On this basis, oxidative reducing reactions and reactions proceeding without changing the oxidation states of chemical elements.
These include many reactions, including all substitution reactions, as well as those reactions of combination and decomposition in which at least one simple substance participates, for example:
As you remember, the coefficients in complex redox reactions are placed using the electronic balance method:
AT organic chemistry a striking example of redox reactions are the properties of aldehydes.
1. They are reduced to the corresponding alcohols:
2. Aldehydes are oxidized to the corresponding acids:
The essence of all the above examples of redox reactions was presented using the well-known to you method of electron balance. It is based on comparing the oxidation states of atoms in the reactants and reaction products and on balancing the number of electrons in the oxidation and reduction processes. This method is used to compile equations for reactions occurring in any phases. This makes it versatile and convenient. But at the same time, it has a serious drawback - when expressing the essence of redox reactions occurring in solutions, particles are indicated that do not really exist.
In this case, it is more convenient to use another method - the method of half-reactions. It is based on the compilation of ion-electronic equations for the processes of oxidation and reduction, taking into account real-life particles and their subsequent summation in general equation. This method does not use the concept of "oxidation state", and the products are determined by deriving the reaction equation.
Let's demonstrate this method using an example: we will make an equation for the redox reaction of zinc with concentrated nitric acid.
1. We write down the ionic scheme of the process, which includes only the reducing agent and its oxidation product, the oxidizing agent and its reduction product:
2. We compose the ion-electronic equation of the oxidation process (this is the 1st half-reaction):
3. We compose the ion-electronic equation of the reduction process (this is the 2nd half-reaction):
Please note: electron-ionic equations are compiled in accordance with the law of conservation of mass and charge.
4. We write down the half-reaction equations so that the number of electrons between the reducing agent and the oxidizing agent is balanced:
5. We sum term by term the equations of half reactions. Compiling a general ionic equation reactions:
We check the correctness of the reaction equation in ionic form:
- Compliance with equality in the number of atoms of elements and in the number of charges
- The number of element atoms must be equal in the left and right parts ionic reaction equation.
- The total charge of the particles on the left and right sides of the ionic equation must be the same.
6. Write down the equation in molecular form. To do this, add to the ions included in the ionic equation, the required number of ions of opposite charge.
Calculation of the degree of oxidation
Summary
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To calculate the oxidation state of an element, the following provisions should be taken into account:
1. The oxidation states of atoms in simple substances are equal to zero (Na 0 ; H 2 0).
2. The algebraic sum of the oxidation states of all atoms that make up the molecule is always zero, and in a complex ion this sum is equal to the charge of the ion.
3. Atoms have a constant oxidation state: alkali metals (+1), alkaline earth metals(+2), hydrogen (+1) (except for hydrides NaH, CaH 2, etc., where the oxidation state of hydrogen is -1), oxygen (-2) (except for F 2 -1 O +2 and peroxides containing the -O group –O–, in which the oxidation state of oxygen is -1).
4. For elements, the positive oxidation state cannot exceed a value equal to the group number of the periodic system.
Examples:
V 2 +5 O 5 -2; Na 2 +1 B 4 +3 O 7 -2; K +1 Cl +7 O 4 -2; N -3 H 3 +1; K 2 +1 H +1 P +5 O 4 -2; Na 2 +1 Cr 2 +6 O 7 -2
There are two types of chemical reactions:
A Reactions in which the degree element oxidation:
Addition reactions
SO 2 + Na 2 O Na 2 SO 3
Decomposition reactions
Cu(OH) 2 - t CuO + H 2 O
Exchange reactions
AgNO 3 + KCl AgCl + KNO 3
NaOH + HNO 3 NaNO 3 + H 2 O
B Reactions in which there is a change in the oxidation states of the atoms of the elements that make up the reacting compounds:
2Mg 0 + O 2 0 2Mg +2 O -2
2KCl +5 O 3 -2 - t 2KCl -1 + 3O 2 0
2KI -1 + Cl 2 0 2KCl -1 + I 2 0
Mn +4 O 2 + 4HCl -1 Mn +2 Cl 2 + Cl 2 0 + 2H 2 O
Such reactions are called redox.
Task number 1
Establish a correspondence between the reaction equation and the property of the nitrogen element that it exhibits in this reaction: for each position indicated by a letter, select the corresponding position indicated by a number.
Answer: 4221
Explanation:
A) NH 4 HCO 3 - salt, which includes the ammonium cation NH 4 +. In the ammonium cation, nitrogen always has an oxidation state of -3. As a result of the reaction, it turns into ammonia NH 3. Hydrogen almost always (except for its compounds with metals) has an oxidation state of +1. Therefore, for the ammonia molecule to be electrically neutral, nitrogen must have an oxidation state of -3. Thus, there is no change in the degree of nitrogen oxidation; it does not exhibit redox properties.
B) As already shown above, nitrogen in ammonia NH 3 has an oxidation state of -3. As a result of the reaction with CuO, ammonia is converted into a simple substance N 2. In any simple substance, the oxidation state of the element with which it is formed is equal to zero. Thus, the nitrogen atom loses its negative charge, and since electrons are responsible for the negative charge, this means that they are lost by the nitrogen atom as a result of the reaction. An element that loses some of its electrons in a reaction is called a reducing agent.
C) As a result of the reaction, NH 3 with an oxidation state of nitrogen equal to -3 turns into nitric oxide NO. Oxygen almost always has an oxidation state of -2. Therefore, in order for the nitric oxide molecule to be electrically neutral, the nitrogen atom must have an oxidation state of +2. This means that the nitrogen atom changed its oxidation state from -3 to +2 as a result of the reaction. This indicates the loss of 5 electrons by the nitrogen atom. That is, nitrogen, as in the case of B, is a reducing agent.
