Engineer Naydenov, who had only a few years left until his retirement, did not ...
We continue to discuss the solution to the problem of the type C1 (No. 30), which is sure to meet everyone who will take the exam in chemistry. In the first part of the article, we presented a general algorithm for solving Problem 30, in the second part we analyzed several rather complex examples.
We begin the third part by discussing typical oxidants and reducing agents and their transformations in various environments.
Fifth step: discussing typical IDS that can be encountered in problem # 30
I would like to recall a few points related to the concept of the oxidation state. We have already noted that a constant oxidation state is characteristic only for a relatively small number of elements (fluorine, oxygen, alkali and alkaline earth metals, etc.). Most of the elements can exhibit different oxidation states. For example, for chlorine, all states from -1 to +7 are possible, although odd values \u200b\u200bare the most stable. Nitrogen exhibits oxidation states from -3 to +5, etc.
There are two important rules to remember clearly.
1. The highest oxidation state of a non-metal element in most cases coincides with the number of the group in which this element is located, and the lowest oxidation state \u003d group number - 8.
For example, chlorine is in group VII, therefore, its highest oxidation state \u003d +7, and the lowest - 7 - 8 \u003d -1. Selenium is in group VI. The highest oxidation state \u003d +6, the lowest - (-2). Silicon is located in group IV; the corresponding values \u200b\u200bare +4 and -4.
Remember that there are exceptions to this rule: the highest oxidation state of oxygen \u003d +2 (and even it manifests itself only in oxygen fluoride), and the highest oxidation state of fluorine \u003d 0 (in a simple substance)!
2. Metals are not capable of exhibiting negative oxidation states. This is quite important, considering that more than 70% of chemical elements belong to metals.
And now the question: "Can Mn (+7) act as a reducing agent in chemical reactions?" Take your time, try to answer yourself.
The correct answer is: "No, it cannot!" This is very easy to explain. Take a look at the position of this element in the periodic table. Mn is in group VII, therefore, its HIGHER oxidation state is +7. If Mn (+7) acted as a reducing agent, its oxidation state would increase (remember the definition of a reducing agent!), Which is impossible, since it already has a maximum value. Conclusion: Mn (+7) can only be an oxidizing agent.
For the same reason, ONLY OXIDIZING properties can show S (+6), N (+5), Cr (+6), V (+5), Pb (+4), etc. Take a look at the position of these elements in the periodic system and see for yourself.
And another question: "Can Se (-2) act as an oxidizing agent in chemical reactions?"
Again, negative answer. You've probably already guessed what this is about. Selenium is in group VI, its LOWEST oxidation state is -2. Se (-2) cannot acquire electrons, that is, it cannot be an oxidizing agent. If Se (-2) participates in OVR, then only in the role of RESTORER.
For a similar reason, the ONLY RESTORER can be N (-3), P (-3), S (-2), Te (-2), I (-1), Br (-1), etc.
The final conclusion: an element in the lowest oxidation state can act in the ORP only as a reducing agent, and an element with the highest oxidation state only as an oxidizing agent.
"What if the element has an intermediate oxidation state?" - you ask. Well, then both its oxidation and its reduction are possible. For example, sulfur in reaction with oxygen is oxidized, and in reaction with sodium it is reduced.
It is probably logical to assume that each element in the highest oxidation state will be a pronounced oxidizing agent, and in the lowest, a strong reducing agent. In most cases, this is true. For example, all compounds Mn (+7), Cr (+6), N (+5) can be classified as strong oxidants. But, for example, P (+5) and C (+4) are restored with difficulty. And it is almost impossible to force Ca (+2) or Na (+1) to act as an oxidizing agent, although, formally speaking, +2 and +1 are also the highest oxidation states.
On the contrary, many chlorine compounds (+1) are powerful oxidizing agents, although the oxidation state +1 in this case is far from the highest.
F (-1) and Cl (-1) are bad recovers, and their counterparts (Br (-1) and I (-1)) are good. Oxygen in the lowest oxidation state (-2) practically does not exhibit reducing properties, and Te (-2) is a powerful reducing agent.
We see that everything is not as obvious as we would like. In some cases, the ability to oxidize - reduction can be easily foreseen, in other cases - you just need to remember that substance X is, say, a good oxidizing agent.
We seem to have finally gotten to a list of typical oxidizing and reducing agents. I would like you not only to "memorize" these formulas (although it will be good too!), But also to be able to explain why this or that substance was included in the corresponding list.
Typical oxidants
- Simple substances - non-metals: F 2, O 2, O 3, Cl 2, Br 2.
- Concentrated sulfuric acid (H 2 SO 4), nitric acid (HNO 3) in any concentration, hypochlorous acid (HClO), perchloric acid (HClO 4).
- Potassium permanganate and potassium manganate (KMnO 4 and K 2 MnO 4), chromates and dichromates (K 2 CrO 4 and K 2 Cr 2 O 7), bismuthates (eg NaBiO 3).
- Oxides of chromium (VI), bismuth (V), lead (IV), manganese (IV).
- Hypochlorites (NaClO), chlorates (NaClO 3) and perchlorates (NaClO 4); nitrates (KNO 3).
- Peroxides, superoxides, ozonides, organic peroxides, peroxoacids, all other substances containing the -O-O- group (eg, hydrogen peroxide - H 2 O 2, sodium peroxide - Na 2 O 2, potassium superoxide - KO 2).
- Metal ions located on the right side of the voltage row: Au 3+, Ag +.
Typical reducing agents
- Simple substances - metals: alkali and alkaline earth, Mg, Al, Zn, Sn.
- Simple substances - non-metals: H 2, C.
- Metal hydrides: LiH, CaH 2, lithium aluminum hydride (LiAlH 4), sodium borohydride (NaBH 4).
