Zinc (Zn) is a chemical element belonging to the group of alkaline earth metals. In the periodic table, Mendeleev is located at number 30, which means that the charge of the atomic nucleus, the number of electrons and protons is also 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 on its valence or external energy level - respectively, 2.
Zinc as a typical alkali metal
Zinc is a typical representative of metals, in the normal state it has a bluish-gray color, it is easily oxidized in air, acquiring an oxide film (ZnO) on the surface.
As a typical amphoteric metal, zinc interacts with atmospheric oxygen: 2Zn + O2 = 2ZnO - without temperature, with the formation of an oxide film. When heated, a white powder forms.
The oxide itself reacts with acids to form salt and water:
2ZnO+2HCl=ZnCl2+H2O.
with acid solutions. If zinc is of ordinary purity, then the reaction equation for HCl Zn is below.
Zn+2HCl= ZnCl2+H2 - molecular reaction equation.
Zn (charge 0) + 2H (charge +) + 2Cl (charge -) = Zn (charge +2) + 2Cl (charge -) + 2H (charge 0) - complete Zn HCl ionic reaction equation.
Zn + 2H(+) = Zn(2+) +H2 - S.I.U. (abbreviated ionic reaction equation).
The reaction of zinc with hydrochloric acid
This HCl Zn reaction equation 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. O. hydrogen before the start of the reaction was "+", and after it became "0". He participated in the reduction process, giving 2 electrons.
Zn is a reducing agent, it participates in oxidation, accepting 2 electrons, increasing the s.d. (degree of oxidation).
It is also a substitution reaction. During it, 2 substances participated, simple Zn and 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 activity series of metals before H2, it displaced it from the substance that reacted with it.
Zinc (Zn) is a chemical element belonging to the group of alkaline earth metals. In the periodic table, Mendeleev is located at number 30, which means that the charge of the atomic nucleus, the number of electrons and protons is also 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 on its valence or external energy level - respectively, 2.
Zinc as a typical alkali metal
Zinc is a typical representative of metals, in the normal state it has a bluish-gray color, it is easily oxidized in air, acquiring an oxide film (ZnO) on the surface.
As a typical amphoteric metal, zinc interacts with atmospheric oxygen: 2Zn + O2 = 2ZnO - without temperature, with the formation of an oxide film. When heated, a white powder forms.
The oxide itself reacts with acids to form salt and water:
2ZnO+2HCl=ZnCl2+H2O.
with acid solutions. If zinc is of ordinary purity, then the reaction equation for HCl Zn is below.
Zn+2HCl= ZnCl2+H2 - molecular reaction equation.
Zn (charge 0) + 2H (charge +) + 2Cl (charge -) = Zn (charge +2) + 2Cl (charge -) + 2H (charge 0) - complete Zn HCl ionic reaction equation.
Zn + 2H(+) = Zn(2+) +H2 - S.I.U. (abbreviated ionic reaction equation).
The reaction of zinc with hydrochloric acid
This HCl Zn reaction equation 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. O. hydrogen before the start of the reaction was "+", and after it became "0". He participated in the reduction process, giving 2 electrons.
Zn is a reducing agent, it participates in oxidation, accepting 2 electrons, increasing the s.d. (degree of oxidation).
It is also a substitution reaction. During it, 2 substances participated, simple Zn and 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 activity series of metals before H2, it displaced it from the substance that reacted with it.
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 on both the right and left sides of the equation. These particles are sometimes called "observer ions"; they do not take part 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 full and short equations can coincide (for more details, see example 9).
Example 5. Write a complete and short ionic equation describing the interaction of silicic acid and potassium hydroxide in an aqueous solution.
Solution. Let's start, of course, with the molecular equation:
H 2 SiO 3 + 2KOH = K 2 SiO 3 + 2H 2 O.
Silicic acid is one of the rare examples of insoluble acids; written in molecular form. KOH and K 2 SiO 3 are written in ionic form. H 2 O, of course, we write in molecular form:
H 2 SiO 3 + 2K++ 2OH - = 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 obtain the desired short ionic equation:
H 2 SiO 3 + 2OH - \u003d SiO 3 2- + 2H 2 O.
As you can see, the process comes down 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 the full and short ionic equations for this reaction.
Solution. Main 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 = Cu 2+ + SO 4 2-+ H2O
2H + + CuO = Cu 2+ + H 2 O
Example 7. Use ionic equations to describe the interaction of zinc with hydrochloric acid.
Solution. Metals standing in series of voltages to the left of hydrogen, they react with acids with the release of hydrogen (we do not discuss the specific properties of oxidizing acids now):
Zn + 2HCl \u003d ZnCl 2 + H 2.
The full ionic equation can be written without difficulty:
Zn+2H++ 2Cl-= Zn2+ + 2Cl-+H2.
Unfortunately, when switching to a short equation in tasks of this type, students often make mistakes. For example, take zinc out of two parts 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! There are even more fantastic options. For example, H+ ions are crossed out on the left side, and H 2 molecules are crossed out on the right. This is motivated by the fact that both are hydrogen. But then, following this logic, one can, for example, consider that H 2 , HCOH and CH 4 are "one and the same", since all these substances contain hydrogen. See how absurd it can get!
