On the eve of the Unified State Examination in Chemistry on June 4, 2018, photos with problems for electrolysis and portions flashed on the Internet. And taken together. Let me remind you that earlier electrolysis tasks were not encountered in real exam tasks.
Leaves look like this:
All this suggests that the tasks real USE again leaked to the Internet before the exam. Well, we'll see tomorrow.
In the meantime, I would like to recall how they solve problems for electrolysis and for portions.
Task 1.
During the electrolysis of 500 g of a 16% copper (II) sulfate solution, the process was stopped when 1.12 liters of gas were released at the anode. A portion weighing 98.4 g was taken from the resulting solution. Calculate the mass of a 20% sodium hydroxide solution necessary for the complete precipitation of copper ions from the selected portion of the solution.
Solution.
First, we compose the reaction equation for the electrolysis of a solution of copper sulfate. How to do this is described in detail in the article.
2CuSO 4 + 2H 2 O → 2Cu + 2H 2 SO 4 + O 2
Find the mass of pure copper sulfate:
m (CuSO 4) \u003d m solution * ω (CuSO 4) \u003d 500 * 0.16 \u003d 80 g
The amount of copper sulfate substance:
ν (CuSO 4) \u003d m / M \u003d 80/160 \u003d 0.5 mol
It can be seen that 0.25 mol of gas, or 5.6 liters, should be released at the anode.
However, the condition says that only 1.12 liters of gas were released. Consequently, copper sulfate did not react completely, but only partially.
We find the amount and mass of oxygen that was released at the anode:
ν (O 2) \u003d V / V m \u003d 1.12 / 22.4 \u003d 0.05 mol,
m(O 2) \u003d ν * M \u003d 0.05 * 32 \u003d 1.6 g.
Consequently, 0.1 mol of copper sulfate entered the electrolysis.
remains in the solution 0.4 mol copper sulfate. Wherein formed 0.1 mol of sulfuric acid weighing 9.8 g and 0.1 mol of copper precipitated (mass of copper 6.4 g).
In this case, the mass of the resulting solution after electrolysis m r-ra2 is equal to:
m solution 2 = 500 - 1.6 - 6.4 = 492 g
A portion weighing 98.4 g was taken from the resulting solution. In this case, the amount of dissolved substances changed. But their mass fraction has not changed. Let's put the mass fraction of copper sulfate ω (CuSO 4) 2 and sulfuric acid ω (H 2 SO 4) in the solution that remained after electrolysis:
m(CuSO 4) rest \u003d ν * M \u003d 0.4 * 160 \u003d 64 g
ω (CuSO 4) 2 \u003d m (CuSO 4) 2 / * m p-ra2 \u003d 64/492 \u003d 0.13 \u003d 13%
ω (H 2 SO 4) \u003d m (H 2 SO 4) / * m p-ra2 \u003d 9.8 / 492 \u003d 0.02 \u003d 2%
Let's find the mass and amount of sulfuric acid and the mass of copper sulfate in a portion with a mass of m p-ra3 = 98.4 g, which we selected:
m (CuSO 4) 3 \u003d ω (CuSO 4) 2 * m solution 3 \u003d 0.13 * 98.4 \u003d 12.79 g
m (H 2 SO 4) 2 \u003d ω (H 2 SO 4) * m p-ra3 \u003d 0.02 * 98.4 \u003d 1.97 g
ν (CuSO 4) \u003d m / M \u003d 12.79 / 160 \u003d 0.08 mol
ν (H 2 SO 4) \u003d m / M \u003d 1.97 / 98 \u003d 0.02 mol
To precipitate copper ions, sodium hydroxide must react with both sulfuric acid in solution and copper sulfate:
H 2 SO 4 + 2NaOH \u003d Na 2 SO 4 + 2H 2 O
CuSO 4 + 2NaOH \u003d Cu (OH) 2 + 2H 2 O
In the first reaction, 0.04 mol of sodium hydroxide is consumed, in the second reaction, 0.16 mol of sodium hydroxide. In total, 0.2 mol of sodium hydroxide is required. Or 8 g of pure NaOH, which corresponds to 40 g of a 20% solution. Thus, the answer on the leaflets distributed on the Internet is slightly wrong.
Other tasks for electrolysis:
2. The electrolysis of 282 g of a 40% solution of copper (II) nitrate was stopped after the mass of the solution decreased by 32 g. 140 g of a 40% solution of sodium hydroxide was added to the resulting solution. Determine the mass fraction of alkali in the resulting solution.
3. During the electrolysis of 340 g of a 20% solution of silver (I) nitrate, the process was stopped when 1.12 l of gas was released at the anode. A portion weighing 79.44 g was taken from the resulting solution. Calculate the mass of a 10% sodium chloride solution necessary for the complete precipitation of silver ions from the selected portion of the solution.
4. During the electrolysis of 312 g of a 15% sodium chloride solution, the process was stopped when 6.72 liters of gas were released at the cathode. A portion weighing 58.02 g was taken from the resulting solution. Calculate the mass of a 20% copper (II) sulfate solution necessary for the complete precipitation of hydroxide ions from the selected portion of the solution.
5. The electrolysis of 640 g of a 15% solution of copper (II) sulfate was stopped after the mass of the solution decreased by 32 g. 400 g of a 20% solution of sodium hydroxide was added to the resulting solution. Determine the mass fraction of alkali in the resulting solution.
6. During the electrolysis of 360 g of 18.75% copper (II) chloride solution, the process was stopped when 4.48 liters of gas were released at the anode. A portion weighing 22.2 g was taken from the resulting solution. Calculate the mass of a 20% sodium hydroxide solution necessary for the complete precipitation of copper ions from the selected portion of the solution.
7. During the electrolysis of 624 g of a 10% solution of barium chloride, the process was stopped when 4.48 liters of gas were released at the cathode. A portion weighing 91.41 g was taken from the resulting solution. Calculate the mass of a 10% sodium carbonate solution necessary for the complete precipitation of barium ions from the selected portion of the solution.
8. When carrying out the electrolysis of 500 g of a 16% solution of copper (II) sulfate, stop the process when 1.12 liters of gas are released at the anode. To the resulting solution was added 53 g of a 10% sodium carbonate solution. Determine the mass fraction of copper (II) sulfate in the resulting solution.
Establish a correspondence between the salt formula and the product formed on an inert anode during its electrolysis aqueous solution: for each position indicated by a letter, select the corresponding position indicated by a number.
| SALT FORMULA | PRODUCT ON ANODE | |
| A | B | AT | G |
Solution.
In the electrolysis of aqueous solutions of salts, alkalis and acids on an inert anode:
Water is discharged and oxygen is released if it is a salt of an oxygen-containing acid or a salt of hydrofluoric acid;
Hydroxide ions are discharged and oxygen is released if it is alkali;
The acid residue that is part of the salt is discharged, and the corresponding simple substance is released if it is a salt of an oxygen-free acid (except for).
The process of electrolysis of salts of carboxylic acids takes place in a special way.
Answer: 3534.
Answer: 3534
Source: Yandex: training work USE in chemistry. Option 1.
Establish a correspondence between the formula of a substance and the product formed on the cathode during the electrolysis of its aqueous solution: for each position indicated by a letter, select the corresponding position indicated by a number.
| SUBSTANCE FORMULA | ELECTROLYSIS PRODUCT, PRODUCED AT THE CATHODE |
|
Write down the numbers in response, arranging them in the order corresponding to the letters:
| A | B | AT | G |
Solution.
During the electrolysis of aqueous solutions of salts, the following is released at the cathode:
Hydrogen, if it is a salt of a metal that is in the series of metal stresses to the left of aluminum;
Metal, if it is a salt of a metal that is in the series of metal voltages to the right of hydrogen;
Metal and hydrogen, if it is a salt of a metal in the series of metal voltages between aluminum and hydrogen.
Answer: 3511.
Answer: 3511
Source: Yandex: Training USE work in chemistry. Option 2.
Establish a correspondence between the salt formula and the product formed on an inert anode during the electrolysis of its aqueous solution: for each position indicated by a letter, select the corresponding position indicated by a number.
| SALT FORMULA | PRODUCT ON ANODE | |
Write down the numbers in response, arranging them in the order corresponding to the letters:
| A | B | AT | G |
Solution.
During the electrolysis of aqueous solutions of salts of oxygen-containing acids and fluorides, oxygen is oxidized from water, so oxygen is released at the anode. During the electrolysis of aqueous solutions of anoxic acids, the acid residue is oxidized.
Answer: 4436.
Answer: 4436
Establish a correspondence between the formula of a substance and the product that is formed on an inert anode as a result of the electrolysis of an aqueous solution of this substance: for each position indicated by a letter, select the corresponding position indicated by a number.
| SUBSTANCE FORMULA | PRODUCT ON ANODE |
2) sulfur oxide (IV) 3) carbon monoxide (IV) 5) oxygen 6) nitric oxide (IV) |
Write down the numbers in response, arranging them in the order corresponding to the letters:
| A | B | AT | G |
What is electrolysis? For a simpler understanding of the answer to this question, let's imagine any source of direct current. For every DC source, you can always find a positive and a negative pole:
Let us connect to it two chemically resistant electrically conductive plates, which we will call electrodes. The plate connected to the positive pole is called the anode, and to the negative pole is called the cathode:
Sodium chloride is an electrolyte; when it melts, it dissociates into sodium cations and chloride ions:
NaCl \u003d Na + + Cl -
It is obvious that the negatively charged chlorine anions will go to the positively charged electrode - the anode, and the positively charged Na + cations will go to the negatively charged electrode - the cathode. As a result of this, both Na + cations and Cl - anions will be discharged, that is, they will become neutral atoms. The discharge occurs through the acquisition of electrons in the case of Na + ions and the loss of electrons in the case of Cl − ions. That is, the process proceeds at the cathode:
Na + + 1e − = Na 0 ,
And on the anode:
Cl − − 1e − = Cl
Since each chlorine atom has an unpaired electron, their single existence is unfavorable and the chlorine atoms combine into a molecule of two chlorine atoms:
Сl∙ + ∙Cl \u003d Cl 2
Thus, in total, the process occurring at the anode is more correctly written as follows:
2Cl - - 2e - = Cl 2
That is, we have:
Cathode: Na + + 1e − = Na 0
Anode: 2Cl - - 2e - = Cl 2
Let's sum up the electronic balance:
Na + + 1e − = Na 0 |∙2
2Cl − − 2e − = Cl 2 |∙1<
Add the left and right sides of both equations half reactions, we get:
2Na + + 2e − + 2Cl − − 2e − = 2Na 0 + Cl 2
We reduce two electrons in the same way as it is done in algebra, we get the ionic equation of electrolysis:
2NaCl (l.) => 2Na + Cl 2
From a theoretical point of view, the case considered above is the simplest, since in the sodium chloride melt, among the positively charged ions, there were only sodium ions, and among the negative ones, only chlorine anions.
In other words, neither Na + cations nor Cl − anions had "competitors" for the cathode and anode.
And what will happen, for example, if instead of a melt of sodium chloride, a current is passed through its aqueous solution? Dissociation of sodium chloride is also observed in this case, but the formation of metallic sodium in an aqueous solution becomes impossible. After all, we know that sodium, a representative of alkali metals, is an extremely active metal that reacts very violently with water. If sodium cannot be reduced under such conditions, then what will be reduced at the cathode?
Let's remember the structure of the water molecule. It is a dipole, that is, it has a negative and a positive pole:
It is due to this property that it is able to “stick around” both the cathode surface and the anode surface:
The following processes may take place:
2H 2 O + 2e - \u003d 2OH - + H 2
2H 2 O - 4e - \u003d O 2 + 4H +
Thus, it turns out that if we consider a solution of any electrolyte, we will see that the cations and anions formed during the dissociation of the electrolyte compete with water molecules for reduction at the cathode and oxidation at the anode.
So what processes will take place at the cathode and at the anode? Discharge of ions formed during the dissociation of the electrolyte or oxidation / reduction of water molecules? Or, perhaps, all of these processes will occur simultaneously?
Depending on the type of electrolyte, a variety of situations are possible during the electrolysis of its aqueous solution. For example, cations of alkali, alkaline earth metals, aluminum and magnesium are simply not able to recover in aquatic environment, since during their reduction, alkali, alkaline earth metals, aluminum or magnesium, i.e., should be obtained, respectively. metals that react with water.
In this case, only the reduction of water molecules at the cathode is possible.
It is possible to remember what process will take place on the cathode during the electrolysis of a solution of any electrolyte, following the following principles:
1) If the electrolyte consists of a metal cation, which in a free state under normal conditions reacts with water, the following process takes place on the cathode:
2H 2 O + 2e - \u003d 2OH - + H 2
This applies to metals that are at the beginning of the Al activity series, inclusive.
2) If the electrolyte consists of a metal cation, which in its free form does not react with water, but reacts with non-oxidizing acids, two processes take place at once, both the reduction of metal cations and water molecules:
Me n+ + ne = Me 0
These metals include those between Al and H in the activity series.
3) If the electrolyte consists of hydrogen cations (acid) or metal cations that do not react with non-oxidizing acids, only electrolyte cations are restored:
2H + + 2e - \u003d H 2 - in the case of acid
Me n + + ne = Me 0 - in the case of salt
At the anode, meanwhile, the situation is as follows:
1) If the electrolyte contains anions of oxygen-free acid residues (except F -), then the process of their oxidation takes place at the anode, water molecules are not oxidized. For example:
2Cl - - 2e \u003d Cl 2
S 2- − 2e = S o
Fluoride ions are not oxidized at the anode because fluorine is not able to form in an aqueous solution (reacts with water)
2) If the electrolyte contains hydroxide ions (alkalis), they are oxidized instead of water molecules:
4OH - - 4e - \u003d 2H 2 O + O 2
3) If the electrolyte contains an oxygen-containing acid residue (except for organic acid residues) or a fluoride ion (F -) on the anode, the process of oxidizing water molecules takes place:
2H 2 O - 4e - \u003d O 2 + 4H +
4) In the case of an acidic residue of a carboxylic acid on the anode, the following process takes place:
2RCOO - - 2e - \u003d R-R + 2CO 2
Let's practice writing electrolysis equations for various situations:
Example #1
Write the equations for the processes occurring at the cathode and anode during the electrolysis of a zinc chloride melt, as well as the general electrolysis equation.
Solution
When zinc chloride is melted, it dissociates:
ZnCl 2 \u003d Zn 2+ + 2Cl -
Further, attention should be paid to the fact that it is the zinc chloride melt that undergoes electrolysis, and not the aqueous solution. In other words, without options, only the reduction of zinc cations can occur at the cathode, and the oxidation of chloride ions at the anode. no water molecules
Cathode: Zn 2+ + 2e − = Zn 0 |∙1
Anode: 2Cl − − 2e − = Cl 2 |∙1
ZnCl 2 \u003d Zn + Cl 2
Example #2
Write the equations for the processes occurring at the cathode and anode during the electrolysis of an aqueous solution of zinc chloride, as well as the general electrolysis equation.
Since in this case, an aqueous solution is subjected to electrolysis, then, theoretically, water molecules can take part in electrolysis. Since zinc is located in the activity series between Al and H, this means that both the reduction of zinc cations and water molecules will occur at the cathode.
2H 2 O + 2e - \u003d 2OH - + H 2
Zn 2+ + 2e − = Zn 0
The chloride ion is the acidic residue of the oxygen-free acid HCl, therefore, in the competition for oxidation at the anode, chloride ions “win” over water molecules:
2Cl - - 2e - = Cl 2
In this particular case, it is impossible to write the overall electrolysis equation, since the ratio between hydrogen and zinc released at the cathode is unknown.
Example #3
Write the equations for the processes occurring at the cathode and anode during the electrolysis of an aqueous solution of copper nitrate, as well as the general electrolysis equation.
Copper nitrate in solution is in a dissociated state:
Cu(NO 3) 2 \u003d Cu 2+ + 2NO 3 -
Copper is in the activity series to the right of hydrogen, that is, copper cations will be reduced at the cathode:
Cu 2+ + 2e − = Cu 0
Nitrate ion NO 3 - is an oxygen-containing acid residue, which means that in oxidation at the anode, nitrate ions “lose” in competition with water molecules:
2H 2 O - 4e - \u003d O 2 + 4H +
In this way:
Cathode: Cu 2+ + 2e − = Cu 0 |∙2
2Cu 2+ + 2H 2 O = 2Cu 0 + O 2 + 4H +
The equation obtained as a result of addition is the ionic equation of electrolysis. To get the complete molecular electrolysis equation, you need to add 4 nitrate ions to the left and right sides of the resulting ionic equation as counterions. Then we will get:
2Cu(NO 3) 2 + 2H 2 O = 2Cu 0 + O 2 + 4HNO 3
Example #4
Write the equations for the processes occurring at the cathode and anode during the electrolysis of an aqueous solution of potassium acetate, as well as the general electrolysis equation.
Solution:
Potassium acetate in an aqueous solution dissociates into potassium cations and acetate ions:
CH 3 COOK \u003d CH 3 COO − + K +
Potassium is an alkali metal, i.e. is in the electrochemical series of voltages at the very beginning. This means that its cations are not capable of being discharged at the cathode. Instead, water molecules will be restored:
2H 2 O + 2e - \u003d 2OH - + H 2
As mentioned above, the acid residues of carboxylic acids “win” in the competition for oxidation from water molecules at the anode:
2CH 3 COO - - 2e - \u003d CH 3 -CH 3 + 2CO 2
Thus, summing up the electronic balance and adding the two equations of half-reactions at the cathode and anode, we obtain:
Cathode: 2H 2 O + 2e − = 2OH − + H 2 |∙1
Anode: 2CH 3 COO - - 2e - \u003d CH 3 -CH 3 + 2CO 2 | ∙ 1
2H 2 O + 2CH 3 COO - \u003d 2OH - + H 2 + CH 3 -CH 3 + 2CO 2
We got complete equation electrolysis in ionic form. By adding two potassium ions to the left and right sides of the equation and adding them with counterions, we get the complete electrolysis equation in molecular form:
2H 2 O + 2CH 3 COOK \u003d 2KOH + H 2 + CH 3 -CH 3 + 2CO 2
Example #5
Write the equations for the processes occurring at the cathode and anode during the electrolysis of an aqueous solution of sulfuric acid, as well as the general electrolysis equation.
Sulfuric acid dissociates into hydrogen cations and sulfate ions:
H 2 SO 4 \u003d 2H + + SO 4 2-
Hydrogen cations H + will be reduced at the cathode, and water molecules will be oxidized at the anode, since sulfate ions are oxygen-containing acid residues:
Cathode: 2Н + + 2e − = H 2 |∙2
Anode: 2H 2 O - 4e - = O 2 + 4H + |∙1
4H + + 2H 2 O \u003d 2H 2 + O 2 + 4H +
Reducing the hydrogen ions in the left and right and left sides of the equation, we obtain the equation for the electrolysis of an aqueous solution of sulfuric acid:
2H 2 O \u003d 2H 2 + O 2
As can be seen, the electrolysis of an aqueous solution of sulfuric acid is reduced to the electrolysis of water.
Example #6
Write the equations for the processes occurring at the cathode and anode during the electrolysis of an aqueous solution of sodium hydroxide, as well as the general electrolysis equation.
Dissociation of sodium hydroxide:
NaOH = Na + + OH -
Only water molecules will be reduced at the cathode, since sodium is a highly active metal, and only hydroxide ions at the anode:
Cathode: 2H 2 O + 2e − = 2OH − + H 2 |∙2
Anode: 4OH − − 4e − = O 2 + 2H 2 O |∙1
4H 2 O + 4OH - \u003d 4OH - + 2H 2 + O 2 + 2H 2 O
Let us reduce two water molecules on the left and on the right and 4 hydroxide ions and come to the conclusion that, as in the case of sulfuric acid, the electrolysis of an aqueous solution of sodium hydroxide is reduced to the electrolysis of water.
