Criteria for the direction of the RIA

The course of redox reactions depends on the nature of the interacting substances and on the conditions of the reactions. 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. The 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 spontaneously proceed always towards the transformation of a strong oxidizing agent into a weak conjugated reducing agent or a strong reducing agent into a weak conjugated oxidizing agent. The direction of the redox reaction can be predicted by analyzing the values of the redox potentials of conjugated pairs.

irreversible sources direct current. Examples

Galvanic cells are classified as chemical current sources in which the direct conversion of chemical energy into electrical energy takes place. Galvanic cells are primary chemical current sources in which chemical reactions are irreversible. In its simplest form, the cell consists of two electrodes made of different metals, immersed in an electrolyte solution. At the same time, one of the electrodes (cathode) undergoes a reaction of dissolution of the electrode material - or oxidation, in which the electrode loses electrons, giving them to an external electrical circuit. On the other electrode (anode) there is a reduction reaction - neutralization of the ions of the material surrounding the electrode, due to the electrons coming from the cathode through the external circuit. Potential difference ( electromotive force) for various elements is in the range 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 is a zinc lining 3 in the form of a cup or cup. Ammonia solution is most often used as an electrolyte. The biscuit element is given a flat shape, convenient for connection into 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. Such a device has, for example, a Krona battery. The film coating isolates individual elements from the sides, prevents electrolyte leakage, but easily passes gases formed inside the element. The mass of activated carbon protrudes slightly from the biscuit for easy contact with another biscuit. In these batteries, the active material is used better and more completely than in cup batteries. Dry galvanic elements are delivered to the consumer in finished form; water-filled before use must be filled with clean water. The voltage of a galvanic cell is always less than the emf it develops, firstly, due to the voltage drop inside the cell on its internal resistance, 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 of the current passing in the circuit. For example, the release 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 effect of polarization on the operation of the element, depolarizers are used - substances that take on hydrogen or oxygen, react with them and thus help to reduce the polarization potential. In galvanic cells with a carbon electrode, manganese dioxide is used as a depolarizer. To in electrical circuit to obtain a voltage exceeding 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 cells:

E about \u003d E 1 + E 2 + E 3 + ... + E n.

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

E about \u003d n * E el.

Galvanic cells must be protected from short circuits and it is not recommended to check "for a spark". Their voltage should be measured under load. In the absence of load, the voltmeter will show emf, which does not characterize the degree of battery usage.

Spontaneous flow criterion chemical processes is the change in Gibbs free energy (ΔG< О). Изменение энергии Гиббса ОВР связано с разностью окислительно-восстановительных (электродных) потенциалов участников окислительно-восстановительного процесса Е:

where F is Faraday's constant; n is the number of electrons involved in the redox process; E is the difference in redox potentials or the electromotive force of the OVR (EMF of a galvanic cell formed by two redox systems):

E \u003d j 0 - j B,

where j 0 is the potential of the oxidizer, j B is the potential of the reductant .

Given the above: OVR flows in the forward direction if its EMF is positive, i.e. E>O; otherwise (E<О) ОВР будет протекать в обратном направлении. The emf calculated for standard conditions is called standard and is denoted by E.

Example 1: Determine if the reaction can proceed in the forward direction at standard conditions:

2Fe 3+ + 2 I D 2Fe 2+ + I 2 .

When the reaction proceeds in the forward direction, Fe3 + ions will be the oxidizing agent, and iodide ions (I) will be the reducing agent. Calculate the standard EMF:

Answer: This reaction can only proceed in the forward direction.

Example 2. Determine the direction of the reaction under standard conditions:

2KCI + 2MnCI 2 + 5CI 2 + 8H 2 O D 2KMnO 4 + 16HCI.

Assume that the reaction proceeds in the forward direction, then

The reaction cannot proceed in the forward direction. It will flow from right to left, in this case.

Answer: This reaction proceeds from right to left.

Thus, the reaction will proceed in the direction in which the EMF is positive. Always systems with a higher redox potential will oxidize systems with a lower value.

Electrochemical processes

The processes of mutual transformation of chemical and electrical forms of energy are called electrochemical processes. Electrochemical processes can be divided into two main groups:

1) the processes of converting chemical energy into electrical energy (in galvanic cells);

2) the processes of converting electrical energy into chemical energy (electrolysis).

An electrochemical system consists of two electrodes and an ionic conductor between them (melt, electrolyte solution or solid electrolytes - conductors of the 2nd kind). Electrodes are called conductors of the first kind, having electronic conductivity and being in contact with the ionic conductor. To ensure the operation of the electrochemical system, the electrodes are connected to each other by a metal conductor, called the external circuit of the electrochemical system.

10.1. Galvanic cells (chemical sources of electric current)

A galvanic cell (GE) is a device in which the chemical energy of a redox reaction is converted into electric current energy. Theoretically, any OVR can be used to produce electrical energy.