- Hydrides of some non-metals: HI, HBr, H 2 S, H 2 Se, H 2 Te, PH 3, silanes and boranes.
- Iodides, bromides, sulfides, selenides, phosphides, nitrides, carbides, nitrites, hypophosphites, sulfites.
- Carbon monoxide (CO).
I would like to emphasize a few points:
- I did not set myself the goal of listing all oxidizing and reducing agents. It is impossible and unnecessary.
- One and the same substance can act in one process as an oxidizing agent, and in another - as an activator.
- No one can guarantee that you will definitely encounter one of these substances in exam problem C1, but the probability of this is very high.
- It is not rote memorization of formulas that is important, but UNDERSTANDING. Try to test yourself: write down a mixture of substances from two lists, and then try to separate them yourself into typical oxidizing and reducing agents. Be guided by the considerations we discussed at the beginning of this article.
And now a little test work. I will offer you some incomplete equations, and you will try to find an oxidizing agent and a reducing agent. It is not necessary to add the right-hand sides of the equations yet.
Example 12... Determine the oxidizing and reducing agent in the ORP:
HNO 3 + Zn \u003d ...
CrO 3 + C 3 H 6 + H 2 SO 4 \u003d ...
Na 2 SO 3 + Na 2 Cr 2 O 7 + H 2 SO 4 \u003d ...
O 3 + Fe (OH) 2 + H 2 O \u003d ...
CaH 2 + F 2 \u003d ...
KMnO 4 + KNO 2 + KOH \u003d ...
H 2 O 2 + K 2 S + KOH \u003d ...
I think you coped with this task without difficulty. If you have problems, read the beginning of this article again, work on a list of typical oxidants.
“All this is wonderful!” The impatient reader exclaims. “But where are the promised problems C1 with incomplete equations? Yes, in example 12 we were able to determine the oxidizer and g-tel, but this is not the main thing. The main thing is to be able to COMPLETE the reaction equation, but can a list of oxidants help us with this? "
Yes, it can, if you understand WHAT HAPPENS with typical oxidants under different conditions. This is exactly what we are going to do now.
Sixth step: transformations of some oxidants in different environments. "Fate" of permanganates, chromates, nitric and sulfuric acids
So, we must not only be able to recognize typical oxidants, but also understand what these substances turn into in the course of ORR. Obviously, without this understanding, we will not be able to correctly solve Problem 30. The situation is complicated by the fact that it is impossible to indicate the products of interaction UNIVERSAL. It makes no sense to ask: "What will the potassium permanganate turn into during the recovery process?" It all depends on many reasons. In the case of KMnO 4, the main one is the acidity (pH) of the medium. In principle, the nature of the recovery products may depend on:
- the reducing agent used during the process,
- acidity of the environment,
- concentrations of reaction participants,
- process temperature.
We will not now talk about the effect of concentration and temperature (although inquisitive young chemists may remember that, for example, chlorine and bromine interact differently with an aqueous solution of alkali in the cold and when heated). Let's focus on the pH of the medium and the strength of the reducing agent.
The information below should simply be remembered. Don't try to analyze the reasons, just REMEMBER the reaction products. I assure you, on the exam in chemistry, this may be useful to you.
Reduction products of potassium permanganate (KMnO 4) in various media
Example 13... Supplement the equations of redox reactions:
KMnO 4 + H 2 SO 4 + K 2 SO 3 \u003d ...
KMnO 4 + H 2 O + K 2 SO 3 \u003d ...
KMnO 4 + KOH + K 2 SO 3 \u003d ...
Decision... Guided by the list of typical oxidizing and reducing agents, we come to the conclusion that potassium permanganate is the oxidizing agent in all these reactions, and potassium sulfite is the reducing agent.
H 2 SO 4, H 2 O and KOH determine the nature of the solution. In the first case, the reaction takes place in an acidic medium, in the second - in a neutral one, in the third - in an alkaline one.
Conclusion: in the first case, permanganate will be reduced to the Mn (II) salt, in the second - to manganese dioxide, in the third - to potassium manganate. Let's supplement the reaction equations:
KMnO 4 + H 2 SO 4 + K 2 SO 3 \u003d MnSO 4 + ...
KMnO 4 + H 2 O + K 2 SO 3 \u003d MnO 2 + ...
KMnO 4 + KOH + K 2 SO 3 \u003d K 2 MnO 4 + ...
And what will potassium sulfite turn into? Well, naturally, in sulfate. It is obvious that K in the composition of K 2 SO 3 is simply nowhere to be oxidized, oxygen oxidation is extremely unlikely (although, in principle, it is possible), but S (+4) easily turns into S (+6). The oxidation product is K 2 SO 4, you can add this formula to the equations:
KMnO 4 + H 2 SO 4 + K 2 SO 3 \u003d MnSO 4 + K 2 SO 4 + ...
KMnO 4 + H 2 O + K 2 SO 3 \u003d MnO 2 + K 2 SO 4 + ...
KMnO 4 + KOH + K 2 SO 3 \u003d K 2 MnO 4 + K 2 SO 4 + ...
Our equations are almost ready. It remains to add substances that are not directly involved in the OVR and set the coefficients. By the way, if you start from the second point, it may be even easier. Let's build, for example, an electronic balance for the last reaction
| Mn (+7) + 1e | = | Mn (+6) | (2) |
| S (+4) - 2e | = | S (+6) | (1) |
We put the coefficient 2 in front of the formulas KMnO 4 and K 2 MnO 4; before the formulas of sulfite and potassium sulfate we mean coeff. 1:
2KMnO 4 + KOH + K 2 SO 3 \u003d 2K 2 MnO 4 + K 2 SO 4 + ...
