The specificity of many OVRs is that, when compiling their equations, the selection of coefficients causes difficulty. To facilitate the selection of coefficients, it is most often used electron balance method and ion-electronic method (half-reaction method). Consider the application of each of these methods with examples.

Electronic balance method

It is based on next rule: the total number of electrons donated by reducing atoms must match the total number of electrons received by oxidizing atoms.

As an example of compiling an OVR, consider the process of interaction of sodium sulfite with potassium permanganate in an acidic environment.

1. First you need to draw up a reaction scheme: write down the substances at the beginning and end of the reaction, given that in an acidic environment MnO 4 - is reduced to Mn 2+ ():

1. Next, we determine which of the compounds are; find their oxidation state at the beginning and end of the reaction:

Na 2 S +4 O 3 + KMn +7 O 4 + H 2 SO 4 = Na 2 S +6 O 4 + Mn +2 SO 4 + K 2 SO 4 + H 2 O

From the above diagram, it is clear that during the reaction, the oxidation state of sulfur increases from +4 to +6, so S +4 donates 2 electrons and is reducing agent. The oxidation state of manganese decreased from +7 to +2, i.e. Mn +7 accepts 5 electrons and is oxidizing agent.

1. We compose electronic equations and find the coefficients for the oxidizing agent and reducing agent.

S +4 - 2e - \u003d S +6 ¦ 5

Mn +7 +5e - = Mn +2 ¦ 2

In order for the number of electrons donated by the reducing agent to be equal to the number of electrons accepted by the reducing agent, it is necessary:

• Put the number of electrons donated by the reducing agent as a factor in front of the oxidizing agent.
• Put the number of electrons accepted by the oxidizing agent as a factor in front of the reducing agent.

Thus, 5 electrons received by the oxidizing agent Mn +7, we put the coefficient in front of the reducing agent, and 2 electrons given away by the reducing agent S +4 as a coefficient in front of the oxidizing agent:

5Na 2 S +4 O 3 + 2KMn +7 O 4 + H 2 SO 4 = 5Na 2 S +6 O 4 + 2Mn +2 SO 4 + K 2 SO 4 + H 2 O

1. Next, you need to equalize the number of atoms of elements that do not change the oxidation state, in the following sequence: the number of metal atoms, acid residues, the number of molecules of the medium (acid or alkali). Lastly, the number of water molecules formed is counted.

So, in our case, the number of metal atoms in the right and left parts are the same.

By the number of acid residues on the right side of the equation, we find the coefficient for the acid.

As a result of the reaction, 8 acid residues SO 4 2- are formed, of which 5 are due to the transformation 5SO 3 2- → 5SO 4 2-, and 3 are due to sulfuric acid molecules 8SO 4 2- - 5SO 4 2- \u003d 3SO 4 2 - .

Thus, sulfuric acid must take 3 molecules:

5Na 2 SO 3 + 2KMnO 4 + 3H 2 SO 4 = 5Na 2 SO 4 + 2MnSO 4 + K 2 SO 4 + H 2 O

1. Similarly, we find the coefficient for water by the number of hydrogen ions, in a given amount of acid

6H + + 3O -2 = 3H 2 O

The final form of the equation is as follows:

A sign that the coefficients are placed correctly is an equal number of atoms of each of the elements in both parts of the equation.

Ion-electronic method (half-reaction method)

Oxidation-reduction reactions, as well as exchange reactions, in electrolyte solutions occur with the participation of ions. That is why the ionic-molecular equations of the OVR more clearly reflect the essence of the redox reactions. When writing ion-molecular equations, strong electrolytes are written as , and weak electrolytes, precipitates and gases are written as molecules (in undissociated form). In the ionic scheme indicate the particles undergoing a change in their oxidation states, as well as characterizing the environment, particles: H + - acidic environment,OH — — alkaline environment and H 2 O - neutral environment.

Consider an example of compiling a reaction equation between sodium sulfite and potassium permanganate in an acidic environment.

1. First you need to draw up a reaction scheme: write down the substances at the beginning and end of the reaction:

Na 2 SO 3 + KMnO 4 + H 2 SO 4 = Na 2 SO 4 + MnSO 4 + K 2 SO 4 + H 2 O

1. We write the equation in ionic form, reducing those ions that do not take part in the redox process:

SO 3 2- + MnO 4 - + 2H + = Mn 2+ + SO 4 2- + H 2 O

1. Next, we define the oxidizing agent and reducing agent and compose the half-reactions of the reduction and oxidation processes.

In the above reaction oxidizing agent - MnO 4- accepts 5 electrons recovering in an acidic environment to Mn 2+. In this case, oxygen is released, which is part of MnO 4 -, which, combining with H +, forms water:

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

Reducing agent SO 3 2-- oxidized to SO 4 2-, giving 2 electrons. As you can see, the resulting SO 4 2- ion contains more oxygen than the original SO 3 2- . The lack of oxygen is replenished by water molecules and as a result, 2H + is released:

SO 3 2- + H 2 O - 2e - \u003d SO 4 2- + 2H +

1. We find the coefficient for the oxidizing agent and reducing agent, considering that the oxidizing agent adds as many electrons as the reducing agent gives up in the oxidation-reduction process:

MnO 4 - + 8H + + 5e - \u003d Mn 2+ + 4H 2 O ¦2 oxidizing agent, reduction process

SO 3 2- + H 2 O - 2e - \u003d SO 4 2- + 2H + ¦5 reducing agent, oxidation process

1. Then it is necessary to sum both half-reactions, preliminarily multiplying by the found coefficients, we obtain:

2MnO 4 - + 16H + + 5SO 3 2- + 5H 2 O \u003d 2Mn 2+ + 8H 2 O + 5SO 4 2- + 10H +

Reducing like terms, we find the ionic equation:

2MnO 4 - + 5SO 3 2- + 6H + = 2Mn 2+ + 5SO 4 2- + 3H 2 O

1. Let's write the molecular equation, which has the following form:

5Na 2 SO 3 + 2KMnO 4 + 3H 2 SO 4 = 5Na 2 SO 4 + 2MnSO 4 + K 2 SO 4 + 3H 2 O

Na 2 SO 3 + KMnO 4 + H 2 O \u003d Na 2 SO 4 + MnO 2 + KOH

AT ionic form the equation becomes:

SO 3 2- + MnO 4 - + H 2 O \u003d MnO 2 + SO 4 2- + OH -

Also, as in the previous example, the oxidizing agent is MnO 4 -, and the reducing agent is SO 3 2-.

In a neutral and slightly alkaline environment, MnO 4 - accepts 3 electrons and is reduced to MnO 2. SO 3 2- - is oxidized to SO 4 2-, giving 2 electrons.

Half reactions have the following form:

MnO 4 - + 2H 2 O + 3e - \u003d MnO 2 + 4OH - ¦2 oxidizing agent, reduction process

SO 3 2- + 2OH - - 2e - \u003d SO 4 2- + H 2 O ¦3 reducing agent, oxidation process

We write the ionic and molecular equations, taking into account the coefficients for the oxidizing agent and reducing agent:

3SO 3 2- + 2MnO 4 - + H 2 O \u003d 2 MnO 2 + 3SO 4 2- + 2OH -

3Na 2 SO 3 + 2KMnO 4 + H 2 O \u003d 2MnO 2 + 3Na 2 SO 4 + 2KOH

And one more example - drawing up an equation for the reaction between sodium sulfite and potassium permanganate in an alkaline medium.

Na 2 SO 3 + KMnO 4 + KOH \u003d Na 2 SO 4 + K 2 MnO 4 + H 2 O

AT ionic form the equation becomes:

SO 3 2- + MnO 4 - + OH - \u003d MnO 2 + SO 4 2- + H 2 O

In an alkaline environment oxidizing agent MnO 4 - accepts 1 electron and is reduced to MnO 4 2-. Reductant SO 3 2- - is oxidized to SO 4 2-, giving 2 electrons.

