Task book on general and inorganic chemistry
2.2. Redox reactions
See tasks >>>Theoretical part
Redox reactions include chemical reactions, which are accompanied by a change in the oxidation states of the elements. In the equations of such reactions, the selection of coefficients is carried out by compiling electronic balance. The method of selecting coefficients using the electronic balance consists of the following steps:
a) write down the formulas of the reactants and products, and then find the elements that increase and decrease their oxidation states, and write them out separately:
MnCO 3 + KClO 3 ® MnO2+ KCl + CO2
Cl V¼ = Cl - I
Mn II¼ = Mn IV
b) compose the equations of half-reactions of reduction and oxidation, observing the laws of conservation of the number of atoms and charge in each half-reaction:
half reaction recovery Cl V + 6 e - = Cl - I
half reaction oxidation Mn II- 2 e - = Mn IV
c) select additional factors for the equation of half-reactions so that the law of conservation of charge is fulfilled for the reaction as a whole, for which the number of accepted electrons in the reduction half-reactions is made equal to the number of electrons donated in the oxidation half-reaction:
Cl V + 6 e - = Cl - I 1
Mn II- 2 e - = Mn IV 3
d) put down (according to the factors found) stoichiometric coefficients in the reaction scheme (coefficient 1 is omitted):
3 MnCO 3 + KClO 3 = 3 MNO 2 + KCl+CO2
d) equalize the number of atoms of those elements that do not change their oxidation state during the course of the reaction (if there are two such elements, then it is enough to equalize the number of atoms of one of them, and check the second one). Get the equation of the chemical reaction:
3 MnCO 3 + KClO 3 = 3 MNO 2 + KCl+ 3CO2
Example 3. Fit Coefficients in Redox Equation
Fe 2 O 3 + CO ® Fe + CO2
Solution
Fe 2 O 3 + 3 CO \u003d 2 Fe + 3 CO 2
Fe III + 3 e - = Fe 0 2
C II - 2 e - = C IV 3
With the simultaneous oxidation (or reduction) of atoms of two elements of one substance, the calculation is carried out for one formula unit of this substance.
Example 4 Fit Coefficients in Redox Equation
Fe(S ) 2 + O 2 = Fe 2 O 3 + SO 2
Solution
4 Fe(S ) 2 + 11 O 2 = 2 Fe 2 O 3 + 8 SO 2
FeII- e - = Fe III
- 11 e - 4
2S - I - 10 e - = 2SIV
O 2 0 + 4 e - = 2O - II + 4 e - 11
In examples 3 and 4, the functions of the oxidizing and reducing agent are divided between different substances, Fe 2 O 3 and O 2 - oxidizing agents, CO and Fe(S)2 - reducing agents; such reactions are intermolecular redox reactions.
When intramolecular oxidation-reduction, when in the same substance the atoms of one element are oxidized, and the atoms of another element are reduced, the calculation is carried out per one formula unit of the substance.
Example 5 Find the coefficients in the equation of the redox reaction
(NH 4) 2 CrO 4 ® Cr 2 O 3 + N 2 + H 2 O + NH 3
Solution
2 (NH 4) 2 CrO 4 \u003d Cr 2 O 3 + N 2 +5 H 2 O + 2 NH 3
Cr VI + 3 e - = Cr III 2
2N - III - 6 e - = N 2 0 1
For reactions dismutations (disproportionation, autoxidation- self-healing), in which atoms of the same element in the reagent are oxidized and reduced, additional factors are put down first on the right side of the equation, and then the coefficient for the reagent is found.
Example 6. Fit Coefficients in Dismutation Reaction Equation
H2O2 ® H 2 O + O 2
Solution
2 H 2 O 2 \u003d 2 H 2 O + O 2
O - I + e - = O - II 2
2O - I - 2 e - = O 2 0 1
For the commutation reaction ( synproportionation), in which the atoms of the same element of different reagents, as a result of their oxidation and reduction, receive the same oxidation state, additional factors are put down first on the left side of the equation.
Example 7 Select the coefficients in the commutation reaction equation:
H 2 S + SO 2 \u003d S + H 2 O
Solution
2 H 2 S + SO 2 \u003d 3 S + 2H 2 O
S - II - 2 e - = S 0 2
SIV+4 e - = S 0 1
To select coefficients in the equations of redox reactions occurring in an aqueous solution with the participation of ions, the method is used electron-ion balance. The method of selection of coefficients using the electron-ion balance consists of the following steps:
a) write down the formulas of the reagents of this redox reaction
K 2 Cr 2 O 7 + H 2 SO 4 + H 2 S
and establish the chemical function of each of them (here K2Cr2O7 - oxidizing agent, H 2 SO 4 - acid reaction medium, H 2 S - reducing agent);
b) write down (on the next line) the formulas of the reagents in ionic form, indicating only those ions (for strong electrolytes), molecules (for weak electrolytes and gases) and formula units (for solids) that will take part in the reaction as an oxidizing agent ( Cr2O72 - ), environments ( H+- more precisely, the oxonium cation H3O+ ) and reducing agent ( H2S):
Cr2O72 - + H + + H 2 S
c) determine the reduced formula of the oxidizing agent and the oxidized form of the reducing agent, which must be known or specified (for example, here the dichromate ion passes chromium cations ( III), and hydrogen sulfide - into sulfur); these data are written on the next two lines, the electron-ion equations of the reduction and oxidation half-reactions are compiled, and additional factors are selected for the half-reaction equations:
half reaction reduction of Cr 2 O 7 2 - + 14 H + + 6 e - \u003d 2 Cr 3+ + 7 H 2 O 1
half reaction H 2 S oxidation - 2 e - = S(t) + 2H + 3
d) by summing up the equations of half-reactions, they compose the ionic equation of this reaction, i.e. supplement entry (b):
Cr2O72 - + 8 H + + 3 H 2 S = 2 Cr 3+ + 7 H 2 O + 3 S ( T )
d) on the basis of the ionic equation make up the molecular equation of this reaction, i.e. supplement the entry (a), and the formulas of cations and anions that are absent in the ionic equation are grouped into formulas of additional products ( K2SO4):
K 2 Cr 2 O 7 + 4H 2 SO 4 + 3H 2 S \u003d Cr 2 (SO 4) 3 + 7H 2 O + 3S ( m) + K 2 SO 4
f) check the selected coefficients by the number of atoms of elements in the left and right parts of the equation (usually it is enough to check only the number of oxygen atoms).
