Oxidation state (S.O.) is the conditional charge of an atom in a compound, calculated based on the proposal that the chemical bond is purely ionic. The oxidation state can have a negative, positive and zero value, which is denoted by Arabic numerals with a “+” or “-” sign and placed above the element symbol, for example: Cl 2 0, K + 2 O -2, H + N +5 O -2
Zero degree oxidation have neutral atoms (for example, Mg, H 2, O 2). For a number of elements S.O. atoms in compounds is constant. These include:
Element S.O.
Li, Na, K, Rb, Cs, Fr, H (except hydrides) +1
Be, Mg, Ca, Sr, Ba +2
Halogens in gallides (MeHx –1) and hydrogen in hydrides (MeHx) -1
O (except peroxides) -2
Using this information, you can calculate S.O. other atoms in compounds, taking into account that THE ALGEBRAIC SUM OF OXIDATION DEGREES OF ALL ATOMS IN A COMPOUND IS ALWAYS EQUAL TO ZERO, AND IN A COMPLEX ION – TO THE CHARGE OF THE ION.
For example, in the compounds FeO, NaFeO 2, K 2 FeO 4 C.O. iron atom will be +2, +3, +6, respectively, because S.O. oxygen atom is -2, sodium and potassium - +1, and algebraic sum S.O. of all atoms is zero:
Fe +2 O -2 Na + Fe +3 O 2 -2 K 2 + Fe +6 O 4 -2
2 +(-2)=0 +1+3+2(-2)=0 +1·2+6+4(-2)=0.
Valence defined as a number chemical bonds formed by a given atom in a compound.
In the simplest cases, the valence of an element’s atom is determined by the number of unpaired electrons in it that go to form common electron pairs. For example, in the compounds HC1, CH 4, H 2 S, the valence is: C1 - I, C - IV, S - II, because communication in molecules is carried out due to one, four and two electron pairs at the corresponding atom (a pair of electrons at graphic representation indicated by a dash).
H―Cl, H―S―H, N
EXERCISE 2. Determine the oxidation state of the atom of the underlined element in the given molecules or ions:
K Mn O4, H N O2, P O 4 3- , Cr 2 O 3 . IN last example Also determine the valency of the underlined element (indicate the number of chemical bonds formed by a given atom of the element in the compound).
ANSWER: To determine S.O. underlined elements in the given examples KMnO 4, HNO 2, Cr 2 O 3, we indicate S.O. all other atoms in the compounds and, by calculating their algebraic sum, we find S.O. the underlined element as a value equal and opposite in sign to the algebraic sum S.O. all other atoms.
In K +1 MnO 4 -2 algebraic sum S.O. is equal to -7, therefore S.O. manganese is +7; in H +1 NO 2 -2, reasoning similarly, S.O. nitrogen is +3; in Cr 2 O 3 -2 S.O. chromium is +3.
In the PO 4 3- ion, it is necessary to determine the CO. phosphorus. The sum of all oxidation states of atoms in an ion must be equal to the charge of the ion. Then, denoting S.O. phosphorus through x and multiplying the oxidation state of oxygen (-2) by the number of its atoms, we create the equation
x + (-2) 4 = -3, hence x = +5.
In the Cr 2 O 3 molecule, chromium is trivalent, i.e. the number of chemical bonds formed by chromium is 3:
In the given graphic formula electron pair covalent bond indicated by a dash.
TASK 2(for self-control) Determine S.O. the underlined element in the following examples:
a) K 2 MnO 4, CrO 2 -, SnCl 4; b) HVO 3, CrO 4 2-, CuI 2;
c) Na 2 CO 3, PO 3 3-, Fe 2 (SO 4) 3; d) K 2 Cr 2 O 7, NO 3 -, NH 4 OH;
e) NH 4 NO 3, CO 3 2-, SnCl 4; e) KNO 2, SO 4 2-, FeCO 3;
g) NiSO 4, AlO 2 -, Fe(OH) 3; h) K 2 SO 3, SnO 3 2-, CaF 2;
i) MnO 2, SnO 2 2-, Fe 2 O 3.
In each last example of the task, also determine the valence of the underlined element (indicate the number of chemical bonds formed by a given atom of the element in the compound).