On the right we see 6 potassium atoms, on the left - so far only 5. We need to correct the situation; we put coefficient 2 in front of the KOH formula:
2KMnO 4 + 2KOH + K 2 SO 3 \u003d 2K 2 MnO 4 + K 2 SO 4 + ...
The final touch: on the left we see hydrogen atoms, on the right they are not. Obviously, we urgently need to find some substance that contains hydrogen in the oxidation state +1. Let's get some water!
2KMnO 4 + 2KOH + K 2 SO 3 \u003d 2K 2 MnO 4 + K 2 SO 4 + H 2 O
We check the equation again. Yes, everything is great!
“An interesting movie!” The alert young chemist remarks. “Why did you add water at the last step? And if I want to add hydrogen peroxide or just H2 or potassium hydride or H2S? You added water, because it’s NEEDED to add or did you just feel like it? "
Well, let's figure it out. Well, first of all, we naturally have no right to add substances to the reaction equation at our will. The reaction goes exactly the way it goes; as nature ordered. Our likes and dislikes cannot influence the course of the process. We can try to change the reaction conditions (increase the temperature, add a catalyst, change the pressure), but if the reaction conditions are specified, its result can no longer depend on our will. Thus, the formula for water in the equation of the last reaction is not my desire, but a fact.
Secondly, you can try to equalize the reaction in cases where the substances you listed are present instead of water. I assure you: in no case will you be able to do this.
Thirdly, options with H 2 O 2, H 2, KH or H 2 S are simply unacceptable in this case for one reason or another. For example, in the first case, the oxidation state of oxygen changes, in the second and third - hydrogen, and we agreed that the oxidation state will change only for Mn and S. In the fourth case, sulfur generally acted as an oxidizing agent, and we agreed that S - reducing agent. In addition, potassium hydride is unlikely to "survive" in an aqueous medium (and the reaction, recall, takes place in an aqueous solution), and H 2 S (even if this substance was formed) will inevitably enter into a solution with KOH. As you can see, knowledge of chemistry allows us to reject these issues.
"But why exactly water?" - you ask.
Yes, because, for example, in this process (as in many others), water acts as a solvent. Because, for example, if you analyze all the reactions that you have written in 4 years of studying chemistry, you will find that H 2 O occurs in almost half of the equations. Water is generally quite a "popular" compound in chemistry.
Understand, I am not saying that every time in Problem 30 you need to "send hydrogen somewhere" or "get oxygen from somewhere," you need to grab onto water. But this will probably be the first substance to think about.
Similar logic is used for the equations of reactions in acidic and neutral media. In the first case, it is necessary to add the formula of water to the right side, in the second - potassium hydroxide:
KMnO 4 + H 2 SO 4 + K 2 SO 3 \u003d MnSO 4 + K 2 SO 4 + H 2 O,
KMnO 4 + H 2 O + K 2 SO 3 \u003d MnO 2 + K 2 SO 4 + KOH.
The arrangement of coefficients for highly experienced young chemists should not cause the slightest difficulty. Final answer:
2KMnO 4 + 3H 2 SO 4 + 5K 2 SO 3 \u003d 2MnSO 4 + 6K 2 SO 4 + 3H 2 O,
2KMnO 4 + H 2 O + 3K 2 SO 3 \u003d 2MnO 2 + 3K 2 SO 4 + 2KOH.
In the next part, we will talk about the reduction products of chromates and dichromates, about nitric and sulfuric acids.
Part I
Problem number 30 on the exam in chemistry is devoted to the topic "Redox reactions". Previously, a job of this type was included in version of the exam under the number C1.
The meaning of task 30: it is necessary to arrange the coefficients in the reaction equation using the electronic balance method. Usually, in the condition of the problem, only the left side of the equation is given, the student must independently complete the right side.
The complete solution of the problem is estimated at 3 points. One point is given for the determination of the oxidizing agent and the reducing agent, another point is given directly for the construction of the electronic balance, the last one is for the correct arrangement of the coefficients in the reaction equation. Note: at the Unified State Exam-2018, the maximum score for solving task 30 will be 2 points.
In my opinion, the most difficult thing in this process is the first step. Not everyone is able to correctly predict the result of the reaction. If the interaction products are indicated correctly, all subsequent stages are already a matter of technology.
First step: remembering oxidation states
We must start with a concept element oxidation state... If you are not yet familiar with this term, refer to the Oxidation State section in a Chemistry Handbook. You must learn to confidently determine the oxidation states of all elements in inorganic compounds and even in the simplest organic substances. Without 100% understanding of this topic, it is pointless to move on.
Step two: oxidizing and reducing agents. Oxidation - reduction reactions
I want to remind you that all chemical reactions in nature can be divided into two types: redox and occurring without changing the oxidation state.
In the course of OVR (we will use just such a reduction below for redox reactions), some elements change their oxidation states.
Element whose oxidation state goes downis called oxidizing agent.
Element whose oxidation state risesis called reducing agent.
The oxidizing agent is reduced during the reaction.
The reducing agent is oxidized during the reaction.
Example 1... Consider the reaction of sulfur with fluorine:
S + 3F 2 \u003d SF 6.
Arrange the oxidation states of all elements yourself. We see that the oxidation state of sulfur increases (from 0 to +6), and the oxidation state of fluorine decreases (from 0 to -1). Conclusion: S is a reducing agent, F 2 is an oxidizing agent. During the process, sulfur is oxidized and fluorine is reduced.
Example 2... Let's discuss the reaction of manganese (IV) oxide with hydrochloric acid:
MnO 2 + 4HCl \u003d MnCl 2 + Cl 2 + 2H 2 O.
During the reaction, the oxidation state of manganese decreases (from +4 to +2), and the oxidation state of chlorine increases (from -1 to 0). Conclusion: manganese (in the composition of MnO 2) - an oxidizing agent, chlorine (in the composition of HCl - a reducing agent). Chlorine is oxidized, manganese is reduced.
