In every redox reaction, including the reaction
Zn + CuSO 4 \u003d ZnSO 4 + Cu (1)
two redox pairs are involved - a reducing agent (Zn) and its oxidized form (Zn 2+); oxidizing agent (Cu 2+) and its reduced form (Cu). A measure of the redox power of a given pair is the redox or electrode potential, which is denoted by , where Ox is the oxidized form, Red is the reduced form (for example, , ). It is impossible to measure the absolute value of the potential, so the measurements are carried out relative to a standard, for example, a standard hydrogen electrode.
Standard hydrogen electrode consists of a platinum plate coated with a thin platinum powder, immersed in a solution of sulfuric acid with a concentration of hydrogen ions equal to 1 mol / l. The electrode is washed with a current of hydrogen gas under a pressure of 1.013 10 5 Pa at a temperature of 298 K. A reversible reaction occurs on the platinum surface, which can be represented as:
2H + + 2 Û H 2 .
The potential of such an electrode taken as zero: 
B (potential dimension - Volt).
Standard potentials have been measured or calculated for a large number of redox pairs (half-reactions) and are listed in tables. For example, 
. How more value, especially strong oxidizing agent is the oxidized form (Ox) of this pair. How smaller potential value, especially strong reducing agent is the reduced form (Red) of the redox pair.
A series of metals arranged in increasing order of their standard electrode potentials is called electrochemical series voltages of metals (near the activity of metals):
Li Ba Sr Ca Na Mg Al Mn Zn Cr Fe Cd Co Ni Sn Pb H Bi Cu Ag Hg Au
E0< 0 E 0 =0 E 0 > 0
The series begins with the most active metals (alkaline), and ends with "noble", i.e. hard-to-oxidize metals. The more to the left the metals are located in the row, the stronger the reducing properties they have, they can displace the metals to the right from salt solutions. Metals located before hydrogen displace it from acid solutions (except for HNO 3 and H 2 SO 4 conc).
In those cases when the system is in non-standard conditions, the value
,
where is the potential of the system under non-standard conditions, V;
is the potential of the system under standard conditions, V;
R is the universal gas constant (8.31 J/mol K);
T—temperature, K;
n is the number of electrons involved in the process;
F is the Faraday number (96500 K/mol);
A, c are the product of concentrations (mol/l) of the oxidized and reduced forms of the process participants raised to the power of stoichiometric coefficients.
The solids and water concentrations are taken as unity.
At a temperature of 298 K, after substituting the numerical values of R and F,
the Nernst equation takes the form:
. (2)
So, for the half-reaction
Û 
Nernst equation
Using the values of the electrode potentials, it is possible to determine the direction of the spontaneous occurrence of the redox reaction. In the course of redox, electrons always move from a pair containing a reducing agent to a pair containing an oxidizing agent. Denote
The electrode potential of a pair containing an oxidizing agent;
By the values of the standard redox potentials (E 0) one can judge the direction of the redox reaction.
for example. For the reaction equation
MnO 4 - + 5Fe 2+ + 8H + → Mn 2+ + 5Fe 3+ + 4H 2 O
spontaneously direct reaction will proceed if the standard potential of the redox couple of the oxidizing agent is greater than the standard potential of the redox couple of the reducing agent, and Δ E> 0 . Table values of standard electrode potentials for the following redox pairs:
E 0 (MnO 4 - / Mn 2+) \u003d 1.51 B; E 0 (Fe 3+ / Fe 2+) = 0.77 B.
The value of 1.51 V> 0.77 V, therefore, upon contact, the permanganate ion MnO 4 - acts as an oxidizing agent, and the iron cation Fe 2+ - a reducing agent, a direct reaction proceeds. Calculate Δ E this reaction:
ΔE \u003d E 0 ok - E 0 restore \u003d 1.51 - 0.77 \u003d 0.74 V.
The value of ΔE is positive, the reaction proceeds spontaneously in the forward direction. If ΔE turns out to be a negative value, then the reaction proceeds in reverse direction under standard conditions.