Consider one of the simplest GEs - copper-zinc, or the Daniel-Jacobi element (Fig. 10.1). Plates of zinc and copper are connected in it with a conductor, while each of the metals is immersed in a solution of the corresponding salt: zinc sulfate and copper (II) sulfate. The half-cells are connected by an electrolytic key1 if they are in different vessels or separated by a porous partition if they are in the same vessel.

Let us first consider the state of this element with an open external circuit - the "idle" mode. As a result of the exchange process, the following equilibria are established on the electrodes, which, under standard conditions, correspond to standard electrode potentials:

Zn 2+ + 2e - D Zn \u003d - 0.76V

Cu 2+ + 2e - D Cu \u003d + 0.34V.

The potential of the zinc electrode has a more negative value than the potential of the copper electrode, therefore, when the external circuit is closed, i.e. when connecting zinc to copper with a metal conductor, electrons will move from zinc to copper. As a result of the transfer of electrons from zinc to copper, the equilibrium on the zinc electrode will shift to the left, so an additional amount of zinc ions will go into solution (dissolution of zinc on the zinc electrode). At the same time, the equilibrium on the copper electrode will shift to the right and a discharge of copper ions will occur (copper precipitation on the copper electrode). These spontaneous processes will continue until the potentials of the electrodes are equalized or all the zinc is dissolved (or all the copper is deposited on the copper electrode).

So, during the operation of the Daniel-Jacobi element (when the internal and external circuits of the GE are closed), the following processes occur:

1) the movement of electrons in the external circuit from the zinc electrode to the copper one, because< ;

2) zinc oxidation reaction: Zn - 2e - = Zn 2+.

Oxidation processes in electrochemistry are called anode processes, and the electrodes on which oxidation processes take place are called anodes; therefore, the zinc electrode is an anode;

3) reduction reaction of copper ions: Cu 2+ + 2e = Cu.

The reduction processes in electrochemistry are called cathodic processes, and the electrodes on which the reduction processes take place are called cathodes; therefore, the copper electrode is the cathode;

4) the movement of ions in solution: anions (SO 4 2-) to the anode, cations (Cu 2+, Zn 2+) to the cathode, closes the electrical circuit of the galvanic cell.

The direction of this movement is determined by the electric field resulting from the occurrence of electrode processes: anions are consumed at the anode, and cations at the cathode;

5) summing up the electrode reactions, we obtain:

Zn + Cu 2+ = Cu + Zn 2+

or in molecular form: Zn + CuSO 4 \u003d Cu + ZnSO 4.

As a result of this chemical reaction in a galvanic cell, the movement of electrons occurs in the external circuit of ions inside the cell, i.e. electricity, therefore, the total chemical reaction occurring in a galvanic cell is called current-generating.

In a schematic notation that replaces the drawing of a galvanic cell, the interface between the conductor of the 1st kind and the conductor of the 2nd kind is indicated by one vertical line, and the interface between the conductors of the 2nd kind is indicated by two lines. The anode - the source of electrons entering the external circuit - is considered to be negative, the cathode - positive. The anode is placed in the circuit on the left. The Daniel-Jacobi GE scheme, for example, is written as:

(-) Zn |ZnSO 4 | |CuSO 4 | Cu(+)

or in ion-molecular form:

(-) Zn |Zn 2+ ||Cu 2+ | Cu (+).

The reason for the occurrence and flow of electric current in a galvanic cell is the difference in redox potentials (electrode potentials 1) private reactions that determine the electromotive force E e of the galvanic cell, and in this case:

In the general case: E e \u003d j k - j a,

where j k is the cathode potential, j a is the anode potential.

E e is always greater than zero (E e > O). If the reaction is carried out under standard conditions, then the EMF observed in this case is called the standard electromotive force of this element. For the Daniel - Jacobi element, the standard EMF \u003d 0.34 - (-0.76) \u003d 1.1 (V).

Make a diagram, write the equations of electrode processes and current-generating reaction for a galvanic cell formed by bismuth and iron, immersed in solutions of their own salts with a concentration of metal ions in a solution of C Bi 3+ = 0.1 mol / l, C Fe 2+ = 0.01 mol/l. Calculate the EMF of this element at 298K.

The concentrations of metal ions in the solution are different from the concentration of 1 mol/l, therefore, it is necessary to calculate the metal potentials using the Nernst equation, compare them and determine the anode and cathode.

j me n + /me = j about me n + /me + lgSme n + ;

j Bi 3+ / Bi \u003d 0.21 + lg10 -1 \u003d 0.19V; j F e 2+ / F e \u003d -0.44 + lg10 -2 \u003d - 0.499V.

The iron electrode is the anode, the bismuth electrode is the cathode. GE scheme:

(-)Fe |Fe(NO 3) 2 ||Bi(NO 3) 3 |Bi(+)

or (-) Fe|Fe 2+ ||Bi 3+ |Bi (+).

Equations of electrode processes and current-generating reaction:

A: Fe - 2 = Fe 2+ 3

K: Bi 3+ + 3 = Bi 2

3 Fe + 2Bi 3+ = 3Fe 2+ + 2 Bi

EMF of this element E e \u003d 0.19 - (-0.499) \u003d 0.689 V.