D) N 2 is a simple substance. In all simple substances, the element that forms them has an oxidation state of 0. As a result of the reaction, nitrogen is converted into lithium nitride Li3N. The only oxidation state of an alkali metal other than zero (any element has an oxidation state of 0) is +1. Thus, for the Li3N structural unit to be electrically neutral, nitrogen must have an oxidation state of -3. It turns out that as a result of the reaction, nitrogen acquired a negative charge, which means the addition of electrons. Nitrogen is the oxidizing agent in this reaction.
Task number 2
Establish a correspondence between the reaction scheme and the property of the phosphorus element that it exhibits in this reaction: for each position indicated by a letter, select the corresponding position indicated by a number.
Write in the table the selected numbers under the corresponding letters.
Answer: 1224
Task number 3
| REACTION EQUATION | |
| A) 4NH 3 + 5O 2 → 4NO + 6H 2 O B) 2Cu(NO 3) 2 → 2CuO + 4NO 2 + O 2 C) 4Zn + 10HNO 3 → NH 4 NO 3 + 4Zn (NO 3) 2 + 3H 2 O D) 3NO 2 + H 2 O → 2HNO 3 + NO |
Write in the table the selected numbers under the corresponding letters.
Answer: 1463
Task number 4
Establish a correspondence between the reaction equation and the change in the oxidation state of the oxidizing agent in it: for each position indicated by a letter, select the corresponding position indicated by a number.
| REACTION EQUATION | CHANGING THE OXIDIZER DEGREE |
| A) SO 2 + NO 2 → SO 3 + NO B) 2NH 3 + 2Na → 2NaNH 2 + H 2 C) 4NO 2 + O 2 + 2H 2 O → 4HNO 3 D) 4NH 3 + 6NO → 5N 2 + 6H 2 O |
Write in the table the selected numbers under the corresponding letters.
Answer: 3425
Task number 5
Establish a correspondence between the reaction scheme and the coefficient in front of the oxidizing agent in it: for each position indicated by a letter, select the corresponding position indicated by a number.
| REACTION SCHEME | COEFFICIENT BEFORE THE OXIDIZER |
| A) NH 3 + O 2 → N 2 + H 2 O B) Cu + HNO 3 (conc.) → Cu(NO 3) 2 + NO 2 + H 2 O C) C + HNO 3 → NO 2 + CO 2 + H 2 O D) S + HNO 3 → H 2 SO 4 + NO |
Write in the table the selected numbers under the corresponding letters.
Answer: 3442
Task number 6
Establish a correspondence between the reaction equation and the change in the oxidation state of the oxidizing agent in it: for each position indicated by a letter, select the corresponding position indicated by a number.
| REACTION EQUATION | CHANGING THE OXIDIZER DEGREE |
| A) 2NH 3 + K → 2KNH 2 + H 2 B) H 2 S + K → K 2 S + H 2 C) 4NH 3 + 6NO → 5N 2 + 6H 2 O D) 2H 2 S + 3O 2 → 2SO 2 + 2H 2 O |
Write in the table the selected numbers under the corresponding letters.
Answer: 4436
Task number 7
Establish a correspondence between the starting materials and the property of copper that this element exhibits in this reaction: for each position indicated by a letter, select the corresponding position indicated by a number.
Write in the table the selected numbers under the corresponding letters.
Answer: 2124
Task number 8
Establish a correspondence between the reaction scheme and the property of sulfur that it exhibits in this reaction: for each position indicated by a letter, select the corresponding position indicated by a number.
Write in the table the selected numbers under the corresponding letters.
Answer: 3224
Task number 9
Establish a correspondence between the reaction scheme and the property of phosphorus that it exhibits in this reaction: for each position indicated by a letter, select the corresponding position indicated by a number.
Write in the table the selected numbers under the corresponding letters.
Answer: 3242
Task number 10
Establish a correspondence between the reaction scheme and the property of nitrogen that it exhibits in this reaction: for each position indicated by a letter, select the corresponding position indicated by a number.
Write in the table the selected numbers under the corresponding letters.
Answer: 2141
Task number 11
Establish a correspondence between the reaction scheme and the property of fluorine that it exhibits in this reaction: for each position indicated by a letter, select the corresponding position indicated by a number.
Write in the table the selected numbers under the corresponding letters.
Answer: 1444
Task number 12
Establish a correspondence between the reaction scheme and the change in the oxidation state of the reducing agent: for each position indicated by a letter, select the corresponding position indicated by a number.
| REACTION SCHEME | |
| A) NaIO → NaI + NaIO 3 B) HI + H 2 O 2 → I 2 + H 2 O C) NaIO 3 → NaI + O 2 D) NaIO 4 → NaI + O 2 | 1) I +5 → I −1 2) O −2 → O 0 3) I +7 →I −1 4) I +1 → I −1 5) I +1 → I +5 6) I −1 → I 0 |
Write in the table the selected numbers under the corresponding letters.
Answer: 5622
Task number 13
Establish a correspondence between the reaction equation and the change in the oxidation state of the reducing agent in this reaction: for each position indicated by a letter, select the corresponding position indicated by a number.
| REACTION EQUATION | CHANGING THE OXIDATION DEGREE OF THE REDUCER |
| A) H 2 S + I 2 → S + 2HI B) Cl 2 + 2HI → I 2 + 2HCl C) 2SO 3 + 2KI → I 2 + SO 2 + K 2 SO 4 D) S + 3NO 2 → SO 3 + 3NO |
Write in the table the selected numbers under the corresponding letters.
Answer: 5331
Task number 14
Establish a correspondence between the redox reaction equation and the change in the oxidation state of sulfur in this reaction: for each position indicated by a letter, select the corresponding position indicated by a number.
| REACTION EQUATION | CHANGES IN THE OXIDATION STATE OF SULFUR |
| A) S + O 2 → SO 2 B) SO 2 + Br 2 + 2H 2 O → H 2 SO 4 + 2HBr C) C + H 2 SO 4 (conc.) → CO 2 + 2SO 2 + 2H 2 O D) 2H 2 S + O 2 → 2H 2 O + 2S |
Write in the table the selected numbers under the corresponding letters.