Half reactions have the following form:

MnO 4 - + e - \u003d MnO 2 ¦2 oxidizing agent, reduction process

SO 3 2- + 2OH - - 2e - \u003d SO 4 2- + H 2 O ¦1 reducing agent, oxidation process

Let's write down the ionic and molecular equations, taking into account the coefficients for the oxidizing agent and reducing agent:

SO 3 2- + 2MnO 4 - + 2OH - \u003d 2MnO 4 2- + SO 4 2- + H 2 O

Na 2 SO 3 + 2KMnO 4 + H 2 O \u003d 2K 2 MnO 4 + 3Na 2 SO 4 + 2KOH

It should be noted that not always in the presence of an oxidizing agent and a reducing agent, spontaneous OVR may occur. Therefore, to quantitatively characterize the strength of the oxidizing agent and reducing agent and to determine the direction of the reaction, the values ​​of redox potentials are used.

Drawing up equations of redox reactions

In order to write the OVR equation, it is necessary, first of all, to know which substances are formed as a result of the reaction. In the general case, this issue is solved experimentally. However, knowledge of the chemical characteristics of various oxidizing and reducing agents often makes it possible to predict quite reliably (although not with a 100% guarantee) the composition of the interaction products.

If the reaction products are known, the stoichiometric coefficients in the reaction equation can be found by equalizing the number of electrons added by oxidant atoms and lost by reducing agent atoms. Two methods of selection of coefficients in the OVR equations are used - the method of electronic balance and the method of ion-electron balance. Let's consider these methods.

The method is based on the principle of conservation of electric charge during a chemical reaction, as a result of which substances react in such a ratio that ensures the equality of the number of electrons given up by all atoms of the reducing agent and attached by all atoms of the oxidizing agent. To select the coefficients, it is advisable to use the following algorithm:

1. Write down the OVR scheme (starting substances and reaction products).

2. Determine the elements whose oxidation state changes during the reaction.

3. Make diagrams of the processes of oxidation and reduction.

4. Find the factors equalizing the number of electrons attached by the oxidizing agent atoms and lost by the reducing agent atoms (balancing factors). To do this, find the least common multiple for electrons attached by one atom of the oxidizing agent and given away by one atom of the reducing agent; the balancing factors will be equal to the least common multiple divided by the number of electrons attached (for an oxidizing agent) and donated electrons (for a reducing agent).

5. Determine and enter into the equation the coefficients for substances containing elements whose oxidation state changes (reference coefficients) by dividing the balancing factors by the number of oxidizing or reducing agent atoms in the formula unit of the substance. If the quotient is not an integer, the balancing factors should be increased by the required number of times.

6. Find and arrange additional coefficients that equalize the number of atoms that have not changed their oxidation state (except for hydrogen and oxygen); at the same time, if the medium is acidic, first equalize the metal atoms, and then the anions of the acids, if the medium is alkaline or neutral, vice versa.

7. Equalize the number of hydrogen atoms, adding, if necessary, water to the right or left side equations.

8. Check whether the coefficients for oxygen are correctly selected.

Consider, as an example, the formulation of the equation for the interaction of potassium permanganate with iron (II) sulfate in a sulfuric acid medium according to the stages of the proposed algorithm:

1. KMnO 4 + FeSO 4 + H 2 SO 4 → MnSO 4 + Fe 2 (SO 4) 3 + K 2 SO 4

2. KMn +7 O 4 + Fe +2 SO 4 + H 2 S0 4 → Mn +2 SO 4 + Fe(SO 4) 3 + K 2 SO 4

3. Fe +2 - 1e - = Fe +3 (oxidation)

Mn +7 +5e - = Mn +2 (recovery)

4. Fe +2 - 1e - = Fe +3 │5 │ 10

Mn +7 + 5e - = Mn +2 │1 │2

5. Reference coefficients: with KMnO 4 - 2:1=2, with FeSO 4 - 10:1=10, with MnSO 4 - 2:1=2, with Fe 2 (SO 4) 3 - 10:2=5.

2KMnO 4 + 10FeSO 4 + H 2 SO 4 → 2MnSO 4 + 5Fe 2 (SO 4) 3 + K 2 SO 4

6. The medium is acidic, so first we equalize the potassium atoms, then the sulfate ions.

2KMnO 4 + 10FeSO 4 + 5H 2 SO 4 → 2MnSO 4 + 5Fe 2 (SO 4) 3 + K 2 SO 4

7. Since the left side of the equation contains 10 hydrogen atoms, we add 5 water molecules to the right side:

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

8. The number of oxygen atoms (excluding oxygen included in sulfate ions) in the right and left parts of the equation is 8. The coefficients are chosen correctly.

When OVR occurs, there may be cases when the oxidizing agent or reducing agent is partially spent on binding the oxidation or reduction products without changing the oxidation state of the corresponding element. In this case, the coefficient for a substance with a dual function is equal to the sum of the reference and additional coefficients and is introduced into the equation after the additional coefficient is found. Thus, the reaction between zinc and very dilute nitric acid proceeds according to the equation

4Zn + 10HNO 3 \u003d 4Zn (NO 3) 2 + NH 4 NO 3 + 3H 2 O

Zn 0 - 2e - = Zn +2 │4

N +5 + 8e - = N -3 │1

As follows from the redox schemes, the oxidation of four zinc atoms requires one nitric acid molecule (the reference factor for HNO 3 is 1); however, the formation of four molecules of zinc nitrate and one molecule of ammonium nitrate requires nine more HNO 3 molecules that react without changing the nitrogen oxidation state (additional coefficient for HNO 3 - 9). Accordingly, the coefficient for nitric acid in the reaction equation will be equal to 10, and 3 water molecules should be entered into the right side of the equation.

If one of the substances simultaneously performs the function of both an oxidizing agent and a reducing agent (disproportionation reactions) or is a product of both oxidation and reduction (counter-disproportionation reactions), then the coefficient for this substance is equal to the sum of the reference coefficients for the oxidizing agent and reducing agent. For example, in the reaction equation for the disproportionation of sulfur in an alkaline medium, the coefficient for sulfur is three.

3S 0 + 6NaOH \u003d Na 2 S +4 O 3 + Na 2 S -2 + 3H 2 O

S - 4e - = S +4 │1

S + 2e - = S -2 │2

Sometimes, during the course of OVR, a change in the oxidation state of more than two elements is observed; in this case, the coefficients of the equation can be uniquely determined if all oxidizing agents or all reducing agents are part of one molecule. In this case, the calculation of donated or attached electrons is rationally carried out for the formula unit of a substance containing these oxidizing or reducing agents. As an example, consider the interaction of arsenic(III) sulfide with nitric acid according to the stages of the above algorithm.

1. As 2 S 3 + HNO 3 → H 3 AsO 4 + H 2 SO 4 + NO

2. AsS+ HN +5 O 3 → H 3 As +5 O 4 + H 2 S +6 O 4 + N +2 O

The reaction involves two reducing agents (As +3 and S -2) and one oxidizing agent (N +5).

3. N +5 + 3e - = N +2 │28

As 2 S 3 - 28e - \u003d 2As +5 + 3S +6 │ 3

4. Least common multiple - 84, balancing factors - 28 and 3.

5. 3As 2 S 3 + 28HNO 3 → 6H 3 AsO 4 + 9H 2 SO 4 + 28NO

6. There are no additional coefficients.

7. Water molecules should be entered on the left side of the equation:

3As 2 S 3 + 28HNO 3 + 4H 2 O \u003d 6H 3 AsO 4 + 9H 2 SO 4 + 28NO

8. The number of oxygen atoms in both the left and right sides of the equation is 88. The coefficients are chosen correctly.

If they participate in the OVR organic matter, then the oxidation states are not determined for them, since in this case each atom can have its own oxidation state value, and often not integer. When drawing up redox schemes for such reactions, the following rules should be followed:

1. the addition of an oxygen atom is identical to the loss of two electrons by a molecule;

2. the loss of an oxygen atom is identical to the addition of two electrons;

3. the addition of a hydrogen atom is identical to the addition of one electron;

4. The loss of a hydrogen atom is identical to the loss of one electron.

Below is an example of the equation for the oxidation reaction ethyl alcohol potassium dichromate:

3C 2 H 5 OH + 2K 2 Cr 2 O 7 + 8H 2 SO 4 \u003d 3CH 3 COOH + 2Cr 2 (SO 4) 3 + 2K 2 SO 4 + 11H 2 O

C 2 H 5 OH + [O] - 2 [H] - 4e - \u003d 3CH 3 COOH │3

Cr +6 + 3e - = Cr +3 │4

The conversion of ethyl alcohol to acetic acid requires the addition of an oxygen atom and the loss of two hydrogen atoms, which corresponds to the loss of four electrons.

