Redox reactions. CC (valuable instructions) H2 o2 h2o redox reaction

Before giving examples of redox reactions with a solution, we highlight the main definitions associated with these transformations.

Those atoms or ions that, during interaction, change their oxidation state with a decrease (accept electrons) are called oxidizing agents. Among the substances with such properties are strong inorganic acids: sulfuric, hydrochloric, nitric.

Oxidizer

Alkali metal permanganates and chromates are also strong oxidizing agents.

The oxidizing agent takes what it needs during the reaction to complete energy level(installation of the completed configuration).

Reducing agent

Any redox reaction scheme involves identifying a reducing agent. It includes ions or neutral atoms that can increase their oxidation state during interaction (they donate electrons to other atoms).

Typical reducing agents include metal atoms.

Processes in OVR

What else are they characterized by a change in the oxidation states of the starting substances.

Oxidation involves the process of releasing negative particles. Reduction involves accepting them from other atoms (ions).

Parsing algorithm

Examples of redox reactions with solutions are offered in various reference materials designed to prepare high school students for final chemistry tests.

In order to successfully cope with those offered in the OGE and Unified State Exam assignments, it is important to master the algorithm for compiling and analyzing redox processes.

1. First of all, charge values ​​are assigned to all elements in the substances proposed in the diagram.
2. Atoms (ions) from the left side of the reaction are written out, which during the interaction changed their indicators.
3. When the oxidation state increases, the sign “-” is used, and when the oxidation state decreases, “+”.
4. The least common multiple (the number by which they are divided without a remainder) is determined between the given and accepted electrons.
5. When dividing NOC by electrons, we obtain stereochemical coefficients.
6. We place them in front of the formulas in the equation.

The first example from the OGE

In the ninth grade, not all students know how to solve redox reactions. That is why they make many mistakes and do not receive high scores for the OGE. The algorithm of actions is given above, now let’s try to work it out using specific examples.

The peculiarity of the tasks concerning the arrangement of coefficients in the proposed reaction, given to graduates of the basic stage of education, is that both the left and right sides of the equation are given.

This greatly simplifies the task, since you do not need to independently invent interaction products or select missing starting substances.

For example, it is proposed to use an electronic balance to identify the coefficients in the reaction:

At first glance, this reaction does not require stereochemical coefficients. But, in order to confirm your point of view, it is necessary for all elements to have charge numbers.

In binary compounds, which include copper oxide (2) and iron oxide (2), the sum of oxidation states is zero, given that for oxygen it is -2, for copper and iron this indicator is +2. Simple substances do not give up (do not accept) electrons, so they are characterized by a zero oxidation state.

Let's draw up an electronic balance, showing with a "+" and "-" sign the number of electrons received and given during the interaction.

Fe 0 -2e=Fe 2+.

Since the number of electrons accepted and donated during the interaction is the same, there is no point in finding the least common multiple, determining stereochemical coefficients, and putting them in the proposed interaction scheme.

In order to get the maximum score for the task, it is necessary not only to write down examples of redox reactions with solutions, but also to write out separately the formula of the oxidizing agent (CuO) and the reducing agent (Fe).

Second example with OGE

Let us give more examples of redox reactions with solutions that may be encountered by ninth-graders who have chosen chemistry as their final exam.

Suppose it is proposed to place the coefficients in the equation:

Na+HCl=NaCl+H2.

In order to cope with the task, it is first important to determine each simple and complex substance indicators of oxidation states. For sodium and hydrogen they will be equal to zero, since they are simple substances.

In hydrochloric acid, hydrogen has a positive oxidation state and chlorine has a negative oxidation state. After arranging the coefficients, we obtain a reaction with coefficients.

The first from the Unified State Exam

How to complement redox reactions? Examples with solutions found on the Unified State Exam (grade 11) require the completion of gaps, as well as the placement of coefficients.

For example, you need to supplement the reaction with an electronic balance:

H 2 S+ HMnO 4 = S+ MnO 2 +…

Identify the reducing agent and oxidizing agent in the proposed scheme.

How to learn to write redox reactions? The sample assumes the use of a specific algorithm.

First, in all substances given according to the conditions of the problem, it is necessary to set the oxidation states.

Next, you need to analyze which substance can become an unknown product in this process. Since there is an oxidizing agent (manganese plays its role) and a reducing agent (sulfur is its role), the oxidation states in the desired product do not change, therefore, it is water.

When discussing how to correctly solve redox reactions, we note that next step there will be a compilation of the electronic ratio:

Mn +7 takes 3 e= Mn +4 ;

S -2 gives 2e= S 0 .

The manganese cation is a reducing agent, and the sulfur anion is a typical oxidizing agent. Since the smallest multiple between the received and donated electrons will be 6, we get the coefficients: 2, 3.

The last step will be to insert the coefficients into the original equation.

3H 2 S+ 2HMnO 4 = 3S+ 2MnO 2 + 4H 2 O.

The second sample of OVR in the Unified State Exam

How to correctly formulate redox reactions? Examples with solutions will help you work out the algorithm of actions.

It is proposed to use the electronic balance method to fill in the gaps in the reaction:

PH 3 + HMnO 4 = MnO 2 +…+…

We arrange the oxidation states of all elements. In this process, oxidizing properties are manifested by manganese, which is part of the composition and the reducing agent must be phosphorus, changing its oxidation state to positive in phosphoric acid.

According to the assumption made, we obtain the reaction scheme, then we compose the electron balance equation.

P -3 gives 8 e and turns into P +5;

Mn +7 takes 3e, becoming Mn +4.

The LOC will be 24, so phosphorus must have a stereometric coefficient of 3, and manganese -8.

We put the coefficients into the resulting process, we get:

3 PH 3 + 8 HMnO 4 = 8 MnO 2 + 4H 2 O+ 3 H 3 PO 4.

Third example from the Unified State Exam

Using electron-ion balance, you need to compose a reaction, indicate the reducing agent and oxidizing agent.

KMnO 4 + MnSO 4 +…= MnO 2 +…+ H2SO 4.

According to the algorithm, we arrange the oxidation states of each element. Next, we determine those substances that are missed in the right and left parts of the process. Here a reducing agent and an oxidizing agent are given, so the oxidation states of the missing compounds do not change. The lost product will be water, and the starting compound will be potassium sulfate. We obtain a reaction scheme for which we will draw up an electronic balance.

Mn +2 -2 e= Mn +4 3 reducing agent;

Mn +7 +3e= Mn +4 2 oxidizing agent.

We write the coefficients into the equation, summing up the manganese atoms on the right side of the process, since it relates to the disproportionation process.

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

Conclusion

Redox reactions have special meaning for the functioning of living organisms. Examples of OVR are the processes of rotting, fermentation, nervous activity, breathing, metabolism.

Oxidation and reduction are relevant for metallurgical and chemical industry Thanks to such processes, it is possible to restore metals from their compounds, protect them from chemical corrosion, and process them.

