Redox reactions. Ion exchange reactions, conditions for their completion (using the example of two reactions)

Ion exchange reactions are reactions in aqueous solutions between electrolytes that occur without changes in the oxidation states of their constituent elements

A necessary condition for the reaction between electrolytes (salts, acids and bases) is the formation of a slightly dissociating substance (water, weak acid, ammonium hydroxide), precipitate or gas.

Let's consider the reaction that results in the formation of water. Such reactions include all reactions between any acid and any base. For example, the reaction of nitric acid with potassium hydroxide:

HNO 3 + KOH = KNO 3 + H 2 O (1)

Starting materials, i.e. nitric acid and potassium hydroxide, as well as one of the products, namely potassium nitrate, are strong electrolytes, i.e. in aqueous solution they exist almost exclusively in the form of ions. The resulting water belongs to weak electrolytes, i.e. practically does not disintegrate into ions. Thus, the equation above can be rewritten more accurately by indicating the real state of substances in an aqueous solution, i.e. in the form of ions:

H + + NO 3 − + K + + OH ‑ = K + + NO 3 − + H 2 O (2)

As can be seen from equation (2), both before and after the reaction, NO 3 − and K + ions are present in the solution. In other words, essentially, nitrate ions and potassium ions did not participate in the reaction at all. The reaction occurred only due to the combination of H + and OH − particles into water molecules. Thus, by performing an algebraic reduction of identical ions in equation (2):

H + + NO 3 − + K + + OH ‑ = K + + NO 3 − + H 2 O

we will get:

H + + OH ‑ = H 2 O (3)

Equations of the form (3) are called abbreviated ionic equations, type (2) - complete ionic equations, and type (1) - molecular reaction equations.

In fact, the ionic equation of a reaction maximally reflects its essence, precisely what makes its occurrence possible. It should be noted that many different reactions can correspond to one abbreviated ionic equation. Indeed, if we take, for example, not nitric acid, but hydrochloric acid, and instead of potassium hydroxide we use, say, barium hydroxide, we have the following molecular equation of the reaction:

2HCl+ Ba(OH) 2 = BaCl 2 + 2H 2 O

Hydrochloric acid, barium hydroxide and barium chloride are strong electrolytes, that is, they exist in solution primarily in the form of ions. Water, as discussed above, is a weak electrolyte, that is, it exists in solution almost only in the form of molecules. Thus, complete ionic equation This reaction will look like this:

2H + + 2Cl − + Ba 2+ + 2OH − = Ba 2+ + 2Cl − + 2H 2 O

Let's cancel the same ions on the left and right and get:

2H + + 2OH − = 2H 2 O

Dividing both the left and right sides by 2, we get:

H + + OH − = H 2 O,

Received abbreviated ionic equation completely coincides with the abbreviated ionic equation for the interaction of nitric acid and potassium hydroxide.

When composing ionic equations in the form of ions, write only the formulas:

1) strong acids (HCl, HBr, HI, H 2 SO 4, HNO 3, HClO 4) (the list of strong acids must be learned!)

2) strong bases (hydroxides of alkali (ALM) and alkaline earth metals (ALM))

3) soluble salts

The formulas are written in molecular form:

1) Water H 2 O

2) Weak acids (H 2 S, H 2 CO 3, HF, HCN, CH 3 COOH (and others, almost all organic))

3) Weak bases (NH 4 OH and almost all metal hydroxides except alkali metal and alkali metal

4) Slightly soluble salts (↓) (“M” or “H” in the solubility table).

5) Oxides (and other substances that are not electrolytes)

Let's try to write down the equation between iron (III) hydroxide and sulfuric acid. In molecular form, the equation of their interaction is written as follows:

2Fe(OH) 3 + 3H 2 SO 4 = Fe 2 (SO 4) 3 + 6H 2 O

Iron (III) hydroxide corresponds to the designation “H” in the solubility table, which tells us about its insolubility, i.e. in the ionic equation it must be written in its entirety, i.e. as Fe(OH) 3 . Sulfuric acid is soluble and belongs to strong electrolytes, that is, it exists in solution mainly in a dissociated state. Iron(III) sulfate, like almost all other salts, is a strong electrolyte, and since it is soluble in water, it must be written as an ion in the ionic equation. Taking into account all of the above, we obtain a complete ionic equation of the following form:

2Fe(OH) 3 + 6H + + 3SO 4 2- = 2Fe 3+ + 3SO 4 2- + 6H 2 O

Reducing the sulfate ions on the left and right, we get:

2Fe(OH) 3 + 6H + = 2Fe 3+ + 6H 2 O

Dividing both sides of the equation by 2 we get the abbreviated ionic equation:

Fe(OH) 3 + 3H + = Fe 3+ + 3H 2 O

Now let's look at the ion exchange reaction that produces a precipitate. For example, the interaction of two soluble salts:

All three salts - sodium carbonate, calcium chloride, sodium chloride and calcium carbonate (yes, that too) - are strong electrolytes and all except calcium carbonate are soluble in water, i.e. are involved in this reaction in the form of ions:

2Na + + CO 3 2- + Ca 2+ + 2Cl − = CaCO 3 ↓+ 2Na + + 2Cl −

By canceling the same ions on the left and right in this equation, we get the abbreviated ionic equation:

CO 3 2- + Ca 2+ = CaCO 3 ↓

The last equation reflects the reason for the interaction of solutions of sodium carbonate and calcium chloride. Calcium ions and carbonate ions combine into neutral calcium carbonate molecules, which, when combined with each other, give rise to small crystals of CaCO 3 precipitate of ionic structure.

Let's consider the third condition for the occurrence of ion exchange reactions - the formation of gas. Strictly speaking, only as a result of ion exchange, gas formation is possible only in rare cases, for example, during the formation of hydrogen sulfide gas:

K 2 S + 2HBr = 2KBr + H 2 S

In most other cases, gas is formed as a result of the decomposition of one of the products of the ion exchange reaction. For example, you need to know for sure as part of the Unified State Examination that with the formation of gas, due to instability, products such as H 2 CO 3, NH 4 OH and H 2 SO 3 decompose:

H 2 CO 3 = H 2 O + CO 2

NH 4 OH = H 2 O + NH 3

H 2 SO 3 = H 2 O + SO 2

In other words, if an ion exchange produces carbonic acid, ammonium hydroxide, or sulfurous acid, the ion exchange reaction proceeds due to the formation of a gaseous product:

Let us write the ionic equations for all the above reactions leading to the formation of gases. 1) For reaction:

K 2 S + 2HBr = 2KBr + H 2 S

Potassium sulfide and potassium bromide will be written in ionic form, because are soluble salts, as well as hydrobromic acid, because refers to strong acids. Hydrogen sulfide, being a poorly soluble gas that dissociates poorly into ions, will be written in molecular form:

2K + + S 2- + 2H + + 2Br — = 2K + + 2Br — + H 2 S

Reducing identical ions we get:

S 2- + 2H + = H 2 S

2) For the equation:

Na 2 CO 3 + H 2 SO 4 = Na 2 SO 4 + H 2 O + CO 2

In ionic form, Na 2 CO 3, Na 2 SO 4 will be written as highly soluble salts and H 2 SO 4 as a strong acid. Water is a poorly dissociating substance, and CO 2 is not an electrolyte at all, so their formulas will be written in molecular form:

2Na + + CO 3 2- + 2H + + SO 4 2- = 2Na + + SO 4 2 + H 2 O + CO 2

CO 3 2- + 2H + = H 2 O + CO 2

3) for the equation:

NH 4 NO 3 + KOH = KNO 3 + H 2 O + NH 3

Molecules of water and ammonia will be written in their entirety, and NH 4 NO 3, KNO 3 and KOH will be written in ionic form, because all nitrates are highly soluble salts, and KOH is an alkali metal hydroxide, i.e. strong base:

NH 4 + + NO 3 − + K + + OH − = K + + NO 3 − + H 2 O + NH 3

NH 4 + + OH − = H 2 O + NH 3

For the equation:

Na 2 SO 3 + 2HCl = 2NaCl + H 2 O + SO 2

The full and abbreviated equation will look like:

2Na + + SO 3 2- + 2H + + 2Cl − = 2Na + + 2Cl − + H 2 O + SO 2

1. Redox reactions. Oxidizing agent and reducing agent (using the example of two reactions).

Redox reactions occur with a change in the oxidation state. Widespread reactions of this type are combustion reactions. This also includes slow oxidation reactions (corrosion of metals, decay of organic substances).

The oxidation state of an element shows the number of electrons displaced (attracted or given up). In simple substances it is equal to zero. In binary compounds (consisting of 2 elements), it is equal to the valency, which is preceded by a sign (therefore it is sometimes called a “conditional charge”).

In substances consisting of 3 or more elements, the oxidation number can be calculated using an equation, taking the unknown oxidation number as “x”, and equating the total amount to zero. For example, in nitric acid HNO 3 the oxidation state of hydrogen is +1, oxygen is −2, we get the equation: +1 + x −2 3 = 0

An element that gains electrons is called oxidizing agent. An element that is an electron donor (giving away electrons) is called reducing agent.

2 e − _ l ↓ Fe 0 + S 0 = Fe +2 S −2

When iron and sulfur powders are heated, iron sulfide is formed. Iron is a reducing agent (oxidizes), sulfur is an oxidizing agent (reduced).