For example. Can chlorine Cl 2 oxidize the bromide ion Br - to bromine Br 2?
Let us write out the values of the standard potentials of redox pairs from the reference table:
E 0 (Cl 2 / 2Cl -) \u003d 1.36 B; E 0 (Br 2 / 2Br -) \u003d 1.07 V.
From the values of the standard potentials it can be seen that the value of 1.36 V > 1.07 V, therefore, chlorine will oxidize the bromide ion to bromine according to the reaction equation:
Cl 2 + 2Br - \u003d 2Cl - + Br 2
Standard redox potentials (E 0)
with respect to the potential of a standard hydrogen electrode at 25 °C
| Element | Highest oxidation state | +ne- | Lowest oxidation state | E 0 , V |
| As | As↓+3H+ As↓ + 3H 2 OH AsO 2 + 3H + H 3 AsO4 + 2H + AsO 4 3- + 2H 2 O |
+3e +3e +3e +2e +2e |
Ash 3 AsH 3 + 3OH - As↓ + 2H 2 OH AsO 2 +2H 2 O AsO 2 - + 4OH - |
-0,60 -1,43 +0,234 +0,56 -0,71 |
| Br | Br2 BrO 3 - + 5H + BrO 3 - + 2H 2 O |
+2e +4e +4e |
2Br- HBrO + 2H 2 O BrO - + 4OH - |
+1,087 +1,45 +0,54 |
| 2BrO 3 - + 12H + 2 BrO 3 - + 6H 2 O BrO 3 - + 6H + BrO 3 - + 3H 2 O |
+10e +10e +6e +6e |
Br 2 + 6H 2 O Br 2 + 12OH - Br - + 3H 2 O Br - + 6OH - |
+1,52 +0,50 +1,45 +0,61 |
|
| C | C 6 H 4 O 2 + 2H + Quinone HCHO+2H+ HCOOH+2H+ CO2+2H+ 2CO2+2H+ |
+2e +2e +2e +2e +2e |
C 6 H 4 (OH) 2 Gdroquinone CH3OH HCHO + H2O HCOOH H 2 C 2 O 4 |
+0,699 +0,19 -0,01 -0,20 -0,49 |
| Cl | Cl2↓ 2ClO 3 - + 12H + ClO 4 - + 2H + 2ClO 4 - + 16H + ClO 4 - + 8H + |
+2e +10e +2e +14e +8e |
2Cl-Cl 2 ↓ + 6H 2 O ClO 3 - + H 2 O Cl 2 ↓ + 8H 2 O Cl - + 4H 2 O |
+1,359 +1,47 +1,19 +1,39 +1,38 |
| Cr | Cr3+ Cr3+ Cr2+ Cr(OH) 3 ↓ Cr 2 O 7 2- + 14H + CrO 4 2- + 4H 2 O |
+e +3e +2e +3e +6e +3e |
Cr2+ Cr↓ Cr↓ Cr↓ + 3OH — 2Cr 3+ + 7H 2 O Cr(OH) 3 ↓ |
-0,41 -0,74 -0,91 -1,3 +1,33 -0,13 |
| Cu | Cu2+ Cu+ Cu2+ CuI↓ Cu(NH 3) 4 2+ |
+2e +e +e +e +2e |
Cu↓ Cu↓ Cu+ Cu↓ + I — Cu↓ + 4NH 3 |
+0,345 +0,531 +0,159 -0,185 -0,07 |
| F | F2 | +2e | 2F- | +2,77 |
| Fe | Fe3+ Fe3+ Fe2+ Fe(CN) 6 3- |
+e +3e +2e +e |
Fe2+ Fe↓ Fe↓ Fe(CN) 6 4- |
+0,771 -0,058 -0,473 +0,364 |
| H | 2H+ 2H+(10-7M) H2 2H2O H 2 O 2 + 2H + |
+2e +2e +2e +2e +2e |
H2 H2 2H- H 2 + 2OH - 2H2O |
0,0000 -0,414 -2,25 -0,828 +1,77 |
| I | I 2 ↓ I 2 I 3 — 2IO 3 - + 12H + 2IO 3 - + 6H 2 O IO 3 - + 6H + IO 3 - + 3H 2 O |
+2e +2e +2e +10e +10e +6e +6e |
2I- 2I- 3I- I 2 ↓ + 6H 2 O I 2 ↓ + 12OH - I - + 3H 2 O I - + 6OH - |
+0,536 +0,621 +0,545 +1,19 +0,21 +1,08 +0,26 |
| K | K+ | +e | K↓ | -2,923 |
| Li | Li+ | +e | Li↓ | -3,04 |
| mg | Mg2+ | +2e | Mg↓ | -2,37 |
| Mn | Mn3+ Mn2+ |
+e +2e |
Mn2+ Mn↓ |
+1,51 -1,17 |
| MnO 2 ↓ + 4H + MnO 4 2- + 2H 2 O MnO4 - MnO 4 - + 4H + MnO 4 - + 2H 2 O MnO 4 - + 8H + |
+2e +2e +e +3e +3e +5e |
Mn 2+ + 2H 2 O MnO 2 ↓ + 4OH - MnO 4 2- MnO 2 ↓ + 2H 2 O MnO 2 ↓ + 4OH - Mn 2+ + 4H 2 O |
+1,23 +0,58 +0,558 +1,69 +0,60 +1,51 |
|
| Mo | Mo 3+ H 2 MoO 4 + 6H + MoO 4 2- + 4H 2 O |
+3e +6e +6e |
Mo↓ Mo↓ + 4H 2 O Mo↓ + 8OH — |
-0,20,0 -1,05 |
| Na | Na+ | +e | Na↓ | -2,713 |
When describing redox reactions, one should not be limited only to the qualitative side of the process, but it is necessary to be able to characterize it quantitatively, which makes it possible to determine the direction of this reaction. For the quantitative assessment of redox reactions, such characteristics as redox potentials and determining the change in the Gibbs energy of the system are important.