To compile a redox process in organic matter, it is necessary to use a certain algorithm of actions. First, in the proposed scheme, the oxidation states are set, then those elements that increased (decreased) the indicator are determined, and the electronic balance is recorded.

If you follow the sequence of actions suggested above, you can easily cope with the tasks offered in the tests.

In addition to the electronic balance method, the arrangement of coefficients is also possible by composing half-reactions.

Task No. 1

Si + HNO 3 + HF → H 2 SiF 6 + NO + …

N +5 + 3e → N +2 │4 reduction reaction

Si 0 − 4e → Si +4 │3 oxidation reaction

N +5 (HNO 3) – oxidizing agent, Si – reducing agent

3Si + 4HNO 3 + 18HF → 3H 2 SiF 6 + 4NO +8H 2 O

Task No. 2

Using the electron balance method, create an equation for the reaction:

B+ HNO 3 + HF → HBF 4 + NO 2 + …

Identify the oxidizing agent and the reducing agent.

N +5 + 1e → N +4 │3 reduction reaction

B 0 -3e → B +3 │1 oxidation reaction

N +5 (HNO 3) – oxidizing agent, B 0 – reducing agent

B+ 3HNO 3 + 4HF → HBF 4 + 3NO 2 + 3H 2 O

Task No. 3

Using the electron balance method, create an equation for the reaction:

K 2 Cr 2 O 7 + HCl → Cl 2 + KCl + … + …

Identify the oxidizing agent and the reducing agent.

2Cl -1 -2e → Cl 2 0 │3 oxidation reaction

Cr +6 (K 2 Cr 2 O 7) – oxidizing agent, Cl -1 (HCl) – reducing agent

K 2 Cr 2 O 7 + 14HCl → 3Cl 2 + 2KCl + 2CrCl 3 + 7H 2 O

Task No. 4

Using the electron balance method, create an equation for the reaction:

Cr 2 (SO 4) 3 + … + NaOH → Na 2 CrO 4 + NaBr + … + H 2 O

Identify the oxidizing agent and the reducing agent.

Br 2 0 + 2e → 2Br -1 │3 reduction reaction

2Cr +3 - 6e → 2Cr +6 │1 oxidation reaction

Br 2 – oxidizing agent, Cr +3 (Cr 2 (SO 4) 3) – reducing agent

Cr 2 (SO 4) 3 + 3Br 2 + 16NaOH → 2Na 2 CrO 4 + 6NaBr + 3Na 2 SO 4 + 8H 2 O

Task No. 5

Using the electron balance method, create an equation for the reaction:

K 2 Cr 2 O 7 + … + H 2 SO 4 → l 2 + Cr 2 (SO 4) 3 + … + H 2 O

Identify the oxidizing agent and the reducing agent.

2Cr +6 + 6e → 2Cr +3 │1 reduction reaction

2I -1 -2e → l 2 0 │3 oxidation reaction

Cr +6 (K 2 Cr 2 O 7) – oxidizing agent, l -1 (Hl) – reducing agent

K 2 Cr 2 O 7 + 6HI + 4H 2 SO 4 → 3l 2 + Cr 2 (SO 4) 3 + K 2 SO 4 + 7H 2 O

Task No. 6

Using the electron balance method, create an equation for the reaction:

H 2 S + HMnO 4 → S + MnO 2 + …

Identify the oxidizing agent and the reducing agent.

3H 2 S + 2HMnO 4 → 3S + 2MnO 2 + 4H 2 O

Task No. 7

Using the electron balance method, create an equation for the reaction:

H 2 S + HClO 3 → S + HCl + …

Identify the oxidizing agent and the reducing agent.

S -2 -2e → S 0 │3 oxidation reaction

Mn +7 (HMnO 4) – oxidizing agent, S -2 (H 2 S) – reducing agent

3H 2 S + HClO 3 → 3S + HCl + 3H 2 O

Task No. 8

Using the electron balance method, create an equation for the reaction:

NO + HClO 4 + … → HNO 3 + HCl

Identify the oxidizing agent and the reducing agent.

Cl +7 + 8e → Cl -1 │3 reduction reaction

N +2 -3e → N +5 │8 oxidation reaction

Cl +7 (HClO 4) – oxidizing agent, N +2 (NO) – reducing agent

8NO + 3HClO4 + 4H2O → 8HNO3 + 3HCl

Task No. 9

Using the electron balance method, create an equation for the reaction:

KMnO 4 + H 2 S + H 2 SO 4 → MnSO 4 + S + … + …

Identify the oxidizing agent and the reducing agent.

S -2 -2e → S 0 │5 oxidation reaction

Mn +7 (KMnO 4) – oxidizing agent, S -2 (H 2 S) – reducing agent

2KMnO 4 + 5H 2 S + 3H 2 SO 4 → 2MnSO 4 + 5S + K 2 SO 4 + 8H 2 O

Task No. 10

Using the electron balance method, create an equation for the reaction:

KMnO 4 + KBr + H 2 SO 4 → MnSO 4 + Br 2 + … + …

Identify the oxidizing agent and the reducing agent.

Mn +7 + 5e → Mn +2 │2 reduction reaction

2Br -1 -2e → Br 2 0 │5 oxidation reaction

Mn +7 (KMnO 4) – oxidizing agent, Br -1 (KBr) – reducing agent

2KMnO 4 + 10KBr + 8H 2 SO 4 → 2MnSO 4 + 5Br 2 + 6K 2 SO 4 + 8H 2 O

Task No. 11

Using the electron balance method, create an equation for the reaction:

PH 3 + HClO 3 → HCl + …

Identify the oxidizing agent and the reducing agent.

Cl +5 + 6e → Cl -1 │4 reduction reaction

Cl +5 (HClO 3) – oxidizing agent, P -3 (H 3 PO 4) – reducing agent

3PH 3 + 4HClO 3 → 4HCl + 3H 3 PO 4

Task No. 12

Using the electron balance method, create an equation for the reaction:

PH 3 + HMnO 4 → MnO 2 + … + …

Identify the oxidizing agent and the reducing agent.

Mn +7 + 3e → Mn +4 │8 reduction reaction

P -3 − 8e → P +5 │3 oxidation reaction

Mn +7 (HMnO 4) – oxidizing agent, P -3 (H 3 PO 4) – reducing agent

3PH 3 + 8HMnO 4 → 8MnO 2 + 3H 3 PO 4 + 4H 2 O

Task No. 13

Using the electron balance method, create an equation for the reaction:

NO + KClO + … → KNO 3 + KCl + …

Identify the oxidizing agent and the reducing agent.

Cl +1 + 2e → Cl -1 │3 reduction reaction

N +2 − 3e → N +5 │2 oxidation reaction

Cl +1 (KClO) – oxidizing agent, N +2 (NO) – reducing agent

2NO + 3KClO + 2KOH → 2KNO 3 + 3KCl + H 2 O

Task No. 14

Using the electron balance method, create an equation for the reaction:

PH 3 + AgNO 3 + … → Ag + … + HNO 3

Identify the oxidizing agent and the reducing agent.