Naturally, in this example, we can (and should!) erase only chloride ions. We get the final answer:
Zn + 2H + = Zn 2+ + H 2 .
Unlike all the examples discussed above, this reaction is redox (during this process, a change occurs oxidation states). For us, however, this is completely unprincipled: 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 occurring in the solution.
Solution. More active metals (those to the left in series of voltages) displace less active ones from solutions of their salts. Copper is in the voltage series to the left of silver, therefore, it displaces Ag from the salt solution:
Сu + 2AgNO 3 \u003d Cu (NO 3) 2 + 2Ag ↓.
The full and short ionic equations are given below:
Cu 0 + 2Ag + + 2NO 3 -= Cu 2+ + 2NO 3 -+ 2Ag↓ 0 ,
Cu 0 + 2Ag + = Cu 2+ + 2Ag↓ 0 .
Example 9. Write ionic equations describing the interaction of aqueous solutions of barium hydroxide and sulfuric acid.
Solution. This is a well-known neutralization reaction, the molecular equation can be written without difficulty:
Ba(OH) 2 + H 2 SO 4 = BaSO 4 ↓ + 2H 2 O.
Full ionic equation:
Ba 2+ + 2OH - + 2H + + SO 4 2- = BaSO 4 ↓ + 2H 2 O.
It's time to make a short equation, and here it turns out an interesting detail: there is, in fact, nothing to reduce. We do not observe identical 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 encountered is atypical, but quite acceptable. There are no observer ions here; all particles participate in the reaction: when barium ions and sulfate anions are combined, a precipitate of barium sulfate is formed, and when H + and OH ions interact, a weak electrolyte (water) is formed.
"But, let me!" - you exclaim. - "How do we write a short ionic equation?"
No way! You can say that the short equation is the same as the full one, you can rewrite the previous equation again, but the meaning of the reaction will not change from this. Let's hope that the compilers of the USE options will save you from such "slippery" questions, but, in principle, you should be prepared 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 (full and short) for the following reactions:
- Ba(OH) 2 + HNO 3 =
- Fe + HBr =
- Zn + CuSO 4 \u003d
- SO 2 + KOH =
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 the assignment on the exam is formulated somewhat ... unusually. You will be presented with a list of several substances. You will have to choose two compounds between which a reaction is possible, make a molecular and ionic equation. For example, a task might be formulated as follows:
Example 10. Aqueous solutions of sodium hydroxide, barium hydroxide, potassium sulfate, sodium chloride and potassium nitrate are available. Choose two substances that can react with each other; write the molecular equation for the reaction, as well as the full and short ionic equations.
Solution. remembering properties of the main classes of inorganic compounds, we conclude that the only possible reaction is the interaction of aqueous solutions of barium hydroxide and potassium sulfate:
Ba(OH) 2 + K 2 SO 4 = BaSO 4 ↓ + 2KOH.
Full ionic equation:
Ba 2+ + 2OH- + 2K++ SO 4 2- = BaSO 4 ↓ + 2K+ + 2OH-.
Brief ionic equation:
Ba 2+ + SO 4 2- \u003d BaSO 4 ↓.
By the way, pay attention to an interesting point: the brief ionic equations turned out to be identical in this example and in example 1 from first part this article. At first glance, this seems strange: completely different substances react, but the result is the same. In fact, there is nothing strange here: 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 make ionic equations. Well, for example, consider the interaction of potassium nitrate and sodium chloride. Let's write the molecular equation:
KNO 3 + NaCl = NaNO 3 + KCl.
So far, everything looks plausible enough, and we move on to the full ionic equation:
K + + NO 3 - + Na + + Cl - \u003d Na + + NO 3 - + K + + Cl -.
We begin to remove the excess and find an unpleasant detail: EVERYTHING in this equation is "superfluous". All the particles that are present on the left side, we find in the right. What does this mean? Is it possible? Yes, perhaps, just no reaction occurs in this case; particles that were originally present in the solution will remain in it. No reaction!
You see, in the molecular equation we quietly wrote nonsense, but we failed to "cheat" the short ionic equation. This is the case when the formulas are 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 made a mistake and are trying to "reduce" something superfluous, 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 can react with each other, write the molecular equation for the reaction, as well as the full and short ionic equations.
Solution. In the above list, there are 4 salts and one acid. Salts can react with each other only if a precipitate is formed during the reaction, but none of the listed salts is able to form a precipitate in the reaction with another salt from this list (verify this fact using solubility table!) An acid can react with a 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, because carbonic acid is extremely unstable even at room temperature and easily decomposes into water and carbon dioxide.
When writing the full 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 excess, we get a short ionic equation:
CO 3 2- + 2H + = H 2 O + CO 2.
Now experiment a little! Try, as we did in the previous problem, to write ionic equations for impossible reactions. Take, for example, sodium carbonate and potassium sulfate, or cesium bromide and sodium nitrate. Make sure the short ionic equation is "empty" again.
- consider 6 more examples of solving USE-31 tasks,
- discuss how to write ionic equations for complex redox reactions,
- we give examples of ionic equations involving organic compounds,
- Let us touch upon the ion exchange reactions occurring in a non-aqueous medium.
