During the lesson, we will study the topic "Redox Reactions". You will learn the definition of these reactions, their differences from reactions of other types. Remember what an oxidation state, an oxidizing agent and a reducing agent are. Learn how to draw electronic balance diagrams for redox reactions, get acquainted with the classification of redox reactions.
Topic: Redox reactions
Lesson: Redox reactions
Reactions that occur with a change in the oxidation states of the atoms that make up the reactants are called redox . The change in oxidation states occurs due to the transfer of electrons from the reducing agent to the oxidizing agent. is the formal charge of the atom, assuming that all bonds in the compound are ionic.
Oxidizer - a substance whose molecules or ions accept electrons. If an element is an oxidizing agent, its oxidation state decreases.
O 0 2 + 4e - → 2O -2 (Oxidizing agent, reduction process)
Process reception substances of electrons is called restoration. The oxidizing agent is reduced during the process.
Restorer - a substance whose molecules or ions donate electrons. The reducing agent has an increased oxidation state.
S 0 -4e - →S +4 (Reductant, oxidation process)
Process returns electrons is called. The reducing agent is oxidized during the process.
Example #1. Obtaining chlorine in the laboratory
In the laboratory, chlorine is obtained from potassium permanganate and concentrated of hydrochloric acid. Potassium permanganate crystals are placed in a Wurtz flask. Close the flask with a stopper with a dropping funnel. Hydrochloric acid is poured into the funnel. Hydrochloric acid is poured from a dropping funnel. Vigorous release of chlorine begins immediately. Through the vent tube, chlorine gradually fills the cylinder, displacing air from it. Rice. 1.
Rice. 1
Using this reaction as an example, let's consider how to draw up an electronic balance.
KMnO 4 + HCI \u003d KCI + MnCI 2 + CI 2 + H 2 O
K + Mn +7 O -2 4 + H + CI - = K + CI - + Mn +2 CI - 2 + CI 0 2 + H + 2 O -2
The oxidation states changed manganese and chlorine.
Mn +7 +5e - = Mn +2 oxidizing agent, reduction process
2 CI - -2e - \u003d CI 0 2 reducing agent, oxidation process
4. Equalize the number of given and received electrons. To do this, we find the least common multiple for the numbers 5 and 2. This is 10. As a result of dividing the least common multiple by the number of given and received electrons, we find the coefficients in front of the oxidizing agent and reducing agent.
Mn +7 +5e - = Mn +2 2
2 CI - -2e - \u003d CI 0 2 5
2KMnO 4 + ? HCI = ?KCI + 2MnCI 2 + 5CI 2 +? H2O
However, no coefficient was placed before the hydrochloric acid formula, since not all chloride ions participated in the redox process. The electron balance method allows you to equalize only the ions involved in the redox process. Therefore, it is necessary to equalize the number of ions not participating in. Namely, potassium cations, hydrogen and chloride anions. The result is the following equation:
2KMnO 4 + 16 HCI = 2KCI + 2MnCI 2 + 5CI 2 + 8H 2 O
Example #2. Interaction of copper with concentrated nitric acid. Rice. 2.
A “copper” coin was placed in a glass with 10 ml of acid. Emission of brown gas quickly began (brown bubbles in a still colorless liquid looked especially impressive). The entire space above the liquid turned brown, brown vapors poured out of the glass. The solution turned into green color. The reaction was constantly accelerating. After about half a minute, the solution turned blue, and after two minutes the reaction began to slow down. The coin did not completely dissolve, but lost a lot in thickness (it could be bent with fingers). The green color of the solution in the initial stage of the reaction is due to the products of the reduction of nitric acid.
Rice. 2
1. Let's write the scheme of this reaction:
Cu + HNO 3 \u003d Cu (NO 3) 2 + NO 2 + H 2 O
2. Let's arrange the oxidation states of all elements in the substances participating in the reaction:
Cu 0 + H + N +5 O -2 3 = Cu +2 (N +5 O -2 3) 2 + N +4 O -2 2 + H + 2 O -2
The oxidation states changed copper and nitrogen.
