR) consists of two coupled processes:

1. Oxidation is the process of electron donation by a particle, which leads to an increase in the CO of the element:

2. Recovery is the process of electron acceptance by a particle, which is accompanied by a decrease in the CO of the element:

Substances that give up their electrons during oxidation are called restorers, and substances that accept electrons during the reduction process are oxidizing agents. If we denote by Oh oxidized form of the substance, and through Red– restored, then any ORR can be represented as the sum of two processes:

Red 1 – n ē® Ox 1 (oxidation);

Ox 2 + n ē ® Red 2 (recovery).

The manifestation of certain redox properties of atoms depends on many factors, the most important of which are the position of the element in the Periodic Table, its CO in the substance, and the nature of the properties exhibited by other participants in the reaction. Based on their redox activity, substances can be divided into three groups:

1. Typical reducing agents- these are simple substances whose atoms have low EO values (for example, metals, hydrogen, carbon), as well as particles in which there are atoms in the minimum (lowest) oxidation state for them (for example, chlorine in the compound).

2. Typical oxidizing agents- these are simple substances whose atoms are characterized by high EO (for example, fluorine and oxygen), as well as particles that contain atoms in the highest (maximum) CO (for example, chromium in the compound).

3. Substances with redox dual properties– these are many non-metals (for example, sulfur and phosphorus), as well as substances containing elements in intermediate CO (for example, manganese in the compound).

Reactions in which the oxidizing and reducing agents are different substances are called intermolecular. For example:

In some reactions, the oxidizing and reducing agents are atoms of elements of the same molecule of different nature; such ORRs are called intramolecular, For example:

Reactions in which the oxidizing and reducing agent is an atom of the same element located in the composition of the same substance are called disproportionation reactions(auto-oxidation-self-healing), For example:

There are several ways to select coefficients in the OVR equations, of which the most common are electronic balance method And method of ion-electronic equations(otherwise half-reaction method). Both methods are based on the implementation of two principles:

1. Principle material balance– the number of atoms of all elements before and after the reaction must be the same;

2. Principle electronic balance– the number of electrons donated by the reducing agent must be equal to the number of electrons accepted by the oxidizing agent.

The electronic balance method is universal, that is, it can be used to equalize OVR occurring in any conditions. The half-reaction method is applicable to compiling equations only for redox processes that occur in solutions. However, it has a number of advantages compared to the electronic balance method. In particular, when using it, there is no need to determine the oxidation states of elements; in addition, the role of the environment and the actual state of particles in solution are taken into account.

The main stages of compiling reaction equations using the electronic balance method are as follows:

It is obvious that CO changes for manganese (decreases) and for iron (increases). Thus, KMnO 4 is an oxidizing agent, and FeSO 4 is a reducing agent.

2. Compose half-reactions of oxidation and reduction:

(recovery)

(oxidation)

3. Balance the number of received and given electrons by transferring the coefficients in front of the electrons in the form of multipliers, swapping them:

½´ 1½´ 2

½´ 5½´ 10

If the coefficients are multiples of each other, they should be reduced by dividing each by the greatest common multiple. If the coefficients are odd, and the formula of at least one substance contains an even number of atoms, then the coefficients should be doubled.

So, in the example under consideration, the coefficients in front of the electrons are odd (1 and 5), and the formula Fe 2 (SO 4) 3 contains two iron atoms, so we double the coefficients.

4. Record the total reaction of the electronic balance. In this case, the number of received and given electrons should be the same and should decrease at this stage of equalization.

5. Arrange the coefficients in the molecular equation of the reaction and add the missing substances. In this case, the coefficients for substances that contain atoms of elements that have changed CO are taken from the total electron balance reaction, and the atoms of the remaining elements are equalized in the usual way, observing the following sequence:

– metal atoms;

– atoms of non-metals (except oxygen and hydrogen);

– hydrogen atoms;

– oxygen atoms.

For this example

2KMnO 4 + 10FeSO 4 + 8H 2 SO 4 =
= 2MnSO 4 + 5Fe 2 (SO 4) 3 + K 2 SO 4 + 8H 2 O.

When equalizing reactions using the method of ion-electronic equations follow the following sequence of actions:

1. Write down the reaction scheme, determine the CO elements, identify the oxidizing agent and the reducing agent. For example:

CO changes for chromium (decreases) and for iron (increases). Thus, K 2 Cr 2 O 7 is an oxidizing agent, and Fe is a reducing agent.