oxidizedand restored forms of oxidizing and reducing agent often differ in oxygen content (compare Cr2O72 - and Cr3+ ). Therefore, when compiling half-reaction equations using the electron-ion balance method, they include H + / H 2 O pairs (for an acidic environment) and OH - / H 2 O (for an alkaline environment). If during the transition from one form to another, the original form (usually - oxidized) loses its oxide ions (shown below in square brackets), then the latter, since they do not exist in a free form, must be combined with hydrogen cations in an acidic environment, and in an alkaline environment - with water molecules, which leads to the formation of water molecules (in an acidic environment) and hydroxide ions (in an alkaline environment):
acid environment[ O2 - ] + 2 H + = H 2 O
alkaline environment [ O 2 - ] + H 2 O \u003d 2 OH -
Lack of oxide ions in their original form (more often- reduced) in comparison with the final form is compensated by the addition of water molecules (in an acid medium) or hydroxide ions (in an alkaline medium):
acidic environment H 2 O \u003d [ O 2 - ] + 2 H +
alkaline environment2 OH - = [ O 2 - ] + H 2 O
Example 8 Select the coefficients using the electron-ion balance method in the redox reaction equation:
® MnSO 4 + H 2 O + Na 2 SO 4 + ¼
Solution
2 KMnO 4 + 3 H 2 SO 4 + 5 Na 2 SO 3 \u003d
2 MnSO 4 + 3 H 2 O + 5 Na 2 SO 4 + + K 2 SO 4
2 MnO 4 - + 6 H + + 5 SO 3 2 - = 2 Mn 2+ + 3 H 2 O + 5 SO 4 2 -
MnO4 - + 8H + + 5 e - = Mn 2+ + 4 H 2 O2
SO 3 2 - + H2O - 2 e - = SO 4 2 - + 2 H + 5
Example 9. Select the coefficients using the electron-ion balance method in the redox reaction equation:
Na 2 SO 3 + KOH + KMnO 4 ® Na 2 SO 4 + H 2 O + K 2 MnO 4
Solution
Na 2 SO 3 + 2 KOH + 2 KMnO 4 = Na 2 SO 4 + H 2 O + 2 K 2 MnO 4
SO 3 2 - + 2OH - + 2 MnO 4 - = SO 4 2 - + H 2 O + 2 MnO 4 2 -
MnO4 - + 1 e - = MnO 4 2 - 2
SO 3 2 - + 2OH - - 2 e - = SO 4 2 - + H 2 O 1
If the permanganate ion is used as an oxidizing agent in a weakly acidic environment, then the reduction half-reaction equation is:
MnO4 - + 4 H + + 3 e - = MnO 2( m) + 2 H 2 O
and if in a weakly alkaline medium, then
MNO 4 - + 2 H 2 O + 3 e - = MnO 2( m) + 4 OH -
Often, a weakly acidic and weakly alkaline medium is conditionally called neutral, while only water molecules are introduced into the half-reaction equations on the left. In this case, when compiling the equation, one should (after selecting additional factors) write an additional equation that reflects the formation of water from H + and OH ions - .
Example 10. Select the coefficients in the equation for the reaction taking place in a neutral medium:
KMnO 4 + H 2 O + Na 2 SO 3 ® Mn O 2( t) + Na 2 SO 4 ¼
Solution
2 KMnO 4 + H 2 O + 3 Na 2 SO 3 \u003d 2 MnO 2( t) + 3 Na 2 SO 4 + 2 KOH
MnO4 - + H 2 O + 3 SO 3 2 - = 2 MnO 2( m) + 3 SO 4 2 - + 2 OH -
MNO 4 - + 2 H 2 O + 3 e - = MnO 2( m) + 4 OH -
SO 3 2 - + H2O - 2 e - = SO 4 2 - +2H+
8OH - + 6 H + = 6 H 2 O + 2 OH -
Thus, if the reaction from example 10 is carried out by simply draining aqueous solutions of potassium permanganate and sodium sulfite, then it proceeds in a conditionally neutral (and in fact, in a slightly alkaline) environment due to the formation of potassium hydroxide. If the solution of potassium permanganate is slightly acidified, then the reaction will proceed in a weakly acidic (conditionally neutral) medium.
Example 11. Select the coefficients in the equation for the reaction taking place in a weakly acidic environment:
KMnO 4 + H 2 SO 4 + Na 2 SO 3 ® Mn O 2( t) + H 2 O + Na 2 SO 4 + ¼
Solution
2KMnO 4 + H 2 SO 4 + 3Na 2 SO 3 \u003d 2Mn O 2( T ) + H 2 O + 3Na 2 SO 4 + K 2 SO 4
2 MnO 4 - + 2 H + + 3 SO 3 2 - = 2 MnO 2( t) + H 2 O + 3 SO 4 2 -
MnO4 - +4H + + 3 e - = Mn O 2( t ) + 2 H 2 O2
SO 3 2 - + H2O - 2 e - = SO 4 2 - + 2 H + 3
Forms of existence of oxidizing agents and reducing agents before and after the reaction, i.e. their oxidized and reduced forms are called redox couples. So, it is known from chemical practice (and this needs to be remembered) that the permanganate ion in an acidic medium forms a manganese cation ( II ) (pair MNO 4 - + H + / Mn 2+ + H 2 O ), in a weakly alkaline medium- manganese(IV) oxide (pair MNO 4 - +H+ ¤ Mn O 2 (t) + H 2 O or MNO 4 - + H 2 O = Mn O 2(t) + OH - ). The composition of the oxidized and reduced forms is determined, therefore, chemical properties of this element in various oxidation states, i.e. unequal stability of specific forms in various media of an aqueous solution. All redox pairs used in this section are given in problems 2.15 and 2.16.
18. Redox reactions (continued 1)
18.5. OVR hydrogen peroxide
In hydrogen peroxide H 2 O 2 molecules, oxygen atoms are in the –I oxidation state. This is an intermediate and not the most stable oxidation state of the atoms of this element, so 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 ones and also depend on the concentration.
