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The direction of the course of the ravine. The direction of the redox reaction. Methodical development for teachers and students

08.08.2020

Oxidation - reduction potential is a particular, narrow case of the concept of electrode potential. Let's take a closer look at these concepts.

V OVR electron transfer reducing agents oxidizing agents occurs with direct contact of particles, and the energy chemical reaction turns into warmth. Energy any OVR flowing in solution can be converted into electrical energy. For example, if the redox processes are separated spatially, i.e. the transfer of electrons by the reducing agent will take place through the conductor of electricity. This is realized in galvanic cells, where electrical energy is obtained from chemical energy.

Let us 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, helps the electrode to acquire an electrical charge. The potential difference arising at the interface between the metal and the electrolyte solution is called electrode potential... Its meaning 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 ones (Zn, Fe, etc.) are charged negatively, and the less active ones (Cu, Ag, Au, etc.) are positively charged.

The result of the connection of the zinc and copper plate with a conductor of electricity is the appearance in the circuit electric current due to flowing from the zinc to the copper plate along 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 the 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 = ZnSO 4 + Cu

Quantitatively characterize 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 a cell, one of the electrodes of which is the metal (or non-metal) under test, and the other is a hydrogen electrode. The found potential difference at the poles of the element is used to determine the normal potential of the metal under study.

Redox potential

The values of the redox potential are used when 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 internal and external circuit. An electric current is generated in the circuit. Iodide ions donate their electrons, which will flow along the conductor to an inert electrode immersed in a solution of Fe 3+ salt:

2I - - 2e - = I 2

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

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

The nature of the substance(oxidizing 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 - gas constant (R = 8.314 J);

T - absolute temperature, K

n - the number of donated or received electrons in the redox process;

F - Faraday constant (F = 96484.56 C / mol);

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

C vos - concentration (activity) of the reduced form.

Substituting the known data into the equation and passing to the decimal logarithm, we get 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 pairs, the oxidized form of which contains oxygen (for example, Cr 2 O 7 2-, CrO 4 2-, MnO 4 -), with a decrease in the pH of the solution, the redox potential increases, i.e. potential grows with increasing H +. Conversely, the redox potential decreases with decreasing H +.
Temperature. As the temperature rises, the redox potential of this pair also increases.

Standard redox potentials are presented in the tables of special reference books. It should be borne in mind that only reactions in aqueous solutions at a temperature of ≈ 25 ° C are considered. Such tables make it possible to draw some conclusions:

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

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

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

This pair will have the greater recovery capacity, the more numerical value its negative potential, and the oxidizing capacity is higher, 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 is positive. The ORR flow in the selected direction is also possible if the potential difference has a positive value. ORR flows towards weaker oxidizing agents and reducing agents from stronger ones, for example, the reaction

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

It 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 °(Sn 4+ / Sn 2+)< 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) = - 0.44 V, i.e. E °(Fe 2+ / Fe)< E °(Cu 2+ / Cu).

In the process of RVR, 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. when chemical equilibrium occurs, the potentials of both processes become equal.

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

So, how to determine the EMF of the reaction?

Let's consider several reactions and define their EMF:

Mg + Fe 2+ = Mg 2+ + Fe
Mg + 2H + = Mg 2+ + H 2
Mg + Cu 2+ = Mg 2+ + Cu

E °(Mg 2+ / Mg) = - 2.36 V

E °(2H + / H 2) = 0.00V

E °(Cu 2+ / Cu) = +0.34 V

E °(Fe 2+ / Fe) = - 0.44 V

To determine the EMF of the reaction, you need to find the difference between the potential of the oxidizing agent and the potential of the reducing agent

EMF = E 0 ok - E 0 recovery

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

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

Equilibrium constant.

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

For example, for the reaction

Zn + Cu 2+ = Zn 2+ + Cu

By applying law of mass action, you can write

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)

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

E 0 Zn / Zn2 + = -0.76 + (0.59 / 2) logC Zn / Zn2 and E 0 Cu / Cu2 + = +0.34 + (0.59 / 2) logC Cu / Cu2 +

In a state of equilibrium E 0 Zn / Zn2 + = E 0 Cu / Cu2 +, i.e.