The electron balance method is a universal method applicable to any OVR occurring in the gas phase, condensed systems, and solutions. The disadvantage of the method is that this technique is formal and operates with particles that do not really exist (Mn +7, N +5, etc.).

Redox reactions.

Redox reactions – these are reactions that occur with a change in the oxidation states of the atoms of the elements that make up the molecules of the reacting substances:

0 0 +2 -2

2Mg + O 2  2MgO,

5 -2 -1 0

2KClO 3 2KCl + 3O 2 .

Recall that oxidation state – this is the conditional charge of an atom in a molecule, arising from the assumption that the electrons are not displaced, but are completely given to an atom of a more electronegative element.

The most electronegative elements in the compound have negative oxidation states, and the atoms of elements with less electronegativity are positive.

The oxidation state is a formal concept; in some cases, the value of the oxidation state of an element does not coincide with its valence.

To find the oxidation state of the atoms of the elements that make up the reactants, the following rules should be borne in mind:

1. The oxidation state of the atoms of elements in the molecules of simple substances is zero.

For example:

Mg0, Cu0.

2. The oxidation state of hydrogen atoms in compounds is usually +1.

For example: +1 +1

HCl, H2S

Exceptions: in hydrides (compounds of hydrogen with metals), the degree of oxidation of hydrogen atoms is –1.

For example:

NaH -1 .

3. The oxidation state of oxygen atoms in compounds is usually -2.

For example:

H 2 O -2, CaO -2.

Exceptions:

 oxidation state of oxygen in oxygen fluoride (OF 2 ) is equal to +2.

 oxidation state of oxygen in peroxides (H 2 O 2, Na 2 O 2 ) containing the –O–O– group is equal to –1.

4. The oxidation state of metals in compounds is usually a positive value.

For example: +2

CuSO 4 .

5. The oxidation state of non-metals can be both negative and positive.

For example: -1 +1

HCl, HClO.

6. The sum of the oxidation states of all atoms in a molecule is zero.

Redox reactions are two interrelated processes - the oxidation process and the reduction process.

Oxidation process – it is the process of donating electrons by an atom, molecule or ion; in this case, the oxidation state increases, and the substance is a reducing agent:

– 2ē  2H + oxidation process,

Fe +2 – ē  Fe +3 oxidation process,

2J – – 2ē  oxidation process.

The reduction process is the process of adding electrons, while the oxidation state decreases, and the substance is an oxidizing agent:

+ 4ē  2O –2 recovery process,

Mn +7 + 5ē  Mn +2 recovery process,

Cu +2 +2ē  Cu 0 recovery process.

Oxidizing agent - a substance that accepts electrons and is reduced in the process (the element's oxidation state is reduced).

Restorer - a substance that donates electrons and is oxidized at the same time (the oxidation state of an element decreases).

It is possible to make a reasonable conclusion about the nature of the behavior of a substance in specific redox reactions based on the value of the redox potential, which is calculated from the value of the standard redox potential. However, in some cases, it is possible, without resorting to calculations, but knowing the general laws, to determine which substance will be an oxidizing agent and which one will be a reducing agent, and make a conclusion about the nature of the redox reaction.

Typical reducing agents are:

 some simple things:

Metals: e.g. Na, Mg, Zn, Al, Fe,

Nonmetals: e.g. H 2, C, S;

 some complex substances: for example, hydrogen sulfide (H 2 S) and sulfides (Na 2 S), sulfites (Na 2 SO 3 ), carbon monoxide (II) (CO), hydrogen halides (HJ, HBr, HCI) and salts of hydrohalic acids (KI, NaBr), ammonia (NH 3 );

 metal cations in lower oxidation states: e.g. SnCl 2 , FeCl 2 , MnSO 4 , Cr 2 (SO 4 ) 3 ;

 cathode in electrolysis.

Typical oxidizers are:

 some simple substances are non-metals: for example, halogens (F 2, CI 2, Br 2, I 2 ), chalcogens (O 2 , O 3 , S);

 some complex substances: for example, nitric acid (HNO 3 ),sulfuric acid(H 2 SO 4 conc. ), potassium premanganate (K 2 MnO 4 ), potassium dichromate (K 2 Cr 2 O 7 ), potassium chromate (K 2 CrO 4 ), manganese(IV) oxide (MnO 2 ), lead(IV) oxide (PbO 2 ), potassium chlorate (KCIO 3 ), hydrogen peroxide (H 2 O 2 );

 electrolysis anode.

When compiling the equations of redox reactions, it should be borne in mind that the number of electrons donated by the reducing agent is equal to the number of electrons accepted by the oxidizing agent.

There are two methods for compiling equations of redox reactions -electron balance method and electron-ion method (half-reaction method).

When compiling the equations of redox reactions by the electron balance method, a certain procedure should be followed. Consider the procedure for compiling equations by this method using the example of the reaction between potassium permanganate and sodium sulfite in an acidic medium.

1. We write down the reaction scheme (indicate the reagents and reaction products):

1. We determine the oxidation state of the atoms of elements that change its value:

7 + 4 + 2 + 6

KMnO 4 + Na 2 SO 3 + H 2 SO 4 → MnSO 4 + Na 2 SO 4 + K 2 SO 4 + H 2 O.

3) We draw up an electronic balance diagram. To do this, we write down the chemical signs of the elements whose atoms change their oxidation state, and determine how many electrons give or add the corresponding atoms or ions.

We indicate the processes of oxidation and reduction, the oxidizing agent and the reducing agent.

We equalize the number of given and received electrons and, thus, determine the coefficients for the reducing agent and oxidizing agent (in this case, they are respectively equal to 5 and 2):

5 S +4 – 2 e- → S +6 oxidation process, reducing agent

2 Mn +7 + 5 e- → Mn +2 reduction process, oxidizing agent.

2KMnO 4 + 5Na 2 SO 3 + 8H 2 SO 4 \u003d 2MnSO 4 + 5Na 2 SO 4 + K 2 SO 4 + 8H 2 O.

5) If hydrogen and oxygen do not change their oxidation states, then their number is counted last and the required number of water molecules is added to the left or right side of the equation.

Redox reactions are divided into three types:intermolecular, intramolecular and self-oxidation reactions - self-healing (disproportionation).

Reactions of intermolecular oxidation - reductionredox reactions are called, in which the oxidizing agent and reducing agent are represented by molecules of different substances.

For example:

0 +3 0 +3

2Al + Fe 2 O 3 \u003d 2Fe + Al 2 O 3,

Al 0 – 3e – → Al +3 oxidation, reducing agent,

Fe +3 +3e – → Fe 0 reduction, oxidizing agent.

In this reaction, the reducing agent (Al) and the oxidizing agent (Fe+3 ) are part of various molecules.

Reactions of intramolecular oxidation– recovery called reactions in which the oxidizing agent and reducing agent are part of the same molecule (and are represented either by different elements, or by one element, but with varying degrees oxidation):

5 –1 0

2 KClO 3 \u003d KCl + 3O 2

2 CI +5 + 6e – → CI –1 reduction, oxidizer

3 2O –2 – 4e – → oxidation, reducing agent

In this reaction, the reducing agent (O–2) and an oxidizing agent (CI +5 ) are part of one molecule and are represented by different elements.

In the reaction of thermal decomposition of ammonium nitrite, the atoms of the same chemical element, nitrogen, which are part of one molecule, change their oxidation states:

3 +3 0

NH 4 NO 2 \u003d N 2 + 2H 2 O

N –3 – 3e – → N 0 reduction, oxidizer

N +3 + 3e – → N 0 oxidation, reducing agent.

Reactions of this type are often called reactions.counterproportionation.

Self-oxidation reactions - self-healing(disproportionation) -These are reactions in the course of which the same element with the same oxidation state itself both increases and lowers its oxidation state.

For example: 0 -1 +1

Cl 2 + H 2 O \u003d HCI + HCIO

CI 0 + 1e – → CI –1 reduction, oxidizer

CI 0 – 1e – → CI +1 oxidation, reducing agent.

Disproportionation reactions are possible when the element in the original substance has an intermediate oxidation state.

The properties of simple substances can be predicted by the position of the atoms of their elements in the periodic system of elements D.I. Mendeleev. So, all metals in redox reactions will be reducing agents. Metal cations can also be oxidizing agents. Non-metals in the form of simple substances can be both oxidizing and reducing agents (excluding fluorine and inert gases).