Hydrogen peroxide is a very weak acid (see Appendix 13), therefore, in strongly alkaline solutions, its molecules are converted into hydroperoxide ions.
Depending on the reaction of the medium and on whether the oxidizing or reducing agent is hydrogen peroxide 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 for redox half-reactions of H 2 O 2 in solutions
Environment reaction |
H 2 O 2 oxidizer |
H 2 O 2 reducing agent |
| Acid | ||
| Neutral | H 2 O 2 + 2e - \u003d 2OH | H 2 O 2 + 2H 2 O - 2e - \u003d O 2 + 2H 3 O |
| alkaline | HO 2 + H 2 O + 2e - \u003d 3OH |
Let us consider examples of OVR 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 \u003d 4H 2 O + I 2
H 2 O 2 + H 2 SO 4 + 2KI \u003d 2H 2 O + I 2 + K 2 SO 4
Example 2. Write an equation for the reaction between potassium permanganate and hydrogen peroxide in aqueous solution acidified with sulfuric acid.
| 2 | MnO 4 + 8H 3 O + 5e - \u003d Mn 2 + 12H 2 O |
| 5 | H 2 O 2 + 2H 2 O - 2e - \u003d 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 - \u003d 3OH |
| 1 | 2 | I + 6OH - 6e - \u003d 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 \u003d 3H 2 O + NaIO 3 (in the presence of NaOH)
The same equation will be obtained if the formation of hydroperoxide ions is not taken into account immediately (at the stage of compiling the balance).
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. Recovering 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 - \u003d O 2 + H 2 O |
PbO 2 + H 2 O + HO 2 \u003d + 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 assignment condition, the added hydrogen peroxide solution was alkaline, then the molecular equation should be written as follows:
PbO 2 + H 2 O + KHO 2 \u003d K + O 2
If in reaction mixture containing alkali, a neutral solution of hydrogen peroxide is added, then the molecular equation can be written without taking into account the formation of potassium hydroperoxide:
PbO 2 + KOH + H 2 O 2 \u003d K + O 2
18.6. OVR dismutations and intramolecular OVR
Among the redox reactions are dismutation reactions (disproportionation, self-oxidation-self-healing).
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 is reduced to the –I oxidation state:
Let us use the electron-ion balance method to 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 - \u003d 2Cl |
| 1 | Cl 2 + 4OH - 2e - \u003d 2ClO + 2H 2 O |
2Cl 2 + 4OH = 2Cl + 2ClO + 2H 2 O
All coefficients in this equation have a common divisor, hence:
Cl 2 + 2OH \u003d Cl + ClO + H 2 O
Cl 2 + 2KOH \u003d KCl + KClO + H 2 O
The dismutation of chlorine in a hot solution proceeds somewhat differently:
| 5 | Cl 2 + 2e - \u003d 2Cl | |
| 1 | Cl 2 + 12OH - 10e - \u003d 2ClO 3 + 6H 2 O |
3Cl 2 + 6OH = 5Cl + ClO 3 + 3H 2 O
3Cl 2 + 6KOH \u003d 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 - \u003d NO 3 + 2H 3 O | NO 2 + 2OH - e - \u003d NO 3 + H 2 O | |||
| NO 2 + H 2 O + e - \u003d HNO 2 + OH | NO 2 + e - \u003d NO 2 | ||||
2NO 2 + 2H 2 O \u003d NO 3 + H 3 O + HNO 2 |
2NO 2 + 2OH \u003d NO 3 + NO 2 + H 2 O |
||||
2NO 2 + H 2 O \u003d HNO 3 + HNO 2 |
2NO 2 + 2NaOH \u003d NaNO 3 + NaNO 2 + H 2 O |
||||
Dismutation reactions occur not only in solutions, but also when solids are heated, for example, potassium chlorate:
4KClO 3 \u003d KCl + 3KClO 4
A characteristic and very effective example of intramolecular OVR 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 decomposes rapidly. In this case, chromium(VI) transforms into chromium(III), the most stable state of chromium, while nitrogen(–III) transforms into nitrogen(0), also the most stable state. Taking into account the number of atoms in the formula unit of the electronic balance equation:
| 2Cr + VI + 6e – = 2Cr + III | |
| 2N -III - 6e - \u003d N 2, |
and the reaction equation itself:
(NH 4) 2 Cr 2 O 7 \u003d Cr 2 O 3 + N 2 + 4H 2 O.
Another important example of intramolecular OVR is the thermal decomposition of potassium perchlorate KClO 4 . In this reaction, chlorine(VII), as always, when it acts as an oxidizing agent, passes into chlorine(–I), oxidizing oxygen(–II) to a simple substance:
| 1 | Cl + VII + 8e – = Cl –I | |
| 2 | 2O -II - 4e - \u003d O 2 |
and hence the reaction equation
KClO 4 \u003d KCl + 2O 2
Similarly, potassium chlorate KClO 3 decomposes when heated, if the decomposition is carried out in the presence of a catalyst (MnO 2): 2KClO 3 \u003d 2KCl + 3O 2
In the absence of a catalyst, the dismutation reaction proceeds.
The group of intramolecular OVR also includes reactions of thermal decomposition of nitrates.
Usually, 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 weak heating, dehydration of nitrate occurs [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 ], if the water is more strongly bound [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 at more high temperature decompose to oxides.
Anhydrous nitrates, when heated, can decompose to nitrites (if they exist and are still stable at this temperature), and nitrites can decompose to oxides. If the heating is carried out to a sufficiently high temperature, or the corresponding oxide is unstable (Ag 2 O, HgO), then the metal (Cu, Cd, Ag, Hg) can also be a product of thermal decomposition.
A somewhat simplified scheme of the thermal decomposition of nitrates is shown in fig. 5.
Examples of successive 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(NO 3) 2 . 6H 2 O Cu(NO 3) 2 CuO Cu 2 O Cu;
Bi(NO3)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 ongoing processes, when answering the question of what will happen when the corresponding anhydrous nitrate is "calcined" (that is, at a temperature of 400 - 500 o C), they are 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 a series of voltages - from magnesium to copper) decompose to oxides;
3) nitrates of the least active metals (to the right of copper in the voltage series) 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.