S 0 + O 2 0 = S +4 O 2 −2

In this reaction, sulfur is a reducing agent, oxygen is an oxidizing agent. Sulfur (IV) oxide is formed

An example involving a complex substance can be given:

Zn 0 + 2H +1 Cl = Zn +2 Cl 2 + H 2 0

zinc is a reducing agent, hydrogen hydrochloric acid is an oxidizing agent.

You can give an example involving a complex substance and create an electronic balance:

Cu 0 + 4HN +5 O 3 = Cu +2 (NO 3) 2 + 2H 2 O + 2N +4 O 2

Ticket number 8

1. Ion exchange reactions, conditions for their completion (using the example of two reactions). The difference between ion exchange reactions and redox reactions.

Exchange reactions in electrolyte solutions are called ion exchange reactions. These reactions proceed to completion in 3 cases:

1. If as a result of the reaction a precipitate forms (an insoluble or slightly soluble substance is formed, which can be determined from the solubility table): CuSO 4 + BaCl 2 = BaSO 4 ↓ + CuCl 2

2. If gas is released (often formed during the decomposition of weak acids): Na 2 CO 3 + 2HCl = 2NaCl + H 2 O + CO 2

3. If a slightly dissociating substance is formed. For example, water, acetic acid: HCl + NaOH = NaCl + H 2 O

This is due to a shift in chemical equilibrium to the right, which is caused by the removal of one of the products from the reaction zone.

Ion exchange reactions are not accompanied by the transfer of electrons and changes in the oxidation state of elements, unlike redox reactions.

If you are asked to write an equation in ionic form, you can check the correct spelling of ions using the solubility table. Don't forget to change indexes to coefficients. We do not separate insoluble substances, released gases, water (and other oxides) into ions.

Cu 2+ + SO 4 2− + Ba 2+ + 2Cl − = BaSO 4 ↓ + Cu 2+ + 2Cl − Cross out the unchanged ions.

96. 161 g of Glauber’s salt Na 2 SO 4 ∙10H 2 O was dissolved in 180 ml of water. What will be the mass fraction of sodium sulfate in the resulting solution? How many ions of each type are there in it?

97. Write down the equations of electrolytic dissociation of substances:

A) lithium hydroxide

B) potassium carbonate

B) barium nitrate

D) sulfurous acid

E) chromium (III) sulfate

E) potassium phosphate

98. Write four equations for the electrolytic dissociation of substances that form only sulfate ions as anions.

99. Write the formulas of substances that dissociate into ions in water:

A) Ba 2+ and Cl ─

B) Fe 3+ and NO 3 ─

B) H + and SO 4 2─

D) K + and OH ─

100. Write down the molecular and ionic equations of practically feasible reactions:

A) Na 2 CO 3 + Ca(NO 3) 2 →

B) Cu(OH) 2 + HCl→

B) K 2 CO 3 + HNO 3 →

D) NaOH + H 3 PO 4 →

D) KNO 3 + Na 2 SO 4 →

E) MgCO 3 + HCl→

G) Fe(NO 3) 3 + KOH→

101. Write two molecular equations, the essence of which is expressed by the ionic equation a) Ba 2+ + SO 4 2─ → BaSO 4 ↓, b) H + + OH - → H 2 O.

102. Complete the reaction equations, indicate their type, name the products. For exchange reactions, write ionic equations.

1) HNO 3 + Li 2 CO 3 →

2) H 2 SO 4 + Al →

3) HCl + Fe 2 O 3 →

4) H 3 PO 4 + KOH→

103. Which of the following substances will a solution of sulfuric acid react with: silicon (IV) oxide, lithium hydroxide, barium nitrate, hydrochloric acid, potassium oxide, sodium silicate, potassium nitrate, iron (II) hydroxide? Write down the equations of possible reactions in molecular and ionic form.

104. What amount of substance and what mass of each product will be obtained when carrying out the following transformations: sulfur → sulfur oxide (IV) → sulfurous acid → barium sulfite, if 16 g of sulfur was taken?

105. Which of the following substances will a barium hydroxide solution react with: nitric acid, sodium oxide, ammonium chloride, potassium hydroxide, sulfur (VI) oxide, copper (II) chloride, sodium nitrate, iron (II) hydroxide, carbon dioxide? Write down the equations of possible reactions in molecular and ionic form.

106. Complete the reaction equations, indicate their type, name the products. Write down the ionic equations.

1) HNO 3 + Al(OH) 3 →

2) LiOH + H 2 SO 4 →

3) KOH + SO 2 →

4) NaOH + FeCl 3 →

107. What mass of each product will be obtained when carrying out the following transformations: calcium → calcium oxide → calcium hydroxide → calcium chloride, if 80 g of calcium was taken?

