Module 2. Basic processes of chemistry and properties of substances

Lab #7

Topic: Electrolysis of aqueous salt solutions

by electrolysis called a redox process that occurs on the electrodes when an electric current passes through a solution or an electrolyte melt.

When a constant electric current is passed through an electrolyte solution or melt, cations move towards the cathode, and anions move towards the anode. Oxidation-reduction processes take place on the electrodes; The cathode is a reducing agent, since it donates electrons to cations, and the anode is an oxidizing agent, since it accepts electrons from anions. The reactions occurring on the electrodes depend on the composition of the electrolyte, the nature of the solvent, the material of the electrodes, and the mode of operation of the cell.

Chemistry of the electrolysis process of calcium chloride melt:

CaCl 2 ↔ Ca 2+ + 2Cl -

at the cathode Ca 2+ + 2e → Ca °

at the anode 2Cl - - 2e → 2C1 ° → C1 2

The electrolysis of a solution of potassium sulfate on an insoluble anode schematically looks like this:

K 2 SO 4 ↔ 2K + + SO 4 2 -

H 2 O ↔ H + + OH -

at the cathode 2Н + + 2е→2Н°→ Н 2 2

at the anode 4OH - 4e → O 2 + 4H + 1

K 2 SO 4 + 4H 2 O 2H 2 + O 2 + 2K0H + H 2 SO 4

Objective: familiarization with the electrolysis of salt solutions.

Devices and equipment: electric current rectifier, electrolyzer, carbon electrodes, sandpaper, cups, washer.

Rice. 1. Device for carrying out

electrolysis

1 - electrolyzer;

2 - electrodes;

3-conductive wires; DC source.

Reagents and solutions: 5% solutions of copper chloride СuС1 2, potassium iodide KI , potassium hydrogen sulfate KHSO 4 , sodium sulfate Na 2 SO 4 , copper sulfate CuSO 4 , zinc sulfate ZnSO 4 , 20% sodium hydroxide solution NaOH, copper and nickel plates, phenolphthalein solution, nitric acid (conc.) HNO 3 , 1% starch solution , neutral litmus paper, 10% sulfuric acid solution H 2 SO 4 .

Experience 1. Electrolysis of copper chloride with insoluble electrodes

Fill the electrolyzer up to half the volume with a 5% copper chloride solution. Lower the graphite rod into both knees of the electrolyzer, fasten them loosely to the segments and the rubber tube. Connect the ends of the electrodes with conductors to direct current sources. If there is a slight smell of chlorine, immediately disconnect the electrolyser from the power source. What happens at the cathode? Make equations of electrode reactions.

Experience 2. Electrolysis of potassium iodide with insoluble electrodes

Fill the electrolytic cell with 5% potassium iodide solution, . add 2 drops of phenolphthalein to each knee. Paste in each knee of the electrolyzer graphite electrodes and connect them to a direct current source.

In which knee and why did the solution turn colored? Add 1 drop of starch paste to each knee. Where and why is iodine released? Make equations of electrode reactions. What is formed in the cathode space?

Experience 3. Electrolysis of sodium sulfate with insoluble electrodes

Fill half of the volume of the electrolyzer with 5% sodium sulfate solution and add 2 drops of methyl orange or litmus to each knee. Insert electrodes into both knees and connect them to a direct current source. Write down your observations. Why did electrolyte solutions turn different colors at different electrodes? Make equations of electrode reactions. What gases and why are released on the electrodes? What is the essence of the process of electrolysis of an aqueous solution of sodium sulfate

ELECTROLYSIS

One of the ways to obtain metals is electrolysis. Active metals occur in nature only in the form of chemical compounds. How to isolate from 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 can occur. Consider what will happen if two metal plates are placed in an electrolyte solution or melt, each of which is connected to one of the poles of the current source. These plates are called electrodes. Electric current is a moving stream of electrons. As a result of the fact that the electrons in the circuit move from one electrode to another, an excess of electrons appears on one of the electrodes. The electrons have a negative charge, so this electrode becomes negatively charged. It is called the cathode. On the other electrode, a lack of electrons is created, and it is 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 a negatively charged electrode - the cathode. Anions are attracted to a positively charged electrode - the anode. On the surface of the electrodes, interaction between ions and electrons can occur.

Electrolysis refers to the processes that occur when an 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.

Melt electrolysis

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, there is a transfer of electrons from the electrode surface to sodium ions. At the same time, ions Na+ are converted into sodium atoms, that is, cations are reduced 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 surface of the electrode. At the same time, negatively charged ions Cl- are converted into chlorine atoms, which immediately combine to form chlorine molecules C l2 :

2C l - -2e - \u003d Cl 2

Chloride ions lose electrons, that is, they are oxidized.

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

Na + + e - = Na

2 C l - -2 e - \u003d Cl 2

One electron is involved in the process of reduction of sodium cations, and 2 electrons are involved in the process of 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, the 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 - \u003d Na 2

2C l - -2e - \u003d Cl 2 1

We add both equations together and get the general equation for the reaction.

2 Na + + 2C l - \u003d 2 Na + Cl 2 (ionic reaction equation), or

2 NaCl \u003d 2 Na + Cl 2 (molecular reaction equation)

So, in the considered example, we see that electrolysis is a redox reaction. At the cathode, the reduction of positively charged ions - cations, at the anode - the oxidation of negatively charged ions - anions. To remember which process happens where, you can use the "T rule":

cathode - cation - reduction.

Example 2Electrolysis of sodium hydroxide melt.

Sodium hydroxide in solution dissociates into cations and hydroxide ions.

