The concept of oxidizing agent and reducing agent. Atoms or ions that gain electrons in a given reaction are oxidizing agents, and those that donate electrons are reducing agents. Oxygen gives or accepts electrons
Redox reactions – reactions that occur with a change in the oxidation state of elements.
Oxidation- the process of giving up electrons by an atom
Recovery- the process of receiving electrons by an atom
Reducing agent- an element that donates electrons
Oxidizer– element that accepts electrons
For a visual, but simplified idea of the reasons for the change in charges of elements, let us turn to the figures:
An atom is an electrically neutral particle. Therefore the number of protons is equal to the number of electrons
If an element gives up an electron, its charge changes. It becomes positively charged (if it accepts, on the contrary, negatively)
That. the charge of an element is influenced by the number of electrons given or received
I. Drawing up equations for redox reactions
1. Write down the reaction scheme
Na + Cl 2 -> NaCl
2. We arrange the oxidation states of the elements:
Na 0 + Cl 2 0 -> Na + Cl -
3. We write down the elements that have changed the oxidation state and determine the number of electrons given/received:
Na 0 -1e -> Na +
Cl 2 +2e ->2Cl -
4. Find the least common multiple of the numbers of donated and attached electrons:
That. we got the necessary coefficients
5. We set the coefficients:
2Na 0 + Cl 2 0 -> 2Na + Cl —
8. Classification of chemical reactions. OVR. Electrolysis
8.3. Redox reactions: general provisions
Redox reactions(ORR) are reactions that occur with a change in the oxidation state of atoms of elements. As a result of these reactions, some atoms give up electrons, while others accept them.
A reducing agent is an atom, ion, molecule or PU that donates electrons, an oxidizing agent is an atom, ion, molecule or PU that accepts electrons:
The process of giving up electrons is called oxidation, and the process of accepting electrons is called restoration. The OVR must contain a reducing substance and an oxidizing substance. There is no oxidation process without a reduction process and there is no reduction process without an oxidation process.
The reducing agent gives up electrons and is oxidized, and the oxidizing agent accepts electrons and is reduced
The reduction process is accompanied by a decrease in the oxidation state of atoms, and the oxidation process is accompanied by an increase in the oxidation state of atoms of elements. It is convenient to illustrate the above with a diagram (CO - oxidation state):
Specific examples of oxidation and reduction processes (electronic balance diagrams) are given in Table. 8.1.
Table 8.1
Examples of electronic balance schemes
| Electronic balance scheme | Process characteristics |
|---|---|
| Oxidation process | |
 | The calcium atom donates electrons, increases the oxidation state, and is a reducing agent. |
| The Cr +2 ion donates electrons, increases the oxidation state, and is a reducing agent | |
 | A chlorine molecule gives up electrons, chlorine atoms increase the oxidation state from 0 to +1, chlorine is a reducing agent |
| Recovery process | |
| Carbon atom accepts electrons, lowers oxidation number, is an oxidizing agent | |
 | The oxygen molecule accepts electrons, the oxygen atoms reduce the oxidation state from 0 to −2, the oxygen molecule is an oxidizing agent |
| The ion accepts electrons, lowers the oxidation number, is an oxidizing agent | |
The most important reducing agents: simple substances metals; hydrogen; carbon in the form of coke; carbon(II) monoxide; compounds containing atoms in the lowest oxidation state (metal hydrides, sulfides, iodides, ammonia); the strongest reducing agent is electric current at the cathode.
The most important oxidizing agents: simple substances - halogens, oxygen, ozone; concentrated sulfuric acid; Nitric acid; a number of salts (KClO 3, KMnO 4, K 2 Cr 2 O 7); hydrogen peroxide H 2 O 2; the most powerful oxidizing agent is electric current at the anode.
According to the period, the oxidizing properties of atoms and simple substances increase: fluorine - the most powerful oxidizing agent of all simple substances. In each period, halogens form simple substances with the most pronounced oxidizing properties.
In groups A, from top to bottom, the oxidizing properties of atoms and simple substances weaken, and the reducing properties increase.
For atoms of the same type, the reducing properties increase with increasing radius; for example, the reducing properties of the anion
I − are more pronounced than the Cl − anion.
For metals, the redox properties of simple substances and ions in an aqueous solution are determined by the position of the metal in the electrochemical series: from left to right (top to bottom), the reducing properties of simple metals weaken: the most powerful reducing agent- lithium.
