Ufa State Petroleum Technical University

Department of General and Analytical Chemistry

Educational and methodological manual

for laboratory work on the topic:

Hydrolysis of salts

Designed for non-chemistry students

and chemical faculties of universities.

Compiled by: Syrkin A.M., prof., candidate of chemical sciences, Rolnik L.Z., associate professor,

Doctor of Chemical Sciences

Reviewer Sergeeva L.G., Associate Professor, Candidate of Chemical Sciences.

© Ufa State

petroleum technical

university, 2002

Hydrolysis of salts

Hydrolysis of salts is the interaction of their ions with water, based on the polarizing effect of ions on water molecules, as a result of which, as a rule, the equality

Characteristic of clean water.

There are 4 groups of salts:

a salt formed by a strong base and a strong acid;

a salt formed by a weak base and a strong acid;

a salt formed by a strong base and a weak acid;

a salt formed by a weak base and a weak acid.

Therefore, we will consider 4 options for the effect of water on salt.

1) This group includes salts such as NaCI, KCI, NaNO 3, Na 2 SO 4, etc. The cations and anions of these salts have small charges and significant sizes. At the same time, their polarizing effect on water molecules is small, that is, the interaction of salt with water practically does not occur. This applies to cations such as K + and Na +, and to such anions as CI - and NO 3 -. Therefore, salts of a strong base and a strong acid do not undergo hydrolysis. In this case, the equilibrium of water dissociation in the presence of salt ions is almost not disturbed.

Therefore, solutions of such salts are practically neutral (pH ≈ 7).

2) If the salt is formed by a weak base cation NH 4 +, AI 3+, Mg 2+, etc. and a strong acid anion (Cl -, NO 3 -, SO 4 2-, etc.), then hydrolysis occurs according to cation(only the salt cation has a polarizing effect). An example is the process:

a) in molecular form

NH 4 CI + H 2 O NH 4 OH + HCI;

b) in ionic-molecular form

NH 4 + + CI - + H 2 O NH 4 OH + H + + CI - ;

c) in a short ion-molecular form

NH 4 + + H 2 O NH 4 OH + H + .

Hydrolysis is due to the formation of a slightly dissociated compound - NH 4 OH. As a result, the equilibrium of electrolytic dissociation of water shifts and an excess of hydrogen ions appears in the solution, so the reaction of the medium is acidic (pH< 7). Очевидно, чем полнее протекает гидролиз, тем более показатель среды отличается от состояния нейтральности.

Let us immediately note that the hydrolysis process can be quantitatively characterized by two quantities: 1) the degree of hydrolysis (h); 2) hydrolysis constant (K g).

Degreehydrolysis is called the ratio of the number of salt molecules that have undergone hydrolysis to the total number of salt molecules in solution; or the degree of hydrolysis is understood as a number indicating what part of the total amount of salt is hydrolyzed, that is, converted by the action of water into the corresponding acid or base (either into acidic or basic salts).

The degree of hydrolysis is calculated based on the equation of the dissociation constant of the corresponding weak base (or acid) and the ionic product of water.

Let's consider these characteristics for the hydrolysis of ammonium chloride salt.

Let us write the hydrolysis equation again in ionic-molecular form:

NH 4 + + H 2 O NH 4 OH + H +

According to the law of mass action, the equilibrium constant of this reaction will have the following form:

K p =
(1)

The concentration of water practically does not change in a salt solution, that is, 0 = equal = const (2)

= K p = K g (3)

The product of two constants K p is a constant quantity and is called constanthydrolysis To Mr.

From the equation of the ionic product of water we have

K H 2 O = (4)

=
(5)

Then equation (1) can be written as follows:

K g =
(6)

Attitude

=, (7)

where K main. – dissociation constant of the weak base NH 4 OH.

Then expression (6) has the form

K g = (8)

The higher the K g, the more the salt undergoes hydrolysis.

From equation (3) the degree of hydrolysis of the salt can be calculated.

K g = =
(9)

Let us assume that the concentration of the original salt is c mol/l, the degree of hydrolysis is h, then ch moles of salt are hydrolyzed, ch moles of NH 4 OH and ch g- ions H + are formed.

At equilibrium, the concentrations will have the following values:

= (c - ch)

Let's substitute these values ​​into equation (5).