3. We draw up a diagram that reflects the process of electron transition:
N +5 + e - \u003d N +4 oxidizing agent, reduction process
Cu 0 -2e - = Cu +2 reducing agent, oxidation process
4. Equalize the number of given and received electrons. To do this, we find the least common multiple for the numbers 1 and 2. This is 2. As a result of dividing the least common multiple by the number of given and received electrons, we find the coefficients in front of the oxidizing agent and reducing agent.
N +5 + e - \u003d N +4 2
Cu 0 -2e - \u003d Cu +2 1
5. We transfer the coefficients to the original scheme and transform the reaction equation.
Cu + ?HNO 3 \u003d Cu (NO 3) 2 + 2NO 2 + 2H 2 O
Nitric acid is involved not only in the redox reaction, so the coefficient is not written at first. As a result, the following equation is finally obtained:
Cu + 4HNO 3 \u003d Cu (NO 3) 2 + 2NO 2 + 2H 2 O
Classification of redox reactions
1. Intermolecular redox reactions .
These are reactions in which the oxidizing and reducing agents are different substances.
H 2 S -2 + Cl 0 2 → S 0 + 2HCl -
2. Intramolecular reactions in which oxidizing and stopping atoms are in the molecules of the same substance, for example:
2H + 2 O -2 → 2H 0 2 + O 0 2
3. Disproportionation (self-oxidation-self-recovery) - reactions in which the same element acts both as an oxidizing agent and as a reducing agent, for example:
Cl 0 2 + H 2 O → HCl + O + HCl -
4. Conproportionation (Reproportionation) - reactions in which one oxidation state is obtained from two different oxidation states of the same element
Homework
1. No. 1-3 (p. 162) Gabrielyan O.S. Chemistry. Grade 11. A basic level of. 2nd ed., ster. - M.: Bustard, 2007. - 220 p.
2. Why does ammonia only show restorative properties, and nitric acid - only oxidizing?
3. Arrange the coefficients in the equation for the reaction of obtaining nitric acid using the electron balance method: ?NO 2 + ?H 2 O + O 2 = ?HNO 3
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 is 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-over" 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 of nitric acid (HNO 3), and in this compound the degree of nitrogen oxidation is (+5), therefore, in its salt - sodium nitrate, the degree of nitrogen oxidation is also (+5). sodium bicarbonate(NaHCO 3) is an 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 power 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 quantityparticles 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
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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 balance methods, but in the latter case, the number of protons and water molecules involved in the redox process is set “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+ | 10 e
S 2- - 8 e→ S 6+
_______________________
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)
_______________________
N 5+ + 3 e→ N 2+ 10 (oxidation process)
_____________________________________
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
___________________________________________________
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 sulfuric 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
________________________________________________
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
_____________________________________
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
________________________________
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
_________________________________
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
________________________________
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 | 1
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 Nitric oxide NO2. It remains only to solve this equation by the balance method:
P - 3 - 8 e= P+5 | 1
N+ 5 + e= N+4 | 8
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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Redox reactions (ORD)- reactions accompanied by the addition or release of electrons, or the redistribution of electron density on atoms (change in the degree of oxidation).
OVR stages
Oxidation- the return of electrons by atoms, molecules or ions. As a result, the oxidation state rises. Reducing agents donate electrons.
Recovery- the addition of electrons. As a result, the oxidation state decreases. Oxidizers accept electrons.
OVR- coupled process: if there is reduction, then there is also oxidation.
OVR rules
Equivalent exchange of electrons and atomic balance.
acid environment
In an acidic environment, the released oxide ions bind with protons to form water molecules; the missing oxide ions are supplied by water molecules, then protons are released from them.
Where there are not enough oxygen atoms, we write as many water molecules as there are not enough oxide ions.
Sulfur in potassium sulfite has an oxidation state of +4, manganese in potassium permanganate has an oxidation state of +7, sulfuric acid is the reaction medium.
Maraganets in the highest degree oxidation - an oxidizing agent, therefore, potassium sulfite is a reducing agent.
Note: +4 is an intermediate oxidation state for sulfur, so it can act as both a reducing agent and an oxidizing agent. With strong oxidizing agents (permanganate, dichromate), sulfite is a reducing agent (oxidized to sulfate), with strong reducing agents (halides, chalcogenides), sulfite is an oxidizing agent (reduced to sulfur or sulfide).
