Engineering Chemistry - Module 3 - Phase Equilibria PDF

ENGINEERING CHEMISTRY Module 3- Phase equilibria Module content: Phase equilibria Gibb’s phase rule Phase diagram of 1-component system Phase diagram of 2-component system Fe-C phase diagram ENGINEERING CHEMISTRY Module 3- Phase equilibria Class content: Free energy in Phase equilibria Chemical potential Phase equilibria Phase Component Degree of freedom ENGINEERING CHEMISTRY Module 3- Phase equilibria Free energy in Phase equilibria All substances have tendency to minimize their Gibbs energy at constant temperature and pressure to attain stable state Phase transformations from one phase to another occur to reduce free energy of the system Gibb’s energy is an extensive property Chemical potential Chemical potential is defined as the partial molar Gibb’s energy for a component i in a mixture , and is denoted by μ G  i  ( ) p ,T ,n j i ni ENGINEERING CHEMISTRY Module 3- Phase equilibria Phase equilibria Phase equilibria between phases exist when chemical potential of a component is equal in all the phases in equilibrium e.g. for water at triple point Solid ⇌ liquid ⇌ vapour The chemical potential of water will be equal in all the three phases For systems not at equilibrium, the chemical potential will point to the direction in which the system can move in order to achieve equilibrium viz. , the system moves from higher chemical potential to lower chemical potential When various phases are in equilibrium with one another in a heterogeneous system, there can be no transfer of energy or mass from one phase to another. For a system at equilibrium, the various phases must have the same temperature and pressure and their respective compositions must remain constant all along ENGINEERING CHEMISTRY Module 3- Phase equilibria Phase A phase is defined as any homogeneous and physically distinct part of a system bounded by a surface and is mechanically separable from other parts of the system. It is denoted by P Gaseous state : P =1 gases are completely miscible Liquid state : P = No. of layers when liquids are immiscible P = 1 when liquids are completely miscible Solid state : Each solid constitutes a separate phase Each polymorphic form constitutes a separate phase P = 1 for solid solution ENGINEERING CHEMISTRY Module 3- Phase equilibria Counting the number of phases 1) Solid ⇌ liquid ⇌ vapour ; P = 3 Ice in the system is a single phase even if it is present as a number of pieces. 2) Calcium carbonate undergoes thermal decomposition CaCO3(s) ⇌ CaO(s) + CO2 (g) P= 3; 2 solid phases, CaCO3 and CaO and one gaseous phase, that of CO2 3) Ammonium chloride undergoes thermal decomposition NH4Cl(s) ⇌ NH3 (g) + HCl (g) P = 2; one solid, NH4Cl and one gaseous, a mixture of NH3 and HCl ENGINEERING CHEMISTRY Module 3- Phase equilibria Components A component is defined as the smallest number of independently varying chemical constituents using which the composition of each and every phase in the system can be expressed When no reaction is taking place in a system, the number of components is the same as the number of constituents While expressing in terms of constituents zero and negative values are allowed ENGINEERING CHEMISTRY Module 3- Phase equilibria Counting the number of components: 1-component system All the different e.g. Pure water ; C = 1 phases can be expressed in terms of solid water ⇌ liquid water the single constituent water 2-component system Salt hydrate system e.g. Na2SO4.10H2O ; C=2 Na2SO4.10H2O = Na2SO4 + 10H2O Na2SO4.7H2O = Na2SO4 + 7H2O The composition of all the Na2SO4 = Na2SO4 + 0H2O phases can be expressed in terms of 2 components H2O = 0 Na2SO4 + H2O ENGINEERING CHEMISTRY Module 3- Phase equilibria Thermal decomposition of solid CaCO3 in a closed container ; C = 2 CaCO3 (s) ⇌ CaO(s) + CO2(g) Though there are 3 species present, the number of components is only two Phases are : CaCO3(s), CaO(s) and CO2(g) Any two of the three constituents may be chosen as the components If CaO and CO2 are chosen, CaCO3(s) = CaO + CO2 CaO(s) = CaO + 0 CO2 CO2(g) = 0 CaO + CO2 If CaCO3 and CO2 are chosen, CaCO3(s) = CaCO3 + 0 CO2 CaO(s) = CaCO3 - CO2 CO2(g) = 0 CaCO3 + CO2 ENGINEERING CHEMISTRY Module 3- Phase equilibria Thermal decomposition of ammonium chloride in a closed system ;C = 1 NH4Cl (s) ⇌ NH3(g) + HCl(g) Phases are : solid phase of NH4Cl(s)and gaseous phase of NH3(g) and HCl(g) solid : NH4Cl (s) = NH4Cl gas : NH3(g) +HCl(g) = NH4Cl The composition of both the solid and gaseous phase can be expressed in terms of NH4Cl. Hence the number of components is one; C=1 If additional HCl (or NH3) were added to the system, then C = 2 The decomposition of NH4Cl would not give the correct composition of the gas phase A second component, HCl (or NH3) would be needed to describe the gas phase, therefore C=2 ENGINEERING CHEMISTRY Module 3- Phase equilibria Degrees of freedom (or variance) The degrees of freedom or variance of a system is defined as the minimum number of intensive variables such as temperature, pressure, concentration, which must be fixed in order to define the system completely; it is denoted by F OR The degree of freedom of a system may also be defined as the number of variables, such as temperature, pressure and concentration that can be varied independently without altering the number of phases. Example : water system Only 1 phase (solid , liquid or gas) Both temperature and pressure need to be mentioned in order to define the system; F = 2 ENGINEERING CHEMISTRY Module 3- Phase equilibria 2 phases in equilibrium, Only one variable, either temperature or pressure need to be specified in order to define the system; F =1 solid water ⇌ liquid water If the pressure on the system is maintained at 1 atm, then the temperature of the system gets automatically fixed at 0oC, the normal melting point of ice 3 phases in equilibrium, No variable can be changed temperature and pressure are fixed, F = 0 solid water ⇌ liquid water ⇌ water vapour Three phases, ice, water, vapour can coexist in equilibrium at triple point of water at 0.0098oC and 4.6mm of Hg pressure only ENGINEERING CHEMISTRY Module 3- Phase equilibria Class content: Gibb’s Phase rule Derivation of Gibb’s Phase rule Phase diagram of a 1-component system ENGINEERING CHEMISTRY Module 3- Phase equilibria Phase rule It was given by Williams Gibbs in 1874 Statement of Gibb’s phase rule Provided the equilibrium in a heterogeneous system is not influenced by external forces (gravity, electrical or magnetic forces ) , the number of degrees of freedom (F) of the system is related to number of components (C) and number of phases (P) existing at equilibrium to one another by the equation F=C–P+2 ENGINEERING CHEMISTRY Module 3- Phase equilibria Derivation of the phase rule A system at equilibrium satisfies the following conditions: Thermal equilibrium – Temperature is constant Mechanical equilibrium – Pressure is constant Chemical or material equilibrium – Chemical potential of a substance is same in all the phases Mathematically, 𝝁𝒊𝜶 = 𝝁𝒊𝜷 = 𝝁𝒊γ = …. The system considered is: All C components distributed between P phases ENGINEERING CHEMISTRY Module 3- Phase equilibria Total number of intensive variables that need to be ascertained to describe the system: Temperature 1 Pressure 1 Composition mole fraction of each component in every phase For each phase, the sum of mole fractions equals unity 𝝌𝟏𝜶 + 𝝌𝟐𝜶+ 𝝌𝟑𝜶+ ….. + 𝝌𝒄𝜶= 1 (C-1) 𝝌𝟏𝜷 + 𝝌𝟐𝜷+ 𝝌𝟑𝜷+ ….. + 𝝌𝒄𝜷= 1 (C-1) 𝝌𝟏𝜸 + 𝝌𝟐𝜸+ 𝝌𝟑𝜸+ ….. + 𝝌𝒄𝜸= 1 (C-1)... 𝝌𝟏𝑷 + 𝝌𝟐𝑷+ 𝝌𝟑𝑷+ ….. + 𝝌𝒄𝑷= 1 (C-1) In each phase (C-1) mole fraction terms need to be defined Number of phases : P Number of composition variables = P(C-1) Total number of intensive variables = P(C-1) + 2 ENGINEERING CHEMISTRY Module 3- Phase equilibria Total number of equations(constraints) : At equilibrium the chemical potential of particular component is same in every phase in a system µ1α = µ1β = µ1γ =.... (P-1) µ2α = µ2β = µ2γ =.... (P-1) µ3α = µ3β = µ3γ =.... (P-1)... µcα = µcβ = µcγ =.... (P-1) For C components C(P-1) Total number of equations or constraints =C(P-1) F = Total number of variables – total number of equations F = P(C-1) + 2 – {C(P-1)} F=C-P+2 which is the Gibb’s phase rule ENGINEERING CHEMISTRY Module 3- Phase equilibria Application of Gibb’s phase rule F = C-P+ 2 to 1-component system: Water system When only 1 phase is present : C = 1, P = 1; F = 2 ;Temperature and Pressure can be varied independently Bivariant system When 2 phases are in equilibrium: C = 1, P = 2; F = 1 ; Temperature or Pressure can be varied independently Univariant system When all 3 phases are in equilibrium: C = 1, P = 3; F = 0 ; Neither Temperature nor Pressure can be varied Invariant system ENGINEERING CHEMISTRY Module 3- Phase equilibria Phase diagram A diagram which represents the conditions under which a substance exists in different phases in a system Phase diagram of a 1-component system F = C- P + 2 For a 1-component system F = 3 – P Single phase : F = 2 ; Area in a diagram Two phases in equilibrium : F = 1 ; line in a diagram Three phases in equilibrium : F = 0 ; point in a diagram ENGINEERING CHEMISTRY Module 3- Phase equilibria High pressure, low temperature: solid phase A R High temperature, low pressure : vapour phase E In between : liquid phase A L solid ⇌ liquid I liquid ⇌ vapour N vapour ⇌ solid E POINT Solid ⇌ liquid ⇌ vapour Source:http://abyss.uoregon.edu/~js/glossary/triple_point.html ENGINEERING CHEMISTRY Module 3- Phase equilibria Class content: Phase diagram of water Reduced phase rule for a 2-component system ENGINEERING CHEMISTRY Module 3- Phase equilibria Phase diagram of water OC : Melting point curve OA : Vaporisation curve OB : Sublimation curve O: Triple Point A: Critical point OA’: Metastable equilibrium Source: https://imbooz.com/engineering-chemistry/phase-diagram-for-water-system/ ENGINEERING CHEMISTRY Module 3- Phase equilibria Equilibrium between solid and liquid (fusion curve OC) ice ⇌ water F=1, monovariant system variation of melting point of ice with pressure slope is negative; as ice melts its volume decreases or density increases Clausius-Clapeyron equation dp H Fusion = = negative dT TV Where, V = decrease in volume as ice melts is -ve; Hfusion = endothermic,+ve ENGINEERING CHEMISTRY Module 3- Phase equilibria Equilibrium between solid and vapour (sublimation curve OB) ice ⇌ water vapour F=1, monovariant system variation of sublimation temperature of ice with pressure slope is positive Clausius Clapeyron equation dp H sub = = positive dT TV Where , V = Increase in volume as ice sublimates ,+ve ;𝛥𝐻𝑠𝑢𝑏𝑙𝑖𝑚𝑎𝑡𝑖𝑜𝑛 = endothermic reaction, +𝑣𝑒 ENGINEERING CHEMISTRY Module 3- Phase equilibria Equilibrium between liquid and vapour (vaporization curve OA) liquid water ⇌ vapour F=1, monovariant system variation of boiling temperature of water with pressure slope is positive Clausius - Clapeyron equation dp H Vapourisation = = positive dT TV Where, V = Increase in volume as liquid water vapourises,+ve;𝛥𝐻𝑉𝑎𝑝𝑜𝑢𝑟𝑖𝑠𝑎𝑡𝑖𝑜𝑛 = endothermic reaction, +𝑣𝑒 ENGINEERING CHEMISTRY Module 3- Phase equilibria Triple point “O”: Represents equilibrium between liquid, vapour and solid water (ice) All three phases are present together F = 0, invariant system Triple point for water lies at 0.0098 0C and 4.58 mmHg Critical point “A”: the interface between liquid water and water vapour vanishes a point above which water does not exist in liquid state Critical point lies at 3740C and 220 atm pressure ENGINEERING CHEMISTRY Module 3- Phase equilibria Metastable equilibrium (OA’): Ice fails to form at the triple point and water continues to exist in liquid phase The vapour pressure of the liquid continues along OA’ This is called super cooled water and represents metastable equilibrium involving liquid and vapour phases. Any disturbance will cause the system to go back to stable equlilibrium (OB) The vapour pressure of the system in the metastable region is more than that of the stable system ice at the same temperature ENGINEERING CHEMISTRY Module 3- Phase equilibria Phase diagram of a two component system The phase rule equation is F = C – P + 2 For a 2 component system the ordinary phase rule cannot be used For a two component system, C = 2 then F = 2 – P + 2 = 4 – P The minimum number of phase is 1; F = 4 – 1 = 3 This requires 3 dimensional space which cannot be explained on the plane of paper One of the three variables is kept constant Measurements in these systems are generally carried out at atmospheric pressure Pressure may be considered constant ; degrees of freedom is reduced by 1, F = C – P + 2 -1 ENGINEERING CHEMISTRY Module 3- Phase equilibria The phase rule takes the form F = C – P + 1 and is known as the reduced phase rule Equilibria such as solid-liquid equilibria are such systems in which the gas phase is absent and hence are hardly affected by small changes in pressure Systems in which the gas phase is absent are called condensed systems F = C – P + 1 is also known as condensed phase rule ENGINEERING CHEMISTRY Module 3- Phase equilibria Class content: Phase diagram of Pb-Ag system Determination of solid-liquid equilibria Pattinson’s process ENGINEERING CHEMISTRY Module 3- Phase equilibria Phase diagram of Pb-Ag system Phase rule for a 2- component system : F = C-P+1 Pressure is constant Plot is between Temperature and Composition Pb and Ag are miscible in all proportions in the liquid (molten) state In solid state they are completely immiscible T T Pure Ag Pure Pb Composition ENGINEERING CHEMISTRY Module 3- Phase equilibria Simple eutectic system When a mixture of the 2 components is heated till the mixture melts and then cooled: First crystal of A if formed when f.pt. of A in the mixture is reached On further cooling more solid A precipitates and liquid melt becomes richer in B With decrease in temperature more and more solid A separates and the liquid melt moves along curve ME Finally when temperature reaches F, solid B also starts precipitating Three phases are present at E: solid A, solid B and the liquid melt Further cooling will just result in cooling of solid A & B Source:https://in.pinterest.com/pin/818107088531422918/ ENGINEERING CHEMISTRY Module 3- Phase equilibria Simple eutectic : Pb-Ag system Pb-Ag system A : 100%Ag, M : 961oC B : 100%Pb, N : 327oC Area above MEN: Liquid melt Area MEF : Solid A + Liquid melt Area NEG : Solid B + Liquid melt Below FEG : Solid A + Solid B Eutectic point – “E” Curves : ME and NE Source:https://in.pinterest.com/pin/818107088531422918/ ENGINEERING CHEMISTRY Module 3- Phase equilibria Eutectic mixture: A mixture of two components which has the lowest freezing point of all the possible mixtures of the components It has a definite composition and a sharp melting point Number of phases at eutectic point = 3 F = C-P+1; C =2, P = 3,so F = 0; invariant point, Eutectic temperature: 303oC Eutectic composition: 97.4 % Pb and 2.6 % Ag ENGINEERING CHEMISTRY Module 3- Phase equilibria CURVE ME : freezing point curve of Ag It shows decrease in freezing point / melting point of Ag due to the addition of Pb to Ag Solid Ag is in equilibrium with liquid melt of Pb in Ag Here C = 2 and P = 2, then the reduced phase rule is F = C – P + 1 = 2 – 2 +1 = 1 Hence the system is univariant. CURVE NE: freezing point curve of Pb It shows decrease in freezing point of Pb due to the addition of Ag to Pb Solid Pb is in equilibrium with liquid melt of Ag in Pb Here C = 2 and P = 2, then the reduced phase rule is F = C – P + 1 = 2 – 2 +1 = 1 Hence the system is univariant ENGINEERING CHEMISTRY Module 3- Phase equilibria Determination of solid-liquid equilibria For the determination of equilibrium conditions between solid and liquid phases – Thermal Analysis Thermal analysis: The study of the cooling curves of various compositions of a system during solidification Cooling curves: Temperature versus time Freezing point and eutectic point can be determined from the cooling curves ENGINEERING CHEMISTRY Module 3- Phase equilibria Construction of phase diagram using cooling curves Source:http://www.mchmultimedia.co m/PhysicalChemistry- help/clientstories/study-tips/digging- into-phase-diagrams-cooling- curves.html For pure solid: When the freezing point is For a mixture of solids: When reached, temperature remains constant crystallisation of one of the until the liquid is fully solidified. components starts ,cooling curve exhibits a break. The temperature decreases continuously until the eutectic point is reached. Now the temperature remains constant, till the completion of solidification. ENGINEERING CHEMISTRY Module 3- Phase equilibria Pattinson’s process for the desilverisation of argentiferous lead The process of heating argentiferous lead containing a very small quantity of silver (~0.1 %) and cooling it to get pure lead and liquid richer in silver Argentiferous lead is heated to a temperature above the melting point of pure lead The melt is allowed to cool Temperature of the melt reaches the freezing curve of Pb where solid lead starts separating As the system further cools, more and more lead separates and the liquid in equilibrium with the solid lead gets richer in silver ENGINEERING CHEMISTRY Module 3- Phase equilibria The lead that separates, floats and is continuously removed by ladles When the temperature of the liquid reaches the eutectic temperature, solid lead is in equilibrium with the liquid having the eutectic composition After removing the lead that separates, the liquid is cooled further when it solidifies to give a mixture of lead and silver having the eutectic composition of 2.6 % of silver This solid mixture of lead and silver is subjected to other processes for the recovery of silver ENGINEERING CHEMISTRY Module 3- Phase equilibria Class content: Fe-C phase diagram ENGINEERING CHEMISTRY Module 3- Phase equilibria Iron– Carbon (Fe–C ) Phase Diagram Fe-C system is a 2-component system; F = C – P + 1 the Fe-C phase diagram is a fairly complex one we consider the part of the diagram, up to around 6.7% Carbon α- ferrite : 0oC – 900oC γ- austenite : 900oC – 1400oC δ- ferrite : 1400oC – 1540oC Beyond 1540oC – Fe melts Cementite(Fe3C) : 6.7% C ENGINEERING CHEMISTRY Module 3- Phase equilibria Phases in Fe–Fe3C Phase Diagram α-ferrite solid solution of C in BCC Fe Stable form of iron at room temperature Transforms to FCC γ-austenite at 900 °C γ-austenite solid solution of C in FCC Fe Transforms to BCC δ-ferrite at 1400 °C δ-ferrite solid solution of C in BCC Fe The same structure as α-ferrite Stable only at high T, above 1400 °C Melts at 1540 °C Fe3C (iron carbide or cementite) This is a intermetallic compound. Fe-C liquid solution ENGINEERING CHEMISTRY Module 3- Phase equilibria Areas in the Fe–Fe3C phase diagram 1 : liquid melt 2 : γ- austenite + liquid melt 3 : cementite + liquid melt 4 : γ- austenite + cementite 5 : α- ferrite + γ- austenite 6 : α- ferrite + cementite 7 : δ- ferrite + liquid melt 8 : δ- ferrite + γ- austenite ENGINEERING CHEMISTRY Module 3- Phase equilibria Eutectic point A : liquid melt of Fe-C transforms into two different solid phases γ-austenite and cementite (Fe3C) on cooling; L ↔ γ + Fe3C corresponds to 4.3 % C, 1130 °C 3 phases are in equilibrium ; γ-austenite, cementite and liquid melt F=0. The system is invariant Eutectoid point B : γ-austenite phase transforms into two different solid phases α-ferrite and cementite (Fe3C) on cooling ; γ ↔ α + Fe3C corresponds to 0.8 % C, 