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

(0.59 / 2) (logC Zn 2 - logC Cu 2+) = 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 proceeds 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 related by the general formula:

logK = (E 1 0 -E 2 0) n / 0.059, where

K - equilibrium constant

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

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

If E 1 0> E 2 0, then logK> 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 necessary conditions are created for this.

Knowing equilibrium constant value without resorting to experimental data, one can judge the depth of the chemical reaction. It should be borne in mind that these values of the standard potentials do not allow determining the rate of establishment of the equilibrium of the reaction.

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

How to make a diagram of a galvanic cell?

EMF element- the value is positive, because work is done in the galvanic cell.
EMF value of 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 on which the recovery process (cathode).

Phase boundary 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 right of it (if necessary, the concentration of solutions is also indicated here). The components of the same phase, in this case, are written separated by commas.

For example, let's make galvanic cell circuit, in which the following reaction is carried out:

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.

Categories ,

By the values of the redox potentials, one can judge the direction of the redox reactions. Based on equation (7.4), which connects the potential difference of the half-reactions with the change in the Gibbs free energy DG, and remembering that the possibility of any chemical process occurring is due to the negative value of DG, we can draw the following conclusion:

The redox reaction will spontaneously proceed in such a direction that a half-reaction with a higher redox potential acts as an oxidizing agent with respect to a half-reaction with a lower potential.

At the same time, it is taken into account that in the tables the values of j 0 are given for half-reactions in one direction - oxidation.

In other words, the redox reaction can proceed if the potential difference of the half-reactions of the corresponding galvanic cell is positive.

As the reaction proceeds, the concentrations of the oxidized and reduced forms in the half-reactions change in such a way that the potential of the oxidizing agent decreases and the potential of the reducing agent increases. As a result, the potential difference decreases, the driving force of the process weakens. The redox reaction will continue until the potentials of the half-reactions become equal. When the potentials are equal, chemical equilibrium is established in the system.

As a first approximation, already by comparing the standard potentials of half-reactions, one can solve the question - which of them is capable of performing the function of an oxidizing agent or a reducing agent in relation to the other. For this, the standard potentials must differ significantly from one another. For example, zinc (j 0 = -0.76 V) will reduce (displace) copper (j 0 = +0.34 V) from aqueous solution its salts at any practicable concentration of this solution. But, if the difference between the standard potentials is small (the standard potentials are close), it is necessary to calculate the actual potentials taking into account the concentrations according to equation (7.2) and only after that decide the question of the direction of flow of this redox
telny reaction.

Example 12. Establish the possibility and direction of the reaction

2KBr + PbO 2 + 2H 2 SO 4 = Br 2 + PbSO 4 + K 2 SO 4 + 2H 2 O.

Solution. To answer the question, let us write down the reaction in ionic form, dividing it into half-reactions, and write down the standard potentials of half-reactions:

2Br - + PbO 2 + 4H + = Pb 2+ + Br 2 + 2H 2 O.

1) Br 2 + 2? = 2Br -; j 0 = 1.09 B,

2) PbO 2 + 4H + + 2? = Pb 2+ + 2H 2 O; j 0 = 1.46 B.

According to the expression (7.4) Dj = j ok-la - j recovery = 1.46 - 1.09 = 0.37 V> 0.

Therefore, DG< 0, т.е. реакция будет протекать слева направо. Из сопоставления потенциалов видно, что PbO 2 в кислой среде имеет больший потенциал, следовательно, он может окислить ионы Br - , которые образуют при этом Br 2 .

Example 13. Determine the possibility of redox reactions

a) HBrO + HIO = HBrO 3 + HI,

b) HBrO + HIO = HBr + HIO 3.

Solution. Let us write down the corresponding half-reactions and the corresponding potentials: for reaction a)

BrO 3 - + 5H + +4? = HBrO + 2H 2 O; j 0 = 1.45 B,

HIO + H + + 2e = I - + H 2 O; j 0 = 0.99 B.

Using equation (7.3), we find the difference of redox potentials

Dj 0 = j 0 ok-la - j 0 rev-la = 0.99 - 1.45 = -0.46 B.