The oxidizing ability of non-metals increases in the period from left to right, and in the group - from bottom to top.

Restorative abilities, on the contrary, decrease from left to right and from bottom to top for both metals and non-metals.

If the redox reaction of metals occurs in solution, then to determine the reducing ability, usea range of standard electrode potentials(activity series of metals). In this series, metals are arranged as the reducing ability of their atoms decreases and the oxidizing ability of their cations increases (see table. 9 applications).

The most active metals, standing in a series of standard electrode potentials up to magnesium, can react with water, displacing hydrogen from it.

For example:

Ca + 2H 2 O \u003d Ca (OH) 2 + H 2

When interacting metals with salt solutions, it should be borne in mind thateach more active metal (not interacting with water) is able to displace (restore) the metal behind it from a solution of its salt.

Thus, iron atoms can reduce copper cations from a solution of copper sulfate (CuSO 4 ):

Fe + CuSO 4 \u003d Cu + FeSO 4

Fe 0 - 2e - \u003d Fe +2 oxidation, reducing agent

Cu +2 + 2e – = Cu 0 reduction, oxidizing agent.

In this reaction, iron (Fe) is located in the activity series before copper (Cu) and is a more active reducing agent.

The reaction, for example, of silver with a solution of zinc chloride will be impossible, since silver is located in the series of standard electrode potentials to the right of zinc and is a less active reducing agent.

Ag + ZnCl 2 ≠

All metals that are in the activity series up to hydrogen can displace hydrogen from solutions of ordinary acids, that is, restore it:

Zn + 2HCl \u003d ZnCI 2 + H 2

Zn 0 – 2e – = Zn +2 oxidation, reducing agent

2H + + 2e – → reduction, oxidizing agent.

Metals that are in the activity series after hydrogen will not reduce hydrogen from solutions of ordinary acids.

Cu + HCI ≠

To determine if there might be oxidizing agent or reducing agentcomplex substance, it is necessary to find the degree of oxidation of the elements that make it up. The elements that are inhighest oxidation state, can only lower it by accepting electrons. Hence,substances whose molecules contain atoms of elements in the highest oxidation state will only be oxidizing agents.

For example, HNO 3, KMnO 4, H 2 SO 4 in redox reactions, they will only function as an oxidizing agent. Nitrogen oxidation states (N+5 ), manganese (Mn +7 ) and sulfur (S +6 ) in these compounds have maximum values ​​(coincide with the group number of the given element).

If the elements in the compounds have the lowest oxidation state, then they can only increase it by donating electrons. At the same time, suchsubstances containing elements in the lowest oxidation state will only function as a reducing agent.

For example, ammonia, hydrogen sulfide and hydrogen chloride (NH 3 , H 2 S, HCI) will only be reducing agents, since the oxidation states of nitrogen (N–3 ), sulfur (S –2 ) and chlorine (Cl –1 ) are the lowest for these elements.

Substances that contain elements with intermediate oxidation states can be both oxidizing and reducing agents., depending on the specific reaction. Thus, they may exhibit redox duality.

Such substances include, for example, hydrogen peroxide (H 2O2 ), an aqueous solution of sulfur oxide (IV) (sulphurous acid), sulfites, etc. Similar substances, depending on environmental conditions and the presence of stronger oxidizing agents (reducing agents), can exhibit oxidizing properties in some cases, and reducing properties in others.

As you know, many elements have a variable degree of oxidation, being part of various compounds. For example, sulfur in compounds H 2 S, H 2 SO 3, H 2 SO 4 and sulfur S in the free state has oxidation states –2, +4, +6 and 0, respectively. Sulfur refers to the elements R -electron family, its valence electrons are located on the last s - and p -sublevels (...3 s 3 p ). The sulfur atom with the oxidation state - 2 valence sublevels is fully equipped. Therefore, a sulfur atom with a minimum oxidation state (–2) can only donate electrons (oxidize) and be only a reducing agent. A sulfur atom with an oxidation state of +6 has lost all of its valence electrons and in this state can only accept electrons (recover). Therefore, the sulfur atom with the maximum oxidation state (+6) can only be an oxidizing agent.

Sulfur atoms with intermediate oxidation states (0, +4) can both lose and gain electrons, that is, they can be both reducing agents and oxidizing agents.

Similar reasoning is valid when considering the redox properties of atoms of other elements.

The nature of the course of the redox reaction is affected by the concentration of substances, the environment of the solution and the strength of the oxidizing agent and reducing agent. Thus, concentrated and dilute nitric acid react differently with active and inactive metals. Nitrogen reduction depth (N+5 ) of nitric acid (oxidizer) will be determined by the activity of the metal (reductant) and the concentration (dilution) of the acid.

4HNO 3 (conc.) + Cu \u003d Cu (NO 3) 2 + 2NO 2 + 2H 2 O,

8HNO 3 (razb.) + 3Cu \u003d 3Cu (NO 3) 2 + 2NO + 4H 2 O,

10HNO 3 (conc.) + 4Mg \u003d 4Mg (NO 3) 2 + N 2 O + 5H 2 O,

10HNO 3 (c. razb.) + 4Mg \u003d 4Mg (NO 3) 2 + NH 4 NO 3 + 3H 2 O.

The reaction of the medium has a significant influence on the course of redox processes.

If potassium permanganate (KMnO 4 ), then, depending on the reaction of the solution medium, Mn+7 will recover in different ways:

in an acidic environment (up to Mn +2 ) the reduction product will be a salt, e.g. MnSO 4 ,

in a neutral environment(up to Mn +4 ) the reduction product will be MnO 2 or MnO (OH) 2,

in an alkaline environment(up to Mn +6 ) the reduction product will be a manganate, for example, K 2 MnO 4 .

For example, when reducing a solution of potassium permanganate with sodium sulfite, depending on the reaction of the medium, the corresponding products will be obtained:

acidic environment -

2KMnO 4 + 5Na 2 SO 3 + 3H 2 SO 4 = 5Na 2 SO 4 + 2MnSO 4 + K 2 SO 4 + H 2 O

neutral environment –

2KMnO 4 + 3Na 2 SO 3 + H 2 O \u003d 3Na 2 SO 4 + 2MnO 2 + 2KOH

alkaline environment -

2KMnO 4 + Na 2 SO 3 + 2NaOH \u003d Na 2 SO 4 + Na 2 MnO 4 + K 2 MnO 4 + H 2 O.

The temperature of the system also affects the course of the redox reaction. So, the products of the interaction of chlorine with an alkali solution will be different depending on the temperature conditions.

When chlorine reacts withcold alkali solutionThe reaction proceeds with the formation of chloride and hypochlorite:

0 -1 +1

Cl 2 + KOH → KCI + KCIO + H 2 O

CI 0 + 1e – → CI –1 reduction, oxidizer

CI 0 – 1e – → CI +1 oxidation, reducing agent.

If you take hot concentrated KOH solution, then as a result of interaction with chlorine we get chloride and chlorate:

0 t ° -1 +5

3CI 2 + 6KOH → 5KCI + KCIO 3 + 3H 2 O

5 │ CI 0 + 1e – → CI –1 reduction, oxidizer

1 │ CI 0 – 5e – → CI +5 oxidation, reducing agent.

Questions for self-control on the topic

"Redox Reactions"

1. What reactions are called redox reactions?

2. What is the oxidation state of an atom? How is it defined?

3. What is the degree of oxidation of atoms in simple substances?

4. What is the sum of the oxidation states of all atoms in a molecule?

5. What process is called the oxidation process?

6. What substances are called oxidizing agents?

7. How does the oxidation state of an oxidizing agent change in redox reactions?

8. Give examples of substances that are only oxidizing agents in redox reactions.

9. What process is called the recovery process?

10. Define the term "reductant".

11. How does the oxidation state of the reducing agent change in redox reactions?

12. What substances can only be reducing agents?

13. What element is an oxidizing agent in the reaction of dilute sulfuric acid with metals?

14. What element is an oxidizing agent in the interaction of concentrated sulfuric acid with metals?

15. What is the function of nitric acid in redox reactions?

16. What compounds can be formed as a result of the reduction of nitric acid in reactions with metals?

17. What element is an oxidizing agent in concentrated, dilute and very dilute nitric acid?

18. What role can hydrogen peroxide play in redox reactions?

19. How are all redox reactions classified?

Tests for self-control of knowledge of the theory on the topic "Oxidation-reduction reactions"

Option number 1

1) CuSO 4 + Zn = ZnSO 4 + Cu,

2) CaCO 3 + CO 2 + H 2 O \u003d Ca (HCO 3) 2,

3) SO 3 + H 2 O \u003d H 2 SO 4,

4) FeCl 3 + 3NaOH \u003d Fe (OH) 3 + 3NaCl,

5) NaHCO 3 + NaOH = Na 2 CO 3 + H 2 O.