Consider typical examples of reactions related to these three types:
KNO 3 KNO 2 + O 2
2KNO 3 \u003d 2KNO 2 + O 2 |
Zn(NO 3) 2 ZnO + NO 2 + O 2
2Zn(NO 3) 2 \u003d 2ZnO + 4NO 2 + O 2 |
AgNO 3 Ag + NO 2 + O 2 18.7. Redox switching reactions These reactions can be both intermolecular and intramolecular. For example, intramolecular OVR occurring during the thermal decomposition of ammonium nitrate and nitrite belong to commutation reactions, since the degree of oxidation of nitrogen atoms is equalized here: NH 4 NO 3 \u003d 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, in both cases water is formed. An example of an intermolecular switching reaction is the reaction that occurs when hot solutions of potassium nitrite and ammonium chloride are poured: NH 4 + NO 2 \u003d N 2 + 2H 2 O NH 4 Cl + KNO 2 \u003d 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 of the reactions will proceed at a temperature above 250 o C depends on the ratio of the reagents. Switching 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 \u003d KCl + 3Cl 2 + 3H 2 O Also, by the switching reaction, sulfur is formed from gaseous hydrogen sulfide and sulfur dioxide: 2H 2 S + SO 2 \u003d 3S + 2H 2 O OVR commutations are quite numerous and varied - they even include some acid-base reactions, for example: NaH + H 2 O \u003d NaOH + H 2. Both the electron-ionic and electronic balances are used to compile the equations of the OVR commutation, depending on whether a given reaction occurs in a solution or not. 18.8. Electrolysis In 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 are usually used made of a material that does not react (graphite, platinum, etc.), but sometimes electrolysis is also carried out with a "soluble" anode. "Soluble" anode is used in those cases when it is necessary to obtain an electrochemical connection of the element from which the anode is made. During electrolysis, it has great importance 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. 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, reactions take place on the cathode and on the anode: K: Na + e - = Na Having thus written the equations of reactions occurring on the electrodes, we obtain half-reactions with which we can act 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 electrolysis equation 2Na + 2Cl 2Na + Cl2 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. In industry, this reaction is used to produce metallic sodium. 2. Electrolysis of K 2 CO 3 melt. The 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 is extremely slow. 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 oxidizing agents than H 3 O ions (see the series of voltages), therefore, copper ions will first of all be discharged at the cathode, and only when there are very few of them left, oxonium ions will be discharged. For anions, you can follow the following rule: |
Task number 1
Si + HNO 3 + HF → H 2 SiF 6 + NO + ...
N +5 + 3e → N +2 │4 reduction reaction
Si 0 - 4e → Si +4 │3 oxidation reaction
N +5 (HNO 3) - oxidizing agent, Si - reducing agent
3Si + 4HNO 3 + 18HF → 3H 2 SiF 6 + 4NO + 8H 2 O
Task number 2
Using the electron balance method, write the equation for the reaction:
B+ HNO 3 + HF → HBF 4 + NO 2 + …
Determine the oxidizing agent and reducing agent.
N +5 + 1e → N +4 │3 reduction reaction
B 0 -3e → B +3 │1 oxidation reaction
N +5 (HNO 3) - oxidizing agent, B 0 - reducing agent
B+ 3HNO 3 + 4HF → HBF 4 + 3NO 2 + 3H 2 O
Task number 3
Using the electron balance method, write the equation for the reaction:
K 2 Cr 2 O 7 + HCl → Cl 2 + KCl + … + …
Determine the oxidizing agent and reducing agent.
2Cl -1 -2e → Cl 2 0 │3 oxidation reaction
Cr +6 (K 2 Cr 2 O 7) - oxidizing agent, Cl -1 (HCl) - reducing agent
K 2 Cr 2 O 7 + 14HCl → 3Cl 2 + 2KCl + 2CrCl 3 + 7H 2 O
Task number 4
Using the electron balance method, write the equation for the reaction:
Cr 2 (SO 4) 3 + ... + NaOH → Na 2 CrO 4 + NaBr + ... + H 2 O
Determine the oxidizing agent and reducing agent.
Br 2 0 + 2e → 2Br -1 │3 reduction reaction
2Cr +3 - 6e → 2Cr +6 │1 oxidation reaction
Br 2 - oxidizing agent, Cr +3 (Cr 2 (SO 4) 3) - reducing agent
Cr 2 (SO 4) 3 + 3Br 2 + 16NaOH → 2Na 2 CrO 4 + 6NaBr + 3Na 2 SO 4 + 8H 2 O
Task number 5
Using the electron balance method, write the equation for the reaction:
K 2 Cr 2 O 7 + ... + H 2 SO 4 → l 2 + Cr 2 (SO 4) 3 + ... + H 2 O
Determine the oxidizing agent and reducing agent.
2Cr +6 + 6e → 2Cr +3 │1 reduction reaction
2I -1 -2e → l 2 0 │3 oxidation reaction
Cr +6 (K 2 Cr 2 O 7) - oxidizing agent, l -1 (Hl) - reducing agent
K 2 Cr 2 O 7 + 6HI + 4H 2 SO 4 → 3l 2 + Cr 2 (SO 4) 3 + K 2 SO 4 + 7H 2 O
Task number 6
Using the electron balance method, write the equation for the reaction:
H 2 S + HMnO 4 → S + MnO 2 + ...
Determine the oxidizing agent and reducing agent.
3H 2 S + 2HMnO 4 → 3S + 2MnO 2 + 4H 2 O
Task number 7
Using the electron balance method, write the equation for the reaction:
H 2 S + HClO 3 → S + HCl + ...
Determine the oxidizing agent and reducing agent.
S -2 -2e → S 0 │3 oxidation reaction
Mn +7 (HMnO 4) - oxidizing agent, S -2 (H 2 S) - reducing agent
3H 2 S + HClO 3 → 3S + HCl + 3H 2 O
Task number 8
Using the electron balance method, write the equation for the reaction:
NO + HClO 4 + ... → HNO 3 + HCl
Determine the oxidizing agent and reducing agent.