Sulfur from the +4 oxidation state goes to +6 - sulfite is oxidized to sulfate. Manganese from the +7 oxidation state goes into +2 (acidic environment) - the permanganate ion is reduced to Mn 2+.
2. Compose half-reactions. Equalizing manganese: 4 oxide ions are released from permanganate, which are bound by hydrogen ions (acidic environment) into water molecules. Thus, 4 oxide ions bind to 8 protons in 4 water molecules.
In other words, 4 oxygen is missing on the right side of the equation, so we write 4 water molecules, on the left side of the equation - 8 protons.
Seven minus two is plus five electrons. You can equalize by the total charge: on the left side of the equation, eight protons minus one permanganate \u003d 7+, on the right side, manganese with a charge of 2+, water is electrically neutral. Seven minus two is plus five electrons. Everything is balanced.
Equalizing sulfur: the missing oxide ion on the left side of the equation is supplied by a water molecule, from which two protons are subsequently released to the right side.
Left charge 2-, right 0 (-2+2). minus two electrons.
Multiply the upper half-reaction by 2, the lower by 5.
We reduce protons and water.
Sulfate ions bind to potassium and manganese ions.
Alkaline environment
In an alkaline environment, the released oxide ions are bound by water molecules, forming hydroxide ions (OH - groups). The missing oxide ions are supplied by hydroxo groups, which must be taken twice as much.
Where there are not enough oxide ions, we write 2 times more hydroxo groups than there are not enough, on the other hand - water.
Example. Using the electron balance method, write the reaction equation, determine the oxidizing agent and reducing agent:
Determine the degree of oxidation:
Bismuth (III) with strong oxidizing agents (for example, Cl 2) in an alkaline medium exhibits reducing properties (oxidizes to bismuth V):
Since on the left side of the equation there are not enough 3 oxygens for balance, we write 6 hydroxo groups, and on the right - 3 waters.
Final reaction equation:
Neutral environment
In a neutral environment, the released oxide ions are bound by water molecules to form hydroxide ions (OH - groups). The missing oxide ions are supplied by water molecules. H + ions are released from them.
Using the electron balance method, write the reaction equation, determine the oxidizing agent and reducing agent:
1. Determine the degree of oxidation: sulfur in potassium persulfate has an oxidation state of +7 (it is an oxidizing agent, because the highest oxidation state), bromine in potassium bromide has an oxidation state of -1 (it is a reducing agent, because the lowest oxidation state), water is the reaction medium.
Sulfur from the +7 oxidation state goes to +6 - persulfate is reduced to sulfate. Bromine goes from oxidation state -1 to 0 - the bromide ion is oxidized to bromine.
2. Compose half-reactions. Equalize sulfur (factor 2 before sulfate). Oxygen is balanced.
On the left side, the charge is 2-, on the right side, the charge is 4-, 2 electrons are attached, so we write +2
Equalize bromine (factor 2 in front of the bromide ion). On the left side, the charge is 2-, on the right side, the charge is 0, 2 electrons are given away, so we write -2
3. The overall equation of the electronic balance.
4. Final reaction equation: Sulfate ions bind with potassium ions to potassium sulfate, a factor of 2 before KBr and before K 2 SO 4 . Water was not needed - we enclose it in square brackets.
OVR classification
- Oxidizing agent and reducing agent- various substances
- Self-oxidizers, self-restorers (disproportionation, dismutation). An element in an intermediate oxidation state.
- Oxidizing or reducing agent - medium for the process
- Intramolecular oxidation-reduction. The composition of the same substance includes an oxidizing agent and a reducing agent.
Solid state, high temperature reactions.
Quantitative characteristics of OVR
Standard redox potential, E 0- electrode potential relative to the standard hydrogen potential. More about .
For OVR to pass, it is necessary that the potential difference be greater than zero, that is, the potential of the oxidizer must be greater than the potential of the reductant:
, 
For example:
The lower the potential, the stronger the reducing agent; the higher the potential, the stronger the oxidizing agent.
Oxidative properties are stronger in an acidic environment, reducing - in an alkaline.
The whole variety of chemical reactions can be reduced to two types. If as a result of the reaction the oxidation states of the elements do not change, then such reactions are called exchange, otherwise - redox reactions.