2. Based on the structure of atoms, determine under what number the formula of the ion is indicated, which can only be an oxidizing agent:

1) Mn , 2) NO 3– , 3) ​​Br – , 4) S 2– , 5) NO 2– ?

3. Under what number is the formula of the substance that is the most powerful reducing agent, from among the following:

1) NO 3–, 2) Сu, 3) Fe, 4) Ca, 5) S?

4. What number indicates the amount of substance KMnO 4 , in moles, which interacts with 10 moles of Na 2 SO 3 in the reaction represented by the following scheme:

KMnO 4 + Na 2 SO 3 + H 2 SO 4 → MnSO 4 + Na 2 SO 4 + K 2 SO 4 + H 2 O?

1) 4, 2) 2, 3) 5, 4) 3, 5) 1.

5. What number is the disproportionation reaction (self-oxidation - self-recovery)?

1) 2H 2 S + H 2 SO 3 \u003d 3S + 3H 2 O,

2) 4KClO 3 \u003d KCl + 3KClO 4,

3) 2F 2 + 2H 2 O \u003d 4HF + O 2.

4) 2Au 2 O 3 \u003d 4Au + 3O 2,

5) 2KClO 3 \u003d 2KCl + 3O 2.

Option number 2

1. Under what number is the equation of the redox reaction given?

1) 4KClO 3 \u003d KCl + 3KClO 4,

2) CaCO 3 \u003d CaO + CO 2,

3) CO 2 + Na 2 O \u003d Na 2 CO 3,

4) CuOHCl + HCl \u003d CuCl 2 + H 2 O,

5) Pb(NO 3 ) 2 + Na 2 SO 4 = PbSO 4 + 2NaNO 3 .

2. Under what number is the formula of a substance that can only be a reducing agent:

1) SO 2, 2) NaClO, 3) KI, 4) NaNO 2, 5) Na 2 SO 3?

3. Under what number is the formula of the substance, which is the most powerful oxidizing agent, among the given:

1) I 2 , 2) S, 3) F 2 , 4) O 2 , 5) Br 2 ?

4. Under what number is the volume of hydrogen in liters under normal conditions, which can be obtained from 9 g of Al as a result of the following redox reaction:

2Al + 6H 2 O \u003d 2Al (OH) 3 + 3H 2

1) 67,2, 2) 44,8, 3) 33,6, 4) 22,4, 5) 11,2?

5. What number is the scheme of the redox reaction that takes place at pH > 7?

1) I 2 + H 2 O → HI + HIO,

2) FeSO 4 + HIO 3 + ... → I 2 + Fe(SO 4 ) 3 + …,

3) KMnO4 + NaNO2 + … → MnSO4 + …,

4) KMnO4 + NaNO2 + … → K2 MNO4 + …,

5) CrCl3 + KMnO4 + … → K2 Cr2 O7 + MnO(OH)2 + … .

Option number 3

1. Under what number is the equation of the redox reaction given?

1) H2 SO4 + Mg → MgSO4 + H2 ,

2) CuSO4 + 2NaOH →Cu(OH)2 + Na2 SO4 ,

3) SO3 + K2 O→K2 SO4 ,

4) CO2 + H2 O→H2 CO3 ,

5)H2 SO4 + 2KOH → K2 SO4 + 2H2 Oh

2. Based on the structure of the atom, determine the number under which the formula of the ion is given, which can be a reducing agent:

1) Ag+ , 2) Al3+, 3) Cl7+, 4) Sn2+ , 5) Zn2+ ?

3. What is the recovery process number?

1) NO2– → NO3– , 2) S2– → S0 , 3) Mn2+ →MnO2 ,

4) 2I– → I2 , 5) → 2Cl– .

4. Under what number is the mass of the reacted iron given, if as a result of the reaction represented by the following scheme:

Fe + HNO3 → Fe(NO3 ) 3 + NO + H2 O

formed 11.2 L NO(n.o.)?

1) 2,8, 2) 7, 3) 14, 4) 56, 5) 28.

5. Under what number is the scheme of the reaction of self-oxidation-self-recovery (dismutation)?

1) HI+H2 SO4 → I2 + H2 S+H2 O

2) FeCl2 + SnCl4 → FeCl3 + SnCl2 ,

3) HNO2 → NO + NO2 + H2 O

4) KClO3 → KCl + O2 ,

5) Hg(NO3 ) 2 → HgO + NO2 + O2 .

See the answers to the test questions on p.

Questions and exercises for self-study

research work on the topic.

1. Indicate the number or sum of the conditional numbers under which the schemes of redox reactions are located:

1) MgCO3 + HCl MgCl2 + CO2 + H2 O

2) FeO + P Fe+P2 O5 ,

4)H2 O2  H2O+O2 , 8) KOH + CO2  KHCO3 .

2. Indicate the number or sum of conditional numbers under which redox processes are located:

1) electrolysis of sodium chloride solution,

2) pyrite firing,

3) hydrolysis of sodium carbonate solution,

4) lime slaking.

3. Indicate the number or sum of conditional numbers under which the names of groups of substances are located, characterized by an increase in oxidizing properties:

1) chlorine, bromine, fluorine,

2) carbon, nitrogen, oxygen,

3) hydrogen, sulfur, oxygen,

4) bromine, fluorine, chlorine.

4. Which of the substances -chlorine, sulfur, aluminum, oxygen– is a stronger reducing agent? In your answer, indicate the value of the molar mass of the selected compound.

5. Indicate the number or sum of conditional numbers under which only oxidizing agents are located:

1) K2 MNO4 , 2) KMnO4 , 4) MnO3 , 8) MnO2 ,

16) K2 Cr2 O7 , 32) K2 SO3 .

6. Indicate the number or sum of conditional numbers under which the formulas of substances with redox duality are located:

1) KI, 2) H2 O2 , 4) Al, 8) SO2 , 16) K2 Cr2 O7 , 32) H2 .

7. Which of the compounds -iron oxide(III)chromium oxide(III)sulfur oxide(IV)nitrogen oxide(II)nitrogen oxide(V) - can only be an oxidizing agent? In your answer, indicate the value of the molar mass of the selected compound.

8. Indicate the number or sum of conditional numbers, under which are the formulas of substances that have an oxygen oxidation state - 2:

1) H2 O, Na2 O, Cl2 O, 2) HPO3 , Fe2 O3 , SO3 ,

4) OF2 , Ba(OH)2 , Al2 O3 , 8) BaO2 , Fe3 O4 , SiO2 .

9. Which of the following compounds can only be an oxidizing agent:sodium nitrite, sulfurous acid, hydrogen sulfide, nitric acid? In your answer, indicate the value of the molar mass of the selected compound.

10. Which of the following nitrogen compounds - NH3 ; HNO3 ; HNO2 ; NO2 - can only be an oxidizing agent? In your answer, write down the value of the relative molecular weight of the selected compound.

11. Under what number, among the names of substances listed below, is the strongest oxidizing agent indicated?

1) concentrated nitric acid,

2) oxygen,

3) electricity at the anode during electrolysis,

4) fluorine.

12. Which of the following nitrogen compounds - HNO3 ; NH3 ; HNO2 ; NO - can only be a reducing agent? In your answer, write down the molar mass of the selected compound.

13. Which of the compounds - Na2 S; K2 Cr2 O7 ; KMnO4 ; NaNO2 ; KClO4 - can be both an oxidizing agent and a reducing agent, depending on the reaction conditions? In your answer, write down the molar mass of the selected compound.

14. Indicate the number or sum of conditional numbers, where ions are indicated that can be reducing agents:

1) (MnO4 ) 2– , 2) (CrO4 ) –2 , 4) Fe+2 , 8) Sn+4 , 16) (ClO4 ) – .

15. Indicate the number or sum of conditional numbers, under which only oxidizing agents are located:

1) K2 MNO4 , 2) HNO3 , 4) MNO3 , 8) MnO2 , 16) K2 CrO4 , 32) H2 O2 .

