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 oxidizer accepts during the reaction what it needs before completing the energy level (establishing 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 the tasks proposed in the OGE and the Unified State Exam, it is important to master the algorithm for compiling and analyzing redox processes.
- First of all, charge values are assigned to all elements in the substances proposed in the diagram.
- Atoms (ions) from the left side of the reaction are written out, which during the interaction changed their indicators.
- When the oxidation state increases, the sign “-” is used, and when the oxidation state decreases, “+”.
- The least common multiple (the number by which they are divided without a remainder) is determined between the given and accepted electrons.
- When dividing NOC by electrons, we obtain stereochemical coefficients.
- 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 the oxidation states of each simple and complex substance. 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.
Discussing how to correctly solve redox reactions, we note that the next step will be to compile an electronic relationship:
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 create 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 are of particular importance for the functioning of living organisms. Examples of OVR are the processes of rotting, fermentation, nervous activity, respiration, and metabolism.
Oxidation and reduction are relevant for the metallurgical and chemical industries; 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.
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:
| Typical oxidizing agents | Typical reducing agents |
| Simple substances consisting of elements with high electronegativity (non-metals): iodine, fluorine, chlorine, bromine, oxygen, ozone, sulfur, etc. | Simple substances consisting of atoms of elements with low electronegativity (metals or non-metals): hydrogen H2, carbon C ( graphite), zinc Zn, aluminum Al, calcium Ca, barium Ba, iron Fe, chromium Cr and so on. |
Molecules or ions containing metal or non-metal atoms with high oxidation states:
|
Molecules or ions containing atoms of metals or non-metals with low oxidation states:
|
| Ionic compounds containing cations of some metals with high CO: Pb3+, Au3+, Ag+, Fe3+ and others. | Organic compounds: alcohols, acids, aldehydes, sugars. |
Based on the periodic law of chemical elements, one can most often 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 of ORR:
- 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 the production of electrical energy through a chemical reaction (such energy sources can be used in machine 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
The equations of most chemical reactions can be 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:
- The reaction equation is written.
- The reference values of the elements are determined.
- Elements that have changed their oxidation states as a result of the reaction are determined. The oxidation and reduction half-reactions are recorded separately.
- 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.
- The selected coefficients are entered into the reaction equation.
- The remaining reaction coefficients are selected.
Using a simple example aluminum interactions with oxygen it is convenient to write the equation step by step:
- Equation: Al + O2 = Al2O3
- The COs of 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 the laboratory using 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 +5e = 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.
We 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 medium
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 pH values:
- Na2SO3 + KMnO4 = Na2SO4 + MnSO4 + K2SO4 (pH<7 кислая среда);
- Na2SO3 + KMnO4 = Na2SO4 + MnO2 + KOH (pH = 7 neutral environment);
- 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 a reaction occurs in an 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 a molecule of an undissolved or weakly dissociating substance in water).
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 canceling 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.
18. Redox reactions (continued 1)
18.5. ORR of hydrogen peroxide
In molecules of hydrogen peroxide H 2 O 2, oxygen atoms are in the oxidation state –I. This is an intermediate and not the most stable oxidation state of the atoms of this element, therefore hydrogen peroxide exhibits both oxidizing and reducing properties.
The redox activity of this substance depends on the concentration. In commonly used solutions with a mass fraction of 20%, hydrogen peroxide is a rather strong oxidizing agent; in dilute solutions its oxidizing activity decreases. The reducing properties of hydrogen peroxide are less characteristic than the oxidizing properties and also depend on the concentration.
Hydrogen peroxide is a very weak acid (see Appendix 13), therefore, in strongly alkaline solutions its molecules transform into hydroperoxide ions.
Depending on the reaction of the medium and whether hydrogen peroxide is the oxidizing or reducing agent in this reaction, the products of the redox interaction will be different. The half-reaction equations for all these cases are given in Table 1.
Table 1
Equations of redox half-reactions of H 2 O 2 in solutions
Environment reaction |
H 2 O 2 oxidizing agent |
H 2 O 2 reducing agent |
| Acidic | ||
| Neutral | H 2 O 2 + 2e – = 2OH | H 2 O 2 + 2H 2 O – 2e – = O 2 + 2H 3 O |
| alkaline | HO 2 + H 2 O + 2e – = 3OH |
Let's consider examples of ORR involving hydrogen peroxide.