Redox potentials. Consider the reaction of the interaction of zinc with a solution of copper sulfate:
At constant temperature and pressure (25°C and 101325 Pa), in accordance with Hess's law, the process will be accompanied by a thermal effect:
Zinc donates electrons as a reducing agent. This process can be represented as a half-reaction
The copper ion accepts electrons as an oxidizing agent, which is expressed by the second half-reaction
Both half-reactions occur at the point of contact between zinc and CuSO 4 solution, and in this case electrons pass directly from zinc to copper ions. This reaction can also be carried out in such a way that the oxidation and reduction half-reactions are spatially separated. Then the electrons will move from the reducing agent to the oxidizing agent along the conductor electric current- external circuit. To do this, a zinc plate is immersed in a ZnS0 4 solution, and a copper plate is immersed in a CuS0 4 solution. If both obtained half-elements are connected by a tube filled with a conductive solution, then a Daniel-Jacobi galvanic cell will be obtained (Fig. 9.2).
Rice. 9.2.
galvanic cell
In the first half-cell on a zinc electrode (anode), zinc is oxidized, and in the second half-cell on a copper electrode (cathode), copper is reduced. Electrons move from the zinc electrode along the external value to the copper one due to the potential difference that has arisen. When summing the equations of both half-reactions, we obtain the equation for the current-generating reaction:
The zinc cations formed during the oxidation process create an excess positive charge in the solution. The solution in which the copper electrode is located is depleted in copper cations, so this solution is characterized by an excess negative charge. In the resulting electric field, copper and zinc cations move from the zinc electrode to the copper one, and S0 4 + anions move from the copper to the zinc electrode, which can be represented by the scheme
The electromotive force (EMF) of a galvanic cell arises due to the occurrence of an oxidizing-reducing reaction. driving force chemical reaction is the loss of the Gibbs energy AG, which determines the maximum work of a chemical reaction. When the external circuit is closed, part of the energy of the system is spent on the release of Joule heat, which is not accompanied by useful work, and the process is irreversible. The galvanic cell performs the maximum work when the reaction is carried out reversibly under equilibrium conditions. This is possible when the EMF of the element is fully compensated by the external EMF (equal in magnitude and opposite in sign). In this case, the change in free energy is determined by the product of the electricity flowing through the galvanic cell and the voltage of the cell E:
where P- number of moles transferred in the process of electrons; F- Faraday's constant (value nF equal to the amount of electricity).