Cl +7 + 8e → Cl -1 │3 reduction reaction
N +2 -3e → N +5 │8 oxidation reaction
Cl +7 (HClO 4) - oxidizing agent, N +2 (NO) - reducing agent
8NO + 3HClO 4 + 4H 2 O → 8HNO 3 + 3HCl
Task number 9
Using the electron balance method, write the equation for the reaction:
KMnO 4 + H 2 S + H 2 SO 4 → MnSO 4 + S + ... + ...
Determine the oxidizing agent and reducing agent.
S -2 -2e → S 0 │5 oxidation reaction
Mn +7 (KMnO 4) - oxidizing agent, S -2 (H 2 S) - reducing agent
2KMnO 4 + 5H 2 S + 3H 2 SO 4 → 2MnSO 4 + 5S + K 2 SO 4 + 8H 2 O
Task number 10
Using the electron balance method, write the equation for the reaction:
KMnO 4 + KBr + H 2 SO 4 → MnSO 4 + Br 2 + ... + ...
Determine the oxidizing agent and reducing agent.
Mn +7 + 5e → Mn +2 │2 reduction reaction
2Br -1 -2e → Br 2 0 │5 oxidation reaction
Mn +7 (KMnO 4) - oxidizing agent, Br -1 (KBr) - reducing agent
2KMnO 4 + 10KBr + 8H 2 SO 4 → 2MnSO 4 + 5Br 2 + 6K 2 SO 4 + 8H 2 O
Task number 11
Using the electron balance method, write the equation for the reaction:
PH 3 + HClO 3 → HCl + ...
Determine the oxidizing agent and reducing agent.
Cl +5 + 6e → Cl -1 │4 reduction reaction
Cl +5 (HClO 3) - oxidizing agent, P -3 (H 3 PO 4) - reducing agent
3PH 3 + 4HClO 3 → 4HCl + 3H 3 PO 4
Task number 12
Using the electron balance method, write the equation for the reaction:
PH 3 + HMnO 4 → MnO 2 + … + …
Determine the oxidizing agent and reducing agent.
Mn +7 + 3e → Mn +4 │8 reduction reaction
P -3 - 8e → P +5 │3 oxidation reaction
Mn +7 (HMnO 4) - oxidizing agent, P -3 (H 3 PO 4) - reducing agent
3PH 3 + 8HMnO 4 → 8MnO 2 + 3H 3 PO 4 + 4H 2 O
Task number 13
Using the electron balance method, write the equation for the reaction:
NO + KClO + … → KNO 3 + KCl + …
Determine the oxidizing agent and reducing agent.
Cl +1 + 2e → Cl -1 │3 reduction reaction
N +2 − 3e → N +5 │2 oxidation reaction
Cl +1 (KClO) - oxidizing agent, N +2 (NO) - reducing agent
2NO + 3KClO + 2KOH → 2KNO 3 + 3KCl + H 2 O
Task number 14
Using the electron balance method, write the equation for the reaction:
PH 3 + AgNO 3 + ... → Ag + ... + HNO 3
Determine the oxidizing agent and reducing agent.
Ag +1 + 1e → Ag 0 │8 reduction reaction
P -3 - 8e → P +5 │1 oxidation reaction
Ag +1 (AgNO 3) - oxidizing agent, P -3 (PH 3) - reducing agent
PH 3 + 8AgNO 3 + 4H 2 O → 8Ag + H 3 PO 4 + 8HNO 3
Task number 15
Using the electron balance method, write the equation for the reaction:
KNO 2 + ... + H 2 SO 4 → I 2 + NO + ... + ...
Determine the oxidizing agent and reducing agent.
N +3 + 1e → N +2 │ 2 reduction reaction
2I -1 - 2e → I 2 0 │ 1 oxidation reaction
N +3 (KNO 2) - oxidizing agent, I -1 (HI) - reducing agent
2KNO 2 + 2HI + H 2 SO 4 → I 2 + 2NO + K 2 SO 4 + 2H 2 O
Task number 16
Using the electron balance method, write the equation for the reaction:
Na 2 SO 3 + Cl 2 + ... → Na 2 SO 4 + ...
Determine the oxidizing agent and reducing agent.
Cl 2 0 + 2e → 2Cl -1 │1 reduction reaction
Cl 2 0 - oxidizing agent, S +4 (Na 2 SO 3) - reducing agent
Na 2 SO 3 + Cl 2 + H 2 O → Na 2 SO 4 + 2HCl
Task number 17
Using the electron balance method, write the equation for the reaction:
KMnO 4 + MnSO 4 + H 2 O → MnO 2 + ... + ...
Determine the oxidizing agent and reducing agent.
Mn +7 + 3e → Mn +4 │2 reduction reaction
Mn +2 − 2e → Mn +4 │3 oxidation reaction
Mn +7 (KMnO 4) - oxidizing agent, Mn +2 (MnSO 4) - reducing agent
2KMnO 4 + 3MnSO 4 + 2H 2 O → 5MnO 2 + K 2 SO 4 + 2H 2 SO 4
Task number 18
Using the electron balance method, write the equation for the reaction:
KNO 2 + ... + H 2 O → MnO 2 + ... + KOH
Determine the oxidizing agent and reducing agent.
Mn +7 + 3e → Mn +4 │2 reduction reaction
N +3 − 2e → N +5 │3 oxidation reaction
Mn +7 (KMnO 4) - oxidizing agent, N +3 (KNO 2) - reducing agent
3KNO 2 + 2KMnO 4 + H 2 O → 2MnO 2 + 3KNO 3 + 2KOH
Task #19
Using the electron balance method, write the equation for the reaction:
Cr 2 O 3 + ... + KOH → KNO 2 + K 2 CrO 4 + ...
Determine the oxidizing agent and reducing agent.
N +5 + 2e → N +3 │3 reduction reaction
2Cr +3 − 6e → 2Cr +6 │1 oxidation reaction
N +5 (KNO 3) - oxidizing agent, Cr +3 (Cr 2 O 3) - reducing agent
Cr 2 O 3 + 3KNO 3 + 4KOH → 3KNO 2 + 2K 2 CrO 4 + 2H 2 O
Task number 20
Using the electron balance method, write the equation for the reaction:
I 2 + K 2 SO 3 + ... → K 2 SO 4 + ... + H 2 O
Determine the oxidizing agent and reducing agent.
