Electrochemistry

Electrochemistry(Types of Electrodes, Applications of EMF, Determination of pH)(Faradays law of Electrolysis, Electrolysis of NaCl, Ionic Strength)

Dr.S.SURESHAssistant Professor

Email:[email protected]

Types of Electrodes

Different Types of Electrodes

• An electrochemical cell consists of two electrodes, positive and negative. Each electrode along with the electrolyte constitutes a half cell. The commonly used electrodes in different electrochemical cells are

• Metal-Metal ion electrodes• Metal-Amalgam electrodes• Metal insoluble metal salt electrodes• Gas electrodes• Oxidation-reduction electrodes

Metal-Metal ion electrodes

An electrode of this type consists of a metal rod (M) dipping into a solution of its metal ions(Mn+). This is represented as

M/Mn+

Example:Cu rod dipping in Copper sulphate solution (Cu/Cu2+)

Metal-Metal ion electrodes

The electrode reaction may be represented asMn+ + ne‒ ⇌ M

If the metal rod behaves as negative electrode (i.e the reaction involves oxidation) the equilibrium will shift to the left and hence the concentration of the Metal ions in the solution will increase. On the other hand, if the metal rod behaves as positive electrode (i.e the reaction involves reduction), and the equilibrium will shift to the right and hence the concentration of the metal ions will decrease.

Metal-Amalgam Electrodes

The activities and the electrode potentials of highly reactive metals such as sodium, potassium, etc. are difficult to measure in aqueous solution. Hence, the activity of the metal is lowered by dilution with mercury.

M(Hg)/Mn+

ExampleNa, Hg (C1)/Na+(C2)

Metal-Insoluble metal salt electrodes

• This type of electrode consists of a metal (M) covered by a layer of sparingly soluble salt (MX) dipping into a solution containing a common anion (X‒). These electrodes are represented as

M/MX // X‒ (a)

Example:

• (a)Calomel Electode: It consists of Mercury-Mercurous chloride in contact with a solution of potassium chloride (Hg/Hg2Cl2, KCl)

Gas Electrodes• In a gas electrode hydrogen gas is continuously

bubbled through a 1M solution of the acid. A inert metal like platinum is used, since it is not attacked by acid. The inert metal in electrode does not participate in the electrode reaction but helps in making electrical contact. Let the gas bubbled be X2, then the electrode is denoted as

Pt, X2 (1atm) /X+(a)

Example:Standard hydrogen electrode: Pt, H2 (1atm)/H+(1M)

Oxidation-Reduction electrodes (or) Red-Ox electrodes

• In this type of electrode the potential is developed due to the presence of ions of the same substance in two different valence(i.e. oxidation) states.

• This electrode is set up by inserting an inert metal like platinum in an appropriate solution containing a mixture of ferrous (Fe2+) and Ferric (Fe3+) ions.

• In this electrode the potential is due to the tendency of the ions to change from one oxidation state to the other more stable oxidation state.These electrodes are represented as

Pt/Mn1

+ (a1), M n2

+ (a2)

Applications of EMF measurements


(i) Determination of valency of ion in doubtful cases

• The valency of mercurous ion can be determined by determining the EMF of a concentration cell of the type given below.

Hg/ Hg2(NO3)2 (C1) // Hg2(NO3)2(C2) / Mercury

• The EMF of the cell, E, is given by the expression

E =

• It was found that when C2/C1 was 10, (i.e. C2 = 1M and C1 = 0.1M) the EMF was 0.0295V. Therefore, the valency of mercurous ion is 2, and it should be represented as .

1

2

C

C log

n

0.059

22Hg


(ii) Determination of solubility product of sparingly soluble salt:The solubility product constant of a sparingly soluble salt is a kind of equilibrium constant. Consider the salt MX in equilibrium with its ions in a saturated solution.

MX(s) +⇌

The solubility product of the salt is given by Ksp = [M+] [X‒]

)(aqM

)(aqX

The cell is represented as M, M+ X‒(sat.sol.) // MX(s), M

R.H.E MX(s) + e‒ M + X⇌ ‒

L.H.E M + e⇌ ‒

Overall reaction MX(s) +⇌

E° = We know that ‒ ∆G° = nFE°And ‒∆G° = 2.303 RT log Ksp

log Ksp =

E° =

=

)(aqM

)(aqM

)(aqX

οL

οR E - E

RT 2.303

nFE

spK log nF

RT 2.303

spK log n

0.059

(iii)Free energy, enthalpy and entropy changes in electrochemical reactions:

The standard free energy can be calculated as follows∆G° = ‒nFE°n= number of electrons, F = 96500 C, E° = EMF of the cell.