If the process proceeds under standard conditions (25 ° C, ion activity is 1), then the cell voltage is indicated E°, and the corresponding energy change is A G°. Equation (9.1) takes the form
The negative sign of the right side of equation (9.2) shows that if the electrochemical element spontaneously creates electrical voltage, then the system must lose energy as a result. Value E°, called the standard electrode (redox) potential, is the potential of a given electrode process, in which the activity of all ions participating in it is equal to 1.
To determine E°, it is necessary to have another electrode system with a known potential. By combining two electrode systems into a galvanic cell, its EMF can be measured. A hydrogen system consisting of platinum black saturated with gaseous hydrogen was adopted as a standard electrode. The values of the standard electrode potential of solutes refer to 1 M solutions, and for gaseous compounds - to 101 325 Pa. As E° adopted for hydrogen electrode zero, then AC° reactions
should also be conditionally taken equal to zero. Therefore, if M "+ (the oxidized form of the M and + / M (t) system) is a better oxidizing agent (electron acceptor) than H 3 0, then the reaction
accompanied by a decrease in free energy. The standard electrode potential of such a reaction is positive (Equation (9.1)). On the other hand, if M is a better reducing agent than hydrogen, then the standard electrode potential of such a reaction will be negative. So, for the system
the equilibrium is shifted towards the formation of the metal, so A G° E° > 0; true value E°= +0.80 V.
For example, in sodium, the equilibrium "metal - metal ion" is shifted towards the ion and E° =-2.71 V.
From these examples, it is easy to understand that reactions involving strongly basic elements (Li, K, Na, Ca, etc.), which easily donate electrons (reducing agents), are characterized by a negative value E°, while reactions involving weakly basic elements that tend to add electrons (oxidizing agents) correspond to positive values E°. A series of elements arranged in ascending order of standard electrode potentials is called an electrochemical series (Table 9.1).
In the processes of formation of cations in solution, an element higher in this row will displace elements lower.
From Table. 9.1 clear that chlorine (E°\u003d +1.36) - a stronger oxidizing agent than bromine ( E°= +1.07), and zinc ( E° =-0.76) is a stronger reducing agent than lead ( E° = -0,13).
Table 9.1
Electrochemical series
If you want to compare two redox systems and find out in which direction the reaction can proceed, then you need to compare their normal potentials. A system characterized by a more positive potential will play the role of an oxidizing agent, i.e. show a tendency to accept electrons. So, if you mix solutions of salts (Sn 4+, Sn 2 ") and (Fe 3+, Fe 2+), then the reaction
will go from left to right, as E p e > E$ p.
Many oxidizing agents are anions of oxygen-containing acids, and their reactions usually proceed in an acidic environment: 
In table. 9.2 shows the values of the reduction potentials of redox systems in aqueous solutions.
Table 9.2
Standard potentials of some redox systems
in aqueous solutions at 25°С
|
E°, AT |
|
Often instead of a symbol E° use the symbol cf°.
Using the data in Table. 9.1 and 9.2, it is possible to determine in which direction the reactions involving FeCl 3 and halides will proceed, which is presented in general form:
For five possible electrode reactions, we find the values of the standard electrode potentials: 
Let's represent the expected reaction in ionic form:
It is known that the independent transition of electrons occurs from an electrochemical system with more low value electrode potential to a system with a higher value. It follows from this that the first of these systems will be a reducing agent, and the second - an oxidizing agent. Difference between values E° in the reaction of FeCl 3 with halides will be respectively equal to: for KF E° = 0.77 - 2.80 = -2.03 V; for KS1 E° = 0.77 - 1.36 = -0.59 V; for KVg E°= 0.77 - 1.07 = -0.30 V; for KI E°= 0.77 - 0.54 = +0.23 V. As you can see, the difference between the standard potentials has a positive value only in the case of KI, therefore, it is in the presence of potassium iodide that the reaction proceeds from left to right, i.e. there will be a reduction of Fe 3 "to Fe 2+.
