Zn ionic. HCl Zn reaction equation, ORP, abbreviated-ionic equation. Reaction of zinc with hydrochloric acid

Zinc (Zn) is a chemical element belonging to the group of alkaline earth metals. In the periodic table of Mendeleev it is located at number 30, which means that the charge of the atomic nucleus, the number of electrons and protons is also equal to 30. Zinc is in the side II group of the IV period. By the group number, you can determine the number of atoms that are at its valence or external energy level - respectively, 2.

Zinc as a typical alkali metal

Zinc is a typical representative of metals; in its normal state it has a bluish-gray color, easily oxidizes in air, acquiring an oxide film (ZnO) on the surface.

As a typical amphoteric metal, zinc interacts with atmospheric oxygen: 2Zn + O2 \u003d 2ZnO - without temperature, with the formation of an oxide film. When heated, it forms a white powder.

The oxide itself reacts with acids to form salt and water:

2ZnO + 2HCl \u003d ZnCl2 + H2O.

With acid solutions. If zinc is of normal purity, then the reaction equation for HCl Zn is below.

Zn + 2HCl \u003d ZnCl2 + H2 is the molecular equation of the reaction.

Zn (charge 0) + 2H (charge +) + 2Cl (charge -) \u003d Zn (charge +2) + 2Cl (charge -) + 2H (charge 0) - complete Zn HCl ionic reaction equation.

Zn + 2H (+) \u003d Zn (2+) + H2 - S.I.U. (abbreviated ionic reaction equation).

Reaction of zinc with hydrochloric acid

This reaction equation for HCl Zn belongs to the redox type. This can be proved by the fact that the charge of Zn and H2 changed during the reaction, a qualitative manifestation of the reaction was observed, and the presence of an oxidizing agent and a reducing agent was also observed.

In this case, H2 is an oxidizing agent, since c. about. hydrogen before the start of the reaction was "+", and after it became "0". He participated in the recovery process, donating 2 electrons.

Zn is a reducing agent; it participates in oxidation, accepting 2 electrons, increasing the S.O. (oxidation state).

It is also a substitution reaction. It involved 2 substances, a simple Zn and a complex HCl. As a result of the reaction, 2 new substances were formed, as well as one simple - H2 and one complex - ZnCl2. Since Zn is located in the range of metal activity up to H2, it displaced it from the substance that reacted with it.

Zinc as a typical alkali metal

Zinc is a typical representative of metals; in its normal state it has a bluish-gray color, easily oxidizes in air, acquiring an oxide film (ZnO) on the surface.

As a typical amphoteric metal, zinc interacts with atmospheric oxygen: 2Zn + O2 \u003d 2ZnO - without temperature, with the formation of an oxide film. When heated, it forms a white powder.

The oxide itself reacts with acids to form salt and water:

2ZnO + 2HCl \u003d ZnCl2 + H2O.

With acid solutions. If zinc is of normal purity, then the reaction equation for HCl Zn is below.

Zn + 2HCl \u003d ZnCl2 + H2 is the molecular equation of the reaction.

Zn (charge 0) + 2H (charge +) + 2Cl (charge -) \u003d Zn (charge +2) + 2Cl (charge -) + 2H (charge 0) - complete Zn HCl ionic reaction equation.

Zn + 2H (+) \u003d Zn (2+) + H2 - S.I.U. (abbreviated ionic reaction equation).

Reaction of zinc with hydrochloric acid

In this case, H2 is an oxidizing agent, since c. about. hydrogen before the start of the reaction was "+", and after it became "0". He participated in the recovery process, donating 2 electrons.

Zn is a reducing agent; it participates in oxidation, accepting 2 electrons, increasing the S.O. (oxidation state).

It's time to move on. As we already know, the complete ionic equation needs to be "cleaned up". It is necessary to remove those particles that are present in both the right and left sides of the equation. These particles are sometimes referred to as "observer ions"; they do not participate in the reaction.

In principle, there is nothing complicated in this part. You just need to be careful and realize that in some cases the complete and short equations may coincide (for more details, see example 9).

Example 5... Write a complete and concise ionic equation describing the interaction of silicic acid and potassium hydroxide in aqueous solution.

Decision... Let's start, naturally, with the molecular equation:

H 2 SiO 3 + 2KOH \u003d K 2 SiO 3 + 2H 2 O.

