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oxides incl. H 2 O, precipitation (solubility table), weakly dissociating compounds: H 2 S; HNO 2, H 2 SO 3 → SO 2 + H 2 O, H 2 CO 3 → CO 2 + H 2 O, NH 4 OH → NH 3 + H 2 O; CH3COOH; HMnO 4 H 2 SiO 3 , H 3 PO 4

The following have a constant oxidation state:

Group I main subgroup +1, Group II main subgroup +2, H +, O –2, OH –, Al 3+, Zn 2+.

Redox reactions(ORR) are reactions in which elements change their oxidation state (CO) due to the transfer of electrons.

Algorithm for solving redox reactions

We indicate the oxidation number (CO) of each element in the reaction.

We find elements that change their oxidation state.

We select ions or molecules that contain elements with a changed oxidation state.

We sign the oxidizing agent, the reducing agent.

Acidic environment: add nH 2 O, where lack of O → 2nH +

Alkaline environment: add nH 2 O, where excess O → 2nOH –

We equalize each half-reaction (the left side of the half-reaction = the right side), and write down the number of electrons given and received.

We equalize the number of accepted and given electrons, set the coefficients before the half-reactions.

We sign the oxidation process and the reduction process.

We write the total ionic equation taking into account the coefficients.

We transfer the coefficients from the ionic equation to the molecular equation and present similar ones (the left side of the reaction = the right side)

http://ru.wikipedia.org/wiki/%DD%EB%E5%EA%F2%F0%EE%EB%E8%E7

According to Faraday's law: m = EIt/96,500, Q = It, Cl (electricity consumed)

where m is the mass of the substance oxidized or reduced at the electrode; E is the equivalent mass of the substance; I – current strength, A; t – duration of electrolysis, s. Ve N 2 = 11.2 l, Ve O 2 = 5.6 l

To remember cathodic and anodic processes in electrochemistry, there is the following mnemonic rule:

At the anode, the anions are oxidized.

At the cathode, cations are reduced.

In the first line, all words begin with a vowel, in the second - with a consonant.

Or simpler:

CAThode - CATIONS (ions at the cathode)

ANode - ANions (ions at the anode)

Definition

Reactions occurring between ions in electrolyte solutions are called ion exchange reactions(RIO).

During RIO, there is no change in the oxidation states of elements, so RIO is not redox.

The criterion for the irreversibility of ion exchange reactions is the formation of a weak electrolyte.

Berthollet's rule

Ion exchange reactions proceed almost irreversibly if one of the resulting reaction products “leaves” the reaction sphere in the form of:

• gas,
• draft
• or a weakly dissociating electrolyte (for example, water).

If there are no ions in the solution that form a weak electrolyte, the reaction is reversible and in this case its equation is not written, putting the sign “$\ne$”

To write ionic equations, molecular (1), full ionic (2) and short ionic forms of equations (3,4) are used:

$2KOH + H_2SO_4 = K_2SO_4 + 2H_2O \hspace(3cm) (1)$

$2K^+ +2OH^- + 2H^+ + SO_4^(2-) = 2K^+ + SO_4^(2-) +2H_2O \hspace(0.2cm) (2)$

$2OH^- + 2H^+ = 2H_2O \hspace(5cm) (3)$

$OH^- + H^+ = H_2O \hspace(5.5cm) (4)$

Please note that in In a short ionic equation, the coefficients should be minimal. Therefore, in equation (3) all coefficients are reduced by 2, and the resulting equation (4) is considered a short ionic equation.