723 °C 3 phases are in equilibrium; γ-austenite, α – ferrite and cementite F=0. The system is invariant ENGINEERING CHEMISTRY Module 3- Phase equilibria Peritectic point C: liquid melt of Fe-C transforms into two different solid phases γ-austenite and cementite (Fe3C) on cooling: δ + L ↔ γ corresponds to 0.16 % C, 1498 °C 3 phases are in equilibrium ; γ-austenite, δ- ferrite and liquid melt F=0. The system is invariant ENGINEERING CHEMISTRY Module 3- Phase equilibria Three types of ferrous alloys of Fe and Carbon Wrought Iron: less than 0.008 % C Steel: 0.008 - 2.14 % C Cast iron: 2.14 - 6.7 wt % C ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Module content: Electrode potential and cell potential Nernst Equation Types of electrodes Reference electrodes Concentration cells Ion-selective electrodes ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Class content: Origin of electrode potential Cell potential Nernst Equation ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Electrochemistry Deals with the inter conversion of chemical energy and electrical energy CHEMICAL ENERGY ELECTRICAL ENERGY Two types of cells: Electrolytic Galvanic cell cell Converts chemical Converts electrical energy to electrical energy to chemical energy energy Batteries and fuel cells Cells used in electroplating ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Electrochemical studies: Redox reaction Electrodes- Anode (oxidation) Cathode (reduction) Electrolytic conductance through electrolyte due to movement of ions Acid, alkali or salt solutions Molten electrolytes Solid electrolytes Electrode potential When a metal rod is dipped in a solution of its own ions, the electrical potential developed at the interface of the metal and its solution It is denoted by E ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Origin of Electrode potential When a metal M is in contact with solution containing its ions Mn+, two reactions are possible: 1. Ionisation (Oxidation) M ⇌ Mn+ + ne- 2. Deposition (Reduction) Mn+ + ne- ⇌ M ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Case I : If ionization is faster than deposition the metal acquires net negative charge, consequently retards the rate of ionization and increases the rate of deposition. This ultimately lead to the establishment of equilibrium the metal electrode gets negatively charged and attracts the layer of positive ions at the interface an electrical double layer is formed at the interface of metal and solution known as Helmholtz electrical double layer M ⇌ Mn+ + ne- ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Case II : deposition is faster than ionization the metal acquires net positive charge, consequently retards the rate of deposition and increases rate of ionization. This ultimately lead to the establishment of equilibrium. The metal electrode gets positively charged and attracts the layer of negative ions at the interface, an electrical double layer is formed at the interface of metal and solution known as Helmholtz electrical double layer Mn+ + ne- ⇌ M ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Standard electrode potential The potential developed at the interface of metal and solution, when the metal is in contact with a solution of its own ions having unit concentration at 298 K In case of gas electrodes the partial pressure of gas is maintained at 1 atmospheric pressure. It is represented as Eo Electrochemical Cell Single electrode potentials cannot be measured hence two electrodes are coupled together to form a cell Cell notation e.g. Daniel cell: Zn(s)/Zn2+ (1M)//Cu2+ (1M) / Cu(s) ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Cell potential The difference in electrode potentials of the electrodes constituting the cell It is denoted by Ecell Standard cell potential Ecell depends on concentration of the ions in the cell, temperature and the partial pressures of any gases involved in the cell reaction. When all the concentrations are 1M, all partial pressures of gases are 1atm and temperature is 298K, the emf is called Standard cell potential, Eocell Calculation of Ecell Ecell = Erhs – Elhs = Ecathode – E anode Ecell represents the driving force for the cell reaction to take place ∆ G = - nFECELL If reaction is spontaneous ∆ G is negative, thus ECELL should be positive If reaction is non spontaneous ∆ G is positive, thus ECELL should be negative ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Electrochemical Series: In order to predict the electrochemical behavior of an electrode – electrolyte system, elements are arranged in the order of their standard reduction potentials. This arrangement is known as electrochemical series. A negative value indicates oxidation tendency while a positive value indicates a reduction tendency. Source:https://www.syedgilanis.co m/2019/04/electrochemicalseries. html ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Nernst equation for a single electrode A quantitative relationship between electrode potential and concentration of species with which the electrode is reversible The reaction at the electrode is Mn+ + ne- ⇌ M The maximum work that can be obtained is -∆G = Wmax For an electrochemical system, maximum work done is Wmax = Total charge available × Energy available per unit charge Total charge available , i.e.,No. of moles of electrons exchanged in redox reaction (n), multiplied by charge carried per mole of electrons ,F(96,500 C/mol) = nF Energy available per unit charge, i.e., electrode potential because electrode potential = energy/unit charge = E Therefore , Wmax = nFE ; ∆ G = - nFE Under standard conditions, ∆ Go = - nFEo ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Mn+ + ne- ⇌ M A thermodynamic equation which relates reaction quotient and decrease in free energy is given by, ∆ G = ∆Go + RTlnQ , where Q is the reaction quotient The reaction quotient for the reaction is, Q= [M]/[Mn+] Substituting for ∆ G, ∆Go and Q , we get Where , Eo = Standard electrode potential, n = number of electrons exchanged in the redox reaction, R = Gas constant. 