Since Dj 0< 0, то DG >0, therefore, the reaction is impossible (it proceeds in the opposite direction).

For reaction b)

HBrO + H + + 2? = Br - + H 2 O; j 0 = 1.33 B,

IO 3 - + 5H + + 4? = HIO + 2H 2 O; j 0 = 1.14.

Dj 0 = j 0 ok-la - j 0 recovery = 1.33 - 1.14 = 0.19 B. Since for this reaction Dj 0> 0, the reaction is possible.

Example 14. Is it possible to oxidize Cr 3+ ions in Cr 2 O 7 2 - by the action of NO 3 - -ion?

Solution. Acting as an oxidizing agent, the NO 3 ion is reduced to NO, so we will compare the potentials of the following half-reactions:

NO 3 - + 4H + +3? = NO + 2H 2 O; j 0 = 0.96 B,

Cr 2 O 7 2 - + 4H + + 6? = 2Cr 3+ + 7H 2 O; j 0 = 1.33 B.

Dj 0 = j 0 ok-la - j 0 recovery = 0.96 - 1.33 = -0.37 V. DG> 0, i.e. NO 3 - -ion cannot be an oxidizing agent in relation to Cr 3+ -ion under standard conditions.

Conversely, dichromic acid H 2 Cr 2 O 7 and its salts ("dichromates") oxidize NO to HNO 3.

= Our Discussions =

OVR flow criteria. Standard conditions and standard potentials

Methodical development for teachers and students

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

analytical chemistry

In this case, it is assumed that the reactions at the interphase boundaries proceed rather quickly, and in the volume of the solution - almost instantaneously. In inorganic chemistry, where reactions often involve ions, this assumption is almost always justified (some examples of thermodynamically possible reactions that do not actually proceed due to kinetic difficulties will be discussed below).

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

First of all, let us ask ourselves the following question: what does the oxidizing ability of an oxidizing agent participating in a particular half-reaction depend on, for example, MnO 4 - + 8H + + 5e = Mn 2+ + 4H 2 O? Although we traditionally use the equal sign and not the reversibility sign when writing the half-reaction equations and the ionic OVR equations as a whole, in reality, all reactions are chemically reversible to one degree or another, and therefore, in a state of equilibrium with equal speeds both direct and reverse reactions always take place.
An increase in the concentration of the oxidizing agent MnO 4 - increases, as is known from school course chemistry, the rate of direct reaction, i.e. leads to a shift in balance to the right; the completeness of the oxidation of the reducing agent increases. The use of Le Chatelier's principle naturally leads to the same conclusion.
Thus, the oxidizing ability of an oxidizing agent always increases with an increase in its concentration.
This conclusion seems to be quite trivial and almost self-evident.
However, reasoning in exactly the same way, we can conclude that the oxidizing ability of the permanganate ion in an acidic medium will increase with an increase in the concentration of hydrogen ions and decrease with an increase in the concentration of the Mn 2+ cation (in particular, with its accumulation in solution as the reaction proceeds) ...
In the general case, the oxidizing capacity of an oxidizing agent depends on the concentrations of all particles appearing in the half-reaction equation. At the same time, its increase, i.e. the oxidizer reduction process is promoted by an increase in the concentration of particles in the left side of the half-reaction; an increase in the concentration of particles in its right side, on the contrary, prevents this process.
Exactly the same conclusions can be drawn with respect to the half-reactions of oxidation of a reducing agent and redox reactions in general (if they are written in ionic form).

Redox potentials

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

Nernst equation

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

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

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

Standard redox potentials

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

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

The table of standard potentials, compiled in descending order, uniquely ranks oxidants (i.e. oxidized forms of various redox pairs) according to their strength. At the same time, the restorers are ranked by strength ( reconstituted steam forms).

Criterion for the direction of the reaction under standard conditions.