Answer: 4123
Task number 15
| CHANGING THE OXIDATION DEGREE | SUBSTANCE FORMULA |
| A) S -2 → S +4 B) S −2 → S +6 C) S +6 → S −2 D) S −2 → S 0 | 1) Cu 2 S and O 2 2) H 2 S and Br 2 (solution) 3) Mg and H 2 SO 4 (conc.) 4) H 2 SO 3 and O 2 5) PbS and HNO 3 (conc.) 6) C and H 2 SO 4 (conc.) |
Write in the table the selected numbers under the corresponding letters.
Answer: 1532
Task number 16
Establish a correspondence between the change in the oxidation state of sulfur in the reaction and the formulas of the starting substances that enter into it: for each position indicated by a letter, select the corresponding position indicated by a number.
| CHANGING THE OXIDATION DEGREE | SUBSTANCE FORMULA |
| A) S 0 → S +4 B) S +4 → S +6 C) S −2 → S 0 D) S +6 → S +4 | 1) Cu and H 2 SO 4 (diff.) 2) H 2 S and O 2 (insufficient) 3) S and H 2 SO 4 (conc.) |
Write in the table the selected numbers under the corresponding letters.
Answer: 3523
Task number 17
Establish a correspondence between the properties of nitrogen and the equation of the redox reaction in which it exhibits these properties: for each position indicated by a letter, select the corresponding position indicated by a number.
Write in the table the selected numbers under the corresponding letters.
Answer: 2143
Task number 18
Establish a correspondence between the change in the oxidation state of chlorine in the reaction and the formulas of the starting substances that enter into it: for each position indicated by a letter, select the corresponding position indicated by a number.
| CHANGING THE OXIDATION DEGREE | FORMULA OF STARTING SUBSTANCES |
| A) Cl 0 → Cl -1 B) Cl -1 → Cl 0 C) Cl +5 → Cl -1 D) Cl 0 → Cl +5 | 1) KClO 3 (heating) 2) Cl 2 and NaOH (hot solution) 3) KCl and H 2 SO 4 (conc.) 6) KClO 4 and H 2 SO 4 (conc.) |
Write in the table the selected numbers under the corresponding letters.
Answer: 2412
Task #19
Establish a correspondence between the formula of an ion and its ability to show oxidative restorative properties: for each position indicated by a letter, select the corresponding position indicated by a number.
Write in the table the selected numbers under the corresponding letters.
Answer: 2332
Task number 20
Match the schema chemical reaction and a change in the degree of oxidation of the oxidizing agent: for each position indicated by a letter, select the corresponding position indicated by a number.
| REACTION SCHEME | CHANGING THE OXIDIZER DEGREE |
| A) MnCO 3 + KClO 3 → MnO 2 + KCl + CO 2 B) Cl 2 + I 2 + H 2 O → HCl + HIO 3 C) H 2 MnO 4 → HMnO 4 + MnO 2 + H 2 O D) Na 2 SO 3 + KMnO 4 + KOH → Na 2 SO 4 + K 2 MnO 4 + H 2 O | 1) Cl 0 → Cl - 2) Mn+6 → Mn+4 3) Cl+5 → Cl- 4) Mn +7 → Mn +6 5) Mn+2 → Mn+4 6) S+4 → S+6 |
Write in the table the selected numbers under the corresponding letters.
Answer: 3124
Task number 21
Establish a correspondence between the reaction scheme and the change in the oxidation state of the reducing agent in this reaction: for each position indicated by a letter, select the corresponding position indicated by a number.
1. How to determine the redox reaction?
There are various classifications of chemical reactions. One of them includes those in which the substances that interact with each other (or the substance itself) change the oxidation states of the elements.
As an example, consider two reactions:
Zn 0 + 2H +1 C1 -1 \u003d Zn +2 Cl 2 -1 + H 2 0 (1)
H +1 Cl -1 + K +1 O -2 H +1 = K +1 Cl -1 + H 2 +1 O -2 (2)
Reaction (1) involves zinc and hydrochloric acid . Zinc and hydrogen change their oxidation states, chlorine leaves its oxidation state unchanged:
Zn 0 - 2e = Zn 2+
2H + 1 + 2e \u003d H 2 0
2Cl -1 \u003d 2 Cl -1
And in reaction (2), ( neutralization reaction), chlorine, hydrogen, potassium, and oxygen do not change their oxidation states: Cl -1 = Cl -1, H +1 = H +1, K +1 = K +1, O -2 = O -2; Reaction (1) belongs to the redox reaction, and reaction (2) belongs to another type.
Chemical reactions that are carried out with a changeoxidation states of elementsare called redox. 
In order to determine the redox reaction, it is necessary to establish steppeno oxidation of elements on the left and right sides of the equation. This requires knowing how to determine the oxidation state of an element.
In the case of reaction (1), the elements Zn and H change their states by losing or gaining electrons. Zinc, giving up 2 electrons, passes into the ionic state - it becomes the Zn 2+ cation. In this case, the process recovery and zinc is oxidized. Hydrogen gains 2 electrons, exhibits oxidative properties, itself in the process of reaction recovering.