The course of chemical reactions is due to the exchange of particles between the reacting substances. For example, in a neutralization reaction, an exchange occurs between cations and anions of an acid and a base, resulting in the formation of a weak electrolyte - water:
Often the exchange is accompanied by the transfer of electrons from one particle to another. So, when copper is displaced by zinc in a solution of copper sulfate (II)
electrons from zinc atoms go to copper ions:
The process of losing electrons by a particle is called oxidation, and the process of acquiring electrons is restoration. Oxidation and reduction proceed simultaneously, therefore, interactions accompanied by the transfer of electrons from one particle to another are called redox reactions.
The transfer of electrons may be incomplete. For example, in the reaction
instead of low-polarity C-H connections strongly polar H-Cl bonds appear. For the convenience of writing redox reactions, the concept of the degree of oxidation is used, which characterizes the state of an element in chemical compound and its behavior in reactions.
Oxidation state- a value numerically equal to the formal charge that can be attributed to an element, based on the assumption that all the electrons of each of its bonds have passed to a more electronegative atom of this compound.
Using the concept of oxidation state, one can give more general definition oxidation and reduction processes. redox called chemical reactions that are accompanied by a change in the oxidation states of the elements of the substances involved in the reaction. When reducing the oxidation state of the element decreases, when oxidized - increases. A substance that contains an element that lowers the oxidation state is called oxidizing agent; a substance that contains an element that increases the oxidation state is called reducing agent.
The oxidation state of an element in a compound is determined according to the following rules:
The oxidation state of an element in a simple substance is zero;
· the algebraic sum of all oxidation states of atoms in a molecule is equal to zero;
the algebraic sum of all oxidation states of atoms in a complex ion, as well as the oxidation state of an element in a simple monatomic ion, is equal to the charge of the ion;
A negative oxidation state is shown in the compound by the atoms of the element that has the highest electronegativity;
The maximum possible (positive) oxidation state of an element corresponds to the number of the group in which the element is located in Periodic system DI. Mendeleev.
The oxidation state of the atoms of the elements in the compound is written above the symbol of this element, indicating first the sign of the oxidation state, and then its numerical value, for example.
A number of elements in compounds exhibit a constant oxidation state, which is used in determining the oxidation states of other elements:
The redox properties of atoms of various elements manifest themselves depending on many factors, the most important of which are electronic structure element, its oxidation state in the substance, the nature of the properties of other participants in the reaction. Compounds that contain atoms of elements with a maximum (positive) oxidation state, for example, can only be reduced, acting as oxidizing agents. Compounds containing elements with a minimum oxidation state, for example, 
can only oxidize and act as reducing agents.
Substances containing elements with intermediate oxidation states, for example, 
possess redox duality. Depending on the reaction partner, such substances are capable of both accepting (when interacting with stronger reducing agents) and donating (when interacting with stronger oxidizing agents) electrons.
The composition of reduction and oxidation products also depends on many factors, including the medium in which the chemical reaction proceeds, the concentration of reagents, and the activity of the partner in the redox process.
To write down the equation of a redox reaction, it is necessary to know how the oxidation states of the elements change and to what others the oxidizing agent and reducing agent pass. Consider brief characteristics most commonly used oxidizing and reducing agents.
The most important oxidizing agents. Among simple substances, oxidizing properties are typical for typical non-metals: fluorine F 2, chlorine Cl 2, bromine Br 2, iodine I 2, oxygen O 2.
Halogens, recovering, they acquire an oxidation state of -1, and from fluorine to iodine their oxidizing properties weaken (F 2 has limited use due to its high aggressiveness):
Oxygen, recovering, acquires an oxidation state of -2:
The most important oxidizing agents among oxygen-containing acids and their salts are nitric acid HNO 3 and its salts, concentrated sulfuric acid H 2 SO 4, oxygen-containing halogen acids HNalO x and their salts, potassium permanganate KMnO 4 and potassium dichromate K 2 Cr 2 O 7 .
Nitric acid exhibits oxidizing properties due to nitrogen in the oxidation state +5. In this case, the formation of various recovery products is possible:
The depth of nitrogen reduction depends on the concentration of the acid, as well as on the activity of the reducing agent, determined by its redox potential:
Fig.1. Depth of nitrogen reduction as a function of acid concentration.