For metal ions in an aqueous solution from left to right in the same row, the oxidizing properties increase accordingly: the most powerful oxidizing agent- Au 3 + ions.
To assign coefficients in ORR, you can use a method based on drawing up diagrams of oxidation and reduction processes. This method is called electronic balance method.
The essence of the electronic balance method is as follows.
1. Draw up a reaction scheme and identify the elements that changed the oxidation state.
2. Compose electronic equations for half-reactions of reduction and oxidation.
3. Since the number of electrons donated by the reducing agent must be equal to the number of electrons accepted by the oxidizing agent, additional factors are found using the least common multiple (LCM) method.
4. Additional factors are placed before the formulas of the corresponding substances (coefficient 1 is omitted).
5. The numbers of atoms of those elements that have not changed the oxidation state are equalized (first - hydrogen in water, and then - the number of oxygen atoms).
An example of drawing up an equation for a redox reaction
electronic balance method.
We find that the carbon and sulfur atoms have changed their oxidation state. We compose the equations for the half-reactions of reduction and oxidation:
For this case, the LOC is 4, and the additional factors are 1 (for carbon) and 2 (for sulfuric acid).
We place the additional factors found on the left and right sides of the reaction diagram before the formulas of substances containing carbon and sulfur:
C + 2H 2 SO 4 → CO 2 + 2SO 2 + H 2 O
We equalize the number of hydrogen atoms by putting a coefficient of 2 in front of the water formula, and make sure that the number of oxygen atoms in both sides of the equation is the same. Therefore, the OVR equation
C + 2H 2 SO 4 = CO 2 + 2SO 2 + 2H 2 O
The question arises: in which part of the OVR circuit should the found additional multipliers be placed - on the left or on the right?
For simple reactions this does not matter. However, it should be kept in mind: if additional factors are defined on the left side of the equation, then the coefficients are also placed in front of the formulas of the substances on the left side; if the calculations were carried out for the right side, then the coefficients are placed on the right side of the equation. For example:
Based on the number of Al atoms on the left side:
Based on the number of Al atoms on the right side:
In the general case, if substances of molecular structure (O 2 , Cl 2 , Br 2 , I 2 , N 2 ) are involved in the reaction, then when selecting the coefficients it is based on the number of atoms in the molecule:
If N 2 O is formed in a reaction involving HNO 3, then it is also better to write the electronic balance diagram for nitrogen based on two nitrogen atoms 
.
In some redox reactions, one of the substances can act as both an oxidizing agent (reducing agent) and a salt former (i.e., participate in the formation of salt).
Such reactions are typical, in particular, for the interaction of metals with oxidizing acids (HNO 3, H 2 SO 4 (conc)), as well as oxidizing salts (KMnO 4, K 2 Cr 2 O 7, KClO 3, Ca(OCl) 2) with hydrochloric acid (due to the Cl anions, hydrochloric acid has reducing properties) and other acids, the anion of which is a reducing agent.
Let's create an equation for the reaction of copper with dilute nitric acid:
We see that part of the nitric acid molecules is spent on the oxidation of copper, being reduced to nitrogen oxide(II), and part is used to bind the resulting Cu 2+ ions into the salt Cu(NO 3) 2 (in the composition of the salt, the oxidation state of the nitrogen atom is the same , as in acid, i.e. does not change). In such reactions, an additional factor for the oxidizing element is always placed on the right side before the formula of the reduction product, in this case before the formula NO, and not HNO 3 or Cu(NO 3) 2.
Before the formula HNO 3 we put the coefficient 8 (two HNO 3 molecules are spent on the oxidation of copper and six on the binding of three Cu 2+ ions into salt), we equalize the numbers of H and O atoms and get
3Cu + 8HNO 3 = 3Cu(NO 3) 2 + 2NO + 4H 2 O.
In other cases, an acid, for example hydrochloric acid, can simultaneously be both a reducing agent and participate in the formation of a salt:
Example 8.5. Calculate what mass of HNO 3 is spent on salt formation when the reaction whose equation is
zinc weighing 1.4 g enters.