2. Definitionoxidation states of elements.
The oxidation state of the elements in its compounds is determined based on the position that the total total charge of the oxidation states of all elements of a given compound zero. For example, in the compound H 3 PO 4, the oxidation state of hydrogen is +1, phosphorus is +5, and oxygen is -2; Having made a mathematical equation, we determine that in the sum number of particles(atoms or ions) will have a charge equal to zero: (+1)x3+(+5)+(-2)x4 = 0
But in this example, the oxidation states of the elements are already set. How can one determine the degree of oxidation of sulfur, for example, in the compound sodium thiosulfate Na 2 S 2 O 3, or manganese in the compound potassium permanganate- KMnO 4 ? For this you need to know constant oxidation states of a number of elements. They have the following meanings:
1) Elements of group I of the periodic system (including hydrogen in combination with non-metals) +1;
2) Elements of group II of the periodic system +2;
3) Elements of group III of the periodic system +3;
4) Oxygen (except in combination with fluorine or in peroxide compounds) -2;
Based on these constant values of oxidation states (for sodium and oxygen), we determine oxidation state sulfur in the Na 2 S 2 O 3 compound. Since the total charge of all oxidation states of the elements whose composition reflects this compound formula, is equal to zero, then denoting the unknown charge of sulfur " 2X”(since there are two sulfur atoms in the formula), we compose the following mathematical equation:
(+1) x 2 + 2X+ (-2) x 3 = 0
Solving this equation for 2 x, we get
2X = (-1) x 2 + (+2) x 3
or
X = [(-2) + (+6)] : 2 = +2;
Therefore, the oxidation state of sulfur in the Na 2 S 2 O 3 compound is (+2). But will it really always be necessary to use such an inconvenient method to determine the oxidation states of certain elements in compounds? Of course not always. For example, for binary compounds: oxides, sulfides, nitrides, etc., you can use the so-called "cross-to-cross" method to determine the oxidation states. Let's say given compound formula:titanium oxide– Ti 2 O 3 . Using a simple mathematical analysis, based on the fact that the oxidation state of oxygen is known to us and is equal to (-2): Ti 2 O 3, it is easy to establish that the oxidation state of titanium will be equal to (+3). Or, for example, in conjunction methane CH 4 it is known that the oxidation state of hydrogen is (+1), then it is not difficult to determine the oxidation state of carbon. It will correspond in the formula of this compound (-4). Also, using the "criss-cross" method, it is not difficult to establish that if the following compound formula Cr 4 Si 3, then the degree of oxidation of chromium into it is (+3), and silicon (-4).
For salts, this is also not difficult. And it doesn't matter if it's given or medium salt or acid salt. In these cases, it is necessary to proceed from the salt-forming acid. For example, given salt sodium nitrate(NaNO3). It is known that it is a derivative nitric acid(HNO 3), and in this compound the oxidation state of nitrogen is (+5), therefore, in its salt - sodium nitrate, the oxidation state of nitrogen is also (+5). sodium bicarbonate(NaHCO 3) is acid salt carbonic acid(H 2 CO 3). Just like in an acid, the oxidation state of carbon in this salt will be (+4).
It should be noted that the oxidation states in compounds: metals and non-metals (when compiling electronic balance equations) are equal to zero: K 0, Ca 0, Al 0, H 2 0, Cl 2 0, N 2 0 As an example, we give the oxidation states of the most typical elements:
Only oxidizing agents are substances that have a maximum, usually positive, oxidation state, for example: KCl +7 O 4, H 2 S +6 O 4, K 2 Cr +6 O 4, HN +5 O 3, KMn +7 O 4 . This is easy to prove. If these compounds could be reducing agents, then in these states they would have to donate electrons:
Cl +7 - e \u003d Cl +8
S +6 - e \u003d S +7
But the elements chlorine and sulfur cannot exist with such oxidation states. Similarly, the only reducing agents are substances that have a minimum, as a rule, negative degree oxidation, for example: H 2 S -2, HJ -, N -3 H 3. In the process of redox reactions, such compounds cannot be oxidizing agents, since they would have to add electrons:
S-2 + e = S-3
J - + e \u003d J -2
But for sulfur and iodine, ions with such degrees of oxidation are not typical. Elements with intermediate oxidation states, for example N +1 , N +4 , S +4 , Cl +3 , C +2 can exhibit both oxidizing and reducing properties.
3
. Types of redox reactions.
There are four types of redox reactions.
1)
Intermolecular redox reactions.
The most common type of reaction. These reactions change oxidation stateselements in different molecules, for example:
2Bi +3 Cl 3 + 3Sn +2 Cl 2 = 2Bi 0 + 3Sn +4 Cl 4
Bi +3 - 3 e= Bi0
sn+2+2 e= Sn+4
2) A kind of intermolecular redox reactions is the reaction proportionate, in which the oxidizing and reducing agents are atoms of the same element: in this reaction, two atoms of the same element with different oxidation states form one atom with a different oxidation state:
SO 2 +4 + 2H 2 S -2 \u003d 3S 0 + 2H 2 O
S-2-2 e= S 0
S+4+4 e= S 0
3) Reactions disproportionation are carried out if the oxidizing and reducing agents are atoms of the same element, or one atom of an element with one oxidation state forms a compound with two oxidation states:
N +4 O 2 + NaOH = NaN +5 O 3 + NaN +3 O 2 + H 2 O
N +4 - e=N+5
N+4+ e= N+3
4) Intramolecular redox reactions occur when the oxidizing atom and the reducing atom are in the same substance, for example:
N -3 H 4 N +5 O 3 \u003d N +1 2 O + 2H 2 O
2N -3 - 8 e=2N+1
2N+5+8 e= 2N+1
4 . The mechanism of redox reactions.