Oxidation-reduction reactions (ORR) are reactions accompanied by a change in the oxidation state of elements due to the displacement or complete transition of electrons from one atom or ion to another. Oxidation – the process of giving up electrons by an atom, molecule or ion; recovery- the process of adding electrons to an atom, molecule or ion. Oxidation and reduction are interrelated processes that occur simultaneously. Oxidizing agents can be atoms and molecules of some non-metals; complex ions and molecules containing atoms of elements in the highest or one of the higher oxidation states MnO 4 -, NO 3 -, SO 4 2-, Cr 2 O 7 2-, ClO 3 -, PbO 2, etc.; positively charged metal ions (Fe 3+, Au 3+, Ag +, Sn 4+, Hg 2+, etc.). Typical reducing agents are almost all metals and some non-metals (C, H 2, etc.) in a free state; negatively charged non-metal ions (S 2-, I -, etc.), cations whose oxidation state may increase (Sn 2+, Fe 2+, Cu +, etc.). If a substance contains an element in an intermediate oxidation state, then, depending on the reaction conditions, it can be both an oxidizing agent and a reducing agent. For example, potassium nitrite in the presence of a strong oxidizing agent exhibits restorative properties, oxidizing to nitrate
3KNO 2 + K 2 Cr 2 O 7 + 4H 2 SO 4 = 3KNO 3 + Cr 2 (SO 4) 3 + K 2 SO 4 + 4H 2 O
When interacting with a reducing agent, on the contrary, it exhibits oxidizing properties
2KNO 2 + 2KI + 2H 2 SO 4 = 2NO + I 2 + 2K 2 SO 4 + 2H 2 O
This redox duality is also characteristic of H 2 O 2, H 2 SO 3 (and its salts), etc.
Drawing up OVR equations.
Method electronic balance is based on a comparison of the oxidation states of atoms in the initial and final substances. It is based on the rule that the number of electrons donated by the reducing agent must be equal to the number of electrons added by the oxidizing agent.
Let's consider this method on the reaction of hydrogen sulfide with potassium permanganate in an acidic environment.
H 2 S + KMnO 4 + H 2 SO 4 = S + MnSO 4 + K 2 SO 4 + H 2 O
Then we determine the change in oxidation states of atoms
H 2 S -2 + KMn +7 O 4 + H 2 SO 4 = S 0 + Mn +2 SO 4 + K 2 SO 4 + H 2 O
This shows that the oxidation state changes for sulfur and manganese
S -2 – 2 e = S 5
Mn +7 + 5 e = Mn +2 2
We find the coefficients for the oxidizing agent and the reducing agent, and then for other reactants. From the electronic equations it is clear that we need to take 5 H 2 S molecules and 2 KMnO 4 molecules, then we get 5 S atoms and 2 MnSO 4 molecules. In addition, from a comparison of atoms on the left and right sides of the equation, we find that 1 molecule of K 2 SO 4 and 8 molecules of water are also formed.
The final reaction equation will be:
5H 2 S + 2KMnO 4 + 3H 2 SO 4 = 5S + 2MnSO 4 + K 2 SO 4 + 8H 2 O
The correctness of the equation is checked by counting the atoms of each element on the left and right sides of the equation.
Laboratory work. Redox reactions Experiment 1. Oxidative properties of kMnO4 in various media.
Place 3 drops of KMnO 4 solution into three conical flasks. Then add 2 drops of 2 N solution to the first one. H 2 SO 4 , in the second - 2 drops of distilled water, in the third - 2 drops of NaOH solution, then add Na 2 SO 3 solution drop by drop into each test tube until the color of the solution changes. How does KMnO 4 behave in acidic, neutral and alkaline environment?
KMnO 4 + Na 2 SO 3 + H 2 SO 4 MnSO 4 + Na 2 SO 4 + K 2 SO 4 + H 2 O
KMnO 4 + Na 2 SO 3 + H 2 O MnO 2 + Na 2 SO 4 + KOH
KMnO 4 + Na 2 SO 3 + NaOH Na 2 MnO 4 + K 2 SO 4 + H 2 O
Experiment 2. Oxidative properties of potassium dichromate .
Pour 3-4 drops of K 2 Cr 2 O 7 solution into two test tubes, add 3-4 drops of 2 N into one of the test tubes. H 2 SO 4 solution, into another - 3-4 drops of 2 N. alkali solution. Notice the change in color of the solution in the second test tube. Add sodium sulfite to all test tubes. Give explanations for the observed phenomena.