For example, the oxidation of zinc (active metal) with nitric acid is accompanied by the formation of various reduction products, but about a concentration of HNO 3 of about 2% (mass.) NH 4 NO 3 is predominantly formed:
at a concentration of HNO 3 approximately 5% (mass.) - N 2 O:
at a concentration of HNO 3 about 30% (mass.) - NO:
and at a concentration of HNO 3 of approximately 60% (mass.) predominantly formed - NO 2:
The oxidative activity of nitric acid increases with increasing concentration, therefore, concentrated HNO 3 oxidizes not only active, but also inactive metals, such as copper and silver, forming mainly nitric oxide (IV):
as well as non-metals, such as sulfur and phosphorus, oxidizing them to acids corresponding to the highest oxidation states:
Salts of nitric acid ( nitrates) can be reduced in acidic, and when interacting with active metals and in alkaline media, as well as in melts:
Aqua regia- a mixture of concentrated and nitric acids, mixed in a ratio of 1:3 by volume. The name of this mixture is due to the fact that it dissolves even noble metals such as gold and platinum:
The course of this reaction is due to the fact that aqua regia releases nitrosyl chloride NOCl and free chlorine Cl 2:
which convert metals to chlorides.
Sulfuric acid exhibits oxidizing properties in a concentrated solution due to sulfur in the oxidation state +6:
The composition of the reduction products is determined mainly by the activity of the reducing agent and the concentration of the acid:
Fig.2. Reducing activity of sulfur depending on
acid concentration.
So, the interaction of concentrated H 2 SO 4 with low-active metals, some non-metals and their compounds leads to the formation of sulfur oxide (IV):
Active metals restore concentrated sulfuric acid to sulfur or hydrogen sulfide:
in this case, H 2 S, S and SO 2 are simultaneously formed in various ratios. However, in this case, the main product of the reduction of H 2 SO 4 is SO 2, since the released S and H 2 S can be oxidized with concentrated sulfuric acid:
and their salts (see Table A.1.1) are often used as oxidizing agents, although many of them exhibit a dual character. As a rule, the reduction products of these compounds are chlorides and bromides (oxidation state -1), as well as iodine (oxidation state 0);
However, even in this case, the composition of the reduction products depends on the reaction conditions, the concentration of the oxidizing agent, and the activity of the reducing agent:
Potassium permanganate exhibits oxidizing properties due to manganese in the oxidation state +7. Depending on the medium in which the reaction takes place, it is reduced to different products: in an acidic medium to manganese (II) salts, in a neutral medium to manganese (IV) oxide in the hydrated form MnO (O) 2 , in an alkaline medium to manganate -and she
acid environment
neutral environment
alkaline environment
Potassium dichromate, whose molecule contains chromium in the +6 oxidation state, is a strong oxidizing agent during sintering and in an acid solution
exhibits oxidizing properties in a neutral environment
In an alkaline environment, the equilibrium between chromate and dichromate ions
is shifted towards the formation, therefore, in an alkaline environment, the oxidizing agent is potassium chromate K 2 CrO 4:
however, K 2 CrO 4 is a weaker oxidizing agent compared to K 2 Cr 2 O 7 .
Among ions, oxidizing properties are exhibited by the hydrogen ion H + and metal ions in the highest oxidation state. hydrogen ion H + acts as an oxidizing agent when active metals interact with dilute acid solutions (with the exception of HNO 3)
metal ions in a relatively high oxidation state, such as Fe 3+, Cu 2+, Hg 2+, being restored, turning into ions of a lower oxidation state
or are isolated from solutions of their salts in the form of metals
The most important reducing agents. Typical reducing agents among simple substances include active metals such as alkali and alkaline earth metals, zinc, aluminum, iron and others, as well as some non-metals (hydrogen, carbon, phosphorus, silicon).
Metals in an acidic environment, they are oxidized to positively charged ions:
In an alkaline environment, metals exhibiting amphoteric properties are oxidized; in this case, negatively charged anions or hydroxocomponents are formed:
non-metals, oxidized, form oxides or the corresponding acids:
Reducing functions have oxygen-free anions, such as Cl - , Br - , I - , S 2- , H - and metal cations in the highest degree of oxidation.