8.314 JK-1 mol-1 , T = temp in Kelvin, F = Faraday 96500 C mol-1 dividing throughout by –nF, ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria since [M] = 1 for pure substances, at 298K, Nernst equation may also be used to calculate. emf of electrochemical cells. For the cell reaction aA + bB ⇌ cC + dD [𝐂]𝐜 [𝐃]𝐝 Q= [𝐀]𝐚 [𝐁 ]𝐛 Nernst equation is n= no. of e -s transferred, Eocell = std. emf of the cell ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Class content: Types of electrodes Metal-metal-ion electrode Metal-insoluble salt –ion electrode Gas electrode Amalgam electrode Redox electrode Ion selective electrode ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Types of electrodes In order to form a cell, 2 half cells or 2 electrodes are required Various types of electrodes are available which are constructed based on the application 1. Metal-metal ion electrode: Metal in contact with a solution of its own ions e.g., Zn/Zn2+, Cu/Cu2+, Ag/Ag+ Mn+ + ne- ⇌ M Nernst equation Source:http://www.valgetal.com/physics/ Batteries/batteries.htm ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria 2. Metal-Metal insoluble salt- ion electrode: These electrodes consist of a metal in contact with a sparingly soluble salt of the same metal dipped in a solution of soluble salt of the same anion e.g.,Calomel electrode Hg/Hg2Cl2/KCl, Ag/AgCl(s)/HCl For silver –silver chloride electrode AgCl + e- ⇌ Ag + Cl- Nernst equation: 0.0591 𝐸𝐴𝑔/𝐴𝑔𝐶𝑙/𝐶𝑙− = 𝐸𝑜𝐴𝑔/𝐴𝑔𝐶𝑙/𝐶𝑙− − log[Cl-] 1 Source:https://www.corrosion-doctors.org/Corrosion- Thermodynamics/Reference-Half-Cells-Silver.htm ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria 3. Gas electrode: It consists of gas bubbling about an inert metal foil, immersed in solution containing ions to which the gas is reversible. The metal provides electrical contact and facilitates the establishment of equilibrium between the gas and its ions e.g.,Hydrogen electrode Pt/H2/H+,Chlorine electrode Pt/Cl2/Cl- For a hydrogen electrode 2H+ + 2e- ⇌ H2 Nernst equation: 0.0591 𝑝𝐻 𝐸𝑃𝑡 𝐻2 𝐻+ = 𝐸 0 𝑃𝑡 𝐻2 𝐻+ − log( 2 ) 2 𝐻+ 2 Source:https://thefactfactor.com/facts/pure_science/chemistry/physical- chemistry/reference-electrodes/5844/ ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria 4. Amalgam electrode: It is similar to metal- metal ion electrode in which metal amalgam is in contact with a solution containing its own ions e.g., Lead amalgam electrode Pb-Hg/Pb2+ For lead amalgam electrode Pb2+ + 2e- ⇌ Pb-Hg Nernst equation: 0.0591 𝑃𝑏 − 𝐻𝑔 𝐸𝑃𝑏2+ /𝑃𝑏−𝐻𝑔 = 𝐸 0 𝑃𝑏2+ /𝑃𝑏−𝐻𝑔 − log( ) 2 [𝑃𝑏2+ ] Source:https://www.semanticscholar.org/paper/Potentiometric-Titration-of- Sulfate-in-Water-and-a-Robbins- Carter/c823ab0578481e876975ee707a5f8adca14c512f ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria 5. Oxidation - reduction electrode : It consists of an inert metal such as platinum immersed in a solution containing an appropriate oxidized and reduced form of redox system. The metal merely acts as electrical contact. The potential arises due to the tendency of one form to change in to other form. e.g.,Pt/Fe2+,Fe3+, Pt/Ce3+,Ce4+ , Pt/Sn2+,Sn4+ For stannous stannic electrode Sn4+ + 2e- ⇌ Sn2+ Nernst equation: 0.0591 [𝑆𝑛2+ ] 𝐸𝑃𝑡/𝑆𝑛4+ /𝑆𝑛2+ = 𝐸0 𝑃𝑡/𝑆𝑛4+ /𝑆𝑛2+ − log( [𝑆𝑛4+ ]) 2 Source:https://slideplayer.com/slide /13860805/ ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Quinhydrone electrode It consists of an inert metal such as platinum immersed in a solution containing quinone and hydroquinone The metal merely acts as electrical contact The potential arises due to the tendency of quinone to change to hydroquinone Pt/Q,QH2 O OH + - + 2H + 2e O OH Nernst equation: 𝐸𝑃𝑡/𝑄/𝑄𝐻2 = 𝐸 𝑜 𝑃𝑡/𝑄/𝑄𝐻2 − 0.0591 log( [𝑄𝐻2 ] ) 2 𝑄 𝐻+ 2 ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria 6. Ion selective electrode:(membrane electrode) It consists of a membrane in contact with a solution, with which it can exchange ions. e.g., glass electrode: selective to H+, Na+, K+ etc. Equation for determining potential for pH sensitive Glass electrode EG = E0G + 0.0591 log10[H+] Source:Analytical Chemistry 2.0, David Harvey, community.asdlib.org/activele...line-textbook/ ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Class content: Reference electrodes Primary reference electrode Standard Hydrogen electrode Secondary reference electrodes Calomel electrode Silver – silver chloride electrode ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Reference electrodes Electrodes whose potentials are accurately known, stable and with reference to which the electrode potential of any electrode can be measured Reference electrode is combined with indicator electrode and emf of the cell is measured Two types of reference electrodes: Primary reference electrodes Standard Hydrogen electrode(SHE) Secondary reference electrodes Calomel electrode Silver-silver chloride electrode ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Primary reference electrode: Standard hydrogen electrode Electrode potential is assigned a value of 0.0 V Gas electrode Pt/H2/H+ 2H+ + 2e- ⇌ H2 Used to measure potential of other electrodes e.g., Zn/Zn2+//H+(1M)/H2(1atm),Pt Ecell = Erhs – Elhs = Ecathode – E anode 0.76 = 0.0 – E Zn/Zn2+ E Zn/Zn2+ = - 0.76 V Pt ,H2(1atm) / H+(1M)// Cu2+/ Cu Ecell = Erhs – Elhs = Ecathode – E anode 0.34 = E Cu/Cu2+ - 0.0 E Cu/Cu2+ = 0.34 V ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Disadvantages of SHE: Maintaining concentration of H+ ions at 1M and pressure of H2 gas at 1 atm is difficult. Platinum is highly susceptible to poisoning by different impurities in gas It cannot be used with oxidizing and reducing environment Secondary reference electrodes: Due to the limitations of standard hydrogen electrode some other electrodes whose electrode potentials are accurately known and remain stable over a long period of time and can be easily assembled. With respect to these electrodes , electrode potentials of other electrodes can be assigned e.g.,Calomel electrode, silver silver chloride electrode ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Secondary Reference electrodes: Calomel electrode Most widely used reference electrode Metal-insoluble salt –ion electrode Construction: A glass tube containing a layer of mercury over which a paste of insoluble salt Hg2Cl2 (calomel) + Hg and the next layer is a solution of KCl A Pt wire dipped in Hg provides electrical https://doubtnut.com/question-answer- chemistry/describe-the-construction-and- contact working-of-the-calomel-electrode- 96607395 Tube is fitted with a side tube to fill KCl solution of known concentration and another side tube which connects to the salt bridge ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Hg/Hg2Cl2(s)/Cl- Working : Can act as anode or cathode depending on the nature of the electrode with which it is coupled As anode: 2Hg ⇌ Hg22+ + 2e- Hg22+ + 2 Cl- ⇌ Hg2Cl2 2Hg + 2 Cl- ⇌ Hg2Cl2 + 2e- As