Ag +1 + 1e → Ag 0 │8 reduction reaction

P -3 - 8e → P +5 │1 oxidation reaction

Ag +1 (AgNO 3) – oxidizing agent, P -3 (PH 3) – reducing agent

PH 3 + 8AgNO 3 + 4H 2 O → 8Ag + H 3 PO 4 + 8HNO 3

Task No. 15

Using the electron balance method, create an equation for the reaction:

KNO 2 + … + H 2 SO 4 → I 2 + NO + … + …

Identify the oxidizing agent and the reducing agent.

N +3 + 1e → N +2 │ 2 reduction reaction

2I -1 − 2e → I 2 0 │ 1 oxidation reaction

N +3 (KNO 2) – oxidizing agent, I -1 (HI) – reducing agent

2KNO 2 + 2HI + H 2 SO 4 → I 2 + 2NO + K 2 SO 4 + 2H 2 O

Task No. 16

Using the electron balance method, create an equation for the reaction:

Na 2 SO 3 + Cl 2 + … → Na 2 SO 4 + …

Identify the oxidizing agent and the reducing agent.

Cl 2 0 + 2e → 2Cl -1 │1 reduction reaction

Cl 2 0 – oxidizing agent, S +4 (Na 2 SO 3) – reducing agent

Na 2 SO 3 + Cl 2 + H 2 O → Na 2 SO 4 + 2HCl

Task No. 17

Using the electron balance method, create an equation for the reaction:

KMnO 4 + MnSO 4 + H 2 O→ MnO 2 + … + …

Identify the oxidizing agent and the reducing agent.

Mn +7 + 3e → Mn +4 │2 reduction reaction

Mn +2 − 2e → Mn +4 │3 oxidation reaction

Mn +7 (KMnO 4) – oxidizing agent, Mn +2 (MnSO 4) – reducing agent

2KMnO 4 + 3MnSO 4 + 2H 2 O → 5MnO 2 + K 2 SO 4 + 2H 2 SO 4

Task No. 18

Using the electron balance method, create an equation for the reaction:

KNO 2 + … + H 2 O → MnO 2 + … + KOH

Identify the oxidizing agent and the reducing agent.

Mn +7 + 3e → Mn +4 │2 reduction reaction

N +3 − 2e → N +5 │3 oxidation reaction

Mn +7 (KMnO 4) – oxidizing agent, N +3 (KNO 2) – reducing agent

3KNO 2 + 2KMnO 4 + H 2 O → 2MnO 2 + 3KNO 3 + 2KOH

Task No. 19

Using the electron balance method, create an equation for the reaction:

Cr 2 O 3 + … + KOH → KNO 2 + K 2 CrO 4 + …

Identify the oxidizing agent and the reducing agent.

N +5 + 2e → N +3 │3 reduction reaction

2Cr +3 − 6e → 2Cr +6 │1 oxidation reaction

N +5 (KNO 3) – oxidizing agent, Cr +3 (Cr 2 O 3) – reducing agent

Cr 2 O 3 + 3KNO 3 + 4KOH → 3KNO 2 +2K 2 CrO 4 + 2H 2 O

Task No. 20

Using the electron balance method, create an equation for the reaction:

I 2 + K 2 SO 3 + … → K 2 SO 4 + … + H 2 O

Identify the oxidizing agent and the reducing agent.

I 2 0 + 2e → 2I -1 │1 reduction reaction

S +4 - 2e → S +6 │1 oxidation reaction

I 2 – oxidizing agent, S +4 (K 2 SO 3) – reducing agent

I 2 + K 2 SO 3 +2KOH → K 2 SO 4 +2KI + H 2 O

Task No. 21

Using the electron balance method, create an equation for the reaction:

KMnO 4 + NH 3 → MnO 2 +N 2 + … + …

Identify the oxidizing agent and the reducing agent.

Mn +7 + 3e → Mn +4 │2 reduction reaction

2N -3 − 6e → N 2 0 │1 oxidation reaction

Mn +7 (KMnO 4) – oxidizing agent, N -3 (NH 3) – reducing agent

2KMnO 4 + 2NH 3 → 2MnO 2 +N 2 + 2KOH + 2H 2 O

Task No. 22

Using the electron balance method, create an equation for the reaction:

NO 2 + P 2 O 3 + … → NO + K 2 HPO 4 + …

Identify the oxidizing agent and the reducing agent.

N +4 + 2e → N +2 │2 reduction reaction

2P +3 - 4e → 2P +5 │1 oxidation reaction

N +4 (NO 2) – oxidizing agent, P +3 (P 2 O 3) – reducing agent

2NO 2 + P 2 O 3 + 4KOH → 2NO + 2K 2 HPO 4 + H 2 O

Task No. 23

Using the electron balance method, create an equation for the reaction:

KI + H 2 SO 4 → I 2 + H 2 S + … + …

Identify the oxidizing agent and the reducing agent.

S +6 + 8e → S -2 │1 reduction reaction

2I -1 − 2e → I 2 0 │4 oxidation reaction

S +6 (H 2 SO 4) – oxidizing agent, I -1 (KI) – reducing agent

8KI + 5H 2 SO 4 → 4I 2 + H 2 S + 4K 2 SO 4 + 4H 2 O

Task No. 24

Using the electron balance method, create an equation for the reaction:

FeSO 4 + ... + H 2 SO 4 → ... + MnSO 4 + K 2 SO 4 + H 2 O

Identify the oxidizing agent and the reducing agent.

Mn +7 + 5e → Mn +2 │2 reduction reaction

2Fe +2 − 2e → 2Fe +3 │5 oxidation reaction

Mn +7 (KMnO 4) – oxidizing agent, Fe +2 (FeSO 4) – reducing agent

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

Task No. 25

Using the electron balance method, create an equation for the reaction:

Na 2 SO 3 + … + KOH → K 2 MnO 4 + … + H 2 O

Identify the oxidizing agent and the reducing agent.

Mn +7 + 1e → Mn +6 │2 reduction reaction

S +4 − 2e → S +6 │1 oxidation reaction

Mn +7 (KMnO 4) – oxidizing agent, S +4 (Na 2 SO 3) – reducing agent

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

Task No. 26

Using the electron balance method, create an equation for the reaction:

H 2 O 2 + … + H 2 SO 4 → O 2 + MnSO 4 + … + …

Identify the oxidizing agent and the reducing agent.