Solution. From the reaction equation we see that out of 8 moles of nitric acid, only 2 moles went to oxidize 3 moles of zinc (the formula of the acid reduction product, NO, is preceded by a coefficient of 2). 6 moles of acid were consumed for salt formation, which can be easily determined by multiplying the coefficient 3 in front of the salt formula Zn(HNO 3) 2 by the number of acid residues in one formula unit of the salt, i.e. on 2.
n(Zn) = 1.4/65 = 0.0215 (mol).
x = 0.043 mol;
m (HNO 3) = n (HNO 3) M (HNO 3) = 0.043 ⋅ 63 = 2.71 (g)
Answer: 2.71 g.
In some ORRs, the oxidation state is changed by the atoms of not two, but three elements.
Example 8.6. Arrange the coefficients in the ORR flowing according to the scheme FeS + O 2 → Fe 2 O 3 + SO 2 using the electronic balance method.
Solution. We see that the oxidation state is changed by the atoms of three elements: Fe, S and O. In such cases, the numbers of electrons donated by atoms of different elements are summed up:
Arranging the stoichiometric coefficients, we get:
4FeS + 7O 2 = 2Fe 2 O 3 + 4SO 2.
Let's look at examples of solving other types of exam tasks on this topic.
Example 8.7. Indicate the number of electrons transferred from the reducing agent to the oxidizing agent during the complete decomposition of copper(II) nitrate, weighing 28.2 g.
Solution. We write down the equation for the reaction of salt decomposition and the diagram of the electronic balance of the ORR; M = 188 g/mol.
We see that 2 moles of O 2 are formed from the decomposition of 4 moles of salt. In this case, 4 moles of electrons pass from the reducing agent atoms (in this case, ions) to the oxidizing agent (i.e., ions): 
. Since the chemical amount of salt n = 28.2/188 = 0.15 (mol), we have:
2 moles of salt - 4 moles of electrons
0.15 mol - x
n (e) = x = 4 ⋅ 0.15/2 = 0.3 (mol),
N (e) = N A n (e) = 6.02 ⋅ 10 23 ⋅ 0.3 = 1.806 ⋅ 10 23 (electrons).
Answer: 1.806 ⋅ 10 23.
Example 8.8. When sulfuric acid reacted with a chemical amount of 0.02 mol with magnesium, the sulfur atoms added 7.224 ⋅ 10 22 electrons. Find the formula of the acid reduction product.
Solution. In general, the schemes for the reduction of sulfur atoms in sulfuric acid can be as follows:
those. 1 mole of sulfur atoms can accept 2, 6 or 8 moles of electrons. Considering that 1 mole of acid contains 1 mole of sulfur atoms, i.e. n (H 2 SO 4) = n (S), we have:
n (e) = N (e)/N A = (7.224 ⋅ 10 22)/(6.02 ⋅ 10 23) = 0.12 (mol).
We calculate the number of electrons accepted by 1 mole of acid:
0.02 mol of acid accepts 0.12 mol of electrons
1 mol - x
n(e) = x = 0.12/0.02 = 6 (mol).
This result corresponds to the process of reducing sulfuric acid to sulfur:
Answer: sulfur.
Example 8.9. The reaction of carbon with concentrated nitric acid produces water and two salt-forming oxides. Find the mass of carbon that reacted if the oxidizing atoms received 0.2 mol of electrons in this process.
Solution. The interaction of substances proceeds according to the reaction scheme
We compose the equations for the half-reactions of oxidation and reduction:
From the electronic balance diagrams we see that if the oxidizing atoms () accept 4 moles of electrons, then 1 mole (12 g) of carbon enters into the reaction. We compose and solve the proportion:
4 moles of electrons - 12 g of carbon
0.2 - x
x = 0.2 ⋅ 12 4 = 0.6 (g).
Answer: 0.6 g.
Classification of redox reactions
There are intermolecular and intramolecular redox reactions.
When intermolecular ORRs oxidizing and reducing atoms are part of different substances and are atoms of different chemical elements.
When intramolecular ORR oxidizing and reducing atoms are part of the same substance. Intramolecular reactions include disproportionation, in which the oxidizing agent and the reducing agent are atoms of the same chemical element in the composition of the same substance. Such reactions are possible for substances containing atoms with an intermediate oxidation state.
