Category: Chemistry

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Enzyme Catalysis

Enzymes are complex protein molecules with three dimensional structures. They catalyse the chemical reaction in living organism. They are often present in colloidal state and extremely specific in catalytic action. Each enzyme produced in a particular living cell can catalyse a particular reaction in the cell.

Some common examples for enzyme catalysis

1. The peptide glycyl L-glutamyl L-tyrosin is hydrolysed by an enzyme called pepsin.

2. The enzyme diastase hydrolyses starch into maltose
2(C₆H₁₀O₅)_n + nH₂O → nC₁₂H₂₂O₁₁

3. The yeast contains the enzyme zymase which converts glucose into ethanol.
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

4. The enzyme micoderma aceti oxidises alcohol into acetic acid.
C₂H₅OH + O₂ → CH₃COOH + H₂O

5. The enzyme urease present in soya beans hydrolyses the urea.
NH₂ – CO – NH₂ + H₂O → 2NH₃ + CO₂

Mechanism of Enzyme Catalysed Reaction

The following mechanism is proposed for the enzyme catalysis
E + S ⇄ ES → P + E

Where E is the enzyme, S the substrate (reactant), Es represents activated complex and P the products.

Enzyme Catalysed Reaction show Certain General Special Characteristics.

(i) Effective and efficient conversion is the special characteristic of enzyme catalysed reactions. An enzyme may transform a million molecules of reactant into product in a minute.
For eg. 2H₂O₂ → 2H₂O + O₂

For this reaction, the activation energy is 18k cal/mole without a catalyst. With colloidal platinum as a catalyst the activation energy is 11.7 kcal/mole.

But with the enzyme catalyst the activation energy of this reaction is less than 2kcal/mole.

(ii) Enzyme catalysis is highly specific in nature.

H₂N-CO-NH₂ + H₂O → 2NH₃ + CO₂

The enzyme urease which catalyses the reaction of urea does not catalyse the following reaction of methyl urea

H₂N-CO-NH-CH₃ + H₂O → No reaction

(iii) Enzyme catalysed reaction has maximum rate at optimum temperature. At first rate of reaction increases with the increase of temperature, but above a particular temperature the activity of enzyme is destroyed. The rate may even drop to zero. The temperature at which enzymic activity is high or maximum is called as optimum temperature.

For Example:

• Enzymes involved in human body have an optimum temperature 37°C/98°F
• During high fever, as body temperature rises the enzymatic activity may collapse and lead to danger.

(iv) The rate of enzyme catalysed reactions varies with the pH of the system. The rate is maximum at a pH called optimum pH.

(v) Enzymes can be inhibited i.e. poisoned. Activity of an enzyme is decreased and destroyed by a poison. The physiological action of drugs is related to their inhibiting action.

Example: Sulpha drugs. Penicillin inhibits the action of bacteria and used for curing diseases like pneumonia, dysentery, cholera and other infectious diseases.

(vi) Catalytic activity of enzymes is increased by coenzymes or activators. A small non protein (vitamin) called a coenzyme promotes the catalytic activity of enzyme.

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Catalysis

In 1836 Berzelius identified certain substances loosen the bond in the reacting molecules and increased the rate of the reaction. But he also found these substances didn’t undergo any change chemically. In order to indicate the property, he gave them the name catalyst. (In greek, kata-wholly, lein-to loosen).

Later it was identified that there were many substances which retarded the speed of a reaction.

Hence a catalyst is defined as a substance which alters the rate of chemical reaction without itself undergoing chemical change. The phenomenon which involves the action of a catalyst is called catalysis.

Positive and Negative Catalysis:

In positive catalysis, the rate of a reaction is increased by the presence of catalyst but in negative catalysis, the rate of reaction is decreased by the presence of a catalyst. The two main types of catalysis

1. Homogeneous Catalysis and
2. Heterogeneous Catalysis

Homogeneous Catalysis

In a homogeneous catalysed reaction, the reactants, products and catalyst are present in the same phase.

Illustration (1):

In this reaction the catalyst NO, reactants, SO₂ and O₂, and product, SO₃ are present in the gaseous form.

Illustration (2):

In the decomposition of acetaldehyde by I₂ catalyst, the reactants and products are all present in the vapour phase.

CH₃CHO_(g) + [I₂]_(g) → CH_4(g) + CO_(g) + [I₂]_(g)

Let us consider some examples in which the reactants, products and catalyst are present in aqueous solution.

(1) Hydrolysis of cane sugar with a mineral acid as catalyst

(2) Ester hydrolysis with acid or alkali as catalyst

Heterogeneous Catalysis

In a reaction, the catalyst is present in a different phase i.e. it is not present in the same phase as that of reactants or products. This is generally referred as contact catalysis and the catalyst present is in the form of finely divided metal or as gauze.

Illustration

(i) In the manufacture of sulphuric acid by contact process SO₃ is prepared by the action of SO₂ and O₂
in the presence of Pt or V₂O₅ as a catalyst.

(ii) In the Haber’s process for the manufacture of ammonia, iron is used as a catalyst for the reaction between Hydrogen and Nitrogen.

(iii) Oxidation of ammonia is carried out in presence of platinum gauze

(iv) The hydrogenation of unsaturated organic compounds is carried out using finely divided nickel as a catalyst.

(v) Decomposition of H₂O₂ occurs in the presence of the Pt catalyst

(vi) In the presence of anhydrous AlCl₃, benzene reacts with ethanoyl chloride to produce acetophenone

Characteristics of Catalysts

1. For a chemical reaction, catalyst is needed in very small quantity. Generally, a pinch of catalyst is enough for a reaction in bulk.
2. There may be some physical changes, but the catalyst remains unchanged in mass and chemical composition in a chemical reaction.
3. A catalyst itself cannot initiate a reaction. It means it can not start a reaction which is not taking place. But, if the reaction is taking place in a slow rate it can increase its rate.
4. A solid catalyst will be more effective if it is taken in a finely divided form.
5. A catalyst can catalyse a particular type of reaction, hence they are said to be specific in nature.
6. In an equilibrium reaction, presence of catalyst reduces the time for attainment of equilibrium and hence it does not affect the position of equilibrium and the value of equilibrium constant.
7. A catalyst is highly effective at a particular temperature called as optimum temperature.
8. Presence of a catalyst generally does not change the nature of products.

For example: 2SO₂ + O₂ → SO₃
This reaction is slow in the absence of a catalyst, but fast in the presence of Pt catalyst

Promoters and Catalyst Poison

1. In a catalysed reaction the presence of a certain substance increases the activity of a catalyst. Such a substance is called a promoter.
2. For example in the Haber’s process of manufacture of ammonia, the activity of the iron catalyst is increased by the presence of molybdenum.
3. Hence molybdenum is called a promoter. In the same way Al₂O₃ can also be used as a promoter to increase the activity of the iron catalyst.

On the other hand, certain substances when added to a catalysed reaction decreases or completely destroys the activity of catalyst and they are often known as catalytic poisons.

Few examples,

In the reaction, 2SO₂ + O₂ → 2SO₃ with a Pt catalyst, the poison is As₂O₃
blocks the activity of the catalyst. So, the activity is lost.

In the Haber’s process of the manufacture of ammonia, the Fe catalyst is poisoned by the presence of H₂S.

In the reaction, 2H₂ + O₂ → 2H₂O,
CO acts as a catalytic poison for Pt – catalyst

Auto Catalysis

In certain reactions one of the products formed acts as a catalyst to the reaction. Initially the rate of reaction will be very slow but with the increase in time the rate of reaction increases.

Auto catalysis is observed in the following reactions.

CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH

Acetic acid acts as the autocatalyst

2AsH₃ → 2As + 3H₂

Arsenic acts as an autocatalyst

Negative Catalysis

In certain reactions, presence of certain substances, decreases the rate of the reaction. Ethanol is a negative catalyst for the following reaction.

(i) 4CHCl₃ + 3O₂ → 4COCl₂ + 2H₂O + 2Cl₂

Ethanol decreases the rate of the reaction

(ii) 2H₂O₂ → 2H₂O + O₂

In the decomposition of hydrogen peroxide, dilute acid or glycerol acts as a negative catalyst.

Theories of Catalysis

For a chemical reaction to occur, the reactants are to be activated to form the activated complex. The energy required for the reactants to reach the activated complex is called the activation energy. The activation energy can be decreased by increasing the reaction temperature. In the presence of a catalyst, the reactants are activated at reduced temperatures in otherwords, the activation energy is lowered.

The catalyst adsorbs the reactants activates them by weakening the bonds and allows them to react to form the products. As activation energy is lowered in presence of a catalyst, more molecules take part in the reaction and hence the rate of the reaction increases.

The action of catalysis in chemical reactions is explained mainly by two important theories. They are

• The intermediate compound formation theory
• The adsorption theory.

The Intermediate Compound Formation Theory

A catalyst acts by providing a new path with low energy of activation. In homogeneous catalysed reactions a catalyst may combine with one or more reactant to form an intermediate which reacts with other reactant or decompose to give products and the catalyst is regenerated.

Consider the reactions:

A + B → AB (1)
A + C → AC (intermediate) (2)
C is the catalyst
AC + B → AB + C (3)

Activation energies for the reactions (2) and (3) are lowered compared to that of (1). Hence the formation and decomposition of the intermediate accelerate the rate of the reaction.

Example 1:

The mechanism of Fridel craft reaction is given below

The action of catalyst is explained as follows

CH₃Cl + AlCl₃ → [CH₃]⁺[AlCl₄]^–

It is an intermediate.

C₆H₆ + [CH₃⁺] [AlCl₄]^– → C₆H₅CH₃ + AlCl₃ + HCl

Example 2:

Thermal decomposition of KClO₃ in presence of MnO₂ proceeds as follows.