In some cases, the electrode metal does not undergo changes during the electrode process, but participates only in the transfer of electrons from the reduced form of the substance to its oxidized form. So, in a galvanic cell

Pt |Fe 2+ , Fe 3+ || MnO , Mn 2+ , H + | Pt

platinum plays the role of inert electrodes. Iron (II) is oxidized on a platinum anode:

Fe 2+ - e - \u003d Fe 3+, ,

and MnO is reduced on the platinum cathode:

MnO 4 - + 8H + + 5e - \u003d Mn 2+ + 4H 2 O,

Current-generating reaction equation:

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

Standard EMF E \u003d 1.51-0.77 \u003d 0.74 V.

A galvanic cell can be composed not only of different, but also of the same electrodes immersed in solutions of the same electrolyte, differing only in concentration (concentration galvanic cells). For example:

(-) Ag |Ag + ||Ag + |Ag (+)

C Ag< C Ag

Electrode reactions: A: Ag – eˉ = Ag + ;

K: Ag + + eˉ = Ag.

Current generating reaction equation: Ag + Ag + = Ag + + Ag.

Lead battery. A ready-to-use lead-acid battery consists of gridded lead plates, some of which are filled with lead dioxide and others filled with spongy lead metal. The plates are immersed in a 35 - 40% solution of H 2 SO 4; at this concentration, the electrical conductivity of the sulfuric acid solution is maximum.

During battery operation - when it is discharged - OVR occurs in it, during which lead (Pb) is oxidized, and lead dioxide is reduced:

(-) Рb|H 2 SO 4 | PbO 2 (+)

A: Pb + SO -2eˉ = PbSO 4

K: РbО 2 + SO + 4Н + + 2еˉ = PbSO 4 + 2H 2 O

Pb + PbO 2 + 4H + + 2SO 4 2- \u003d 2PbSO 4 + 2H 2 O (current-forming reaction). .

In the internal circuit (in H 2 SO 4 solution), when the battery is operating, ions are transferred: SO 4 2- ions move towards the anode, and H + cations move towards the cathode. The direction of this movement is determined by the electric field resulting from the occurrence of electrode processes: anions are consumed at the anode, and cations are consumed at the cathode. As a result, the solution remains electrically neutral.

To charge the battery, they are connected to an external DC source (“+” to “+”, “–“ to “–“). In this case, the current flows through the battery in the opposite direction, opposite to that in which it passed when the battery was discharged; electrolysis takes place in an electrochemical system (see p. 10.2). As a result, the electrochemical processes on the electrodes are "reversed". The lead electrode now undergoes a reduction process (the electrode becomes the cathode):

PbSO 4 + 2eˉ \u003d Pb + SO 4 2-.

On the PbO 2 electrode, when charging, the oxidation process takes place (the electrode becomes the anode):

PbSO 4 + 2H 2 O - 2eˉ \u003d PbO 2 + 4H + + SO 4 2-.

Summary Equation:

2PbSO 4 + 2H 2 O \u003d Pb + PbO 2 + 4H + + 2SO 4 2-.

It is easy to see that this process is the opposite of that which occurs during the operation of the battery: when the battery is charged, the substances necessary for its operation are again obtained in it.

Electrolysis

Electrolysis is a redox reaction occurring on electrodes in an electrolyte solution or melt under the action of a direct electric current supplied from an external source. Electrolysis converts electrical energy into chemical energy. The device in which electrolysis is carried out is called an electrolyser. On the negative electrode of the electrolyzer (cathode), the reduction process occurs - the oxidizer attaches electrons coming from the electrical circuit, and on the positive electrode (anode) - the oxidation process - the transfer of electrons from the reducing agent to the electrical circuit.

In this way, the distribution of signs of the charge of the electrodes is opposite to that which exists during the operation of the galvanic cell. The reason for this is that the processes occurring during electrolysis are, in principle, the reverse of the processes occurring during the operation of a galvanic cell. During electrolysis, the processes are carried out due to the energy of an electric current supplied from the outside, while during the operation of a galvanic cell, the energy of a spontaneous chemical reaction occurring in it is converted into electrical energy. For electrolysis processes DG>0, i.e. under standard conditions, they do not spontaneously go.

Electrolysis of melts. Consider the electrolysis of a melt of sodium chloride (Fig. 10.2). This is the simplest case of electrolysis, when the electrolyte consists of one type of cations (Na +) and one type of anions (Cl) and there are no other particles that can participate in electrolysis. The process of electrolysis of the NaCl melt proceeds as follows. Using an external current source, electrons are brought to one of the electrodes, imparting a negative charge to it. Na + cations under the action electric field move to the negative electrode, interacting with the electrons coming through the external circuit. This electrode is the cathode, and the process of reduction of Na + cations takes place on it. Cl anions move towards the positive electrode and, having donated electrons to the anode, are oxidized. The electrolysis process is visually depicted by a diagram that shows the dissociation of the electrolyte, the direction of movement of the ions, the processes on the electrodes and the released substances. . The electrolysis scheme of a sodium chloride melt looks like this:

NaCl = Na + + Cl

(-) Cathode: Na + Anode (+): Cl

Na + + e - = Na 2Cl - 2eˉ = Cl 2

Summary Equation:

2Na + + 2Cl electrolysis 2Na + Cl 2

or in molecular form

2NaCl ELECTROLYSIS 2Na + Cl 2

This reaction is a redox reaction: the oxidation process occurs at the anode, and the reduction process occurs at the cathode.