Cathode (-)<-- Na + + OH - à Анод (+)

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

cathode (-) Na + +e à Na

Hydroxide ions are oxidized on the anode surface, while 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 should be the same. So 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

Putting both equations together, we get the equation for the electrolysis reaction:

4 NaOH à 4 Na + 2 H 2 O + O 2

Example 3Consider 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, lowering the melting point to 800-900º C. In the melt, aluminum oxide dissociates into ions Al 3+ and O 2-. H cations are reduced at the cathode Al 3+ , turning into aluminum atoms:

Al +3 e a Al

Anions are oxidized at the anode O 2- turning into oxygen atoms. Oxygen atoms immediately combine into O 2 molecules:

2 O 2- – 4 e à O 2

The number of electrons involved in the reduction of aluminum cations and the oxidation of oxygen ions must be equal, so we 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- a 4 Al 0 +3 O 2 0 (ionic reaction equation)

2 Al 2 O 3 à 4 Al + 3 O 2

Solution electrolysis

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

δ+

H-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 a positively charged pole to a negatively charged electrode - the cathode, and by a negative pole - to a positively charged electrode - anode. At the cathode, water molecules can be reduced, and hydrogen is released:

Oxidation of water molecules can occur at the anode with the release of oxygen:

2 H 2 O - 4e - \u003d 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 takes place 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 series 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 are evolved simultaneously. In general, this is a rather complicated case, much depends on the reaction conditions: the concentration of the solution, the current strength, 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 takes place depends on the nature of the anion. During the electrolysis of salts of anoxic acids or the acids themselves, anions are oxidized at the anode. The only exception is the fluoride ion F- . In the case of oxygen-containing acids, water molecules are oxidized at the anode and oxygen is released.

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

In an aqueous solution of sodium chloride there will be sodium cations Na + , chlorine anions Cl - and water molecules.

2 NaCl a 2 Na + + 2 Cl -

2Н 2 О а 2 H + + 2 OH -

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

anode (+) 2 Cl - ; 2OH-; 2 Cl - – 2e a 2 Cl 0

2NaCl + 2H 2 O à H 2 + Cl 2 + 2NaOH

Chemical activity anions hardly decreases.

Example 2What if the salt contains SO 4 2- ? 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 between metal ions Al 3+ and Pb 2+ , occupying a middle position in the voltage series, the recovery process at the cathode occurs according to both schemes:

2 H 2 O + 2e - \u003d H 2 + 2OH -

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

anode e à O 2 + 4H +

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

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

2 H 2 O + 2e - \u003d H 2 + 2OH -

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

4 electrons are involved in the reduction processes, and 4 electrons are also involved in the oxidation process. Putting these equations together, we 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 simultaneously H + ions 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- we 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 full ionic equation, you need to add to both parts of the sulfate ion 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 is released at the anode.

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

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

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

2Н 2 О а 2 H + + 2 OH -

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

At the anode, electrochemical oxidation of water will occur, leading to the release of oxygen.

2 H 2 O - 4e à O 2 + 4 H +

because corresponding to this system standard electrode potential (1.23 V) is significantly lower than the standard electrode potential (2.01 V) that characterizes the system

2 SO 4 2- + 2 e \u003d S 2 O 8 2-.

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

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

6 H 2 O \u003d 2 H 2 + 4 OH - + O 2 + 4 H +

Taking into account that ions are simultaneously accumulated in the cathode space and ions in the anode space, the overall process equation can be written in the following form:

6H 2 O + 2Na 2 SO 4 \u003d 2H 2 + 4Na + + 4OH - + O 2 + 4H + + 2SO 4 2-

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

Example 4Electrolysis of copper sulfate solution ( II) CuSO4.

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 and SO 4 2- , since 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 H2

anode(+) 2H 2 O - 4eà O2+4H+

2H+ + 2eà H 2 2

2H2O-4eà O 2 + 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 water molecules will be reduced:

2H2O + 2eà H 2 + 2OH - 4OH - -4eà 2H 2 O + O 2

cathode(-)2H2O+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 (+)

2H2O + 2eà H 2 + 2OH - 2H 2 O - 4eà O2+4H+

cathode(-)2H2O+2eà H 2 + 2OH-2

anode(+) 2H 2 O - 4eà O 2 + 4H+1

4H2O + 2H2Oà 2H2+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 that are in the voltage series of metals, to the left of aluminum, water electrolysis 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 reattach electrons, yielding in this respect to ionsH + (cm. Cation activity series). In the electrolysis of aqueous solutions of compounds containing these cations, the function of an oxidizing agent on the cathode is performed by ionsH + , while 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 attach electrons than ions. During the electrolysis of aqueous solutions of their salts, these cations emit the function of an oxidizing agent on the cathode, while being reduced to a metal according to the scheme, for example:

Cu 2+ +2 eà Cu 0

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

4. In aqueous solutions of electrolytes, monatomic anions (Cl - , Br - , J - ), oxygen-containing anions (NO 3 - , SO 4 2- , PO 4 3- and others), as well as hydroxyl ions of water. Of these, halide ions have the stronger reducing properties, with the exception ofF. ionsOhoccupy an intermediate position between them and polyatomic anions. Therefore, during the electrolysis of aqueous solutionsHCl, HBr, HJor their salts on the anode, halide ions are oxidized according to the scheme:

2 X - -2 eà X 2 0

During the electrolysis of aqueous solutions of sulfates, nitrates, phosphates, etc. the function of the reducing agent is performed by ions, while being oxidized according to the scheme:

4 HOH – 4 eà 2 H 2 O + O 2 + 4 H +

.

Tasks.

W a dacha 1. During the electrolysis of a solution of copper sulfate, 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 neither ionsC 2+ andS0 4 2 ".

CuS0 4 \u003d 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, electrolysis of water occurs at the anode:

Cu 2+ + 2e- \u003d Cu12

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

General electrolysis equation:

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

Add to both sides of the equation 2 sulfate ions each, which are formed during the dissociation of copper sulfate, 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, we calculate the amount of copper substance:

According to the reaction equation, when 2 mol of copper is released from the anode, 1 mol of oxygen is released. 0.75 mol of copper was released at the cathode, let x mol 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.