I 2 0 + 2e → 2I -1 │1 reduction reaction
S +4 - 2e → S +6 │1 oxidation reaction
I 2 - oxidizing agent, S +4 (K 2 SO 3) - reducing agent
I 2 + K 2 SO 3 + 2KOH → K 2 SO 4 + 2KI + H 2 O
Task number 21
Using the electron balance method, write the equation for the reaction:
KMnO 4 + NH 3 → MnO 2 +N 2 + ... + ...
Determine the oxidizing agent and reducing agent.
Mn +7 + 3e → Mn +4 │2 reduction reaction
2N -3 - 6e → N 2 0 │1 oxidation reaction
Mn +7 (KMnO 4) - oxidizing agent, N -3 (NH 3) - reducing agent
2KMnO 4 + 2NH 3 → 2MnO 2 + N 2 + 2KOH + 2H 2 O
Task #22
Using the electron balance method, write the equation for the reaction:
NO 2 + P 2 O 3 + ... → NO + K 2 HPO 4 + ...
Determine the oxidizing agent and reducing agent.
N +4 + 2e → N +2 │2 reduction reaction
2P +3 - 4e → 2P +5 │1 oxidation reaction
N +4 (NO 2) - oxidizing agent, P +3 (P 2 O 3) - reducing agent
2NO 2 + P 2 O 3 + 4KOH → 2NO + 2K 2 HPO 4 + H 2 O
Task #23
Using the electron balance method, write the equation for the reaction:
KI + H 2 SO 4 → I 2 + H 2 S + … + …
Determine the oxidizing agent and reducing agent.
S +6 + 8e → S -2 │1 reduction reaction
2I -1 - 2e → I 2 0 │4 oxidation reaction
S +6 (H 2 SO 4) - oxidizing agent, I -1 (KI) - reducing agent
8KI + 5H 2 SO 4 → 4I 2 + H 2 S + 4K 2 SO 4 + 4H 2 O
Task #24
Using the electron balance method, write the equation for the reaction:
FeSO 4 + ... + H 2 SO 4 → ... + MnSO 4 + K 2 SO 4 + H 2 O
Determine the oxidizing agent and reducing agent.
Mn +7 + 5e → Mn +2 │2 reduction reaction
2Fe +2 − 2e → 2Fe +3 │5 oxidation reaction
Mn +7 (KMnO 4) - oxidizing agent, Fe +2 (FeSO 4) - reducing agent
10FeSO 4 + 2KMnO 4 + 8H 2 SO 4 → 5Fe 2 (SO 4) 3 + 2MnSO 4 + K 2 SO 4 + 8H 2 O
Task #25
Using the electron balance method, write the equation for the reaction:
Na 2 SO 3 + ... + KOH → K 2 MnO 4 + ... + H 2 O
Determine the oxidizing agent and reducing agent.
Mn +7 + 1e → Mn +6 │2 reduction reaction
S +4 − 2e → S +6 │1 oxidation reaction
Mn +7 (KMnO 4) - oxidizing agent, S +4 (Na 2 SO 3) - reducing agent
Na 2 SO 3 + 2KMnO 4 + 2KOH → 2K 2 MnO 4 + Na 2 SO 4 + H 2 O
Task #26
Using the electron balance method, write the equation for the reaction:
H 2 O 2 + ... + H 2 SO 4 → O 2 + MnSO 4 + ... + ...
Determine the oxidizing agent and reducing agent.
Mn +7 + 5e → Mn +2 │2 reduction reaction
2O -1 - 2e → O 2 0 │5 oxidation reaction
Mn +7 (KMnO 4) - oxidizing agent, O -1 (H 2 O 2) - reducing agent
5H 2 O 2 + 2KMnO 4 + 3H 2 SO 4 → 5O 2 + 2MnSO 4 + K 2 SO 4 + 8H 2 O
Task number 27
Using the electron balance method, write the equation for the reaction:
K 2 Cr 2 O 7 + H 2 S + H 2 SO 4 → Cr 2 (SO 4) 3 + K 2 SO 4 + ... + ...
Determine the oxidizing agent and reducing agent.
2Cr +6 + 6e → 2Cr +3 │1 reduction reaction
S -2 - 2e → S 0 │3 oxidation reaction
Cr +6 (K 2 Cr 2 O 7) - oxidizing agent, S -2 (H 2 S) - reducing agent
K 2 Cr 2 O 7 + 3H 2 S + 4H 2 SO 4 → Cr 2 (SO 4) 3 + K 2 SO 4 + 3S + 7H 2 O
Task #28
Using the electron balance method, write the equation for the reaction:
KMnO 4 + HCl → MnCl 2 + Cl 2 + ... + ...
Determine the oxidizing agent and reducing agent.
Mn +7 + 5e → Mn +2 │2 reduction reaction
2Cl -1 - 2e → Cl 2 0 │5 oxidation reaction
Mn +7 (KMnO 4) - oxidizing agent, Cl -1 (HCl) - reducing agent
2KMnO 4 + 16HCl → 2MnCl 2 + 5Cl 2 + 2KCl + 8H 2 O
Task #29
Using the electron balance method, write the equation for the reaction:
CrCl 2 + K 2 Cr 2 O 7 + ... → CrCl 3 + ... + H 2 O
Determine the oxidizing agent and reducing agent.
2Cr +6 + 6e → 2Cr +3 │1 reduction reaction
Cr +2 − 1e → Cr +3 │6 oxidation reaction
Cr +6 (K 2 Cr 2 O 7) - oxidizing agent, Cr +2 (CrCl 2) - reducing agent
6CrCl 2 + K 2 Cr 2 O 7 + 14HCl → 8CrCl 3 + 2KCl + 7H 2 O
Task number 30
Using the electron balance method, write the equation for the reaction:
K 2 CrO 4 + HCl → CrCl 3 + ... + ... + H 2 O
Determine the oxidizing agent and reducing agent.