In the processes of electrolysis of electrolyte solutions, water molecules can participate and polarization of the electrodes takes place.

Polarization and overvoltage. Electrode potentials determined in electrolyte solutions in the absence of electric current in the circuit are called equilibrium potentials (under standard conditions - standard electrode potentials). With the passage of electric current, the potentials of the electrodes change . The change in the potential of the electrode during the passage of current is called polarization:

Dj \u003d j i - j p,

where Dj - polarization;

j i is the potential of the electrode during the passage of current;

j p is the equilibrium potential of the electrode.

When the cause of the change in potential during the passage of current is known, instead of the term "polarization", use the term "overvoltage". It is also related to some specific processes, such as cathodic hydrogen evolution (hydrogen surge).

To experimentally determine the polarization, a curve of dependence of the electrode potential on the current density flowing through the electrode is built. Since the electrodes can be different in area, depending on the area of the electrode at the same potential, there can be different currents; therefore, the current is usually referred to a unit surface area. The ratio of the current I to the electrode area S is called the current density I:

The graphical dependence of the potential on the current density is called the polarization curve(Fig. 10.3). With the passage of current, the potentials of the electrodes of the electrolyzer change, i.e. electrode polarization occurs. Due to cathodic polarization (Dj k), the cathode potential becomes more negative, and due to anodic polarization (Dj a), the anode potential becomes more positive.

The sequence of electrode processes in the electrolysis of electrolyte solutions. Water molecules, H + and OH ions can participate in the processes of electrolysis of electrolyte solutions, depending on the nature of the medium. When determining the products of electrolysis of aqueous solutions of electrolytes, in the simplest cases, one can be guided by the following considerations:

1. Cathodic processes.

1.1. At the cathode, the processes characterized by the highest electrode potential, i.e. the strongest oxidizing agents are reduced first.

1.2. Metal cations that have a standard electrode potential greater than that of hydrogen (Cu 2+ , Ag + , Hg 2+ , Au 3+ and other low-active metal cations - see p.11.2) are almost completely reduced on the cathode during electrolysis:

Me n + + neˉ "Me.

1.3. Metal cations, the potential of which is much lower than that of hydrogen (those in the “Row of voltages” from Li + to Al 3+ inclusive, i.e. active metal cations), are not reduced at the cathode, since water molecules are reduced at the cathode:

2H 2 O + 2eˉ ® H 2 + 2OH.

The electrochemical release of hydrogen from acidic solutions occurs due to the discharge of hydrogen ions:

2Н + + 2еˉ " Н 2 .

1.4. Metal cations that have a standard electrode potential are less than that of hydrogen, but more than that of aluminum (standing in the "Row of voltages" from Al 3+ to 2H + - metal cations of medium activity), during electrolysis at the cathode, they are reduced simultaneously with water molecules:

Me n + + neˉ ® Me

2H 2 O + 2eˉ ® H 2 + 2OH.

This group includes ions Sn 2+ , Pb 2+ , Ni 2+ , Co 2+ , Zn 2+ , Cd 2+ , etc. When comparing the standard potentials of these metal ions and hydrogen, one could conclude that precipitation of metals at the cathode. However, you should take into account:

· the standard potential of the hydrogen electrode refers to a n+ [H + ] 1 mol/l., i.е. pH=0; with an increase in pH, the potential of the hydrogen electrode decreases, becomes more negative ( ; see section 10.3); at the same time, the potentials of metals in the region where their insoluble hydroxides do not precipitate do not depend on pH;

the polarization of the hydrogen reduction process is greater than the polarization of the discharge of metal ions of this group (or in other words, hydrogen evolution at the cathode occurs with a higher overvoltage compared to the overvoltage of the discharge of many metal ions of this group); example: polarization curves of the cathodic release of hydrogen and zinc (Fig. 10.4).

As can be seen from this figure, the equilibrium potential of the zinc electrode is less than the potential of the hydrogen electrode; at low current densities, only hydrogen is released at the cathode. But the hydrogen overvoltage of the electrode is greater than the overvoltage of the zinc electrode, therefore, with an increase in current density, zinc also begins to be released on the electrode. At the potential φ 1, the current densities of the evolution of hydrogen and zinc are the same, and at the potential φ 2, i.e. zinc is released mainly on the electrode.

2. Anode processes.

2.1. At the anode, the processes characterized by the lowest electrode potential occur first, i.e. strong reducing agents are oxidized first.