16. Indicate the number or sum of conditional numbers, under which are located only the names of substances between which redox reactions are not possible:

1) carbon and sulfuric acid,

2) sulfuric acid and sodium sulfate,

4) hydrogen sulfide and hydrogen iodide,

8) sulfur oxide (IV) and hydrogen sulfide.

17. Indicate the number or sum of conditional numbers under which the oxidation processes are located:

1) S+6  S–2 , 2) Mn+2  Mn+7 , 4) S–2  S+4 ,

8) Mn+6  Mn+4 , 16) O2  2O–2 , 32) S+4  S+6 .

18. Indicate the number or sum of conditional numbers under which the recovery processes are located:

1) 2I–1  I2 , 2) 2N+3  N2 , 4) S–2  S+4 ,

8) Mn+6  Mn+2 , 16) Fe+3  Fe0 , 32) S0  S+6 .

19. Specify the number or sum of conditional numbers under which the recovery processes are located:

1) C0  CO2 , 2) Fe+2  Fe+3 ,

4) (SO3 ) 2–  (SO4 ) 2– , 8) MnO2  Mn+2 .

20. Indicate the number or sum of conditional numbers under which the recovery processes are located:

1) Mn+2  MNO2 , 2) (IO3 ) –  (IO4 ) – ,

4) (NO2 ) –  (NO3 ) – , 8) MnO2  Mn+2 .

21. Indicate the number or the sum of conditional numbers under which the ions that are reducing agents are located.

1) Ca+2 , 2) Al+3 , 4) K+ , 8) S–2 , 16) Zn+2 , 32) (SO3 ) 2– .

22. Under what number is the formula of a substance, in the interaction with which hydrogen acts as an oxidizing agent?

1) O2 , 2) Na, 3) S, 4) FeO.

23. Under what number is the reaction equation in which the reducing properties of the chloride ion appear?

1) MNO2 + 4HCl = MnCl2 +Cl2 + 2H2 O,

2) CuO + 2HCl = CuCl2 + H2 O

3) Zn + 2HCl = ZnCl2 + H2 ,

4) AgNO3 + HCl = AgCl + HNO3 .

24. When interacting with which of the following substances - O2 , NaOH, H2 S - sulfur oxide (IV) exhibits the properties of an oxidizing agent? Write the equation of the corresponding reaction and in the answer indicate the sum of the coefficients of the starting substances.

25. Indicate the number or sum of conditional numbers under which the disproportionation reaction schemes are located:

1) NH4 NO3  N2 O+H2 O, 2) NH4 NO2  N2 + H2 O

4) KClO3  KClO4 + KCl, 8) KClO3  KCl+O2 .

26. Draw an electronic balance diagram and indicate how much potassium permanganate is involved in the reaction with ten moles of sulfur oxide (IV). The reaction proceeds according to the scheme:

KMnO4 + SO2  MnSO4 + K2 SO4 + SO3 .

27. Draw an electronic balance diagram and indicate how much potassium sulfide substance interacts with six moles of potassium permanganate in the reaction:

K2 S+KMnO4 + H2 O MNO2 + S + KOH.

28. Draw an electronic balance diagram and indicate how much potassium permanganate substance interacts with ten moles of iron (II) sulfate in the reaction:

KMnO4 + FeSO4 + H2 SO4  MnSO4 + Fe2 (SO4 ) 3 + K2 SO4 + H2 Oh

29. Draw an electronic balance diagram and indicate how much potassium chromite (KCrO2 ) reacts with six moles of bromine in the reaction:

KCrO2 +Br2 +KOH K2 CrO4 + KBr + H2 Oh

30. Draw an electronic balance diagram and indicate how much of the manganese (IV) oxide substance interacts with six moles of lead (IV) oxide in the reaction:

MNO2 + PbO2 + HNO3  HMnO4 + Pb(NO3 ) 2 + H2 Oh

31. Write the reaction equation:

KMnO4 + NaI + H2 SO4 I2 + K2 SO4 +MnSO4 + Na2 SO4 + H2 Oh

32. Write the reaction equation:

KMnO4 + NaNO2 + H2 O MNO2 + NaNO3 + KOH.

In your answer, indicate the sum of the stoichiometric coefficients in the reaction equation.

33. Write the reaction equation:

K2 Cr2 O7 + HClconc. KCl + CrCl3 +Cl2 + H2 Oh

In your answer, indicate the sum of the stoichiometric coefficients in the reaction equation.

34. Draw an electronic balance diagram and indicate how much sodium nitrite (NaNO2 ) reacts with four moles of potassium permanganate in the reaction:

KMnO4 + NaNO2 + H2 SO4  MnSO4 + NaNO3 + K2 SO4 + H2 Oh

35. Draw an electronic balance diagram and indicate how much hydrogen sulfide substance interacts with six moles of potassium permanganate in the reaction:

KMnO4 + H2 S+H2 SO4  S+MnSO4 + K2 SO4 + H2 Oh

36. What amount of iron substance in moles will be oxidized by oxygen with a volume of 33.6 liters (n.o.) in the reaction proceeding according to the scheme below?

Fe+H2 O+O2  Fe(OH)3 .

37. Which of the following metals - Zn, Rb, Ag, Fe, Mg - does not dissolve in dilute sulfuric acid? In your answer, indicate the value of the relative atomic mass of this metal.

38. Which of the following metals - Zn, Rb, Ag, Fe, Mg - does not dissolve in concentrated sulfuric acid? In your answer, indicate the ordinal number of the element in the periodic system of D.I. Mendeleev.

39. Indicate the number or sum of conditional numbers under which the metals are passivated in concentrated solutions of oxidizing acids.

1) Zn, 2) Cu, 4) Au, 8) Fe, 16) Mg, 32) Cr.

40. Indicate the number or sum of conditional numbers under which there are chemical signs of metals that do not displace hydrogen from a dilute solution of sulfuric acid, but displace mercury from solutions of Hg salts2+ :

1) Fe, 2) Zn, 4) Au, 8) Ag, 16) Cu.

41. Under what number are the chemical signs of metals, each of which does not react with nitric acid?

1) Zn, Ag; 2) Pt, Au; 3) Cu, Zn; 4) Ag, Hg.

42. Under what number is the method of obtaining chlorine in industry indicated?

1) electrolysis of sodium chloride solution;

2) the action of manganese oxide (1V) on hydrochloric acid;

3) thermal decomposition of natural chlorine compounds;

4) the action of fluorine on chlorides.

43. Under what number is the chemical formula of the gas that is predominantly released during the action of a concentrated solution of nitric acid on copper?

1) N2 , 2) NO2 , 3) NO, 4) H2 .

44. Under what number are the formulas of the reaction products of the combustion of hydrogen sulfide in air with a lack of oxygen?

1) SO2 + H2 O, 2) S + H2 O

3) SO3 + H2 O, 4) SO2 + H2 .

Give the number of the correct answer.

45. Write an equation for the reaction of the interaction of concentrated sulfuric acid with copper. In your answer, indicate the sum of the coefficients in the reaction equation.

8. Classification of chemical reactions. OVR. Electrolysis

8.3. Redox reactions: general provisions

redox reactions(OVR) are called reactions that occur with a change in the oxidation state of the atoms of the elements. As a result of these reactions, some atoms donate electrons, while others accept them.

A reducing agent is an atom, ion, molecule or FE that donates electrons, an oxidizing agent is an atom, ion, molecule or FE that accepts electrons:

The process of giving off electrons is called oxidation, and the process of accepting - restoration. In OVR, there must be a reducing agent and an oxidizing agent. There is no oxidation process without a reduction process and there is no reduction process without an oxidation process.

The reducing agent donates electrons and is oxidized, while the oxidizing agent accepts electrons and is reduced.

The reduction process is accompanied by a decrease in the degree of oxidation of atoms, and the oxidation process is accompanied by an increase in the degree of oxidation of atoms of elements. It is convenient to illustrate the above with a diagram (CO - oxidation state):

Specific examples oxidation and reduction processes (electronic balance schemes) are given in Table. 8.1.