If the reaction mixture contains both the starting materials and the reaction products formed by them during the course of ORP, or, in other words, two oxidizing agents and two reducing agents, then the direction of the reaction is determined by which of the oxidizing agents under these conditions will be stronger in accordance with the Nernst equation.
It is especially easy to determine the direction of the reaction under standard conditions, when all the substances (particles) participating in it are in their standard states. Obviously, the oxidizer of the pair with the higher standard potential turns out to be stronger under these conditions.
Although the direction of the reaction under standard conditions is uniquely determined by this, we, without knowing it in advance, can write the reaction equation or right(the reaction under standard conditions really goes in the accepted by us, i.e. in the forward direction) or not right(the reaction is going in the opposite direction we have taken).
Any record of the OVR equation presupposes a certain choice of oxidizer on the left side of the equation. If under standard conditions this oxidizing agent is stronger, the reaction will proceed to direct direction, if not - in reverse.
The standard potential of a redox pair, in which the oxidizing agent we have chosen is the oxidized form, will be called oxidizer potentialφ о OK, and the standard potential of another pair, in which the reduced form is the reducing agent we have chosen, is reducing agent potentialφ about Sun

The quantity Δφ о = φ о ОК - φ о С is called standard redox potential difference.
After the introduction of these designations reaction direction criterion under standard conditions, you can give a simple look:

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

The criterion for the completeness of the ORR (or the criterion for the chemical irreversibility of the ORR)

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

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

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

Kinetic Difficulties in Ion Interaction

As already noted, reactions with the participation of ions that take place in the entire volume of the solution almost always proceed very quickly. However, there are also exceptions. Thus, the oxidation reaction of the ammonium cation in an acidic medium by the iron (III) cation 6Fe 3+ + 2NH 4 + ≠ N 2 + 6Fe 2+ + 8H + is thermodynamically possible (Δφ о = 0.499 V), but in fact it does not go.

The reason kinetic difficulties here is the Coulomb repulsion of the cations of the oxidizing agent and the reducing agent, which prevents them from approaching each other at a distance at which an electronic transition is possible. For a similar reason (but already due to the Coulomb repulsion of anions), the oxidation of the iodide ion by the nitrate ion does not occur in an acidic medium, although for this reaction Δφ o = 0.420 V.
After adding zinc, neutral molecules appear in the system nitrous acid, which nothing prevents from oxidizing iodide ions.

PRACTICAL EXERCISES ON THE TOPIC

"OXIDATIVE-REDUCING
REACTIONS AND ELECTROCHEMICAL
PROCESSES "

FOR THE DISCIPLINE "CHEMISTRY"

Study guide

CHEREPOVETS

Practical classes on the topic "Redox reactions and electrochemical processes" in the discipline "Chemistry": Textbook. Method. allowance. Cherepovets: GOU VPO ChGU, 2005.45 p.

Considered at a meeting of the Department of Chemistry, Protocol No. 11 dated 09.06.2004.

Approved by the Editorial and Publishing Commission of the Institute of Metallurgy and Chemistry of the State Educational Institution of Higher Professional Education, ChGU, protocol No. 6 of June 21, 2004.

Compilers: O.A. Calco - Cand. tech. Sciences, Associate Professor; N.V. Kunina

Reviewers: T.A. Okuneva, Associate Professor (GOU VPO ChSU);

G.V. Kozlova, Cand. chem. Sci., Associate Professor (GOU VPO ChSU)

Scientific editor: G.V. Kozlova - Cand. chem. Sciences, Associate Professor

Vienna University, 2005

INTRODUCTION

The manual includes brief theoretical information, examples of problem solving and options test assignments on the topic "Redox reactions and electrochemical processes" of the course of general chemistry. The content of the training manual corresponds to the state standard of the discipline "Chemistry" for chemical and engineering specialties.

OXIDATION-REDUCTION REACTIONS

Reactions proceeding with a change in the oxidation states of elements in substances are called redox.

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

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

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

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

- for fluorine - "-1";

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

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

4. The algebraic sum of CO of all elements in a neutral molecule is zero, and in an ion - the charge of the ion.

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

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

1. Oxidation Is the process of giving up electrons by a particle, which leads to an increase in the CO of an element:

2. Recovery Is the process of accepting electrons by a particle, which is accompanied by a decrease in the CO of an element:

Substances that donate their electrons during oxidation are called restorers, and the substances that receive electrons in the reduction process are oxidants... If we denote by Oh the oxidized form of the substance, and through Red- restored, then any OVR can be represented as the sum of two processes:

Red 1 – n ē® Ox 1 (oxidation);

Ox 2 + n ē ® Red 2 (recovery).

