The oxidation state of hno3 is equal. Redox reactions. Oxidative properties of nitric acid

Let's consider the oxidation states of all elements in nitric acid. Oxygen in complex compounds is almost always in the -2 oxidation state (with the exception of peroxides, superoxides, oxygen fluoride, etc.). The hydrogen atom, which is necessarily included in protic acids, has an oxidation state of +1. To determine the oxidation state of a nitrogen atom, you need to solve a simple equation. Let x be the oxidation state of nitrogen, then, according to the principle of electrical neutrality of the molecule, 1 + x + 3 * (-2) = 0, whence x = 5. Answer: the oxidation states of elements in nitric acid are +1, +5, -2 for hydrogen, nitrogen and oxygen, respectively.

In this task you need to determine the oxidation state of the following compound:

Determine the sequence for completing this task

• Write down what oxidation number means;
• Determine the oxidation degree of nitric acid;
• Write down a description.

The oxidation state in this compound is as follows

Oxidation state - auxiliary conventional value for recording the processes of oxidation, reduction and oxidative - reduction reactions. It indicates the oxidation state of an individual atom of a molecule and is only a convenient method of accounting for electron transfer: it is not the true charge of an atom in the molecule.

Ideas about the oxidation state of elements form the basis and are used in classification chemical substances, description of their properties, compilation of formulas of compounds and their international names (nomenclature). But it is especially widely used in the study of redox reactions.

Concept oxidation state often used in inorganic chemistry instead of the concept of valence.

The oxidation number is indicated above the element symbol. Unlike indicating the charge of an ion, when indicating the oxidation state, the sign is given first, and then the numerical value, and not vice versa.

Oxidation number (as opposed to valency) can have zero, negative and positive values, which are usually placed above the element symbol at the top.

The oxidation state of nitric acid is as follows:

HNO3 - oxidation state of hydrogen + 1, oxidation state of nitrogen + 5, oxidation state of oxygen - 2.

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 F 2, oxygen O 2, chlorine Cl 2);
• 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 of oxygen-free acids (K 2 S, NaI);
• some compounds containing cations with minimal positive oxidation state(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.

Several methods for composing redox reactions (ORR) are used: the electron balance method, the electron-ion balance method (half-reaction method) and others.

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 the TOTAL number of electrons given and received is 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 and basic oxides cannot be present among the reaction products, because they react with acid. And, conversely, in an alkaline environment the formation of acid and acid oxide is excluded. 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 nitric acid HNO 3 interacts with reducing agents, a pattern is observed - 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 by degree of grinding of solids . For example, powdered zinc with nitric acid forms some products, while granulated zinc forms completely different ones. The greater the degree of grinding of 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, 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. Under 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 fact is that 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 a chemical reaction is just a conventional notation that does not indicate each process that occurs, but is merely a reflection 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 negative oxidation state to simple substances (with oxidation state 0), exceptions — phosphorus, arsenic - up to +5 ;
• nonmetals with intermediate oxidation state to the highest degree of 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 valence 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), insoluble in a neutral environment amphoteric chromium (III) hydroxide 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 in a negative oxidation state 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, various products are formed during the thermal (temperature) decomposition of nitrates: 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.

At higher temperatures, nitrogen 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 it decomposes to iron (II) oxide; when heated, iron oxidizes to the 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 inactive 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 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 the lower the degree of dilution of the acid, the more we shift to the left.

For example , concentrated acid and inactive metal copper Cu interact. 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 an ordinary 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 is 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

The essence electronic balance method is:

• Calculating the change in oxidation state for each of the elements included in the chemical reaction equation
• Elements whose oxidation state does not change as a result of the reaction are not taken into account
• Of the remaining elements, the oxidation state of which has changed, a balance is drawn up, which consists of calculating the number of electrons acquired or lost
• For all elements that have lost or gained electrons (the number of which differs for each element), find the least common multiple
• The found value is the base coefficients for composing the equation.

Visually, the algorithm for solving the problem using electronic balance method presented in the diagram.

What this looks like in practice is discussed using the example of tasks step by step.

A) Ag + HNO 3 → AgNO 3 + NO + H 2 O
b) Ca + H 2 SO 4 → CaSO 4 + H 2 S + H 2 O
c) Be + HNO 3 → Be(NO 3) 2 + NO + H 2 O

Solution.
To solve this problem, we will use the rules for determining the oxidation state.

Applying the electronic balance method step by step. Example "a"

Let's compose electronic balance for each element of the oxidation reaction Ag + HNO 3 → AgNO 3 + NO + H 2 O.