Mn +7 + 5e → Mn +2 │2 reduction reaction

2O -1 − 2e → O 2 0 │5 oxidation reaction

Mn +7 (KMnO 4) – oxidizing agent, O -1 (H 2 O 2) – reducing agent

5H 2 O 2 + 2KMnO 4 + 3H 2 SO 4 → 5O 2 + 2MnSO 4 + K 2 SO 4 + 8H 2 O

Task No. 27

Using the electron balance method, create an equation for the reaction:

K 2 Cr 2 O 7 + H 2 S + H 2 SO 4 → Cr 2 (SO 4) 3 + K 2 SO 4 + … + …

Identify the oxidizing agent and the reducing agent.

2Cr +6 + 6e → 2Cr +3 │1 reduction reaction

S -2 − 2e → S 0 │3 oxidation reaction

Cr +6 (K 2 Cr 2 O 7) – oxidizing agent, S -2 (H 2 S) – reducing agent

K 2 Cr 2 O 7 + 3H 2 S + 4H 2 SO 4 → Cr 2 (SO 4) 3 + K 2 SO 4 + 3S + 7H 2 O

Task No. 28

Using the electron balance method, create an equation for the reaction:

KMnO 4 + HCl → MnCl 2 + Cl 2 + … + …

Identify the oxidizing agent and the reducing agent.

Mn +7 + 5e → Mn +2 │2 reduction reaction

2Cl -1 − 2e → Cl 2 0 │5 oxidation reaction

Mn +7 (KMnO 4) – oxidizing agent, Cl -1 (HCl) – reducing agent

2KMnO 4 + 16HCl → 2MnCl 2 + 5Cl 2 + 2KCl + 8H 2 O

Task No. 29

Using the electron balance method, create an equation for the reaction:

CrCl 2 + K 2 Cr 2 O 7 + … → CrCl 3 + … + H 2 O

Identify the oxidizing agent and the reducing agent.

2Cr +6 + 6e → 2Cr +3 │1 reduction reaction

Cr +2 − 1e → Cr +3 │6 oxidation reaction

Cr +6 (K 2 Cr 2 O 7) – oxidizing agent, Cr +2 (CrCl 2) – reducing agent

6CrCl 2 + K 2 Cr 2 O 7 + 14HCl → 8CrCl 3 + 2KCl + 7H 2 O

Task No. 30

Using the electron balance method, create an equation for the reaction:

K 2 CrO 4 + HCl → CrCl 3 + … + … + H 2 O

Identify the oxidizing agent and the reducing agent.

Cr +6 + 3e → Cr +3 │2 reduction reaction

2Cl -1 − 2e → Cl 2 0 │3 oxidation reaction

Cr +6 (K 2 CrO 4) – oxidizing agent, Cl -1 (HCl) – reducing agent

2K 2 CrO 4 + 16HCl → 2CrCl 3 + 3Cl 2 + 4KCl + 8H 2 O

Task No. 31

Using the electron balance method, create an equation for the reaction:

KI + … + H 2 SO 4 → I 2 + MnSO 4 + … + H 2 O

Identify the oxidizing agent and the reducing agent.

Mn +7 + 5e → Mn +2 │2 reduction reaction

2l -1 − 2e → l 2 0 │5 oxidation reaction

Mn +7 (KMnO 4) – oxidizing agent, l -1 (Kl) – reducing agent

10KI + 2KMnO 4 + 8H 2 SO 4 → 5I 2 + 2MnSO 4 + 6K 2 SO 4 + 8H 2 O

Task No. 32

Using the electron balance method, create an equation for the reaction:

FeSO 4 + KClO 3 + KOH → K 2 FeO 4 + KCl + K 2 SO 4 + H 2 O

Identify the oxidizing agent and the reducing agent.

Cl +5 + 6e → Cl -1 │2 reduction reaction

Fe +2 − 4e → Fe +6 │3 oxidation reaction

3FeSO 4 + 2KClO 3 + 12KOH → 3K 2 FeO 4 + 2KCl + 3K 2 SO 4 + 6H 2 O

Task No. 33

Using the electron balance method, create an equation for the reaction:

FeSO 4 + KClO 3 + … → Fe 2 (SO 4) 3 + … + H 2 O

Identify the oxidizing agent and the reducing agent.

Cl +5 + 6e → Cl -1 │1 reduction reaction

2Fe +2 − 2e → 2Fe +3 │3 oxidation reaction

Cl +5 (KClO 3) – oxidizing agent, Fe +2 (FeSO 4) – reducing agent

6FeSO 4 + KClO 3 + 3H 2 SO 4 → 3Fe 2 (SO 4) 3 + KCl + 3H 2 O

Task No. 34

Using the electron balance method, create an equation for the reaction.

Problem book on general and inorganic chemistry

2.2. Redox reactions

Theoretical part

Redox reactions include chemical reactions that are accompanied by a change in the oxidation states of elements. In the equations of such reactions, the selection of coefficients is carried out by compiling electronic balance. The method for selecting odds using an electronic balance consists of the following steps:

a) write down the formulas of the reagents and products, and then find the elements that increase and decrease their oxidation states and write them out separately:

MnCO 3 + KClO 3 ® MnO2+ KCl + CO2

Cl V¼ = Cl - I

Mn II¼ = Mn IV

b) compose equations for half-reactions of reduction and oxidation, observing the laws of conservation of the number of atoms and charge in each half-reaction:

half-reaction recovery Cl V + 6 e - = Cl - I

half-reaction oxidation Mn II- 2 e - = Mn IV

c) additional factors are selected for the equation of half-reactions so that the law of conservation of charge is satisfied for the reaction as a whole, for which the number of accepted electrons in the reduction half-reactions is made equal to the number of electrons donated in the oxidation half-reaction:

Cl V + 6 e - = Cl - I 1

Mn II- 2 e - = Mn IV 3

d) insert (using the found factors) stoichiometric coefficients into the reaction scheme (coefficient 1 is omitted):

3 MnCO 3 + KClO 3 = 3 MnO 2 + KCl+CO2

d) equalize the number of atoms of those elements that do not change their oxidation state during the reaction (if there are two such elements, then it is enough to equalize the number of atoms of one of them, and check for the second). The equation for the chemical reaction is obtained:

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

Example 3. Select the coefficients in the equation of the redox reaction

Fe 2 O 3 + CO ® Fe + CO 2

Solution

Fe 2 O 3 + 3 CO = 2 Fe +3 CO 2

Fe III + 3 e - = Fe 0 2

C II - 2 e - = C IV 3

With the simultaneous oxidation (or reduction) of atoms of two elements of one substance, the calculation is carried out for one formula unit of this substance.

Example 4. Select the coefficients in the equation of the redox reaction

Fe(S ) 2 + O 2 = Fe 2 O 3 + SO 2

Solution

4Fe(S ) 2 + 11 O 2 = 2 Fe 2 O 3 + 8 SO 2

Fe II- e - = Fe III

- 11 e - 4

2S - I - 10 e - = 2S IV

O 2 0 + 4 e - = 2O - II+4 e - 11

In examples 3 and 4, the functions of the oxidizing and reducing agent are divided between different substances, Fe 2 O 3 and O 2 - oxidizing agents, CO and Fe(S)2 - reducing agents; Such reactions are classified as intermolecular redox reactions.