Redox reactions are carried out due to the transfer of electrons from the atoms of one element to another. If an atom or molecule loses electrons, then this process is called oxidation, and this atom is a reducing agent, for example:
Al 0 - 3 e=Al3+
2Cl - - 2 e= Cl 2 0
Fe 2+ - e= Fe3+
In these examples, Al 0 , Cl - , Fe 2+ are reducing agents, and the processes of their transformation into compounds Al 3+ , Cl 2 0 , Fe 3+ are called oxidative. If an atom or molecule acquires electrons, then such a process is called reduction, and this atom is an oxidizing agent, for example:
Ca 2+ + 2 e= Ca0
Cl 2 0 + 2 e= 2Cl -
Fe3+ + e= Fe 2+
Oxidizing agents, as a rule, are non-metals (S, Cl 2, F 2, O 2) or metal compounds with a maximum oxidation state (Mn +7, Cr +6, Fe +3). Reducing agents are metals (K, Ca, Al) or non-metal compounds having a minimum oxidation state (S -2, Cl -1, N -3, P -3);
Redox equations differ from molecular equations other reactions by the difficulty of selecting coefficients in front of the reactants and reaction products. For this use electronic balance method, or method of electron-ion equations(sometimes the latter is called " half-reaction method"). As an example of compiling equations for redox reactions, consider a process in which concentrated sulfuric acid(H 2 SO 4) will react with hydrogen iodide (HJ):
H 2 SO 4 (conc.) + HJ → H 2 S + J 2 + H 2 O
First of all, let us establish that oxidation state iodine in hydrogen iodide is (-1), and sulfur in sulfuric acid: (+6). During the reaction, iodine (-1) will be oxidized to a molecular state, and sulfur (+6) will be reduced to the oxidation state (-2) - hydrogen sulfide:
J - → J 0 2
S+6 → S-2
In order to make it necessary to take into account that amountparticles atoms in the left and right parts of the half-reactions should be the same
2J - - 2 e→ J 0 2
S+6+8 e→S-2
By setting the vertical line on the right of this half-reaction scheme, we determine the reaction coefficients:
2J - - 2 e→ J 0 2 |8
S+6+8 e→ S-2 |2
Reducing by "2", we get the final values of the coefficients:
2J - - 2 e→ J 0 2 |4
S+6+8 e→ S-2 |1
Let's sum up under this scheme half reactions horizontal line and summarize the reaction number of particles atoms:
2J - - 2 e→ J 0 2 |4
S+6+8 e→ S-2 |1
____________________
8J - + S +6 → 4 J 0 2 + S -2
After that it is necessary. Substituting the obtained values of the coefficients in molecular equation, we bring it to this form:
8HJ + H 2 SO 4 \u003d 4J 2 + H 2 S + H 2 O
Having counted the number of hydrogen atoms in the left and right parts of the equation, we will make sure that the coefficient “4” in front of water needs to be corrected, we get the complete equation:
8HJ + H 2 SO 4 \u003d 4J 2 + H 2 S + 4H 2 O
This equation can be written using method of electronicion balance. In this case, there is no need to correct the coefficient in front of water molecules. The equation is compiled on the basis of the dissociation of ions of the compounds participating in the reaction: For example, dissociation of sulfuric acid leads to the formation of two hydrogen protons and a sulfate anion:
H 2 SO 4 ↔ 2H + + SO 4 2-
Similarly, the dissociation of hydrogen iodide and hydrogen sulfide can be written:
HJ ↔ H + + J -
H 2 S ↔ 2H + + S 2-
J 2 does not dissociate. It also practically does not dissociate H 2 O. Compilation half-reaction equations for iodine remains the same:
2J - - 2 e→ J 0 2
The half-reaction for sulfur atoms will take the following form:
SO 4 -2 → S -2
Since four oxygen atoms are missing on the right side of the half-reaction, this amount must be balanced with water:
SO 4 -2 → S -2 + 4H 2 O
Then, in the left part of the half-reaction, it is necessary to compensate for hydrogen atoms due to protons (since the reaction of the medium is acidic):
SO 4 2- + 8H + → S -2 + 4H 2 O
Having counted the number of passing electrons, we obtain a complete record of the equation in terms of half-reaction method:
SO 4 2- + 8H + + 8 e→ S -2 + 4H 2 O
Summing up both half-reactions, we get electronic balance equation:
2J - - 2 e→ J 0 2 |8 4
SO 4 2- + 8H + + 8 e→ S -2 + 4H 2 O | 2 1
8J - + SO 4 2- + 8Н + → 4J 2 0 + S 0 + 4H 2 O
From this entry it follows that the method electron-ion equation gives a more complete picture of the redox reaction than electronic balance method. The number of electrons involved in the process is the same for both methods of balance, but in the latter case, the number of protons and water molecules involved in the redox process is set as if “automatically”.
Let us analyze several specific cases of redox reactions that can be compiled by the method electron-ion balance. Some redox processes are carried out with the participation of an alkaline environment, for example:
KCrO 2 + Br 2 + KOH → KBr + K 2 CrO 4 + H 2 O
In this reaction, the reducing agent is chromite ion (CrO 2 -), which is oxidized to chromate ion (CrO -2 4). Oxidizing agent - bromine (Br 0 2) is reduced to bromide ion (Br -):
CrO 2 - → CrO 4 2-
Br 0 2 → 2 Br -
Since the reaction occurs in an alkaline medium, the first half-reaction must be composed taking into account hydroxide ions (OH -):
CrO 2 - + 4OH - - 3 e\u003d CrO 2- 4 + 2H 2 O
We compose the second half-reaction in the already known way:
CrO 2 - + 4OH - -3 e\u003d CrO 4 2 - + 2H 2 O | 2
Br 0 2 + 2 e= Br - |3
__________
2CrO 2 - + 3Br 2 0 + 8OH - \u003d 2CrO 2- 4 + 6Br - + 4H 2 O
After this, it is necessary to arrange the coefficients in the reaction equation and completely molecular equation of this redox process will take the form:
2KCrO 2 + 3Br 2 + 8KOH = 2K 2 CrO 4 + 6KBr + 4H 2 O.