Example 1. Write an equation for the reaction that occurs when a solution of potassium iodide is added to a solution of hydrogen peroxide acidified with sulfuric acid.
| 1 | H 2 O 2 + 2H 3 O + 2e – = 4H 2 O |
| 1 | 2I – 2e – = I 2 |
H 2 O 2 + 2H 3 O +2I = 4H 2 O + I 2
H 2 O 2 + H 2 SO 4 + 2KI = 2H 2 O + I 2 + K 2 SO 4
Example 2. Write an equation for the reaction between potassium permanganate and hydrogen peroxide in an aqueous solution acidified with sulfuric acid.
| 2 | MnO 4 + 8H 3 O + 5e – = Mn 2 + 12H 2 O |
| 5 | H 2 O 2 + 2H 2 O – 2e – = O 2 + 2H 3 O |
2MnO 4 + 6H 3 O+ + 5H 2 O 2 = 2Mn 2 + 14H 2 O + 5O 2
2KMnO 4 + 3H 2 SO 4 + 5H 2 O 2 = 2MnSO 4 + 8H 2 O + 5O 2 + K 2 SO 4
Example 3. Write an equation for the reaction of hydrogen peroxide with sodium iodide in solution in the presence of sodium hydroxide.
| 3 | 6 | HO 2 + H 2 O + 2e – = 3OH |
| 1 | 2 | I + 6OH – 6e – = IO 3 + 3H 2 O |
3HO 2 + I = 3OH + IO 3
3NaHO 2 + NaI = 3NaOH + NaIO 3
Without taking into account the neutralization reaction between sodium hydroxide and hydrogen peroxide, this equation is often written as follows:
3H 2 O 2 + NaI = 3H 2 O + NaIO 3 (in the presence of NaOH)
The same equation will be obtained if we do not immediately (at the stage of drawing up the balance) take into account the formation of hydroperoxide ions.
Example 4. Write an equation for the reaction that occurs when lead dioxide is added to a solution of hydrogen peroxide in the presence of potassium hydroxide.
Lead dioxide PbO 2 is a very strong oxidizing agent, especially in an acidic environment. Reducing under these conditions, it forms Pb 2 ions. In an alkaline environment, when PbO 2 is reduced, ions are formed.
| 1 | PbO 2 + 2H 2 O + 2e – = + OH |
| 1 | HO 2 + OH – 2e – = O 2 + H 2 O |
PbO 2 + H 2 O + HO 2 = + O 2
Without taking into account the formation of hydroperoxide ions, the equation is written as follows:
PbO 2 + H 2 O 2 + OH = + O 2 + 2H 2 O
If, according to the conditions of the task, the added solution of hydrogen peroxide was alkaline, then the molecular equation should be written as follows:
PbO 2 + H 2 O + KHO 2 = K + O 2
If a neutral solution of hydrogen peroxide is added to a reaction mixture containing an alkali, then the molecular equation can be written without taking into account the formation of potassium hydroperoxide:
PbO 2 + KOH + H 2 O 2 = K + O 2
18.6. ORR dismutation and intramolecular ORR
Among the redox reactions there are dismutation reactions (disproportionation, self-oxidation-self-reduction).
An example of a dismutation reaction known to you is the reaction of chlorine with water:
Cl 2 + H 2 O HCl + HClO
In this reaction, half of the chlorine(0) atoms are oxidized to the +I oxidation state, and the other half are reduced to the –I oxidation state:
Using the electron-ion balance method, let’s compose an equation for a similar reaction that occurs when chlorine is passed through a cold alkali solution, for example KOH:
| 1 | Cl 2 + 2e – = 2Cl |
| 1 | Cl 2 + 4OH – 2e – = 2ClO + 2H 2 O |
2Cl 2 + 4OH = 2Cl + 2ClO + 2H 2 O
All coefficients in this equation have a common divisor, therefore:
Cl 2 + 2OH = Cl + ClO + H 2 O
Cl 2 + 2KOH = KCl + KClO + H 2 O
The dismutation of chlorine in a hot solution proceeds somewhat differently:
| 5 | Cl 2 + 2e – = 2Cl | |
| 1 | Cl 2 + 12OH – 10e – = 2ClO 3 + 6H 2 O |
3Cl 2 + 6OH = 5Cl + ClO 3 + 3H 2 O
3Cl 2 + 6KOH = 5KCl + KClO 3 + 3H 2 O
Of great practical importance is the dismutation of nitrogen dioxide during its reaction with water ( A) and with alkali solutions ( b):
| A) | NO 2 + 3H 2 O – e – = NO 3 + 2H 3 O | NO 2 + 2OH – e – = NO 3 + H 2 O | |||
| NO 2 + H 2 O + e – = HNO 2 + OH | NO 2 + e – = NO 2 | ||||
2NO 2 + 2H 2 O = NO 3 + H 3 O + HNO 2 |
2NO 2 + 2OH = NO 3 + NO 2 + H 2 O |
||||
2NO 2 + H 2 O = HNO 3 + HNO 2 |
2NO 2 + 2NaOH = NaNO 3 + NaNO 2 + H 2 O |
||||
Dismutation reactions occur not only in solutions, but also when heating solids, for example, potassium chlorate:
4KClO 3 = KCl + 3KClO 4
A typical and very effective example of intramolecular ORR is the reaction of thermal decomposition of ammonium dichromate (NH 4) 2 Cr 2 O 7. In this substance, nitrogen atoms are in their lowest oxidation state (–III), and chromium atoms are in their highest (+VI). At room temperature this compound is quite stable, but when heated it intensively decomposes. In this case, chromium(VI) transforms into chromium(III) - the most stable state of chromium, and nitrogen(–III) - into nitrogen(0) - also the most stable state. Taking into account the number of atoms in the formula unit of the electron balance equation:
| 2Cr +VI + 6e – = 2Cr +III | |
| 2N –III – 6e – = N 2, |
and the reaction equation itself:
(NH 4) 2 Cr 2 O 7 = Cr 2 O 3 + N 2 + 4H 2 O.
Another important example of intramolecular ORR is the thermal decomposition of potassium perchlorate KClO 4 . In this reaction, chlorine(VII), as always when it acts as an oxidizing agent, turns into chlorine(–I), oxidizing oxygen(–II) to a simple substance:
| 1 | Cl +VII + 8e – = Cl –I | |
| 2 | 2O –II – 4e – = O 2 |
and therefore the reaction equation
KClO 4 = KCl + 2O 2
Potassium chlorate KClO 3 decomposes similarly when heated, if the decomposition is carried out in the presence of a catalyst (MnO 2): 2KClO 3 = 2KCl + 3O 2
In the absence of a catalyst, a dismutation reaction occurs.
The group of intramolecular redox reactions also includes reactions of thermal decomposition of nitrates.
Typically, the processes that occur when nitrates are heated are quite complex, especially in the case of crystalline hydrates. If water molecules are weakly retained in the crystalline hydrate, then with low heating the nitrate dehydrates [for example, LiNO 3. 3H 2 O and Ca(NO 3) 2 4H 2 O are dehydrated to LiNO 3 and Ca(NO 3) 2 ], but if water is bound more tightly [as, for example, in Mg(NO 3) 2. 6H 2 O and Bi(NO 3) 3. 5H 2 O], then a kind of “intramolecular hydrolysis” reaction occurs with the formation of basic salts - hydroxide nitrates, which upon further heating can turn into oxide nitrates (and (NO 3) 6), the latter decompose to oxides at a higher temperature .
When heated, anhydrous nitrates can decompose into nitrites (if they exist and are still stable at this temperature), and nitrites can decompose into oxides. If heating is carried out to a sufficiently high temperature, or the corresponding oxide is unstable (Ag 2 O, HgO), then the product of thermal decomposition can also be a metal (Cu, Cd, Ag, Hg).
A somewhat simplified diagram of the thermal decomposition of nitrates is shown in Fig. 5.
Examples of sequential transformations that occur when certain nitrates are heated (temperatures are given in degrees Celsius):
KNO 3 KNO 2 K 2 O;
Ca(NO3)2. 4H 2 O Ca(NO 3) 2 Ca(NO 2) 2 CaO;
Mg(NO3)2. 6H 2 O Mg(NO 3)(OH) MgO;
Cu(NO3)2. 6H 2 O Cu(NO 3) 2 CuO Cu 2 O Cu;
Bi(NO 3) 3 . 5H 2 O Bi(NO 3) 2 (OH) Bi(NO 3)(OH) 2 (NO 3) 6 Bi 2 O 3.
Despite the complexity of the processes taking place, when answering the question of what happens when the corresponding anhydrous nitrate is “calcined” (that is, at a temperature of 400 – 500 o C), one is usually guided by the following extremely simplified rules:
1) nitrates of the most active metals (in the series of voltages - to the left of magnesium) decompose to nitrites;
2) nitrates of less active metals (in the range of voltages - from magnesium to copper) decompose to oxides;
3) nitrates of the least active metals (in the series of voltages - to the right of copper) decompose to metal.
When using these rules, it should be remembered that in such conditions
LiNO 3 decomposes to oxide,
Be(NO 3) 2 decomposes to oxide at a higher temperature,
from Ni(NO 3) 2, in addition to NiO, Ni(NO 2) 2 can also be obtained,
Mn(NO 3) 2 decomposes to Mn 2 O 3,
Fe(NO 3) 2 decomposes to Fe 2 O 3;
from Hg(NO 3) 2, in addition to mercury, its oxide can also be obtained.