When intramolecular oxidation-reduction, when in the same substance the atoms of one element are oxidized and the atoms of another element are reduced, the calculation is carried out per one formula unit of the substance.

Example 5. Select the coefficients in the oxidation-reduction reaction equation

(NH 4) 2 CrO 4 ® Cr 2 O 3 + N 2 + H 2 O + NH 3

Solution

2 (NH 4) 2 CrO 4 = Cr 2 O 3 + N 2 +5 H 2 O + 2 NH 3

CrVI + 3 e - = Cr III 2

2N - III - 6 e - = N 2 0 1

For reactions dismutation (disproportionation, autoxidation- self-healing), in which atoms of the same element in the reagent are oxidized and reduced, additional factors are first added to the right side of the equation, and then the coefficient for the reagent is found.

Example 6. Select the coefficients in the dismutation reaction equation

H2O2 ® H2O+O2

Solution

2 H 2 O 2 = 2 H 2 O + O 2

O - I+ e - = O - II 2

2O - I - 2 e - = O 2 0 1

For the commutation reaction ( synproportionation), in which atoms of the same element of different reagents, as a result of their oxidation and reduction, receive the same oxidation state, additional factors are first put in left side equations

Example 7. Select the coefficients in the commutation reaction equation:

H 2 S + SO 2 = S + H 2 O

Solution

2H2S + SO2 = 3S + 2H2O

S - II - 2 e - = S 0 2

SIV+4 e - = S 0 1

To select coefficients in the equations of redox reactions occurring in an aqueous solution with the participation of ions, the method is used electron-ion balance. The method for selecting coefficients using electron-ion balance consists of the following steps:

a) write down the formulas of the reagents of this redox reaction

K 2 Cr 2 O 7 + H 2 SO 4 + H 2 S

and establish the chemical function of each of them (here K2Cr2O7 - oxidizing agent, H 2 SO 4 - acidic reaction medium, H2S - reducing agent);

b) write down (on the next line) the formulas of the reagents in ionic form, indicating only those ions (for strong electrolytes), molecules (for weak electrolytes and gases) and formula units (for solids), which will take part in the reaction as an oxidizing agent ( Cr2O72 - ), environment ( H+- more precisely, oxonium cation H3O+ ) and reducing agent ( H2S):

Cr2O72 - +H++H2S

c) determine the reduced formula of the oxidizing agent and the oxidized form of the reducing agent, which must be known or specified (for example, here the dichromate ion passes chromium cations ( III), and hydrogen sulfide - into sulfur); This data is written down on the next two lines, the electron-ion equations for the reduction and oxidation half-reactions are drawn up, and additional factors are selected for the half-reaction equations:

half-reaction reduction of Cr 2 O 7 2 - + 14 H + + 6 e - = 2 Cr 3+ + 7 H 2 O 1

half-reaction oxidation of H 2 S - 2 e - = S (t) + 2 H + 3

d) compose, by summing up the half-reaction equations, the ionic equation of a given reaction, i.e. supplement entry (b):

Cr2O72 - + 8 H + + 3 H 2 S = 2 Cr 3+ + 7 H 2 O + 3 S ( T )

d) based on the ionic equation, make up the molecular equation of this reaction, i.e. supplement entry (a), and the formulas of cations and anions that are missing in the ionic equation are grouped into the formulas of additional products ( K2SO4):

K 2 Cr 2 O 7 + 4H 2 SO 4 + 3H 2 S = Cr 2 (SO 4) 3 + 7H 2 O + 3S ( t ) + K 2 SO 4

f) check the selected coefficients by the number of atoms of the elements on the left and right sides of the equation (usually it is enough to only check the number of oxygen atoms).

OxidizedAnd restored The oxidizing and reducing forms often differ in oxygen content (compare Cr2O72 - and Cr 3+ ). Therefore, when compiling half-reaction equations using the electron-ion balance method, they include the pairs H + / H 2 O (for an acidic medium) and OH - / H 2 O (for alkaline environment). If, when moving from one form to another, the original form (usually - oxidized) loses its oxide ions (shown below in square brackets), then the latter, since they do not exist in free form, must be combined with hydrogen cations in an acidic environment, and in an alkaline environment - with water molecules, which leads to the formation of water molecules (in an acidic environment) and hydroxide ions (in an alkaline environment):

acidic environment[ O2 - ] + 2 H + = H 2 O

alkaline environment[ O 2 - ] + H 2 O = 2 OH -

Lack of oxide ions in their original form (usually- in reduced) compared to the final form is compensated by the addition of water molecules (in an acidic environment) or hydroxide ions (in an alkaline environment):

acidic environment H 2 O = [ O 2 - ] + 2 H +

alkaline environment2 OH - = [ O 2 - ] + H 2 O

Example 8. Select the coefficients using the electron-ion balance method in the equation of the redox reaction:

® MnSO 4 + H 2 O + Na 2 SO 4 + ¼

Solution

2 KMnO 4 + 3 H 2 SO 4 + 5 Na 2 SO 3 =

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

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

MnO4 - + 8H + + 5 e - = Mn 2+ + 4 H 2 O2

SO 3 2 - +H2O - 2 e - = SO 4 2 - + 2 H + 5

Example 9. Select the coefficients using the electron-ion balance method in the equation of the redox reaction:

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

Solution

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

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

MnO4 - + 1 e - = MnO 4 2 - 2

SO 3 2 - + 2 OH - - 2 e - = SO 4 2 - + H 2 O 1

If the permanganate ion is used as an oxidizing agent in a weakly acidic environment, then the equation for the reduction half-reaction is:

MnO4 - + 4 H + + 3 e - = MnO 2( t) + 2 H 2 O

and if in a slightly alkaline environment, then

MnO 4 - + 2 H 2 O + 3 e - = MnO 2( t) + 4 OH -

Often, a weakly acidic and slightly alkaline medium is conventionally called neutral, and only water molecules are introduced into the half-reaction equations on the left. In this case, when composing the equation, you should (after selecting additional factors) write down an additional equation reflecting the formation of water from H + and OH ions - .

Example 10. Select the coefficients in the equation of the reaction occurring in a neutral medium:

KMnO 4 + H 2 O + Na 2 SO 3 ® Mn ABOUT 2( t) + Na 2 SO 4 ¼

Solution

2 KMnO 4 + H 2 O + 3 Na 2 SO 3 = 2 MnO 2( t) + 3 Na 2 SO 4 + 2 KOH

MnO4 - + H 2 O + 3 SO 3 2 - = 2 MnO 2( t ) + 3 SO 4 2 - + 2 OH -

MnO 4 - + 2 H 2 O + 3 e - = MnO 2( t) + 4 OH -

SO 3 2 - +H2O - 2 e - = SO 4 2 - +2H+

8OH - + 6 H + = 6 H 2 O + 2 OH -

Thus, if the reaction from example 10 is carried out by simply combining aqueous solutions of potassium permanganate and sodium sulfite, then it proceeds in a conditionally neutral (and in fact, slightly alkaline) environment due to the formation of potassium hydroxide. If the potassium permanganate solution is slightly acidified, the reaction will proceed in a weakly acidic (conditionally neutral) environment.