Step 1. Let's calculate the oxidation states for each element involved in a chemical reaction.

Ag. Silver is initially neutral, that is, it has an oxidation state of zero.

For HNO 3 we determine the oxidation state as the sum of the oxidation states of each element.

The oxidation state of hydrogen is +1, oxygen is -2, therefore, the oxidation state of nitrogen is:

0 - (+1) - (-2)*3 = +5

(in total, again, we get zero, as it should be)

Now let's move on to the second part equations

For AgNO 3, the oxidation state of silver is +1 oxygen -2, therefore the oxidation state of nitrogen is equal to:

0 - (+1) - (-2)*3 = +5

For NO, the oxidation state of oxygen is -2, therefore nitrogen is +2

For H 2 O, the oxidation state of hydrogen is +1, oxygen -2

Step 2. Write the equation in a new form, indicating the oxidation state of each of the elements involved in the chemical reaction.

Ag 0 + H +1 N +5 O -2 3 → Ag +1 N +5 O -2 3 + N +2 O -2 + H +1 2 O -2

From the resulting equation with the indicated oxidation states, we see an imbalance in the sum of positive and negative oxidation states individual elements.

Step 3. Let us write them separately in the form electronic balance- which element and how many electrons it loses or gains:
(It is necessary to take into account that elements whose oxidation state has not changed are not included in this calculation)

Ag 0 - 1e = Ag +1
N +5 +3e = N +2

Silver loses one electron, nitrogen gains three. Thus, we see that for balancing we need to apply a factor of 3 for silver and 1 for nitrogen. Then the number of electrons lost and acquired will be equal.

Step 4. Now, based on the obtained coefficient “3” for silver, we begin to balance the entire equation taking into account the number of atoms participating in the chemical reaction.

• In the initial equation we put a three in front of Ag, which will require the same coefficient in front of AgNO 3
• Now we have an imbalance in the number of nitrogen atoms. There are four of them on the right side, one on the left. Therefore, we put a coefficient of 4 in front of HNO 3
• Now it remains to equalize 4 hydrogen atoms on the left and two on the right. We solve this by applying a factor of 2 in front of H 2 O

Answer:
3Ag + 4HNO3 = 3AgNO3 + NO + 2H2O

Example "b"

Let's compose electronic balance for each element of the oxidation reaction Ca + H 2 SO 4 → CaSO 4 + H 2 S + H 2 O

For H 2 SO 4, the oxidation state of hydrogen is +1 of oxygen -2, whence the oxidation state of sulfur is 0 - (+1)*2 - (-2)*4 = +6

For CaSO 4, the oxidation state of calcium is +2 of oxygen -2, whence the oxidation state of sulfur is 0 - (+2) - (-2)*4 = +6

For H 2 S, the oxidation state of hydrogen is +1, respectively, of sulfur -2

Ca 0 +H +1 2 S +6 O -2 4 → Ca +2 S +6 O -2 4 + H +1 2 S -2 + H +1 2 O -2
Ca 0 - 2e = Ca +2 (factor 4)
S +6 + 8e = S -2

4Ca + 5H 2 SO 4 = 4CaSO 4 + H 2 S + 4H 2 O

Under normal conditions, nitric acid is a colorless liquid (density 1.52 g/cm 3), boiling at 82.6 o C, and at a temperature (-41.6 o C) it solidifies into a transparent crystalline mass. Gross formula - HNO 3. Molar mass - 93 g/mol. The structure of the nitric acid molecule is shown in Fig. 1.

Nitric acid mixes with water in any ratio. It is a strong electrolyte, i.e. V aqueous solution almost completely dissociates into ions. In ORR it manifests itself as an oxidizing agent.

Rice. 1. The structure of the nitric acid molecule, indicating the bond angles between bonds and the lengths of chemical bonds.

HNO3, oxidation states of elements in it

To determine the oxidation states of the elements that make up nitric acid, you first need to understand for which elements this value is precisely known.

The oxidation states of hydrogen and oxygen in the composition of inorganic acids are always equal to (+1) and (-2), respectively. To find the oxidation state of nitrogen, we take its value as “x” and determine it using the electrical neutrality equation:

(+1) + x + 3×(-2) = 0;

1 + x - 6 = 0;

This means the oxidation state of nitrogen in nitric acid is (+5):

H +1 N +5 O -2 3 .

Examples of problem solving

EXAMPLE 1

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