cathode: Hg22+ + 2e- ⇌ 2Hg Hg2Cl2 ⇌ Hg22+ + 2Cl- Hg2Cl2 + 2e- ⇌ 2Hg + 2 Cl- ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Applying Nernst’s equation Hg2Cl2 + 2e- ⇌ 2Hg + 2 Cl- E = Eo - 2.303RT/2F log [Cl-]2 at 298K E = Eo - 0.0591 log [Cl-] Electrode is reversible to chloride ions Electrode potential depends on chloride ion concentration Types of calomel electrodes: [KCl] Name Electrode potential at 298K 0.1M Decinormal electrode 0.3358 V 1M Normal electrode 0.2824 V Saturated solution of KCl Saturated Calomel 0.2422 V Electrode(SCE) ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Silver-silver chloride electrode Widely used as reference electrode Metal-insoluble salt –ion electrode Construction: It has a silver wire or a silver coated platinum wire, coated electrolytically with a thin layer of silver chloride which is dipped in a solution of KCl or HCl of known concentration Source:https://www.corrosion-doctors.org/Corrosion- Ag/AgCl/Cl- Thermodynamics/Reference-Half-Cells-Silver.htm ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Working : Can act as anode or cathode depending on the nature of the electrode with which it is coupled As anode: Ag ⇌ Ag+ + e- Ag+ + Cl- ⇌ AgCl Ag + Cl- ⇌ AgCl + e- As cathode: Ag+ + e- ⇌Ag AgCl ⇌ Ag+ + Cl- AgCl + e- ⇌ Ag + Cl- ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Applying Nernst’s equation AgCl + e- ⇌ Ag + Cl- E = Eo - 2.303RT/F log [Cl-] at 298K E = Eo - 0.0591 log [Cl-] Electrode is reversible to chloride ions Electrode potential depends on chloride ion concentration Types of silver - silver chloride electrodes : [KCl] Name Electrode potential at 298K 0.1N Decinormal electrode 0.289 V 1N Normal electrode 0.223 V Saturated solution of KCl Saturated silver-silver 0.199 V chloride Electrode ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Class content: Concentration cells Types of concentration cells Electrolyte concentration cells Electrode concentration cells Ion- selective electrodes Types of ion – selective electrodes Electrode potential for an ion-selective electrode ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Concentration cells: An electrochemical cell in which identical electrodes are in contact with a solution of identical species but of different concentration https://chemdemos.uoregon.edu/demos/Voltaic-Cell- CuCu-concentration-cell-Demonstration ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria In this cell 2 copper electrodes are immersed in copper sulphate solutions of concentration c1 & c2 , such that c2> c1 An electrolyte has spontaneous tendency to diffuse from a solution of higher concentration to a solution of lower concentration which is the driving force for development of potential Oxidation takes place at anode and reduction takes place at cathode e.g.,Cu/Cu2+(c1)//Cu2+(c2)/Cu ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Reactions : At anode: Cu → Cu2+ (c1)+2e- At cathode: Cu2+ (c2)+ 2e-→ Cu Expression for cell potential: The emf of the cell = E cathode – E anode 𝟐. 𝟑𝟎𝟑𝑹𝑻 𝑬𝒄𝒂𝒕𝒉𝒐𝒅𝒆 = 𝑬𝒐 + 𝐥𝐨𝐠 𝒄𝟐 𝒏𝑭 𝟐. 𝟑𝟎𝟑𝐑𝐓 𝐄𝐚𝐧𝐨𝐝𝐞 = 𝐄 𝐨 + 𝐥𝐨𝐠 𝐜𝟏 𝐧𝐅 𝟐. 𝟑𝟎𝟑𝑹𝑻 𝟐. 𝟑𝟎𝟑𝑹𝑻 𝑬𝒄𝒆𝒍𝒍 = 𝑬𝒐 + 𝐥𝐨𝐠 𝒄𝟐 − 𝑬𝒐 + 𝒍𝒐𝒈 𝒄𝟏 𝒏𝑭 𝒏𝑭 2.303RT 𝒄𝟐 𝑬cell = log( ) nF 𝒄𝟏 𝟎.𝟎𝟓𝟗𝟏 𝒄 At 298K, 𝑬cell = log( 𝟐 ) 𝒏 𝒄𝟏 ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria The emf of the cell is positive only if c2 > c1 i.e., conc of metal ion at cathode > conc. of metal ion at anode The emf of the cell depends upon the ratio c2/c1 When c2 = c1, the emf of the cell becomes zero During working of the cell, concentration of ions increases at anode decreases at cathode When current is drawn from the cell c1 increases and c2 decreases The cell can operate only as long as the concentration terms are different ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Types of concentration cells: Electrolyte concentration cell Electrode concentration cell Electrolyte concentration cell: Electrolyte concentration cell consists of two same electrodes that are dipped in the same electrolyte but with different concentrations of electrolytes Cu/Cu2+(c1)//Cu2+(c2)/Cu 2.303RT 𝒄𝟐 𝐄cell = log( ) Cell potential is given by nF 𝒄𝟏 Electrode concentration cell Electrode concentration cell consists of two identical electrodes of different activity which are dipped in the same solution of electrolyte Na-Hg(c1)/Na+/Na-Hg(c2) 𝟐. 𝟑𝟎𝟑𝐑𝐓 Na−Hg(c1) Cell potential is given by 𝐄𝐜𝐞𝐥𝐥 = 𝐥𝐨𝐠 𝐧𝐅 Na−Hg(c2) ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Na-Hg(c1)/Na+/Na-Hg(c2) : Reactions are : At anode: Na-Hg(c1) → Na+ + e- At cathode: Na+ + e-→ Na-Hg(c2) Cell potential = E cathode – E anode 𝟐. 𝟑𝟎𝟑𝑹𝑻 Na−Hg(c2) 𝑬𝒄𝒂𝒕𝒉𝒐𝒅𝒆 = 𝑬𝒐 − 𝐥𝐨𝐠 𝒏𝑭 Na+ 𝟐. 𝟑𝟎𝟑𝐑𝐓 Na−Hg(c1) 𝐄𝐚𝐧𝐨𝐝𝐞 = 𝐄𝐨 − 𝐥𝐨𝐠 𝐧𝐅 Na+ 𝟐. 𝟑𝟎𝟑𝐑𝐓 Na−Hg(c1) 𝐄𝐜𝐞𝐥𝐥 = 𝐥𝐨𝐠 𝐧𝐅 Na−Hg(c2) ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Pt/H2(p1 atm)/H+/H2(p2 atm)/Pt : Nernst Equation: 𝟐. 𝟑𝟎𝟑𝐑𝐓 𝐩𝟏 𝐄𝐜𝐞𝐥𝐥 = 𝐥𝐨𝐠 𝐧𝐅 𝐩𝟐 Pt/Cl2(p1 atm)/Cl-/Cl2(p2 atm)/Pt : Nernst Equation: 𝟐. 𝟑𝟎𝟑𝐑𝐓 𝐩𝟐 𝐄𝐜𝐞𝐥𝐥 = 𝐥𝐨𝐠 𝐧𝐅 𝐩𝟏 ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Ion selective electrodes (ISE) Selectively respond to a specific ion in a mixture Potential developed is a function of concentration of that ion Have a membrane which is capable of exchanging the specific ion with solution with which it is in contact Membrane electrodes e.g., glass electrode ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Types of Ion selective electrodes : Electrodes are classified based on the membrane material Crystalline / solid state membrane electrodes: Single crystal LaF3 selective to F- Polycrystalline such as Ag2S selective to S2- Non-crystalline membrane electrodes: e.g., Glass membrane selective to H+ , Na+ Liquid membrane electrodes: An ion-exchanger is dissolved in a viscous organic liquid membrane; used for Ca+, K+ Immobilised liquid in a rigid polymer: e.g., immobilized ion exchanger in PVC matrix ; used for Ca+, NO3- ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Electrode potential of an ion-selective electrode Schematic representation: analyte solution reference solution inner reference electrode [Mn+] = C1 [Mn+] = C2 boundary potential is 2.303RT 𝐂𝟏 𝐄𝐣 = log( ) nF 𝐂𝟐 since concentration of reference solution C2 is constant 2.303RT where 2.303RT 𝐄𝐣 = logC𝟏 +K K= − logC𝟐 nF nF ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Overall potential of the membrane electrode is given by EM = Ej + Eref 2.303RT since 𝐄𝐣 = logC𝟏 +K nF 2.303RT 𝐄𝐌 = logC𝟏 +K +Eref nF 2.303RT where EoM = K + Eref 𝐄𝐌 =E𝐨 𝐌+ logC𝟏 nF 𝟎. 