Example 11. Select the coefficients in the equation of the reaction occurring in a weakly acidic environment:

KMnO 4 + H 2 SO 4 + Na 2 SO 3 ® Mn ABOUT 2( t) + H 2 O + Na 2 SO 4 + ¼

Solution

2KMnO 4 + H 2 SO 4 + 3Na 2 SO 3 = 2Mn O 2( T ) + H 2 O + 3Na 2 SO 4 + K 2 SO 4

2 MnO 4 - + 2 H + + 3 SO 3 2 - = 2 MnO 2( t ) + H 2 O + 3 SO 4 2 -

MnO4 - + 4 H + + 3 e - = Mn O 2( t ) + 2 H 2 O2

SO 3 2 - +H2O - 2 e - = SO 4 2 - + 2 H + 3

Forms of existence of oxidizing agents and reducing agents before and after the reaction, i.e. their oxidized and reduced forms are called redox couples. Thus, from chemical practice it is known (and this must be remembered) that the permanganate ion in an acidic environment forms a manganese cation ( II) (pair MnO 4 - +H+/ Mn 2+ + H 2 O ), in a slightly alkaline environment- manganese(IV) oxide (pair MnO 4 - +H+ ¤ Mn O 2(t) + H 2 O or MnO 4 - + H 2 O = Mn O 2(t) + OH - ). The composition of oxidized and reduced forms is determined, therefore, chemical properties of this element in different oxidation states, i.e. unequal stability of specific forms in different environments of aqueous solution. All redox couples used in this section are given in problems 2.15 and 2.16.

Reactions, which are called redox reactions (ORR), occur with a change in the oxidation states of the atoms contained in the reagent molecules. These changes occur due to the transfer of electrons from atoms of one element to another.

Processes occurring in nature and carried out by humans mostly represent OVR. Such important processes as respiration, metabolism, photosynthesis (6CO2 + H2O = C6H12O6 + 6O2) are all OVR.

In industry, with the help of ORR, sulfuric acid, hydrochloric acid and much more are produced.

The recovery of metals from ores - in fact, the basis of the entire metallurgical industry - is also an oxidation-reduction process. For example, the reaction for producing iron from hematite: 2Fe2O3 + 3C = 4Fe+3CO2.

Oxidizing agents and reducing agents: characteristics

Atoms that donate electrons during a chemical transformation are called reducing agents, and their oxidation state (CO) increases as a result. Atoms that accept electrons are called oxidizing agents and their CO decreases.

They say that oxidizing agents are reduced by accepting electrons, and reducing agents are oxidized by losing electrons.

The most important representatives of oxidizing and reducing agents are presented in the following table:

Based periodic law chemical elements Most often, one can assume the redox abilities of the atoms of a particular element. From the reaction equation it is also easy to understand which atoms are the oxidizing agent and the reducing agent.

How to determine whether an atom is an oxidizing agent or a reducing agent: it is enough to write down CO and understand which atoms increased it during the reaction (reducing agents) and which decreased it (oxidizing agents).

Substances with dual nature

Atoms with intermediate COs are capable of both accepting and donating electrons; as a result, substances containing such atoms in their composition will have the opportunity to act as both an oxidizing agent and a reducing agent.

An example would be hydrogen peroxide. The oxygen contained in CO -1 can either accept an electron or give it away.

When interacting with a reducing agent, peroxide exhibits oxidizing properties, and when interacting with an oxidizing agent, it exhibits reducing properties.

You can take a closer look using the following examples:

• reduction (peroxide acts as an oxidizing agent) when interacting with a reducing agent;

SO2 + H2O2 = H2SO4

O -1 +1e = O -2

• oxidation (peroxide is a reducing agent in this case) when interacting with an oxidizing agent.

2KMnO4 + 5H2O2 + 3H2SO4 = 2MnSO4 + 5O2 + K2SO4 + 8H2O

2О -1 -2е = О2 0

OVR classification: examples

The following types of redox reactions are distinguished:

• intermolecular oxidation-reduction (the oxidizing agent and the reducing agent are contained in different molecules);
• intramolecular oxidation-reduction (the oxidizing agent is part of the same molecule as the reducing agent);
• disproportionation (the oxidizing agent and the reducing agent are an atom of the same element);
• reproportionation (the oxidizing agent and the reducing agent form one product as a result of the reaction).

Examples of chemical transformations related to various types OVR:

• Intramolecular ORRs are most often reactions of thermal decomposition of a substance:

2KCLO3 = 2KCl + 3O2

(NH4)2Cr2O7 = N2 + Cr2O3 + 4H2O

2NaNO3 = 2NaNO2 + O2

• Intermolecular OVR:

3Cu + 8HNO3 = 3Cu(NO3)2 + 2NO + 4H2O

2Al + Fe2O3 = Al2O3 + 2Fe

• Disproportionation reactions:

3Br2 + 6KOH = 5KBr + KBrO3 + 6H2O

3HNO2 = HNO3 + 2NO + H2O

2NO2 + H2O = HNO3 + HNO2

4KClO3 = KCl + 3KClO4

• Reproportionation reactions:

2H2S + SO2 = 3S + 2H2O

HOCl + HCl = H2O + Cl2

Current and non-current OVR

Redox reactions are also divided into current and non-current.

The first case is receiving electrical energy due to a chemical reaction (such energy sources can be used in car engines, in radio devices, control devices), or electrolysis, that is, a chemical reaction, on the contrary, occurs due to electricity (with the help of electrolysis, you can obtain various substances, treat the surfaces of metals and products made from them).

Examples currentless OVR we can name the processes of combustion, corrosion of metals, respiration and photosynthesis, etc.

Electron balance method of ORR in chemistry

Majority equations chemical reactions equalized by simple selection stoichiometric coefficients. However, when selecting coefficients for ORR, you may encounter a situation where the number of atoms of some elements cannot be equalized without violating the equality of the numbers of atoms of others. In the equations of such reactions, coefficients are selected using the electronic balance method.

The method is based on the fact that the sum of electrons accepted by the oxidizing agent and the number given off by the reducing agent is brought to equilibrium.

The method consists of several stages:

1. The reaction equation is written.
2. The reference values ​​of the elements are determined.
3. Elements that have changed their oxidation states as a result of the reaction are determined. The oxidation and reduction half-reactions are recorded separately.
4. The factors for the half-reaction equations are selected so as to equalize the electrons accepted in the reduction half-reaction and the electrons donated in the oxidation half-reaction.
5. The selected coefficients are entered into the reaction equation.
6. The remaining reaction coefficients are selected.