Let's look at typical examples of reactions belonging to these three types:
KNO 3 KNO 2 + O 2
2KNO 3 = 2KNO 2 + O 2 |
Zn(NO 3) 2 ZnO + NO 2 + O 2
2Zn(NO 3) 2 = 2ZnO + 4NO 2 + O 2 |
AgNO 3 Ag + NO 2 + O 2 18.7. Redox commutation reactions These reactions can be either intermolecular or intramolecular. For example, intramolecular ORRs that occur during the thermal decomposition of ammonium nitrate and nitrite belong to commutation reactions, since here the oxidation state of nitrogen atoms is equalized: NH 4 NO 3 = N 2 O + 2H 2 O (about 200 o C) At a higher temperature (250 - 300 o C) ammonium nitrate decomposes to N 2 and NO, and at an even higher temperature (above 300 o C) - to nitrogen and oxygen, and in both cases water is formed. An example of an intermolecular commutation reaction is the reaction that occurs when hot solutions of potassium nitrite and ammonium chloride are combined: NH 4 + NO 2 = N 2 + 2H 2 O NH 4 Cl + KNO 2 = KCl + N 2 + 2H 2 O If a similar reaction is carried out by heating a mixture of crystalline ammonium sulfate and calcium nitrate, then, depending on the conditions, the reaction can proceed in different ways: (NH 4) 2 SO 4 + Ca(NO 3) 2 = 2N 2 O + 4H 2 O + CaSO 4 (t< 250 o C) The first and third of these reactions are commutation reactions, the second is a more complex reaction, including both the commutation of nitrogen atoms and the oxidation of oxygen atoms. Which reaction will occur at temperatures above 250 o C depends on the ratio of the reagents. Conversion reactions leading to the formation of chlorine occur when salts of oxygen-containing chlorine acids are treated with hydrochloric acid, for example: 6HCl + KClO 3 = KCl + 3Cl 2 + 3H 2 O Also, by the commutation reaction, sulfur is formed from gaseous hydrogen sulfide and sulfur dioxide: 2H 2 S + SO 2 = 3S + 2H 2 O OVR commutations are quite numerous and varied - they even include some acid-base reactions, for example: NaH + H 2 O = NaOH + H 2. To compile ORR commutation equations, both electron-ion and electron balances are used, depending on whether the reaction occurs in solution or not. 18.8. Electrolysis While studying Chapter IX, you became acquainted with the electrolysis of melts of various substances. Since mobile ions are also present in solutions, solutions of various electrolytes can also be subjected to electrolysis. Both in the electrolysis of melts and in the electrolysis of solutions, electrodes made of non-reactive material (graphite, platinum, etc.) are usually used, but sometimes electrolysis is carried out with a “soluble” anode. A “soluble” anode is used in cases where it is necessary to obtain an electrochemical connection of the element from which the anode is made. During electrolysis, it is of great importance whether the anode and cathode spaces are separated, or the electrolyte is mixed during the reaction - the reaction products in these cases may turn out to be different. Let us consider the most important cases of electrolysis. 1. Electrolysis of NaCl melt. The electrodes are inert (graphite), the anode and cathode spaces are separated. As you already know, in this case the following reactions occur at the cathode and anode: K: Na + e – = Na Having written the equations for the reactions occurring on the electrodes in this way, we obtain half-reactions, which we can deal with in exactly the same way as in the case of using the electron-ion balance method:
Adding these half-reaction equations, we obtain the ionic equation of electrolysis 2Na + 2Cl 2Na + Cl 2 and then molecular 2NaCl 2Na + Cl 2 In this case, the cathode and anode spaces must be separated so that the reaction products do not react with each other. Industrially, this reaction is used to produce sodium metal. 2. Electrolysis of the K 2 CO 3 melt. Electrodes are inert (platinum). The cathode and anode spaces are separated.
4K+ + 2CO 3 2 4K + 2CO 2 + O 2 3. Electrolysis of water (H 2 O). The electrodes are inert.
4H 3 O + 4OH 2H 2 + O 2 + 6H 2 O 2H 2 O 2H 2 + O 2 Water is a very weak electrolyte, it contains very few ions, so the electrolysis of pure water proceeds extremely slowly. 4. Electrolysis of CuCl 2 solution. Graphite electrodes. The system contains Cu 2 and H 3 O cations, as well as Cl and OH anions. Cu 2 ions are stronger oxidizers than H 3 O ions (see voltage series), therefore copper ions will be discharged at the cathode first, and only when there are very few of them left will oxonium ions be discharged. For anions, you can follow the following rule: |