When drawing up the RIO, it should be remembered that

• water, metals, oxides, gases, precipitation do not disintegrate into ions and are written in all equations in molecular form;
• $H_2SO_3$, $H_2CO_3$, $NH_4OH$, $AgOH$ are unstable and decompose almost instantly when formed:

  $H_2SO_3 = H_2O + SO_2 \uparrow$

  $H_2CO_3 = H_2O + CO_2 \uparrow$

  $NH_4OH = H_2O + NH_3 \uparrow$

  $2AgOH = Ag_2O \downarrow + H_2O$

Algorithm for composing ion exchange reactions

1. Write down the molecular equation and assign the coefficients. When recording chemical formulas reaction products, it is important to remember that the sum of the charges in the molecule must be equal to zero.
2. A complete ionic equation is drawn up, which takes into account the result of dissociation of both the starting substances and the products of the exchange reaction. All soluble compounds are recorded in the form of ions (indicated in the solubility table by the letter “P” (highly soluble in water), with the exception of calcium hydroxide). The formulas of insoluble substances, gases, oxides, and water are written in molecular form. Counting total reaction coefficient, for which we add up all the coefficients on the right and left sides of the equation.
3. To obtain the abbreviated ionic form of the equation, similar ones are given, that is, identical ions are abbreviated before and after the equal sign in the equation. The coefficients must be minimal, and the sums of charges on the left and right sides of the equation must be the same. The total coefficient is calculated in the abbreviated form (similar to the full form).
4. The abbreviated ionic form of the equation reflects the essence of the chemical reaction that took place.

Interaction of basic oxides with acids. Write down the molecular, short and complete ionic equations for the interaction of calcium oxide and hydrochloric acid. Calculate the total coefficients in full and abbreviated form.

Solution

1. Molecular equation:

$CaO + 2HCl = CaCl_2 + H_2O$

2. Complete ionic equation:

$CaO + 2H^+ + \underline(2Cl^-) = Ca^(2+) + \underline(2Cl^-) + H_2O$

The sum of the coefficients is (1+2+2+1+2+1)=9.

3. Abbreviated ionic equation:

$CaO + 2H^+ = Ca^(2+) + H_2O$

The total coefficient is (1+2+1+1)=5.

4. A short ionic equation shows that when calcium oxide interacts with strong acids ($H^+$), the reaction is almost irreversible, resulting in the formation of a soluble calcium salt and a low-dissociating substance (water)

Interaction of salts with acids. Write down the molecular, short and complete ionic equations for the interaction of potassium carbonate and nitric acid. Calculate the total coefficients in full and abbreviated form.

Solution

1. Molecular equation:

$K_2CO_3 + 2HNO_3 = 2KNO_3 + CO_2\uparrow + H_2O$

2. Complete ionic equation:

$\underline(2K^+) + CO_3^(2-) + 2H^+ + \underline(2NO_3^-) = \underline(2K^+) + \underline(2NO_3^-) + CO_2\uparrow + H_2O$

The sum of the coefficients is (2+1+2+2+2+2+1+1)=13.

3. Brief ionic equation:

$ CO_3^(2-) + 2H^+ = CO_2\uparrow + H_2O$

The sum of the coefficients is (1+2+1+1)=5.

4. A short ionic equation shows that when soluble carbonates (alkali metals) interact with strong acids ($H^+$), the reaction is almost irreversible, resulting in the always formation of carbon dioxide ($CO_2\uparrow$) and a poorly dissociating substance (water )

Instructions

Before you begin ionic equations, you need to understand some rules. Insoluble in water, gaseous and poorly dissociating substances (for example, water) do not disintegrate into ions, which means write them in molecular form. This also includes weak electrolytes such as H2S, H2CO3, H2SO3, NH4OH. The solubility of compounds can be determined from the solubility table, which is the permitted reference material on all types of control. All charges that are inherent in cations and anions are also indicated there. To fully complete the task, you need to write molecular, complete and ionic abbreviated equations.