Calculate the volume of released oxygen:

V(O2) \u003d v (O2) "VM \u003d 0.375 mol" 22.4 l / mol \u003d 8.4 l

According to the reaction equation, when 2 mol of copper is released at the cathode, 2 mol of sulfuric acid is formed in the solution, which means that if 0.75 mol of copper is released at the cathode, then 0.75 mol of sulfuric acid is formed in the solution, v (H2SO4) = 0.75 mol . Calculate the molar mass of sulfuric acid:

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

Calculate the mass of sulfuric acid:

m (H2S04) \u003d v (H2S04> M (H2S04) \u003d \u003d 0.75 mol \u003d 98 g / mol \u003d 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

Task 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 is formed in solution? Find its mass.

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

2KS1 \u003d K + + Cl

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

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

2SG-2e "= C12|1

General electrolysis equation:

2CHl + 2H2O \u003d H2 + 2OH "+ C12 (short ionic equation) The solution also contains K + ions, which were formed during the dissociation of potassium chloride and do not participate in the reaction:

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

Let's rewrite the equation in molecular form:

2KS1 + 2H2O = H2 + C12 + 2KOH

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

The solution contained 111.75 g of potassium chloride.

Calculate the molar mass of potassium chloride:

M(KC1) = 39+35.5 = 74.5 g/mol

Calculate the amount of potassium chloride substance:

According to the reaction equation, electrolysis of 2 mol of potassium chloride releases 1 mol of chlorine. Let the electrolysis of 1.5 mol of potassium chloride release 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) \u003d 0.75 mol. According to the reaction equation, when 1 mol of chlorine is released at the anode, 1 mol 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 us calculate the volume of chlorine released at the anode:

V (C12) \u003d v (Cl2) -VM \u003d 0.75 mol \u003d 22.4 l / mol \u003d 16.8 l.

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

Y (H2) \u003d Y (C12) \u003d 16.8 l.

According to the reaction equation, during the electrolysis of 2 mol of potassium chloride, 2 mol of potassium hydroxide is formed, which means that during the electrolysis of 0.75 mol of potassium chloride, 0.75 mol of potassium hydroxide is formed. Calculate the molar mass of potassium hydroxide:

M (KOH) \u003d 39 + 16 + 1 - 56 g / mol.

Calculate the mass of potassium hydroxide:

m(KOH) \u003d v (KOH> M (KOH) \u003d 0.75 mol-56 g / mol \u003d 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 formed in the solution.

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

We 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, water molecules are reduced:

2H20 + 2e- = H2 + 2OH|1

2Cl -2e "= C12! 1

General electrolysis equation:

2Cl + 2H2O \u003d H2 + 2OH "+ C12 (short ionic equation)

The solution also contains M2+ ions, which do not change during the reaction. We write the full ionic reaction equation:

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

Let's rewrite the reaction equation in molecular form:

MS12 + 2H2O - H2 + M(OH)2 + C12

Find the amount of chlorine released at the anode:

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

The molar mass of chloride of an unknown metal 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 that 0.4 mol of hydrogen was released at the cathode. Calculate the volume of hydrogen:

V (H2) \u003d v (H2> VM \u003d 0.4 mol \u003d 22.4 l / mol \u003d 8.96 l.

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

*Problem 4. During the electrolysis of 200 g of a solution of potassium sulfate 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 the electrolysis.

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

2H20 + 2e "= H2 + 20H-|2

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

Let's put both equations together:

6H2O \u003d 2H2 + 4OH "+ 4H + + O2, or

6H2O \u003d 2H2 + 4H2O + O2, or

2H2O = 2H2 + 02

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

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

C=m(solute) 100% / m(solution)

To find the concentration of the potassium sulfate solution at the end of the electrolysis, it is necessary to know the mass of potassium sulfate and the mass of the solution. The mass of potassium sulfate does not change during the reaction. Calculate the mass of potassium sulfate in the initial 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. Calculate the amount of released oxygen:

(O 2) \u003d V (O2) / Vm \u003d 14.56 l / 22.4 l / mol \u003d 0.65 mol

According to the reaction equation, 1 mole of oxygen is formed from 2 moles of water. 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.

Calculate the molar mass of water:

M(H2O) \u003d 1-2 + 16 \u003d 18 g / mol.

Calculate the mass of decomposed water:

m(H2O) \u003d v (H2O> M (H2O) \u003d 1.3 mol * 18 g / mol \u003d 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 the electrolysis:

С2 (K2 SO4)=m(K2 SO4) 100% / m(solution)=30g 100% / 176.6g=17%

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

* Problem 5. 188.3 g of a mixture of sodium and potassium chlorides were 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 in percent 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. Chloride ions are oxidized at the anode and chlorine is released:

Let's rewrite the equations in molecular form:

2KS1 + 2H20 = H2 + C12 + 2KOH

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

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

Let's make the equation: 0.5x + 0.5y \u003d 1.5

Calculate the molar masses of potassium and sodium chlorides:

M(KC1) = 39+35.5 = 74.5 g/mol

M (NaCl) \u003d 23 + 35.5 \u003d 58.5 g / mol

Mass x mole of potassium chloride is:

m (KCl) \u003d v (KCl) -M (KCl) \u003d x mol-74.5 g / mol \u003d 74.5 x g.

The mass of a mole of sodium chloride is:

m (KCl) \u003d v (KCl) -M (KCl) \u003d y mol-74.5 g / mol \u003d 58.5 u g.