Silicic acid is one of the rare examples of insoluble acids; written in molecular form. We write KOH and K 2 SiO 3 in ionic form. H 2 O, naturally, we write in molecular form:

H 2 SiO 3 + 2K + + 2OH - \u003d 2K + + SiO 3 2- + 2H 2 O.

We see that potassium ions do not change during the reaction. These particles do not take part in the process, we must remove them from the equation. We get the desired short ionic equation:

H 2 SiO 3 + 2OH - \u003d SiO 3 2- + 2H 2 O.

As you can see, the process is reduced to the interaction of silicic acid with OH - ions. Potassium ions do not play any role in this case: we could replace KOH with sodium hydroxide or cesium hydroxide, and the same process would take place in the reaction flask.

Example 6... Copper (II) oxide was dissolved in sulfuric acid. Write a complete and concise ionic equation for this reaction.

Decision... Basic oxides react with acids to form salt and water:

H 2 SO 4 + CuO \u003d CuSO 4 + H 2 O.

The corresponding ionic equations are given below. I think it is unnecessary to comment on anything in this case.

2H + + SO 4 2- + CuO \u003d Cu 2+ + SO 4 2- + H 2 O

2H + + CuO \u003d Cu 2+ + H 2 O

Example 7... Using ionic equations, describe the interaction of zinc with hydrochloric acid.

Decision... Metals in the series of voltages to the left of hydrogen react with acids with the evolution of hydrogen (we are not discussing the specific properties of oxidizing acids now):

Zn + 2HCl \u003d ZnCl 2 + H 2.

The complete ionic equation can be easily written:

Zn + 2H + + 2Cl - \u003d Zn 2+ + 2Cl - + H 2.

Unfortunately, when switching to a short equation in tasks of this type, schoolchildren often make mistakes. For example, removing zinc from two sides of the equation. This is a gross mistake! On the left side there is a simple substance, uncharged zinc atoms. On the right side, we see zinc ions. These are completely different objects! Even more fantastic options come across. For example, H + ions are crossed out on the left side, and H 2 molecules on the right. This is motivated by the fact that both are hydrogen. But then, following this logic, one can, for example, assume that H 2, HCOH and CH 4 are "one and the same", since all these substances contain hydrogen. You see how absurd you can get!

Naturally, in this example we can (and should!) Erase only chlorine ions. We get the final answer:

Zn + 2H + \u003d Zn 2+ + H 2.

Unlike all the examples discussed above, this reaction is redox (during this process, the oxidation states change). For us, however, this is completely unimportant: the general algorithm for writing ionic equations continues to work here as well.

Example 8... Copper was placed in an aqueous solution of silver nitrate. Describe the processes taking place in the solution.

Decision... More active metals (standing to the left in the series of voltages) displace less active metals from solutions of their salts. Copper is in the series of voltages to the left of silver, therefore, displaces Ag from the salt solution:

Сu + 2AgNO 3 \u003d Cu (NO 3) 2 + 2Ag ↓.

The complete and concise ionic equations are given below:

Cu 0 + 2Ag + + 2NO 3 - \u003d Cu 2+ + 2NO 3 - + 2Ag ↓ 0,

Cu 0 + 2Ag + \u003d Cu 2+ + 2Ag ↓ 0.

Example 9... Write ionic equations describing the interaction of aqueous solutions of barium hydroxide and sulfuric acid.

Decision... We are talking about the well-known neutralization reaction, the molecular equation can be written without difficulty:

Ba (OH) 2 + H 2 SO 4 \u003d BaSO 4 ↓ + 2H 2 O.

Complete ionic equation:

Ba 2+ + 2OH - + 2H + + SO 4 2- \u003d BaSO 4 ↓ + 2H 2 O.

It's time to make a short equation, and here an interesting detail becomes clear: there is, in fact, nothing to reduce. We don't see the same particles on the right and left sides of the equation. What to do? Looking for a bug? No, there is no mistake here. The situation we have encountered is not typical, but quite acceptable. There are no observer ions here; All particles participate in the reaction: when barium ions and sulfate anion combine, a precipitate of barium sulfate is formed, and when H + and OH - ions interact, a weak electrolyte (water) is formed.

"But, excuse me!" you exclaim. - "How do we make a short ionic equation?"

No way! You can say that the short equation is the same as the complete one, you can rewrite the previous equation again, but the meaning of the reaction will not change. Let's hope that the compilers of the USE versions will save you from such "slippery" questions, but, in principle, you should be ready for any scenario.