Table 8.1

Examples of electronic balance schemes

Scheme of electronic balance	Process characteristic
Oxidation process
	The calcium atom donates electrons, increases the degree of oxidation, is a reducing agent
	Ion Cr +2 donates electrons, increases the degree of oxidation, is a reducing agent
	The chlorine molecule donates electrons, chlorine atoms increase the oxidation state from 0 to +1, chlorine is a reducing agent
Recovery process
	The carbon atom accepts electrons, lowers the oxidation state, is an oxidizing agent
	The oxygen molecule accepts electrons, the oxygen atoms lower their oxidation state from 0 to -2, the oxygen molecule is an oxidizing agent
	The ion accepts electrons, lowers the oxidation state, is an oxidizing agent

The most important reducing agents: simple substances metals; hydrogen; carbon in the form of coke; carbon monoxide(II); compounds containing atoms in the lowest oxidation state (metal hydrides,, sulfides, iodides, ammonia); the strongest reducing agent is the electric current at the cathode.

The most important oxidizers: simple substances - halogens, oxygen, ozone; concentrated sulfuric acid; Nitric acid; a number of salts (KClO 3 , KMnO 4 , K 2 Cr 2 O 7); hydrogen peroxide H 2 O 2 ; the strongest oxidizing agent is an electric current at the anode.

Over the period, the oxidizing properties of atoms and simple substances are enhanced: fluorine - the strongest oxidizing agent of all simple substances. In each period, halogens form simple substances with the most pronounced oxidizing properties.

In groups A, from top to bottom, the oxidizing properties of atoms and simple substances weaken, while the reducing properties increase.

For atoms of the same type, the reducing properties increase with an increase in their radius; for example, the reducing properties of the anion
I - are more pronounced than the anion Cl - .

For metals, the redox properties of simple substances and ions in an aqueous solution are determined by the position of the metal in the electrochemical series: from left to right (top to bottom), the reducing properties of simple metals weaken: the strongest reducing agent- lithium.

For metal ions in an aqueous solution, from left to right in the same row, respectively, the oxidizing properties are enhanced: most powerful oxidizing agent- Au 3 + ions.

To arrange the coefficients in the OVR, you can use a method based on the mapping of oxidation and reduction processes. This method is called electronic balance method.

The essence of the electronic balance method is as follows.

1. Draw up a reaction scheme and determine the elements that have changed the oxidation state.

2. Compose electronic equations for half-reactions of reduction and oxidation.

3. Since the number of electrons donated by the reducing agent must be equal to the number of electrons accepted by the oxidizing agent, additional factors are found using the least common multiple (LCM) method.

4. Additional multipliers are put down before the formulas of the corresponding substances (coefficient 1 is omitted).

5. Equalize the number of atoms of those elements that have not changed the degree of oxidation (first - hydrogen in water, and then - the number of oxygen atoms).

An example of compiling an equation for a redox reaction

electronic balance method.

We find that the carbon and sulfur atoms have changed their oxidation state. We compose the equations of half-reactions of reduction and oxidation:

For this case, the LCM is 4, and the additional factors are 1 (for carbon) and 2 (for sulfuric acid).

We put down the additional factors found in the left and right parts of the reaction scheme in front of the formulas of substances containing carbon and sulfur:

C + 2H 2 SO 4 → CO 2 + 2SO 2 + H 2 O

We equalize the number of hydrogen atoms by putting a factor of 2 in front of the water formula, and we make sure that the number of oxygen atoms in both parts of the equation is the same. Therefore, the OVR equation

C + 2H 2 SO 4 \u003d CO 2 + 2SO 2 + 2H 2 O

The question arises: in which part of the OVR scheme should the found additional factors be placed - on the left or on the right?

For simple reactions, this does not matter. However, it should be borne in mind: if additional factors are defined on the left side of the equation, then the coefficients are put down before the formulas of substances on the left side; if the calculations were carried out for the right side, then the coefficients are put on the right side of the equation. For example:

According to the number of Al atoms on the left side:

According to the number of Al atoms on the right side:

In the general case, if substances of a molecular structure participate in the reaction (O 2, Cl 2, Br 2, I 2, N 2), then when selecting the coefficients, they proceed precisely from the number of atoms in the molecule:

If N 2 O is formed in a reaction involving HNO 3, then it is also better to write the electron balance scheme for nitrogen based on two nitrogen atoms .

In some redox reactions, one of the substances can perform the function of both an oxidizing agent (reducing agent) and a salt former (i.e., participate in the formation of salt).

Such reactions are typical, in particular, for the interaction of metals with oxidizing acids (HNO 3, H 2 SO 4 (conc)), as well as oxidizing salts (KMnO 4 , K 2 Cr 2 O 7 , KClO 3 , Ca (OCl) 2) with hydrochloric acid (due to the anions Cl − hydrochloric acid has reducing properties) and other acids, the anion of which is a reducing agent.

Let's make an equation for the reaction of copper with dilute nitric acid:

We see that part of the nitric acid molecules is spent on the oxidation of copper, while being reduced to nitric oxide (II), and part is used to bind the formed Cu 2+ ions to the salt Cu (NO 3) 2 (in the composition of the salt, the degree of oxidation of the nitrogen atom is the same , as in acid, i.e. does not change). In such reactions, an additional factor for the oxidizing element is always placed on the right side before the reduction product formula, in this case, before the NO formula, and not HNO 3 or Cu(NO 3) 2 .

Before the HNO 3 formula, we put a coefficient of 8 (two HNO 3 molecules are spent on the oxidation of copper and six on the binding of three Cu 2+ ions into a salt), we equalize the numbers of H and O atoms and get

3Cu + 8HNO 3 \u003d 3Cu (NO 3) 2 + 2NO + 4H 2 O.

In other cases, an acid, such as hydrochloric acid, can simultaneously be both a reducing agent and participate in the formation of a salt:

Example 8.5. Calculate what mass of HNO 3 is spent on salt formation, when in the reaction, the equation of which

zinc enters with a mass of 1.4 g.

Decision. From the reaction equation, we see that out of 8 moles of nitric acid, only 2 moles went to the oxidation of 3 moles of zinc (there is a factor of 2 in front of the formula for the acid reduction product, NO). Salt formation consumed 6 mol of acid, which is easy to determine by multiplying the coefficient 3 in front of the salt formula Zn(HNO 3) 2 by the number of acid residues in one formula unit of the salt, i.e. on 2.

n (Zn) \u003d 1.4 / 65 \u003d 0.0215 (mol).

x = 0.043 mol;

m (HNO 3) \u003d n (HNO 3) M (HNO 3) \u003d 0.043 ⋅ 63 \u003d 2.71 (g)

Answer: 2.71 g.

In some OVR, the oxidation state is changed by the atoms of not two, but three elements.

Example 8.6. Arrange the coefficients in the OVR flowing according to the scheme FeS + O 2 → Fe 2 O 3 + SO 2 using the electron balance method.

Decision. We see that the oxidation state is changed by the atoms of three elements: Fe, S and O. In such cases, the numbers of electrons donated by atoms of different elements are summed up:

Having placed the stoichiometric coefficients, we obtain:

4FeS + 7O 2 \u003d 2Fe 2 O 3 + 4SO 2.

Consider examples of solving other types of examination tasks on this topic.

Example 8.7. Indicate the number of electrons passing from the reducing agent to the oxidizing agent during the complete decomposition of copper(II) nitrate, with a mass of 28.2 g.

Decision. We write down the reaction equation for the decomposition of salt and the scheme of the electronic balance of the OVR; M = 188 g/mol.

We see that 2 mol O 2 is formed during the decomposition of 4 mol salt. At the same time, 4 mol of electrons pass from the atoms of the reducing agent ( in this case, these are ions) to the oxidizing agent ( i.e. to ions): . Since the chemical amount of salt is n = 28.2/188 = = 0.15 (mol), we have:

2 moles of salt - 4 moles of electrons

0.15 mol - x

n (e) \u003d x \u003d 4 ⋅ 0.15 / 2 \u003d 0.3 (mol),

N (e) \u003d N A n (e) \u003d 6.02 ⋅ 10 23 ⋅ 0.3 \u003d 1.806 ⋅ 10 23 (electrons).

Answer: 1.806 ⋅ 10 23 .

Example 8.8. During the interaction of sulfuric acid with a chemical amount of 0.02 mol with magnesium, sulfur atoms added 7.224 ⋅ 10 22 electrons. Find the formula for the acid recovery product.