The mass of the mixture is 188.3 g, we make 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 \u003d 1.5 / 0.5 \u003d 3,

x = 3-y

Substituting this value of x into the second equation, we get:

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

223.5-74.5y + 58.5y = 188.3

-16y = -35.2

y \u003d 2.2 100% / 188.3g \u003d 31.65%

Compute mass fraction sodium chloride:

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

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

Solving chemical problems
aware of Faraday's law
high school

Author's development

Among the great variety of various chemical problems, as the practice of teaching at school shows, the greatest difficulties are caused by problems for the solution of which, in addition to solid chemical knowledge, it is required to have a good command of the material of the physics course. And although far from every secondary school pays attention to solving at least the simplest problems using the knowledge of two courses - chemistry and physics, problems of this type are sometimes found at entrance exams in universities where chemistry is a major discipline. And therefore, without examining tasks of this type in the classroom, a teacher can unintentionally deprive his student of the chance to enter a university in a chemical specialty.
This author's development contains over twenty tasks, one way or another related to the topic "Electrolysis". To solve problems of this type, it is necessary not only to have a good knowledge of the topic "Electrolysis" school course chemistry, but also to know Faraday's law, which is studied in the school physics course.
Perhaps this selection of tasks will not be of interest to absolutely all students in the class or is available to everyone. Nevertheless, tasks of this type are recommended to be analyzed with a group of interested students in a circle or optional class. It can be noted with confidence that tasks of this type are complicated and at least not typical for a school chemistry course (we are talking about an average general education school), and therefore problems of this type can be safely included in the variants of a school or district Chemistry Olympiad for 10th or 11th grades.
Having a detailed solution for each problem makes development a valuable tool, especially for beginning teachers. Having analyzed several tasks with students in an optional lesson or a circle lesson, a creatively working teacher will certainly set several tasks of the same type at home and use this development in the process of checking homework, which will significantly save valuable teacher time.

Theoretical information on the problem

chemical reactions, flowing under the action of an electric current on electrodes placed in a solution or melt of an electrolyte, is called electrolysis. Consider an example.

In a glass at a temperature of about 700 ° C there is a melt of sodium chloride NaCl, electrodes are immersed in it. Before passing an electric current through the melt, Na + and Cl - ions move randomly, however, when an electric current is applied, the movement of these particles becomes ordered: Na + ions rush to the negatively charged electrode, and Cl - ions - to the positively charged electrode.

And he A charged atom or group of atoms that has a charge.

Cation is a positively charged ion.

Anion is a negatively charged ion.

Cathode- a negatively charged electrode (positively charged ions - cations) move towards it.

Anode- a positively charged electrode (negatively charged ions - anions) move towards it.

Electrolysis of sodium chloride melt on platinum electrodes

Total reaction:

Electrolysis of an aqueous solution of sodium chloride on carbon electrodes

Total reaction:

or in molecular form:

Electrolysis of an aqueous solution of copper(II) chloride on carbon electrodes

Total reaction:

AT electrochemical series activity of metals, copper is located to the right of hydrogen, so copper will be reduced at the cathode, and chlorine will be oxidized at the anode.

Electrolysis of an aqueous solution of sodium sulfate on platinum electrodes

Total reaction:

Similarly, the electrolysis of an aqueous solution of potassium nitrate occurs (platinum electrodes).

Electrolysis of an aqueous solution of zinc sulfate on graphite electrodes

Total reaction:

Electrolysis of an aqueous solution of iron(III) nitrate on platinum electrodes

Total reaction:

Electrolysis of an aqueous solution of silver nitrate on platinum electrodes

Total reaction:

Electrolysis of an aqueous solution of aluminum sulfate on platinum electrodes

Total reaction:

Electrolysis of an aqueous solution of copper sulfate on copper electrodes - electrochemical refining

The concentration of CuSO 4 in the solution remains constant, the process is reduced to the transfer of the anode material to the cathode. This is the essence of the process of electrochemical refining (obtaining pure metal).

When drawing up schemes for the electrolysis of a particular salt, it must be remembered that:

– metal cations having a higher standard electrode potential (SEP) than hydrogen (from copper to gold inclusive) are almost completely reduced at the cathode during electrolysis;

– metal cations with small SEP values ​​(from lithium to aluminum inclusive) are not reduced at the cathode, but instead water molecules are reduced to hydrogen;

– metal cations, whose SEC values ​​are less than those of hydrogen, but greater than those of aluminum (from aluminum to hydrogen), are reduced simultaneously with water during electrolysis at the cathode;

- if the aqueous solution contains a mixture of cations of various metals, for example, Ag +, Cu 2+, Fe 2+, then silver will be the first to be reduced in this mixture, then copper, and the last iron;

- on an insoluble anode during electrolysis, anions or water molecules are oxidized, and anions S 2–, I –, Br – , Cl – are easily oxidized;

– if the solution contains anions of oxygen-containing acids , , , , then water molecules are oxidized to oxygen at the anode;

- if the anode is soluble, then during electrolysis it itself undergoes oxidation, i.e. it sends electrons to the external circuit: when electrons are released, the balance between the electrode and the solution is shifted and the anode dissolves.

If from the whole series of electrode processes we single out only those that correspond to the general equation

M z+ + ze=M,

then we get metal stress range. Hydrogen is also always placed in this row, which makes it possible to see which metals are able to displace hydrogen from aqueous solutions of acids, and which are not (table).

Table

A range of stress metals

The equation electrode process	Standard electrode potential at 25 °С, V	The equation electrode process	Standard electrode potential at 25 °C, V
Li + + 1 e= Li0	–3,045	Co2+ + 2 e= Co0	–0,277
Rb + + 1 e= Rb0	–2,925	Ni 2+ + 2 e= Ni0	–0,250
K++1 e= K0	–2,925	Sn 2+ + 2 e= Sn0	–0,136
Cs + + 1 e= Cs 0	–2,923	Pb 2+ + 2 e= Pb 0	–0,126
Ca 2+ + 2 e= Ca0	–2,866	Fe 3+ + 3 e= Fe0	–0,036
Na + + 1 e= Na 0	–2,714	2H++2 e=H2	0
Mg 2+ + 2 e=Mg0	–2,363	Bi 3+ + 3 e= Bi0	0,215
Al 3+ + 3 e=Al0	–1,662	Cu 2+ + 2 e= Cu 0	0,337
Ti 2+ + 2 e= Ti0	–1,628	Cu + +1 e= Cu 0	0,521
Mn 2+ + 2 e=Mn0	–1,180	Hg 2 2+ + 2 e= 2Hg0	0,788
Cr 2+ + 2 e=Cr0	–0,913	Ag + + 1 e= Ag0	0,799
Zn 2+ + 2 e= Zn0	–0,763	Hg 2+ + 2 e= Hg0	0,854
Cr 3+ + 3 e=Cr0	–0,744	Pt 2+ + 2 e= Pt0	1,2
Fe 2+ + 2 e= Fe0	–0,440	Au 3+ + 3 e= Au 0	1,498
CD 2+ + 2 e= CD 0	–0,403	Au++1 e= Au 0	1,691

In a simpler form, a series of metal stresses can be represented as follows:

To solve most electrolysis problems, knowledge of Faraday's law is required, the formula expression of which is given below:

m = M I t/(z F),

where m is the mass of the substance released on the electrode, F- Faraday number, equal to 96 485 A s / mol, or 26.8 A h / mol, M is the molar mass of the element that is reduced during electrolysis, t– the time of the electrolysis process (in seconds), I- current strength (in amperes), z is the number of electrons involved in the process.