In a row halide ions, which, when oxidized, usually form halogens:
reducing properties are enhanced from Cl - to I - .
hydrides metals exhibit reducing properties due to the oxidation of hydrogen bound (oxidation state -1) to free hydrogen:
metal cations in the lowest degree of oxidation, such as Sn 2+, Fe 2+, Cu +, Hg 2 2+ and others, when interacting with oxidizing agents, an increase in the degree of oxidation is characteristic:
Redox duality. Among simple substances, redox duality is characteristic of elements VIIA, VIA and VA subgroups, which can both increase and decrease their oxidation state.
Often used as oxidizers halogens under the action of stronger oxidizing agents, they exhibit reducing properties (with the exception of fluorine). Their oxidizing abilities decrease, and their reducing properties increase from Cl 2 to I 2:
Fig.3. Redox ability of halogens.
This feature is illustrated by the reaction of the oxidation of iodine with chlorine in an aqueous solution:
The composition of oxygen-containing compounds that exhibit dual behavior in redox reactions also includes elements in an intermediate oxidation state. Oxygen-containing acids of halogens and their salts, whose molecules include halogen in an intermediate oxidation state, can be both oxidizing agents
and restorers
Hydrogen peroxide, containing oxygen in the -1 oxidation state, exhibits oxidizing properties in the presence of typical reducing agents, since the oxygen oxidation state can drop to -2:
The latter reaction is used in the restoration of paintings by old masters, the paints of which, containing white lead, turn black due to interaction with air hydrogen sulfide.
When interacting with strong oxidizing agents, the oxidation state of oxygen, which is part of hydrogen peroxide, rises to 0, H 2 O 2 exhibits the properties of a reducing agent:
Nitrous acid And nitrites, which include nitrogen in the +3 oxidation state, and can also act as oxidizing agents
and as a restorer
Classification. There are four types of redox reactions.
1. If the oxidizing agent and reducing agent are different substances, then such reactions are related to intermolecular. All of the above reactions are examples.
2. During the thermal decomposition of complex compounds, which include an oxidizing agent and a reducing agent in the form of atoms of different elements, redox reactions occur, called intramolecular:
3. Reactions disproportionation (dismutations or, according to outdated terminology, self-oxidation - self-healing) can occur if compounds containing elements in intermediate oxidation states are exposed to conditions where they are unstable (for example, at elevated temperatures). The oxidation state of this element both rises and falls:
4. Reactions counterproportionation (switching) are the processes of interaction of an oxidizing agent and a reducing agent, which include the same element with varying degrees oxidation. As a result, the product of oxidation and reduction is a substance with an intermediate oxidation state of atoms of a given element:
There are also mixed reactions. For example, an intramolecular counterproportionation reaction includes the decomposition of ammonium nitrate
Drawing up equations.
The equations of redox reactions are based on the principles of equality of the number of the same atoms before and after the reaction, and also taking into account the equality of the number of electrons donated by the reducing agent and the number of electrons accepted by the oxidizing agent, i.e. electroneutrality of molecules. The reaction is presented as a system of two half-reactions - oxidation and reduction, the summation of which, taking into account the indicated principles, leads to the compilation of the general equation of the process.
To compile the equations of redox reactions, the method of electron-ion half-reactions and the method of electronic balance are most often used.
Method of electron-ion half-reactions used in the preparation of reaction equations occurring in an aqueous solution, as well as reactions involving substances whose oxidation state of elements is difficult to determine (for example, KNCS, CH 3 CH 2 OH).
According to this method, the following main stages of compiling the reaction equation are distinguished.
a) write down the general molecular scheme of the process indicating the reducing agent, oxidizing agent and the medium in which the reaction proceeds (acidic, neutral or alkaline). For example
b) taking into account the dissociation of electrolytes in an aqueous solution, this scheme is presented in the form of molecular-ion interaction. Ions whose oxidation states of atoms do not change are not indicated in the scheme, with the exception of medium ions (H +, OH -):
c) determine the oxidation states of the reducing agent and oxidizing agent, as well as the products of their interaction:
f) add ions that did not participate in the oxidation-reduction process, equalize their number on the left and right, and write down the molecular reaction equation
The greatest difficulties arise in compiling the material balance of the oxidation and reduction half-reactions, when the number of oxygen atoms that make up the particles of the oxidizing agent and reducing agent changes. It should be taken into account that in aqueous solutions the binding or addition of oxygen occurs with the participation of water molecules and ions of the environment.