The manifestation of certain redox properties of atoms depends on many factors, the most important of which is the position of the element in Periodic table, its CO in the substance, the nature of the properties exhibited by other participants in the reaction. According to the redox activity, substances can be conditionally divided into three groups:

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

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

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

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

In some reactions, the oxidizing and reducing agents are atoms of elements of the same molecule that are different in nature, such ORP are called intramolecular, for example:

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

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

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

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

The electronic balance method is universal, that is, it can be used to equalize RVRs occurring in any conditions. The half-reaction method is applicable for drawing up equations only for such redox processes that occur in solutions. However, it has several advantages over the electronic balance method. In particular, when using it, there is no need to determine the oxidation states of elements; in addition, the role of the medium and the real state of particles in solution are taken into account.

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

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

2. Make up half-reactions of oxidation and reduction:

(recovery)

(oxidation)

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

½´ 1½´ 2

½´ 5½´ 10

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

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

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

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

- metal atoms;

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

- hydrogen atoms;

- oxygen atoms.

For the considered example

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

When equalizing reactions by the method of ion-electronic equations observe the following sequence of actions:

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

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

2. Record the ionic scheme of the reaction. Wherein strong electrolytes are recorded in the form of ions, and weak electrolytes, insoluble and slightly soluble substances, as well as gases, are left in molecular form. For the process under consideration

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

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

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

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

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

Cr 2 O + 14Н + ® 2Cr 3+ + 7Н 2 О

Then the charges are equalized by adding or subtracting a certain number of electrons:

12+ + 6 ē ® 6+

Fe 0 - 2 ē ® Fe 2+

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

12+ + 6 ē ® 6+ ½´2½´1

Fe 0 - 2 ē ® Fe 2+ ½´6½´3

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

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

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

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

Calculation of molar equivalent masses MNS oxidizing agent or reducing agent in the OVR should be carried out according to the formula

M E =, (1)

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

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

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

Solution

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

V this process oxidizing agent - НNO 3, reducing agent - As 2 S 3.

We draw up an ionic reaction scheme

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

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

0 – 28ē ® 28+ ½´3

3+ + 3ē ® 0 ½´28

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

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

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

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

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

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

M e, oxidizer = g / mol;

M uh, reducing agent = g / mol.

MULTI-VARIANT PROBLEM No. 1

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

DIRECTION OF OVR FLOW

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

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

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

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

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

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

Fe 3+ + Cl - "Fe 2+ + Cl 2

Solution

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

Fe 3+ + 1 ē = Fe 2+, E= 0.77 V;

Cl 2 + 2 ē = 2Cl -, E= 1.36 V.

Using the formula (3), we calculate the EMF:

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

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

Solution

Let's draw up an ionic reaction scheme:

Fe 3+ + H 2 S ® Fe 2+ + S + H +

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

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

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

MULTI-VARIANT PROBLEM No. 2

Equalize one of the redox reactions using the ion-electronic equation method. Using the standard redox potential table, calculate the EMF and D G reaction, and also indicate the direction of the flow of this ORP:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

GALVANIC ELEMENTS

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

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

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

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

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

UHE is a gas electrode that consists of platinum in contact with hydrogen gas ( R= 1 atm) and a solution in which the activity of hydrogen ions a= 1 mol / dm 3. Equilibrium in the hydrogen electrode is reflected by the equation

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

2) for a hydrogen electrode

where pH is the pH of water.

In a galvanic cell, an electrode with a lower potential value is called anode and is denoted by a "-" sign. Reducing agent particles are oxidized at the anode. An electrode with high potential is called cathode and is indicated by a "+" sign. Reduction of oxidant particles occurs at the cathode. The transition of electrons from a reducing agent to an oxidizing agent occurs along a metal conductor, which is called external circuit... OVR, which underlies the operation of a galvanic cell, is called current-forming reaction.