In a number of cases, non-dissociable substances simultaneously participate in the redox reaction. For example:
AsH 3 + HNO 3 \u003d H 3 AsO 4 + NO 2 + 4H 2 O
Then half-reaction method is compiled taking into account this process:
AsH 3 + 4H 2 O - 8 e\u003d AsO 4 3- + 11H + | 1
NO 3 + 2H + + e= NO 2 + H 2 O | 8
AsH 3 + 8NO 3 + 4H 2 O + 2H + = AsO 4 3- + 8NO 2 + 11H + O
molecular equation will take the form:
AsH 3 + 8HNO 3 \u003d H 3 AsO 4 + 8NO 2 + 4H 2 O.
Redox reactions are sometimes accompanied by a simultaneous oxidation-reduction process of several substances. For example, in the reaction with copper sulfide interacts concentrated nitric acid:
Cu 2 S + HNO 3 \u003d Cu (NO 3) 2 + H 2 SO 4 + NO + H 2 O
The redox process involves the atoms of copper, sulfur and nitrogen. When compiling the equation half-reaction method the following steps must be taken into account:
Cu + → Cu 2+
S 2- → S +6
N5+ → N+2
In this situation, it is necessary to combine the oxidation and reduction processes in one stage:
2Cu + - 2 e→ 2Cu 2+ | ten e
S 2- - 8 e→ S 6+
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N 5+ + 3 e→ N 2+ | 3 e
At which the redox half-reaction will take the form:
2Cu + - 2 e→ 2Cu 2+
S 2- - 8 e→ S 6+ 3 ( recovery processes)
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N 5+ + 3 e→ N 2+ 10 (oxidation process)
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6Cu + + 3S 2- + 10N 5+ → 6Cu 2+ + 3S 6+ + 10N 2+
Eventually molecular reaction equation will take the form:
3Cu 2 S + 22HNO 3 \u003d 6Cu (NO 3) 2 + 3H 2 SO 4 + 10NO + 8H 2 O.
Particular attention should be paid to redox reactions involving organic matter. For example, when glucose is oxidized potassium permanganate in an acidic environment, the following reaction occurs:
C 6 H 12 O 6 + KMnO 4 + H 2 SO 4 > CO 2 + MnSO 4 + K 2 SO 4 + H 2 O
When drawing up a balance sheet half-reaction method The conversion of glucose takes into account the absence of its dissociation, but the correction of the number of hydrogen atoms is carried out due to protons and water molecules:
C 6 H 12 O 6 + 6H 2 O - 24 e\u003d 6CO 2 + 24H +
Half-reaction involving potassium permanganate will take the form:
MnO 4 - + 8H + + 5 e\u003d Mn 2+ + 4H 2 O
As a result, we obtain the following scheme of the redox process:
C 6 H 12 O 6 + 6H 2 O - 24 e= 6CO 2 + 24H + | 5
MnO 4 - + 8H + + 5 e= Mn +2 + 4H 2 O | 24
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5C 6 H 12 O 6 + 30H 2 O + 24MnО 4 - + 192H + = 30CO 2 + 120H + + 24Mn 2+ + 96H 2 O
By reducing the number of protons and water molecules on the left and right sides half reactions, we get the final molecular equation:
5C 6 H 12 O 6 + 24KMnO 4 + 36H 2 SO 4 = 30CO 2 + 24MnSO 4 + 12K 2 SO 4 + 66H 2 O
5. Influence of the environment on the nature of the course of redox reactions.
Depending on the medium (excess H +, neutral, excess OH -), the nature of the reaction between the same substances may change. To create an acidic environment is usually used sulphuric acid(H 2 SO 4), Nitric acid(HNO 3), hydrochloric acid (HCl), as an OH medium, sodium hydroxide (NaOH) or potassium hydroxide (KOH) is used. For example, we will show how the environment affects potassium permanganate(KMnO 4). and its reaction products:
For example, let's take Na 2 SO 3 as a reducing agent, KMnO 4 as an oxidizing agent
In an acidic environment:
5Na 2 SO 3 + 2KMnO 4 + 3H 2 SO 4 → 5Na 2 SO 4 + 2MnSO 4 + K 2 SO 4 + 3H 2 O
SO 3 2- + H 2 O - 2 e→ SO 4 2- + 2H + |5
MnO 4 - + 8H + + 5 e→ Mn 2+ + 4H 2 O |2
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5SO 3 2- + 2MnO 4 - + 6H + → 5SO 4 2- + 2Mn 2+ + 3H 2 O
In neutral (or slightly alkaline):
3Na 2 SO 3 + 2KMnO 4 + H 2 O → 3Na 2 SO 4 + 2MnO 2 + 2KOH
SO 3 2- + H 2 O - 2 e→ SO 4 2- + 2H + |3
MnO 4 - + 2H 2 O + 3 e→ MnO 2 + 4OH | 2
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3SO 3 2- + 2 MnO 4 - + H 2 O → 3SO 4 2- + 2MnO 2 + 2OH
In a highly alkaline environment:
Na 2 SO 3 + 2KMnO 4 + 2NaOH → Na 2 SO 4 + K 2 MnO 4 + Na 2 MnO + H 2 O
SO 3 2- + 2 OH - - 2 e→ SO 4 2- + H 2 O | 1
MnO4 - + e→ MnO 4 2 |2
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SO 3 2- + 2 MnO 4 - + 2OH → SO 4 2- + 2MnO 4 2- + H 2 O
Hydrogen peroxide(H 2 O 2), depending on the environment, is restored according to the scheme:
1) Acidic environment (H +) H 2 O 2 + 2H + + 2 e→ 2H2O
2) Neutral medium (H 2 O) H 2 O 2 + 2 e→ 2OH
3) Alkaline medium (OH -) H 2 O 2 + 2 e→ 2OH
Hydrogen peroxide(H 2 O 2) acts as an oxidizing agent:
2FeSO 4 + H 2 O 2 + H 2 SO 4 → Fe 2 (SO 4) 3 + 2H 2 O
Fe 2+ - e= Fe3+ |2
H 2 O 2 + 2H + + 2 e\u003d 2H 2 O | 1
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2Fe 2+ + H 2 O 2 + 2H + → 2Fe 3+ + 2 H 2 O
However, when meeting with very strong oxidizing agents (KMnO 4) Hydrogen peroxide(H 2 O 2) acts as a reducing agent:
5H 2 O 2 + 2KMnO 4 + 3H 2 SO 4 → 5O 2 + 2MnSO 4 + K 2 SO 4 + 8H 2 O
H 2 O 2 - 2 e→ O 2 + 2H + |5
MnO 4 - + 8H + + 5 e→ Mn 2+ + 4H 2 O |2
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5H 2 O + 2 MnO 4 - + 6H + → 5O 2 + 2Mn 2+ + 8H 2 O
6.