Example 8.10. Specify the OVR disproportionation scheme:
1) MnO 2 + HCl → MnCl 2 + Cl 2 + H 2 O
2) Zn + H 2 SO 4 → ZnSO 4 + H 2
3) KI + Cl 2 → KCl + I 2
4) Cl 2 + KOH → KCl + KClO + H 2 O
Solution . Reactions 1)–3) are intermolecular ORRs:
The disproportionation reaction is reaction 4), since in it the chlorine atom is both an oxidizing agent and a reducing agent:
Answer: 4).
The redox properties of substances can be qualitatively assessed based on an analysis of the oxidation states of atoms in the composition of the substance:
1) if the atom responsible for the redox properties is in the highest oxidation state, then this atom can no longer give up electrons, but can only accept them. Therefore, in the OVR this substance will exhibit only oxidizing properties. Examples of such substances (the formulas indicate the oxidation state of the atom responsible for redox properties):
2) if the atom responsible for the redox properties is in the lowest oxidation state, then this substance will exhibit only restorative properties(this atom can no longer accept electrons, it can only give them away). Examples of such substances: , . Therefore, only reducing properties in ORR are exhibited by all halogen anions (with the exception of F−, for the oxidation of which electric current is used at the anode), the sulfide ion S2−, the nitrogen atom in the ammonia molecule, and the hydride ion H−. Metals (Na, K, Fe) have only reducing properties;
3) if an element’s atom is in an intermediate oxidation state (the oxidation state is greater than the minimum but less than the maximum), then the corresponding substance (ion) will, depending on the conditions, exhibit dual oxidative-restorative properties: stronger oxidizing agents will oxidize these substances (ions), and stronger reducing agents will reduce them. Examples of such substances: sulfur, since the highest oxidation state of the sulfur atom is +6, and the lowest is −2, sulfur oxide(IV), nitrogen oxide(III) (the highest oxidation state of the nitrogen atom is +5, and the lowest is −3), hydrogen peroxide ( the highest oxidation state of the oxygen atom is +2, and the lowest is −2). Metal ions in intermediate oxidation states exhibit dual redox properties: Fe 2+, Mn +4, Cr +3, etc.
Example 8.11. An oxidation-reduction reaction cannot occur, the scheme of which is:
1) Cl 2 + KOH → KCl + KClO 3 + H 2 O
2) S + NaOH → Na 2 S + Na 2 SO 3 + H 2 O
3) KClO → KClO 3 + KClO 4
4) KBr + Cl 2 → KCl + Br
Solution. The reaction whose scheme is indicated under number 3) cannot occur, since it contains a reducing agent, but no oxidizing agent:
Answer: 3).
For some substances, redox duality is due to the presence in their composition of various atoms in both the lowest and highest oxidation states; for example, hydrochloric acid (HCl), due to the hydrogen atom (highest oxidation state equal to +1), is an oxidizing agent, and due to the Cl − anion, it is a reducing agent (lowest oxidation state).
ORR is not possible between substances that exhibit only oxidizing (HNO 3 and H 2 SO 4, KMnO 4 and K 2 CrO 7) or only reducing properties (HCl and HBr, HI and H 2 S)
OVRs are extremely common in nature (metabolism in living organisms, photosynthesis, respiration, decay, combustion), and are widely used by humans for various purposes (obtaining metals from ores, acids, alkalis, ammonia and halogens, creating chemical current sources, obtaining heat and energy when burning various substances). Let us note that OVRs often complicate our lives (spoilage of food, fruits and vegetables, corrosion of metals - all this is associated with the occurrence of various redox processes).
An oxidizing agent and a reducing agent are used to formulate a reaction in organic and inorganic chemistry. Let us consider the main characteristics of such interactions, identify the algorithm for composing the equation and arranging the coefficients.
Definitions
An oxidizing agent is an atom or ion that, when interacting with other elements, accepts electrons. The process of accepting electrons is called reduction, and it is associated with a decrease in the oxidation state.
In the course of inorganic chemistry, two main methods of arranging coefficients are discussed. The reducing agent and oxidizing agent in reactions are determined by drawing up an electronic balance or by the method of half-reactions. Let's take a closer look at the first method of arranging coefficients in the OVR.
Oxidation states
Before determining the oxidizing agent in a reaction, it is necessary to determine the oxidation states of all elements in the substances involved in the transformation. It represents the charge of an atom of an element, calculated according to certain rules. In complex substances, the sum of all positive and negative oxidation states must be equal to zero. For metals of the main subgroups, it corresponds to valence and has a positive value.