K 2 Cr 2 O 7 + H 2 SO 4 + Na 2 SO 3 Cr 2 (SO 4) 3 + K 2 SO 4 + Na 2 SO 4 + H 2 O
Determine the oxidation state for all elements, arrange the coefficients in the reaction equations.
Experiment 3. Reduction of potassium dichromate .
Pour 5-6 drops of potassium dichromate solution into a test tube, add 2-3 drops of sulfuric acid and add a few crystals of potassium sulfide. Shake the contents of the test tube. Observe the color change.
Experiment 4. Redox properties of iron compounds ( III )
Pour 4-5 drops of KMnO 4 solution and 1-2 drops of H 2 SO 4 into a test tube, add iron (II) sulfate solution drop by drop until the solution becomes discolored.
Pour 4-5 drops of ferric chloride solution and 1-2 drops of potassium iodide solution into a test tube. Note the change in color of the solution. Add 1-2 drops of the resulting solution into a test tube with 7-8 drops of starch. Determine the oxidation state for all elements, arrange the coefficients in the reaction equations.
KMnO 4 + FeSO 4 + H 2 SO 4 MnSO 4 + Fe 2 (SO 4) 3 + K 2 SO 4 + H 2 O
FeCl 3 +KI FeCl 2 +KCl+I 2
Experiment 5. Autoxidation and self-healing (disproportionation) of sodium sulfite.
Place 2-3 Na 2 SO 3 crystals into two cylindrical test tubes. Leave one test tube as a control. Secure the second one in a tripod and heat for 5-6 minutes. Allow the test tube to cool. Add 2-3 ml of distilled water to both test tubes and stir with glass rods until the salts in the test tubes dissolve. Add 2-3 ml of CuSO 4 solution to each test tube. Note the color of the sediments in the test tubes. How to explain the difference in color? The black precipitate obtained in the second test tube is copper sulfide. What product of calcination of sodium sulfite with copper sulfate gave this precipitate? Write an equation for the decomposition reaction of sodium sulfite, taking into account that the second product of calcination is sodium sulfite.
Na 2 SO 3 + H 2 O + CuSO 4 H 2 SO 4 + Cu 2 O + NaOH
Determine the degree of oxidation for all elements, arrange the coefficients in the reaction equations.
Experiment 6. Oxidative properties of hydrogen peroxide .
Add 3-4 drops of 2 N H 2 SO 4 to a test tube with 5-6 drops of KI solution and then add H 2 O 2 solution drop by drop until a yellow color appears. To detect iodine in a solution, add a few drops of chloroform or benzene to the test tube. Write an equation for the reaction.
KI + H 2 O 2 + H 2 SO 4 I 2 + H 2 O + K 2 SO 4
Determine the degree of oxidation for all elements, arrange the coefficients in the reaction equations.
Experiment 7. Reductive properties of hydrogen peroxide.
Add 3-4 drops of 2 N H 2 SO 4 and 5-6 drops of hydrogen peroxide to a test tube with 5-6 drops of KMnO 4 and heat. What's happening? Write an equation for the reaction, taking into account that hydrogen peroxide is oxidized to oxygen.
KMnO 4 + H 2 O 2 + H 2 SO 4 MnSO 4 + O 2 + K 2 SO 4 + H 2 O
Determine the degree of oxidation for all elements, arrange the coefficients in the reaction equations.
Experiment 8. Oxidation of copper with nitric acid.
Place a piece of copper wire into the test tube and add 5-6 drops of 0.2 N HNO 3 . Note the evolution of gas, dissolution of copper, and change in color of the solution. Write electronic equations for the reaction, indicating the oxidizing agent and reducing agent. Determine the degree of oxidation for all elements, arrange the coefficients in the reaction equations.
Cu + HNO 3 Cu(NO 3) 2 + NO + H 2 O
Control questions
1.Which of the following reactions are redox reactions:
a) Na 2 CO 3 + SiO 2 = Na 2 SiO 3 + CO 2 b) Fe 2 O 3 + CO = 2FeO + CO 2
c) K 2 Cr 2 O 7 + 2KOH = 2K 2 CrO 4 + H 2 O
2. Determine the oxidizing agent and reducing agent and select the coefficients in the following oxidation-reduction reactions:
a) Na 2 SO 3 + I 2 + H 2 O = Na 2 SO 4 + HI b) S + HNO 3 = H 2 SO 4 + NO
3. Metals have oxidation state +2 in compounds: Cu, Al, Zn, Sn, Pb, Cr, Fe, Mn
4. Metals have an oxidation state of +3 in compounds: Cu, Al, Zn, Sn, Pb, Cr, Fe, Mn
5. Metals have an oxidation state of +1: Cu, Al, Zn, Sn, Pb, Cr, Fe, Mn, Na, Ca, Ag
Oxidizing agents are particles (atoms, molecules or ions) that accept electrons during a chemical reaction. In this case, the oxidation state of the oxidizing agent goes down. Oxidizing agents are being restored.