Task Conditions

1. What mass of nickel will be released during the electrolysis of a nickel nitrate solution for 1 hour at a current of 20 A?

2. At what current strength is it necessary to carry out the process of electrolysis of a solution of silver nitrate in order to obtain 0.005 kg of pure metal within 10 hours?

3. What mass of copper will be released during the electrolysis of a copper (II) chloride melt for 2 hours at a current of 50 A?

4. How long does it take to electrolyze an aqueous solution of zinc sulfate at a current of 120 A in order to obtain 3.5 g of zinc?

5. What mass of iron will be released during the electrolysis of an iron(III) sulfate solution at a current of 200 A for 2 hours?

6. At what current strength is it necessary to carry out the process of electrolysis of a solution of copper (II) nitrate in order to obtain 200 g of pure metal within 15 hours?

7. During what time is it necessary to carry out the process of electrolysis of a melt of iron (II) chloride at a current of 30 A in order to obtain 20 g of pure iron?

8. At what current strength is it necessary to carry out the process of electrolysis of a solution of mercury (II) nitrate in order to obtain 0.5 kg of pure metal within 1.5 hours?

9. At what current strength is it necessary to carry out the process of electrolysis of a sodium chloride melt in order to obtain 100 g of pure metal within 1.5 hours?

10. The potassium chloride melt was subjected to electrolysis for 2 hours at a current of 5 A. The resulting metal reacted with water weighing 2 kg. What concentration of alkali solution was obtained in this case?

11. How many grams of a 30% hydrochloric acid solution will be required for complete interaction with iron obtained by electrolysis of an iron (III) sulfate solution for 0.5 h at current strength
10 A?

12. In the process of electrolysis of a melt of aluminum chloride, carried out for 245 min at a current of 15 A, pure aluminum was obtained. How many grams of iron can be obtained by the aluminothermic method when a given mass of aluminum interacts with iron(III) oxide?

13. How many milliliters of a 12% solution of KOH with a density of 1.111 g / ml will be required to react with aluminum (with the formation of potassium tetrahydroxyaluminate) obtained by electrolysis of an aluminum sulfate solution for 300 minutes at a current of 25 A?

14. How many milliliters of a 20% sulfuric acid solution with a density of 1.139 g / ml will be required to interact with zinc obtained by electrolysis of a zinc sulfate solution for 100 minutes at a current of 55 A?

15. What volume of nitric oxide (IV) (n.o.) will be obtained when an excess of hot concentrated nitric acid reacts with chromium obtained by electrolysis of a solution of chromium (III) sulfate for 100 minutes at a current of 75 A?

16. What volume of nitric oxide (II) (n.o.) will be obtained when an excess of nitric acid solution reacts with copper obtained by electrolysis of a copper(II) chloride melt for 50 minutes at a current strength of 10.5 A?

17. During what time is it necessary to carry out the electrolysis of a melt of iron (II) chloride at a current of 30 A in order to obtain the iron necessary for complete interaction with 100 g of a 30% hydrochloric acid solution?

18. How long does it take to electrolyze a nickel nitrate solution at a current of 15 A in order to obtain the nickel necessary for complete interaction with 200 g of a 35% sulfuric acid solution when heated?

19. The sodium chloride melt was electrolyzed at a current of 20 A for 30 minutes, and the potassium chloride melt was electrolyzed for 80 minutes at a current of 18 A. Both metals were dissolved in 1 kg of water. Find the concentration of alkalis in the resulting solution.

20. Magnesium obtained by electrolysis of a magnesium chloride melt for 200 min at current strength
10 A, dissolved in 1.5 l of a 25% sulfuric acid solution with a density of 1.178 g / ml. Find the concentration of magnesium sulfate in the resulting solution.

21. Zinc obtained by electrolysis of a solution of zinc sulfate for 100 min at current strength

17 A, was dissolved in 1 l of a 10% sulfuric acid solution with a density of 1.066 g/ml. Find the concentration of zinc sulfate in the resulting solution.

22. Iron obtained by electrolysis of a melt of iron(III) chloride for 70 min at a current of 11 A was powdered and immersed in 300 g of an 18% copper(II) sulfate solution. Find the mass of copper precipitated.

23. Magnesium obtained by electrolysis of a magnesium chloride melt for 90 minutes at current strength
17 A, were immersed in an excess of hydrochloric acid. Find the volume and amount of hydrogen released (n.o.s.).

24. A solution of aluminum sulfate was subjected to electrolysis for 1 hour at a current of 20 A. How many grams of a 15% hydrochloric acid solution would be required for complete interaction with the resulting aluminum?

25. How many liters of oxygen and air (N.O.) will be required for the complete combustion of magnesium obtained by electrolysis of a magnesium chloride melt for 35 minutes at a current of 22 A?

See the following numbers for answers and solutions

Which flows under the action of an electric current on electrodes immersed in a solution or molten electrolyte.

There are two types of electrodes.

Anode oxidation.

Cathode is the electrode at which recovery. Anions tend to the anode because it has a positive charge. Cations tend to the cathode, because it is negatively charged and, according to the laws of physics, opposite charges attract. In any electrochemical process, both electrodes are present. The device in which electrolysis is carried out is called an electrolyzer. Rice. one.