For nonmetals, which are located at the end of the formula, the degree is determined by subtracting the group number from eight and has a negative value.
For simple substances it is zero, since there is no process of accepting or giving up electrons.
For complex compounds consisting of several chemical elements, mathematical calculations are used to determine oxidation states.
So, an oxidizing agent is an atom that, in the process of interaction, lowers its oxidation state, and a reducing agent, on the contrary, increases its value.
Examples of OVR
The main feature of tasks related to the arrangement of coefficients in redox reactions is the identification of missing substances and the compilation of their formulas. An oxidizing agent is an element that will accept electrons, but in addition to it, a reducing agent must also participate in the reaction and donate them.
Let us present a generalized algorithm by which you can complete the tasks offered to high school graduates on the unified state exam. Let's look at a few specific examples to understand that an oxidizing agent is not only an element in a complex substance, but also a simple substance.
First, you need to assign oxidation states for each element using certain rules.
Next, you need to analyze the elements that did not participate in the formation of substances and create formulas for them. After all gaps have been eliminated, you can proceed to the process of drawing up an electronic balance between the oxidizing agent and the reducing agent. The resulting coefficients are put into the equation, if necessary adding them in front of those substances that are not included in the balance.
For example, using the electronic balance method, it is necessary to complete the proposed equation and place the necessary coefficients in front of the formulas.
H 2 O 2 + H 2 SO 4 + KMnO 4 = MnSO 4 + O 2 + …+…
To begin with, we determine the values of oxidation states for each, we get
H 2+ O 2 - + H 2+ S +6 O 4 -2 +K + Mn +7 O 4 -2 = Mn +2 S +6 O 4 -2 + O 2 0 + …+…
In the proposed scheme, they change for oxygen, as well as for manganese in potassium permanganate. Thus, we have found a reducing agent and an oxidizing agent. On the right side there is no substance that would contain potassium, so instead of the gaps we will create the formula for its sulfate.
The last action in this task will be the placement of coefficients.
5H 2 O 2 + 3H 2 SO 4 + 2KMnO 4 = 2Mn SO 4 + 5O 2 + 8H 2 O + K 2 SO 4
Acids, potassium permanganate, and hydrogen peroxide can be considered as strong oxidizing agents. All metals exhibit reducing properties, transforming into cations with a positive charge during reactions.
Conclusion
Processes involving the acceptance and donation of negative electrons occur not only in inorganic chemistry. The metabolism that occurs in living organisms is a clear example of the occurrence of redox reactions in organic chemistry. This confirms the significance of the processes considered, their relevance for living and inanimate nature.
Oxidation-reduction reactions, or ORR for short, are one of the fundamentals of the subject of chemistry, as they describe the interaction of individual chemical elements with each other. As the name of these reactions suggests, they involve at least two different chemicals, one of which acts as an oxidizing agent, and the other as a reducing agent. Obviously, it is very important to be able to distinguish and identify them in various chemical reactions.
How to determine an oxidizing agent and a reducing agentThe main difficulty in determining the oxidizing agent and the reducing agent in chemical reactions is that the same substances in different cases can be both oxidizing agents and reducing agents. To learn how to correctly determine the role of a specific chemical element in a reaction, you need to clearly understand the following basic concepts.
- Oxidation is the process of losing electrons from the outer electron layer of a chemical element. In its turn oxidizing agent there will be an atom, molecule or ion that accepts electrons and thereby lowers its oxidation state, which is are being restored . After a chemical reaction of interaction with another substance, the oxidizing agent always acquires a positive charge.
- Recovery is the process of adding electrons to the outer electron layer of a chemical element. Restorer there will be an atom, molecule or ion that donates its electrons and thereby increases its oxidation state, that is oxidize . After a chemical reaction of interaction with another substance, a reducing agent always acquires a positive charge.
- Simply put, an oxidizing agent is a substance that “takes” electrons, and a reducing agent is a substance that gives them to the oxidizing agent. It is possible to determine who in a redox reaction plays the role of an oxidizing agent, who is a reducing agent, and in what cases the oxidizing agent becomes a reducing agent and vice versa, by knowing the typical behavior of individual elements in chemical reactions.