Decision. In the general case, the schemes for the processes of reduction of sulfur atoms in the composition of sulfuric acid can be as follows:

those. 1 mole of sulfur atoms can accept 2, 6 or 8 moles of electrons. Given that 1 mol of acid contains 1 mol of sulfur atoms, i.e. n (H 2 SO 4) = n (S), we have:

n (e) \u003d N (e) / N A \u003d (7.224 ⋅ 10 22) / (6.02 ⋅ 10 23) \u003d 0.12 (mol).

We calculate the number of electrons accepted by 1 mol of acid:

0.02 mole of acid accept 0.12 mole of electrons

1 mol - x

n (e) \u003d x \u003d 0.12 / 0.02 \u003d 6 (mol).

This result corresponds to the process of reducing sulfuric acid to sulfur:

Answer: sulfur.

Example 8.9. In the reaction of carbon with concentrated nitric acid, water and two salt-forming oxides are formed. Find the mass of carbon that reacted if the atoms of the oxidizing agent took 0.2 mol of electrons in this process.

Decision. The interaction of substances proceeds according to the reaction scheme

We compose the equations for the half-reactions of oxidation and reduction:

From the schemes of the electronic balance, we see that if the atoms of the oxidizing agent () accept 4 mol of electrons, then 1 mol (12 g) of carbon enters into the reaction. Compose and solve the proportion:

4 moles of electrons - 12 g of carbon

0.2 - x

x = 0.2 ⋅ 12 4 = 0.6 (d).

Answer: 0.6 g.

Classification of redox reactions

There are intermolecular and intramolecular redox reactions.

When intermolecular OVR the atoms of the oxidizing agent and the reducing agent are part of different substances and are atoms of different chemical elements.

When intramolecular OVR The oxidizing and reducing atoms are in the same substance. Intramolecular reactions are disproportionation, in which the oxidizing agent and reducing agent are atoms of the same chemical element in the composition of the same substance. Such reactions are possible for substances containing atoms with an intermediate oxidation state.

Example 8.10. Specify the OVR disproportionation scheme:

1) MnO 2 + HCl → MnCl 2 + Cl 2 + H 2 O

2) Zn + H 2 SO 4 → ZnSO 4 + H 2

3) KI + Cl 2 → KCl + I 2

4) Cl 2 + KOH → KCl + KClO + H 2 O

Decision . Reactions 1)–3) are intermolecular OVR:

The disproportionation reaction is reaction 4), since it contains a chlorine atom and an oxidizing agent and a reducing agent:

Answer: 4).

It is possible to qualitatively assess the redox properties of substances based on the analysis of the oxidation states of atoms in the composition of the substance:

1) if the atom responsible for the redox properties is in the highest degree of oxidation, then this atom can no longer donate electrons, but can only accept them. Therefore, in OVR, this substance will exhibit only oxidizing properties. Examples of such substances (in the formulas the oxidation state of the atom responsible for the redox properties is indicated):

2) if the atom responsible for the redox properties is in the lowest oxidation state, then this substance in the OVR will show only restorative properties(A given atom can no longer accept electrons, it can only give them away). Examples of such substances:,. Therefore, all halogen anions (with the exception of F - for the oxidation of which an electric current at the anode is used), the sulfide ion S 2-, the nitrogen atom in the ammonia molecule, and the hydride ion H - show only reducing properties in the OVR. Metals (Na, K, Fe) have only reducing properties;

3) if an atom of an element is in an intermediate oxidation state (the oxidation state is greater than the minimum, but less than the maximum), then the corresponding substance (ion) will, depending on the conditions, exhibit dual oxidation-restorative properties: stronger oxidizing agents will oxidize these substances (ions), and stronger reducing agents will reduce them. Examples of such substances: sulfur, since the highest oxidation state of the sulfur atom is +6, and the lowest is -2, sulfur oxide (IV), nitric oxide (III) (the highest oxidation state of the nitrogen atom is +5, and the lowest is -3), hydrogen peroxide ( The highest oxidation state of the oxygen atom is +2, and the lowest is -2). Dual redox properties are exhibited by metal ions in an intermediate oxidation state: Fe 2+, Mn +4, Cr +3, etc.

Example 8.11. A redox reaction cannot proceed, the scheme of which is:

1) Cl 2 + KOH → KCl + KClO 3 + H 2 O

2) S + NaOH → Na 2 S + Na 2 SO 3 + H 2 O

3) KClO → KClO 3 + KClO 4

4) KBr + Cl 2 → KCl + Br

Decision. The reaction, the scheme of which is indicated at number 3), cannot proceed, since it contains a reducing agent, but no oxidizing agent:

Answer: 3).

For some substances, redox duality is due to the presence in their composition of various atoms in both the lowest and highest oxidation states; for example, hydrochloric acid (HCl) due to the hydrogen atom (the highest oxidation state, equal to +1) is an oxidizing agent, and due to the Cl − anion, it is a reducing agent (lower oxidation state).

OVR is impossible between substances that exhibit only oxidizing (HNO 3 and H 2 SO 4, KMnO 4 and K 2 CrO 7) or only reducing properties (HCl and HBr, HI and H 2 S)

OVR are extremely common in nature (metabolism in living organisms, photosynthesis, respiration, decay, combustion), are widely used by humans for various purposes (obtaining metals from ores, acids, alkalis, ammonia and halogens, creating chemical current sources, obtaining heat and energy during combustion of various substances). Note that OVR often complicate our lives (spoilage of food, fruits and vegetables, corrosion of metals - all this is associated with the occurrence of various redox processes).

Ion-electronic method (half-reaction method)

When compiling the OVR equations flowing in aqueous solutions, it is preferable to select the coefficients using the half-reaction method.

The procedure for selecting coefficients using the half-reaction method:

1. Write down the reaction scheme in molecular and ionic-molecular forms and determine the ions and molecules that change the oxidation state.

2. Determine the environment in which the reaction proceeds (H + - acidic; OH - alkaline; H 2 O - neutral)

3. Make up an ion-molecular equation for each half-reaction and equalize the number of atoms of all elements.

1. The number of oxygen atoms is equalized using water molecules or OH - ions.
2. If the original ion or molecule contains more oxygen atoms than the reaction product, then

• an excess of oxygen atoms in an acidic environment binds with H + ions into water molecules
• in neutral and alkaline environment excess oxygen atoms are bound by water molecules into OH groups -
  

1. If the original ion or molecule contains fewer oxygen atoms than the reaction product, then

· the lack of oxygen atoms in acidic and neutral solutions is compensated by water molecules

· in alkaline solutions - due to OH - ions.

4. Compose electron-ion equations of half-reactions.

To do this, electrons are added (or subtracted) to the left side of each half-reaction in such a way that the total charge on the left and right sides of the equations becomes the same. We multiply the resulting equations by the smallest factors, for the balance of electrons.

5. Summarize the resulting electron-ion equations. Cancel like terms and get the ion-molecular OVR equation

6. According to the obtained ion-molecular equation, a molecular equation is made.

Example :

1 . Na 2 SO 3 + KMnO 4 + H 2 SO 4 → Na 2 SO 4 + MnSO 4 + K 2 SO 4 + H 2 O

2Na + +SO 3 2- +K + +MnO 4 - +2H + +SO 4 2- →2Na + +SO 4 2- +Mn 2+ +SO 4 2- +2K + +SO 4 2- +H 2 O

SO 3 2- → SO 4 2-

MNO 4 - → Mn 2+

2 . Acidic environment - H +

3 .

MnO 4 - + 8 H + → Mn 2+ + 4 H 2 O

SO 3 2- + H 2 O → SO 4 2- + 2 H +

4 .

MnO 4 - + 8 H + + 5ē → Mn 2+ + 4 H 2 O│ X2

SO 3 2- + H 2 O - 2ē → SO 4 2- + 2 H + │ X5

5 .

2MnO 4 - + 16 H + + 10ē →2Mn 2+ + 8 H 2 O

5SO 3 2- + 5H 2 O - 10ē → 5SO 4 2- + 10 H +

2MnO 4 - + 16 H + + 5SO 3 2- + 5H 2 O →2Mn 2+ + 8 H 2 O + 5SO 4 2- + 10 H +

2MnO 4 - + 6 H + + 5SO 3 2- →2Mn 2+ + 3 H 2 O + 5SO 4 2-

6 . 5Na 2 SO 3 + 2KMnO 4 + 3H 2 SO 4 → 5Na 2 SO 4 + 2MnSO 4 + K 2 SO 4 + 3H 2 O

Reminder!