Steps in the reaction
2KClO₃ → 2KCl + 3O₂ can be given as
It is an intermediate.
6MnO₃ → 6MnO₂ + 3O₂

Example 3:

Formation of water due to the reaction of H₂ and O₂ in the presence of Cu proceeds as follows.
Steps in the reaction H₂ + \(\frac{1}{2}\)O₂ → H₂O can be given as

2Cu + \(\frac{1}{2}\)O₂ → Cu₂O

It is an intermediate.

Cu₂O + H₂ → H₂O + 2Cu

Example 4:

Oxidation of HCl by air in presence of CuCl₂ proceeds as follows. Steps in the reaction

4HCl + O₂ → 2H₂O + 2Cl₂ can be given as

2CuCl₂ → Cl₂ + Cu₂Cl₂
2Cu₂Cl₂ + O₂ → 2Cu₂OCl₂

It is an intermediate.

2Cu₂OCl₂ + 4HCl → 2H₂O + 4CuCl₂

This theory describes

• The specificity of a catalyst and
• The increase in the rate of the reaction with increase in the concentration of a catalyst.

Limitations

• The intermediate compound theory fails to explain the action of catalytic poison and activators (promoters).
• This theory is unable to explain the mechanism of heterogeneous catalysed reactions.

Adsorption Theory

Langmuir explained the action of catalyst in heterogeneous catalysed reactions based on adsorption. The reactant molecules are adsorbed on the catalyst surfaces, so this can also be called as contact catalysis. According to this theory, the reactants are adsorbed on the catalyst surface to form an activated complex which subsequently decomposes and gives the product.

The various steps involved in a heterogeneous catalysed reaction are given as follows:

1. Reactant molecules diffuse from bulk to the catalyst surface.
2. The reactant molecules are adsorbed on the surface of the catalyst.
3. The adsorbed reactant molecules are activated and form activated complex which is decomposed to form the products.
4. The product molecules are desorbed.
5. The product diffuse away from the surface of the catalyst.

Active Centres

The surface of a catalyst is not smooth. It bears steps, cracks and corners. Hence the atoms on such locations of the surface are co-ordinatively unsaturated. So, they have much residual force of attraction. Such sites are called active centres. So, the surface carries high surface free energy.

The presence of such active centres increases the rate of reaction by adsorbing and activating the reactants. The adsorption theory explains the following

1. Increase in the surface area of metals and metal oxides by reducing the particle size increases acting of the catalyst and hence the rate of the reaction.

2. The action of catalytic poison occurs when the poison blocks the active centres of the catalyst.

3. A promoter or activator increases the number of active centres on the surfaces.

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Adsorption and Absorption

Solid surfaces have the ability to attract the contacting species due to free valency or residual force on them.

For example:
charcoal adsorbs ammonia, silica gel adsorbs water., charcoal adsorbs colorants from sugar.

These examples prove that adsorption is a surface phenomenon. In contrast to adsorption, absorption is a bulk phenomenon i.e. the adsorbate molecules are distributed throughout the adsorbent.

1. Adsorbent is the material on which adsorption takes place.
2. Adsorbed substance is called an adsorbate.
3. The surface of separation of the two phases where the concentration of adsorbed molecule is high is known as interface.
4. In adsorption, if the concentration of a substance in the interface is high, then it is called positive adsorption.
5. If it is less, then it is called negative adsorption.
6. The process of removing an adsorbed substance from the surface is called desorption.
7. The gaseous molecules like He, Ne, O₂, SO₂ and NH₃ and solutions of NaCl or KCl can be adsorbed by suitable adsorbents. These are referred as adsorbates.
8. Silica gel and metals like Ni, Cu, Pt, Ag and Pd and certain colloids can act as adsorbents.

Characteristics of Adsorption

1. Adsorption can occur in all interfacial surfaces i.e. the adsorption can occur in between gas-solid, liquid solid, liquid-liquid, solid-solid and gas-liquid.

2. Adsorption is a spontaneous process and it is always accompanied by decrease in free energy. When ∆G reaches zero, the equilibrium is attained. We know,

∆G = ∆H – T ∆S where ∆G is Change in Free energy.
∆H is Change in enthalpy and ∆S = Change in entropy.

3. When molecules are adsorbed, there is always a decrease in randomness of the molecules. ie., ∆S < 0, and T∆S is negative. Hence, adsorption is exothermic. Adsorption is a quick process whereas absorption is a slow process.

Types of Adsorption

Adsorption is classifid as physical adsorption and chemical adsorption, depending on the nature of forces acting between adsorbent and adsorbate. In chemical adsorption, gas molecules are held to the surface by formation of chemical bonds. Since strong bond is formed, nearly 400 KJ / mole is given out as heat of adsorption.

Examples

(a) Adsorption of O₂ on tungsten, Adsorption of H₂ on nickel, Adsorption of ethyl alcohol vapours on nickel.

In physical adsorption, physical forces such as van der waals force of attraction, dipole – dipole interaction, dispersion forces etc., exist between adsorbent and adsorbate. As these forces are weak, heat of adsorption is low, hence physical adsorption occurs at low temperatures.

Examples

(a) Adsorption of N₂ on mica.
(b) Adsorption of gases on charcoal.

The following table 10.1 illustrates the distinction between chemical and physical adsorption.

Distinction Between Chemical and Physical Adsorption

Chemical adsorption or Chemisorption or Activated adsorption	Physical adsorption or van der waals adsorption or Physisorption
1. It is very slow	1. It is instantaneous
2. It is very specific depends on nature of adsorbent and adsorbate.	2. It is non-specific
3. Chemical adsorption is fast with increase pressure, it can not alter the amount	3. In Physisorption, when pressure increases the extent of adsorption increases
4. When temperature is raised chemisorption first increases and then decreases	4. Physisorption decreases with increase in temperature
5. Chemisorption involves transfer of electrons between the adsorbent and adsorbate	5. No transfer of electrons
6. Heat of adsorption is high i.e., from 40- 400kJ/mole	6. Heat of adsorption is low in the order of 40kJ/mole.
7. Monolayer of the adsorbate is formed	7. Multilayer of the adsorbate is formed on the adsorbent
8. Adsorption occurs at fixed sites called active centres. It depends on surface area	8. It occurs on all sides
9. Chemisorption involves the formation of activated complex with appreciable activation energy	9. Activation energy is insignifiant

Factors Affecting Adsorption

The adsorption is well understood by considering the various factors affecting it. Qualitatively, the extent of surface adsorption depends on the following factors.

1. Nature of adsorbent
2. Nature of adsorbate
3. Pressure
4. Concentration at a given temperature.

1. Surface Area of Adsorbent:

As the adsorption is a surface phenomenon it depends on the surface area of adsorbent. i.e., higher the surface area, higher is the amount adsorbed.

2. Nature of Adsorbate

The nature of adsorbate can inflence the adsorption. Gases like SO₂, NH₃, HCl and CO₂ are easily liquefiable as they have greater van der waal’s force of attraction. On the other hand, permanent gases like H₂, N₂ and O₂ can not be liquefied easily. These permanent gases are having low critical temperature and adsorbed slowly, while gases with high critical temperature are adsorbed readily.

3. Effect of Temperature

When temperature is raised chemisorption fist increases and then decreases. whereas physisorption decreases with increase in temperature.

4. Effect of Pressure:

chemical adsorption is fast with increase in pressure, it can not alter the amount of adsorption. In Physisorption the extent of adsorption increases with increase in pressure.

Adsorption Isotherms and Isobars

Adsorption isotherms represents the variation of adsorption at constant temperature. When amount of adsorption is plotted versus temperature at constant pressure it is called adsorption isobar. Adsorption isobars of physisorption and chemisorption are different as represented in the graphs.

In physical adsorption, \(\frac{x}{m}\) decreases with increase in Temprature, But in chemical adsorption, \(\frac{x}{m}\) increases with rise in temperature and then decreases.

The increase illustrates the requirement of activation of the surface for adsorption is due to fact that formation of activated complex requires certain energy. The decrease at high temperature is due to desorption, as the kinetic energy of the adsorbate increases.

Adsorption Isotherms

Adsorption isotherm can be studied quantitatively. A plot between the amount of adsorbate adsorbed and pressure (or concentration of adsorbate) at constant temperature is called adsorption isotherms. In order to explain these isotherms various equations were suggested as follows:

(i) Freundlich adsorption isotherm.
According to Freundlich,
\(\frac{x}{m}\) = kp^1/n

where x is the amount of adsorbate, adsorbed on ‘m’ gm of adsorbent at a pressure of p. K and n are constant introduced by freundlich. Value n is always less than unity.

This equation is applicable for adsorption of gases on solid surfaces. The same equation becomes \(\frac{x}{m}\) = K c^1/n, when used for adsorption in solutions with c as concentration.

This equation quantitively predicts the effect of pressure(or concentration) on the adsorption of gases(or adsorbates) at constant temperature.

Taking log on both sides of equation \(\frac{x}{m}\) = Kp^1/n
log \(\frac{x}{m}\) = logK + \(\frac{1}{n}\)log P

Hence the intercept represents the value of log k and the slope gives \(\frac{1}{n}\).

This equation explains the increase of \(\frac{x}{m}\) with increase in pressure. But experimental values show the deviation at low pressure.

Limitations

This equation is purely empirical and valid over a limited pressure range. The values of constants k and n also found vary with temperature. No theoretical explanations were given.

Applications of Adsorption

Though we have innumerable applications for adsorption, we consider few of them

1. Gas Masks:

During world war I charcoal gas mask was employed by both the British and American. Activated charcoal was found to be one of the best adsorbents.

2. To create high vacuum in vessels, Tail and Dewar used activated charcoal. For dehydration and also purification of gases like CO₂, N₂, Cl₂, O₂ and He, alumina and silica are employed. In the blast furnace silica gel is also used for drying air.

3. One of the highly important use of adsorption is the softening of hardwater. Permutit is employed for this process which adsorbs Ca²⁺ and Mg²⁺ ions in its surface, there is an ion exchange as shown below it occurs on the surface.

Na₂Al₂Si₄O₁₂ + CaCl₂ → CaAl₂Si₄O₁₂ + 2NaCl

Exhausted permutit is regenerated by adding a solution of common salt.

CaAl₂Si₄O₁₂ + 2NaCl → Na₂Al₂Si₄O₁₂ + CaCl₂

4. Ion Exchange Resins

Ion exchange resins are working only based on the process of adsorption. Ion exchange resins are used to demineralise water. This process is carried out by passing water through two columns of cation and anion exchange resins.

5. Petroleum Refining and Refining of Vegetable Oil

Fuller’s earth and silica gel are used for refining process.

6. Decolourisation of Sugar:

Sugar prepared from molasses is decolourised to remove coloured impurities by adding animal charcoal which acts as decolourising material.

7. Chromatography

The chromatographic technique is applied for separation of components in a mixture. It is mainly based on adsorption of components on the surface of adsorbents. This method is very effective and used for identification, detection and estimation of many substances even if they are contained in micro quantities.

8. Catalysed Reaction

Catalysis is an important branch of surface chemistry which is based on the phenomenon of adsorption of materials on the catalyst surface.

Examples:

In the Haber’s process, ammonia is manufactured from N₂ and H₂ as shown by the following reactions.

In this process, Fe is the catalyst and Mo is a promoter. The surface of the Fe catalyses the reaction.

In the hydrogenation of oils to obtain vanaspathi, Nickel is used as a catalyst. Nickel surface catalyses the reaction.

9. Qualitative Analysis

When blue litmus solution is added to Al³⁺ ion, a red coloration is seen due to the acidic nature of the solution. Addition of ammonium hydroxide to it gives a blue lake. This is due to the adsorption of blue colour litmus compound on the surface of Al (OH)₃ Which is formed during the addition of NH₄OH.

10. Medicine:

Drugs cure diseases by adsorption on body tissues.

11. Concentration of Ores of Metals

Sulphides ores are concentrated by a process called froth fltation in which light ore particles are wetted by pine oil.

12. Mordants and Dyes

Most of the dyes are adsorbed on the surface of the fabrics. Mordants are the substances used for fining dyes onto the fabric.

13. Adsorption Indicators

In the precipitation titrations, the end point is indicated by an external indicator which changes its colour after getting adsorbed on precipitate. It is used to indicate the end point of the titration.

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Thermodynamics of Cell Reactions

We have just learnt that in a galvanic cell, the chemical energy is converted into electrical energy. The electrical energy produced by the cell is equal to the product of the total charge of electrons and the emf of the cell which drives these electrons between the electrodes.

If ‘n’ is the number of moles of electrons exchanged between the oxidising and reducing agent in the overall cell reaction, then the electrical energy produced by the cell is given as below.

Electrical energy = Charge of ‘n’ mole of electrons × E_cell ……. (9.20)
Charge of 1 mole of electrons = one Faraday (1F)
∴ Charge of ‘n’ mole of electrons = nF
Equation (9.20) ⇒ Electrical energy = nFE_cell ……. (9.21)

∴ Charge of one elctron = 1.602 × 10^-19C
∴ Charge of one mole of elctron = 6.023 × 10²³ × 1.602 × 10^-19C
= 96488 C
i.e., 1F = 96500 C

This energy is used to do the electric work. Therefore the maximum work that can be obtained from a galvanic cell is

(W_max)_cell = – nFE_cell …………… (9.22)

Here the (-) sign is introduced to indicate that the work is done by the system on the surroundings.

We know from the Second Law of thermodynamics that the maximum work done by the system is equal to the change in the Gibbs free energy of the system.

i.e., W_max = ∆G …………… (9.23)

From (9.22) and (9.23),

∆G = – nFE_cell ……………. (9.24)

For a spontaneous cell reaction, the DG should be negative. The above expression (9.24) indicates that E_cell should be positive to get a negative ∆G value.

When all the cell components are in their standard state, the equation (9.24) becomes

∆G° = – nFE°_cell ………….. (9.25)

We know that the standard free energy change is related to the equilibrium constant as per the following expression.

∆G° = – RTlnK_eq …………… (9.26)
Comparing (9.25) and (9.26),
⇒ E°_cell = \(\frac{2.303RT}{nF}\)log K_eq ………….. (9.27)

Nernst Equation

Nernst equation is the one which relates the cell potential and the concentration of the species involved in an electrochemical reaction. Let us consider an electrochemical cell for which the overall redox reaction is,
xA + yB ⇄ lC + mD. The reaction quotient Q for the above reaction is given below

……….. (9.28)

We have already learnt that,
∆G = ∆G° + RT lnQ

The Gibbs free energy can be related to the cell emf as follows [∴equation (9.24) and (9.25)]
∆G = – nFE _cell; ∆G° = – nFE°_cell
Substitute these values and Q from (9.28) in the equation (9.29)

………… (9.30)

Divide the whole equation (9.30) by (-nF)

………… (9.31)

The above equation (9.31) is called the Nernst equation. At 25°C (298K), the above equation (9.31) becomes,

……………… (9.32)

Let us calculate the emf of the following cell at 25°C using Nernst equation

Cu (s)|Cu²⁺(0.25 aq, M)|Fe²⁺(0.1 aq M)|Pt(s)
Given: (E°)_{Fe²⁺|Fe²⁺} = 0.77V and (E°)_Cu²⁺|Cu = 0.34 V

Half reactions are
Cu (s) → Cu²⁺(aq) + 2e^– …………… (1)
2 Fe³⁺(aq) + 2e^– → 2Fe²⁺(aq) …………. (2)

the overall reaction is

Cu (s) + 2Fe³⁺(aq) → Cu²⁺(aq) + 2Fe²⁺ (aq), and n = 2
Apply Nernst equation at 25°C.

Given standard reduction potential of Cu²⁺|Cu is 0.34V

Electrolytic Cell and Electrolysis

Electrolysis is a process in which the electrical energy is used to cause a non-spontaneous chemical reaction to occur; the energy is often used to decompose a compound into its elements. The device which is used to carry out the electrolysis is called the electrolytic cell.

The electrochemical process occurring in the electrolytic cell and galvanic cell are the reverse of each other. Let us understand the function of a electrolytic cell by considering the electrolysis of molten sodium chloride.

The electrolytic cell consists of two iron electrodes dipped in molten sodium chloride and they are connected to an external DC power supply via a key as shown in the figure (9.8). The electrode which is attached to the negative end of the power supply is called the cathode, and the one which attached to the positive end is called the anode. Once the key is closed, the external DC power supply drives the electrons to the cathode and at the same time pull the electrons from the anode.

Cell Reactions

Na⁺ ions are attracted towards cathode, where they combine with the electrons and reduced to liquid sodium.

Cathode (Reduction)

Na⁺(l) + e^– → Na(l) E° = – 2.71
Similarly, Cl^– ions are attracted towards anode where they lose their electrons and oxidised to chlorine gas.

Anode (Oxidation)

2Cl^–(l) → Cl₂(g) + 2e^– E° = – 1.36V

The overall reaction is,
2Na⁺(l) + 2Cl^–(l) → 2Na(l) + Cl₂(g) E° = – 4.07V

The negative E° value shows that the above reaction is a non spontaneous one. Hence, we have to supply a voltage greater than 4.07V to cause the electrolysis of molten NaCl. In electrolytic cell, oxidation occurs at the anode and reduction occur at the cathode as in a galvanic cell, but the sign of the electrodes is the reverse i.e., in the electrolytic cell cathode is – ve and anode is + ve.

Faraday’s Laws of Electrolysis
First Law

The mass of the substance (m) liberated at an electrode during electrolysis is directly proportional to the quantity of charge (Q) passed through the cell.

i.e m α Q

We know that the charge is related to the current by the equation I = \(\frac{Q}{t}\) ⇒ Q = It
∴ m α It
(or)
m = Z It …………. (9.33)

Where is Z is known as the electro chemical equivalent of the substance produced of the electrode. When, I = 1A and t = 1sec, Q = 1C, in such case the equation (9.32) becomes, (9.33)
⇒ m = Z …………….. (9.34)

Thus, the electrochemical equivalent is defined as the amount of substance deposited or liberated at the electrode by a charge of 1 coulomb.

Electro Chemical Equivalent and Molar Mass

Consider the following general electrochemical redox reaction
Mⁿ⁺(aq) + ne^– → M(s)
We can infer from the above equation that ‘n’ moles of electrons are required to precipitate 1 mole of Mⁿ⁺ as M(s).

The quantity of charge required to precipitate one mole of Mⁿ⁺ = Charge of ‘n’ moles of electrons
= nF

In other words, the mass of substance deposited by one coulomb of charge

…………… (9.35)

Second Law

When the same quantity of charge is passed through the solutions of different electrolytes, the amount of substances liberated at the respective electrodes are directly proportional to their electrochemical equivalents.

Let us consider three electrolytic cells connected in series to the same DC electrical source as shown in the figure 9.9. Each cell is filled with a different electrolytes namely NiSO₄, CuSO₄ and CoSO₄, respectively.

When Q coulomb charge is passed through the electrolytic cells the masses of Nickel, copper and cobalt deposited at the respective electrodes be m_Ni, m_Cu and m_Co, respectively.

According to Faraday’s second Law,

………….. (9.36)

Batteries

Batteries are indispensable in the modern electronic world. For example, Li – ion batteries are used in cell phones, dry cell in flashlight etc. These batteries are used as a source of direct current at a constant voltage. We can classify them into primary batteries (non – rechargeable) and secondary batteries (rechargeable). In this section, we will briefly discuss the electrochemistry of some batteries.

Leclanche Cell

Anode: Zinc container
Cathode: Graphite rod in contact with MnO₂
Electrolyte: ammonium chloride and zinc chloride in water

Emf of the cell is about 1.5V
Cell reaction

Oxidation at Anode
Zn (s) → Zn²⁺(aq) + 2e^– ……… (1)

Reduction at Cathode
2 NH₄⁺(aq) + 2e^– → 2NH₃(aq) + H₂(g) ………… (2)

The hydrogen gas is oxidised to water by MnO₂
H₂ (g) + 2 MnO₂ (s) → Mn₂O₃(s) + H₂O(l)
Equation (1) + (2)+(3) gives the overall redox reaction

Zn (s) + 2NH₄⁺(aq) + 2 MnO₂(s) → Zn²⁺(aq) + Mn₂O₃ (s) + H₂O(l) + 2NH₃ ………… (4)

Ammonia produced at the cathode combines with Zn²⁺ to form a complex ion [Zn(NH₃)₄]²⁺(aq). As the reaction proceeds, the concentration of NH₄⁺ will decrease and the aqueous NH₃ will increase which lead to the decrease in the emf of cell.

Mercury Button Cell

Anode: zinc amalgamated with mercury
Cathode: HgO mixed with graphite
Electrolyte: Paste of KOH and ZnO

Overall reaction: Zn (s) + HgO (s) → ZnO (s) + Hg (l)
Cell emf: about 1.35V.
Uses: It has higher capacity and longer life. Used in pacemakers, electronic watches, cameras etc.

Secondary Batteries

We have already learnt that the electrochemical reactions which take place in a galvanic cell may be reversed by applying a potential slightly greater than the emf generated by the cell. This principle is used in secondary batteries to regenerate the original reactants. Let us understand the function of secondary cell by considering the lead storage battery as an example.

Lead Storage Battery

Anode: Spongy lead
Cathode: lead plate bearing PbO₂
Electrolyte: 38% by mass of H₂SO₄ with density 1.2g / mL.

Oxidation occurs at the anode

Pb(s) → Pb²⁺(aq) + 2e^– ……….. (1)
The Pb²⁺ ions combine with SO₄^2-(aq) → PbSO₄ (s) …………. (2)

Reduction occurs at the cathode

PbO₂ (s) + 4 H⁺(aq) + 2e^– → Pb²⁺(aq) + 2H₂O(l) ………. (3)
The Pb²⁺ ions also combine with SO₄^2- → Pb²⁺(aq) + 2H₂O(l)

The emf of a single cell is about 2V. Usually six such cells are combined in series to produce 12 volt

The emf of the cell depends on the concentration of H₂SO₄. As the cell reaction uses SO₄^2-
ions, the concentration H₂SO₄ decreases. When the cell potential falls to about 1.8V, the cell has to be recharged.

Recharge of the Cell

As said earlier, a potential greater than 2V is applied across the electrodes, the cell reactions that take place during the discharge process are reversed. During recharge process, the role of anode and cathode is reversed and H₂SO₄ is regenerated.

Oxidation occurs at the cathode (now act as anode)

image

Reduction occurs at the anode (now act as cathode) PbSO₄(s) + 2e^– → Pb(s) + SO₄^2-(aq)
Overall reaction
2PbSO₄(s) + 2H₂O(l) → Pb(s) + PbO₂(s) + 4H⁺(aq) + 2SO₄^2-(aq)

Thus, the overall cell reaction is exactly the reverse of the redox reaction which takes place while discharging.

Uses:
Used in automobiles, trains, inverters etc.

The lithium – ion Battery

Anode: Porus graphite
Cathode: Transition metal oxide such as CoO₂.

Electrolyte: Lithium salt in an organic solvent
At the anode oxidation occurs
Li (s) → Li⁺(aq) + e^–

At the cathode reduction occurs
Li⁺ + CoO₂(s) + e^– → LiCoO₂(s)

Overall reactions
Li (s) + CoO₂ → LiCoO₂ (s)

Both electrodes allow Li⁺ ions to move in and out of their structures.

During discharge, the Li⁺ ions produced at the anode move towards cathode through the non – aqueous electrolyte. When a potential greater than the emf produced by the cell is applied across the electrode, the cell reaction is reversed and now the Li⁺ ions move from cathode to anode where they become embedded on the porous graphite electrode. This is known as intercalation.

Uses:

Used in cellular phones, laptops, computers, digital cameras, etc.

Fuel Cell:

The galvanic cell in which the energy of combustion of fuels is directly converted into electrical energy is called the fuel cell. It requires a continuous supply of reactant to keep functioning. The general representation of a fuel cell is follows.

Fuel | Electrode | Electrolyte | Electrode | Oxidant

Let us understand the function of fuel cell by considering hydrogen – oxygen fuel cell. In this case, hydrogen act as a fuel and oxygen as an oxidant and the electrolyte is aqueous KOH maintained at 200°C and 20 – 40 atm. Porous graphite electrode containing Ni and NiO serves as the inert electrodes.

Hydrogen and oxygen gases are bubbled through the anode and cathode, respectively.

Oxidation occurs at the anode:
2H₂g + 4 OH^–(aq) → 4H₂O(l) + 4e^–
Reduction occurs at the cathode O₂(g) + 2H₂O(l) + 4e^– → 4 OH^–(aq)

The overall reaction is 2H₂ (g) + O₂ (g) → 2H₂O(l)

The above reaction is the same as the hydrogen combustion reaction, however, they do not react directly ie., the oxidation and reduction reactions take place separately at the anode and cathode respectively like H₂ – O₂ fuel cell. Other fuel cells like propane – O₂ and methane O₂ have also been developed.

Corrosion

We are familiar with the rusting of iron. Have you ever noticed a green film formed on copper and brass vessels?. In both, the metal is oxidised by oxygen in presence of moisture. This redox process which causes the deterioration of metal is called corrosion.

As the corrosion of iron causes damages to our buildings, bridges etc… it is important to know the chemistry of rusting and how to prevent it. Rusting of iron is an electrochemical process.

Electrochemical Mechanism of Corrosion

The formation of rust requires both oxygen and water. Since it is an electrochemical redox process, it requires an anode and cathode in different places on the surface of iron. The iron surface and a droplet of water on the surface as shown in figure (9.15) form a tiny galvanic cell. The region enclosed by water is exposed to low amount of oxygen and it acts as the anode.

The remaining area has high amount of oxygen and it acts as cathode. So based on the oxygen content, an electro chemical cell is formed corrosion occurs at the anode i,e,. in the region enclosed by the water as discussed below.

At Anode (Oxidation):

Iron dissolves in the anode region

2Fe (s) → 2Fe²⁺ (aq) + 4e^– E° = 1.23V

The electrons move through the iron metal from the anode to the cathode area where the oxygen dissolved in water, is reduced to water.

At Cathode (Reduction):

The reaction of atmospheric carbon dioxide with water gives carbonic acid which furnishes the H⁺ ions for reduction.

O₂(g) + 4H⁺(aq) + 4e^– → 2H₂O(l) E° = 0.44 + 1.23 = 1.67V

The positive emf value indicates that the reaction is spontaneous.

Fe²⁺(aq) + O₂(g) + 4H⁺(aq) → 4Fe³⁺(aq) + 2H₂O(l)
4Fe³⁺(aq) + 4H₂O(l) → Fe₂O₃.H₂O(s) + 6H⁺(aq)

Other metals such as aluminium, copper and silver also undergo corrosion, but at a slower rate than iron. For example, let us consider the reduction of aluminium,
Al(s) → Al³⁺(aq) + 3e^–

Al³⁺, which reacts with oxygen in air to forms a protective coating of Al₂O₃. This coating act as a protective film for the inner surface. So, further corrosion is prevented.

Protection of Metals form Corrosion

This can be achieved by the following methods.

• Coating metal surface by paint.
• Galvanizing – by coating with another metal such as zinc. zinc is stronger reducing agent than iron and hence it can be more easily corroded than iron. i.e., instead of iron, the zinc is oxidised.
• Cathodic protection – In this technique, unlike galvanising the entire surface of the metal to be protected need not be covered with a protecting metal.

Instead, metals such as Mg or zinc which is corroded more easily than iron can be used as a sacrificial anode and the iron material acts as a cathode. So iron is protected, but Mg or Zn is corroded.

Passivation:

The metal is treated with strong oxidising agents such as concentrated HNO₃. As a result, a protective oxide layer is formed on the surface of metal.

Alloy Formation:

The oxidising tendency of iron can be reduced by forming its alloy with other more anodic metals.
Example, stainless steel – an alloy of Fe and Cr.

Electrochemical Series

We have already learnt that the standard single electrode potentials are measured using standard hydrogen electrode. The standard electrode potential at 298K for various metal metal ion electrodes are arranged in the decreasing order of their standard reduction potential values as shown in the figure. This series is called electrochemical series.

The standard reduction potential (E°) is a measure of the oxidising tendency of the species. The greater the E° value, greater is the tendency shown by the species to accept electrons and undergo reduction. So higher the (E°) Value, lesser is the tendency to undergo corrosion.

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Electrochemical Cell

Electrochemical cell is a device which converts chemical energy into electrical energy and vice versa. It consists of two separate electrodes which are in contact with an electrolyte solution. Electrochemical cells are mainly classified into the following two types.

1. Galvanic Cell (Voltaic Cell):

It is a device in which a spontaneous chemical reaction generates an electric current i.e., it converts chemical energy into electrical energy. It is commonly known as a battery.

2. Electrolytic Cell:

It is a device in which an electric current from an external source drives a nonspontaneous reaction i.e., it converts electrical energy into chemical energy.

Galvanic Cell

We have already learnt in XI standard that when a zinc metal strip is placed in a copper sulphate solution, the blue colour of the solution fades and the copper is deposited on the zinc strip as red – brown crust due to the following spontaneous chemical reaction.

Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s)

The energy produced in the above reaction is lost to the surroundings as heat.

In the above redox reaction, Zinc is oxidised to Zn²⁺ ions and the Cu²⁺ ions are reduced to metallic copper.
The half reactions are represented as below.

Zn(s) → Zn²⁺(aq) + 2e^– (oxidation)
Cu²⁺(aq) + 2e^– → Cu(s) (reduction)

If we perform the above two half reactions separately in an apparatus as shown in figure 9.5, some of the energy produced in the reaction will be converted into electrical energy. Let us understand the function of a galvanic cell by considering Daniel cell as an example. It uses the above reaction for generation of electrical energy. The separation of half reaction is the basis for the construction of Daniel cell. It consists of two half cells.

Oxidation Half Cell

A metallic zinc strip that dips into an aqueous solution of zinc sulphate taken in a beaker, as shown in Figure 9.5.

Reduction Half Cell

A copper strip that dips into an aqueous solution of copper sulphate taken in a beaker, as shown in Figure 9.5.

Joining the Half Cells

The zinc and copper strips are externally connected using a wire through a switch (k) and a load (example: volt meter). The electrolytic solution present in the cathodic and anodic compartment are connected using an inverted U tube containing a agar-agar gel mixed with an inert electrolytes such as KCl, Na₂SO₄ etc., The ions of inert electrolyte do not react with other ions present in the half cells and they are not either oxidised (or) reduced at the electrodes. The solution in the salt bridge cannot get poured out, but through which the ions can move into (or) out of the half cells.

When the switch (k) closes the circuit, the electrons flows from zinc strip to copper strip. Ths is due to the following redox reactions which are taking place at the respective electrodes.

Anodic Oxidation

The electrode at which the oxidation occurs is called the anode. In Daniel cell, the oxidation take place at zinc electrode, i.e., zinc is oxidised to Zn²⁺ ions by loosing its electrons. The Zn²⁺ ions enter the solution and the electrons enter the zinc metal, then flow through the external wire and then enter the copper strip. Electrons are liberated at zinc electrode and hence it is negative (- ve).

Zn(s) → Zn²⁺(aq) + 2e^– (loss of electron-oxidation)

Cathodic Reduction

As discussed earlier, the electrons flow through the circuit from zinc to copper, where the Cu²⁺ ions in the solution accept the electrons, get reduced to copper and the same get deposited on the electrode. Here, the electrons are consumed and hence it is positive (+ ve).

Cu²⁺ + 2e^– → Cu (s) (gain of electron-reduction)

Salt Bridge

The electrolytes present in two half cells are connected using a salt bridge. We have learnt that the anodic oxidation of zinc electrodes results in the increase in concentration of Zn²⁺ in solution. i.e., the solution contains more number of Zn²⁺ ions as compared to SO₄^2- and hence the solution in the anodic compartment would become positively charged.

Similarly, the solution in the cathodic compartment would become negatively charged as the Cu²⁺ ions are reduced to copper i.e., the cathodic solution contain more number of SO₄^2- ions compared to Cu²⁺.

To maintain the electrical neutrality in both the compartments, the non reactive anions Cl^– (from KCl taken in the salt bridge) move from the salt bridge and enter into the anodic compartment, at the same time some of the K⁺ ions move from the salt bridge into the cathodic compartment.

Completion of Circuit

Electrons flow from the negatively charged zinc anode into the positively charged copper cathode through the external wire, at the same time, anions move towards anode and cations are move towards the cathode compartment. This completes the circuit.

Consumption of Electrodes

As the Daniel cell operates, the mass of zinc electrode gradually decreases while the mass of the copper electrode increases and hence the cell will function until the entire metallic zinc electrode is converted in to Zn²⁺ or the entire Cu²⁺ ions are converted in to metallic copper.

Unlike Daniel cell, in certain cases, the reactants (or) products cannot serve as electrodes and in such cases insert electrode such as graphite (or) platinum is used which conducts current in the external circuit.

Galvanic Cell Notation

The galvanic cell is represented by a cell diagram, for example, Daniel cell is represented as

Zn (s)|Zn²⁺(aq)||Cu²⁺(aq)|Cu (s)

In the above notation, a single vertical bar (|) represents a phase boundary and the double vertical bar (||) represents the salt bridge. The anode half cell is written on the left side of the salt bridge and the cathode half cell on the right side.

The anode and cathode are written on the extreme left and extreme right, respectively. The emf of the cell is written on the right side after cell diagram.

Example

The net redox reaction of a galvanic cell is given below
2 Cr (s) + 3Cu²⁺(aq) → 2Cr³⁺ (aq) + 3Cu (s)

Write the half reactions and describe the cell using cell notation.
Anodic oxidation 2Cr (s) → 2Cr³⁺ (aq) + 6e^– ….. (1)
Cathodic reduction : 3Cu²⁺ (aq) + 6e^– → 3 Cu (s) ….. (2)

Cell Notation is
Cr (s)|Cr ³⁺(aq)||Cu²⁺(aq)|Cu(s)

Emf of a Cell

We have learnt that when two half cells of a Daniel cell are connected, a spontaneous redox reaction will take place which results in the flow of electrons from anode to cathode. The force that pushes the electrons away from the anode and pulls them towards cathode is called the electromotive force (emf) (or) the cell potential. The SI unit of cell potential is the volt (v).

When there is one volt difference in electrical potential between the anode and cathode, one joule of energy is released for each columb of charge that moves between them.

i.e., 1J = 1C × 1V ….. (9.18)

The cell voltage depends on the nature of the electrodes, the concentration of the electrolytes and the temperature at which the cell is operated. For example

At, 25°C , The emf of the below mentioned Daniel cell is 1.107 Volts

Zn (s)|Zn²⁺(aq, 1M)||Cu²⁺(aq, 1M)|Cu (s) E° = 1.107 V0lts

Measurement of Electrode Potential

The overall redox reaction can be considered as the sum of two half reactions i.e., oxidation and reduction. Similarly, the emf of a cell can be considered as the sum of the electrode potentials at the cathode and anode,

E_cell = (E_ox)_anode + (E_red)_cathode ….. (9.19)

Here, (E_ox)_anode represents the oxidation potential at anode and (E_red)_cathode represents the reduction potential at cathode. It is impossible to measure the emf of a single electrode, but we can measure the potential difference between the two electrodes (E_cell) using a voltmeter.

If we know the emf of any one of the electrodes which constitute the cell, we can calculate the emf of the other electrode from the measured emf of the cell using the expression (9.19). Hence, we need a reference
electrode whose emf is known.

For that purpose, Standard Hydrogen Electrode (SHE) is used as the reference electrode. It has been assigned an arbitrary emf of exactly zero volt. It consists of a platinum electrode in contact with 1M HCl solution and 1 atm hydrogen gas. The hydrogen gas is bubbled through the solution at 25°C as shown in the figure 9.6. SHE can act as a cathode as well as an anode.

The Half cell reactions are given below.

If SHE is used as a cathode, the reduction reaction is
2H⁺(aq, 1M) + 2e^– → H₂(g, 1 atm) E° = 0 volt

If SHE is used as an anode, the oxidation reaction is
H₂(g, 1 atm) → 2H⁺(aq, 1M) + 2e^– E° = 0 volt

Illustration

Let us calculate the reduction potential of zinc electrode dipped in zinc sulphate solution using SHE.

Step: 1

The following galvanic cell is constructed using SHE
Zn (s) | Zn²⁺(aq, 1M)|| H⁺(aq, 1M)| H₂ (g, 1atm)| Pt (s)

Step: 2

The emf of the above galvanic cell is measured using a volt meter. In this case, the measured emf of the above galvanic cell is 0.76V.

Calculation
We know that,

E°_cell = (E°_ox)_zn|zn²⁺+ (E°_red)_SHE [From equation (9.19)]
E°_cell = 0.76 and (E°_red)_SHE = 0V . Substitute these values in the above equation

⇒ 0.76V = (E°_ox)_zn|zn²⁺ + 0V
⇒ (E°_ox)_zn|zn²⁺ = 0.76V

This oxidation potential corresponds to the below mentioned half cell reaction which takes place at the cathode.

Zn → Zn²⁺+ 2e^– (Oxidation)

The emf for the reverse reaction will give the reduction potential
Zn²⁺+2e^– → Z ; E° = – 0.76V
∴ (E^o°_red)_{zn²⁺|zn = – 0.76V.}

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Variation of Molar Conductivity With Concentration

Friedrich Kohlraush studied the molar conductance of different electrolytes at different concentrations. He observed that, increase of the molar conductance of an electrolytic solution with the increase in the dilution. One such experimental results is given in the following table for better understanding.

Based on the above such results, Kohlraush deduced the following empirical relationship between the molar conductance (Λ_m) and the concentration of the electrolyte (C).

Λ_m = ^° – k\(\sqrt{C}\) ………… (9.11)

The above equation represents a straight line of the form y = mx + c. Hence, the plot of Λ_m Vs \(\sqrt{C}\) gives a straight line with a negative slope of – k and the y intercept. Λ^°_m. Where Λ^°_m is called the limiting molar value in very dilute solutions.

For strong electrolytes such as KCl, NaCl etc., the plot, Λ_m Vs \(\sqrt{C}\), gives a straight line as shown in the graph (9.4). It is also observed that the plot is not a linear one for weak electrolytes.

For a strong electrolyte, at high concentration, the number of constituent ions of the electrolyte in a given volume is high and hence the attractive force between the oppositely charged ions is also high. Moreover the ions also experience a viscous drag due to greater solvation.

These factors attribute for the low molar conductivity at high concentration. When the dilution increases, the ions are far apart and the attractive forces decrease. At infinite dilution the ions are so far apart, the interaction between them becomes insignificant and hence, the molar conductivity increases and reaches a maximum value at infinite dilution.

For a weak electrolyte, at high concentration, the plot is almost parallel to concentration axis with slight increase in conductivity as the dilution increases. When the concentration approaches zero, there is a sudden increase in the molar conductance and the curve is almost parallel to Λ_m axis.

This is due to the fact that the dissociation of the weak electrolyte increases with the increase in dilution (Ostwald dilution law). Λ^°_m values for strong electrolytes can be obtained by extrapolating the straight line, as shown in figure (9.4). But the same procedure is not applicable for weak electrolytes, as the plot is not a linear one, Λ^°_m values of the weak electrolytes can be determined using Kohlraush’s law.

Debye – Huckel and Onsager Equation

We have learnt that at infinite dilution, the interaction between the ions in the electrolyte solution is negligible. Except this condition, electrostatic interaction between the ions alters the properties of the solution from those expected from the free – ions value. The influence of ion-ion interactions on the conductivity of strong electrolytes was studied by Debye and Huckel.

They considered that each ion is surrounded by an ionic atmosphere of opposite sign, and derived an expression relating the molar conductance of strong electrolytes with the concentration by assuming complete dissociation. Later, the equation was further developed by Onsager. For a uni – univalent electrolyte the Debye Huckel and Onsager equation is given below.

Λ_m = Λ^°_m – (A + BΛ^°_m)\(\sqrt{C}\) ………. (9.12)

Where A and B are the constants which depend only on the nature of the solvent and temperature. The expression for A and B are

Here, D is the dielectric constant of the medium, η the viscosity of the medium and T the temperature in Kelvin.

Kohlraush’s Law

The limiting molar conductance Λ^°_m is the basis for kohlraush law. At infinite dilution, the limiting molar conductivity of an electrolyte is equal to the sum of the limiting molar conductivities of its constituent ions. i.e., the molar conductivity is due to the independent migration of cations in one direction and anions in the opposite direction.

(Λ^°_m)_NaCl = (λ^°_m)_Na + (λ^°_m)_Cl^–

In general, according to Kohlraush’s law, the molar conductivity at infinite dilution for a electrolyte represented by the formula A_xB_y, is given below.

(Λ^°_m)A_xB_y = x(λ^°_m)A^y+ + y(λ^°_m)B^x- ………… (9.13)

Kohlraush arrived the above mentioned relationship based on the experimental observations such as the one as shown in the table. These result show that at infinite dilution each constituent ion of the electrolyte makes a definite contribution towards the molar conductance of the electrolyte irrespective of nature of other ion with which it is associated i.e;

(Λ^°_m)KCl – (Λ^°_m)_NaCl = 149.86 – 126.45
{(λ^°_K)K + (λ^°_m)Cl^–} – {(λ^°_m)_Cl^–} = 23.41
(λ^°_m)_K⁺ – (λ^°_m)_Na⁺ = 23.41

Similarly, we can conclude that (λ^°_m)_Br^– – (λ^°_m)_Cl^– = 2.06

Applications of Kohlrausch’s Law

1. Calculation of Molar Conductance at Infinite Dilution of a Weak Electrolyte.

It is impossible to determine the molar conductance at infinite dilution for weak electrolytes experimentally. However, the same can be calculated using Kohlraush’s Law.

For example, the molar conductance of CH₃COOH, can be calculated using the experimentally determined molar conductivities of strong electrolytes HCl, NaCl and CH₃COONa.

Equation (1) + Equation (2) – Equation (3) gives,

2. Calculation of Degree of Dissociation of Weak Electrolytes

The degree of dissociation of weak electrolyte can be calculated from the molar conductivity at a given concentration and the molar conductivity at infinite dilution using the following expression

………… (9.14)

Calculation of dissociation constant using Λ_m values. According to Ostwald dilution Law,

K_a = \(\frac{α²C}{(1-α}\) ……….. (9.15)

Substitute α value in the above expression (9.15)

3. Calculation of Solubility of Sparingly Soluble Salts

Substances like AgCl, PbSO₄etc., are sparingly soluble in water. The solubility product of such substances can be determined using conductivity measurements.

Let us consider AgCl as an example

AgCl(s) ⇄ Ag⁺ + Cl^–
K_sp = [Ag⁺] [Cl^–]

Let the concentration of [Ag⁺] be ‘C’ molL^-1.
As per the stoichiometry, if [Ag⁺] = C, then [Cl^–] also equal to ‘C’ mol L^-1.

K_sp = C.C
⇒ K_sp = C²
We know that the concentration (in mol dm^-3) is related to the molar and specific conductance by the following expressions

Substitute the concentration value in the relation K_sp = C²

………… (9.17)

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Conductivity of Electronic Solution

We have already learnt that when an electrolyte such as sodium chloride, potassium chloride etc is dissolved in a solvent like water, the electrolyte is completely dissociated to give its constituent ions (namely cations and anions). When an electric field is applied to such an electrolytic solution, the ions present in the solution carry charge from one electrode to another electrode and thereby they conduct electricity. The conductivity of the electrolytic solution is measured using a conductivity cell. (Fig 9.1)

A conductivity cell consists of two electrodes immersed in an electrolytic solution. It obeys Ohm’s law like metallic conductor. i.e., at a constant temperature, the current flowing through the cell (I) is directly proportional to the voltage across the cell (V). i.e., I α V (or) I = \(\frac{V}{R}\)
⇒ V = IR …………….. (9.1)

Where ‘R’ is the resistance of the solution in ohm (Ω). Here the resistance is the opposition that a cell offers to the flow of electric current through it.

Resistivity (ρ)

Let us consider a conductivity cell in which the electrolytic solution is confied between the two electrodes having cross sectional area (A) and are separated by a distance ‘l’. Like the metallic conductor, the resistance of such an electrolytic solution is also directly proportional to the length (l) and inversely proportional to the cross sectional area (A).

R α \(\frac{l}{A}\)
R = ρ\(\frac{l}{A}\) (9.2)

Where ρ (rho) is called the specific resistance or resistivity, which depends on the nature of the electrolyte.

If \(\frac{l}{A}\) = 1 m^-1, then, ρ = R. Hence the resistivity is defined as the resistance of an electrolyte confined between two electrodes having unit cross sectional area and are separated by a unit distance. The ratio (\(\frac{l}{A}\)) is called the cell constant, Unit of resitivity is ohm metre (Ωm).

Conductivity

It is more convenient to use conductance rather than resistance. The reciprocal of the resistance (\(\frac{1}{R}\)) gives the conductance of an electrolytic solution. The SI unit of conductance is Siemen (S).

C = \(\frac{1}{R}\) ………….. (9.3)

Substitute (R) from (9.2) in (9.3)

⇒ i.e., C = \(\frac{1}{ρ}\).\(\frac{A}{l}\) ………….. (9.4)

The reciprocal of the specific resistance (\(\frac{1}{ρ}\)) is called the specific conductance (or) conductivity. It is represented by the symbol κappa(κ).

Substitute \(\frac{1}{ρ}\) = κ in equation (9.4) and rearranging

⇒ κ = C.(\(\frac{l}{A}\)) ……………. (9.5)

If A = 1m² and l = 1m; then κ = C.

The specific conductance is defined as the conductance of a cube of an electrolytic solution of unit dimensions (Fig 9.2). The SI unit of specific conductance is Sm^-1.

Example

A conductivity cell has two platinum electrodes separated by a distance 1.5 cm and the cross sectional area of each electrode is 4.5 sq cm. Using this cell, the resistance of 0.5 N electrolytic solution was measured as 15 Ω. Find the specific conductance of the solution.

Solution

Molar Conductivity (Λ_b)

Solutions of different concentrations have different number of electrolytic ions in a given volume of solution and hence they have different specific conductance. Therefore a new quantity called molar conductance (Λ_m) was introduced.

Let us imagine a conductivity cell in which the electrodes are separated by 1m and having V m³ of electrolytic solution which contains 1 mole of electrolyte. The conductance of such a system is called the molar conductance (Λ_m)

We have just learnt that the conductance of 1 m³ electrolytic solution is called the specific conductance (κ). Therefore, the conductance of the above mentioned V m³ solution (Λ_m) is given by the following expression.

(Λ_m) = κ × V ……….. (9.6)

Therefore, Volume of the solution containing one mole of solute = \(\frac{1}{M}\)(mol^-1L)

The above relation defines the molar conductance in terms of the specific conductance and the concentration of the electrolyte.

Example

Calculate the molar conductance of 0.025 M aqueous solution of calcium chloride at 25°C. The specific conductance of calcium chloride is 12.04 × 10^-2Sm^-1.

Equivalent Conductance (Λ)

Equivalent conductance is defined as the conductance of ‘V’ m³ of electrolytic solution containing one gram equivalent of electrolyte in a conductivity cell in which the electrodes are one metre apart. The relation between the equivalent conductance and the specific conductance is given below.

………….. (9.9)

Where κ the specific conductance and N is the concentration of the electrolytic solution expressed in normality.

Factors Affecting Electrolytic Conductance

If the interionic attraction between the oppositely charged ions of solutes increases, the conductance will decrease.

Solvent of higher dielectric constant show high conductance in solution. Conductance is inversely proportional to the Viscosity of the medium. i.e., conductivity increases with the decrease in viscosity.

If the temperature of the electrolytic solution increases, conductance also increases. Increase in temperature increases the kinetic energy of the ions and decreases the attractive force between the oppositely charged ions and hence conductivity increases.

Molar conductance of a solution increases with increase in dilution. This is because, for a strong electrolyte, interionic forces of attraction decrease with dilution. For a weak electrolyte, degree of dissociation increases with dilution.

Measurement of Conductivity of Ionic Solutions

We have already learnt to measure the specific resistance of a metallic wire using a metre bridge in your physics practical experiment. We know that it works on the principle of wheatstone bridge. Similarly, the conductivity of an electrolytic solution is determined by using a wheatstone bridge arrangement in which one resistance is replaced by a conductivity cell filled with the electrolytic solution of unknown conductivity.

In the measurement of specific resistance of a metallic wire, a DC power supply is used. Here, if we apply DC current through the conductivity cell, it will lead to the electrolysis of the solution taken in the cell. So, AC current is used for this measurement to prevent electrolysis.

A wheatstone bridge is constituted using known resistances P, Q, a variable resistance S and conductivity cell (Let the resistance of the electrolytic solution taken in it be R) as shown in the figure 9.3. An AC source (550 Hz to 5 KHz) is connected between the junctions A and C.

Connect a suitable detector E (Such as the telephone ear piece detector) between the junctions ‘B’ and ‘D’. The variable resistance ‘S’ is adjusted until the bridge is balanced and in this conditions there is no current flow through the detector.

Under Balanced Condition,

\(\frac{P}{Q}\) = \(\frac{R}{S}\)
∴R = \(\frac{P}{Q}\) × S ………….. (9.10)

The resistance of the electrolytic solution (R) is calculated from the known resistance values P, Q and the measured ‘S’ value under balanced condition using the above expression (9.10).

Conductivity Calculation

Specific conductance (or) conductivity of an electrolyte can be calculated from the resistance value using the following expression.

κ = \(\frac{1}{R}\)(\(\frac{l}{A}\)) [∵equation 9.5]

The value of the cell constant \(\frac{l}{A}\) is usually provided by the cell manufacturer. Alternatively the cell constant may be determined using KCl solution whose concentration and specific conductance are known.

Example

The resistance of a conductivity cell is measured as 190 Ω using 0.1M KCl solution (specific conductance of 0.1M KCl is 1.3 Sm^-1). When the same cell is filled with 0.003M sodium chloride solution, the measured resistance is 6.3KΩ. Both these measurements are made at a particular temperature. Calculate the specific and molar conductance of NaCl solution.

Given that

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Solubility Product

We have come across many precipitation reactions in inorganic qualitative analysis. For example, dil HCl is used to precipitate Pb²⁺ ions as PbCl₂ which is sparingly soluble in water.

Kidney stones are developed over a period of time due to the precipitation of Ca²⁺ (as calcium oxalate etc.). To understand the precipitation, let us consider the solubility equilibria that exist between the undissociated sparingly soluble salt and its constituent ions in solution.

For a general salt X_mY_m

The equilibrium constant for the above is

In solubility equilibria, the equilibrium constant is referred as solubility product constant (or) Solubility product. In such heterogeneous equilibria, the concentration of the solid is a constant and is omitted in the above expression

K_{sp = [Xⁿ⁺]^m[Y^m]ⁿ}

The solubility product of a compound is defined as the product of the molar concentration of the constituent ions, each raised to the power of its stoichiometric co – efficient in a balanced equilibrium equation.

Solubility product finds useful to decide whether an ionic compound gets precipitated when solution that contains the constituent ions are mixed.

When the product of molar concentration of the constituent ions i.e., ionic product, exceeds the solubility product then the compound gets precipitated.

The expression for the solubility product and the ionic product appears to be the same but in the solubility product expression, the molar concentration represents the equilibrium concentration and in ionic product, the initial concentration (or) concentration at a given time ‘t’ is used.

In general we can summarise as,

Ionic product > K_sp, precipitation will occur and the solution is super saturated. Ionic product < K_sp, no precipitation and the solution is unsaturated.

Ionic product = K_sp, equilibrium exist and the solution is saturated.

Determination of Solubility Product from Molar Solubility

Solubility product can be calculated from the molar solubility i.e., the maximum number of moles of solute that can be dissolved in one litre of the solution. For a solute X_mY_n,

X_mY_n(s) ⇄ mXⁿ⁺(aq) + nY^m-(aq)

From the above stoichiometrically balanced equation we have come to know that 1 mole of X_mY_n(s) dissociated to furnish ‘m’ moles of Xⁿ⁺ and ‘n’ moles of Y^m- is ‘s’ is molar solubility of X_mY_n, then

[Xⁿ⁺]=ms and [Y^m+]=ns
∴ K_sp = [Xⁿ⁺]^m[Y^m+]ⁿ

K_sp=(ms)^m(ns)ⁿ
K_sp=(m)^m(n)ⁿ(s)^m+n

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Salt Hydrolysis

When an acid reacts with a base, a salt and water are formed and the reaction is called neutralization. Salts completely dissociate in aqueous solutions to give their constituent ions. The ions so produced are hydrated in water. In certain cases, the cation, anion or both react with water and the reaction is called salt hydrolysis. Hence, salt hydrolysis is the reverse of neutralization reaction.

Salts of Strong Acid and a Strong Base

Let us consider the reaction between NaOH and nitric acid to give sodium nitrate and water.
NaOH(aq) + HNO₃(aq) → NaNO₃(aq) + H₂O(l)

The salt NaNO₃ completely dissociates in water to produce Na⁺ and NO₃^– ions.
NaNO₃(aq) → Na⁺(aq) + NO₃^–(aq)

Water dissociates to a small extent as
H₂O(l) ⇄ H⁺(aq) + OH^–(aq)

Since [H⁺] = [OH^–], water is neutral
NO₃^– ion is the conjugate base of the strong acid HNO₃ and hence it has no tendency to react

with H⁺. Similarly, Na⁺ is the conjugate acid of the strong base NaOH and it has no tendency to react with OH^–.

It means that there is no hydrolysis. In such cases [H⁺] = [OH^–] pH is maintained and, therefore, the solution is neutral.

Hydrolysis of Salt of Strong Base and Weak Acid (Anionic Hydrolysis)

Let us consider the reactions between sodium hydroxide and acetic acid to give sodium acetate and water.
NaOH (aq) + CH₃COOH(aq) ⇄ CH₃COONa(aq) + H₂O(l)

In aqueous solution, CH₃COONa is completely dissociated as below

CH₃COONa (aq) → CH₃COO^–(aq) + Na⁺(aq)

CH₃COO^– is a conjugate base of the weak acid CH₃COOH and it has a tendency to react with H⁺
from water to produce unionised acid.

There is no such tendency for Na⁺ to react with OH^–.

CH₃COO^–(aq) + H₂O(l) ⇄ CH₃COOH(aq) + OH^–(aq) and therefore [OH^–]>[H⁺], in such cases, the solution is basic due to hydrolysis and the pH is greater than 7. Let us find a relation between the equilibrium constant for the hydrolysis reaction (hydrolysis constant) and the dissociation constant of the acid.

K_h value in terms of degree of hydrolysis (h) and the concentration of salt (C) for the equilibrium can be obtained as in the case of ostwald’s dilution law. K_h = h²C and i.e [OH^–] = \(\sqrt{\mathrm{K}_{\mathrm{h}} \cdot \mathrm{C}}\) and i.e [OH^–] = \(\sqrt{\mathrm{K}_{\mathrm{h}} \cdot \mathrm{C}}\).

pH of salt solution in terms of K_a and the concentration of the electrolyte

pH + pOH = 14
pH = 14 – p OH = 14 – {- log [OH^–]}
= 14 + log[OH^–]

Hydrolysis of Salt of Strong Acid and Weak Base (Cationic Hydrolysis)

Let us consider the reactions between a strong acid, HCl, and a weak base, NH₄OH, to produce a salt, NH₄Cl, and water.

HCl (aq) + NH₄OH(aq) ⇄ NH₄Cl(aq) + H₂O(l)
NH₄Cl(aq) → NH₄⁺ + Cl^–(aq)

NH₄⁺ is a strong conjugate acid of the weak base NH₄OH and it has a tendency to react with with OH^– from water to produce unionised NH₄OH shown below.

NH₄⁺(aq) + H₂O(l) ⇄ NH₄OH(aq) + H⁺(aq)

There is no such tendency shown by Cl^– and therefore [H⁺]>[OH^–]; the solution is acidic and the pH
is less than 7.

As discussed in the salt hydrolysis of strong base and weak acid. In this case also, we can establish a relationship between the K_h and k_b as K_h.K_b = K_w

Let us calculate the K_h value in terms of degree of hydrolysis (h) and the concentration of salt

Hydrolysis of Salt of Weak Acid and Weak Base (Anionic & Cationic Hydrolysis)

Let us consider the hydrolysis of ammonium acetate.
CH₃COONH₄(aq) → CH₃COO^–(aq) + NH₄⁺(aq)

In this case, both the cation (NH₄⁺) and anion (CH₃COO^–) have the tendency to
react with water

CH₃COO^– + H₂O ⇄ CH₃COOH + OH^–
NH₄⁺ + H₂O ⇄ NH₄OH + H⁺

The nature of the solution depends on the strength of acid (or) base i.e, if K_a > K_b; then the solution is acidic and pH < 7, if K_a < K_b; then the solution is acidic and pH < 7, if K_a < K_b; then the solution is basic and pH > 7, if K_a = K_b; then the solution is basic and pH > 7, if K_a = K_b; then the solution is neutral.

The relation between the dissociation constant (K_a, K_b) and the hydrolysis constant is given by the following
expression.

K_a.K_b.K_h = K_w

pH of the Solution

pH of the solution can be calculated using the following expression,
pH = 7 + 1/2 pK_a – 1/2 pK_b.

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Buffer Solution

Do you know that our blood maintains a constant pH, irrespective of a number of cellular acid – base reactions. Is it possible to maintain a constant hydronium ion concentration in such reactions? Yes, it is possible due to buffer action.

Buffer is a solution which consists of a mixture of a weak acid and its conjugate base (or) a weak base and its conjugate acid. This buffer solution resists drastic changes in its pH upon addition of a small quantities of acids (or) bases, and this ability is called buffer action. The buffer containing carbonic acid (H₂CO₃) and its conjugate base HCO^–₃ is present in our blood. There are two types of buffer solutions.

1. Acidic Buffer Solution:

A solution containing a weak acid and its salt.
Example: Solution containing acetic acid and sodium acetate

2. Basic Buffer Solution:

A solution containing a weak base and its salt.
Example: Solution containing NH₄OH and NH₄Cl

Buffer Action

To resist changes in its pH on the addition of an acid (or) a base, the buffer solution should contain both acidic as well as basic components so as to neutralize the effect of added acid (or) base and at the same time, these components should not consume each other. Let us explain the buffer action in a solution containing CH₃COOH and CH₃COONa. The dissociation of the buffer components occurs as below.

If an acid is added to this mixture, it will be consumed by the conjugate base CH₃COO^– to form the undissociated weak acid i.e, the increase in the concentration of H⁺ does not reduce the pH significantly.

CH₃COO^–(aq) + H⁺(aq) → CH₃COOH(aq)

If a base is added, it will be neutralized by H₃O⁺, and the acetic acid is dissociated to maintain the equlibrium. Hence the pH is not significantly altered.

These neutralization reactions are identical to those reactions that we have already discussed in common ion effect.

Les us analyse the effect of the addition of 0.01 mol of solid sodium hydroxide to one litre of a buffer solution containing 0.8 M CH₃COOH and 0.8 M CH₃COONa. Assume that the volume change due to the addition of NaOH is negligible. (Given: K₄ for CH₃COOH is 1.8 × 10^-5).

The dissociation constant for CH₃COOH is given by

The above expression shows that the concentration of H⁺ is directly proportional to

Let the degree of dissociation of CH₃COOH be α then,
[CH₃OOH] = 0.8 – α and [CH₃COO^–]
= α + 0.8

Given that

K_afor CH₃COOH is 1.8 × 10^-5
∴ [H⁺] = 1.8 × 10^-5; pH = – log (1.8 × 10^-5)
= 5 – log 1.8
= 5 – 0.26
pH = 4.74

Calculation of pH after adding 0.01 mol NaOH to 1 litre of buffer.

Given that the volume change due to the addition of NaOH is negligible
∴[OH^–] = 1.8 × 10^-5; pH = – log (1.8 × 10^-5)
= 5 – log 1.8
= 5 – 0.26
pH = 4.74

Caluculation of pH after adding 0.01 mol NaOH to 1 litre of buffer.

Given that the volume change due to the addition of NaOH is negligible
∴ [OH^–] = 0.01 M.

The consumption of OH^– are expressed by the following equations.

The addition of a strong base (0.01 M NaOH) increased the pH only slightly ie., from 4.74 to 4.75 . So, the buffer action is verified.

Buffer Capacity and Buffer Index

The buffering ability of a solution can be measured in terms of buffer capacity. Vanslyke introduced a quantity called buffer index, β , as a quantitative measure of the buffer capacity. It is defined as the number of gram equivalents of acid or base added to 1 litre of the buffer solution to change its pH by unity.

Here, β = \(\frac{dB}{d(pH)}\) …………. (8.19)
dB = number of gram equivalents of acid / base added to one litre of buffer solution.
d(pH) = The change in the pH after the addition of acid / base.

Henderson – Hassel Balch Equation

We have already learnt that the concentration of hydronium ion in an acidic buffer solution depends on the ratio of the concentration of the weak acid to the concentration of its conjugate base present in the solution i.e.,

………….. (8.20)

The weak acid is dissociated only to a small extent. Moreover, due to common ion effect, the dissociation is further suppressed and hence the equilibrium concentration of the acid is nearly equal to the initial concentration of the unionised acid. Similarly, the concentration of the conjugate base is nearly equal to the initial concentration of the added salt.

…………… (8.21)

Here [acid] and [salt] represent the initial concentration of the acid and salt, respectively used to prepare the buffer solution. Taking logarithm on both sides of the equation

……………. (8.22)

Reverse the sign on both sides

…………….. (8.23)

We know that

pH = – log [H₃O⁺] and pK_a = – logK_a

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Common ion Effect

When a salt of a weak acid is added to the acid itself, the dissociation of the weak acid is suppressed further. For example, the addition of sodium acetate to acetic acid solution leads to the suppression in the dissociation of acetic acid which is already weakly dissociated. In this case, CH₃COOH and CH₃COONa have the common ion, CH₃COO^–.

Let us analyse why this happens. Acetic acid is a weak acid. It is not completely dissociated in aqueous solution and hence the following equilibrium exists. CH₃COOH(aq) ⇄ H⁺(aq) + CH₃COO^–(aq).

However, the added salt, sodium acetate, completely dissociates to produce Na⁺ and CH₃COO^– ion.

CH₃COONa(aq) → Na⁺(aq) + CH₃COO^–(aq)

Hence, the overall concentration of CH₃COO^– is increased, and the acid dissociation equilibrium is disturbed. We know from Le chatelier’s principle that when a stress is applied to a system at equilibrium, the system adjusts itself to nullify the effect produced by that stress.

So, in order to maintain the equilibrium, the excess CH₃COO^– ions combines with H⁺ ions to produce much more unionized CH₃COOH i.e, the equilibrium will shift towards the left. In other words, the dissociation of a weak acid (CH₃COOH) is suppressed. Thus, the dissociation of a weak acid (CH₃COOH) is suppressed in the presence of a salt (CH₃COONa) containing an ion common to the weak electrolyte. It is called the common ion effect.

The common-ion effect refers to the decrease in solubility of an ionic precipitate by the addition to the solution of a soluble compound with an ion in common with the precipitate. This behaviour is a consequence of Le Chatelier’s principle for the equilibrium reaction of the ionic association/dissociation.

The common ion effect is the phenomenon in which the addition of an ion common to two solutes causes precipitation or reduces ionization. An example of the common ion effect is when sodium chloride (NaCl) is added to a solution of HCl and water.

The common ion effect suppresses the ionization of a weak acid by adding more of an ion that is a product of this equilibrium. The common ion effect suppresses the ionization of a weak base by adding more of an ion that is a product of this equilibrium. The reaction is put out of balance, or equilibrium.

Aquifers which contain chalk and limestone, a common ion effect is used in these to obtain drinking water. Calcium carbonate is sparingly soluble in water, and it can be precipitated out by adding sodium chloride in the solution. In this way, the common ion effect is used in treatment of water.

According to Le Chatelier’s principle, addition of more ions alters the equilibrium and shifts the reaction to favor the solid or deionized form. In the case of an an acidic buffer, the hydrogen ion concentration decreases, and the resulting solution is less acidic than a solution containing the pure weak acid.

Le Chatelier’s principle is an observation about chemical equilibria of reactions. It states that changes in the temperature, pressure, volume, or concentration of a system will result in predictable and opposing changes in the system in order to achieve a new equilibrium state.

At a given temperature, the product of water concentrations and ions is known as the ionic product of water. With the rise in temperature, the value increases i.e. the concentration of H⁺ and OH^– ions increases with temperature increases.

If you have a solution and solute in equilibrium, adding a common ion (an ion that is common with the dissolving solid) decreases the solubility of the solute. This is because Le Chatelier’s principle states the reaction will shift toward the left (toward the reactants) to relieve the stress of the excess product.

pH gives us the measure of acid/base strength of any solution. pH scale ranges from 0-14, 7 is neutral, below 7 it represents acidic nature i.e. the compound is acidic and above 7 it represents basic nature i.e. the compound is basic i.e. pH is the negative logarithm of hydrogen ion.

The solubility of slightly soluble substances can be decreased by the presence of a common ion. This effect is known as common ion effect. Addition of a common ion to a slightly soluble salt solution will add up to the concentration of the common ion.

Le Chatelier’s principle (also known as “Chatelier’s principle” or “The Equilibrium Law”) states that when a system experiences a disturbance (such as concentration, temperature, or pressure changes), it will respond to restore a new equilibrium state.

Explanation:

This phenomenon is called the common-ion effect. When a compound dissolves in water it dissociates into ions. Increasing the concentration of one of these ions will shift the equilibrium towards the compound, thereby making it hard for the compound to dissolve in water (decreases solubility of compound).

When there is an increase in pressure, the equilibrium will shift towards the side of the reaction with fewer moles of gas. When there is a decrease in pressure, the equilibrium will shift towards the side of the reaction with more moles of gas.

The classic example of the practical use of the Le Chatelier principle is the Haber-Bosch process for the synthesis of ammonia, in which a balance between low temperature and high pressure must be found.

Le Chatelier’s Principle helps to predict what effect a change in temperature, concentration or pressure will have on the position of the equilibrium in a chemical reaction. This is very important, particularly in industrial applications, where yields must be accurately predicted and maximised.