𝟎𝟓𝟗𝟏 At 298K, 𝐄𝐌 =E𝐨 𝐌 + logC𝟏 𝐧 Membrane electrode is coupled with an external reference electrode ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria External ref. electrode/Analyte/membrane/ ref. solution/Internal ref. electrode Cell potential = E cathode – E anode E cell = E membrane – E ext.ref.electrode E cell can be measured, E ext.ref.electrode is known E membrane can be determined 𝟎. 𝟎𝟓𝟗𝟏 Since 𝐄𝐌 =E 𝐨 𝐌 + logC𝟏 , C1 can be determined 𝐧 The disadvantage of an ion-selective electrode is that the membrane offers very high resistance so ordinary potentiometers cannot be used; special type of potentiometers have to be used. Applications : Used to determine concentration of number of cations and anions such as H+, Li+,Na+,K+,Pb2+,Cu2+,Mg2+,CN-, NO3-, F-etc ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Class content: Glass electrode Construction Working Determination of pH ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Glass electrode Ion-selective electrode Responds to Hydrogen ion pH sensitive; can be used to determine pH of a solution Consists of a glass membrane which is capable of exchanging H+ ions ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Construction: Glass tube , the end of which is a bulb of very thin glass membrane Glass bulb is made up of special type of glass, CORNING 015 The glass bulb is filled with solution of known pH which is the reference solution A silver - silver chloride electrode is dipped inside the reference solution serves as internal reference electrode and also provides external electrical contact The electrode is immersed in a solution containing H+ which is the analyte Ag/AgCl/HCl/glass Source:https://glossary.periodni.com/glossary. php?en=glass+electrode ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Working: analyte solution reference solution Ag-AgCl electrode [H+] = C1 [H+] = 0.1M HCl = C2 H+ solution + Na+Gl- ⇌ Na+solution + H+Gl- The inner and outer surfaces of the glass membrane can exchange H+ ions with the solution they are in contact with Source:http://www.metrohmsiam.com/t eachingresearch/TRL_25/TRL25_955207 _80155013.pdf The hydrated glass membrane brings about ion exchange reaction between singly charged cations in the interstices of glass lattice and protons from the solution A potential is developed, which is a function of H+ of the solution ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Electrode potential of a glass electrode : analyte solution reference solution Ag-AgCl electrode [H+] = C1 [H+] = 0.1M HCl = C2 2.303RT 𝐂 boundary potential is 𝐄𝒃 = log( 𝟏 ) nF 𝐂𝟐 since concentration of reference solution, C2 is constant 2.303RT 2.303RT 𝐄𝒃 = L′ + logC𝟏 where L′= − logC𝟐 nF nF 0.0591 At 298K, 𝐄𝒃 = L′+ log[H+] since for H+, n = 1 n 𝐄𝒃 = L′ − 𝟎. 𝟎𝟓𝟗𝟏pH ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria The glass electrode potential has 3 components 1. The boundary potential 2. The potential of internal reference electrode 3. Asymmetric potential EG = Eb +Eref + Easymmetric Asymmetric potential arises due to difference in responses of inner and outer surfaces of the glass bulb, due to differing conditions of stress on two glass surfaces 0.0591 EG = Eb + Eref + Eassymmetric ; 𝑬𝒃 = L′+ log[H+] n 0.0591 = L′+ n log[H+] + Eref + Eassymmetric = E0G + 0.0591log [H+] where E0G = L’ + Eref + Easymmetric EG = E0G - 0.0591pH ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Determination of pH using glass electrode: Glass electrode is combined with an external reference electrode Hg/Hg2Cl2/Cl-//analyte solution/glass/0.1N HCl/AgCl/Ag Source:https://utkarshiniedu.wordpress.com/201 6/12/22/lecture-1-108-ion-selective-electrodes/ ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Determination of pH using glass electrode: The emf of the cell is determined potentiometrically Ecell = EG - Ecalomel ; EG = E0G - 0.0591pH = E0G - 0.0591pH - Ecalomel 𝐄𝟎𝐆− 𝐄𝐜𝐚𝐥𝐨𝐦𝐞𝐥 − 𝐄𝐜𝐞𝐥𝐥 pH = − 𝟎.𝟎𝟓𝟗𝟏 To evaluate E0G the glass electrode is dipped in a solution of known pH(buffer solution) and combined with calomel electrode, the emf of the cell is measured from which E0G can be evaluated ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Applications of glass electrode: Used extensively in chemical, industrial, agricultural and biological labs Advantages of glass electrode : Can be used in oxidizing and reducing environments and metal ions Does not get poisoned Can be used for very small volumes Accurate results can be obtained between pH 1 to 9 by ordinary electrodes. However by using special glass electrodes pH 1 to 14 can be measured with accuracy Simple to operate and can be used with portable instruments ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Disadvantages of glass electrode: Because of high resistance of glass, a simple potentiometer cannot be used. It requires sensitive potentiometer for emf measurements Glass membrane is very delicate, hence has to be handled carefully At very high pH levels usually over a pH of 9 , Alkaline error is observed H+ solution + Na+Gl- ⇌ Na+solution + H+Gl- When the Sodium ion level is relatively high, some of the H+ ions in the gel layer around the sensitive electrode membrane are replaced by Na+ ions The electrode may eventually respond to Na+ instead of H+ ions, giving a false lower pH value than the actual value ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Class content: Fluoride - ion electrode Construction Working Determination of [F-] ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Fluoride-ion electrode Ion-selective electrode Responds to Fluoride ion Example: lanthanum fluoride electrode Consists of a lanthanum fluoride (LaF3) membrane which is capable of exchanging F- ions ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Construction: A tube to which the sensing membrane is bonded at one end Membrane is a single crystal of Lanthanum fluoride(LaF3) doped with Europium fluoride (EuF2) to improve conductivity Doping creates fluoride ion vacancies which allows F- ions to migrate across the membrane The tube is filled with the reference solution containing 0.1 M NaF and 0.1M NaCl A silver - silver chloride electrode is placed in the reference solution which acts as the internal reference Source:https://slideplayer.com/ electrode and also serves for external electrical contact slide/14040224/ The electrode is immersed in a solution containing F- ions which is the analyte Ag/AgCl/NaCl,NaF/LaF3 ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Working: analyte solution reference solution Ag-AgCl electrode [F-] = C1 [F-] = C2 LaF3 ⇌ LaF2+ + F-solution The inner and outer surfaces of the membrane can exchange F- ions with the solution they are in contact with The F- ions migrate through the membrane by jumping through the lattice vacancies As transference of charge through the crystal is almost exclusively due to fluoride ion, the electrode is highly Source:https://slideplayer.com/sl specific to fluoride ion ide/14040224/ The only ion which significantly interferes is hydroxide (OH−) ion ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Electrode potential of a Fluoride ion electrode : analyte solution reference solution Ag-AgCl electrode [F-] = C1 [F-] = 0.1M NaF= C2 LaF3 ⇌ LaF2+ + F-solution 2.303RT 𝐂 boundary potential is 𝐄𝒃 = nF log(𝐂𝟏) 𝟐 2.303RT 𝐂𝟐 Since n is -1 for F- ion, 𝐄𝒃 = log( ) F 𝐂𝟏 since concentration of reference solution, C2 is constant 2.303RT 2.303RT 𝐄𝒃 = K′ − logC𝟏 where K′= logC𝟐 F F At 298K, 𝐄𝒃 = K′ − 0.0591 log[F−] 𝐄𝒃 = K′ + 𝟎. 𝟎𝟓𝟗𝟏pF ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria The fluoride ion electrode potential is given by 1. The boundary potential 2. The potential of internal reference electrode EF = Eb +Eref 𝑬𝒃 = K′ − 0.0591 log[F−] since EF = K′ − 0.0591 log[F−] + Eref = E0F - 0.0591log[F−] where E0F = K’ + Eref EF = E0F + 0.0591pF ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Determination of pF using Fluoride ion electrode: Fluoride ion electrode is combined with an external reference electrode Hg/Hg2Cl2/Cl-//analyte solution/LaF3/NaF,NaCl/AgCl/Ag Source:http://elchem.kaist.ac.kr/vt/chem-ed/echem/ise.htm ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Determination of pF using Fluoride ion electrode: The emf of the cell is determined potentiometrically Ecell = EF - Ecalomel ; EF = E0F + 0.0591pF = E0F + 0.0591pF - Ecalomel 𝐄 𝐜𝐞𝐥𝐥− 𝐄𝟎𝐅+ 𝐄𝐜𝐚𝐥𝐨𝐦𝐞𝐥 pF = − 𝟎.𝟎𝟓𝟗𝟏 To evaluate E0F the fluoride ion electrode is dipped in a solution of known F- ion concentration and combined with calomel electrode, the emf of the cell is measured from which E0F can be evaluated ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria Applications of Fluoride ion electrode: To determine concentration of F-ion in water pollution studies To determine concentration of F- ion in toothpaste Advantages of Fluoride ion electrode : Can be used for very small volumes Very sensitive ; can determine [F-] upto 10-6M Disadvantages of Fluoride ion electrode: Because of high resistance of the membrane, a simple potentiometer cannot be used. It requires sensitive potentiometer for emf measurements The presence of OH- ions interferes with measurement of F- ions ;shows best results between pH 5 and 8.Hence the [OH-] is kept constant using a buffer ENGINEERING CHEMISTRY Module 4- Electrochemical equilibria End of Module 4- Electrochemical equilibria End of Unit II ENGINEERING CHEMISTRY Module 4 – Electrochemical Equilibria Class content: Numericals on electrochemistry Nernst equation Ion selective electrode ENGINEERING CHEMISTRY Module 4 – Electrochemical Equilibria 1. For the given cell: Fe/Fe 2+ (0.05M)//Ag+ (0.1M)/Ag (i) Write the overall cell reaction (ii) Calculate E0cell and E cell at 250C (Given : E0Fe+2/ Fe= - 0.44V ; E0Ag+/Ag = 0.80V) Sol. Anode : Fe → Fe2+ + 2e- Cathode : 2Ag+ + 2e- → 2Ag Overall reaction : Fe + 2Ag+ → Fe2+ + 2Ag E0cell= EoC-EoA = 0.80 + 0.44 =1.24V 0.0591 [0.05] Ecell = 1.24 − log( ) 2 [0.1]2 Ecell =1.2193V ENGINEERING CHEMISTRY Module 4 – Electrochemical Equilibria 2. For the following concentration cell: Pt/H2 (8atm)/HCl(0.3M)/H2(2atm)/Pt Calculate potential of the cell at 250C. Sol. 0.0591 pH 2 ( anode ) Ecell = log n pH 2 ( cathode ) 0.0591 8 Ecell = log n 2 Ecell = 0.01779V ENGINEERING CHEMISTRY Module 4 – Electrochemical Equilibria 3. A decinormal calomel electrode as cathode is coupled with a saturated calomel electrode as anode to form a cell. Write the cell representation and calculate the concentration of Cl- ion in the saturated calomel electrode, if the cell potential measured is 0.0988 V at 250C. Sol. Pt/Hg/Hg2Cl2/Cl-(x)//Cl-(0.1M)/Hg2Cl2/Hg/Pt Ecell = ER-EL = [E0 - 0.0591log (0.1)] –[ E0 - 0.0591log(x)] 0.0988 x = log 0.0591 0.1 1.6717 - 1= log(x) x = Antilog(0.6717) x = 4.69M ENGINEERING CHEMISTRY Module 4 – Electrochemical Equilibria 4. For the following cell: Ag/AgCl/Cl-(0.1M)// Fe2+ (0.29M), Fe3+ (0.18M)/Pt (i) Write the half cell reactions and overall cell reaction. (ii) Calculate EoCell and ECell at 298 K (Given:E0Fe3+/Fe2+=0.77 V , E0Calomel=0.222 V, R = 8.314 J/K/mol, F =96500 C/mol) Sol. (i) Anode: Ag + Cl- → AgCl + e- Cathode : Fe3+ + e- → Fe2+ Overall : Ag + Cl- + Fe3+→ AgCl + Fe2+ (ii) E0cell = EoC-EoA = 0.77-0.222 = 0.548V 0.0591 ( [𝐹𝑒 2+ ] ൯ 𝐸𝑐𝑒𝑙𝑙 = 𝐸𝑜 𝑐𝑒𝑙𝑙 − log[ 3+ − ሻ] 𝑛 ([𝐹𝑒 ][Cl ] ECell= 0.4767V ENGINEERING CHEMISTRY Module 4 – Electrochemical Equilibria 5. Calculate the EMF of the following cell at 250C. Au/Au3+ (0.05M) // Au3+ (0.12 M)/ Au (Given : R = 8.314 J/K/mol, F = 96500 C/mol) Sol. 0.0591 [M𝑛+ (𝑐𝑎𝑡ℎ𝑜𝑑𝑒ሻ] 𝐸𝑐𝑒𝑙𝑙 = 𝑙𝑜𝑔 𝑛 [𝑀𝑛+ (𝑎𝑛𝑜𝑑𝑒ሻ] n=3, Ecell = 0.00749V ENGINEERING CHEMISTRY Module 4 – Electrochemical Equilibria 6. A decinormal calomel electrode is used to determine the potential of the following redox electrode : Pt/Cu2+(0.58 M),Cu+(0.08M) (i) Write cell representation. (ii) Write the reactions at the electrodes (iii) Calculate E0cell and Ecell at 298 K. (Given : EoHg/Hg2Cl2/Cl- = 0.281V , E Cu2+/Cu = 0.153 V) Sol. (i) Pt/Cu2+(0.58 M),Cu+(0.08M)//Cl-(0.1 M)/Hg2Cl2/Hg (ii) (iii) E0cell = EoC-EoA = 0.281 -0.153 = 0.128 V 0.0591 ([𝐶𝑙 − ]2 [𝐶𝑢 2+ ]2 ሻ 𝐸𝑐𝑒𝑙𝑙 = 𝐸 𝑜 𝑐𝑒𝑙𝑙 − log[ ] 𝑛 ([𝐶𝑢+ ]2 ሻ Ecell = 0.1362 V ENGINEERING CHEMISTRY Module 4 – Electrochemical Equilibria 7. For the following cell: Fe/Fe+2(0.1M)|| Au+3( 0.5M) /Au i. Write the half cell reactions. ii. Calculate E0cell and Ecell at 298K. (Given E0 Au+3/Au =1.52V, E0 Fe+2/Fe = -0.44V, R = 8.314 J/K/mol, F = 96500C/mol) Sol. (ii) E0cell = EoC-EoA = 1.52 +0.44 = 1.96 V 3 0.0591 0.1 Ecell= 1.96 − log[ ] 2 6 0.5 Ecell = 1.9836 V ENGINEERING CHEMISTRY Module 4 – Electrochemical Equilibria 8. A glass electrode is coupled with saturated calomel electrode to measure unknown pH. The cell potentials measured are 0.215V and 0.385V in contact with a solution of pH = 7 and with solution of unknown pH respectively. Calculate the pH of unknown solution. Given ESCE=0.244V Sol. EoG = Ecell + 0.0591pH + ESCE = 0.215 + 0.0591 X 7 + 0.244 = 0.8727 V pH = 4.12

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