Cr +6 + 3e → Cr +3 │2 reduction reaction
2Cl -1 - 2e → Cl 2 0 │3 oxidation reaction
Cr +6 (K 2 CrO 4) - oxidizing agent, Cl -1 (HCl) - reducing agent
2K 2 CrO 4 + 16HCl → 2CrCl 3 + 3Cl 2 + 4KCl + 8H 2 O
Task number 31
Using the electron balance method, write the equation for the reaction:
KI + ... + H 2 SO 4 → I 2 + MnSO 4 + ... + H 2 O
Determine the oxidizing agent and reducing agent.
Mn +7 + 5e → Mn +2 │2 reduction reaction
2l -1 − 2e → l 2 0 │5 oxidation reaction
Mn +7 (KMnO 4) - oxidizing agent, l -1 (Kl) - reducing agent
10KI + 2KMnO 4 + 8H 2 SO 4 → 5I 2 + 2MnSO 4 + 6K 2 SO 4 + 8H 2 O
Task #32
Using the electron balance method, write the equation for the reaction:
FeSO 4 + KClO 3 + KOH → K 2 FeO 4 + KCl + K 2 SO 4 + H 2 O
Determine the oxidizing agent and reducing agent.
Cl +5 + 6e → Cl -1 │2 reduction reaction
Fe +2 − 4e → Fe +6 │3 oxidation reaction
3FeSO 4 + 2KClO 3 + 12KOH → 3K 2 FeO 4 + 2KCl + 3K 2 SO 4 + 6H 2 O
Task number 33
Using the electron balance method, write the equation for the reaction:
FeSO 4 + KClO 3 + ... → Fe 2 (SO 4) 3 + ... + H 2 O
Determine the oxidizing agent and reducing agent.
Cl +5 + 6e → Cl -1 │1 reduction reaction
2Fe +2 − 2e → 2Fe +3 │3 oxidation reaction
Cl +5 (KClO 3) - oxidizing agent, Fe +2 (FeSO 4) - reducing agent
6FeSO 4 + KClO 3 + 3H 2 SO 4 → 3Fe 2 (SO 4) 3 + KCl + 3H 2 O
Task number 34
Using the electron balance method, write an equation for the reaction.
With an increase in the degree of oxidation an oxidation process takes place, and the substance itself is a reducing agent. When the oxidation state decreases, the reduction process proceeds, and the substance itself is an oxidizing agent.
The described method of OVR equalization is called the “oxidation state balance method”.
Stated in most chemistry textbooks and widely used in practice electronic balance method for equalization, OVR can be used with the caveat that the oxidation state is not equal to the charge.
2. Method of half-reactions.
In those cases, when the reaction proceeds in an aqueous solution (melt), when drawing up equations, they proceed not from a change in the oxidation state of the atoms that make up the reactants, but from a change in the charges of real particles, that is, they take into account the form of existence of substances in a solution (simple or complex ion, atom or a molecule of an undissolved or weakly dissociating substance in water).
In this case when compiling ionic equations of redox reactions, one should adhere to the same form of notation that is adopted for ionic equations of an exchange nature, namely: poorly soluble, poorly 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 as separate half-reactions. Having equalized them according to the number of atoms of each type, the half-reactions are added, multiplying each by a coefficient that equalizes the change in the charge of the oxidizing agent and reducing agent.
The half-reaction method more accurately reflects the true changes in substances in the process of redox reactions and facilitates the formulation of equations for these processes in ionic-molecular form.
Insofar as 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 (that is, an H + ion or an OH ion -, or an 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, given 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). It can be seen from the ionic diagram that the oxidizing agent permanganate ion(MnO 4 -) is converted 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 products and reagents: Dq = +2-7 = -5 (the "-" sign indicates that the reduction process is taking place and 5 is attached to the reagents). For the second process, the conversion of NO 2 - to NO 3 - , the missing oxygen comes from the water to the reducing agent, and as a result, an excess of H + ions is formed, while 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 get the ion-molecular equation for this reaction:
2MnO 4 - + 16H + + 5NO 2 - + 5H 2 O \u003d 2Mn 2+ + 8H 2 O + 5NO 3 - + 10H +.
Having canceled identical particles on the left and right sides of the equation, we finally obtain the ion-molecular equation:
2MnO 4 - + 5NO 2 - + 6H + = 2Mn 2+ + 5NO 3 - + 3H 2 O.
According to the ionic equation, we compose 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 with the formation of one molecule water in an alkaline environment, and in a neutral one it comes from water with the release of 2 H + ions.
If involved in redox reactions hydrogen peroxide(H 2 O 2), it is necessary to take into account the role of H 2 O 2 in a particular reaction. In H 2 O 2, oxygen is in an intermediate oxidation state (-1), therefore, hydrogen peroxide in redox reactions exhibits redox duality. 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 (acid medium);
H 2 O 2 +2? ® 2OH - (neutral and alkaline environments).
If hydrogen peroxide is reducing agent:
H 2 O 2 - 2? ® O 2 + 2H + (acid medium);
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, given that H 2 O 2 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.
Final equation: 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 the atoms of the elements that make up different substances change. The reactions discussed in examples 2-6 are of this type.
2 . Intramolecular redox reactions in which the oxidation state is changed by atoms of different elements of the same substance. According to this mechanism, reactions of thermal decomposition of compounds proceed. 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 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) The method of balance of oxidation states.
Let us determine the oxidation states 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.
It follows from a comparison of oxidation states that manganese simultaneously participates in the oxidation process, increasing the oxidation state from +6 to +7, and in the reduction process, lowering 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 same substance (K 2 MnO 4) acts as an oxidizing and reducing agent, the coefficients in front of it are summed up. We write down the equation:
3K 2 MnO 4 + 2H 2 O = 2KMnO 4 + MnO 2 + 4KOH.
b) Method of half-reactions.
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 an oxide that is poorly soluble 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 - (recovery).
We multiply by the coefficients and add both half-reactions, we get the total ionic equation:
3MnO 4 2 - + 2H 2 O \u003d 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 aligned (that is, the reverse of those previously considered), are processes counterdisproportionation(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 are redox reactions in which atoms or ions of not one, but two or more elements are simultaneously oxidized or reduced.