2.2. Usually anodes are divided into inert (insoluble) and active (soluble). The first are made from coal, graphite, titanium, platinum metals, which have a significant positive electrode potential or are covered with a stable protective film, serving only as conductors of electrons. The second ones are from metals whose ions are present in the electrolyte solution - from copper, zinc, silver, nickel, etc.

2.3. On an inert anode during the electrolysis of aqueous solutions of alkalis, oxygen-containing acids and their salts, as well as HF and its salts (fluorides), electrochemical oxidation of hydroxide ions occurs with the release of oxygen. Depending on the pH of the solution, this process proceeds differently and can be written with various equations:

a) in acidic and neutral environments

2 H 2 O - 4eˉ \u003d O 2 + 4 H +;

b) in an alkaline environment

4OH - 4eˉ \u003d O 2 + 2H 2 O.

The oxidation potential of hydroxide ions (oxygen electrode potential) is calculated using the equation (see section 10.3):

Oxygen-containing anions SO, SO, NO, CO, PO, etc. or are not able to oxidize, or their oxidation occurs at very high potentials, for example: 2SO - 2eˉ \u003d S 2 O \u003d 2.01 V.

2.4. During the electrolysis of aqueous solutions of anoxic acids and their salts (except for HF and its salts), their anions are discharged at an inert anode.

Note that the release of chlorine (Cl 2) during the electrolysis of a solution of HCl and its salts, the release of bromine (Br 2) during the electrolysis of a solution of HBr and its salts contradicts the mutual position of the systems.

2Cl - 2eˉ \u003d Cl 2 \u003d 1.356 V

2Br - 2eˉ \u003d Br 2 \u003d 1.087 V

2H 2 O - 4eˉ \u003d O 2 + 4 H + \u003d 0.82 V (pH \u003d 7)

This anomaly is associated with the anodic polarization of processes (Fig. 10.5). As can be seen, the equilibrium potential of the oxygen electrode (oxidation potential of hydroxide ions from water) is less than the equilibrium potential of the chloride electrode (oxidation potential of chloride ions). Therefore, at low current densities, only oxygen is released. However, the evolution of oxygen proceeds with a higher polarization than the evolution of chlorine, therefore, at a potential, the currents for the evolution of chlorine and oxygen are equal, and at a potential (high current density), chlorine is mainly released.

2.5. If the potential of the metal anode is less than the potential of OH ions or other substances present in the solution or on the electrode, then electrolysis proceeds with an active anode. The active anode is oxidized, dissolving: Me - neˉ ® Me n + .

current output . If the potentials of two or more electrode reactions are equal, then these reactions proceed at the electrode simultaneously. In this case, the electricity passed through the electrode is consumed in all these reactions. The fraction of the amount of electricity spent on the transformation of one of the substances (B j) is called current output of this substance:

(Bj) % = (Qj /Q) . 100,

where Q j is the amount of electricity spent on the transformation of the j-th substance; Q is the total amount of electricity passed through the electrode.

For example, from fig. 10.4 it follows that the current efficiency of zinc increases with increasing cathodic polarization. For this example, a high hydrogen overvoltage is a positive phenomenon. As a result, manganese, zinc, chromium, iron, cobalt, nickel, and other metals can be isolated from aqueous solutions at the cathode.

Faraday's law. The theoretical relationship between the amount of electricity passed and the amount of substance oxidized or reduced at the electrode is determined by Faraday's law, according to which the mass of the electrolyte that has undergone a chemical transformation, as well as the mass of substances released on the electrodes, are directly proportional to the amount of electricity passed through the electrolyte and the molar masses of the equivalents of the substances: m \u003d M e It / F,

where m is the mass of electrolyte subjected to chemical transformation,

or the mass of substances - products of electrolysis, released on the electrodes, g; M e is the molar mass of the equivalent of a substance, g / mol; I - current strength, A; t is the duration of electrolysis, s; F - Faraday number - 96480 C / mol.

Example 1 How does the electrolysis of an aqueous solution of sodium sulfate proceed with a carbon (inert) anode?

Na 2 SO 4 \u003d 2Na + + SO

H 2 O D H + + OH

Summary Equation:

6H 2 O \u003d 2H 2 + O 2 + 4OH + 4H +

or in molecular form

6H 2 O + 2Na 2 SO 4 \u003d 2H 2 + O 2 + 4NaOH + 2H 2 SO 4.

Na + ions and OH - ions accumulate in the cathode space; an alkali is formed, and near the anode the environment becomes acidic due to the formation of sulfuric acid. If the cathode and anode spaces are not separated by a partition, then the H + and OH ions form water, and the equation takes the form

The basis for determining the direction of the spontaneous occurrence of redox reactions is the following rule:

This rule is similar to the rule that determines the direction of acid-base transformations.

A quantitative measure of the redox ability of a given conjugated redox pair is its value recovery potential f, which depends on:

The nature of the oxidized and reduced form of a given conjugate pair;

The concentration ratios of the oxidized and reduced forms of a given conjugated pair;

Temperatures.

In cases where H + or OH- ions are involved in the process of converting an oxidizing agent or reducing agent, (p also depends on the pH of the solution. The value that f takes under standard conditions: the concentration of all components involved in the reaction, including water ions H + (in an acidic environment) and OH- (in an alkaline environment), equal to 1 mol / l, temperature 298 K, - is called standard recovery potential and is denoted by (f°. The value of f° is a quantitative characteristic of the redox properties of a given conjugated redox pair under standard conditions.

There is no way to determine the absolute value of potentials for conjugated redox pairs. Therefore, they use relative values (Sec. 25.2), characterizing the potentials of conjugated pairs relative to the reference pair, the potential of which, under standard conditions, is assumed to be conditionally equal to zero

A positive value of f° are redox pairs in which the oxidized form adds electrons more easily than the hydrogen cation in the reference pair. Negative value f° have redox pairs, in which the oxidized form attaches electrons more difficult than H + in the reference pair. Consequently, the greater (i.e., more positive) the value of f ° of a given conjugated redox pair, the more pronounced its oxidizing properties, and the reducing properties, respectively, are weaker.

In table. 9.1 shows the standard values of the potentials of some conjugated redox pairs.

Under conditions other than standard, the value of φ is calculated according to the Nernst equation (sections 25.2, 25.3).

The essence of redox reactions is the competition for the addition of an electron between the participating oxidizing agents. In this case, the electron is attached to that conjugated pair, the oxidized form of which holds it stronger. This is reflected in the following diagram: *

Comparing the potentials of conjugated pairs involved in the redox reaction, it is possible to determine in advance the direction in which this or that reaction will spontaneously proceed.

In the interaction of two conjugated redox pairs, the oxidizing agent will always be the oxidized form of the pair whose potential has a more positive value.

Example. The reaction mixture contains two conjugated redox pairs:

Since the first pair contains a stronger oxidizing agent (I2) than the second pair (S), then under standard conditions a reaction will spontaneously proceed in which I2 will be the oxidizing agent, and the reducing agent

To determine the direction of the redox reaction, you can also use the value of its EMF.

The EMF of the redox reaction under standard conditions (E°) is numerically equal to the difference in the standard potentials of the conjugated redox pairs involved in the reaction:

The condition for the spontaneous occurrence of a redox reaction is the positive value of its EMF, i.e.

Given this condition, for a spontaneously occurring redox reaction, the value f of the redox pair acting as an oxidizing agent must be greater than f of the second redox pair playing the role of a reducing agent in this reaction. So, in the example above:
If a E°= 0, then the redox reaction is equally likely to occur both in the forward and reverse directions, and this is the condition for the occurrence of chemical equilibrium for the redox process. A quantitative characteristic of the course of any reversible processes is the equilibrium constant TO, which is related to the change in the standard Gibbs energy (Sec. 5.5) as follows:

On the other hand, the change in the standard Gibbs energy is related to the EMF of the redox reaction by the relation:

where F= 96 500 C/mol; z- the number of electrons involved in the elementary process.

From these two equations follows:

Using these expressions, it is possible to calculate the equilibrium constant of any redox reaction, but it will only have a real value for those reactions whose EMF is less than 0.35 V, since at high EMF the reactions are considered practically irreversible. Since the EMF of individual stages of redox reactions occurring in living systems usually does not exceed 0.35 V (| E°| < 0,35 В), то большинство из них практически обратимы, причем обратимость процесса выражена тем сильнее, чем величина | E°| closer to zero.

Redox reactions underlie the metabolism of any organisms. When aerobic metabolism the main oxidizing agent is molecular oxygen supplied during respiration, and the reducing agent is organic compounds supplied with food. At anaerobic metabolism it is based mainly on redox reactions, in which organic compounds are both oxidizing and reducing agents.

Redox potential is a special, narrow case of the concept of electrode potential. Let's take a closer look at these concepts.

AT OVR electron transfer reducing agents oxidizing agents occurs when the particles are in direct contact, and the energy of a chemical reaction is converted into heat. Any energy OVR flowing in solution can be converted into electrical energy. For example, if redox processes are separated spatially, i.e. the transfer of electrons by the reducing agent will occur through a conductor of electricity. This is implemented in galvanic cells, where Electric Energy obtained from chemical energy.

Consider, in which the left vessel is filled with a solution of zinc sulfate ZnSO 4, with a zinc plate lowered into it, and the right vessel is filled with a solution of copper sulfate CuSO 4, with a copper plate lowered into it.

Interaction between the solution and the plate, which acts as an electrode, causes the electrode to acquire an electrical charge. The potential difference that occurs at the interface of the metal-electrolyte solution is called electrode potential. Its value and sign (+ or -) are determined by the nature of the solution and the metal in it. When metals are immersed in solutions of their salts, the more active of them (Zn, Fe, etc.) are charged negatively, and the less active ones (Cu, Ag, Au, etc.) are positively charged.

The result of the connection of a zinc and copper plate with a conductor of electricity is the occurrence of an electric current in the circuit due to the flow from the zinc to the copper plate through the conductor.

In this case, there is a decrease in the number of electrons in zinc, which is compensated by the transition of Zn 2+ into solution, i.e. the zinc electrode dissolves anode (oxidation process).

Zn - 2e - = Zn 2+

In turn, the increase in the number of electrons in copper is compensated by the discharge of copper ions contained in the solution, which leads to the accumulation of copper on the copper electrode - cathode (recovery process):

Cu 2+ + 2e - = Cu

Thus, the following reaction occurs in the element:

Zn + Cu 2+ = Zn 2+ + Cu

Zn + CuSO 4 \u003d ZnSO 4 + Cu

Quantify redox processes allow electrode potentials measured relative to a normal hydrogen electrode (its potential is assumed to be zero).

To determine standard electrode potentials use an element, one of the electrodes of which is the tested metal (or non-metal), and the other is the hydrogen electrode. Based on the potential difference found at the poles of the element, the normal potential of the metal under study is determined.

Redox potential

The values of the redox potential are used if it is necessary to determine the direction of the reaction in aqueous or other solutions.

Let's carry out the reaction

2Fe 3+ + 2I - = 2Fe 2+ + I 2

so that iodide ions and iron ions exchanged their electrons through a conductor. In vessels containing solutions of Fe 3+ and I - , we place inert (platinum or carbon) electrodes and close the inner and outer circuit. An electric current is generated in the circuit. Iodide ions give up their electrons, which will flow through the conductor to an inert electrode immersed in a Fe 3+ salt solution:

2I - - 2e - = I 2

2Fe 3+ + 2e - = 2Fe 2+

Oxidation-reduction processes occur at the surface of inert electrodes. The potential that occurs at the boundary of an inert electrode - a solution and contains both an oxidized and a reduced form of a substance is called equilibrium redox potential. The value of the redox potential depends on many factors, including such as:

The nature of matter(oxidizing agent and reducing agent)
Concentration of oxidized and reduced forms. At a temperature of 25°C and a pressure of 1 atm. the value of the redox potential is calculated using Nernst equations:

E=E° + (RT/nF) ln (C ok / C sun), where

E is the redox potential of this pair;

E°- standard potential (measured at C ok =C sun);

R is the gas constant (R = 8.314 J);

T is absolute temperature, K

n is the number of given or received electrons in the redox process;

F is Faraday's constant (F = 96484.56 C/mol);

C ok - concentration (activity) of the oxidized form;

Cvos is the concentration (activity) of the reduced form.

Substituting the known data into the equation and passing to the decimal logarithm, we obtain the following form of the equation:

E=E° + (0,059/ n) lg (C ok /C sun)

At C ok >C sun, E>E° and vice versa if C ok< C sun, then E< E°

The acidity of the solution. For couples whose oxidized form contains oxygen (for example, Cr 2 O 7 2- , CrO 4 2- , MnO 4 —), as the pH of the solution decreases, the redox potential increases, i.e. the potential grows with increasing H + . Conversely, the redox potential decreases with decreasing H + .
Temperature. As the temperature increases, the redox potential of this pair also increases.

Standard redox potentials are presented in the tables of special reference books. It should be noted that only reactions in aqueous solutions at a temperature of ≈ 25°С. Such tables allow us to draw some conclusions:

The value and sign of standard redox potentials, allow you to predict what properties (oxidizing or reducing) atoms, ions or molecules will exhibit in chemical reactions, for example

E°(F 2 / 2F -) \u003d +2.87 V - the strongest oxidizing agent

E°(K + /K) \u003d - 2.924 V - the strongest reducing agent

This pair will have the greater reducing ability, the greater the numerical value of its negative potential, and the higher the oxidizing ability, the greater the positive potential.

It is possible to determine which of the compounds of one element will have the strongest oxidizing or reducing properties.
It is possible to predict the direction of the OVR. It is known that the operation of a galvanic cell takes place provided that the potential difference has a positive value. OVR flow in the chosen direction is also possible if the potential difference has a positive value. OVR proceeds towards weaker oxidizing agents and reducing agents from stronger ones, for example, the reaction

Sn 2+ + 2Fe 3+ = Sn 4+ + 2Fe 2+

Practically flows in the forward direction, because

E°(Sn 4+ / Sn 2+) = +0.15 V, and E°(Fe 3+ /Fe 2+) = +0.77 V, i.e. E°(Sn4+ /Sn2+)< E°(Fe 3+ /Fe 2+).

Cu + Fe 2+ = Cu 2+ + Fe

is impossible in the forward direction and flows only from right to left, because

E°(Cu 2+ /Cu) = +0.34 V, and E°(Fe 2+ / Fe) \u003d - 0.44 V, i.e. E°(Fe 2+ /Fe)< E°(Cu 2+ /Cu).