In the process of oxidation, one molecule of water is consumed per one oxygen atom attached to the reducing agent particle in acidic and neutral media and two H + ions are formed; in an alkaline environment, two OH hydroxide ions are consumed and one water molecule is formed (Table 1.1).
To bind one oxygen atom of an oxidizing agent in an acidic medium, two H + ions are consumed during the reduction process and one water molecule is formed; in neutral and alkaline environments, one molecule of H 2 O is consumed, two OH ions are formed - (Table 1, 2).
Table 1
Attachment of oxygen atoms to a reducing agent during oxidation
table 2
Binding of oxygen atoms of an oxidizing agent during reduction
The advantages of the method of electron-ionic half-reactions are that when compiling the equations of redox reactions, the real states of particles in solution and the role of the medium in the course of processes are taken into account, there is no need to use the formal concept of the degree of oxidation.
Electronic balance method, based on taking into account changes in the oxidation state and the principle of electroneutrality of the molecule, is universal. It is commonly used to formulate equations for redox reactions occurring between gases, solids and in melts.
The sequence of operations, according to the method, is as follows:
1) write down the formulas of the reactants and reaction products in molecular form:
2) determine the oxidation states of atoms that change it during the reaction:
3) by changing the oxidation states, the number of electrons donated by the reducing agent and the number of electrons accepted by the oxidizing agent are determined, and an electronic balance is drawn up, taking into account the principle of equality of the number of given and received electrons:
4) electronic balance factors are written into the redox reaction equation as the main stoichiometric coefficients:
5) select the stoichiometric coefficients of the remaining participants in the reaction:
When compiling equations, it should be taken into account that the oxidizing agent (or reducing agent) can be consumed not only in the main redox reaction, but also when binding the resulting reaction products, that is, it can act as a medium and a salt former.
An example, when an oxidizing agent plays the role of a medium, is the oxidation reaction of a metal in nitric acid, compiled by the method of electron-ion half-reactions:
An example, when the reducing agent is the medium in which the reaction proceeds, is the oxidation of hydrochloric acid with potassium dichromate, compiled by the electron balance method:
When calculating the quantitative, mass and volume ratios of the participants in redox reactions, the basic stoichiometric laws of chemistry and, in particular, the law of equivalents are used. To determine the direction and completeness of the course of redox processes, the values of the thermodynamic parameters of these systems are used, and when reactions occur in aqueous solutions, the values of the corresponding electrode potentials are used.
How to find out where chemical reaction oxidizing agent and where is the reducing agent? and got the best answer
Reply from str.[active]
if after the reaction (after the equal sign) the substance acquires a positive charge, then it is a reducing agent
and if it acquires a negative charge, it means an oxidizing agent
For example
H2 + O2 = H2O
before the reaction, both hydrogen and oxygen have zero charge
after reaction
hydrogen acquires a charge of +1 and oxygen -2 means hydrogen is a reducing agent
and oxygen is an oxidizing agent!
Source: =)) if something is not clear, write)
Answer from 2 answers[guru]
Hello! Here is a selection of topics with answers to your question: How to find out where in a chemical reaction the oxidizing agent and where is the reducing agent?
Answer from BeardMax[guru]
To do this, you need to know what the degree of oxidation is.
Learn to determine the oxidation state of any atom in a chemical compound.
Next, look at which atoms CO increases in the reaction, and which decreases. The former are reducing agents, the latter are oxidizing agents.
IN general chemistry didn't have to walk.
Answer from OOO[newbie]
A reducing agent is a substance that donates electrons. H-r, Ca (2+) - 2e \u003d Ca (0)
An oxidizing agent is a substance that accepts electrons.
Answer from Mariska[newbie]
To find out, you need to look at what are reagents and what is added as a medium. For example, if there is Mn (+4) and water in the starting materials, then Mn will change its oxidation state to (+6), if I'm not mistaken. In addition, you can see what degree of oxidation the elements are in (suddenly somewhere it is minimal or vice versa maximum).