The quantitative characteristics of electrolysis are expressed by two Faraday's laws:

1) The mass of the substance released on the electrode is directly proportional to the amount of electricity that has passed through the electrolyte.

2) During the electrolysis of various chemical compounds, the same amounts of electricity emit masses of substances on the electrodes, proportional to their electrochemical equivalents.

These two laws can be combined in one equation:

where m is the mass of the released substance, g;

n is the number of electrons transferred in the electrode process;

F is the Faraday number ( F=96485 C/mol)

I– current strength, A;

t– time, s;

M is the molar mass of the released substance, g/mol.

With electrolysis aqueous solutions electrode processes are complicated due to the competition of ions (water molecules can also participate in electrolysis). Recovery at the cathode is due to the position of the metal in a series of standard electrode potentials.

Metal cations, which have a standard electrode potential greater than that of hydrogen (from Cu2+ to Au3+), are almost completely reduced at the cathode during electrolysis. Me n+ + nē →Me Metal cations with a low standard electrode potential (Li2+ up to Al3+ inclusive) are not reduced at the cathode, but water molecules are reduced instead. 2H2O + 2ē → H2 + 2OH- Metal cations that have a standard electrode potential less than that of hydrogen, but greater than that of aluminum (from Mn2+ to H), are reduced simultaneously with water molecules during electrolysis at the cathode. Me n+ + nē → Me 2H2O + 2ē → H2 + 2OH- In the presence of several cations in the solution, the cations of the least active metal are reduced first of all on the cathode.

Example sodium sulfate (Na2SO4)

Na2SO4↔ 2Na++ SO42-

cathode: 2H2O + 2e → H2 + 2OH-

anode: 2H2O - 4e → O2 + 4H+

4OH-- 4H+→ 4H2O

by electrolysis melts many reactive metals are obtained. During the dissociation of the sodium sulfate melt, sodium ions and sulfate ions are formed.

Na2SO4 → 2Na+ + SO42−

- sodium is released at the cathode:

Na+ + 1e− → Na

– oxygen and sulfur oxide (VI) are released at the anode:

2SO42− − 4 e− → 2SO3 + О2

- the total ionic equation of the reaction (the equation of the cathode process was multiplied by 4)

4 Na+ + 2SO42− → 4 Na 0 + 2SO3 + O2

- total reaction:

4 Na2SO44 Na 0 + 2SO3 + O2

Electrolysis of molten salts

To obtain highly active metals (sodium, aluminum, magnesium, calcium, etc.), which easily interact with water, electrolysis of molten salts or oxides is used:

1. Electrolysis of copper (II) chloride melt.

Electrode processes can be expressed as half-reactions:

at the cathode K(-): Сu 2+ + 2e = Cu 0 - cathodic reduction

at the anode A (+): 2Cl - - 2e \u003d Cl 2 - anodic oxidation

The overall reaction of the electrochemical decomposition of a substance is the sum of two electrode half-reactions, and for copper chloride it is expressed by the equation:

Cu 2+ + 2 Cl - \u003d Cu + Cl 2

During the electrolysis of alkalis and salts of oxo acids, oxygen is released at the anode:

4OH - - 4e \u003d 2H 2 O + O 2

2SO 4 2– - 4e \u003d 2SO 3 + O 2

2. Potassium chloride melt electrolysis:

Solution electrolysis

The combination of redox reactions that occur on electrodes in electrolyte solutions or melts when an electric current is passed through them is called electrolysis.

On the cathode "-" of the current source, the process of transferring electrons to cations from a solution or melt occurs, therefore the cathode is a "reducing agent".

At the “+” anode, electrons are given off by anions, so the anode is an “oxidizing agent”.

During electrolysis, competing processes can occur both at the anode and at the cathode.

When electrolysis is carried out using an inert (non-consumable) anode (for example, graphite or platinum), as a rule, two oxidative and two reduction processes are competing:
at the anode - oxidation of anions and hydroxide ions,
at the cathode - reduction of cations and hydrogen ions.

When electrolysis is carried out using an active (consumable) anode, the process becomes more complicated and the competing reactions on the electrodes are:
at the anode - oxidation of anions and hydroxide ions, anodic dissolution of the metal - the material of the anode;
at the cathode - the reduction of the salt cation and hydrogen ions, the reduction of metal cations obtained by dissolving the anode.

When choosing the most probable process at the anode and cathode, one should proceed from the position that the reaction that requires the least energy consumption will proceed. In addition, to select the most probable process at the anode and cathode during the electrolysis of salt solutions with an inert electrode, the following rules are used:

1. The following products may form at the anode:

a) during the electrolysis of solutions containing anions SO 4 2-, NO - 3, PO 4 3-, as well as alkali solutions on the anode, water is oxidized and oxygen is released;

A + 2H 2 O - 4e - \u003d 4H + + O 2

b) during the oxidation of anions Cl - , Br - , I - chlorine, bromine, iodine are released, respectively;

A + Cl - + e - \u003d Cl 0

2. The following products can form on the cathode:

a) during the electrolysis of salt solutions containing ions located in a series of voltages to the left of Al 3+, water is reduced on the cathode and hydrogen is released;

K - 2H 2 O + 2e - \u003d H 2 + 2OH -

b) if the metal ion is located in the voltage series to the right of hydrogen, then metal is released at the cathode.

K - Me n + + ne - \u003d Me 0

c) during the electrolysis of salt solutions containing ions located in a series of voltages between Al + and H + , competing processes of both cation reduction and hydrogen evolution can occur at the cathode.

Example: Electrolysis of an aqueous solution of silver nitrate on inert electrodes

Dissociation of silver nitrate:

AgNO 3 \u003d Ag + + NO 3 -

During the electrolysis of an aqueous solution of AgNO 3, Ag + ions are reduced at the cathode, and water molecules are oxidized at the anode:

Cathode: Ag + + e = A g

Anode: 2H 2 O - 4e \u003d 4H + + O 2

Summary Equation:______________________________________________

4AgNO 3 + 2H 2 O \u003d 4Ag + 4HNO 3 + O 2

Make schemes for the electrolysis of aqueous solutions: a) copper sulfate; b) magnesium chloride; c) potassium sulfate.