On simple example aluminum interactions with oxygen it is convenient to write the equation step by step:

• Equation: Al + O2 = Al2O3
• CO at atoms in simple substances aluminum and oxygen are equal to 0.

Al 0 + O2 0 = Al +3 2O -2 3

• Let's compose the half-reactions:

Al 0 -3e = Al +3;

O2 0 +4e = 2O -2

• We select coefficients, when multiplied by which the number of electrons received and the number of electrons given will be equal:

Al 0 -3е = Al +3 coefficient 4;

O2 0 +4e = 2O -2 coefficient 3.

• We put the coefficients in the reaction diagram:

4 Al+ 3 O2 = Al2O3

• It can be seen that to equalize the entire reaction, it is enough to put a coefficient in front of the reaction product:

4Al + 3O2 = 2 Al2O3

Examples of tasks for preparing an electronic balance

The following may occur adjustment tasks OVR:

• The interaction of potassium permanganate with potassium chloride in an acidic environment with the release of chlorine gas.

Potassium permanganate KMnO4 (potassium permanganate, “potassium permanganate”) is a strong oxidizing agent due to the fact that in KMnO4 the oxidation state of Mn is +7. It is often used to produce chlorine gas in laboratory conditions according to the following reaction:

KCl + KMnO4 + H2SO4 = Cl2 + MnSO4 + K2SO4 + H2O

K +1 Cl -1 + K +1 Mn +7 O4 -2 + H2 +1 S +6 O4 -2 = Cl2 0 + Mn +2 S +6 O4 -2 + K2 +1 S +6 O4 -2 + H2 +1 O -2

Electronic balance:

As can be seen after the arrangement of CO, chlorine atoms give up electrons, increasing their CO to 0, and manganese atoms accept electrons:

Mn +7 +5е = Mn +2 factor two;

2Cl -1 -2е = Cl2 0 multiplier five.

We enter the coefficients into the equation in accordance with the selected factors:

10 K +1 Cl -1 + 2 K +1 Mn +7 O4 -2 +H2SO4 = 5 Cl2 0 + 2 Mn +2 S +6 O4 -2 + K2SO4 + H2O

We equalize the number of remaining elements:

10KCl + 2KMnO4 + 8 H2SO4 = 5Cl2 + 2MnSO4 + 6 K2SO4+ 8 H2O

• The interaction of copper (Cu) with concentrated nitric acid (HNO3) with the release of gaseous nitric oxide (NO2):

Cu + HNO3(conc.) = NO2 + Cu(NO3)2 + 2H2O

Cu 0 + H +1 N +5 O3 -2 = N +4 O2 + Cu +2 (N +5 O3 -2)2 + H2 +1 O -2

Electronic balance:

As you can see, copper atoms increase their CO from zero to two, and nitrogen atoms decrease from +5 to +4

Cu 0 -2e = Cu +2 factor one;

N +5 +1e = N +4 factor two.

We put the coefficients into the equation:

Cu 0 + 4 H +1 N +5 O3 -2 = 2 N +4 O2 + Cu +2 (N +5 O3 -2)2 + H2 +1 O -2

Cu+ 4 HNO3(conc.) = 2 NO2 + Cu (NO3)2 + 2 H2O

• Interaction of potassium dichromate with H2S in an acidic environment:

Let's write down the reaction scheme and arrange the COs:

K2 +1 Cr2 +6 O7 -2 + H2 +1 S -2 + H2 +1 S +6 O4 -2 = S 0 + Cr2 +3 (S +6 O4 -2)3 + K2 +1 S +6 O4 -2 + H2O

S -2 –2e = S 0 coefficient 3;

2Cr +6 +6e = 2Cr +3 coefficient 1.

Let's substitute:

К2Сr2О7 + 3Н2S + Н2SO4 = 3S + Сr2(SO4)3 + K2SO4 + Н2О

Let's equalize the remaining elements:

К2Сr2О7 + 3Н2S + 4Н2SO4 = 3S + Сr2(SO4)3 + K2SO4 + 7Н2О

Influence of the reaction environment

The nature of the environment influences the course of certain OVRs. The role of the reaction medium can be seen using the example of the interaction of potassium permanganate (KMnO4) and sodium sulfite (Na2SO3) at different meanings pH:

1. Na2SO3 + KMnO4 = Na2SO4 + MnSO4 + K2SO4 (pH<7 кислая среда);
2. Na2SO3 + KMnO4 = Na2SO4 + MnO2 + KOH (pH = 7 neutral environment);
3. Na2SO3 + KMnO4 = Na2SO4 + K2MnO4 + H2O (pH >7 alkaline environment).

It can be seen that a change in the acidity of the medium leads to the formation of different products of the interaction of the same substances. When the acidity of the medium changes, they also occur for other reagents entering the ORR. Similar to the examples shown above, reactions involving the dichromate ion Cr2O7 2- will occur with the formation of different reaction products in different environments:

in an acidic environment the product will be Cr 3+;

in alkaline - CrO2 - , CrO3 3+ ;

in neutral - Cr2O3.

With increasing oxidation state an oxidation process occurs, and the substance itself is a reducing agent. When the oxidation state decreases, a reduction process occurs, and the substance itself is an oxidizing agent.

The described method for equalizing ORR is called the “method of balance by oxidation states.”

Presented in most chemistry textbooks and widely used in practice electronic balance method to equalize ORR can be used with the caveats that the oxidation state is not equal to the charge.

2. Half-reaction method.

In those cases when the reaction occurs in aqueous solution(melt), when drawing up equations, they do not proceed from changes in the oxidation state of the atoms that make up the reacting substances, but from changes in the charges of real particles, that is, they take into account the form of existence of substances in solution (simple or complex ion, atom or molecule of undissolved or weakly dissociated in water substances).

In this case when drawing up ionic equations of redox reactions, one should adhere to the same form of writing that is accepted for ionic equations of an exchange nature, namely: poorly soluble, slightly dissociated and gaseous compounds should be written in molecular form, and ions that do not change their state should be excluded from the equation . In this case, the processes of oxidation and reduction are recorded in the form of separate half-reactions. Having equalized them by the number of atoms of each type, the half-reactions are added, multiplying each by a coefficient that equalizes the change in charge of the oxidizing agent and the reducing agent.

The half-reaction method more accurately reflects the true changes in substances during redox reactions and facilitates the compilation of equations for these processes in ion-molecular form.

Because the from the same reagents different products can be obtained depending on the nature of the medium (acidic, alkaline, neutral); for such reactions in the ionic scheme, in addition to particles that perform the functions of an oxidizing agent and a reducing agent, a particle characterizing the reaction of the medium must be indicated (that is, the H + ion or OH ion - , or H 2 O molecule).

Example 5. Using the half-reaction method, arrange the coefficients in the reaction:

KMnO 4 + KNO 2 + H 2 SO 4 ® MnSO 4 + KNO 3 + K 2 SO 4 + H 2 O.