Example 8 Equalize the reaction using the above methods:
3 -2 +5 +5 +6 +2
As 2 S 3 + HNO 3 ® H 3 AsO 4 + H 2 SO 4 + NO.
Before giving examples of redox reactions with a solution, let us single out the main definitions associated with these transformations.
Those atoms or ions that, during the interaction, change their oxidation state with a decrease (accept electrons) are called oxidizing agents. Among the substances with such properties, strong inorganic acids can be noted: sulfuric, hydrochloric, nitric.
Oxidizing agent
Alkali metal permanganates and chromates are also strong oxidizing agents.
The oxidizing agent takes in the course of the reaction that it needs to complete the energy level (establishment of the completed configuration).
Reducing agent
Any redox reaction scheme involves the identification of a reducing agent. It includes ions or neutral atoms that can increase the oxidation state during the interaction (give electrons to other atoms).
Metal atoms can be cited as typical reducing agents.
Processes in OVR
What else is characterized by a change in the oxidation states of the starting substances.
Oxidation involves the process of giving off negative particles. Restoration involves taking them from other atoms (ions).
Parsing algorithm
Examples of redox reactions with a solution are offered in various reference materials designed to prepare high school students for graduate tests in chemistry.
In order to successfully cope with the proposed in the OGE and USE assignments, it is important to know the algorithm for compiling and analyzing redox processes.
- First of all, the charge values \u200b\u200bof all the elements in the substances proposed in the scheme are put down.
- Atoms (ions) are written out from the left side of the reaction, which, during the interaction, changed indicators.
- With an increase in the degree of oxidation, the sign "-" is used, and with a decrease in "+".
- Between the given and received electrons, the least common multiple is determined (the number by which they are divided without a remainder).
- When dividing LCM into 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, do not get high scores for the OGE. The algorithm of actions is given above, now let's try to work it out on concrete examples.
The peculiarity of the tasks concerning the placement of coefficients in the proposed reaction, issued to graduates of the main stage of education, is that both the left and right parts of the equation are given.
This greatly simplifies the task, since there is no need to independently invent interaction products, select the missing starting materials.
For example, it is proposed to use the electronic balance to identify the coefficients in the reaction:
At first glance, this reaction does not require stereochemical coefficients. But, in order to confirm his 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 the oxidation states is zero, given that for oxygen it is -2, for copper and iron this indicator is +2. Simple substances do not give (do not accept) electrons, therefore they are characterized by a zero value of the oxidation state.
Let's make an electronic balance, showing the sign "+" and "-" the number of received and given in the course of the interaction of electrons.
Fe 0 -2e \u003d Fe 2+.
Since the number of electrons received and given away during the interaction is the same, it makes no sense to find the least common multiple, determine the stereochemical coefficients, and put 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 a solution, but also to write out the formula of the oxidizing agent (CuO) and reducing agent (Fe) separately.
The second example with the OGE
Let us give more examples of redox reactions with a solution that may be encountered by ninth-graders who have chosen chemistry as their final exam.
Suppose it is proposed to arrange the coefficients in the equation:
Na+HCl=NaCl+H2.
In order to cope with the task, it is first important to determine the indicators of oxidation states for 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, and chlorine has a negative oxidation state. After placing the coefficients, we get the reaction with the coefficients.
The first of the exam
How to supplement redox reactions? Examples with a solution found in the USE (Grade 11) involve the addition 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 +…
Determine the reducing agent and oxidizing agent in the proposed scheme.
How to learn to compose redox reactions? The sample assumes the use of a specific algorithm.
First, in all substances given by the condition 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 an oxidizing agent is present here (manganese plays its role), a reducing agent (it is sulfur), the oxidation states do not change in the desired product, therefore, it is water.
Arguing about how to correctly solve redox reactions, we note that the next step will be to draw up an electronic ratio:
Mn +7 takes 3 e= Mn +4 ;
S -2 gives 2e= S 0 .
The manganese cation is a reducing agent, while the sulfur anion is a typical oxidizing agent. Since the smallest multiple between the received and given electrons will be 6, we get the coefficients: 2, 3.
The last step will be setting the coefficients in the original equation.
3H 2 S+ 2HMnO 4 = 3S+ 2MnO 2 + 4H 2 O.
The second sample of the OVR in the exam
How to write redox reactions correctly? Examples with a solution will help to 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, the oxidizing properties are manifested by manganese, which is part of the composition and the reducing agent should 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 electronic balance equation.
P -3 gives 8 e and turns into P +5 ;
Mn +7 takes 3e, going to Mn +4 .
The LCM will be 24, so phosphorus should have a stereometric coefficient of 3, and manganese -8.
We put the coefficients in the resulting process, we get:
3 PH 3 + 8 HMnO 4 = 8 MnO 2 + 4H 2 O+ 3 H 3 PO 4 .
The third example from the exam
Using the electron-ion balance, you need to compose a reaction, indicate the reducing agent and oxidizing agent.
KMnO 4 + MnSO 4 +…= MnO 2 +…+ H2SO 4 .
According to the algorithm, we place oxidation states for each element. Next, we determine those substances that are omitted in the right and left parts of the process. A reducing agent and an oxidizing agent are given here, so oxidation states do not change in the omitted compounds. The lost product will be water, and the starting compound will be potassium sulfate. We get the reaction scheme for which we will make an electronic balance.
Mn +2 -2 e= Mn +4 3 reducing agent;
Mn +7 +3e= Mn +4 2 oxidizing agent.
We write the coefficients in the equation, summing up the manganese atoms on the right side of the process, since it belongs to the disproportionation process.
2KMnO 4 + 3MnSO 4 + 2H 2 O \u003d 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 putrefaction, fermentation, nervous activity, respiration, and metabolism.
Oxidation and reduction are relevant for the metallurgical and chemical industries, thanks to such processes, metals can be restored from their compounds, protected from chemical corrosion, and processed.
To draw up a redox process in organic or it is necessary to use a certain algorithm of actions. First, in the proposed scheme, the oxidation states are arranged, then those elements that increased (lowered) the indicator are determined, and the electronic balance is recorded.
If you follow the sequence of actions proposed above, you can easily cope with the tasks offered in the tests.
In addition to the electronic balance method, the placement of coefficients is also possible by compiling half-reactions.