It's time to start working on your own. I suggest you complete the following tasks:

Exercise 6... Write molecular and ionic equations (complete and short) for the following reactions:

Ba (OH) 2 + HNO 3 \u003d
Fe + HBr \u003d
Zn + CuSO 4 \u003d
SO 2 + KOH \u003d

How to solve task 31 on the exam in chemistry

In principle, we have already analyzed the algorithm for solving this problem. The only problem is that on the exam the task is formulated somewhat ... unusual. You will be presented with a list of several substances. You will have to choose two compounds between which a reaction is possible, make up the molecular and ionic equations. For example, a task can be formulated as follows:

Example 10... There are aqueous solutions of sodium hydroxide, barium hydroxide, potassium sulfate, sodium chloride and potassium nitrate at your disposal. Choose two substances that can react with each other; write the molecular reaction equation as well as the complete and concise ionic equations.

Decision... Recalling the properties of the main classes of inorganic compounds, we come to the conclusion that the only possible reaction is the interaction of aqueous solutions of barium hydroxide and potassium sulfate:

Ba (OH) 2 + K 2 SO 4 \u003d BaSO 4 ↓ + 2KOH.

Complete ionic equation:

Ba 2+ + 2OH - + 2K + + SO 4 2- \u003d BaSO 4 ↓ + 2K + + 2OH -.

Short ionic equation:

Ba 2+ + SO 4 2- \u003d BaSO 4 ↓.

By the way, pay attention to an interesting point: the short ionic equations turned out to be identical in this example and in example 1 from the first part of this article. At first glance, this seems strange: completely different substances react, and the result is the same. In fact, there is nothing strange here: the ionic equations help to see the essence of the reaction, which can be hidden under different shells.

And one moment. Let's try to take other substances from the proposed list and create ionic equations. Well, for example, consider the interaction of potassium nitrate and sodium chloride. Let's write the molecular equation:

KNO 3 + NaCl \u003d NaNO 3 + KCl.

So far, everything looks quite plausible, and we move on to the full ionic equation:

K + + NO 3 - + Na + + Cl - \u003d Na + + NO 3 - + K + + Cl -.

We start to remove the unnecessary and find an unpleasant detail: EVERYTHING in this equation is "redundant". All the particles present on the left side, we find in the right. What does this mean? Is this possible? Yes, perhaps there is simply no reaction in this case; the particles that were originally present in the solution will remain in it. No reaction!

You see, we quietly wrote nonsense in the molecular equation, but we failed to "cheat" the short ionic equation. This is the case when formulas turn out to be smarter than us! Remember: if, when writing a short ionic equation, you come to the need to remove all substances, this means that either you are mistaken and are trying to "reduce" something unnecessary, or this reaction is generally impossible.

Example 11... Sodium carbonate, potassium sulfate, cesium bromide, hydrochloric acid, sodium nitrate. From the proposed list, select two substances that are capable of reacting with each other, write the molecular equation of the reaction, as well as the complete and short ionic equations.

Decision... This list contains 4 salts and one acid. Salts are capable of reacting with each other only if a precipitate forms during the reaction, but none of the listed salts is capable of forming a precipitate in reaction with another salt from this list (check this fact using the solubility table!) The acid is able to react with salt only when the salt is formed by a weaker acid. Sulfuric, nitric and hydrobromic acids cannot be displaced by the action of HCl. The only reasonable option is the interaction of hydrochloric acid with sodium carbonate.

Na 2 CO 3 + 2HCl \u003d 2NaCl + H 2 O + CO 2

Please note: instead of the formula H 2 CO 3, which, in theory, should have been formed during the reaction, we write H 2 O and CO 2. This is correct, since carbonic acid is extremely unstable even at room temperature and easily decomposes into water and carbon dioxide.

When writing the complete ionic equation, we take into account that carbon dioxide is not an electrolyte:

2Na + + CO 3 2- + 2H + + 2Cl - \u003d 2Na + + 2Cl - + H 2 O + CO 2.

We remove the unnecessary, we get a short ionic equation:

CO 3 2- + 2H + \u003d H 2 O + CO 2.

Now experiment a little! Try, as we did in the previous problem, to compose ionic equations for unrealizable reactions. Take sodium carbonate and potassium sulfate, or cesium bromide and sodium nitrate, for example. Make sure that the concise ionic equation is "empty" again.

consider 6 more examples of solving the USE-31 tasks,
we will discuss how to compose ionic equations in the case of complex redox reactions,
we will give examples of ionic equations involving organic compounds,
let us touch upon the reactions of ion exchange that take place in a non-aqueous medium.