Determination of products of redox reactions. 
In the practical part of this topic, redox processes are considered, indicating only the initial reagents. Reaction products usually need to be determined. For example, the reaction involves ferric chloride(FeCl 3) and potassium iodide(KJ):
FeCl 3 + KJ = A + B + C
required to install compound formulas A, B, C, formed as a result of the redox process.
The initial oxidation states of the reagents are as follows: Fe 3+ , Cl - , K + , J - . It is easy to assume that Fe 3+, being an oxidizing agent (has a maximum oxidation state), can only reduce its oxidation state to Fe 2+:
Fe3+ + e= Fe 2+
Chloride ion and potassium ion do not change their oxidation state in the reaction, and iodide ion can only increase its oxidation state, i.e. go to state J 2 0:
2J - - 2 e= J 2 0
As a result of the reaction, in addition to the redox process, there will be exchange reaction between FeCl 3 and KJ, but taking into account the change in oxidation states, the reaction is not determined according to this scheme:
FeCl 3 + KJ = FeJ 3 + KCl,
but will take the form
FeCl 3 + KJ = FeJ 2 + KCl,
where the product C is the compound J 2 0:
FeCl 3 + 6KJ = 2FeJ 2 + 6KJ + J 2
Fe3+ + e═> Fe2+ |2
2J - - 2 e═> J 2 0 |1
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2Fe +3 + 2J - = 2Fe 2+ + J 2 0
In the future, when determining the products of the redox process, you can use the so-called "elevator system". Its principle is that any redox reaction can be represented as the movement of elevators in a multi-storey building in two mutually opposite directions. Moreover, the "floors" will be oxidation states relevant elements. Since any of the two half-reactions in a redox process is accompanied by either a decrease or an increase oxidation states of this or that element, then by simple reasoning one can assume about their possible oxidation states in the resulting reaction products.
As an example, consider a reaction in which sulfur reacts with concentrated sodium hydroxide solution ( NaOH):
S + NaOH (conc) = (A) + (B) + H 2 O
Since in this reaction changes will occur only with the oxidation states of sulfur, for clarity, we will draw up a diagram of its possible states:
Compounds (A) and (B) cannot simultaneously be the sulfur states S +4 and S +6, since in this case the process would occur only with the release of electrons, i.e. would be restorative:
S 0 - 4 e=S+4
S 0 - 6 e=S+6
But this would be contrary to the principle of redox processes. Then it should be assumed that in one case the process should proceed with the release of electrons, and in the other case it should move in the opposite direction, i.e. be oxidative:
S 0 - 4 e=S+4
S 0 + 2 e=S-2
On the other hand, how likely is it that the recovery process will be carried out to state S +4 or to S +6? Since the reaction proceeds in an alkaline, and not in an acidic environment, its oxidizing ability is much lower, therefore the formation of the S +4 compound in this reaction is preferable than S +6. Therefore, the final reaction will take the form:
4S + 6NaOH (conc) = Na 2 SO 3 + 2Na 2 S + 3H 2 O
S 0 +2 e= S - 2 | 4 | 2
S 0 + 6OH - - 4 e= SO 3 2 - + 3H 2 O | 2 | one
3S 0 + 6OH - \u003d 2S - 2 + SO 3 2 - + 3H 2 O
As another example, consider the following reaction between phosphine and concentrated nitric acid(HNO3) :
PH 3 + HNO 3 \u003d (A) + (B) + H 2 O
In this case, we have varying degrees of oxidation of phosphorus and nitrogen. For clarity, we present diagrams of the state of their oxidation states.
Phosphorus in the oxidation state (-3) will only exhibit reducing properties, so in the reaction it will increase its oxidation state. Nitric acid itself is a strong oxidizing agent and creates an acidic environment, so phosphorus from the state (-3) will reach its maximum oxidation state (+5).
In contrast, nitrogen will lower its oxidation state. In reactions of this type, usually up to the state (+4).
Further, it is not difficult to assume that phosphorus in the state (+5), being the product (A), can only be phosphoric acid H 3 PO 4, since the reaction medium is strongly acidic. Nitrogen in such cases usually takes the oxidation state (+2) or (+4), more often (+4). Therefore, the product (B) will be nitrogen oxide NO2. It remains only to solve this equation by the balance method:
P - 3 - 8 e= P+5 | one
N+ 5 + e= N+4 | eight
P - 3 + 8N +5 = P +5 + 8N +4
PH 3 + 8HNO 3 \u003d H 3 PO 4 + 8NO 2 + 4H 2 O
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According to the change in the oxidation state, all chemical reactions can be divided into two types:
I. Reactions occurring without changing the degree of oxidation of the elements that make up the reactants. Such reactions are referred to as ion exchange reactions.
Na 2 CO 3 + H 2 SO 4 = Na 2 SO 4 + CO 2 + H 2 O.
II. Reactions that occur with a change in the oxidation state of elements,
included in the reactants. Such reactions are referred to as redox reactions.