, (10)

K g = (11)

Since h is an insignificant value (h ≤ 0.01), we can assume that (1 -h) ≈ 1

K g =
; (12)

h = =
. (13)

From the resulting equation it follows that the degree of hydrolysis (h) is the greater:

the more K H 2 O, that is, the higher the temperature (the ionic product of water K H 2 O depends on the temperature in direct proportion);

the less K base, that is, the weaker the base formed as a result of hydrolysis;

the lower the salt concentration, that is, the more diluted the solution.

Thus, to increase the degree of hydrolysis it is necessary to dilute the solution and raise the temperature. We considered the 2nd option of hydrolysis of a salt formed by a weak base and a strong acid. Copper (II) chloride also belongs to this type of salt. This salt is formed by a diacid base Cu(OH) 2 and a monobasic acid. In this case, the hydrolysis process occurs in stages. At room temperature, mainly 1 stage of hydrolysis is carried out. Let us write down the 1st stage of hydrolysis of copper (II) chloride salt in 3 forms:

in molecular form

CuCI 2 + H 2 O CuOHCI + HCI;

in ionic-molecular form

Cu 2+ + 2CI - + H 2 O (CuOH) + + CI - + H + + CI - ;

in short ion-molecular form

Cu 2+ + H 2 O (CuOH) + + H +

Hydrolysis is due to the formation of slightly dissociating particles (CuOH) +. As a result, the equilibrium of electrolytic dissociation of water shifts, an excess of hydrogen ions appears in the solution, the pH reaction of the environment< 7. Гидролиз протекает by cation.

The basic salt formed as a result of the first stage of hydrolysis can be subjected to further interaction with water. However, the second stage of hydrolysis is less pronounced. This is due to a decrease in Kbas. when moving from K main 1 to K main 2, etc. For example, since (CuOH) + ions dissociate more weakly than Cu(OH) 2, it is formed primarily during the hydrolysis of CuCI 2.

The second stage of hydrolysis of copper (II) chloride can be represented as follows:

in molecular form

CuOHCI + H 2 O Cu(OH) 2  + HCI;

(CuOH) + + CI - +H 2 O Cu(OH) 2  + H + + CI - ;

in short ion-molecular form

(CuOH) + + H 2 O Cu(OH) 2 + H + .

"Copper chemistry" - Enrichment. Norilsk enrichment plant. Nickel shot. Products. Plan. Device design for the electrolysis of aqueous salt solutions. Color. Copper nuggets. Copper rod (fig.) Copper cathode. Converter span. Norilsk is the largest center of copper-nickel production in the country. Ni Al Cu Mg Li.

“Metal copper” - The ingress of copper salts into the body leads to various human diseases. Density 8.92 g/cm3, melting point 1083.4 °C, boiling point 2567 °C. COPPER (lat. In total, the average person’s body (body weight 70 kg) contains 72 mg of copper. Due to its high thermal conductivity, copper is an irreplaceable material for various heat exchangers and refrigeration equipment.

“Salt hydrolysis” - Hydrolysis of salts. Control test. Hydrolysis of binary compounds. Blood contains: NaHCO3, Na2H2PO4. Strong bases (Alkalis) LiOH NaOH KOH RbOH CsOH Ca(OH)2 Sr(OH)2 Ba(OH)2. Changing the direction of hydrolysis. Suppression. Hydrolysis by anion (salt is formed by a strong base and a weak acid).

“Examples of salt hydrolysis” - Determine the pH of the solution. Make up ionic and molecular equations of the processes occurring. Degree of hydrolysis? g (proportion of hydrolyzed units) Hydrolysis constant - Kg. Example: Hydrolysis of a cation. M+ + n2o?moh + n+. In many cases it is necessary to prevent hydrolysis. The process of exchange decomposition of water by salt ions is called hydrolysis.

“Hydrolysis of salt solutions” - Problem B9: the mass of acetic acid contained in 0.5 liters of CH3COOH solution with a mass fraction of 80% (density 1.1 g/ml) is equal to ____________. 1) The reaction equation is written: H2SO4 + 2KOH ??? K2SO4 + 2H2O. Unified State Exam in CHEMISTRY (consultation 3). As a result of the reaction, the thermochemical equation of which is C + O2 = CO2 + 393.5 kJ, 1967.5 kJ of heat was released.

“Chemistry Hydrolysis” - The influence of hydrolysis on the process of geological, chemical and biological evolution of the planet. Connection of the topic with everyday life. Lesson content. Familiarization with types of knowledge control. Ways to motivate learning. Methods of activity of the teacher and students. Setting the goals and objectives of the lesson. Tomilova Natalya Vladimirovna.