In all cases, electrolysis is carried out using carbon electrodes.

Example: Electrolysis of an aqueous solution of copper chloride on inert electrodes

Dissociation of copper chloride:

CuCl 2 ↔ Сu 2+ + 2Cl -

The solution contains Cu 2+ and 2Cl - ions, which, under the action of an electric current, are directed to the corresponding electrodes:

Cathode - Cu 2+ + 2e = Cu 0

Anode + 2Cl - - 2e = Cl 2

_______________________________

CuCl 2 \u003d Cu + Cl 2

Metallic copper is released at the cathode, and chlorine gas is released at the anode.

If, in the considered example of the electrolysis of a CuCl 2 solution, a copper plate is taken as the anode, then copper is released at the cathode, and at the anode, where oxidation processes occur, instead of discharging Cl 0 ions and releasing chlorine, the anode (copper) is oxidized.

In this case, the anode itself dissolves, and in the form of Cu 2+ ions it goes into solution.

The electrolysis of CuCl 2 with a soluble anode can be written as follows:

The electrolysis of salt solutions with a soluble anode is reduced to the oxidation of the anode material (its dissolution) and is accompanied by the transfer of metal from the anode to the cathode. This property is widely used in the refining (purification) of metals from contamination.

Example: Electrolysis of an aqueous solution of magnesium chloride on inert electrodes

Dissociation of magnesium chloride in aqueous solution:

MgCl 2 ↔ Mg 2+ + 2Cl -

Magnesium ions cannot be reduced in an aqueous solution (water is being reduced), chloride ions are oxidized.

Electrolysis scheme:

Example: Electrolysis of an aqueous solution of copper sulfate on inert electrodes

In solution, copper sulfate dissociates into ions:

CuSO 4 \u003d Cu 2+ + SO 4 2-

Copper ions can be reduced at the cathode in an aqueous solution.

Sulfate ions in an aqueous solution are not oxidized, so water will be oxidized at the anode.

Electrolysis scheme:

Electrolysis of an aqueous solution of an active metal salt and an oxygen-containing acid (K 2 SO 4) on inert electrodes

Example: Dissociation of potassium sulfate in aqueous solution:

K 2 SO 4 \u003d 2K + + SO 4 2-

Potassium ions and sulfate ions cannot be discharged at the electrodes in an aqueous solution, therefore, reduction will occur at the cathode, and water will be oxidized at the anode.

Electrolysis scheme:

or, given that 4H + + 4OH - \u003d 4H 2 O (carried out with stirring),

H 2 O 2H 2 + O 2

If an electric current is passed through an aqueous solution of an active metal salt and an oxygen-containing acid, then neither the metal cations nor the ions of the acid residue are discharged.

Hydrogen is released at the cathode, and oxygen is released at the anode, and electrolysis is reduced to the electrolytic decomposition of water.

Electrolysis of sodium hydroxide melt

The electrolysis of water is always carried out in the presence of an inert electrolyte (to increase the electrical conductivity of a very weak electrolyte - water):

Faraday's law

The dependence of the amount of a substance formed under the action of an electric current on time, current strength and the nature of the electrolyte can be established on the basis of the generalized Faraday's law:

where m is the mass of the substance formed during electrolysis (g);

E - equivalent mass of a substance (g / mol);

M is the molar mass of the substance (g/mol);

n is the number of given or received electrons;

I - current strength (A); t is the duration of the process (s);

F - Faraday's constant characterizing the amount of electricity required to release 1 equivalent mass of a substance (F = 96,500 C/mol = 26.8 Ah/mol).

Hydrolysis of inorganic compounds

The interaction of salt ions with water, leading to the formation of weak electrolyte molecules, is called salt hydrolysis.

If we consider a salt as a product of neutralization of a base with an acid, then salts can be divided into four groups, for each of which the hydrolysis will proceed in its own way.

1. A salt formed by a strong base and a strong acid KBr, NaCl, NaNO 3) will not undergo hydrolysis, since in this case a weak electrolyte is not formed. The reaction of the medium remains neutral.

2. In a salt formed by a weak base and a strong acid FeCl 2, NH 4 Cl, Al 2 (SO 4) 3, MgSO 4), the cation undergoes hydrolysis:

FeCl 2 + HOH → Fe(OH)Cl + HCl

Fe 2+ + 2Cl - + H + + OH - → FeOH + + 2Cl - + H +

As a result of hydrolysis, a weak electrolyte, H + ion and other ions are formed. solution pH< 7 (раствор приобретает кислую реакцию).

3. A salt formed by a strong base and a weak acid (KClO, K 2 SiO 3, Na 2 CO 3, CH 3 COONa) undergoes anion hydrolysis, resulting in the formation of a weak electrolyte, hydroxide ion and other ions.

K 2 SiO 3 + HOH → KHSiO 3 + KOH

2K + +SiO 3 2- + H + + OH - → HSiO 3 - + 2K + + OH -

The pH of such solutions is > 7 (the solution acquires an alkaline reaction).

4. A salt formed by a weak base and a weak acid (CH 3 COONH 4, (NH 4) 2 CO 3, Al 2 S 3) is hydrolyzed both by the cation and by the anion. As a result, low-dissociating base and acid are formed. The pH of solutions of such salts depends on the relative strength of the acid and base.

Algorithm for writing equations for the reactions of hydrolysis of a salt of a weak acid and a strong base

There are several options for the hydrolysis of salts:

1. Hydrolysis of a salt of a weak acid and a strong base: (CH 3 COONa, KCN, Na 2 CO 3).

Example 1 Hydrolysis of sodium acetate.

or CH 3 COO - + Na + + H 2 O ↔ CH 3 COOH + Na + + OH -

CH 3 COO - + H 2 O ↔ CH 3 COOH + OH -

Since acetic acid dissociates weakly, the acetate ion binds the H + ion, and the water dissociation equilibrium shifts to the right according to Le Chatelier's principle.