2. Write down the ionic scheme of the reaction. In this case, strong electrolytes are written in the form of ions, and weak electrolytes, insoluble and slightly soluble substances, as well as gases are left in molecular form. For the process under consideration

K + + Cr 2 O + Fe + H + + SO ® Cr 3+ + SO + Fe 2+ + H 2 O

3. Compose equations for ionic half-reactions. To do this, first equalize the number of particles containing atoms of elements that have changed their CO:

a) in acidic environments H 2 O and (or) H +;

b) in neutral environments H 2 O and H + (or H 2 O and OH -);

c) in alkaline media H 2 O and (or) OH - .

Cr 2 O + 14H + ® 2Cr 3+ + 7H 2 O

The charges are then equalized by adding or subtracting a certain number of electrons:

12+ + 6 ē ® 6+

Fe 0 – 2 ē ® Fe 2+

4. Balance the number of received and given electrons as described in the electronic balance method

12+ + 6 ē ® 6+ ½´2½´1

Fe 0 – 2 ē ® Fe 2+ ½´6½´3

5. Record the total reaction of the ion-electron balance:

Cr 2 O + 14H + + 6 ē + 3Fe – 6 ē ® 2Cr 3+ + 7H 2 O + 3Fe 2+

6. Arrange the coefficients in the molecular equation of the reaction:

K 2 Cr 2 O 7 + 3Fe + 7H 2 SO 4 = Cr 2 (SO 4) 3 + 3FeSO 4 + K 2 SO 4 + 7H 2 O

Calculation of molar equivalent masses ME oxidizer or reducer in ORR should be carried out according to the formula

M E = , (1)

Where M– molar mass of the substance, g/mol; N ē– the number of electrons involved in the process of oxidation or reduction.

Example: Equalize the reaction using the ion-electron balance method, calculate the molar equivalent masses of the oxidizing agent and reducing agent

As 2 S 3 + HNO 3 ® H 3 AsO 4 + H 2 SO 4 + NO

Solution

We determine the oxidation states of elements, identify the oxidizing agent and the reducing agent

In this process, the oxidizing agent is HNO 3, the reducing agent is As 2 S 3.

Making an ionic reaction scheme

As 2 S 3 ¯ + H + + NO ® H + + AsO + SO + NO

We write down the ion-electronic half-reactions and balance the number of electrons received and given up:

0 – 28ē ® 28+ ½´3

3+ + 3ē ® 0 ½´28

We add up the half-reactions and simplify the overall scheme:

3As 2 S 3 + 60H 2 O + 28NO + 112H + ®
® 6AsO + 9SO + 120H + + 28NO + 56H 2 O

3As 2 S 3 + 4H 2 O + 28NO ® 6AsO + 9SO + 8H + + 28NO

We transfer the coefficients into the molecular equation and equalize the number of atoms of each element:

3As 2 S 3 + 28HNO 3 + 4H 2 O = 6H 3 AsO 4 + 9H 2 SO 4 + 28NO

We calculate the molar equivalent masses of the oxidizing agent and reducing agent using formula (1):

M uh, oxidizing agent = g/mol;

M uh, reducing agent = g/mol.

MULTI-CHOICE TASK No. 1

According to one of the options, equalize the ORR using the method of ion-electronic equations. Determine the type of reaction and calculate the molar equivalent masses of the oxidizing agent and reducing agent:

1. Zn + HNO 3 = Zn(NO 3) 2 + NH 4 NO 3 + H 2 O

2. FeSO 4 + KClO 3 + H 2 SO 4 = Fe 2 (SO 4) 3 + KCl + H 2 O

3. Al + Na 2 MoO 4 + HCl = MoCl 3 + AlCl 3 + NaCl + H 2 O

4. Sb 2 O 3 + HBrO 3 = Sb 2 O 5 + HBr

5. Fe + HNO 3 = Fe(NO 3) 3 + NO + H 2 O

6. Fe + HNO 3 = Fe(NO 3) 2 + NO 2 + H 2 O

7. Zn + H 2 SO 4 = ZnSO 4 + H 2 S + H 2 O

8. Zn + HNO 3 = Zn(NO 3) 2 + NO + H 2 O

9. C + H 2 SO 4 = CO + SO 2 + H 2 O

10. P + HNO 3 + H 2 O = H 3 PO 4 + NO

11. Pb + PbO 2 + H 2 SO 4 = PbSO 4 + H 2 O

12. Zn + H 2 SO 4 = ZnSO 4 + SO 2 + H 2 O

13. C + HNO 3 = CO 2 + NO + H 2 O

14. Na 2 S + HNO 3 = S + NaNO 3 + NO + H 2 O

15. KMnO 4 + HCl = MnCl 2 + Cl 2 + KCl + H 2 O

16. KIO 3 + KI + H 2 SO 4 = I 2 + K 2 SO 4 + H 2 O

17. S + HNO 3 = H 2 SO 4 + NO 2 + H 2 O

18. Al + H 2 SO 4 = Al 2 (SO 4) 3 + SO 2 + H 2 O

19. FeSO 4 + K 2 Cr 2 O 7 + H 2 SO 4 = Fe 2 (SO 4) 3 + Cr 2 (SO 4) 3 + K 2 SO 4 + + H 2 O

20. K 2 Cr 2 O 7 + HCl = CrCl 3 + Cl 2 + KCl + H 2 O

21. Zn + HNO 3 = Zn(NO 3) 2 + N 2 O + H 2 O

22. K 2 SO 3 + Br 2 + H 2 O = K 2 SO 4 + HBr

23. K 2 Cr 2 O 7 + KI + H 2 SO 4 = Cr 2 (SO 4) 3 + I 2 + K 2 SO 4 + H 2 O

24. Zn + H 3 AsO 3 + HCl = AsH 3 + ZnCl 2 + H 2 O

25. HI + H 2 SO 4 = I 2 + H 2 S + H 2 O

26. Cr 2 (SO 4) 3 + K 2 SO 4 + I 2 + H 2 O = K 2 Cr 2 O 7 + KI + H 2 SO 4

27. MnO 2 + KBr + H 2 SO 4 = Br 2 + MnSO 4 + H 2 O

28. HClO + FeSO 4 + H 2 SO 4 = Fe 2 (SO 4) 3 + Cl 2 + H 2 O

29. KMnO 4 + K 2 S + H 2 SO 4 = S + MnSO 4 + K 2 SO 4 + H 2 O

30. CuCl + K 2 Cr 2 O 7 + HCl = CuCl 2 + CrCl 3 + KCl + H 2 O

DIRECTION OF OVR FLOW

The possibility and completeness of the spontaneous occurrence of ORR under isobaric-isothermal conditions, like any chemical process, can be assessed by the sign of the change in the Gibbs free energy of the system D G during the process. Spontaneously when P, T = = const reactions can occur in the forward direction for which D G < 0.

The change in the Gibbs energy of the redox process is also equal to the electrical work done by the system to move electrons from the reducing agent to the oxidizing agent, that is

where D E– EMF of the redox process, V; F– Faraday constant ( F= 96,485 » 96,500 C/mol); n ē– the number of electrons participating in this process.

From equation (2) it follows that the condition for the spontaneous occurrence of redox reaction in the forward direction is a positive value of the emf of the redox process (D E> 0). The calculation of the EMF of the ORR under standard conditions should be carried out according to the equation

where are the standard redox potentials of systems. Their values are determined experimentally and are given in reference literature (for some systems, redox potentials are given in Table 1 of the Appendix).

Example 1. Determine the direction of flow of ORR, the ionic scheme of which is as follows:

Fe 3+ + Cl - « Fe 2+ + Cl 2

Solution

In this process, the Fe 3+ ion is an oxidizing agent, and the Cl – ion is a reducing agent. In table 1 application we find the potentials of half-reactions:

Fe 3+ + 1 ē = Fe 2+, E= 0.77 V;

Cl2+2 ē = 2Cl - , E= 1.36 V.

Using formula (3) we calculate the EMF:

Since the value of D E 0 < 0, то реакция идет самопроизвольно в обратном направлении.

Example 2. Is it possible to oxidize H 2 S to elemental sulfur using FeCl 3?

Solution

Let's make an ionic reaction scheme:

Fe 3+ + H 2 S ® Fe 2+ + S + H +

The Fe 3+ ion in this process acts as an oxidizing agent, and the H 2 S molecule acts as a reducing agent.

We find the redox potentials of the corresponding half-reactions: E= 0.77 V; E= 0.17 V.

The potential of the oxidizing agent is greater than the potential of the reducing agent; therefore, hydrogen sulfide can be oxidized using iron (III) chloride.