The polarization interaction of cations and anions with highly polar water molecules leads to a special chemical ion exchange reaction called hydrolysis of salts .

It is convenient to consider the qualitative and quantitative aspects of hydrolysis from the perspective of the concept of strong and weak electrolytes (unassociated and associated). Almost all electrolytes classified as weak in aqueous solutions (see section 3.2) are characterized by the fact that the equilibrium of their dissociation is shifted to the left, towards undissociated particles. In other words, they are not characterized by dissociation, but, on the contrary, association, that is, the binding of protons by the corresponding anions and OH ions by cations into undissociated particles. And H + and OH – ions are always present in water due to its slight dissociation. Let us consider in more detail the processes taking place using the examples of two salts - CuCl 2 and Na 2 CO 3.

Copper (II) chloride is a strong electrolyte, therefore in an aqueous solution it completely dissociates into ions:

Copper (II) hydroxide is a weak electrolyte (see section 3.2), in other words, the Cu 2+ cation, in the presence of OH – ions in the solution, will actively bind them into a slightly dissociated CuOH + particle, thereby disturbing the equilibrium of water dissociation:

As a result, according to Le Chatelier's principle, the dissociation of water will increase and the concentration of hydrogen ions in the solution will increase compared to what was in the water. The solution becomes acidic, its pH<7, подобная ситуация называется hydrolysis at the cation .

Of course, the hydrolysis of copper chloride can go further, in the second stage:

However, taking into account that the hydrolysis products of the first stage suppress the second stage and that the polarization interaction of the Cu 2+ ion with water molecules is incomparably stronger than the CuOH + ion, we come to the following important conclusion. If there is the possibility of stepwise hydrolysis, this process actually occurs only in the first step.

A similar situation arises in a solution of Na 2 CO 3. As a result of the complete dissociation of this salt in the solution, CO 3 2– ions are formed, which are anions of weak carbonic acid. This ion, if there are protons in the solution, will actively bind them into a slightly dissociated HCO 3 – particle, thereby disturbing the equilibrium of water dissociation:

As a result, the dissociation of water will increase and the concentration of OH ions in the solution will increase compared to what was in the water. The solution has become alkaline, its pH > 7, in this case they say hydrolysis by anion .

To be fair, it should be noted that the actual mechanism of hydrolysis is somewhat different. Any ions in an aqueous solution are hydrated and polarization interaction occurs between the ion and the water molecules that make up its hydration shell, for example:

This clarification in no way changes the conclusions drawn above and does not affect further quantitative calculations.

Thus, either salts containing cations of weak bases (cation hydrolysis) or salts containing anions of weak acids (anion hydrolysis) undergo hydrolysis. If the cation and anion in the salt molecule are ions of the corresponding strong base
and a strong acid, then there is no hydrolysis in the solution of such a salt, its pH is 7.

If the salt contains a cation of a weak base and an anion of a weak acid, then hydrolysis in this case occurs in two directions and, as a rule, deeply. As for the acidity of such a solution, it will be determined by the direction of preferential hydrolysis.

Ways to enhance salt hydrolysis:

1) diluting the salt solution;

2) heating the solution, since the enthalpies of hydrolysis are positive;

3) adding alkali to the solution to enhance hydrolysis of the cation, adding acid to the solution to enhance hydrolysis of the anion.

Methods to suppress hydrolysis:

1) cooling the solution,

2) adding acid to the solution to suppress hydrolysis of the cation, adding alkali to the solution to suppress hydrolysis
by anion.

Let us consider the quantitative characteristics of hydrolysis. These are, first of all, the degree and constant of hydrolysis. The degree of hydrolysis ( h) similar to the degree of dissociation, the proportion of hydrolyzed molecules relative to the total number of molecules is called. The hydrolysis constant is the equilibrium constant of the hydrolysis process. It was shown above that hydrolysis occurs only in the first stage. The first stage of hydrolysis by cation can be written in general form:

K equal = K hydr = . (3.23)

We multiply the numerator and denominator of this expression by the concentration of the OH ion - and we get:

K hydr = = (3.24)

Thus, the hydrolysis constant for a cation is equal to the ratio of the ionic product of water to the dissociation constant of the weakest base whose salt is hydrolyzed, or to the dissociation constant of the base at the corresponding stage.