Note that in the last example, not all chlorine atoms have changed their oxidation states. This did not affect our conclusions in any way.
Example 3... Thermal decomposition of ammonium dichromate:
(NH 4) 2 Cr 2 O 7 \u003d Cr 2 O 3 + N 2 + 4H 2 O.
We see that both the oxidizing agent and the reducing agent are in the composition of one "molecule": chromium changes the oxidation state from +6 to +3 (i.e., it is an oxidizing agent), and nitrogen - from -3 to 0 (therefore, nitrogen - reducing agent).
Example 4... Interaction of nitrogen dioxide with an aqueous solution of alkali:
2NO 2 + 2NaOH \u003d NaNO 3 + NaNO 2 + H 2 O.
By arranging the oxidation states (I hope you do this without difficulty!), We find a strange picture: the oxidation state of only one element, nitrogen, changes. Some of the N atoms increase their oxidation state (from +4 to +5), some decrease (from +4 to +3). In fact, there is nothing strange about this! In this process, N (+4) is both an oxidizing agent and a reducing agent.
Let's talk a little about the classification of redox reactions. Let me remind you that all OVRs are divided into three types:
- 1) intermolecular OVR (oxidizing agent and reducing agent are contained in different molecules);
- 2) intramolecular OVR (oxidizing agent and reducing agent are in one molecule);
- 3) disproportionation reactions (oxidizing agent and reducing agent are atoms of one element with the same initial oxidation state in the composition of one molecule).
I think that based on these definitions, you will easily understand that the reactions from examples 1 and 2 refer to intermolecular ORP, the decomposition of ammonium dichromate is an example of an intramolecular ORP, and the interaction of NO 2 with alkali is an example of a disproportionation reaction.
Step three: we begin to master the method of electronic balance
To check how well you have mastered the previous material, I will ask you a simple question: "Is it possible to give an example of a reaction in which oxidation occurs, but there is no reduction, or, conversely, there is oxidation but no reduction?"
The correct answer is: "No, you can't!"
Indeed, let the oxidation state of element X increase during the reaction. This means that X donates electrons... But to whom? After all, electrons cannot simply evaporate, disappear without a trace! There is some other element Y whose atoms will accept these electrons. Electrons have a negative charge, therefore, the oxidation state of Y will decrease.
Conclusion: if there is a reducing agent X, then there will certainly be an oxidizing agent Y! Moreover, the number of electrons donated by one element will be exactly equal to the number of electrons donated by another element.
It is on this fact that electronic balance methodused in task C1.
Let's start to master this method with examples.
Example 4
C + HNO 3 \u003d CO 2 + NO 2 + H 2 O
electronic balance method.
Decision... Let's start by determining the oxidation states (do it yourself!). We see that during the process, two elements change their oxidation states: C (from 0 to +4) and N (from +5 to +4).
It is obvious that carbon is a reducing agent (is oxidized), and nitrogen (+5) (in the composition of nitric acid) is an oxidizing agent (is reduced). By the way, if you correctly identified the oxidizer and v-tel, you are already guaranteed 1 point for problem N 30!
Now the fun begins. Let's write the so-called. half-reactions of oxidation and reduction:
The carbon atom parts with 4 electrons, the nitrogen atom takes 1 unit. The number of donated electrons is not equal to the number of electrons received. This is bad! The situation needs to be corrected.
Let's "multiply" the first half-reaction by 1, and the second - by 4.
| C (0) - 4e | = | C (+4) | (1) |
| N (+5) + 1e | = | N (+4) | (4) |
Now everything is fine: for one carbon atom (giving up 4 e) there are 4 nitrogen atoms (each of which takes one e). The number of electrons donated is equal to the number of electrons received!
What we have now written is actually called electronic balance... If you write this balance correctly on the real USE in chemistry, you are guaranteed another 1 point for task C1.
The last stage: it remains to transfer the obtained coefficients into the reaction equation. We do not change anything in front of the C and CO 2 formulas (since the coefficient 1 is not put in the equation), before the HNO 3 and NO 2 formulas we put a four (since the number of nitrogen atoms in the left and right sides of the equation should be 4) :
C + 4HNO 3 \u003d CO 2 + 4NO 2 + H 2 O.
It remains to do the last check: we see that the number of nitrogen atoms is the same on the left and right, the same applies to C atoms, but there are still problems with hydrogen and oxygen. But everything is easy to fix: we put the coefficient 2 in front of the formula H 2 O and we get the final answer:
C + 4HNO 3 \u003d CO 2 + 4NO 2 + 2H 2 O.
That's all! The problem is solved, the coefficients are arranged, and we got another point for the correct equation. Result: 3 points for a perfectly solved problem 30. Congratulations on that!
Example 5... Place the coefficients in the reaction equation
NaI + H 2 SO 4 \u003d Na 2 SO 4 + H 2 S + I 2 + H 2 O
electronic balance method.
Decision... Arrange the oxidation states of all elements yourself. We see that during the process, two elements change their oxidation states: S (from +6 to -2) and I (from -1 to 0).
Sulfur (+6) (in sulfuric acid) is an oxidizing agent, and iodine (-1) in NaI is a reducing agent. During the reaction, I (-1) is oxidized, S (+6) is reduced.
We write down the half-reactions of oxidation and reduction:
pay attention to important point: there are two atoms in the iodine molecule. "Half" of the molecule cannot participate in the reaction, therefore, in the corresponding equation we write not I, but I 2.
Let's "multiply" the first half-reaction by 4, and the second by 1.
| 2I (-1) - 2e | = | I 2 (0) | (4) |
| S (+6) + 8e | = | S (-2) | (1) |
The balance is built, for 8 donated electrons there are 8 received.