Restorers are particles (atoms, molecules or ions) that donate electrons during a chemical reaction. In this case, the oxidation state of the reducing agent rises. Reductants in this case oxidize.
Chemicals can be divided into typical oxidizing agents, typical reducing agents, and substances that may exhibit both oxidizing and reducing properties. Some substances exhibit virtually no redox activity.
TO typical oxidizing agents include:
- simple substances-non-metals with the strongest oxidizing properties(fluorine F2, oxygen O2, chlorine Cl2);
- ionsmetals or non-metals With high positive (usually higher) oxidation states : acids (HN +5 O 3, HCl +7 O 4), salts (KN +5 O 3, KMn +7 O 4), oxides (S +6 O 3, Cr +6 O 3)
- compounds containing some metal cations having high oxidation states: Pb 4+, Fe 3+, Au 3+, etc.
Typical reducing agents - this is, as a rule:
- simple substances - metals(the reducing abilities of metals are determined by a number of electrochemical activities);
- complex substances that contain atoms or ions of nonmetals with a negative (usually lowest) oxidation state: binary hydrogen compounds (H 2 S, HBr), salts oxygen-free acids(K 2 S, NaI);
- some compounds containing cations with minimal positive degree oxidation(Sn 2+, Fe 2+, Cr 2+), which, giving up electrons, can increase their oxidation state;
- compounds containing complex ions consisting of nonmetals with an intermediate positive oxidation state(S +4 O 3) 2–, (НР +3 O 3) 2–, in which elements can, by donating electrons, increase its positive oxidation state.
Most other substances may exhibit both oxidizing and reducing properties.
Typical oxidizing and reducing agents are given in the table.
IN laboratory practice the most commonly used are the following oxidizing agents :
potassium permanganate (KMnO 4);
potassium dichromate (K 2 Cr 2 O 7);
nitric acid (HNO 3);
concentrated sulfuric acid(H 2 SO 4);
hydrogen peroxide (H 2 O 2);
oxides of manganese (IV) and lead (IV) (MnO 2, PbO 2);
molten potassium nitrate (KNO 3) and melts of some other nitrates.
TO restoration workers , which apply V laboratory practice relate:
- magnesium (Mg), aluminum (Al), zinc (Zn) and other active metals;
- hydrogen (H 2) and carbon (C);
- potassium iodide (KI);
- sodium sulfide (Na 2 S) and hydrogen sulfide (H 2 S);
- sodium sulfite (Na 2 SO 3);
- tin chloride (SnCl 2).
Classification of redox reactions
Redox reactions are usually divided into four types: intermolecular, intramolecular, disproportionation (auto-oxidation-self-reduction) reactions, and counter-disproportionation reactions.
Intermolecular reactions occur with a change in the oxidation state different elements from different reagents. In this case, various oxidation and reduction products .
2Al 0 + Fe +3 2 O 3 → Al +3 2 O 3 + 2Fe 0,
C 0 + 4HN +5 O 3 (conc) = C +4 O 2 + 4N +4 O 2 + 2H 2 O.
Intramolecular reactions - these are reactions in which different elements from one reagent go to different products, for example:
(N -3 H 4) 2 Cr +6 2 O 7 → N 2 0 + Cr +3 2 O 3 + 4 H 2 O,
2 NaN +5 O -2 3 → 2 NaN +3 O 2 + O 0 2 .
Disproportionation reactions (auto-oxidation-self-healing) are reactions in which the oxidizing agent and the reducing agent are the same element of the same reagent, which then turns into different products:
3Br 2 + 6 KOH → 5KBr + KBrO 3 + 3 H 2 O,
Reproportionation (comproportionation, counter-disproportionation ) are reactions in which the oxidizing agent and the reducing agent are the same element, Which one of different reagents goes into one product. The reaction is the opposite of disproportionation.
2H 2 S -2 + S +4 O 2 = 3S + 2H 2 O
Basic rules for composing redox reactions
Redox reactions are accompanied by oxidation and reduction processes:
Oxidation is the process of donating electrons by a reducing agent.
Recovery is the process of gaining electrons by an oxidizing agent.
Oxidizer is being restored, and the reducing agent oxidizes .
In redox reactions it is observed electronic balance: The number of electrons that the reducing agent gives up is equal to the number of electrons that the oxidizing agent gains.
If the balance sheet is drawn up incorrectly, you will not be able to create complex OVRs.
Let's take a closer look electronic balance method .
It is quite easy to “identify” ORR - it is enough to arrange the oxidation states in all compounds and determine that the atoms change the oxidation state:
K + 2 S -2 + 2K + Mn +7 O -2 4 = 2K + 2 Mn +6 O -2 4 + S 0
We write out separately the atoms of elements that change the oxidation state, in the state BEFORE the reaction and AFTER the reaction.
The oxidation state is changed by manganese and sulfur atoms:
S -2 -2e = S 0
Mn +7 + 1e = Mn +6
Manganese absorbs 1 electron, sulfur gives up 2 electrons. In this case, it is necessary to comply electronic balance. Therefore, it is necessary to double the number of manganese atoms, and leave the number of sulfur atoms unchanged. We indicate balance coefficients both before the reagents and before the products!
Scheme for compiling OVR equations using the electronic balance method:
Attention! There may be several oxidizing or reducing agents in a reaction. The balance must be drawn up so that Total number the electrons given and received were the same.
General patterns of redox reactions
The products of redox reactions often depend on conditions for the process. Let's consider main factors influencing the course of redox reactions.
The most obvious determining factor is reaction solution environment — . Typically (but not necessarily), the substance defining the medium is listed among the reagents. The following options are possible:
- oxidative activity is enhanced in a more acidic environment and the oxidizing agent is reduced more deeply(for example, potassium permanganate, KMnO 4, where Mn +7 in an acidic environment is reduced to Mn +2, and in an alkaline environment - to Mn +6);
- oxidative activity increases in a more alkaline environment, and the oxidizing agent is reduced deeper (for example, potassium nitrate KNO 3, where N +5, when interacting with a reducing agent in an alkaline environment, is reduced to N -3);
- or the oxidizing agent is practically not subject to changes in the environment.
The reaction environment makes it possible to determine the composition and form of existence of the remaining OVR products. The basic principle is that products are formed that do not interact with reagents!
Note! E If the solution medium is acidic, then bases cannot be present among the reaction products and basic oxides, because they react with acid. And, conversely, in an alkaline environment, the formation of acid and acid oxide. This is one of the most common and most serious mistakes.
The direction of the flow of OVR is also affected by nature of the reacting substances. For example, when interacting nitric acid HNO 3 with reducing agents there is a pattern - the greater the activity of the reducing agent, the more nitrogen N +5 is reduced.
When increasing temperature Most ODD tends to be more intense and deeper.
IN heterogeneous reactions the composition of products is often influenced degree of grinding solid . For example, powdered zinc with nitric acid forms some products, while granulated zinc forms completely different ones. How more degree grinding the reagent, the greater its activity, usually.
Let's look at the most typical laboratory oxidizing agents.
Basic schemes of redox reactions
Permanganate recovery scheme
Permanganates contain a powerful oxidizing agent - manganese in oxidation state +7. Manganese salts +7 color the solution in violet color.
Permanganates, depending on the environment of the reaction solution, are reduced in different ways.
IN acidic environment recovery occurs more deeply, up to Mn 2+. Manganese oxide in the +2 oxidation state exhibits basic properties, therefore in acidic environment salt is formed. Manganese salts +2 colorless. IN neutral solution manganese is reduced to oxidation state +4 , with education amphoteric oxide MnO 2 — brown precipitate insoluble in acids and alkalis. IN alkaline environment, manganese is restored minimally - to the nearest oxidation states +6 . Manganese compounds +6 exhibit acidic properties and form salts in an alkaline environment - manganates. Manganates impart to the solution green color .
Let's consider the interaction of potassium permanganate KMnO 4 with potassium sulfide in acidic, neutral and alkaline media. In these reactions, the oxidation product of the sulfide ion is S0.
5 K 2 S + 2 KMnO 4 + 8 H 2 SO 4 = 5 S + 2 MnSO 4 + 6 K 2 SO 4 + 8 H 2 O,
3 K 2 S + 2 KMnO 4 + 4 H 2 O = 2 MnO 2 ↓ + 3 S↓ + 8 KOH,
A common mistake in this reaction is to indicate the interaction of sulfur and alkali in the reaction products. However, sulfur interacts with alkali under rather harsh conditions (elevated temperature), which does not correspond to the conditions of this reaction. At normal conditions It would be correct to indicate molecular sulfur and alkali separately, and not the products of their interaction.
K 2 S + 2 KMnO 4 –(KOH)= 2 K 2 MnO 4 + S↓
Difficulties also arise when composing this reaction. The point is that in in this case writing a molecule of the medium (KOH or another alkali) in the reagents is not required to equalize the reaction. The alkali takes part in the reaction and determines the product of the reduction of potassium permanganate, but the reagents and products are equalized without its participation. This seemingly paradox can be easily resolved if we remember that chemical reaction- this is just a conditional record that does not indicate each ongoing process, but is simply a display of the sum of all processes. How to determine this yourself? If you follow the classical scheme - balance - balance coefficients - metal equalization, then you will see that the metals are equalized by balance coefficients, and the presence of alkali on the left side of the reaction equation will be superfluous.
Permanganates oxidize:
- nonmetals with a negative oxidation state to simple substances (with oxidation state 0), exceptions — phosphorus, arsenic - up to +5 ;
- nonmetals with intermediate oxidation state before highest degree oxidation;
- active metals stable positive degree of oxidation of the metal.
KMnO 4 + neMe (lowest d.o.) = neMe 0 + other products
KMnO 4 + neMe (intermediate d.o.) = neMe (higher d.o.) + other products
KMnO 4 + Me 0 = Me (stable s.o.) + other products
KMnO 4 + P -3 , As -3 = P +5 , As +5 + other products
Chromate/bichromate recovery scheme
A special feature of chromium with valency VI is that it forms 2 types of salts in aqueous solutions: chromates and dichromates, depending on the solution environment. Active metal chromates (for example, K 2 CrO 4) are salts that are stable in alkaline environment. Dichromates (bichromates) of active metals (for example, K 2 Cr 2 O 7) - salts, stable in an acidic environment .
Chromium(VI) compounds are reduced to chromium(III) compounds . Chromium compounds Cr +3 are amphoteric, and depending on the solution environment they exist in solution in various forms: in an acidic environment in the form salts(amphoteric compounds form salts when interacting with acids), in neutral environment- insoluble amphoteric hydroxide chromium (III) Cr(OH) 3 , and in an alkaline environment chromium (III) compounds form complex salt, For example, potassium hexahydroxochromate (III) K 3 .
Chromium VI compounds oxidize:
- nonmetals V negative degree oxidation to simple substances (with oxidation state 0), exceptions — phosphorus, arsenic – up to +5;
- nonmetals in intermediate oxidation state to the highest degree of oxidation;
- active metals from simple substances (oxidation stage 0) to compounds with stable positive degree of oxidation of the metal.
Chromate/bichromate + NeMe (negative d.o.) = NeMe 0 + other products
Chromate/bichromate + neMe (intermediate positive d.o.) = neMe (higher d.o.) + other products
Chromate/bichromate + Me 0 = Me (stable d.o.) + other products
Chromate/bichromate + P, As (negative d.o.) = P, As +5 + other products
Nitrate decomposition
Nitrate salts contain nitrogen in oxidation state +5 - strong oxidizer. Such nitrogen can oxidize oxygen (O -2). This occurs when nitrates are heated. In most cases, oxygen is oxidized to oxidation state 0, i.e. before molecular oxygen O2 .
Depending on the type of metal forming the salt, the thermal (temperature) decomposition of nitrates produces various products: If active metal(in the series of electrochemical activity there are to magnesium), then nitrogen is reduced to the oxidation state +3, and during decomposition nitrite salts and molecular oxygen are formed .
For example:
2NaNO 3 → 2NaNO 2 + O 2 .
Active metals occur in nature in the form of salts (KCl, NaCl).
If a metal is in the series of electrochemical activity to the right of magnesium and to the left of copper (including magnesium and copper) , then upon decomposition it is formed metal oxide in a stable oxidation state, nitric oxide (IV)(brown gas) and oxygen. Metal oxide also forms during decomposition lithium nitrate .