Let's return to relation (3.23). Let the total concentration of the hydrolyzing salt in the solution be equal to With mol/l, and the degree of its hydrolysis is h. Then, given that = and h= /With, we obtain from relation (3.23):

K hydr = . (3.25)

Relationship (3.25) coincides in form with the expression of the Ostwald dilution law (3.8), which once again reminds us of the genetic connection between the processes of hydrolysis and dissociation.

The first stage of hydrolysis at the anion can be written in general form

in the following way:

The equilibrium constant of this process, the hydrolysis constant, is equal to:

K equal = K hydr = . (3.26)

We multiply the numerator and denominator of this expression by the concentration of the H + ion and get:

To hydr = = . (3.27)

Thus, the hydrolysis constant for an anion is equal to the ratio of the ionic product of water to the dissociation constant of the weak acid whose salt is hydrolyzed, or to the dissociation constant of the acid at the corresponding stage. Let us turn again to expression (3.26). Let's transform it, assuming that the total concentration of salt in the solution is equal to With mol/l and, given that = ; h = / c, we get:

K hydr = . (3.28)

Expressions (3.23), (3.24) and (3.27), (3.28) are sufficient to find the equilibrium concentrations of ions, constants and degrees of hydrolysis in aqueous solutions of hydrolyzing salts.

It is not difficult to guess that the hydrolysis constant of a salt undergoing cation and anion hydrolysis at the same time is equal to the ratio of the ionic product of water to the product of the dissociation constants of a weak base and an acid or the product of the dissociation constants of the corresponding stages. Indeed, the hydrolysis of a salt by cation and anion simultaneously can be represented in general form as follows:

The hydrolysis constant has the form:

K hydr = . (3.29)

We multiply the numerator and denominator of relation (3.29) by K W and get:

K hydr = . (3.30)

Let the total concentration of salt hydrolyzed simultaneously into the cation and anion be equal to c mol/l, the degree of hydrolysis is h. It is obvious that ==hc; ==c–hc. We substitute these relations into expression (3.29):

K hydr = . (3.31)

An interesting result was obtained - the concentration is not explicitly included in the expression of the hydrolysis constant, in other words, the degree of hydrolysis of a salt that undergoes cation and anion hydrolysis at the same time will be the same for any salt concentration in the solution.

Let us find an expression for the pH of the salt solution under consideration. To do this, multiply the numerator and denominator of relation (3.29) by the concentration of the H + ion and transform the resulting expression:

K hydr = 3.32)

Finally we get:

K diss.k-you × . (3.33)

Let us now dwell on the connection between the characteristics of hydrolysis and dissociation in the case of stepwise hydrolysis. As an example, consider the hydrolysis of the already mentioned sodium carbonate. The equilibrium of hydrolysis of Na 2 CO 3 in steps and the corresponding equilibrium constants are given below:

K hydr (1) = = = = ;

K hydr (2) = = = .

Thus, the first stage of hydrolysis corresponds to the last stage of dissociation of the corresponding weak electrolyte, and vice versa - the last stage of hydrolysis corresponds to the first stage of dissociation of the electrolyte. When analyzing the issue of hydrolysis of acid salts, it is necessary to compare the values ​​of the hydrolysis constants and the dissociation constants of anions. If the hydrolysis constant is greater than the dissociation constant of the acidic anion, then hydrolysis of the anion takes place and the solution is characterized by pH > 7. If the hydrolysis constant is less than the dissociation constant of the corresponding acidic anion, then hydrolysis is suppressed, only dissociation of the acidic anion actually occurs and the salt solution has a pH< 7.

It was noted above that the simplest way to enhance the hydrolysis of a salt into a cation is to introduce an alkali into such a solution. Similarly, to enhance the hydrolysis of the salt at the anion, it is necessary to introduce an acid into the solution. What happens when you merge solutions of two salts, one of which is hydrolyzed by the cation and the other by the anion, for example, solutions of Na 2 CO 3 and CuCl 2? Hydrolysis equilibria in these solutions:

As can be seen, hydrolysis of the first salt will enhance the hydrolysis of the second and vice versa. In this case, they speak of mutual enhancement of hydrolysis. It is clear that in such a situation the formation of an exchange reaction product is impossible; hydrolysis products must be formed. Their composition depends on a large number of factors: the concentrations of the solutions being drained, the order of mixing, the degree of mixing, etc.

In the system under consideration (and similar ones) basic carbonates are formed; to some approximation, their composition can be considered as ECO 3 ×E(OH) 2 = (EOH) 2 CO 3 .