MULTI-CHOICE TASK No. 2

Balance one of the redox reactions using the method of ion-electronic equations. Using the table of standard redox potentials, calculate the emf and D G reaction, and also indicate the direction of occurrence of this ORR:

1. CuS + H 2 O 2 + HCl = CuCl 2 + S + H 2 O

2. HIO 3 + H 2 O 2 = I 2 + O 2 + H 2 O

3. I 2 + H 2 O 2 = HIO 3 + H 2 O

4. Cr 2 (SO 4) 3 + Br 2 + NaOH = Na 2 CrO 4 + NaBr + Na 2 SO 4 + H 2 O

5. H 2 S + Cl 2 + H 2 O = H 2 SO 4 + HCl

6. I 2 + NaOH = NaI + NaIO + H 2 O

7. Na 2 Cr 2 O 7 + H 2 SO 4 + Na 2 SO 3 = Cr 2 (SO 4) 3 + Na 2 SO 4 + H 2 O

8. H 2 S + SO 2 = S + H 2 O

9. I 2 + NaOH = NaI + NaIO 3 + H 2 O

10. MnCO 3 + KClO 3 = MnO 2 + KCl + CO 2

11. Na 2 S + O 2 + H 2 O = S + NaOH

12. PbO 2 + HNO 3 + H 2 O 2 = Pb(NO 3) 2 + O 2 + H 2 O

13. P + H 2 O + AgNO 3 = H 3 PO 4 + Ag + HNO 3

14. P + HNO 3 = H 3 PO 4 + NO 2 + H 2 O

15. HNO 2 + H 2 O 2 = HNO 3 + H 2 O

16. Bi(NO 3) 3 + NaClO + NaOH = NaBiO 3 + NaNO 3 + NaCl + H 2 O

17. KMnO 4 + HBr + H 2 SO 4 = MnSO 4 + HBrO + K 2 SO 4 + H 2 O

18. H 2 SO 3 + H 2 S = S + SO 2 + H 2 O

19. NaCrO 2 + PbO 2 + NaOH = Na 2 CrO 4 + Na 2 PbO 2 + H 2 O

20. NaSeO 3 + KNO 3 = Na 2 SeO 4 + KNO 2

21. KMnO 4 + KOH = K 2 MnO 4 + O 2 + H 2 O

22. Pb + NaOH + H 2 O = Na 2 + H 2

23. PbO 2 + HNO 3 + Mn (NO 3) 2 = Pb (NO 3) 2 + HMnO 4 + H 2 O

24. MnO 2 K + 2 SO 4 + KOH = KMnO 4 + K 2 SO 3 + H 2 O

25. NO + H 2 O + HClO = HNO 3 + HCl

26. NO + H 2 SO 4 + CrO 3 = HNO 3 + Cr 2 (SO 4) 3 + H 2 O

27. MnCl 2 + KBrO 3 + KOH = MnO 2 + KBr + KCl + H 2 O

28. Cl 2 + KOH = KClO + KCl + H 2 O

29. CrCl 3 + NaClO + NaOH = Na 2 CrO 4 + NaCl + H 2 O

30. H 3 PO 4 + HI = H 3 PO 3 + I 2 + H 2 O

GALVANIC CELLS

The processes of converting the energy of a chemical reaction into electrical energy are the basis of the work chemical current sources(HIT). HCI include galvanic cells, batteries and fuel cells.

Galvanic cell is a device for directly converting the energy of a chemical reaction into an electrical one, in which the reagents (oxidizing agent and reducing agent) are included directly in the composition of the element and are consumed during its operation. After the reagents are consumed, the element can no longer work, that is, it is a one-time HIT.

If the oxidizing agent and the reducing agent are stored outside the cell and during its operation are supplied to electrodes that are not consumed, then such a cell can operate for a long time and is called fuel cell.

The operation of batteries is based on reversible OVR. Under the influence of an external current source, the ORR flows in the opposite direction, while the device accumulates (accumulates) chemical energy. This process is called battery charge. The battery can then convert the stored chemical energy into electrical energy (a process battery discharge). The processes of charging and discharging the battery are carried out repeatedly, that is, it is a reusable HIT.

A galvanic cell consists of two half-cells (redox systems) connected to each other by a metal conductor. Half-element (aka electrode) is most often a metal placed in a solution containing ions that can be reduced or oxidized. Each electrode is characterized by a certain value conditional electrode potential E, which under standard conditions is determined experimentally relative to the potential standard hydrogen electrode(SVE).

SHE is a gas electrode that consists of platinum in contact with hydrogen gas ( R= 1 atm) and a solution in which the activity of hydrogen ions A= 1 mol/dm3. Equilibrium in the hydrogen electrode is reflected by the equation

When calculating the potentials of metal electrodes, the activity of metal ions can be considered approximately equal to their molar concentration A» [Me n + ];

2) for hydrogen electrode

where pH is the pH value of water.

In a galvanic cell, the electrode having a lower potential value is called anode and is indicated by a “–” sign. At the anode, oxidation of reducing agent particles occurs. An electrode with a high potential is called cathode and is indicated by a “+” sign. At the cathode, the oxidizing agent particles are reduced. The transfer of electrons from a reducing agent to an oxidizing agent occurs through a metal conductor called external circuit. The ORR underlying the operation of a galvanic cell is called current-generating reaction.