For example, decomposition zinc nitrate:
2Zn(NO 3) 2 → 2ZnО + 4NO 2 + O 2.
Metals of intermediate activity are most often found in nature in the form of oxides (Fe 2 O 3, Al 2 O 3, etc.).
Ions metals, located in the series of electrochemical activity to the right of copper are strong oxidizing agents. At decomposition of nitrates they, like N +5, participate in the oxidation of oxygen and are reduced to simple substances, i.e. metal is formed and gases are released - nitric oxide (IV) and oxygen .
For example, decomposition silver nitrate:
2AgNO3 → 2Ag + 2NO2 + O2.
Inactive metals occur in nature as simple substances.
Some exceptions!
Decomposition ammonium nitrate :
The ammonium nitrate molecule contains both an oxidizing agent and a reducing agent: nitrogen in the -3 oxidation state exhibits only reducing properties, while nitrogen in the +5 oxidation state exhibits only oxidative properties.
When heated, ammonium nitrate decomposes. At temperatures up to 270 o C, it forms nitric oxide (I)(“laughing gas”) and water:
NH 4 NO 3 → N 2 O + 2H 2 O
This is an example of a reaction counter-disproportionation .
The resulting oxidation state of nitrogen is the arithmetic mean of the oxidation state of nitrogen atoms in the original molecule.
With more high temperature Nitric oxide (I) decomposes into simple substances - nitrogen And oxygen:
2NH 4 NO 3 → 2N 2 + O 2 + 4H 2 O
At decomposition ammonium nitrite NH4NO2 counter-disproportionation also occurs.
The resulting oxidation state of nitrogen is also equal to the arithmetic mean of the oxidation states of the initial nitrogen atoms - oxidizing agent N +3 and reducing agent N -3
NH 4 NO 2 → N 2 + 2H 2 O
Thermal decomposition manganese(II) nitrate accompanied by metal oxidation:
Mn(NO 3) 2 = MnO 2 + 2NO 2
Iron(II) nitrate at low temperatures decomposes to iron (II) oxide; when heated, iron oxidizes to oxidation state +3:
2Fe(NO 3) 2 → 2FeO + 4NO 2 + O 2 at 60°C
4Fe(NO 3) 2 → 2Fe 2 O 3 + 8NO 2 + O 2 at >60°C
Nickel(II) nitrate
decomposes to nitrite when heated.
Oxidative properties of nitric acid
Nitric acid HNO 3 when interacting with metals is practically never produces hydrogen , unlike most mineral acids.
This is due to the fact that the acid contains a very strong oxidizing agent - nitrogen in the oxidation state +5. When interacting with reducing agents - metals, various nitrogen reduction products are formed.
Nitric acid + metal = metal salt + nitrogen reduction product + H 2 O
Nitric acid upon reduction can transform into nitrogen oxide (IV) NO 2 (N +4); nitric oxide (II) NO (N +2); nitric oxide (I) N 2 O (“laughing gas”); molecular nitrogen N 2; ammonium nitrate NH 4 NO 3. As a rule, a mixture of products is formed with a predominance of one of them. Nitrogen is reduced to oxidation states from +4 to −3. The depth of restoration depends primarily by nature of a reducing agent And on the concentration of nitric acid . The rule works: the lower the acid concentration and the higher the activity of the metal, the more electrons nitrogen receives, and the more reduced products are formed.
Some regularities will allow you to correctly determine the main product of the reduction of nitric acid by metals in the reaction:
- upon action very dilute nitric acid on metals is usually formed ammonium nitrate NH 4 NO 3;
For example, reaction of zinc with very dilute nitric acid:
4Zn + 10HNO 3 = 4Zn(NO 3) 2 + NH 4 NO 3 + 3H 2 O
- concentrated nitric acid in the cold passivates some metals - chromium Cr, aluminum Al and iron Fe . When the solution is heated or diluted, the reaction occurs;
metal passivation - this is the transfer of the metal surface into an inactive state due to the formation on the metal surface of thin layers of inert compounds, in this case mainly metal oxides that do not react with concentrated nitric acid
- Nitric acid does not react with metals of the platinum subgroup — gold Au, platinum Pt, and palladium Pd;
- when interacting concentrated acid with no active metals And medium activity metals nitrogen acid is reduced to nitric oxide (IV) NO 2 ;
For example, oxidation of copper with concentrated nitric acid:
Cu+ 4HNO 3 = Cu(NO 3) 2 + 2NO 2 + 2H 2 O
- when interacting concentrated nitric acid with active metals is formed Nitric oxide (I)N2O ;
For example, oxidation sodium concentrated nitric acid:
Na+ 10HNO 3 = 8NaNO 3 + N 2 O + 5H 2 O
- when interacting dilute nitric acid with inactive metals (in the activity series to the right of hydrogen) the acid is reduced to nitric oxide (II) NO ;
- when interacting dilute nitric acid with medium activity metals is formed either nitric oxide (II) NO, or nitric oxide N 2 O, or molecular nitrogen N 2 - depending on the additional factors(metal activity, degree of metal grinding, degree of acid dilution, temperature).