Equation of the ongoing process:

2CuCl 2 + 2 Na 2 CO 3 +H 2 O = (CuOH) 2 CO 3 ¯ + CO 2 + 4 NaCl.

Similar poorly soluble compounds will be obtained by the interaction of soluble carbonates with salts of any divalent metals, hydrolyzed into the cation. If the salts are not hydrolyzed, then the usual metabolic process occurs, for example:

BaCl 2 + Na 2 CO 3 = BaCO 3 ¯ + 2 NaCl.

In general, Me 3+ salts are more hydrolyzed than Me 2+ salts, therefore, if in the process under discussion CuCl 2 is replaced by Me 3+ salt, then a stronger mutual enhancement of hydrolysis should be expected. Indeed, when merging solutions of Fe 3+, Al 3+, Cr 3+ salts with a solution of Na 2 CO 3, the release of carbon dioxide and the precipitation of metal hydroxide are observed. In other words, in this case the mutual enhancement of hydrolysis leads to complete (irreversible) hydrolysis, for example:

2FeCl 3 + 3Na 2 CO 3 + 3H 2 O = 2Fe(OH) 3 ¯ + 6NaCl + 3CO 2.

Similar processes will be observed when mixing solutions of Me 3+ salts with solutions of other salts hydrolyzed by the anion, for example:

2AlCl 3 + 3Na 2 SO 4 + 3H 2 O = 2Al(OH) 3 ¯ + 3SO 2 + 6NaCl

Cr 2 (SO 4) 3 + 3Na 2 S + 6H 2 O = 2Cr(OH) 3 ¯ + 3H 2 S + 3Na 2 SO 4.

Unlike salts, the hydrolysis of acid derivatives - acid halides, thioanhydrides - proceeds deeply and, often, completely (irreversibly), for example:

SO 2 Cl 2 + 2H 2 O = H 2 SO 4 + 2HCl;

SOCl 2 + H 2 O = SO 2 + 2HCl;

COCl 2 + H 2 O = CO 2 + 2HCl;

BCl 3 + 3H 2 O = H 3 BO 3 + 3HCl;

PCl 3 + 3H 2 O = H 3 PO 3 + 3HCl;

CrO 2 Cl 2 + 2H 2 O = H 2 CrO 4 + 2HCl;

CS 2 + 2H 2 O = CO 2 + 2H 2 S.

Finally, we note the special case of hydrolysis of Bi(III), Sb(III) compounds, salts of d-elements - with the formation of oxo compounds, for example:

SbCl 3 + H 2 O = SbOCl + 2HCl;

Bi(NO 3) 3 + H 2 O = BiONO 3 + 2HNO 3;

Ti(SO 4) 2 + H 2 O = TiOSO 4 + H 2 SO 4.

The hydrolysis constant, like any other equilibrium constant, can be calculated based on thermodynamic data.

TICKET No. 23

1. Which of the following salts undergoes hydrolysis:CuCl 2 , Na 2 SO 4 , Sa(NOz) 2 ? Calculate the pH of the solution if the concentration of this salt is 0.5 mol/l and the dissociation constant of the base isKb2= 2.19 * .

Solution:

Sodium sulfate Na 2 SO 4 and calcium nitrate Ca (NO3) 2 are salts of strong bases and strong acids, therefore they are not subject to hydrolysis.

Copper chloride will be hydrolyzed (2)CuCl 2 – salt of a weak base (Cu(OH) 2) and a strong acid (HCl). Hydrolysis proceeds by cation, mainly at stage I. The environment is acidic.

Dissociation of copper chloride (2):

CuCl 2 = Cu 2+ + 2Cl -

Hydrolysis of copper chloride in the first stage:

Cu 2+ + H 2 O ↔ CuOH + + H +

CuCl 2 + H 2 O ↔ Cu(OH)Cl + HCl

= √(K g *s)

For salts hydrolyzing at the cation, the hydrolysis constant K g is equal to:

K g = K w /K b, where K w = 10 -14 is the ionic product of water, K b is the dissociation constant of the base.