5NaNO 2 + 2KMnO 4 + 3H 2 SO 4 = 5NaNO 3 + 2MnSO 4 + K 2 SO 4 + 3H 2 O.
Oxidation state(oxidation) - a characteristic of the state of the atoms of the elements in the composition of the molecule. It characterizes the uneven distribution of electrons between the atoms of elements and corresponds to the charge that an atom of an element would acquire if all the common electron pairs of its chemical bonds shifted towards the more electronegative element. Depending on the relative electronegativity of the elements that form a bond, an electron pair can be shifted to one of the atoms or symmetrically located relative to the nuclei of atoms. Therefore, the oxidation state of elements can be negative, positive, or zero.
Elements whose atoms accept electrons from other atoms have a negative oxidation state. Elements whose atoms donate their electrons to other atoms have a positive oxidation state. Atoms in the molecules of simple substances have a zero oxidation state, as well as if the substance is in the atomic state.
The oxidation state is denoted +1, +2.
Ion charge 1+, 2+.
The oxidation state of an element in a compound is determined by the rules:
1. The degree of oxidation of an element in simple substances is zero.
2. Some elements in almost all of their compounds exhibit a constant oxidation state. These elements include:
It has an oxidation state of +1 (with the exception of metal hydrides).
O has an oxidation state of -2 (with the exception of fluorides).
3. Elements of groups I, II and III of the main subgroups Periodic system DIMendeleev's elements have a constant oxidation state equal to the group number.
Elements Na, Ba, Al: oxidation state +1, +2, +3, respectively.
4. For elements that have a variable oxidation state, there is the concept of higher and lower oxidation states.
The highest oxidation state of an element is equal to the group number of the Periodic Table of Elements of D.I. Mendeleev, in which the element is located.
Elements N,Cl: highest degree oxidation +5, +7, respectively.
The lowest oxidation state of an element is equal to the group number of the Periodic Table of Elements of D.I. Mendeleev, in which the element is located minus eight.
Elements N, Cl: the lowest oxidation state is -3, -1, respectively.
5. In single-element ions, the oxidation state of the element is equal to the charge of the ion.
Fe 3+ - the oxidation state is +3; S 2- - the oxidation state is -2.
6. The sum of the oxidation states of all atoms of elements in a molecule is zero.
KNO 3 ; (+1) + X+ 3 (-2) = 0; X = +5. The oxidation state of nitrogen is +5.
7. The sum of the oxidation states of all atoms of elements in an ion is equal to the charge of the ion.
SO 4 2- ; X+ 4 (-2) = -2; X= +6. The oxidation state of sulfur is +6.
8. In compounds consisting of two elements, the element that is written on the right always has lowest degree oxidation.
Reactions in which the oxidation state of elements changes are referred to as redox reactions /ORD/. These reactions consist of oxidation and reduction processes.
Oxidation The process of donating electrons by an element that is part of an atom, molecule or ion is called.
Al 0 - 3e \u003d Al 3+
H 2 - 2e \u003d 2H +
Fe 2+ - e \u003d Fe 3+
2Cl - - 2e \u003d Cl 2
When oxidized, the oxidation state of an element increases. A substance (atom, molecule, or ion) that contains an element that donates electrons is called a reducing agent. Al, H 2 , Fe 2+ , Cl - - reducing agents. The reducing agent is oxidized.
Recovery The process of adding electrons to an element that is part of an atom, molecule or ion is called.
Cl 2 + 2e \u003d 2Cl -
Fe 3+ + e \u003d Fe 2+
When reduced, the oxidation state of an element decreases. A substance (atom, molecule, or ion) that contains an element that accepts electrons is called an oxidizing agent. S, Fe 3+, Cl 2 are oxidizing agents. The oxidant is restored.
Total number electrons in the system during a chemical reaction does not change. The number of electrons donated by the reducing agent is equal to the number of electrons attached by the oxidizing agent.
To compile the equation of the redox reaction (ORR) in solutions, the ion-electronic method (half-reaction method) is used.
OVR can occur in acidic, neutral or alkaline environments. The reaction equations take into account the possible participation of water molecules (HOH) and those contained in the solution, depending on the nature of the medium, an excess of H + or OH - ions:
in an acidic environment - HOH and H + ions;
in a neutral environment - only HOH;
in an alkaline environment - HOH and OH - ions.
When compiling the OVR equations, it is necessary to adhere to a certain sequence:
1. Write a reaction scheme.
2. Determine the elements that have changed their oxidation state.
3. Write a diagram in a short ion-molecular form: strong electrolytes in the form of ions, weak electrolytes in the form of molecules.
4. Compose equations for the processes of oxidation and reduction (equation of half-reactions). To do this, write down the elements that change the degree of oxidation in the form of real particles (ions, atoms, molecules) and equalize the number of each element in the left and right parts of the half-reaction.
Note:
If the original substance contains fewer oxygen atoms than the products (P PO 4 3-), then the lack of oxygen is supplied by the environment.
If the original substance contains more oxygen atoms than the products (SO 4 2-SO 2), then the released oxygen is bound by the medium.
5. Equalize the left and right parts of the equations by the number of charges. To do this, add or subtract the required number of electrons.
6. Select factors for the oxidation and reduction half-reactions so that the number of electrons during oxidation is equal to the number of electrons during reduction.
7. Summarize the half-reactions of oxidation and reduction, taking into account the found factors.
8. Write down the resulting ion-molecular equation in molecular form.
9. Carry out an oxygen test.
There are three types of redox reactions:
a) Intermolecular - reactions in which the oxidation state changes for the elements that make up different molecules.
2KMnO 4 + 5NaNO 2 + 3H 2 SO 4 2MnSO 4 + 5NaNO 3 + K 2 SO 4 + 3H 2 O
b) Intramolecular - reactions in which the oxidation state changes for the elements that make up one molecule.