Solution. We write the reaction in ionic form, taking into account that all substances except water dissociate into ions:

MnO 4 - + NO 2 - + 2H + ® Mn 2+ + NO 3 - + H 2 O

(K + and SO 4 2 - remain unchanged, therefore they are not indicated in the ionic scheme). From the ionic diagram it is clear that the oxidizing agent permanganate ion(MnO 4 -) turns into Mn 2+ ion and four oxygen atoms are released.

In an acidic environment Each oxygen atom released by the oxidizing agent binds to 2H + to form a water molecule.

this implies: MnO 4 - + 8H + + 5® Mn 2+ + 4H 2 O.

We find the difference in the charges of the products and reagents: Dq = +2-7 = -5 (the “-” sign indicates that the reduction process is occurring and 5 is added to the reagents). For the second process, the conversion of NO 2 - into NO 3 -, the missing oxygen comes from the water to the reducing agent, and as a result, an excess of H + ions is formed, in this case the reagents lose 2 :

NO 2 - + H 2 O - 2® NO 3 - + 2H + .

Thus we get:

2 | MnO 4 - + 8H + + 5® Mn 2+ + 4H 2 O (reduction),

5 | NO 2 - + H 2 O - 2® NO 3 - + 2H + (oxidation).

Multiplying the terms of the first equation by 2, and the second by 5 and adding them, we obtain the ionic-molecular equation of this reaction:

2MnO 4 - + 16H + + 5NO 2 - + 5H 2 O = 2Mn 2+ + 8H 2 O + 5NO 3 - + 10H +.

By reducing identical particles on the left and right sides of the equation, we finally obtain the ionic-molecular equation:

2MnO 4 - + 5NO 2 - + 6H + = 2Mn 2+ + 5NO 3 - + 3H 2 O.

Using the ionic equation, we create a molecular equation:

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

In alkaline and neutral environments you can be guided by the following rules: in an alkaline and neutral environment, each oxygen atom released by the oxidizing agent combines with one molecule of water, forming two hydroxide ions (2OH -), and each missing one goes to the reducing agent from 2 OH - ions to form one molecule water in an alkaline environment, and in a neutral environment it comes from water with the release of 2 H + ions.

If participates in the redox reaction hydrogen peroxide(H 2 O 2), the role of H 2 O 2 in a specific reaction must be taken into account. In H 2 O 2 oxygen is in an intermediate oxidation state (-1), therefore hydrogen peroxide exhibits redox duality in redox reactions. In cases where H 2 O 2 is oxidizing agent, the half-reactions have the following form:

H 2 O 2 + 2H + + 2? ® 2H 2 O (acidic environment);

H 2 O 2 +2? ® 2OH - (neutral and alkaline environments).

If hydrogen peroxide is reducing agent:

H 2 O 2 - 2? ® O 2 + 2H + (acidic environment);

H 2 O 2 + 2OH - - 2? ® O 2 + 2H 2 O (alkaline and neutral).

Example 6. Equalize the reaction: KI + H 2 O 2 + H 2 SO 4 ® I 2 + K 2 SO 4 + H 2 O.

Solution. We write the reaction in ionic form:

I - + H 2 O 2 + 2H + ® I 2 + SO 4 2 - + H 2 O.

We compose half-reactions, taking into account that H2O2 in this reaction is an oxidizing agent and the reaction proceeds in an acidic environment:

1 2I - - 2= I 2 ,

1 H 2 O 2 + 2H + + 2® 2H 2 O.

The final equation is: 2KI + H 2 O 2 + H 2 SO 4 ® I 2 + K 2 SO 4 + 2H 2 O.

There are four types of redox reactions:

1 . Intermolecular redox reactions in which the oxidation states of atoms of elements that make up different substances change. The reactions discussed in examples 2-6 belong to this type.

2 . Intramolecular redox reactions in which the oxidation state changes the atoms of different elements of the same substance. Reactions of thermal decomposition of compounds proceed through this mechanism. For example, in the reaction

Pb(NO 3) 2 ® PbO + NO 2 + O 2

changes the oxidation state of nitrogen (N +5 ® N +4) and the oxygen atom (O - 2 ® O 2 0) located inside the Pb(NO 3) 2 molecule.

3. Self-oxidation-self-healing reactions(disproportionation, dismutation). In this case, the oxidation state of the same element both increases and decreases. Disproportionation reactions are characteristic of compounds or elements of substances corresponding to one of the intermediate oxidation states of the element.

Example 7. Using all the above methods, equalize the reaction:

Solution.

A) Oxidation state balance method.

Let us determine the oxidation degrees of the elements involved in the redox process before and after the reaction:

K 2 MnO 4 + H 2 O ® KMnO 4 + MnO 2 + KOH.

From a comparison of oxidation states, it follows that manganese simultaneously participates in the oxidation process, increasing the oxidation state from +6 to +7, and in the reduction process, decreasing the oxidation state from +6 to +4.2 Mn +6 ® Mn +7; Dw = 7-6 = +1 (oxidation process, reducing agent),

1 Mn +6 ® Mn +4 ; Dw = 4-6 = -2 (reduction process, oxidizing agent).

Since in this reaction the oxidizing agent and the reducing agent are the same substance (K 2 MnO 4), the coefficients in front of it are summed up. We write the equation:

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

b) Half-reaction method.

The reaction takes place in a neutral environment. We draw up an ionic reaction scheme, taking into account that H 2 O is a weak electrolyte, and MnO 2 is a poorly soluble oxide in water:

MnO 4 2 - + H 2 O ® MnO 4 - + ¯MnO 2 + OH - .

We write down the half-reactions:

2 MnO 4 2 - - ? ® MnO 4 - (oxidation),

1 MnO 4 2 - + 2H 2 O + 2? ® MnO 2 + 4OH - (reduction).

We multiply by the coefficients and add both half-reactions, we obtain the total ionic equation:

3MnO 4 2 - + 2H 2 O = 2MnO 4 - + MnO 2 + 4OH -.

Molecular equation: 3K 2 MnO 4 + 2H 2 O = 2KMnO 4 + MnO 2 + 4KOH.

In this case, K 2 MnO 4 is both an oxidizing agent and a reducing agent.

4. Intramolecular oxidation-reduction reactions, in which the oxidation states of atoms of the same element are equalized (that is, the reverse of those previously discussed), are processes counter-disproportionation(switching), for example

NH 4 NO 2 ® N 2 + 2H 2 O.

1 2N - 3 - 6? ® N 2 0 (oxidation process, reducing agent),

1 2N +3 + 6?® N 2 0 (reduction process, oxidizing agent).

The most difficult ones are redox reactions in which atoms or ions of not one, but two or more elements are simultaneously oxidized or reduced.

Example 8. Using the above methods, equalize the reaction:

3 -2 +5 +5 +6 +2

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