The main characteristic of the element’s operation is its emf D E, which is calculated as the difference between the cathode and anode potentials

D E = E cathode – E anode. (6)

Since the cathode potential is always greater than the anode potential, it follows from formula (6) that in a working galvanic cell D E > 0.

Galvanic cells are usually written in the form of diagrams in which one vertical line represents the phase boundary (metal - solution), and two vertical lines represent the boundary between two solutions. In practice, electrical contact between solutions is ensured by salt bridge– U-shaped tube with electrolyte solution.

Example 1. Determine the potential of a nickel electrode if the concentration of Ni 2+ ions in the solution is 0.02 N.

Solution

We determine the molar concentration of nickel ions in solution:

= mol/dm 3,

Where z= 2 – equivalence number of Ni 2+ ions.

E= - 0.250 V. Using formula (4), we calculate the potential of the nickel electrode

Example 2. Determine the concentration of OH - ions in a solution if the potential of the hydrogen electrode placed in this solution is -0.786 V.

Solution

From formula (5) we determine the pH of the solution:

Then the hydroxyl index of water is equal to

R OH = 14 – R N = 14 – 13.32 = 0.68.

Hence the concentration of OH ions is equal to

Mol/dm 3 .

Example 3. Make a diagram, write the equations of electrode processes and calculate the EMF of a galvanic cell composed of lead and copper electrodes immersed in solutions with concentrations of Pb 2+ and Cu 2+ ions equal to 0.1 M and 0.05 M, respectively.

Solution

From the table Select 2 applications E 0 of these metals and using formula (5) we calculate their potentials:

The potential of the copper electrode is greater than the potential of the lead electrode, which means that Pb is the anode, and Cu is the cathode. Consequently, the following processes occur in the element:

on the anode: Pb – 2 ® Pb 2+ ;

at the cathode: Cu 2+ + 2 ® Cu;

current-generating reaction: Pb + Cu 2+ = Pb 2+ + Cu;

element diagram: (-) Pb½Pb 2+ ½½Cu 2+ ½Cu (+).

Using formula (6), we determine the EMF of a given galvanic cell:

D E= 0.298 – (-0.156) = 0.454 V.

MULTI-CHOICE TASK No. 3

Draw up a diagram, write equations for electrode processes and calculate the emf of a galvanic cell composed of the first and second metals immersed in solutions with the indicated concentration of metal ions (Table 1). Calculate D G current-forming reaction.

Table 1

Table of initial data for multivariate task No. 3

To establish the possibility of spontaneous occurrence of ORR under standard conditions, in addition to calculating the EMF, you can use the determination of ∆G 0 298 for this reaction. The negative value of ∆G 0 298, as well as the positive value of the emf, indicates that under standard conditions at 298 K this reaction can occur spontaneously, without supplying energy from the outside.

With the reversible implementation of OVR under conditions p = const and V = const, the change in Gibbs energy will be equal to the electrical work Ael performed by the system.

∆G 0 = - Ael.

The potential of the OB pair is calculated using the Nerist equation, which can be represented as

E=E 0 + 0.059 a oxid.

N lg a recovery

EMF = E 0 oxid. – E 0 restore ; From ref. / From cont. = 1/K

In a state of equilibrium, the activities of the reactants and products will acquire equal values and the emf will become equal to zero, and the expression will be under the sign of the logarithm 1/K, then ln1/K = ln1 - lnK; ln1=0

O = ΔE 0 +(R T/ n F) ln

ΔE 0 - (R T/n F) lnK

or +n ∙ F Δ E 0 = RT lnK

- n · F · ΔE 0 = Δ G

In accordance with the II law of thermodynamics, only processes that occur spontaneously Δ G<0, то реакция ОВР идет слева направо, только если ЭДС>0.

ΔGp = -RT lnK = -nF ΔE 0

If a strong oxidizing agent interacts with a strong reducing agent, then a one-way process occurs. It practically flows to the end, because... its products are usually compounds with weakly expressed redox properties. If there is a small difference in the redox activity of the starting substances and reaction products, the process is two-way. To quantitatively assess the direction of processes, the G 0 values of the reagents and reaction products are used, and for the special case of ORR occurring in dilute aqueous solutions at t = 25 0 C, 101 kPa, the values of normal electrode potentials can be used.

By comparing the electrode potentials of the corresponding systems, it is possible to determine in advance the direction in which the ORR will proceed.

Standard e.m.f. E 0 of a galvanic cell is related to the standard Gibbs energy ∆ G 0 occurring in the reaction element by the relation.

∆G 0 =-nF· ∆E 0

On the other hand, ∆G 0 is related to the equilibrium constant K of the reaction by the equation.

n F ΔE 0 = 2.3 RT logK

logK = n ∙F Δ E 0 / 2.3 RT

logK = n ∙ Δ E 0 / 0.059

This work can be provided as the amount of electricity variable during this process nF (n is the number of electrons passing from the reducing agent to the oxidizing agent in an elementary reaction, F is Faraday’s constant), multiplied by the potential difference E between the electrodes of the corresponding galvanic cell.

ΔG= - n · F · Δ E

For ORR at T = 298 ΔG 0 = - n · F · ΔE 0

Under conditions other than standard for the system

Ox+ne↔Red electrode potential is determined by the Nernst equation

E o x /Red = E 0 0 x /Red - (RT/n F) ln(C(Red)/С(Ox))

E ox / Red and E 0 ox / Red - electrode and standard potentials of the system.

n is the number of electrons participating in the process.

C(Red) and C ox are the molar concentrations of reduced and oxidized forms.

E ox/Red = E 0 ox/Red – (0.059/n) log C(Red)/С(Ox)

For example:

for the system MnO 4 - + 8H + + 5e ↔ Mn 2+ + 4H 2 O

E MnO - 4 / Mn 2+ = 1.51 – (0.059/5) log (C Mn 2+ / C MnO - 4 C (H+) 8)

Select potassium halide as a reducing agent for FeCl 3

2KG - + 2 Fe 3+ Cl 3(P) = G 0 2 + 2KCl (P) + 2 Fe 2+ Cl 2(P)

(G = F -, Cl -, Br -, I -)

According to the table: E 0 Fe 3+ / Fe +2 = +0.77B

E 0 F2/2F- = + 2.86B

E 0 Cl2/2Cl- = + 1.36B

E 0 Br2/2Br- = +1.07B

E 0 I 2/2 I - = +0.54V< 0.77 B

Let's calculate the emf. involving hologenids for KI

ΔE 0 298 = E 0 oxid. – E 0 restore = 0.77 – 0.54 = 0.23V > 0. Only potassium iodide will

change in Gibbs energy of a reaction

ΔGreaction = ∑ΔG0298f(final) - ∑ΔG0298f(initial)

Using the example of the reaction 2NaOH + H2SO4 = Na2SO4 + H2O

Gibbs energy education

ΔGreaction=[ΔGNa2SO4+ΔGH2O] - [ΔGH2SO4+ΔGNaOH*2]

Galvanic cells are widely used in telephones, toys, alarm systems, etc. Since the range of devices that use dry cells is very wide and, in addition, they require periodic replacement

Electrolysis is a set of processes that take place on electrodes when an electric current passes through a solution or melt of an electrolyte. Electrolytes are conductors of the second kind. During electrolysis, the cathode serves as a reducing agent (donates electrons to cations), and the anode serves as an oxidizing agent (receives electrons from anions). The essence of electrolysis is the implementation of chemical reactions using electrical energy - reduction at the cathode (K) and oxidation at the anode (A). These processes are called processes (reactions) of electroreduction and electrooxidation. The reducing and oxidizing effects of electric current are many times stronger than the effects of chemical oxidizing and reducing agents. A distinction is made between electrolysis of melts and electrolyte solutions.

Electrolysis of molten salts

1) All metal cations are reduced at the cathode:

K(-): Zn2+ + 2e- → Zn0; Na+ + 1e- → Na0

2) Anions of oxygen-free acids are oxidized at the anode:

A(+): 2Cl¯ - 2e-→Cl2

3) Anions of oxygen-containing acids form the corresponding acid oxide and oxygen:

A(+): 2SO42ˉ - 4e- → 2SO3 + O2

Electrolysis of aqueous electrolyte solutions

The course of the electrolysis process and the nature of the final products are greatly influenced by the nature of the solvent, the material of the electrodes, the current density on them and other factors. In aqueous solutions of electrolytes, in addition to hydrated cations and anions, there are water molecules, which can also undergo electrochemical oxidation and reduction. Which electrochemical processes will occur on the electrodes during electrolysis depends on the value of the electrode potentials of the corresponding electrochemical systems.

Anodic processes.

In the electrolysis of substances, inert anodes (graphite, platinum) and soluble anodes that do not change during the electrolysis process are used, which are oxidized during the electrolysis process more easily than anions (from zinc, nickel, silver, copper and other metals).