- Typical reducing agents are metals and hydrogen: Fe, K, Ca, Cu, Mg, Na, Zn, H). The less ionized they are, the greater their reducing properties. For example, partially oxidized iron, which has given up one electron and has a charge of +1, will be able to give up one electron less compared to “pure” iron. Also, reducing agents can be compounds of chemical elements in the lowest oxidation state, in which all free orbitals are filled and which can only donate electrons, for example, ammonia NH 3, hydrogen sulfide H 2 S, hydrogen bromide HBr, hydrogen iodide HI, hydrogen chloride HCl.
- Typical oxidizing agents are many nonmetals (F, Cl, I, O, Br). Also, metals with a high degree of oxidation (Fe +3, Sn +4, Mn +4), as well as some compounds of elements with a high degree of oxidation: potassium permanganate KMnO4, sulfuric acid H2SO4, nitric acid HNO3, can act as oxidizing agents. copper oxide CuO, iron chloride FeCl 3.
- Chemical compounds in incomplete or intermediate oxidation states, for example monobasic nitric acid HNO 2, hydrogen peroxide H 2 O 2, sulfurous acid H 2 SO 3 can exhibit both oxidizing and reducing properties depending on the redox properties of the second reagent involved in the interaction .
As follows from this example, one sodium atom gives its electron to one oxygen atom. Therefore, sodium is a reducing agent and oxygen is an oxidizing agent. In this case, sodium will be completely oxidized, since it will give up the maximum possible number of electrons, and the oxygen atom will not be completely reduced, since it will be able to accept another electron from another oxygen atom.
Description
During the redox reaction, the reducing agent gives up electrons, that is, oxidizes; The oxidizing agent gains electrons, that is, is being restored. Moreover, any redox reaction represents the unity of two opposite transformations - oxidation and reduction, occurring simultaneously and without separating one from the other.
Oxidation
Oxidation is the process of losing electrons, with an increase in the degree of oxidation.
When a substance is oxidized, its oxidation state increases as a result of the loss of electrons. The atoms of the substance being oxidized are called electron donors, and the atoms of the oxidizing agent are called electron acceptors.
In some cases, during oxidation, the molecule of the parent substance can become unstable and break down into more stable and smaller constituent parts (see Free radicals). In this case, some of the atoms of the resulting molecules have a higher oxidation state than the same atoms in the original molecule.
The oxidizing agent, accepting electrons, acquires reducing properties, turning into a conjugate reducing agent:
oxidizing agent + e − ↔ conjugate reducing agent.
Recovery
During reduction, atoms or ions gain electrons. In this case, the oxidation state of the element decreases. Examples: reduction of metal oxides to free metals using hydrogen, carbon, and other substances; reduction of organic acids to aldehydes and alcohols; hydrogenation of fats, etc.
The reducing agent, donating electrons, acquires oxidizing properties, turning into a conjugate oxidizing agent:
reducing agent - e − ↔ conjugate oxidizer.
An unbound, free electron is the strongest reducing agent.
Redox couple
An oxidizing agent and its reduced form, or a reducing agent and its oxidized form are conjugate redox couple, and their interconversions are redox half-reactions.
In any redox reaction, two conjugated redox pairs take part, between which there is competition for electrons, as a result of which two half-reactions occur: one is associated with the addition of electrons, i.e. reduction, the other - with the release of electrons, i.e. oxidation.
Types of redox reactions
Intermolecular - reactions in which oxidizing and reducing atoms are located in molecules of different substances, for example:
H 2 S + Cl 2 → S + 2HCl
Intramolecular - reactions in which oxidizing and reducing atoms are located in molecules of the same substance, for example:
2H 2 O → 2H 2 + O 2
Disproportionation (auto-oxidation-self-healing) - reactions in which the same element acts as both an oxidizing agent and a reducing agent, for example:
Cl 2 + H 2 O → HClO + HCl
Reproportionation (conproportionation) - reactions in which one oxidation state is obtained from two different oxidation states of the same element, for example:
NH 4 NO 3 → N 2 O + 2H 2 O
Examples
Redox reaction between hydrogen and fluorine
Divides into two half-reactions:
1) Oxidation:
2) Recovery:
Oxidation, reduction
In redox reactions, electrons are transferred from one atom, molecule, or ion to another. The process of losing electrons is oxidation. During oxidation, the oxidation state increases: