Electrolysis of hydrochloric acid equation. Electrolysis of melts and electrolyte solutions

What is electrolysis? To understand the answer to this question more simply, let's imagine any DC source. For every DC source you can always find a positive and a negative pole:

Let us connect two chemically resistant electrically conductive plates to it, which we will call electrodes. We will call the plate connected to the positive pole an anode, and to the negative pole a cathode:

Sodium chloride is an electrolyte; when it melts, it dissociates into sodium cations and chloride ions:

NaCl = Na + + Cl −

Obviously, negatively charged chlorine anions will go to the positively charged electrode - the anode, and 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. 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 occurs 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 disadvantageous and the chlorine atoms combine into a molecule of two chlorine atoms:

Сl∙ + ∙Cl = 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<

Let's add the left and right sides of both equations half-reactions, we get:

2Na + + 2e − + 2Cl − − 2e − = 2Na 0 + Cl 2

Let's reduce two electrons in the same way as is done in algebra, and we get the ionic equation of electrolysis:

2NaCl (liquid) => 2Na + Cl 2

The case considered above is from a theoretical point of view the simplest, since in the melt of sodium chloride there were only sodium ions among the positively charged ions, and only chlorine anions among the negative ones.

In other words, neither Na + cations nor Cl − anions had “competitors” for the cathode and anode.

What will happen, for example, if instead of molten 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 the alkali metals, is an extremely active metal that reacts very violently with water. If sodium is not able to be reduced under such conditions, what then will be reduced at the cathode?

Let's remember the structure of the water molecule. It is a dipole, that is, it has negative and positive poles:

It is thanks to this property that it is able to “stick” to both the surface of the cathode and the surface of the anode:

In this case, the following processes may occur:

2H 2 O + 2e − = 2OH − + H 2

2H 2 O – 4e − = O 2 + 4H +

Thus, it turns out that if we consider a solution of any electrolyte, we will see that 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 occur at the cathode and anode? Discharge of ions formed during electrolyte dissociation 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 be reduced in an aqueous environment, since their reduction would have to produce alkali, alkaline earth metals, aluminum or magnesium, respectively, i.e. metals that react with water.

In this case, only the reduction of water molecules at the cathode is possible.

You can remember what process will occur at the cathode during the electrolysis of a solution of any electrolyte by 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 process occurs at the cathode:

2H 2 O + 2e − = 2OH − + H 2

This applies to metals located 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 occur simultaneously, both the reduction of metal cations and water molecules:

Me n+ + ne = Me 0

These metals include metals located 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 the electrolyte cations are reduced:

2Н + + 2е − = Н 2 – in 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 acidic residues (except F −), then the process of their oxidation occurs at the anode; water molecules are not oxidized. For example:

2Сl − − 2e = 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 (alkalies), they are oxidized instead of water molecules:

4OH − − 4e − = 2H 2 O + O 2

3) If the electrolyte contains an oxygen-containing acid residue (except for residues organic acids) or fluoride ion (F −) at the anode the process of oxidation of water molecules occurs:

2H 2 O – 4e − = O 2 + 4H +

4) In the case of an acidic residue of a carboxylic acid at the anode, the process occurs:

2RCOO − − 2e − = R-R + 2CO 2

Let's practice writing electrolysis equations for various situations:

Example No. 1

Write the equations for the processes occurring at the cathode and anode during the electrolysis of molten zinc chloride, as well as general equation electrolysis.

Solution

When zinc chloride melts, it dissociates:

ZnCl 2 = Zn 2+ + 2Cl −

Next, you should pay attention to the fact that it is the zinc chloride melt that undergoes electrolysis, and not an aqueous solution. In other words, without options, only the reduction of zinc cations can occur at the cathode, and oxidation of chloride ions at the anode because no water molecules:

Cathode: Zn 2+ + 2e − = Zn 0 |∙1

Anode: 2Cl − − 2e − = Cl 2 |∙1

ZnCl 2 = Zn + Cl 2

Example No. 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 equation for electrolysis.

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 − = 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 down the overall electrolysis equation, since the relationship between the hydrogen and zinc released at the cathode is unknown.

Example No. 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 equation for electrolysis.

Copper nitrate in solution is in a dissociated state:

Cu(NO 3) 2 = 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 acidic residue, which means that in oxidation at the anode, nitrate ions “lose” in competition with water molecules:

2H 2 O – 4e − = O 2 + 4H +

Thus:

Cathode: Cu 2+ + 2e − = Cu 0 |∙2

2Cu 2+ + 2H 2 O = 2Cu 0 + O 2 + 4H +

The resulting equation is the ionic equation of electrolysis. To obtain the complete molecular equation of electrolysis, you need to add 4 nitrate ions to the left and right sides of the resulting ionic equation as counterions. Then we get:

2Cu(NO 3) 2 + 2H 2 O = 2Cu 0 + O 2 + 4HNO 3

Example No. 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 equation for electrolysis.

Solution:

Potassium acetate in an aqueous solution dissociates into potassium cations and acetate ions:

CH 3 COOK = CH 3 COO − + K +

Potassium is an alkali metal, i.e. is in the electrochemical voltage series at the very beginning. This means that its cations are not able to discharge at the cathode. Instead, water molecules will be restored:

2H 2 O + 2e − = 2OH − + H 2

As mentioned above, the acidic residues of carboxylic acids “win” in the competition for oxidation with water molecules at the anode:

2CH 3 COO − − 2e − = CH 3 −CH 3 + 2CO 2

Thus, by 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 − = CH 3 −CH 3 + 2CO 2 |∙1

2H 2 O + 2CH 3 COO − = 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 obtain the complete electrolysis equation in molecular form:

2H 2 O + 2CH 3 COOK = 2KOH + H 2 + CH 3 −CH 3 + 2CO 2

Example No. 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 equation for electrolysis.

Sulfuric acid dissociates into hydrogen cations and sulfate ions:

H 2 SO 4 = 2H + + SO 4 2-

At the cathode, reduction of hydrogen cations H + will occur, and at the anode, oxidation of water molecules, since sulfate ions are oxygen-containing acidic residues:

Cathode: 2Н + + 2e − = H 2 |∙2

Anode: 2H 2 O – 4e − = O 2 + 4H + |∙1

4H + + 2H 2 O = 2H 2 + O 2 + 4H +

By reducing the hydrogen ions on 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 = 2H 2 + O 2

As you can see, the electrolysis of an aqueous solution of sulfuric acid comes down to the electrolysis of water.

Example No. 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 equation for electrolysis.

Dissociation of sodium hydroxide:

NaOH = Na + + OH −

At the cathode, only water molecules will be reduced, since sodium is a highly active metal; at the anode, only hydroxide ions:

Cathode: 2H 2 O + 2e − = 2OH − + H 2 |∙2

Anode: 4OH − − 4e − = O 2 + 2H 2 O |∙1

4H 2 O + 4OH − = 4OH − + 2H 2 + O 2 + 2H 2 O

Let us reduce two water molecules on the left and right and 4 hydroxide ions and we 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.

ELECTROLYSIS

melts and solutions of electrolytes

Electrolysis is a set of chemical reactions that occur during the passage direct electric current through an electrochemical system consisting of two electrodes and a melt or electrolyte solution.

The chemical essence of electrolysis lies in the fact that it is a redox reaction that occurs under the influence of a direct electric current, and the processes of oxidation and reduction are spatially separated.

Cathode – an electrode on which cations or water are reduced. It is negatively charged.

Anode – an electrode on which anions or water are oxidized. It is positively charged.

1. Electrolysis of molten salts and bases.

During the electrolysis of melts at the cathode, metal cations are always reduced.

К(-): Мen+ + nē → Me0

The anodic process is determined by the composition of the anion:

a) If the anion is an oxygen-free acid (Cl-, Br-, I-, S2-), then this anion undergoes anodic oxidation and a simple substance is formed:

A(+): 2Cl - - 2ē → Cl2 or A(+): S2- - 2ē → S0

b) If an oxygen-containing anion (SO42-, SiO32-, HO-, etc.) undergoes anodic oxidation, then the nonmetal forms an oxide (without changing its oxidation state) and oxygen is released.

A(+): 2SiO32-- 4ē → 2SiO2 + O2

A(+): 2SO32-- 4ē → 2SO2 + O2

A(+): 4РО43-- 12ē → 2Р2O5 + 3О2

A(+): 4NO3-- 4ē → 2N2O5 + O2

A(+): 4HO-- 4ē → 2H2O + O2

Example 1.1. Melt salt ZnCl2

ZnCl2 Û Zn2+ + 2Cl-

S: ZnCl2 electrolysis Zn + Cl2

Example 1.2. Melt alkali NaOH

NaOH Û Na+ + OH-

The overall electrolysis equation is obtained by adding the right and left sides of the equations, provided that the electrons participating in the cathode and anodic processes are equal.

https://pandia.ru/text/80/299/images/image006_58.gif" width="70" height="12">4 Na+ + 4 ē + 4 OH - - 4 ē electrolysis 4 Na0 + O2 + 2H2O

4 Na+ + 4 OH - electrolysis 4 Na0 + O2 + 2H2O - ionic equation

4NaOH electrolysis 4Na + 2H2O + O2 - molecular equation

Example 1.3. Melt salt Na2SO4

Na2SO4 Û 2Na+ + SO42-

K(-): Na+ + 1 ē Þ Nao *4

A(+): 2SO42- - 4 ē Þ O2 + 2SO3

4Na+ +2SO42- Þ 2Nao + O2 + 2SO3 – ionic equation of electrolysis

2Na2SO4 electrolysis 4Nao + O2 + 2SO3 – molecular equation

K A

Example 1.4. Melt salt AgNO3

AgNO3 Û Ag+ + NO3-

K(-): Ag+ + 1 ē Þ Ago *4

A(+): 4NO3- - 4 ē Þ 2N2O5 + 2O2 *1

4Ag+ + 4NO3- electrolysis 4Ag + 2N2O5 + 2O2

4AgNO3 electrolysis 4Ag + 2N2O5 +2O2

K A

Tasks for independent work . Draw up equations for the electrolysis of melts of the following salts: AlCl3, Cr2(SO4)3, Na2SiO3, K2CO3.

2. Electrolysis of solutions of salts, hydroxides and acids.

Electrolysis aqueous solutions is complicated by the fact that water can take part in the oxidation and reduction processes.

Cathode processes determined by the electrochemical activity of the salt cation. The further to the left a metal is in the voltage series, the more difficult it is for its cations to be reduced at the cathode:

Li K Ca Na Mg Al Mn Zn Cr Te Ni Sn Pb H2 Cu Hg Ag Pt Au

I groupIIgroupIIIgroup

For metal cations up to and including Al (group I), the cathodic process is the reduction of hydrogen from water:

(-)K: 2H2O + 2ē → H2 + 2HO-

For metal cations after hydrogen (group III), the cathodic process is their reduction to metal:

(-) K: Мen+ + nē → Me0

For metal cations in the voltage series from Mn to H2 (group II), there are parallel competitive processes of reduction of metal cations and hydrogen from water:

(-) K: Мen+ + nē → Me0

2H2O + 2ē → H2+ 2HO-

Which of these processes will prevail depends on a number of factors: Me activity, solution pH, salt concentration, applied voltage and electrolysis conditions.

Anodic processes determined by the composition of salt anions:

A) If the anion is an oxygen-free acid (Cl-, Br-, I-, S2-, etc.), then it is oxidized to simple substances (with the exception of F-):

A(+): S2- - 2ē → S0

b) In the presence of an oxygen-containing anion (SO42-, CO32-, etc. or OH-), only water undergoes anodic oxidation:

A(+): 2H2O - 4ē → O2 + 4H+

Let's look at examples to illustrate all possible options:

Example 2.1 . KCl salt solution

K(-): 2H2O + 2e - Þ H2 + 2OH-

A(+): 2Cl - - 2e - Þ Cl2

å: 2H2O + 2Cl - electrolysis H2 + 2OH - + Cl2 – ionic equation of electrolysis

2KCl + 2H2O electrolysis H2 + 2KOH + Cl2 – molecular equation of electrolysis

K A

Example 2.2 . CuCl2 salt solution

CuCl2 Û Cu2+ + 2Cl-

K(-): Cu2+ + 2e - Þ Cuo

A(+): 2Cl- -2e - Þ Cl2

å: CuCl2 electrolysis Cu + Cl2

Example 2.3. FeCl2 salt solution

FeCl2 Û Fe2+ + 2Cl-

Iron belongs to group II metals, so two parallel processes will occur at the cathode:

1st process:

(-) K: Fe2+ + 2ē → Fe0

(+)A: 2Cl - - 2ē → Cl2

Fe2+ ​​+ 2Cl - el-z Fe0 + Cl2 - ionic equation of the process

FeCl2 el-z Fe0 + Cl2 - molecular equation of the process

2nd process:

(-)K: 2H2O + 2ē → H2+ 2OH-

(+)A: 2Cl - - 2ē → Cl2

2H2O + 2Cl - → H2+ 2OH - + Cl2 - ionic equation of the process

2H2O + FeCl2 electrolysis H2+ Fe(OH)2 + Cl2 - molecular equation.

Thus, in the cathode space Fe, H2 and Fe(OH)2 will be formed in different proportions depending on the electrolysis conditions.

Example 2.4 . Na2SO4 salt solution.

Na2SO4 Û 2Na+ + SO42-

K(-) 2H2O + 2e - Þ H2 + 2OH - *2

A(+) 2H2O – 4e - Þ O2 + 4H+

å: 6H2O electrolysis 2H2 + 4OH - + O2 + 4H+

å: 6H2O + 2Na2SO4 electrolysis 2H2 + 4 NaOH + O2 + 2H2SO4

in the cathode space in the anode space

When the electric current is turned off and the contents of the cathode and anode spaces are mixed, the final result of electrolysis can be represented by the following diagram:

2H2O el-z 2H2 + O2,

since an alkali will react with an acid to form 2 mol of salt and 4 mol of water.

Example 2.5 . Electrolysis of CuSO4 solution.

CuSO4 Û Cu2+ + SO42-

K(-): Cu2+ + 2e - Þ Cuo

A(+): 2H2O – 4e - Þ O2 + 4H+

å: 2Cu2+ + 2H2O electrolysis 2Cuo + O2 + 4H+

å: CuSO4 + 2H2O electrolysis 2Cuo + O2 + 2H2SO4

Example 2.6. Electrolysis of FeSO4 solution

Since iron belongs to group II of metals, two competitive processes will occur in parallel at the cathode (see example 2.3), and water will oxidize at the anode (see example 2.4):

1st process:

https://pandia.ru/text/80/299/images/image043_10.gif" width="41" height="12">2Fe2+ + 2Н2О el-z 2 Fe + O2 + 4H+ - ionic equation of the process

2FeSO4 + 2H2O el-z 2 Fe + O2 + 2H2SO4– molecular equation

2nd process:

K(+): 2H2O + 2ē → H2+ 2OH - *2

A(-): 2H2O - 4ē → O2+ 4H+

6H2O electrolysis 2H2+ 4ОH - + O2 + 4H+

6H2O + 2FeSO4electrolysis 2H2+ 2Fe(OH)2 + O2 + 2H2SO4 - molecular

https://pandia.ru/text/80/299/images/image051_9.gif" width="21" height="50">And only if the processes of cathodic reduction of metal and hydrogen cations from water occur in equal proportions , we can write the overall final reaction equation:

(-) K: Fe2+ + 2ē → Fe0

2H2O + 2ē → H2+ 2HO - only 4 electrons

(+)A: 2H2O - 4ē → O2 + 4H+

Fe2+ ​​+ 2H2O + 2H2O → Fe + H2+ 2HO - + O2 + 4H+

2FeSO4 + 4H2O el-z Fe + H2+ Fe(OH)2 + O2 + 2H2SO4

cathode anode

After turning off the current and mixing the solutions, the final equation will be as follows:

· Create equations for the electrolysis of solutions of K2CO3, ZnSO4, AgNO3, NiI2, CoCl2.

· Solve a problem. To analyze the content of NaCl impurities in technical NaOH, 40 g of the drug was dissolved in water and subjected to electrolysis until the chlorine ions were completely oxidized. In this case, 601 ml of Cl2 was released at the anode at a temperature of 200C and normal pressure. Calculate the mass fraction of NaCl impurity in NaOH.

3. Electrolysis c soluble s m anode

Above, examples of the electrolysis of aqueous solutions of salts with an inert anode, i.e., one that does not take a chemical part in the anodic process, were considered. Such electrodes are made from inactive noble metals, for example Pt, Ir or carbon electrodes are used. If soluble anodes are used, for example, Cu-anode, Zn-anode, then the anodic process is significantly modified, since the anode itself is oxidized. At the anode, out of 2 competitive processes, a process with a lower potential occurs: for the oxidation of copper E0 = - 0.34 V, for the oxidation of zinc E0 = - 0.76 V, and for the oxidation of the Cl anion E0 = + 1.36 V.

Example 3.1. Electrolysis of an aqueous solution of CuCl2 salt with a soluble anode:

Cathode (-): Cu anode (+):

Cu2+ + 2ē → Cu0 Cu0 - 2ē → Cu2+

Thus, a kind of refining of the copper anode occurs: it dissolves, impurities remain in the anode space, and pure copper is deposited on the cathode. In this case, the chlorine anion is not oxidized, but accumulates in the anode space.

Example 3.2. Electrolysis of an aqueous solution of KCl salt with a Cu anode:

Cu anode (+): Cu0 - 2ē → Cu2+

At the cathode, hydrogen initially begins to be reduced from water, but the appearance of Cu2+ in the solution makes two cathodic reduction reactions competitive:

K(-): 2H2O + 2ē → H2+ 2NO - E0 = - 0.828 V

Cu2+ + 2ē → Cu0 E0 = + 0.34 V

As a result, the one that is characterized by a higher potential proceeds predominantly, i.e., the reduction of Cu2+ to Cu0.

Thus, in this case, the Cu anode will dissolve: Cu0 - 2ē → Cu2+, and at the cathode the formed copper cations will be reduced: Cu2+ + 2ē → Cu0. The KCl salt is needed only to increase the electrical conductivity of the solution, and it does not directly participate in redox processes.

Assignment for independent work. Consider the electrolysis of CuSO4 with a Cu anode, Na2SO4 with a Cu anode.

Electrolysis of solutions
and molten salts (2 hours)

Classes elective course"Electrochemistry"

Objectives of the first lesson:

FIRST LESSON PLAN

1. Repetition of studied methods for obtaining metals.

2. Explanation of new material.

3. Solving problems from the textbook by G.E. Rudzitis, F.G. Feldman “Chemistry-9” (M.: Prosveshchenie, 2002), p. 120, no. 1, 2.

4. Testing knowledge acquisition on test tasks.

5. Report on the use of electrolysis.

Objectives of the first lesson: teach how to write diagrams for the electrolysis of solutions and molten salts and apply the acquired knowledge to solve calculation problems; continue to develop skills in working with the textbook and test materials; discuss the use of electrolysis in national economy.

PROGRESS OF THE FIRST LESSON

Repetition of learned methods obtaining metals using the example of copper production from copper(II) oxide.

Writing the equations of the corresponding reactions:

Another way to obtain metals from solutions and melts of their salts is electrochemical, or electrolysis.

Electrolysis is a redox process that occurs at the electrodes when an electric current is passed through a melt or electrolyte solution.

Electrolysis of sodium chloride melt:

NaCl Na + + Cl – ;

cathode (–) (Na+): Na++ e=Na0,

anode (–) (Cl –): Cl – – e= Cl 0, 2Cl 0 = Cl 2;

2NaCl = 2Na + Cl2.

Electrolysis of sodium chloride solution:

NaCl Na + + Cl – ,

H 2 O H + + OH – ;

cathode (–) (Na + ; H +): H + + e= H 0 , 2H 0 = H 2

(2H 2 O + 2 e= H 2 + 2OH –),

anode (+) (Cl – ; OН –): Cl – – e= Cl 0, 2Cl 0 = Cl 2;

2NaCl + 2H 2 O = 2NaOH + Cl 2 + H 2.

Electrolysis of copper(II) nitrate solution:

Cu(NO 3) 2 Cu 2+ +

H 2 O H + + OH – ;

cathode (–) (Cu 2+ ; H +): Cu 2+ + 2 e= Cu 0 ,

anode (+) (OH –): OH – – e=OH 0,

4H 0 = O 2 + 2H 2 O;

2Cu(NO 3) 2 + 2H 2 O = 2Cu + O 2 + 4HNO 3.

These three examples show why electrolysis is more profitable than other methods of producing metals: metals, hydroxides, acids, and gases are obtained.

We wrote electrolysis diagrams, and now let’s try to write electrolysis equations right away, without referring to the diagrams, but only using the ion activity scale:

Examples of electrolysis equations:

2HgSO 4 + 2H 2 O = 2Hg + O 2 + 2H 2 SO 4;

Na 2 SO 4 + 2H 2 O = Na 2 SO 4 + 2H 2 + O 2;

2LiCl + 2H 2 O = 2LiOH + H 2 + Cl 2.

Problem solving from the textbook by G.E. Rudzitis and F.G. Feldman (9th grade, p. 120, No. 1, 2).

Task 1. During electrolysis of a solution of copper(II) chloride, the mass of the cathode increased by 8 g. What gas was released, what is its mass?

Solution

CuCl 2 + H 2 O = Cu + Cl 2 + H 2 O,

(Cu) = 8/64 = 0.125 mol,

(Cu) = (Cl 2) = 0.125 mol,

m(Cl 2) = 0.125 71 = 8.875 g.

Answer. Gas – chlorine weighing 8.875 g.

Task 2. During electrolysis of an aqueous solution of silver nitrate, 5.6 liters of gas were released. How many grams of metal are deposited on the cathode?

Solution

4AgNO 3 + 2H 2 O = 4Ag + O 2 + 4HNO 3,

(O 2) = 5.6/22.4 = 0.25 mol,

(Ag) = 4(O 2) = 4 25 = 1 mol,

m(Ag) = 1,107 = 107 g.

Answer. 107 g silver.

Testing

Option 1

1. During electrolysis of a solution of potassium hydroxide at the cathode, the following is released:

a) hydrogen; b) oxygen; c) potassium.

2. During the electrolysis of a solution of copper(II) sulfate in solution, the following is formed:

a) copper(II) hydroxide;

b) sulfuric acid;

3. During electrolysis of a barium chloride solution at the anode, the following is released:

a) hydrogen; b) chlorine; c) oxygen.

4. During the electrolysis of molten aluminum chloride at the cathode, the following is released:

a) aluminum; b) chlorine;

c) electrolysis is impossible.

5. Electrolysis of a silver nitrate solution proceeds according to the following scheme:

a) AgNO 3 + H 2 O Ag + H 2 + HNO 3;

b) AgNO 3 + H 2 O Ag + O 2 + HNO 3;

c) AgNO 3 + H 2 O AgNO 3 + H 2 + O 2.

Option 2

1. During electrolysis of a sodium hydroxide solution at the anode, the following is released:

a) sodium; b) oxygen; c) hydrogen.

2. During the electrolysis of a solution of sodium sulfide in solution, the following is formed:

a) hydrosulfide acid;

b) sodium hydroxide;

3. During the electrolysis of a melt of mercury(II) chloride at the cathode, the following is released:

a) mercury; b) chlorine; c) electrolysis is impossible.

4.

5. Electrolysis of a solution of mercury(II) nitrate proceeds according to the following scheme:

a) Hg(NO 3) 2 + H 2 O Hg + H 2 + HNO 3;

b) Hg(NO 3) 2 + H 2 O Hg + O 2 + HNO 3;

c) Hg(NO 3) 2 + H 2 O Hg(NO 3) 2 + H 2 + O 2.

Option 3

1. During electrolysis of a solution of copper(II) nitrate at the cathode, the following is released:

a) copper; b) oxygen; c) hydrogen.

2. During the electrolysis of a solution of lithium bromide in solution, the following is formed:

b) hydrobromic acid;

c) lithium hydroxide.

3. During the electrolysis of molten silver chloride at the cathode, the following is released:

a) silver; b) chlorine; c) electrolysis is impossible.

4. During the electrolysis of an aluminum chloride solution, aluminum is released into:

a) cathode; b) anode; c) remains in solution.

5. Electrolysis of a barium bromide solution proceeds according to the following scheme:

a) BaBr 2 + H 2 O Br 2 + H 2 + Ba(OH) 2;

b) BaBr 2 + H 2 O Br 2 + Ba + H 2 O;

c) BaBr 2 + H 2 O Br 2 + O 2 + Ba(OH) 2.

Option 4

1. During the electrolysis of a barium hydroxide solution at the anode, the following is released:

a) hydrogen; b) oxygen; c) barium.

2. During the electrolysis of a solution of potassium iodide in solution, the following is formed:

a) hydroiodic acid;

b) water; c) potassium hydroxide.

3. During the electrolysis of molten lead(II) chloride at the cathode, the following is released:

a) lead; b) chlorine; c) electrolysis is impossible.

4. During electrolysis of a silver nitrate solution at the cathode, the following is released:

a) silver; b) hydrogen; c) oxygen.

5. Electrolysis of a sodium sulfide solution proceeds according to the following scheme:

a) Na 2 S + H 2 O S + H 2 + NaOH;

b) Na 2 S + H 2 O H 2 + O 2 + Na 2 S;

c) Na 2 S + H 2 O H 2 + Na 2 S + NaOH.

Answers

Option	Question 1	Question 2	Question 3	Question 4	Question 5
1	A	b	b	A	b
2	b	b	A	A	b
3	A	V	A	V	A
4	b	V	A	A	A

Application of electrolysis in the national economy

1. To protect metal products from corrosion, a thin layer of another metal is applied to their surface: chromium, silver, gold, nickel, etc. Sometimes, in order not to waste expensive metals, a multilayer coating is produced. For example, the exterior parts of a car are first coated with a thin layer of copper, a thin layer of nickel is applied to the copper, and a layer of chrome is applied to it.

When coatings are applied to metal by electrolysis, they are even in thickness and durable. In this way, you can coat products of any shape. This branch of applied electrochemistry is called electroplating.

2. In addition to protecting against corrosion, galvanic coatings give a beautiful decorative appearance to products.

3. Another branch of electrochemistry, similar in principle to electroplating, is called galvanoplasty. It is the process of making exact replicas of various items. To do this, the object is coated with wax and a matrix is ​​obtained. All recesses of the copied object on the matrix will be bulges. The surface of the wax matrix is ​​coated with a thin layer of graphite, making it conductive to electrical current.

The resulting graphite electrode is immersed in a bath of copper sulfate solution. The anode is copper. During electrolysis, the copper anode dissolves and copper is deposited on the graphite cathode. This way an exact copper copy is obtained.

Electroplating is used to make printing clichés, gramophone records, and metallize various objects. Galvanoplasty was discovered by the Russian scientist B.S. Jacobi (1838).

Making record stamps involves applying a thin silver coating to a plastic record to make it electrically conductive. An electrolytic nickel coating is then applied to the plate.

What should the plate in the electrolytic bath be made of - anode or cathode?

(O t v e t. Cathode.)

4. Electrolysis is used to produce many metals: alkali, alkaline earth, aluminum, lanthanides, etc.

5. To purify some metals from impurities, the metal with impurities is connected to the anode. The metal dissolves during electrolysis and is released at the metal cathode, while the impurity remains in solution.

6. Electrolysis is widely used to obtain complex substances(alkalis, oxygen-containing acids), halogens.

Practical work
(second lesson)

Lesson objectives. Conduct electrolysis of water, demonstrate galvanostegy in practice, and consolidate the knowledge acquired in the first lesson.

Equipment.On students' desks: flat battery, two wires with terminals, two graphite electrodes, a beaker, test tubes, a stand with two legs, a 3% sodium sulfate solution, an alcohol lamp, matches, a torch.

On the teacher's desk: the same + solution of copper sulfate, brass key, copper tube (piece of copper).

Instructing students

1. Attach the wires with terminals to the electrodes.

2. Place the electrodes in a glass so that they do not touch.

3. Pour electrolyte solution (sodium sulfate) into a glass.

4. Pour water into the test tubes and, lowering them upside down into a glass with electrolyte, put them on the graphite electrodes one by one, securing the upper edge of the test tube in the tripod leg.

5. After the device is mounted, attach the ends of the wires to the battery.

6. Observe the release of gas bubbles: less of them are released at the anode than at the cathode. After almost all the water in one test tube is replaced by the released gas, and in the other - half, disconnect the wires from the battery.

7. Light the alcohol lamp, carefully remove the test tube, where the water has almost completely been displaced, and bring it to the alcohol lamp - a characteristic pop of gas will be heard.

8. Light a torch. Remove the second test tube and check the gas with a smoldering torch.

Student assignments

1. Sketch the device.

2. Write an equation for the electrolysis of water and explain why it was necessary to carry out electrolysis in a solution of sodium sulfate.

3. Write reaction equations reflecting the release of gases at the electrodes.

Teacher demonstration experiment
(can perform best students class
if appropriate equipment is available)

1. Connect the wire terminals to the copper tube and brass wrench.

2. Place the tube and key into a glass with a solution of copper(II) sulfate.

3. Connect the other ends of the wires to the battery: “minus” of the battery to the copper tube, “plus” to the key!

4. Observe the release of copper on the surface of the key.

5. After completing the experiment, first disconnect the terminals from the battery, then remove the key from the solution.

6. Disassemble the electrolysis circuit with a soluble electrode:

CuSO 4 = Cu 2+ +

anode (+): Cu 0 – 2 e= Cu 2+ ,

cathode (–): Cu 2+ + 2 e= Cu 0 .

The overall equation for electrolysis with a soluble anode cannot be written.

Electrolysis was carried out in a solution of copper(II) sulfate because:

a) an electrolyte solution is needed for electric current to flow, because water is a weak electrolyte;

b) no reaction by-products will be released, but only copper at the cathode.

7. To consolidate what has been learned, write a diagram of the electrolysis of zinc chloride with carbon electrodes:

ZnCl 2 = Zn 2+ + 2Cl – ,

cathode (–): Zn 2+ + 2 e= Zn 0 ,

2H2O+2 e= H 2 + 2OH – ,

anode (+): 2Cl – – 2 e=Cl2.

The overall reaction equation cannot be written in this case, because it is unknown what part of the total amount of electricity goes to the restoration of water, and what part goes to the restoration of zinc ions.

Scheme of the demonstration experiment

Homework

1. Write the equation electrolysis of solution, containing a mixture of copper(II) nitrate and silver nitrate, with inert electrodes.

2. Write the equation for the electrolysis of sodium hydroxide solution.

3. To clean a copper coin, it must be suspended on a copper wire connected to the negative pole of the battery and immersed in a 2.5% NaOH solution, where a graphite electrode connected to the positive pole of the battery should also be immersed. Explain how the coin becomes clean. ( Answer. The reduction of hydrogen ions occurs at the cathode:

2H + + 2 e= N 2.

Hydrogen reacts with copper oxide located on the surface of the coin:

CuO + H 2 = Cu + H 2 O.

This method is better than cleaning with powder, because... the coin is not erased.)

ELECTROLYSIS

One of the methods for producing metals is electrolysis. Active metals occur in nature only in the form of chemical compounds. How to isolate these compounds in the free state?

Solutions and melts of electrolytes conduct electric current. However, when current is passed through an electrolyte solution, chemical reactions. Let's consider what will happen if two metal plates are placed in a solution or melt of an electrolyte, each of which is connected to one of the poles of a current source. These plates are called electrodes. Electric current is a moving flow of electrons. As electrons in the circuit move from one electrode to another, an excess of electrons appears at one of the electrodes. Electrons have a negative charge, so this electrode is negatively charged. It is called the cathode. A deficiency of electrons is created at the other electrode and it becomes positively charged. This electrode is called the anode. An electrolyte in a solution or melt dissociates into positively charged ions - cations and negatively charged ions - anions. Cations are attracted to the negatively charged electrode - the cathode. Anions are attracted to a positively charged electrode - the anode. At the surface of the electrodes, interactions between ions and electrons can occur.

Electrolysis refers to processes that occur when electric current is passed through solutions or melts of electrolytes.

The processes occurring during the electrolysis of solutions and melts of electrolytes are quite different. Let's consider both of these cases in detail.

Electrolysis of melts

As an example, consider the electrolysis of a sodium chloride melt. In the melt, sodium chloride dissociates into ions Na+
and Cl - : NaCl = Na + + Cl -

Sodium cations move to the surface of a negatively charged electrode - the cathode. There is an excess of electrons on the cathode surface. Therefore, electrons are transferred from the electrode surface to sodium ions. In this case, the ions Na+ transform into sodium atoms, that is, reduction of cations occurs Na+ . Process equation:

Na + + e - = Na

Chloride ions Cl - move to the surface of a positively charged electrode - the anode. A lack of electrons is created on the anode surface and electrons are transferred from anions Cl- to the electrode surface. At the same time, negatively charged ions Cl- are converted into chlorine atoms, which immediately combine to form chlorine molecules C l 2 :

2С l - -2е - = Cl 2

Chloride ions lose electrons, that is, they oxidize.

Let us write down together the equations of the processes occurring at the cathode and anode

Na + + e - = Na

2 C l - -2 e - = Cl 2

One electron is involved in the reduction of sodium cations, and 2 electrons are involved in the oxidation of chlorine ions. However, the law of conservation of electric charge must be observed, that is, the total charge of all particles in the solution must be constant. Therefore, number of electrons, involved in the reduction of sodium cations, must be equal to the number of electrons involved in the oxidation of chloride ions. Therefore, we multiply the first equation by 2:

Na + + e - = Na 2

2С l - -2е - = Cl 2 1

Let's add both equations together and get the general reaction equation.

2 Na + + 2С l - = 2 Na + Cl 2 (ionic reaction equation), or

2 NaCl = 2 Na + Cl 2 (molecular reaction equation)

So, in the example considered, we see that electrolysis is a redox reaction. At the cathode, the reduction of positively charged ions - cations - occurs, and at the anode, the oxidation of negatively charged ions - anions. You can remember which process occurs where using the “T rule”:

cathode - cation - reduction.

Example 2.Electrolysis of molten sodium hydroxide.

Sodium hydroxide in solution dissociates into cations and hydroxide ions.

Cathode (-)<-- Na + + OH - à Анод (+)

On the surface of the cathode, sodium cations are reduced, and sodium atoms are formed:

cathode (-) Na + +e à Na

On the surface of the anode, hydroxide ions are oxidized, oxygen is released and water molecules are formed:

cathode (-) Na + + e à Na

anode (+)4 OH - – 4 e à 2 H 2 O + O 2

The number of electrons involved in the reduction reaction of sodium cations and in the oxidation reaction of hydroxide ions must be the same. Therefore, let's multiply the first equation by 4:

cathode (-) Na + + e à Na 4

anode (+)4 OH - – 4 e à 2 H 2 O + O 2 1

Let's add both equations together and get the electrolysis reaction equation:

4 NaOH à 4 Na + 2 H 2 O + O 2

Example 3.Consider the electrolysis of the melt Al2O3

Using this reaction, aluminum is obtained from bauxite, a natural compound that contains a lot of aluminum oxide. The melting point of aluminum oxide is very high (more than 2000º C), so special additives are added to it to lower the melting point to 800-900º C. In the melt, aluminum oxide dissociates into ions Al 3+ and O 2- . H and cations are reduced at the cathode Al 3+ , turning into aluminum atoms:

Al +3 e à Al

Anions are oxidized at the anode O2- , turning into oxygen atoms. Oxygen atoms immediately combine into O2 molecules:

2 O 2- – 4 e à O 2

The number of electrons involved in the processes of reduction of aluminum cations and oxidation of oxygen ions must be equal, so let’s multiply the first equation by 4, and the second by 3:

Al 3+ +3 e à Al 0 4

2 O 2- – 4 e à O 2 3

Let's add both equations and get

4 Al 3+ + 6 O 2- à 4 Al 0 +3 O 2 0 (ionic reaction equation)

2 Al 2 O 3 à 4 Al + 3 O 2

Electrolysis of solutions

In the case of passing an electric current through an aqueous electrolyte solution, the matter is complicated by the fact that the solution contains water molecules, which can also interact with electrons. Recall that in a water molecule, the hydrogen and oxygen atoms are connected by a polar covalent bond. The electronegativity of oxygen is greater than that of hydrogen, so the shared electron pairs are biased toward the oxygen atom. A partial negative charge arises on the oxygen atom, denoted δ-, and a partial positive charge arises on the hydrogen atoms, denoted δ+.

δ+

N-O δ-

│

H δ+

Due to this shift of charges, the water molecule has positive and negative “poles”. Therefore, water molecules can be attracted by the positively charged pole to the negatively charged electrode - the cathode, and by the negative pole - to the positively charged electrode - the anode. At the cathode, reduction of water molecules can occur, and hydrogen is released:

At the anode, oxidation of water molecules can occur, releasing oxygen:

2 H 2 O - 4e - = 4H + + O 2

Therefore, either electrolyte cations or water molecules can be reduced at the cathode. These two processes seem to compete with each other. What process actually occurs at the cathode depends on the nature of the metal. Whether metal cations or water molecules will be reduced at the cathode depends on the position of the metal in range of metal stresses .

Li K Na Ca Mg Al ¦¦ Zn Fe Ni Sn Pb (H 2) ¦¦ Cu Hg Ag Au

If the metal is in the voltage series to the right of hydrogen, metal cations are reduced at the cathode and free metal is released. If the metal is in the voltage series to the left of aluminum, water molecules are reduced at the cathode and hydrogen is released. Finally, in the case of metal cations from zinc to lead, either metal evolution or hydrogen evolution can occur, and sometimes both hydrogen and metal evolution can occur simultaneously. In general, this is a rather complicated case; a lot depends on the reaction conditions: solution concentration, electric current, and others.

One of two processes can also occur at the anode - either the oxidation of electrolyte anions or the oxidation of water molecules. Which process actually occurs depends on the nature of the anion. During the electrolysis of salts of oxygen-free acids or the acids themselves, anions are oxidized at the anode. The only exception is fluoride ion F- . In the case of oxygen-containing acids, water molecules are oxidized at the anode and oxygen is released.

Example 1.Let's look at the electrolysis of an aqueous solution of sodium chloride.

An aqueous solution of sodium chloride will contain sodium cations Na +, chlorine anions Cl - and water molecules.

2 NaCl à 2 Na + + 2 Cl -

2H 2 O à 2 H + + 2 OH -

cathode (-) 2 Na + ; 2H+; 2Н + + 2е à Н 0 2

anode (+) 2 Cl - ; 2 OH - ; 2 Cl - – 2е à 2 Cl 0

2NaCl + 2H 2 O à H 2 + Cl 2 + 2NaOH

Chemical activity anions are unlikely decreases.

Example 2.And if the salt contains SO 4 2- ? Let us consider the electrolysis of a nickel sulfate solution ( II ). Nickel sulfate ( II ) dissociates into ions Ni 2+ and SO 4 2-:

NiSO 4 à Ni 2+ + SO 4 2-

H 2 O à H + + OH -

Nickel cations are located between metal ions Al 3+ and Pb 2+ , occupying a middle position in the voltage series, the reduction process at the cathode occurs according to both schemes:

2 H 2 O + 2e - = H 2 + 2OH -

Anions of oxygen-containing acids are not oxidized at the anode ( anion activity series ), oxidation of water molecules occurs:

anode e à O 2 + 4H +

Let us write down together the equations of the processes occurring at the cathode and anode:

cathode (-) Ni 2+ ; H+; Ni 2+ + 2е à Ni 0

2 H 2 O + 2e - = H 2 + 2OH -

anode (+) SO 4 2- ; OH - ;2H 2 O – 4 e à O 2 + 4H +

4 electrons are involved in reduction processes and 4 electrons are also involved in oxidation processes. Let's add these equations together and get the general reaction equation:

Ni 2+ +2 H 2 O + 2 H 2 O à Ni 0 + H 2 + 2OH - + O 2 + 4 H +

On the right side of the equation there are both H + and OH- , which combine to form water molecules:

H + + OH - à H 2 O

Therefore, on the right side of the equation, instead of 4 H + ions and 2 ions OH- Let's write 2 water molecules and 2 H + ions:

Ni 2+ +2 H 2 O + 2 H 2 O à Ni 0 + H 2 +2 H 2 O + O 2 + 2 H +

Let's reduce two water molecules on both sides of the equation:

Ni 2+ +2 H 2 O à Ni 0 + H 2 + O 2 + 2 H +

This is a short ionic equation. To get the complete ionic equation, you need to add a sulfate ion to both sides SO 4 2- , formed during the dissociation of nickel sulfate ( II ) and not participating in the reaction:

Ni 2+ + SO 4 2- +2H 2 O à Ni 0 + H 2 + O 2 + 2H + + SO 4 2-

Thus, during the electrolysis of a solution of nickel sulfate ( II ) hydrogen and nickel are released at the cathode, and oxygen at the anode.

NiSO 4 + 2H 2 O à Ni + H 2 + H 2 SO 4 + O 2

Example 3. Write equations for the processes occurring during the electrolysis of an aqueous solution of sodium sulfate with an inert anode.

Standard electrode system potential Na + + e = Na 0 is significantly more negative than the potential of the aqueous electrode in a neutral aqueous medium (-0.41 V). Therefore, electrochemical reduction of water will occur at the cathode, accompanied by the release of hydrogen

2H 2 O à 2 H + + 2 OH -

and Na ions + coming to the cathode will accumulate in the part of the solution adjacent to it (cathode space).

Electrochemical oxidation of water will occur at the anode, leading to the release of oxygen

2 H 2 O – 4e à O 2 + 4 H +

since corresponding to this system standard electrode potential (1.23 V) is significantly lower than the standard electrode potential (2.01 V) characterizing the system

2 SO 4 2- + 2 e = S 2 O 8 2- .

SO 4 2- ions moving towards the anode during electrolysis will accumulate in the anode space.

Multiplying the equation of the cathodic process by two and adding it with the equation of the anodic process, we obtain the total equation of the electrolysis process:

6 H 2 O = 2 H 2 + 4 OH - + O 2 + 4 H +

Taking into account that simultaneous accumulation of ions in the cathode space and ions in the anode space occurs, the overall equation of the process can be written in the following form:

6H 2 O + 2Na 2 SO 4 = 2H 2 + 4Na + + 4OH - + O 2 + 4H + + 2SO 4 2-

Thus, simultaneously with the release of hydrogen and oxygen, sodium hydroxide (in the cathode space) and sulfuric acid (in the anode space) are formed.

Example 4.Electrolysis of copper sulfate solution ( II) CuSO 4 .

Cathode (-)<-- Cu 2+ + SO 4 2- à анод (+)

cathode (-) Cu 2+ + 2e à Cu 0 2

anode (+) 2H 2 O – 4 e à O 2 + 4H + 1

H+ ions remain in the solution SO 4 2- , because sulfuric acid accumulates.

2CuSO 4 + 2H 2 O à 2Cu + 2H 2 SO 4 + O 2

Example 5. Electrolysis of copper chloride solution ( II) CuCl 2.

Cathode (-)<-- Cu 2+ + 2Cl - à анод (+)

cathode (-) Cu 2+ + 2e à Cu 0

anode (+) 2Cl - – 2e à Cl 0 2

Both equations involve two electrons.

Cu 2+ + 2e à Cu 0 1

2Cl - --– 2e à Cl 2 1

Cu 2+ + 2 Cl - à Cu 0 + Cl 2 (ionic equation)

CuCl 2 à Cu + Cl 2 (molecular equation)

Example 6. Electrolysis of silver nitrate solution AgNO3.

Cathode (-)<-- Ag + + NO 3 - à Анод (+)

cathode (-) Ag + + e à Ag 0

anode (+) 2H 2 O – 4 e à O 2 + 4H +

Ag + + e à Ag 0 4

2H 2 O – 4 e à O 2 + 4H + 1

4 Ag + + 2 H 2 O à 4 Ag 0 + 4 H + + O 2 (ionic equation)

4 Ag + + 2 H 2 Oà 4 Ag 0 + 4 H + + O 2 + 4 NO 3 - (full ionic equation)

4 AgNO 3 + 2 H 2 Oà 4 Ag 0 + 4 HNO 3 + O 2 (molecular equation)

Example 7. Electrolysis of hydrochloric acid solutionHCl.

Cathode (-)<-- H + + Cl - à anode (+)

cathode (-) 2H + + 2 eà H 2

anode (+) 2Cl - – 2 eà Cl 2

2 H + + 2 Cl - à H 2 + Cl 2 (ionic equation)

2 HClà H 2 + Cl 2 (molecular equation)

Example 8. Electrolysis of sulfuric acid solutionH 2 SO 4 .

Cathode (-) <-- 2H + + SO 4 2- à anode (+)

cathode (-)2H+ + 2eà H 2

anode(+) 2H 2 O – 4eà O2 + 4H+

2H+ + 2eà H 2 2

2H 2 O – 4eà O2 + 4H+1

4H+ + 2H2Oà 2H 2 + 4H+ +O 2

2H2Oà 2H2 + O2

Example 9. Electrolysis of potassium hydroxide solutionKOH.

Cathode (-)<-- K + + OH - à anode (+)

Potassium cations will not be reduced at the cathode, since potassium is in the voltage series of metals to the left of aluminum; instead, reduction of water molecules will occur:

2H 2 O + 2eà H 2 +2OH - 4OH - -4eà 2H 2 O +O 2

cathode(-) 2H 2 O + 2eà H 2 +2OH - 2

anode(+) 4OH - - 4eà 2H 2 O +O 2 1

4H 2 O + 4OH -à 2H 2 + 4OH - + 2H 2 O + O 2

2 H 2 Oà 2 H 2 + O 2

Example 10. Electrolysis of potassium nitrate solutionKNO 3 .

Cathode (-) <-- K + + NO 3 - à anode (+)

2H 2 O + 2eà H 2 +2OH - 2H 2 O – 4eà O2+4H+

cathode(-) 2H 2 O + 2eà H2+2OH-2

anode(+) 2H 2 O – 4eà O2 + 4H+1

4H 2 O + 2H 2 Oà 2H 2 + 4OH - + 4H ++ O2

2H2Oà 2H2 + O2

When an electric current is passed through solutions of oxygen-containing acids, alkalis and salts of oxygen-containing acids with metals located in the voltage series of metals to the left of aluminum, electrolysis of water practically occurs. In this case, hydrogen is released at the cathode, and oxygen at the anode.

Conclusions. When determining the products of electrolysis of aqueous solutions of electrolytes, in the simplest cases one can be guided by the following considerations:

1.Metal ions with a small algebraic value of the standard potential - fromLi + beforeAl 3+ inclusive - have a very weak tendency to re-add electrons, being inferior in this regard to ionsH + (cm. Cation activity series). During the electrolysis of aqueous solutions of compounds containing these cations, ions perform the function of an oxidizing agent at the cathodeH + , restoring according to the scheme:

2 H 2 O+ 2 eà H 2 + 2OH -

2. Metal cations with positive values ​​of standard potentials (Cu 2+ , Ag + , Hg 2+ etc.) have a greater tendency to add electrons compared to ions. During the electrolysis of aqueous solutions of their salts, the function of the oxidizing agent at the cathode is released by these cations, while being reduced to metal according to the scheme, for example:

Cu 2+ +2 eà Cu 0

3. During electrolysis of aqueous solutions of metal saltsZn, Fe, Cd, Nietc., occupying a middle position in the voltage series between the listed groups, the reduction process at the cathode occurs according to both schemes. The mass of the released metal in these cases does not correspond to the amount of electric current flowing, part of which is spent on the formation of hydrogen.

4. In aqueous solutions of electrolytes, monoatomic anions (Cl - , Br - , J - ), oxygen-containing anions (NO 3 - , SO 4 2- , P.O. 4 3- and others), as well as hydroxyl ions of water. Of these, halide ions have stronger reducing properties, with the exception ofF. IonsOHoccupy an intermediate position between them and polyatomic anions. Therefore, during the electrolysis of aqueous solutionsHCl, HBr, H.J.or their salts at the anode, oxidation of halide ions occurs according to the following scheme:

2 X - -2 eà X 2 0

During the electrolysis of aqueous solutions of sulfates, nitrates, phosphates, etc. The function of a reducing agent is performed by ions, oxidizing according to the following scheme:

4 HOH – 4 eà 2 H 2 O + O 2 + 4 H +

.

Tasks.

Z A cottage 1. During the electrolysis of a copper sulfate solution, 48 g of copper was released at the cathode. Find the volume of gas released at the anode and the mass of sulfuric acid formed in the solution.

Copper sulfate in solution dissociates no ionsC 2+ andS0 4 2 ".

CuS0 4 = Cu 2+ + S0 4 2 "

Let us write down the equations of the processes occurring at the cathode and anode. Cu cations are reduced at the cathode, and water electrolysis occurs at the anode:

Cu 2+ +2e- = Cu12

2H 2 0-4e- = 4H + + 0 2 |1

The general equation for electrolysis is:

2Cu2+ + 2H2O = 2Cu + 4H+ + O2 (short ionic equation)

Let's add 2 sulfate ions to both sides of the equation, which are formed during the dissociation of copper sulfate, and we get the complete ionic equation:

2Cu2+ + 2S042" + 2H20 = 2Cu + 4H+ + 2SO4 2" + O2

2CuSO4 + 2H2O = 2Cu + 2H2SO4 + O2

The gas released at the anode is oxygen. Sulfuric acid is formed in the solution.

The molar mass of copper is 64 g/mol, let’s calculate the amount of copper substance:

According to the reaction equation, when 2 moles of copper are released at the cathode, 1 mole of oxygen is released at the anode. 0.75 moles of copper are released at the cathode, let x moles of oxygen be released at the anode. Let's make a proportion:

2/1=0.75/x, x=0.75*1/2=0.375mol

0.375 mol of oxygen was released at the anode,

v(O2) = 0.375 mol.

Let's calculate the volume of oxygen released:

V(O2) = v(O2) «VM = 0.375 mol «22.4 l/mol = 8.4 l

According to the reaction equation, when 2 moles of copper are released at the cathode, 2 moles of sulfuric acid are formed in the solution, which means that if 0.75 moles of copper are released at the cathode, then 0.75 moles of sulfuric acid are formed in the solution, v(H2SO4) = 0.75 moles . Let's calculate the molar mass of sulfuric acid:

M(H2SO4) = 2-1+32+16-4 = 98 g/mol.

Let's calculate the mass of sulfuric acid:

m(H2S04) = v(H2S04>M(H2S04) = = 0.75 mol «98 g/mol = 73.5 g.

Answer: 8.4 liters of oxygen were released at the anode; 73.5 g of sulfuric acid was formed in the solution

Problem 2. Find the volume of gases released at the cathode and anode during the electrolysis of an aqueous solution containing 111.75 g of potassium chloride. What substance was formed in the solution? Find its mass.

Potassium chloride in solution dissociates into K+ and Cl ions:

2КС1 =К+ + Сl

Potassium ions are not reduced at the cathode; instead, water molecules are reduced. At the anode, chloride ions are oxidized and chlorine is released:

2H2O + 2e" = H2 + 20H-|1

2SG-2e" = C12|1

The general equation for electrolysis is:

2СГl+ 2Н2О = Н2 + 2ОН" + С12 (short ionic equation) The solution also contains K+ ions formed during the dissociation of potassium chloride and not participating in the reaction:

2K+ + 2Cl + 2H20 = H2 + 2K+ + 2OH" + C12

Let's rewrite the equation in molecular form:

2KS1 + 2H2O = H2 + C12 + 2KON

Hydrogen is released at the cathode, chlorine at the anode, and potassium hydroxide is formed in the solution.

The solution contained 111.75 g of potassium chloride.

Let's calculate the molar mass of potassium chloride:

M(KS1) = 39+35.5 = 74.5 g/mol

Let's calculate the amount of potassium chloride:

According to the reaction equation, during the electrolysis of 2 moles of potassium chloride, 1 mole of chlorine is released. Let the electrolysis of 1.5 mol of potassium chloride produce x mol of chlorine. Let's make a proportion:

2/1=1.5/x, x=1.5 /2=0.75 mol

0.75 mol of chlorine will be released, v(C!2) = 0.75 mol. According to the reaction equation, when 1 mole of chlorine is released at the anode, 1 mole of hydrogen is released at the cathode. Therefore, if 0.75 mol of chlorine is released at the anode, then 0.75 mol of hydrogen is released at the cathode, v(H2) = 0.75 mol.

Let's calculate the volume of chlorine released at the anode:

V(C12) = v(Cl2)-VM = 0.75 mol «22.4 l/mol = 16.8 l.

The volume of hydrogen is equal to the volume of chlorine:

Y(H2) = Y(C12) = 16.8l.

According to the reaction equation, the electrolysis of 2 mol of potassium chloride produces 2 mol of potassium hydroxide, which means that the electrolysis of 0.75 mol of potassium chloride produces 0.75 mol of potassium hydroxide. Let's calculate the molar mass of potassium hydroxide:

M(KOH) = 39+16+1 - 56 g/mol.

Let's calculate the mass of potassium hydroxide:

m(KOH) = v(KOH>M(KOH) = 0.75 mol-56 g/mol = 42 g.

Answer: 16.8 liters of hydrogen were released at the cathode, 16.8 liters of chlorine were released at the anode, and 42 g of potassium hydroxide were formed in the solution.

Problem 3. During the electrolysis of a solution of 19 g of divalent metal chloride, 8.96 liters of chlorine were released at the anode. Determine which metal chloride was subjected to electrolysis. Calculate the volume of hydrogen released at the cathode.

Let's denote the unknown metal M, the formula of its chloride is MC12. At the anode, chloride ions are oxidized and chlorine is released. The condition says that hydrogen is released at the cathode, therefore, the reduction of water molecules occurs:

2Н20 + 2е- = Н2 + 2ОH|1

2Cl -2e" = C12! 1

The general equation for electrolysis is:

2Cl + 2H2O = H2 + 2OH" + C12 (short ionic equation)

The solution also contains M2+ ions, which do not change during the reaction. Let us write the complete ionic equation of the reaction:

2SG + M2+ + 2H2O = H2 + M2+ + 2OH- + C12

Let's rewrite the reaction equation in molecular form:

MC12 + 2H2O - H2 + M(OH)2 + C12

Let's find the amount of chlorine released at the anode:

According to the reaction equation, during the electrolysis of 1 mole of chloride of an unknown metal, 1 mole of chlorine is released. If 0.4 mol of chlorine was released, then 0.4 mol of metal chloride was subjected to electrolysis. Let's calculate the molar mass of the metal chloride:

The molar mass of the unknown metal chloride is 95 g/mol. There are 35.5"2 = 71 g/mol per two chlorine atoms. Therefore, the molar mass of the metal is 95-71 = 24 g/mol. Magnesium corresponds to this molar mass.

According to the reaction equation, for 1 mole of chlorine released at the anode, there is 1 mole of hydrogen released at the cathode. In our case, 0.4 mol of chlorine was released at the anode, which means 0.4 mol of hydrogen was released at the cathode. Let's calculate the volume of hydrogen:

V(H2) = v(H2>VM = 0.4 mol «22.4 l/mol = 8.96 l.

Answer: a solution of magnesium chloride was subjected to electrolysis; 8.96 liters of hydrogen were released at the cathode.

*Problem 4. During the electrolysis of 200 g of potassium sulfate solution with a concentration of 15%, 14.56 liters of oxygen were released at the anode. Calculate the concentration of the solution at the end of electrolysis.

In a solution of potassium sulfate, water molecules react at both the cathode and anode:

2Н20 + 2е" = Н2 + 20Н-|2

2H2O - 4e" = 4H+ + O2! 1

Let's add both equations together:

6H2O = 2H2 + 4OH" + 4H+ + O2, or

6H2O = 2H2 + 4H2O + O2, or

2H2O = 2H2 + 02

In fact, when electrolysis of a solution of potassium sulfate occurs, the electrolysis of water occurs.

The concentration of a solute in a solution is determined by the formula:

С=m(solute) 100% / m(solution)

To find the concentration of the potassium sulfate solution at the end of electrolysis, you need to know the mass of potassium sulfate and the mass of the solution. The mass of potassium sulfate does not change during the reaction. Let's calculate the mass of potassium sulfate in the original solution. Let us denote the concentration of the initial solution as C

m(K2S04) = C2 (K2S04) m(solution) = 0.15 200 g = 30 g.

The mass of the solution changes during electrolysis as part of the water is converted into hydrogen and oxygen. Let's calculate the amount of oxygen released:

(O 2)=V(O2) / Vm =14.56l / 22.4l/mol=0.65mol

According to the reaction equation, 2 moles of water produce 1 mole of oxygen. Let 0.65 mol of oxygen be released during the decomposition of x mol of water. Let's make a proportion:

1.3 mol of water decomposed, v(H2O) = 1.3 mol.

Let's calculate the molar mass of water:

M(H2O) = 1-2 + 16 = 18 g/mol.

Let's calculate the mass of decomposed water:

m(H2O) = v(H2O>M(H2O) = 1.3 mol* 18 g/mol = 23.4 g.

The mass of the potassium sulfate solution decreased by 23.4 g and became equal to 200-23.4 = 176.6 g. Let us now calculate the concentration of the potassium sulfate solution at the end of electrolysis:

C2 (K2 SO4)=m(K2 SO4) 100% / m(solution)=30g 100% / 176.6g=17%

Answer: the concentration of the solution at the end of electrolysis is 17%.

*Task 5. 188.3 g of a mixture of sodium and potassium chlorides was dissolved in water and an electric current was passed through the resulting solution. During electrolysis, 33.6 liters of hydrogen were released at the cathode. Calculate the composition of the mixture as a percentage by weight.

After dissolving a mixture of potassium and sodium chlorides in water, the solution contains K+, Na+ and Cl- ions. Neither potassium ions nor sodium ions are reduced at the cathode; water molecules are reduced. At the anode, chloride ions are oxidized and chlorine is released:

Let's rewrite the equations in molecular form:

2KS1 + 2N20 = N2 + C12 + 2KON

2NaCl + 2H2O = H2 + C12 + 2NaOH

Let us denote the amount of potassium chloride contained in the mixture by x mol, and the amount of sodium chloride by mol. According to the reaction equation, during the electrolysis of 2 moles of sodium or potassium chloride, 1 mole of hydrogen is released. Therefore, during the electrolysis of x mole of potassium chloride, x/2 or 0.5x mole of hydrogen is formed, and during the electrolysis of x mole of sodium chloride, 0.5y mole of hydrogen is formed. Let's find the amount of hydrogen released during electrolysis of the mixture:

Let's make the equation: 0.5x + 0.5y = 1.5

Let's calculate molar masses potassium and sodium chlorides:

M(KS1) = 39+35.5 = 74.5 g/mol

M(NaCl) = 23+35.5 = 58.5 g/mol

Mass x mole of potassium chloride is equal to:

m(KCl) = v(KCl)-M(KCl) = x mol-74.5 g/mol = 74.5x g.

The mass of a mole of sodium chloride is:

m(KCl) = v(KCl)-M(KCl) = y mol-74.5 g/mol = 58.5y g.

The mass of the mixture is 188.3 g, let’s create the second equation:

74.5x + 58.5y= 188.3

So, we solve a system of two equations with two unknowns:

0.5(x + y)= 1.5

74.5x + 58.5y=188.3g

From the first equation we express x:

x + y = 1.5/0.5 = 3,

x = 3-y

Let's substitute this value of x into second equation, we get:

74.5-(3-y) + 58.5y= 188.3

223.5-74.5y + 58.5y= 188.3

-16у = -35.2

y = 2.2 100% / 188.3g = 31.65%

Let's calculate the mass fraction of sodium chloride:

w(NaCl) = 100% - w(KCl) = 68.35%

Answer: the mixture contains 31.65% potassium chloride and 68.35% sodium chloride.

When considering the electrolysis of aqueous solutions, it is necessary to keep in mind that, in addition to electrolyte ions, in any aqueous solution there are also ions that are products of the dissociation of water H + and OH –.

In an electric field, hydrogen ions move to the cathode, and OH ions move to the anode. Thus, both electrolyte cations and hydrogen cations can be discharged at the cathode. Similarly, at the anode, both electrolyte anions and hydroxide ions can be discharged. In addition, water molecules can also undergo electrochemical oxidation or reduction.

What kind of electrochemical processes will occur at the electrodes during electrolysis will primarily depend on the relative values ​​of the electrode potentials of the corresponding electrochemical systems. Of several possible processes, the one whose implementation involves minimal energy consumption will proceed. This means that the oxidized forms of electrochemical systems with the highest electrode potential will be reduced at the cathode, and the reduced forms of systems with the lowest electrode potential will be oxidized at the anode. In general, those atoms, molecules and ions whose potentials are the lowest under given conditions are more easily oxidized at the anode, and those ions, molecules, and atoms whose potentials are the highest are restored more easily at the cathode. Let us consider the cathodic processes occurring during the electrolysis of aqueous salt solutions. Here it is necessary to take into account the magnitude of the electrode potential of the process of reduction of hydrogen ions, which depends on the concentration of hydrogen ions. We know the general equation of the electrode potential for a hydrogen electrode (Section 2.3).

In the case of neutral solutions (pH=7), the value of the electrode potential of the process of reduction of hydrogen ions is

φ = –0,059 . 7 = –0.41 V.

1) during the electrolysis of salt solutions containing metal cations, the electrode potential of which is significantly more positive than –0.41 V, the metal will be reduced from a neutral solution of such an electrolyte at the cathode. Such metals are found in the voltage series near hydrogen (starting approximately from and after tin);

2) during the electrolysis of salt solutions containing metal cations, the electrode potential of which is significantly more negative than – 0.41 V, the metal will not be reduced at the cathode, but hydrogen will be released. Such metals include alkali, alkaline earth, magnesium, aluminum, approximately up to titanium;

3) during the electrolysis of salt solutions containing metal cations, the electrode potential of which is close to –0.41 V (metals in the middle part of the series - Zn, Cr, Fe, Cd, Ni), then depending on the concentration of the salt solution and electrolysis conditions ( current density, temperature, solution composition), both metal reduction and hydrogen evolution are possible; Sometimes a joint release of metal and hydrogen is observed.

The electrochemical evolution of hydrogen from acidic solutions occurs due to the discharge of hydrogen ions:

2H + 2ē → 2Н 0

2H 0 = N 2 .

In the case of neutral or alkaline media, the evolution of hydrogen occurs as a result of the electrochemical reduction of water:

НН + ē → Н 0 + HE –

N 0 + N 0 = N 2 ,

Then 2НН + 2ē → Н 2 + 2OH –

Thus, the nature of the cathodic process during the electrolysis of aqueous solutions is determined primarily by the position of the corresponding metal in the series of standard electrode potentials of metals.

If an aqueous solution containing cations of various metals is subjected to electrolysis, then their release at the cathode, as a rule, will proceed in the order of decreasing algebraic value of the electrode potential of the metal. For example, from a mixture of cations Ag +, Cu 2+ and Zn 2+, with sufficient voltage at the electrolyzer terminals, silver cations will first be reduced (φ 0 = +0.8 V), then copper (φ 0 = +0.34 V) and finally, zinc (φ 0 = –0.76 V).

Electrochemical separation of metals from a mixture of cations is used in technology and in quantitative analysis. In general, the ability of metal ions to discharge (gain electrons) is determined by the position of the metals in the series of standard electrode potentials. The further to the left a metal is in the voltage series, the greater its negative potential or the less positive potential, the more difficult it is for its ions to discharge. Thus, of the metal ions standing in the voltage series, trivalent gold ions are most easily discharged (at the lowest electric current voltages), then silver ions, etc. Potassium ions are most difficult to discharge (at the highest electric voltage). But the potential of a metal, as is known, varies depending on the concentration of its ions in solution; in the same way, the ease of discharge of the ions of each metal changes depending on their concentration: an increase in concentration facilitates the discharge of ions, a decrease makes it more difficult. Therefore, during the electrolysis of a solution containing ions of several metals, it may be that the release of more active metal will occur earlier than the release of a less active one (if the concentration of the first metal ion is significant, and the second is very small).

Let us consider the anodic processes occurring during the electrolysis of aqueous salt solutions. The nature of the reactions occurring at the anode depends both on the presence of water molecules and on the substance from which the anode is made. It should be borne in mind that the anode material may oxidize during electrolysis. In this regard, a distinction is made between electrolysis with an inert (insoluble) anode and electrolysis with an active (soluble) anode. Insoluble anodes are made from coal, graphite, platinum, iridium; soluble anodes - made of copper, silver, zinc, cadmium, nickel and other metals. At the insoluble anode, during the electrolysis process, oxidation of anions or water molecules occurs. During the electrolysis of aqueous solutions of oxygen-free acids HI, HBr, HCl, H 2 Si and their salts (except HF and fluorides), anions are discharged at the anode and the corresponding halogen is released. Note that the release of chlorine during the electrolysis of HCl and its salts contradicts the relative position of the systems

2Cl – – 2ē →Cl 2 (φ 0 = +1.36 V)

2 H 2 O– 4ē →O 2 + 4 H + (φ 0 = +1.23 V)

in the range of standard electrode potentials. This anomaly is associated with a significant overvoltage of the second of these two electrode processes - the anode material has an inhibitory effect on the process of oxygen release.

During the electrolysis of aqueous solutions of salts containing the anions SO 4 2-, SO 3 2-, NO 3 -, PO 4 3-, etc., as well as hydrogen fluoride and fluorides, electrochemical oxidation of water occurs. Depending on the pH of the solution, this process occurs differently and can be written by different equations. In an alkaline medium, the equation has the form

4OH – – 4ē → 2H 2 O+O 2 , (pH > 7)

and in acidic or neutral media we have

HOH– 2ē →O 0 + 2 H + (pH ≤ 7)

2 O 0 = O 2 ,

Then 2H 2 О – 4ē → 4Н + + 2О 2 .

In the cases under consideration, electrochemical oxidation of water is the most energetically favorable process. Oxidation of oxygen-containing anions occurs at very high potentials. For example, the standard oxidation potential of the SO 4 2- ion – 2ē →S 2 O 8 2- is 2.01 V, which is significantly higher than the standard oxidation potential of water of 1.228 V.

2H 2 O – 4ē → O 2 + 4H + (φ 0 = 1.228 V).

The standard oxidation potential of the F ion is even more important

2F – – 2ē →F 2 (φ 0 = 2 ,87 IN).

In general, during the electrolysis of aqueous solutions of salts, metal and hydrogen cations simultaneously approach the cathode of the electrolyzer, and each of them “claims” to be reduced by electrons coming from the cathode. How will the reduction process actually proceed at the cathode? The answer can be obtained based on a number of metal stresses. Moreover, the lower the algebraic value of the standard electrode potential of a metal, the weaker the electron acceptors their cations are and the more difficult it is to restore them at the cathode. In this regard, three groups of cations are distinguished according to their relation to electroreduction.

1. Cations characterized by high electron-withdrawing activity (Cu 2+, Hg 2+, Ag+, Au 3+, Pt 2+, Pt 4+). During the electrolysis of salts of these cations, the metal cations are almost completely reduced; current output is 100% or a value close to it.

2. Cations characterized by average values ​​of electron-withdrawing ability (Mn 2+, Zn 2+, Cr 3+, Fe 2+, Ni 2+, Sn 2+, Pb 2+). During electrolysis at the cathode, cations of both metal and water molecules are simultaneously reduced, which leads to a decrease in the current output of the metal.

3. Cations exhibiting low electron-withdrawing ability (K + , Ca 2+ , Mg 2+ , Al 3+ ). In this case, the electron acceptors at the cathode are not cations of the group under consideration, but water molecules. In this case, the cations themselves remain unchanged in the aqueous solution, and the current efficiency approaches zero.

Relation of various anions to electrooxidation at the anode

Anions of oxygen-free acids and their salts (Cl¯,Br¯,J¯,S2-,CN¯, etc.) retain their electrons weaker than water molecules. Therefore, during the electrolysis of aqueous solutions of compounds containing these anions, the latter will play the role of electron donors; they will oxidize and transfer their electrons to the external circuit of the electrolyzer.

Anions of oxygen acids (NO 3 ¯, SO 4 2-, PO 4 3-, etc.) are able to hold their electrons more firmly than water molecules. In this case, water is oxidized at the anode, but the anions themselves remain unchanged.

In the case of a soluble anode, the number of oxidation processes increases to three:

1) electrochemical oxidation of water with the release of oxygen; 2) discharge of the anion (i.e. its oxidation); 3) electrochemical oxidation of the anode metal (anodic dissolution of the metal).

Of the possible processes, the one that is energetically most favorable will take place. If the anode metal is located in a series of standard potentials earlier than both other electrochemical systems, then anodic dissolution of the metal will be observed. Otherwise, oxygen will be released or anion will be discharged. No close sequence has been established for the discharge of anions. By decreasing ability to donate electrons, the most common anions are arranged as follows: S 2-,J ¯,Br ¯,Cl ¯,OH¯,H 2 O,SO 4 2-,NO 3 ¯,CO 3 2-,PO 4 3- .

Let us consider several typical cases of electrolysis of aqueous solutions.

Electrolysis of a CuCl 2 solution with an insoluble anode

In the voltage series, copper is located after hydrogen, so Cu 2+ will be discharged at the cathode and metallic copper will be released, and chloride ions will be oxidized at the anode to molecular chlorine Cl 2 .

	Cathode (–)
	Cu 2+ + 2ē → Cu 0
	2Cl – – 2ē → Cl 2

Cu 2+ + 2 Cl – → Cu 0 + Cl 2

CuCl 2 → Cu 0 + Cl 2

Metal current yield (95-100%).

Electrolysis of NaNO 3 solution

Since sodium is much earlier than hydrogen in the voltage series, water will be discharged at the cathode. Water will also discharge at the anode.

Cathode (–)

2 H 2 O+ 2ē →H 2 + 2 OH –

2H 2 O – 4ē → 4H + +O 2 .

Thus, hydrogen is released at the cathode and an alkaline environment is created, oxygen is released at the anode and an acidic environment is created near the anode. If the anode and cathode spaces are not separated from each other, then the solution in all its parts will remain electrically neutral.

	Cathode (–)
	2 H 2 O+ 2ē →H 2 + 2 OH –
	2H 2 O – 4ē → 4H + +O 2 .

6H 2 O → 2H 2 + 4OH – + 4H + +O 2

6H 2 O → 2H 2 +O 2 + 4H 2 O

2 H 2 O → 2 H 2 + O 2

The current output of the metal is zero.

Therefore, during electrolysis of a NaNO 3 solution, electrolysis of water will occur. The role of NaNO 3 salt is reduced to increasing the electrical conductivity of the solution.

Electrolysis of FeSO 4 solution

Reactions at the cathode (–) (reduction):

A) Fe 2+ + 2ē → Fe 0	simultaneous reactions
b) 2 H 2 O+ 2ē →H 2 + 2 OH – .

Reaction at the anode (+) (oxidation):

2H 2 O – 4ē → 4H + +O 2 .

The metal current output is average.

Electrolysis of KJ solution with insoluble anode

	Cathode (–)
	2 H 2 O+ 2ē →H 2 + 2 OH –
	2J – – 2ē → J 2

2 H 2 O + 2J – → H 2 + 2 OH – +J 2 .

The final equation for the electrolysis reaction of the KJ solution is:

2KJ+2H 2 O → H 2 +J 2 +2KOH.

Electrolysis of a CuSO 4 solution with a copper (soluble) anode.

The standard potential of copper is +0.337 V, which is significantly higher than -0.41 V; therefore, during the electrolysis of a CuSO 4 solution at the cathode, Cu 2+ ions are discharged and metallic copper is released. The opposite process occurs at the anode - oxidation of the metal, since the potential of copper is much less than the oxidation potential of water (+1.228 V), and even more so - the oxidation potential of the SO 4 2- ion (+2.01 V). Consequently, in this case, electrolysis comes down to dissolving the metal (copper) of the anode and separating it at the cathode.

Scheme of electrolysis of copper sulfate solution:

Cathode (–)

Cu 2+ + 2ē → Cu 0

Cu 0 – 2ē → Cu 2+ .

This process is used for the electrical refining of metals (called electrolytic refining).