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

D E = E cathode - E anode. (6)

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

It is customary to write galvanic cells in the form of diagrams in which one vertical line represents the phase boundary (metal - solution), and two vertical lines represent the boundary between two solutions. In practice, electrical contact between solutions is provided by salt bridge- U-shaped tube with electrolyte solution.

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

Solution

Determine the molar concentration of nickel ions in the solution:

= mol / dm 3,

where z= 2 is the equivalence number of Ni 2+ ions.

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

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

Solution

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

Then the hydroxyl index of water is

R OH = 14 - R H = 14 - 13.32 = 0.68.

Hence, the concentration of OH ions is equal to

Mole / dm 3.

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

Solution

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

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

at the anode: Pb - 2 ® Pb 2+;

at the cathode: Cu 2+ + 2 ® Cu;

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

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

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

D E= 0.298 - (-0.156) = 0.454 V.

MULTI-VARIANT PROBLEM No. 3

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

Table 1

Initial data table for multivariate task No. 3

Consider the processes that will be observed if a metal plate (electrode) is immersed in water. Since all substances are to some extent soluble, in such a system the process of transition of metal cations to a solution with their subsequent hydration will begin. The electrons released in this case will remain on the electrode, imparting a negative charge to it. A negatively charged electrode will attract metal cations from the solution, as a result of which an equilibrium will be established in the system:

M M n + + ne -,

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

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

electrode - solution.

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

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

OF + ne - WF

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

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

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

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

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

Н 2 ¾ 2Н ¾® 2Н + + 2е -

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

Rice. 27. Diagram of the structure of the hydrogen electrode.

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

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

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

Direction of redox reactions. Since the electrode potential is associated with a change in the Gibbs free energy by the ratio:

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

Let the redox reactions correspond to the half-reaction:

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

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

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

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

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

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

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

Let us write down the equations for the transition of two oxidized forms into reduced ones and, using the reference tables, we find the corresponding values of the electrode potentials:

Fe 3+ + 1e - = Fe 2+ │5 E ° 1 = 0.77 B

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

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

Galvanic cell

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

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

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

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

Zn 2+ + 2e - = Zn 0; E ° 1 = -0.76 B

Cu 2+ + 2e - = Cu 0; E ° 2 = 0.34 B

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

Zn + Cu 2+ = Zn 2+ + Cu

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

ΔE = E cathode - E anode

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

To record a diagram of galvanic cells, use the form below:

Anode │ Anode solution ││ Cathode solution │ Cathode

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

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

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

Fe 3+ + 1e - = Fe 2+

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

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

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

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

Electrolysis

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

In electrolysis, the anode is the positive electrode, on which the oxidation process takes place, and the cathode is the negative electrode, on which the reduction process takes place. The names "anode" and "cathode", therefore, are not related to the charge of the electrode: during electrolysis, the anode is positive, and the cathode is negative, and vice versa during the operation of a galvanic cell. In the process of electrolysis, the anode is an oxidizing agent, and the cathode is a reducing agent. The electrolysis process is quantitatively described by the laws of M. Faraday (1833):

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

2. The same amount of electricity is consumed to release one mole of an equivalent of any substance on the electrode.

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

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

As already mentioned, both solutions and molten electrolytes are subjected to electrolysis. The most simple process is the electrolysis of melts. In this case, the cation is reduced at the cathode, and the electrolyte anion is oxidized at the anode. For example, electrolysis of sodium chloride melt proceeds according to the equations:

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

Electrolysis equation: 2NaCl = 2Na + Cl 2

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

1. Water recovery (cathodic process):

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

2. Oxidation of water (anodic process):

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

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

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

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

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

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

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

F 2 + 2e - = 2F -

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

2Cl - - 2e - = Cl 2; E ° = 1.36V

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

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

Let's consider some examples of electrolysis of solutions.

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

CuSO 4 = Cu 2+ + SO

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

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

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

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

Cathodic process: Сu 2+ + 2e - = Cu

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

Electrolysis is reduced to the transfer of copper from the anode to the cathode.

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

Na 2 SO 4 = 2Na + + SO

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

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

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

Electrolysis is reduced to the decomposition of water.

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

NaCl = Na + + Cl -

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

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

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

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

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

Colloidal solutions

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