Because hydrolysis of copper chloride (2) occurs predominantly in the first stage, then for the calculation we use the hydrolysis constant in the first stage, which is equal to: K g (1) = K w / K b (2)

So, the pH for this solution is:

pH =- log = - log√(K g(1) *s) = - log√(K w *s/K b (2)) = - log√(10 -14 *0.5/2.19*10 - 7) = 3,82

2. Determine the pH of a 0.1 M solution of hydrofluoric acid (HF), the dissociation constant of which is K a = 6.67* .

Solution:

Hydrofluoric acid is a weak electrolyte. For weak acids, the concentration of hydrogen ions in solution is calculated using the formula:

[ H + ] = √( K A * c M ) = √(6.67*10 -4 * 0.1) = 8.17*10 -3 (mol/l)

pH = -lg= - log 8.17*10 -3 = 2,09

3. In which direction will the equilibrium of the reaction 2CO + shift?O 2 <=>2CO 2 a) with increasing temperature (∆Н<0); б) при увеличении общего давления в системе?

Solution:

According to Le Chatelier's principle, if any external influence is exerted on a system that is in equilibrium, then it favors the occurrence of whichever of the two opposite reactions decreases this influence.

A) As the temperature rises, the equilibrium shifts towards an endothermic reaction that occurs with heat absorption, i.e. left: direct reaction is exothermic, reverse reaction is endothermic;

B) When the total pressure in the system increases, the equilibrium shifts in the direction of decreasing volume, i.e. right(since pV = const)

Solution:

Dissociation of calcium carbonate:

CaCO 3 ↔ Ca 2+ + CO 3 2-

According to the salt dissociation equation,

ETC ( CaCO 3 ) = * = 2

= √PR = √(4.4*10 -9) = 6.63*10 -5 (mol/l)

Let's convert the concentration of calcium ions to g/l:

C = M*M V = 6.63*10 -5 mol/l * 40 g/mol = 2.652*10 -3 g/l

(C is the concentration expressed in grams of dissolved substance per liter of solution, M is the molar concentration of the solution, M B is the molar mass of the calcium cation)

General information about the hydrolysis of copper (II) chloride

DEFINITION

Copper(II) chloride– a medium salt formed by a weak base – copper (II) hydroxide (Cu(OH) 2) and a strong acid – hydrochloric (hydrochloric) (HCl). Formula - CuCl 2.

Represents crystals of yellow-brown (dark brown) color; in the form of crystalline hydrates - green. Molar mass – 134 g/mol.

Rice. 1. Copper(II) chloride. Appearance.

Hydrolysis of copper(II) chloride

Hydrolyzes at the cation. The nature of the environment is acidic. Theoretically, a second stage is possible. The hydrolysis equation is as follows:

First stage:

CuCl 2 ↔ Cu 2+ + 2Cl - (salt dissociation);

Cu 2+ + HOH ↔ CuOH + + H + (hydrolysis by cation);

Cu 2+ + 2Cl - + HOH ↔ CuOH + + 2Cl - + H + (ionic equation);

CuCl 2 + H 2 O ↔ Cu(OH)Cl +HCl (molecular equation).

Second stage:

Cu(OH)Cl ↔ CuOH + + Cl - (salt dissociation);

CuOH + + HOH ↔ Cu(OH) 2 ↓ + H + (hydrolysis by cation);

CuOH + + Cl - + HOH ↔ Cu(OH) 2 ↓ + Cl - + H + (ionic equation);

Cu(OH)Cl + H 2 O ↔ Cu(OH) 2 ↓ + HCl (molecular equation).

Examples of problem solving

EXAMPLE 1

EXAMPLE 2

Exercise	Write down the equation for the electrolysis of a solution of copper (II) chloride. What mass of substance will be released at the cathode if 5 g of copper (II) chloride is subjected to electrolysis?
Solution	Let us write the dissociation equation for copper (II) chloride in an aqueous solution: CuCl 2 ↔ Cu 2+ +2Cl - . Let us conventionally write down the electrolysis scheme: (-) Cathode: Cu 2+, H 2 O. (+) Anode: Cl - , H 2 O. Cu 2+ +2e → Cu o ; 2Cl - -2e → Cl 2. Then, the electrolysis equation for an aqueous solution of copper (II) chloride will look like this: CuCl 2 = Cu + Cl 2. Let's calculate the amount of copper (II) chloride using the data specified in the problem statement (molar mass - 134 g/mol): υ(CuCl 2) = m(CuCl 2)/M(CuCl 2) = 5/134 = 0.04 mol. According to the reaction equation υ(CuCl 2) = υ(Cu) =0.04 mol. Then we calculate the mass of copper released at the cathode (molar mass – 64 g/mol): m(Cu)= υ(Cu)×M(Cu)= 0.04 ×64 = 2.56 g.
Answer	The mass of copper released at the cathode is 2.56 g.