1) Anions of oxygen-free acids (S2ˉ, I¯, Br¯, Cl¯) at sufficient concentrations are easily oxidized to the corresponding simple substances.

2) During the electrolysis of aqueous solutions of alkalis, oxygen-containing acids and their salts, as well as hydrofluoric acid and fluorides, electrochemical oxidation of water occurs with the release of oxygen:

in alkaline solutions: 4OH¯ - 4e- → O2 + 2H2O

in acidic and neutral solutions: 2H2O - 4e- → O2 + 4H+

Application of electrolysis. Obtaining target products by electrolysis makes it possible to relatively simply (by adjusting the current strength) control the speed and direction of the process, thanks to which it is possible to carry out processes both in the “softest” and in the extremely “hard” conditions of oxidation or reduction, obtaining the strongest oxidizing agents and reducing agents. By electrolysis, H2 and O2 are produced from water, C12 from aqueous solutions of NaCl, F2 from the KF melt in KH2F3.

Hydroelectrometallurgy is an important branch of non-ferrous metallurgy (Cu, Bi, Sb, Sn, Pb, Ni, Co, Cd, Zn); it is also used to obtain noble and trace metals, Mn, Cr. Electrolysis is used directly for cathodic separation of metal after it has been transferred from ore to solution, and the solution has been purified. This process is called electroextraction. Electrolysis is also used for metal purification - electrolytic. refining (electrorefining). This process consists of anodic dissolution of the contaminated metal and its subsequent cathodic deposition. Refining and electroextraction are carried out with liquid electrodes made of mercury and amalgams (amalgam metallurgy) and with electrodes made of solid metals.

Electrolysis of electrolyte melts is an important method for the production of many. metals So, for example, raw aluminum is obtained by electrolysis of cryolite-alumina melt (Na3AlF6 + A12O3), the raw material is purified electrolytically. refining. In this case, the anode is melt A1, containing up to 35% Cu (for weighting) and therefore located at the bottom of the electrolyzer bath. The middle liquid layer of the bath contains BaCl2, AlF3 and NaF, and the upper one contains molten refiner. A1 serves as the cathode.

Electrolysis of molten magnesium chloride or dehydrated carnallite - max. a common method for obtaining Mg. In prom. scale, electrolysis of melts is used to produce alkaline and alkaline-earth. metals, Be, Ti, W, Mo, Zr, U, etc.

To electrolytic Methods for producing metals also include the reduction of metal ions to other, more electron-negative ones. metal. The isolation of metals by their reduction with hydrogen also often includes the stages of electrolysis - electrochemical. ionization of hydrogen and deposition of metal ions due to the electrons released during this process. The processes of joint release or dissolution of several play an important role. metals, joint release of metals and mol. hydrogen on the cathode and adsorption of solution components on the electrodes. Electrolysis is used to prepare metallic materials. powders with specified characteristics.

Electric BATTERIES (from Latin accumulator - collector, storage device), chemical. multiple current sources. When charging from external source of electricity current in the battery accumulates energy, edges during discharge due to chemical. r-tion directly transforming. again into the electric one and is released into the external. chain. According to the principle of operation and basic The design elements of batteries do not differ from galvanic cells, but the electrode circuits, as well as the total current-generating circuit in batteries, are reversible. Therefore, after a battery has been discharged, it can be recharged by passing current in the opposite direction: positive. In this case, an oxidizing agent is formed on the electrode, and a reducing agent is formed on the negative electrode.

Naib. Lead-acid batteries are common, often called also acidic. Their action is based on the following:

Corrosion is the process of spontaneous destruction of metal as a result of physical-chemical. interactions with the environment. Mechanism: chemical and electrochemical.

Chemical - destruction of a metal during its oxidation without the generation of an electric current. Electrochemical - destruction of metal in an electrolyte environment with the appearance of an electric current inside the system.

Corrosion affects parts and components of engines, gas turbines, and rocket launchers. Chemical corrosion occurs during metal processing at high temperatures and stress.

Electrochemical corrosion can occur: in electrolytes, in soil, in an atmosphere of any moist gas.