In the redox process, the amount of initial substances decreases, as a result of which the E of the oxidizing agent decreases, and the E of the reducing agent increases. At the end of the reaction, i.e. at the onset of chemical equilibrium, the potentials of both processes are equalized.

If, under these conditions, several OVRs are possible, then the reaction with the largest difference in redox potentials will proceed first.
Using the reference data, you can determine the EMF of the reaction.

So, how to determine the EMF of the reaction?

Consider several reactions and determine their EMF:

Mg + Fe 2+ \u003d Mg 2+ + Fe
Mg + 2H + = Mg 2+ + H 2
Mg + Cu 2+ \u003d Mg 2+ + Cu

E°(Mg 2+ / Mg) \u003d - 2.36 V

E°(2H + /H 2) = 0.00V

E°(Cu 2+ / Cu) = +0.34 V

E°(Fe 2+ / Fe) \u003d - 0.44 V

To determine the EMF of a reaction, find the difference between the potential of the oxidizing agent and the potential of the reducing agent

EMF \u003d E 0 ok - E 0 restore

EMF \u003d - 0.44 - (- 2.36) \u003d 1.92 V
EMF = 0.00 - (- 2.36) = 2.36 V
EMF \u003d + 0.34 - (- 2.36) \u003d 2.70 V

All of the above reactions can proceed in the forward direction, because their EMF > 0.

Equilibrium constant.

If it becomes necessary to determine the degree of reaction, then you can use equilibrium constant.

For example, for the reaction

Zn + Cu 2+ = Zn 2+ + Cu

Applying law of mass action, can be written

K = C Zn 2+ /C Cu 2+

Here equilibrium constant K shows the equilibrium ratio of the concentrations of zinc and copper ions.

The value of the equilibrium constant can be calculated by applying Nernst equation

E=E° + (0,059/ n) lg (C ok /C sun)

Let us substitute the values of the standard potentials of the Zn/Zn 2+ and Cu/Cu 2+ pairs into the equation, we find

E0 Zn/ Zn2+ = -0.76 + (0.59/2)lgC Zn / Zn2 and E0 Cu/Cu2+ = +0.34 + (0.59/2)lgC Cu / Cu2+

In a state of equilibrium E0 Zn/Zn2+ = E0 Cu/Cu2+, i.e.

0.76 + (0.59/2)lgC Zn 2 = +0.34 + (0.59/2)lgC Cu 2+ , whence we obtain

(0.59 / 2) (logC Zn 2 - logC Cu 2+) \u003d 0.34 - (-0.76)

logK = log (C Zn2+ /C Cu2+) = 2(0.34 - (-0.76)) / 0.059 = 37.7

The value of the equilibrium constant shows that the reaction goes almost to the end, i.e. until the concentration of copper ions becomes 10 37.7 times less than the concentration of zinc ions.

Equilibrium constant and redox potential are related by the general formula:

lgK \u003d (E 1 0 -E 2 0) n / 0.059, where

K is the equilibrium constant

E 1 0 and E 2 0 are the standard potentials of the oxidizing agent and reducing agent, respectively

n is the number of electrons donated by the reducing agent or received by the oxidizing agent.

If a E 1 0 > E 2 0 , then lgK > 0 and K > 1. Consequently, the reaction proceeds in the forward direction (from left to right) and if the difference (E 1 0 - E 2 0) is large enough, then it goes almost to the end.

On the contrary, if E 1 0< E 2 0 , то K будет очень мала . The reaction proceeds in the opposite direction, because the balance is strongly shifted to the left. If the difference (E 1 0 - E 2 0) is insignificant, then K ≈ 1 and this reaction does not go to the end, unless the conditions necessary for this are created.

Knowing the value of the equilibrium constant, without resorting to experimental data, one can judge the depth of the course of a chemical reaction. It should be borne in mind that the given values of standard potentials do not allow one to determine the rate at which the equilibrium of the reaction is established.

According to the tables of redox potentials, it is possible to find the values of the equilibrium constants for approximately 85,000 reactions.

How to draw a diagram of a galvanic cell?

element emf is a positive value, since work is done in the galvanic cell.
EMF value of a galvanic circuit is the sum of potential jumps at the interfaces of all phases, but, given that oxidation occurs at the anode, the value of the anode potential is subtracted from the value of the cathode potential.

Thus, when drawing up a circuit of a galvanic cell left record the electrode on which the oxidation process (anode), a on right- the electrode at which recovery process (cathode).

interface denoted by one line - |
electrolyte bridge on the border of two conductors is indicated by two lines - ||
Solutions in which the electrolyte bridge is immersed are written to the left and to the right of it (if necessary, the concentration of the solutions is also indicated here). Components of one phase, in this case, are written separated by commas.

For example, let's make galvanic cell diagram, in which the following reaction takes place:

Fe 0 + Cd 2+ = Fe 2+ + Cd 0

In a galvanic cell, the anode is an iron electrode, and the cathode is a cadmium electrode.

Anode Fe 0 |Fe 2+ || Cd 2+ |Cd 0 Cathode

You will find typical problems with solutions.

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