We transfer the coefficients to the reaction equation. Before the formula I 2 we put 4, before the formula H 2 S - we mean the coefficient 1 - this, I think, is obvious.
NaI + H 2 SO 4 \u003d Na 2 SO 4 + H 2 S + 4I 2 + H 2 O
But further questions may arise. First, it would be wrong to put a four in front of the NaI formula. Indeed, already in the oxidation half-reaction itself, before the symbol I, there is a coefficient 2. Therefore, not 4, but 8 should be written on the left side of the equation!
8NaI + H 2 SO 4 \u003d Na 2 SO 4 + H 2 S + 4I 2 + H 2 O
Secondly, in such a situation, graduates often put a factor of 1 in front of the sulfuric acid formula. They argue as follows: "In the half-reaction of reduction, a coefficient of 1 was found, this coefficient refers to S, which means that there should be one before the sulfuric acid formula."
This reasoning is wrong! Not all sulfur atoms changed the oxidation state, some of them (in the composition of Na 2 SO 4) retained the +6 oxidation state. These atoms are not included in the electronic balance and the coefficient 1 has nothing to do with them.
All this, however, will not prevent us from completing the decision. It is only important to understand that in further considerations we no longer rely on electronic balance, but simply on common sense. So, let me remind you that the coefficients in front of H 2 S, NaI and I 2 are "frozen" and cannot be changed. But the rest are possible and necessary.
On the left side of the equation there are 8 sodium atoms (in the composition of NaI), on the right - so far only 2 atoms. We put a factor of 4 in front of the sodium sulfate formula:
8NaI + H 2 SO 4 \u003d 4Na 2 SO 4 + H 2 S + 4I 2 + H 2 O.
Only now can the number of S atoms be equalized. There are 5 of them on the right, therefore, a factor of 5 must be put in front of the sulfuric acid formula:
8NaI + 5H 2 SO 4 \u003d 4Na 2 SO 4 + H 2 S + 4I 2 + H 2 O.
The last problem: hydrogen and oxygen. Well, I think you yourself guessed that there is not enough coefficient 4 in front of the water formula on the right side:
8NaI + 5H 2 SO 4 \u003d 4Na 2 SO 4 + H 2 S + 4I 2 + 4H 2 O.
We check everything carefully again. Yes everything is correct! The problem is solved, we got our legal 3 points.
So in examples 4 and 5 we discussed in detail algorithm for solving problem C1 (30)... In your solution to a real exam problem, the following points must be present:
- 1) the oxidation state of ALL elements;
- 2) an indication of the oxidizing agent and the reducing agent;
- 3) electronic balance scheme;
- 4) the final reaction equation with coefficients.
A few comments on the algorithm.
1. The oxidation states of all elements on the left and right sides of the equation should be indicated. Everyone, not just an oxidizing agent and a reducing agent!
2. The oxidizing agent and the reducing agent must be marked clearly and clearly: the element X (+ ...) in the composition ... is an oxidizing agent, is reduced; element Y (...) in the composition ... is a reducing agent, it is oxidized. Not everyone can decipher the inscription in small handwriting "ok. V-sya" under the formula of sulfuric acid as "sulfur (+6) in the composition of sulfuric acid - an oxidizing agent, is reduced".
Don't spare letters! You are not giving an advertisement to the newspaper: "Sd. Room with all. Ud."
3. The electronic balance scheme is just a scheme: two half-reactions and the corresponding coefficients.
4. Detailed explanations of exactly how you set the coefficients in the equation are not needed by anyone on the exam. It is only necessary that all the numbers are correct, and the entry itself is made in legible handwriting. Be sure to check yourself several times!
And once again about the assessment of task C1 on the exam in chemistry:
- 1) determination of the oxidizing agent (oxidants) and reducing agent (reducing agents) - 1 point;
- 2) electronic balance scheme with correct coefficients - 1 point;
- 3) the basic reaction equation with all coefficients - 1 point.
Result: 3 points for the complete solution of problem N 30.
Note: I remind you once again that at the Unified State Exam-2018 the maximum score for solving problem No. 30 will be 2 points.
I'm sure you get the idea of \u200b\u200bthe Electronic Balance Method. We understood in general terms how the solution to Example No. 30 is constructed. In principle, everything is not so difficult!
Unfortunately, on the real USE in chemistry, the following problem arises: the reaction equation itself is not completely given. That is, the left side of the equation is present, and the right side either contains nothing at all, or the formula of one substance is indicated. You will have to yourself, relying on your knowledge, to supplement the equation, and only then start placing the coefficients.
This can be tricky. There are no universal recipes for writing equations. In the next part we will discuss this issue in more detail and look at more complex examples.
How to solve problems C1 (36) on the exam in chemistry. Part I
Task No. 36 on the exam in chemistry is devoted to the topic "Redox reactions". Previously, the task of this type was included in the version of the exam under the number C1.
The meaning of the task C1: it is necessary to arrange the coefficients in the reaction equation using the electronic balance method. Usually, in the condition of the problem, only the left side of the equation is given, the student must independently complete the right side.
The complete solution to the problem is estimated at 3 points. One point is given for the determination of the oxidizing agent and the reducing agent, one more point is given directly for the construction of the electronic balance, the last point is for the correct arrangement of the coefficients in the reaction equation.
In my opinion, the most difficult thing in this process is the first step. Not everyone is able to correctly predict the result of the reaction. If the interaction products are indicated correctly, all subsequent stages are already a matter of technology.
First step: remembering oxidation states
We must start with a concept element oxidation state... If you are not yet familiar with this term, refer to the Oxidation State section in a Chemistry Handbook. You must learn to confidently determine the oxidation states of all elements in inorganic compounds and even in the simplest organic substances. Without 100% understanding of this topic, it is pointless to move on.
Step two: oxidizing and reducing agents. Oxidation - reduction reactions
I want to remind you that all chemical reactions in nature can be divided into two types: redox and occurring without changing the oxidation state.
In the course of OVR (we will use just such a reduction below for redox reactions), some elements change their oxidation states.
Example 1... Consider the reaction of sulfur with fluorine:
S + 3F 2 \u003d SF 6.
Arrange the oxidation states of all elements yourself. We see that the oxidation state of sulfur increases (from 0 to +6), and the oxidation state of fluorine decreases (from 0 to -1). Conclusion: S is a reducing agent, F 2 is an oxidizing agent. During the process, sulfur is oxidized and fluorine is reduced.
Example 2... Let's discuss the reaction of manganese (IV) oxide with hydrochloric acid:
MnO 2 + 4HCl \u003d MnCl 2 + Cl 2 + 2H 2 O.
During the reaction, the oxidation state of manganese decreases (from +4 to +2), and the oxidation state of chlorine increases (from -1 to 0). Conclusion: manganese (in the composition of MnO 2) - an oxidizing agent, chlorine (in the composition of HCl - a reducing agent). Chlorine is oxidized, manganese is reduced.
Note that in the last example, not all chlorine atoms have changed their oxidation states. This did not affect our conclusions in any way.
Example 3... Thermal decomposition of ammonium dichromate:
(NH 4) 2 Cr 2 O 7 \u003d Cr 2 O 3 + N 2 + 4H 2 O.
We see that both the oxidizing agent and the reducing agent are in the composition of one "molecule": chromium changes the oxidation state from +6 to +3 (i.e., it is an oxidizing agent), and nitrogen - from -3 to 0 (therefore, nitrogen - reducing agent).
Example 4... Interaction of nitrogen dioxide with an aqueous solution of alkali:
2NO 2 + 2NaOH \u003d NaNO 3 + NaNO 2 + H 2 O.
By arranging the oxidation states (I hope you do this without difficulty!), We find a strange picture: the oxidation state of only one element, nitrogen, changes. Some of the N atoms increase their oxidation state (from +4 to +5), some decrease (from +4 to +3). In fact, there is nothing strange about this! In this process, N (+4) is both an oxidizing agent and a reducing agent.
Let's talk a little about the classification of redox reactions. Let me remind you that all OVRs are divided into three types:
- 1) intermolecular OVR (oxidizing agent and reducing agent are contained in different molecules);
- 2) intramolecular OVR (oxidizing agent and reducing agent are in one molecule);
- 3) disproportionation reactions (oxidizing agent and reducing agent are atoms of one element with the same initial oxidation state in the composition of one molecule).
I think that based on these definitions, you will easily understand that the reactions from examples 1 and 2 refer to intermolecular ORP, the decomposition of ammonium dichromate is an example of an intramolecular ORP, and the interaction of NO 2 with alkali is an example of a disproportionation reaction.
Step three: begin to master the method of electronic balance
To check how well you have mastered the previous material, I will ask you a simple question: "Is it possible to give an example of a reaction in which oxidation occurs, but there is no reduction, or, conversely, there is oxidation but no reduction?"
The correct answer is: "No, you can't!"
Indeed, let the oxidation state of element X increase during the reaction. This means that X donates electrons... But to whom? After all, electrons cannot simply evaporate, disappear without a trace! There is some other element Y whose atoms will accept these electrons. Electrons have a negative charge, therefore, the oxidation state of Y will decrease.
Conclusion: if there is a reducing agent X, then there will certainly be an oxidizing agent Y! Moreover, the number of electrons donated by one element will be exactly equal to the number of electrons donated by another element.
It is on this fact that electronic balance methodused in task C1.
Let's start to master this method with examples.
Example 4
C + HNO 3 \u003d CO 2 + NO 2 + H 2 O
electronic balance method.
Decision... Let's start by determining the oxidation states (do it yourself!). We see that during the process, two elements change their oxidation states: C (from 0 to +4) and N (from +5 to +4).
It is obvious that carbon is a reducing agent (is oxidized), and nitrogen (+5) (in the composition of nitric acid) is an oxidizing agent (is reduced). By the way, if you correctly identified the oxidizer and v-tel, you are already guaranteed 1 point for problem N 36!
Now the fun begins. Let's write the so-called. half-reactions of oxidation and reduction:
The carbon atom parts with 4 electrons, the nitrogen atom takes 1 unit. The number of donated electrons is not equal to the number of electrons received. This is bad! The situation needs to be corrected.
Let's "multiply" the first half-reaction by 1, and the second - by 4.
| C (0) - 4e | = | C (+4) | (1) |
| N (+5) + 1e | = | N (+4) | (4) |
Now everything is fine: for one carbon atom (giving up 4 e) there are 4 nitrogen atoms (each of which takes one e). The number of electrons donated is equal to the number of electrons received!
What we have now written is actually called electronic balance... If you write this balance correctly on the real USE in chemistry, you are guaranteed another 1 point for task C1.
The last stage: it remains to transfer the obtained coefficients into the reaction equation. We do not change anything in front of the C and CO 2 formulas (since the coefficient 1 is not put in the equation), before the HNO 3 and NO 2 formulas we put a four (since the number of nitrogen atoms in the left and right sides of the equation should be 4) :
C + 4HNO 3 \u003d CO 2 + 4NO 2 + H 2 O.
It remains to do the last check: we see that the number of nitrogen atoms is the same on the left and right, the same applies to C atoms, but there are still problems with hydrogen and oxygen. But everything is easy to fix: we put the coefficient 2 in front of the formula H 2 O and we get the final answer:
C + 4HNO 3 \u003d CO 2 + 4NO 2 + 2H 2 O.
That's all! The problem is solved, the coefficients are arranged, and we got another point for the correct equation. Result: 3 points for the ideally solved problem C 1. Congratulations on that!
Example 5... Place the coefficients in the reaction equation
NaI + H 2 SO 4 \u003d Na 2 SO 4 + H 2 S + I 2 + H 2 O
electronic balance method.
Decision... Arrange the oxidation states of all elements yourself. We see that during the process, two elements change their oxidation states: S (from +6 to -2) and I (from -1 to 0).
Sulfur (+6) (in sulfuric acid) is an oxidizing agent, and iodine (-1) in NaI is a reducing agent. During the reaction, I (-1) is oxidized, S (+6) is reduced.
We write down the half-reactions of oxidation and reduction:
Pay attention to an important point: there are two atoms in the iodine molecule. "Half" of the molecule cannot participate in the reaction, therefore, in the corresponding equation we write not I, but I 2.
Let's "multiply" the first half-reaction by 4, and the second by 1.
| 2I (-1) - 2e | = | I 2 (0) | (4) |
| S (+6) + 8e | = | S (-2) | (1) |
The balance is built, for 8 donated electrons there are 8 received.
We transfer the coefficients to the reaction equation. Before the formula I 2 we put 4, before the formula H 2 S - we mean the coefficient 1 - this, I think, is obvious.
NaI + H 2 SO 4 \u003d Na 2 SO 4 + H 2 S + 4I 2 + H 2 O
But further questions may arise. First, it would be wrong to put a four in front of the NaI formula. Indeed, already in the oxidation half-reaction itself, before the symbol I, there is a coefficient 2. Therefore, not 4, but 8 should be written on the left side of the equation!
8NaI + H 2 SO 4 \u003d Na 2 SO 4 + H 2 S + 4I 2 + H 2 O
Secondly, in such a situation, graduates often put a factor of 1 in front of the sulfuric acid formula. They argue as follows: "In the half-reaction of reduction, a coefficient of 1 was found, this coefficient refers to S, which means that there should be one before the sulfuric acid formula."
This reasoning is wrong! Not all sulfur atoms changed the oxidation state, some of them (in the composition of Na 2 SO 4) retained the +6 oxidation state. These atoms are not included in the electronic balance and the coefficient 1 has nothing to do with them.
All this, however, will not prevent us from completing the decision. It is only important to understand that in further considerations we no longer rely on electronic balance, but simply on common sense. So, let me remind you that the coefficients in front of H 2 S, NaI and I 2 are "frozen" and cannot be changed. But the rest are possible and necessary.
On the left side of the equation there are 8 sodium atoms (in the composition of NaI), on the right - so far only 2 atoms. We put a factor of 4 in front of the sodium sulfate formula:
8NaI + H 2 SO 4 \u003d 4Na 2 SO 4 + H 2 S + 4I 2 + H 2 O.
Only now can the number of S atoms be equalized. There are 5 of them on the right, therefore, a factor of 5 must be put in front of the sulfuric acid formula:
8NaI + 5H 2 SO 4 \u003d 4Na 2 SO 4 + H 2 S + 4I 2 + H 2 O.
The last problem: hydrogen and oxygen. Well, I think you yourself guessed that there is not enough coefficient 4 in front of the water formula on the right side:
8NaI + 5H 2 SO 4 \u003d 4Na 2 SO 4 + H 2 S + 4I 2 + 4H 2 O.
We check everything carefully again. Yes everything is correct! The problem is solved, we got our legal 3 points.
So in examples 4 and 5 we discussed in detail algorithm for solving problem C1... In your solution to a real exam problem, the following points must be present:
- 1) the oxidation state of ALL elements;
- 2) an indication of the oxidizing agent and the reducing agent;
- 3) electronic balance scheme;
- 4) the final reaction equation with coefficients.
A few comments on the algorithm.
1. The oxidation states of all elements on the left and right sides of the equation should be indicated. Everyone, not just an oxidizing agent and a reducing agent!
2. The oxidizing agent and the reducing agent must be marked clearly and clearly: the element X (+ ...) in the composition ... is an oxidizing agent, is reduced; element Y (...) in the composition ... is a reducing agent, it is oxidized. Not everyone can decipher the inscription in small handwriting "ok. V-sya" under the formula of sulfuric acid as "sulfur (+6) in the composition of sulfuric acid - an oxidizing agent, is reduced".
Don't spare letters! You are not giving an advertisement to the newspaper: "Sd. Room with all. Ud."
3. The electronic balance scheme is just a scheme: two half-reactions and the corresponding coefficients.
4. Detailed explanations of exactly how you set the coefficients in the equation are not needed by anyone on the exam. It is only necessary that all the numbers are correct, and the entry itself is made in legible handwriting. Be sure to check yourself several times!
And once again about the assessment of the C1 task on the exam in chemistry:
- 1) determination of the oxidizing agent (oxidants) and reducing agent (reducing agents) - 1 point;
- 2) electronic balance scheme with correct coefficients - 1 point;
- 3) the basic reaction equation with all coefficients - 1 point.
Result: 3 points for the complete solution of problem N 36.
I'm sure you get the idea of \u200b\u200bthe Electronic Balance Method. Understood in general terms how the solution of example C1 is built. In principle, everything is not so difficult!
Unfortunately, on the real USE in chemistry, the following problem arises: the reaction equation itself is not completely given. That is, the left side of the equation is present, and the right side either contains nothing at all, or the formula of one substance is indicated. You will have to yourself, relying on your knowledge, to supplement the equation, and only then start placing the coefficients.
This can be tricky. There are no universal recipes for writing equations. In the next part we will discuss this issue in more detail and look at more complex examples.
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Redox reactions
For the correct task you will receive 2 points... Approximately 10-15 minutes.
To complete task 30 in chemistry, you must:
- know what is
- be able to draw up equations for redox reactions
UMK Kuznetsova line. Chemistry (10-11) (D)
UMK Kuznetsova line. Chemistry (10-11) (B)
UMK line N. E. Kuznetsova. Chemistry (10-11) (basic)
Organization of preparation for the exam in chemistry: redox reactions
How should the work in the lesson be organized so that the students achieve good results on the exam?The material was prepared based on the materials of the webinar "Organization of preparation for the exam in chemistry: redox reactions"
“We are considering organizing preparation for the successful completion of tasks related to redox reactions. If we look at the specification and demo version, then such reactions are directly related to tasks # 10 and # 30, but this is the key topic of the school chemistry course. It touches upon a variety of issues, a variety of properties of chemicals. It is very extensive, ”emphasizes Lydia Asanova, host of the webinar, candidate of pedagogical sciences, author of teaching aids.
Task number 30, which considers redox reactions, is a task of a high level of complexity. To get the highest score (3) for its completion, the student's answer should be:
- determination of the oxidation state of elements that are oxidizing and reducing agents;
- oxidizing agent and reducing agent (elements or substances);
- oxidation and reduction processes, and on their basis the compiled electronic (electronic-ionic) balance;
- determination of substances missing in the reaction equation.
However, students often skip, do not assign coefficients, do not indicate the oxidizing agent and reducing agent, the oxidation state. How should you organize the work in the lesson in order to achieve good results on the exam?
Particular attention in the textbook by OS Gabrielyan for grade 10, intended for studying the subject in the amount of 3-4 hours a week, is given to applied topics: the manual covers issues of ecology, medicine, biology and culture related to chemistry. In the 11th grade, the course is completed and generalized.
1. Preparation for the exam should be carried out in the process of teaching the subject and should not be limited to training only in performing tasks similar to tasks examination work... Such "coaching" does not develop thinking, does not deepen understanding. But, by the way, in the examination task it is indicated that other formulations of the answer are allowed that do not distort its meaning. This means that by creatively, with understanding, approaching the solution of the task, you can get the highest score for performance, even if the answer is formulated differently.
The main task of preparing for the exam is purposeful work on repetition, systematization and generalization of the studied material, on bringing the key concepts of the chemistry course into the knowledge system. Of course, experience in conducting a real chemical experiment is required.
2. There is a list of topics and concepts that students should not forget at all. Among them:
- rules for determining the oxidation states of atoms (in simple substances, the oxidation state of elements is zero, the highest (maximum) oxidation state of elements of groups II-VII, as a rule, is equal to the number of the group in which the element is located in the periodic table, the lowest (minimum) oxidation state of metals equal to zero, etc.);
- the most important oxidants and reducing agents, as well as the fact that the oxidation process is always accompanied by a reduction process;
- redox duality;
- types of OVR (intermolecular, intramolecular, counterproportionation reactions, disproportionation reactions (self-oxidation-self-healing)).
The table contains the types of redox reactions, factors affecting the course of reactions (photo pages). Examples are analyzed in detail, and, in addition, there are assignments on the topic "OVR" in the format of the exam.
For instance:
“Using the electronic balance method, write the equation for the chemical reaction:
N 2 O + KMnO 4 +… \u003d NO 2 +… + K 2 SO 4 + H 2 O
Indicate oxidant and reductant. "
However, a variety of examples are given to practice solving problems. For example, in the textbook “Chemistry. Advanced level. Grade 11. Test papers" there are such:
“Based on the theory of redox processes, indicate the schemes of impossible reactions.
SO 2 + H 2 S → S + H 2 O
S + H 2 SO 4 → SO 2 + H 2 O
S + H 2 SO 4 → H 2 S + H 2 O
K 2 SO 3 + K 2 Cr 2 O 7 + H 2 SO 4 → K 2 SO 4 + K 2 CrO 4 + H 2 O
KMnO 4 + HCl → Cl2 + MnCl 2 + KCl + H 2 O
I 2 + SO 2 + H 2 O → HIO 3 + H 2 SO 4
Justify the answer. Convert diagrams of possible processes into reaction equations. Specify oxidant and reductant "
"Make up the reaction equations in accordance with the scheme for changing the oxidation states of carbon atoms: С 0 → С - 4 → С –4 → С +4 → С +2 → С –2."
“Substances are given: carbon, nitrogen oxide (IV), sulfur oxide (IV), water solution potassium hydroxide. Write the equations of four possible reactions between these substances, without repeating the pair of reagents. "
All this allows you to fully study the topic of redox reactions and work out the solution of a variety of problems.
* Since May 2017, the joint publishing group DROFA-VENTANA is a part of the Russian Textbook Corporation. The corporation also includes the Astrel publishing house and the LECTA digital educational platform. Alexander Brychkin, a graduate of Financial Academy under the Government of the Russian Federation, candidate of economic sciences, head of innovative projects of the DROFA publishing house in the field of digital education (electronic forms of textbooks, Russian electronic school, digital educational platform LECTA). Prior to joining the DROFA publishing house, he held the position of Vice President for Strategic Development and Investments of the EKSMO-AST Publishing Holding. Today the publishing corporation "Russian Textbook" has the largest portfolio of textbooks included in the Federal List - 485 titles (approximately 40%, excluding textbooks for a special school). The corporation's publishing houses own the most popular Russian schools sets of textbooks on physics, drawing, biology, chemistry, technology, geography, astronomy - areas of knowledge that are needed to develop the country's production potential. The corporation's portfolio includes textbooks and tutorials for primary school, awarded the Presidential Prize in Education. These are textbooks and manuals on subject areas that are necessary for the development of scientific, technical and production potential of Russia.