OH - ions accumulate in the solution (pH > 7)

If the salt is formed by a polybasic acid, then the hydrolysis proceeds in steps.

For example, carbonate hydrolysis: Na 2 CO 3

Stage I: CO 3 2– + H 2 O ↔ HCO 3 – + OH –

Stage II: HCO 3 - + H 2 O ↔ H 2 CO 3 + OH -

Na 2 CO 3 + H 2 O \u003d NaHCO 3 + NaOH

Of practical importance is usually only the process going through the first stage, which, as a rule, is limited when evaluating the hydrolysis of salts.

The equilibrium of hydrolysis in the second stage is significantly shifted to the left compared to the equilibrium of the first stage, since a weaker electrolyte (HCO 3 -) is formed in the first stage than in the second (H 2 CO 3)

Example 2 . Hydrolysis of rubidium orthophosphate.

1. Determine the type of hydrolysis:

Rb3PO4 ↔ 3Rb + + PO 4 3–

Rubidium is an alkali metal, its hydroxide is a strong base, phosphoric acid, especially in its third stage of dissociation, corresponding to the formation of phosphates, is a weak acid.

Anion hydrolysis occurs.

PO 3- 4 + H–OH ↔ HPO 2- 4 + OH – .

Products - hydrophosphate and hydroxide ions, medium - alkaline.

3. We compose a molecular equation:

Rb 3 PO 4 + H 2 O ↔ Rb 2 HPO 4 + RbOH.

We got an acid salt - rubidium hydrogen phosphate.

Algorithm for writing equations for the reactions of hydrolysis of a salt of a strong acid and a weak base

2. Hydrolysis of a salt of a strong acid and a weak base: NH 4 NO 3, AlCl 3, Fe 2 (SO 4) 3.

Example 1. Hydrolysis of ammonium nitrate.

NH 4 + + NO 3 - + H 2 O ↔ NH 4 OH + NO 3 - + H +

NH 4 + + H 2 O ↔ NH 4 OH + H +

In the case of a multiply charged cation, hydrolysis proceeds in steps, for example:

Stage I: Cu 2+ + HOH ↔ CuOH + + H +

Stage II: CuOH + + HOH ↔ Cu(OH) 2 + H +

CuCl 2 + H 2 O \u003d CuOHCl + HCl

In this case, the concentration of hydrogen ions and the pH of the medium in the solution are also determined mainly by the first stage of hydrolysis.

Example 2 Hydrolysis of Copper(II) Sulfate

1. Determine the type of hydrolysis. At this stage, it is necessary to write the salt dissociation equation:

CuSO4 ↔ Cu 2+ + SO2-4.

A salt is formed by a cation of a weak base (underlined) and an anion of a strong acid. Hydrolysis occurs at the cation.

2. We write the ionic hydrolysis equation, determine the environment:

Cu 2+ + H-OH ↔ CuOH + + H + .

A hydroxomeper(II) cation and a hydrogen ion are formed, the medium is acidic.

3. We make a molecular equation.

It should be taken into account that the compilation of such an equation is a certain formal task. From positive and negative particles in solution, we make up neutral particles that exist only on paper. In this case, we can make the formula (CuOH) 2 SO 4, but for this we must mentally multiply our ionic equation by two.

We get:

2CuSO 4 + 2H 2 O ↔ (CuOH) 2 SO 4 + H 2 SO 4.

Please note that the reaction product belongs to the group of basic salts. The names of the basic salts, as well as the names of the middle salts, should be composed of the names of the anion and cation, in this case we will call the salt “hydroxomeper (II) sulfate”.

Algorithm for writing equations for the reactions of hydrolysis of a salt of a weak acid and a weak base

3. Hydrolysis of a salt of a weak acid and a weak base:

Example 1 Hydrolysis of ammonium acetate.

CH 3 COO - + NH 4 + + H 2 O ↔ CH 3 COOH + NH 4 OH

In this case, two slightly dissociated compounds are formed, and the pH of the solution depends on the relative strength of the acid and base.

If the hydrolysis products can be removed from the solution, for example, in the form of a precipitate or a gaseous substance, then the hydrolysis proceeds to completion.

Example 2 Hydrolysis of aluminum sulfide.

Al 2 S 3 + 6H 2 O \u003d 2Al (OH) 3 + 3H 2 S

2A l 3+ + 3 S 2- + 6H 2 O \u003d 2Al (OH) 3 (precipitate) + ZN 2 S (gas)

Example 3 Hydrolysis of Aluminum Acetate

1. Determine the type of hydrolysis:

Al(CH 3 COO) 3 = Al 3+ + 3CH 3 COO – .

A salt is formed by a cation of a weak base and anions of a weak acid.

2. We write the ionic hydrolysis equations, determine the environment:

Al 3+ + H–OH ↔ AlOH 2+ + H + ,

CH 3 COO - + H-OH ↔ CH 3 COOH + OH - .

Considering that aluminum hydroxide is a very weak base, we assume that hydrolysis at the cation will proceed to a greater extent than at the anion. Therefore, there will be an excess of hydrogen ions in the solution, and the environment will be acidic.

Do not try to make here the total equation of the reaction. Both reactions are reversible, in no way connected with each other, and such a summation is meaningless.

3 . We compose the molecular equation:

Al (CH 3 COO) 3 + H 2 O \u003d AlOH (CH 3 COO) 2 + CH 3 COOH.

This is also a formal exercise to train in formulating salts and their nomenclature. The resulting salt will be called hydroxoaluminum acetate.

Algorithm for writing equations for the reactions of hydrolysis of a salt of a strong acid and a strong base

4. Salts formed by a strong acid and a strong base do not undergo hydrolysis, because the only low-dissociating compound is H 2 O.

The salt of a strong acid and a strong base does not undergo hydrolysis and the solution is neutral.