Example No. 1. neutralization reaction between sulfuric acid and potassium hydroxide, consider it from the point of view of TED (theory electrolytic dissociation). First, write down the reaction equation in molecular form and .H2SO4 + 2KOH = K2SO4 + 2H2O Analyze the resulting substances for their solubility and dissociation. All compounds are soluble in water, which means they are ions. The only exception is water, which does not disintegrate into ions and therefore remains in molecular form. Write the complete ionic equation, find the same ions on the left and right sides and . To cancel identical ions, cross them out.2H+ +SO4 2- +2K+ +2OH- = 2K+ +SO4 2- + 2H2OThe result is an ionic abbreviation equation:2H+ +2OH- = 2H2OCoefficients in the form of twos can also be abbreviated:H+ +OH- = H2O

Example No. 2. Write the exchange reaction between copper chloride and sodium hydroxide, consider it from the point of view of TED. Write the reaction equation in molecular form and assign the coefficients. As a result, the resulting copper hydroxide precipitated blue color. CuCl2 + 2NaOH = Cu(OH) 2↓ + 2NaCl Analyze all substances for their solubility in water - everything is soluble except copper hydroxide, which will not dissociate into ions. Write down the complete ionic equation, underline and abbreviate the identical ions: Cu2+ +2Cl- + 2Na+ +2OH- = Cu(OH) 2↓+2Na+ +2Cl- The ionic abbreviated equation remains: Cu2+ +2OH- = Cu(OH) 2↓

Example No. 3. Write the exchange reaction between sodium carbonate and hydrochloric acid, consider it from the point of view of TED. Write the reaction equation in molecular form and assign the coefficients. As a result of the reaction, sodium chloride is formed and released gaseous substance CO2 (carbon dioxide or carbon monoxide (IV)). It is formed due to the decomposition of weak carbonic acid, which breaks down into oxide and water. Na2CO3 + 2HCl = 2NaCl + CO2+H2OAnalyze all substances for their solubility in water and dissociation. Carbon dioxide leaves the system as a gaseous compound, water is a poorly dissociating substance. All other substances disintegrate into ions. Write down the complete ionic equation, underline and abbreviate the identical ions: 2Na+ +CO3 2- +2H+ +2Cl- =2Na+ +2Cl- +CO2+H2O The ionic abbreviated equation remains: CO3 2- +2H+ =CO2+H2O

Subject: Chemical bond. Electrolytic dissociation

Lesson: Writing Equations for Ion Exchange Reactions

Let's create an equation for the reaction between iron (III) hydroxide and nitric acid.

Fe(OH) 3 + 3HNO 3 = Fe(NO 3) 3 + 3H 2 O

(Iron (III) hydroxide is an insoluble base, therefore it is not subjected to. Water is a poorly dissociated substance; it is practically not dissociated into ions in solution.)

Fe(OH) 3 + 3H + + 3NO 3 - = Fe 3+ + 3NO 3 - + 3H 2 O

Cross out the same number of nitrate anions on the left and right and write the abbreviated ionic equation:

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

This reaction proceeds to completion, because a slightly dissociable substance is formed - water.

Let's write an equation for the reaction between sodium carbonate and magnesium nitrate.

Na 2 CO 3 + Mg(NO 3) 2 = 2NaNO 3 + MgCO 3 ↓

Let's write this equation in ionic form:

(Magnesium carbonate is insoluble in water and therefore does not break down into ions.)

2Na + + CO 3 2- + Mg 2+ + 2NO 3 - = 2Na + + 2NO 3 - + MgCO 3 ↓

Let's cross out the same number of nitrate anions and sodium cations on the left and right, and write the abbreviated ionic equation:

CO 3 2- + Mg 2+ = MgCO 3 ↓

This reaction proceeds to completion, because a precipitate is formed - magnesium carbonate.

Let's write an equation for the reaction between sodium carbonate and nitric acid.

Na 2 CO 3 + 2HNO 3 = 2NaNO 3 + CO 2 + H 2 O

(Carbon dioxide and water are products of the decomposition of the resulting weak carbonic acid.)

2Na + + CO 3 2- + 2H + + 2NO 3 - = 2Na + + 2NO 3 - + CO 2 + H 2 O

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

This reaction proceeds to completion, because As a result, gas is released and water is formed.

Let's create two molecular reaction equations, which correspond to the following abbreviated ionic equation: Ca 2+ + CO 3 2- = CaCO 3 .