- when interacting dilute nitric acid with active metals is formed molecular nitrogen N 2 .
To approximately determine the reduction products of nitric acid when interacting with different metals, I propose to use the pendulum principle. The main factors that shift the position of the pendulum are: acid concentration and metal activity. To simplify, we use 3 types of acid concentrations: concentrated (more than 30%), dilute (30% or less), very dilute (less than 5%). We divide metals according to activity into active (before aluminum), medium activity (from aluminum to hydrogen) and inactive (after hydrogen). We arrange the reduction products of nitric acid in descending order of oxidation state:
NO2; NO; N2O; N 2; NH4NO3
The more active the metal, the more we move to the right. The higher the concentration or less degree dilution of the acid, the more we move to the left.
For example , interact concentrated acid and the inactive metal copper Cu. Consequently, we shift to the extreme left position, nitrogen oxide (IV), copper nitrate and water are formed.
Reaction of metals with sulfuric acid
Dilute sulfuric acid interacts with metals like normal mineral acid. Those. interacts with metals that are located in the series of electrochemical voltages up to hydrogen. The oxidizing agent here is H + ions, which are reduced to molecular hydrogen H 2 . In this case, metals are oxidized, as a rule, to minimum degree of oxidation.
For example:
Fe + H 2 SO 4 (dil) = FeSO 4 + H 2
interacts with metals in the voltage range both before and after hydrogen.
H 2 SO 4 (conc) + metal = metal salt + sulfur reduction product (SO 2, S, H 2 S) + water
When concentrated sulfuric acid interacts with metals, a metal salt (in a stable oxidation state), water and a sulfur reduction product are formed - sulfur dioxide S +4 O 2, molecular sulfur S or hydrogen sulfide H 2 S -2, depending on the degree of concentration, activity of the metal, degree of its grinding, temperature, etc. When concentrated sulfuric acid reacts with metals molecular hydrogen not formed!
Basic principles of interaction of concentrated sulfuric acid with metals:
1. Concentrated sulfuric acid passivates aluminum, chrome, iron at room temperature, or in the cold;
2. Concentrated sulfuric acid doesn't interact With gold, platinum and palladium ;
3. WITH inactive metals concentrated sulfuric acid restored to sulfur(IV) oxide.
For example, copper is oxidized by concentrated sulfuric acid:
Cu 0 + 2H 2 S +6 O 4 (conc) = Cu +2 SO 4 + S +4 O 2 + 2H 2 O
4. When interacting with active metals and zinc concentrated sulfuric acid formssulfur S or hydrogen sulfide H 2 S 2- (depending on temperature, degree of grinding and activity of the metal).
For example , interaction of concentrated sulfuric acid with zinc:
8Na 0 + 5H 2 S +6 O 4 (conc) → 4Na 2 + SO 4 + H 2 S — 2 + 4H 2 O
Hydrogen peroxide
Hydrogen peroxide H 2 O 2 contains oxygen in the oxidation state -1. Such oxygen can both increase and decrease the oxidation state. Thus, hydrogen peroxide exhibits both oxidizing and reducing properties.
When interacting with reducing agents, hydrogen peroxide exhibits the properties of an oxidizing agent and is reduced to an oxidation state of -2. Typically, the product of hydrogen peroxide reduction is water or hydroxide ion, depending on the reaction conditions. For example:
S +4 O 2 + H 2 O 2 -1 → H 2 S +6 O 4 -2
When interacting with oxidizing agents, peroxide is oxidized to molecular oxygen (oxidation state 0): O 2 . For example :
2KMn +7 O 4 + 5H 2 O 2 -1 + 3H 2 SO 4 → 5O 2 0 + 2Mn +2 SO 4 + K 2 SO 4 + 8H 2 O