• By knowing the standard EMF of a cell we can calculate the entropy change using the equation

∆S° = n F

• Then the enthalpy change can be calculated using the equation ∆G° = ∆H° ‒ T∆S°

pT

E

Determination of pH

Determination of pH

By using a glass electrode:

It consists of thin walled glass bulb (made of special type of glass having low melting point and high electrical conductivity) containing a Pt wire in 0.1M HCl. The thin walled glass bulb functions as an ion-exchange resin, and an equilibrium is set up between the sodium ions of glass and hydrogen ions in solution. The potential difference varies with the hydrogen ion concentration, and it given by

pH] 0592.0 [E0G

Determination of pH by using a glass electrode

The cell may be represented as Pt, 0.1M HCl/ Glass// KCl(Satd. Soln.)/ Hg2Cl2(s). Hg

The EMF of the cell is measured byE°Cell =

E°Cell =

= 0.2422 V ‒ = 0.2422 V ‒ pH =

οL

οR E - E

GlassCalomel E - E

pH] 0592.0 [E0G

pH 0592.0 E0G

V 0.0592

E - E - V 0.2422 Cell0G

Faraday’s Law of Electrolysis


First Law: According to it “during electrolysis, the amount of any substance deposited or evolved at any electrode is proportional to the quantity of electricity passed”The quantity of electricity (Q) is equal to the product of the current strength and the time for which it is passed.

Q = current strength × time


If W is the weight of a substance liberated/deposited at an electrode during electrolysis, then from first law, we get:

W α QBut Q = it

W α itW = Zit


Second Law: According to it “the weights of different substance evolved/deposited by the passage of same quantity of electricity are proportional to their chemical equivalent weights

W α EWhere W = Weight of the substance liberated or deposited.E = Chemical equivalent weight of the substance liberated/deposited.

Electrolysis of aqueous sodium chloride

When sodium chloride is dissolved in water, it ionises as

NaCl Na⇌ + + Cl‒

Water also dissociates asH2O H⇌ + + OH‒

When electric current is passed through aqueous sodium chloride solution using platinum electrodes


H+ ions move towards the cathode. The H+ ions gain electrons and change into neutral atoms. Hydrogen atom is unstable and combines with another atom to form stable hydrogen molecule.Hydrogen atom is unstable and combines with another atom to form stable hydrogen molecule

H+ + e‒ H (atom unstable)H + H H2 (stable)


Cl‒ ions move towards anode. These Cl‒ ions lose electrons and change into neutral atoms, chlorine atom is unstable and combines with another atom to form stable chlorine molecule.At Anode:

Cl‒ ‒ e‒ Cl atom (unstable)Cl + Cl Cl2 (stable)

Hence in the electrolysis of aqueous solution of sodium chloride, hydrogen is liberated at cathode while chlorine is liberated at anode.

Electrolysis of aqueous copper sulphate solution

(using pt electrode)

When copper sulphate is dissolved in water, it ionizes as CuSO4 Cu⇌ 2+ + SO4

2‒

H2O H⇌ + + OH‒ (slightly ionized)

When electric current is passed through copper sulphate solution using platinum (Pt) electrodes:(a) Cu2+ ions move towards cathode. These Cu2+ ions gain electrons and change into neutral atoms and get deposited at cathode.

Cu2+ + 2e‒ Cu (deposited)

At CathodeOH‒ ions move towards anode. These OH‒ ions lose electrons and change into neutral hydroxyl groups.At Anode:

2OH‒ ‒ e 2OH (neutral)The neutral hydroxyl groups being unstable react with other neutral OH‒ groups to form water and oxygen.

2OH H2O + O

O + O O2

Hence during electrolysis of copper sulphate solution using platinum electrodes, copper and oxygen are liberated.

Ionic Strength

Ionic strength of solutions

Ionic strength is a measure of the concentration of ions in that solution. Ionic strength may be expressed as

2iiZC

2

1 μ

Ionic strength

Calculate the ionic strength of 0.1M solution of NaCl

For Na+ C = 0.1 and Z = 1For Cl‒ C = 0.1 and Z = 1

2iiZC

2

1 μ

0.1 )11.011.0(2

1 μ 22

Ionic strength

Calculate the ionic strength of 0.1M solution of Na2SO4, MgCl2 and MgSO4 respectively.

Na2SO4

MgCl2

MgSO4

0.3 )2 1.012.0(2

1 μ 22

0.3 )12.021.0(2

1 μ 22

0.4 )21.021.0(2

1 μ 22

Ionic strength

• The ionic strength of a solution containing more than one electrolyte is the sum of the ionic strength of the individual salts comprising the solution.

• Hence, the ionic strength of a solution containing Na2SO4, MgCl2 and MgSO4 each at a concentration of 0.1M is 1.0.

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Transcript of Electrochemistry