The abbreviated ionic equation shows the essence of the ion exchange reaction. IN in this case we can say that to obtain calcium carbonate, it is necessary that the composition of the first substance include calcium cations, and the composition of the second - carbonate anions. Let's create molecular equations for reactions that satisfy this condition:

CaCl 2 + K 2 CO 3 = CaCO 3 ↓ + 2KCl

Ca(NO 3) 2 + Na 2 CO 3 = CaCO 3 ↓ + 2NaNO 3

1. Orzhekovsky P.A. Chemistry: 9th grade: textbook. for general education establishment / P.A. Orzhekovsky, L.M. Meshcheryakova, L.S. Pontak. - M.: AST: Astrel, 2007. (§17)

2. Orzhekovsky P.A. Chemistry: 9th grade: general education. establishment / P.A. Orzhekovsky, L.M. Meshcheryakova, M.M. Shalashova. - M.: Astrel, 2013. (§9)

3. Rudzitis G.E. Chemistry: inorganic. chemistry. Organ. chemistry: textbook. for 9th grade. / G.E. Rudzitis, F.G. Feldman. - M.: Education, OJSC “Moscow Textbooks”, 2009.

4. Khomchenko I.D. Collection of problems and exercises in chemistry for high school. - M.: RIA “New Wave”: Publisher Umerenkov, 2008.

5. Encyclopedia for children. Volume 17. Chemistry / Chapter. ed. V.A. Volodin, Ved. scientific ed. I. Leenson. - M.: Avanta+, 2003.

Additional web resources

1. A unified collection of digital educational resources (video experiences on the topic): ().

2. Electronic version of the journal “Chemistry and Life”: ().

Homework

1. In the table, mark with a plus sign the pairs of substances between which ion exchange reactions are possible and proceed to completion. Write reaction equations in molecular, full and reduced ionic form.

2. p. 67 No. 10,13 from the textbook P.A. Orzhekovsky “Chemistry: 9th grade” / P.A. Orzhekovsky, L.M. Meshcheryakova, M.M. Shalashova. - M.: Astrel, 2013.

Ion exchange reactions are reactions in aqueous solutions between electrolytes that occur without changes in the oxidation states of the elements that form them.

A necessary condition 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 It 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 the complete ionic equation the following type:

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.

Important Note passing the Unified State Exam in chemistry

In order for the reaction of salt1 with salt2 to proceed, in addition to the basic requirements for the occurrence of ionic reactions (gas, sediment or water in the reaction products), such reactions are subject to another requirement - the initial salts must be soluble. That is, for example,

CuS + Fe(NO 3) 2 ≠ FeS + Cu(NO 3) 2

the reaction does not proceed, although FeS could potentially give a precipitate, because insoluble. The reason that the reaction does not proceed is the insolubility of one of the starting salts (CuS).

But, for example,

Na 2 CO 3 + CaCl 2 = CaCO 3 ↓+ 2NaCl

occurs because calcium carbonate is insoluble and the starting salts are soluble.

The same applies to the interaction of salts with bases. In addition to the basic requirements for the occurrence of ion exchange reactions, in order for a salt to react with a base, the solubility of both of them is necessary. Thus:

Cu(OH) 2 + Na 2 S – does not leak,

because Cu(OH)2 is insoluble, although the potential product CuS would be a precipitate.

But the reaction between NaOH and Cu(NO 3) 2 proceeds, so both starting substances are soluble and give a precipitate of Cu(OH) 2:

2NaOH + Cu(NO 3) 2 = Cu(OH) 2 ↓+ 2NaNO 3

Attention! In no case should you extend the requirement of solubility of starting substances beyond the reactions salt1 + salt2 and salt + base.

For example, with acids this requirement is not necessary. In particular, all soluble acids react well with all carbonates, including insoluble ones.

In other words:

1) Salt1 + salt2 - the reaction occurs if the original salts are soluble, but there is a precipitate in the products

2) Salt + metal hydroxide - the reaction occurs if the starting substances are soluble and there is a precipitate or ammonium hydroxide in the products.

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 down 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