As protection against corrosion, application of some kind of coating can be used that prevents the formation of a corrosive element (passive method)

Paint coating, polymer coating and enameling must, first of all, prevent the access of oxygen and moisture. Coating, for example, of steel with other metals such as zinc, tin, chromium, and nickel is often also used. The zinc coating protects the steel even when the coating is partially destroyed. Zinc has a more negative potential and corrodes first. Zn2+ ions are toxic. In the manufacture of cans, tin coated with a layer of tin is used. Unlike galvanized sheet, when the tin layer is destroyed, the iron begins to corrode, and more intensely, since tin has a more positive potential. Another way to protect metal from corrosion is to use a protective electrode with a high negative potential, for example, made of zinc or magnesium. For this purpose, a corrosion element is specially created. The protected metal acts as a cathode, and this type of protection is called cathodic protection. The dissolving electrode is called, accordingly, a sacrificial protection anode. This method is used to protect sea vessels, bridges, boiler plants, and underground pipes from corrosion. To protect the ship's hull, zinc plates are attached to the outside of the hull.

If you compare the potentials of zinc and magnesium with iron, they have more negative potentials. Nevertheless, they corrode more slowly due to the formation of a protective oxide film on the surface, which protects the metal from further corrosion. The formation of such a film is called metal passivation. In aluminum it is enhanced by anodic oxidation (anodizing). When a small amount of chromium is added to steel, an oxide film forms on the surface of the metal. The chromium content in stainless steel is more than 12 percent.

Cathodic corrosion protection

Cathodic electrochemical corrosion protection is used when the metal being protected is not prone to passivation. This is one of the main types of protection of metals from corrosion. The essence of cathodic protection is the application of an external current to the product from the negative pole, which polarizes the cathode sections of the corrosive elements, bringing the potential value closer to the anodic ones. The positive pole of the current source is connected to the anode. In this case, corrosion of the protected structure is almost reduced to zero. The anode gradually deteriorates and must be replaced periodically.

There are several options for cathodic protection: polarization from an external source of electric current; reducing the rate of the cathodic process (for example, deaeration of the electrolyte); contact with a metal whose free corrosion potential in a given environment is more electronegative (so-called sacrificial protection).

Polarization from an external source of electric current is used very often to protect structures located in soil, water (bottoms of ships, etc.). In addition, this type of corrosion protection is used for zinc, tin, aluminum and its alloys, titanium, copper and its alloys, lead, as well as high-chromium, carbon, alloy (both low and high alloy) steels.

The external current source is cathodic protection stations, which consist of a rectifier (converter), a current supply to the protected structure, anode grounding conductors, a reference electrode and an anode cable.

Cathodic protection Both independent and additional types of corrosion protection are used.

The main criterion by which one can judge the effectiveness of cathodic protection is the protective potential. Protective potential is the potential at which the corrosion rate of a metal under certain environmental conditions takes on the lowest (as far as possible) value.

There are disadvantages to using cathodic protection. One of them is the danger of overprotection. Overprotection is observed with a large shift in the potential of the protected object in the negative direction. At the same time it stands out. The result is the destruction of protective coatings, hydrogen embrittlement of the metal, and corrosion cracking.

Tread protection(use of protector)

A type of cathodic protection is sacrificial. When using sacrificial protection, a metal with a more electronegative potential is connected to the protected object. In this case, it is not the structure that is destroyed, but the tread. Over time, the protector corrodes and must be replaced with a new one.

Tread protection is effective in cases where there is a small transition resistance between the protector and the environment.

Each protector has its own radius of protective action, which is determined by the maximum possible distance to which the protector can be removed without losing the protective effect. Protective protection is most often used when it is impossible or difficult and expensive to supply current to the structure.

Protectors are used to protect structures in neutral environments (sea or river water, air, soil, etc.).

The following metals are used to make protectors: magnesium, zinc, iron, aluminum. Pure metals do not fully perform their protective functions, so they are additionally alloyed during the manufacture of protectors.

Iron protectors are made from carbon steel or pure iron.

Corrosion inhibitors, chemical compounds or their compositions, the presence of which in small quantities in an aggressive environment slows down the corrosion of metals. The protective effect of corrosion inhibitors is due to a change in the state of the metal surface due to adsorption (adsorptive corrosion inhibitors) or the formation of sparingly soluble compounds with metal ions. The latter form a film on the surface that is significantly thinner than conversion protective coatings (see Corrosion Protection). The slowdown in corrosion occurs due to a decrease in the active surface area of the metal and a change in the activation energy of electrode reactions, which limit the complex corrosion process. Corrosion inhibitors can inhibit anodic dissolution and cause passivation of the metal (anodic corrosion inhibitors), reduce the rate of the cathodic process (cathodic corrosion inhibitors), or slow down both of these processes (mixed corrosion inhibitors).

Direction of flow. Possibility and direction of ORR occurrence How can you change the direction of the redox reaction

OVR direction criterion

Irreversible DC sources. Examples

= Our discussions =

Criteria for the occurrence of OVR. Standard conditions and standard potentials

Methodological development for teachers and students