108. Indicate the nature of the oxide and create the formula of the corresponding hydroxide (base or acid): Na 2 O, N 2 O 5, Mn 2 O 7, CuO, SO 2, SO 3, FeO, P 2 O 5, CaO.

109. Write down equations for practically feasible reactions and indicate their type. For exchange reactions, write ionic equations.

1) K 2 O + H 2 O →

2) CO 2 + HNO 3 →

3) Fe 2 O 3 + H 2 SO 4 →

4) SO 3 + H 2 O →

5) FeO + H 2 O →

6) SO 2 + KOH →

7) CuO + Ca(OH) 2 →

8) P 2 O 5 + CaO →

9) SiO 2 + Cl 2 O 7 →

110. What mass of salt can be obtained by dissolving magnesium oxide in 100 g of a 10% solution of nitric acid?

111. Complete the reaction equations in molecular and ionic form:

1) CuCl 2 + Al→

2) LiOH + FeSO 4 →

3) Ba(NO 3) 2 + Na 2 SO 4 →

4) CaCO 3 + HNO 3 →

5) FeCl 3 + KOH →

6) K 2 SiO 3 + HCl→

112. Complete the equations of possible reactions, indicate their type, name the products. For exchange reactions, write ionic equations.

1) Na 3 PO 4 + AgNO 3 →

2) K 2 SO 4 + NaCl→

3) BaCO 3 + HCl→

4) Cu(NO 3) 2 + Zn →

5) NaCl + Ca(OH) 2 →

6) Fe(NO 3) 2 + KOH →

113. Write down the equations of all possible reactions that can be used to obtain the salt a) copper (II) chloride, b) iron (II) sulfate.

114. Make genetic series of metals a) sodium, b) magnesium.

115. Make up genetic series of non-metals a) sulfur, b) silicon, c) phosphorus.

116. Solve the chain of transformations, indicate the type of reactions, the conditions for their occurrence, name the products:

A) Ca → CaO → Ca(OH) 2 → CaCO 3 → CO 2 → Na 2 CO 3 → MgCO 3

B) S → SO 2 → SO 3 → H 2 SO 4 → K 2 SO 4 → BaSO 4 .

117. Complete the reaction equations and characterize them according to all known characteristics:

1) Na 2 SO 4 + BaCl 2 →

2) Al + CuCl 2 →

4) CH 4 + O 2 →

118. Arrange the coefficients using the electronic balance method, indicate the oxidizing agent, reducing agent, oxidation and reduction processes:

1) NH 3 + O 2 → NO + H 2 O

2) Al + I 2 → AlI 3

3) CO 2 + Mg → MgO + C

4) HNO 3 + P + H 2 O → H 3 PO 4 + NO 2

5) HCl + KMnO 4 → Cl 2 + KCl + MnCl 2 + H 2 O

119. Complete the equations of exchange reactions and create ionic equations for them:

1) FeO + HNO 3 →

2) MgCO 3 + HCl →

3) Fe 2 (SO 4) 3 + KOH →

120. Complete the abbreviated ionic equations and propose molecular equations for them:

1) OH ─ + H + →

2) SiO 3 2─ + 2 H + →

121. Solve the chain of transformations, indicate the type of reactions, the conditions for their occurrence, name the products: Cu → CuO → CuSO 4 → Cu(OH) 2 → CuO → Cu.

122. Complete the reaction equations and determine their type. Indicate the oxidation states and indicate which of the reactions are redox:

1) Al + CuSO 4 →

3) Fe + Cl 2 →

4) P 2 O 5 + H 2 O →

6) NaCl + AgNO 3 →

7) Zn + H 2 SO 4 →

123. Arrange the coefficients using the electronic balance method:

1) Zn + HCl → ZnCl 2 + H 2

2) NH 3 + O 2 → NO + H 2 O

3) Al + I 2 → AlI 3

4) CO 2 + Mg → MgO + C

5) HNO 3 + P + H 2 O → H 3 PO 4 + NO 2

6) HCl + KMnO 4 → Cl 2 + KCl + MgCl 2 + H 2 O

7) Cu + HNO 3 → Cu(NO 3) 2 + NO 2 + H 2 O

8) K + H 2 SO 4 → K 2 SO 4 + S + H 2 O

9) K 2 S + KMnO 4 + H 2 SO 4 → S + MnSO 4 + K 2 SO 4 + H 2 O

10) Na 2 SO 3 + KIO 3 + H 2 SO 4 → I 2 + Na 2 SO 4 + K 2 SO 4 + H 2 O