Let us consider the possibility of NBr oxidation with aqueous solutions of KMnO 4 or K 2 Cr 2 O 7 .
For systems 
values E° respectively equal to +1.51 V and +1.33 V.
For the reaction Br 2 + 2e~-» 2Вг; E° =+1.07 V.
Consequently, both solutions will oxidize HBr.
Sometimes you have to deal with such redox reactions, when the values E° for the oxidizing agent and reducing agent are close to each other. In such cases, in order to decide the direction of the reaction, it is necessary to take into account the influence of the concentrations of the oxidized and reduced forms of the corresponding substances on the electrode potentials. The equation relating the magnitude of the electrode potential to the concentrations of the oxidized and reduced forms of the substance and temperature has the form
where R- molar gas constant; P - the number of moles of electrons transferred in the process; T - absolute temperature; F- Faraday's constant; [ok| - concentration of the oxidized form; [vo| - concentration of the restored form; tc - coefficients in the reaction equation. Using relation (9.3), called the Nernst equation, we consider the reaction
which spontaneously can go in both directions depending on the concentrations of iron and mercury ions. In this case, two electrochemical systems take place: 
According to formula (9.3), the following potentials correspond to each electrode process: 
Let us assume that = = 10 1 mol / 1000 g H 2 0, a = 10 4 mol / 1000 g H 2 0. Substituting these data into the ratios for?, and E v we get
These data show that E (> E 2 . Therefore, the reaction proceeds from left to right.
If we imagine the inverse ratio of concentrations, i.e. | Hg 2 + ] \u003d \u003d | Fe 2+ ] \u003d 10 "4, a \u003d 10" 1 mol / 1000 g H 2 0, then
We have ?, E v so the reaction proceeds from right to left.
The direction and completeness of the redox reaction can be determined based on the equilibrium constant. Thus, for a process expressed by the following two half-reactions:
where mwq- stoichiometric factors of the redox reaction, the product mwq is equal to the number of electrons transferred during the reaction.
Since at the moment of equilibrium the potentials of the oxidizer and reductant are equal, then, using equation (9.3), we can calculate TO? . Denote E ok = E v a E vos=?. ; , i.e. E x = E.;, then, in accordance with the Nernst equation, we have
Substituting the values R>T and F, for example, at 25°C, we get 
whence it follows that
As 
, then
Hence,
where P is the number of electrons transferred in the process. For example, for the reaction
If a TO?> 1, then the reaction proceeds in the direction of the formation of its products, i.e. from left to right. When TO? 1, then the reaction is biased towards the starting materials.
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 - redox potential difference or OVR electromotive force (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 under 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. electric current, 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 observed emf is called standard electromotive force 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 concentration of metal ions in the solution is different from the concentration of 1 mol/l, so you need to calculate the potentials of the metals according to 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.
Connect to an external power source to charge the battery direct current(“+” 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 electrolyzer. At the negative electrode of the electrolyzer (cathode), a reduction process occurs - the addition of electrons coming from the oxidizer to the oxidizer. electrical circuit, and on the positive electrode (anode) - the oxidation process - the transfer of electrons from the reducing agent to the electrical circuit.
Thus, 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 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).
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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 former are made from coal, graphite, titanium, platinum metals that have a significant positive electrode potential or are coated 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 mutual position 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 - molar mass substance equivalent, g/mol; I - current strength, A; t is the duration of electrolysis, s; F - Faraday number - 96480 C / mol.
Example 1 How does electrolysis work? aqueous solution sodium sulfate 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:
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.
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 
Redox pairs have a positive value of f°, 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 flow 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 coming with food. At anaerobic metabolism it is based mainly on redox reactions, in which organic compounds are both oxidizing and reducing agents.