ELECTROCHEMISTRY - Wits Structural Chemistry

Example 0AgCl(s) → Ag + (aq) + Cl - (aq)Solubility of AgCl(s) in water at 25 o C is 1.274 x 10 -5 mol kg -1 . = [Ag + ] = [Cl - ]Calculate the solubility of AgCl(s) in 0.010 mol kg -1 Na 2 SO 4 (aq).Assume K sp for ideal and non-ideal solutions are the same.Consider ideal vs no longer ideal solutionsEQUILIBRIUM ELECTROCHEMISTRYREVISIONThe Cell PotentialNernst EquationTypes of electrodesStandard Electrode PotentialsThermodynamic Functions from the Cell PotentialApplications of Standard PotentialsPart of Chapter 6w max = ∆GBefore EquilibriumThe Cell PotentialConsider a galvanic cell reaction:The cell can do electrical work by drivinge − ’s through the external circuit.The larger the potential difference between thetwo electrodes, the more work can be done byeach e−.Measure E cell by applying an opposingpotential s.t. the reaction occurs reversiblyand the composition is constant (no nettcurrent flows).∆G = −nFE cellfor electrical work for areversible reactionAt EquilibriumNo more work can be doneE cell = 0 VThe composition in electrode compartment is expressed by:the reaction quotient, Qthe equilibrium constant, KDerive Nernst equation:E = EApplies to:∆G = −nFEo −RTln QnFHalf-reactionsRTEnFandCellsRTnFoo= Ered− lnQredE = Ecell− lnQrxnHalf-cell reaction written as areduction reaction.Nernst Equation∆G= ∆Go + RT ln Q2.303 RTE = Eo − log QnFEocell= Eocathode− EoanodeNOTE: E o used in the calculations are always for the reduction reactions.log xelog10 x =log 10e2 .303 log x = ln x0.05916E = Eo − log Q at 25°CnE o = standard cell potential all reactants and products in their standard states,thus all activities = 1∆G = −nFEDo Self-test 6.7Standard Reduction Potentials in H 2 O at 25 o CThe cell reaction is spontaneousif E cell > 0At equilibrium:∆G = 0 E cell = 0 and Q = KStrong oxidisingagentGibbs free energy, G∆G < 0E > 0∆G = 0E = 0∆G > 0E < 0RTE = E − lnQnFRT0 E − lnKnF= nF ln K = ERTE > 0 spontaneousreductionreactantsExtent of reactionproductsThe steeper the slope, the greaterthe driving power of the cell.Recall:∆G°= –RT ln K ∆G°= –nFE°E < 0 nonspontaneousreductionStrong reducingagent3

Self-test 6.7Express the formation of H 2 O from H 2 and O 2 in acidic solution (a redox reaction)as the difference of two reduction reactions.(You may need to consult the full table of standard reduction potentials in the Appendix)Example 1Consider the Daniell cell:Using the standard reduction potentials, calculate the equilibrium constant at 25 °C.Standard reduction potentials at 25 °C:Cu 2+ + 2e - → Cu(s)Zn 2+ + 2e - → Zn(s)Zn(s)ZnSO 4 (aq) CuSO 4 (aq) Cu(s)E o = +0.34 VE o = −0.76 VReduction rxnOxidation rxnDo Self-test 6.8Metal/metal ion:e.g. Zn(s) in zinc ion solutionM(s) | M n+ (aq)Metal-insoluble salt: M(s) | MX(s) | X - (aq)e.g. silver-silver chloride electrode−Ag(s)AgCl(s) Cl (aCl)AgCl(s) + e - Ag(s) + Cl - (aq)Inert electrode: Pt(s) | M n+ (aq), M p+ (aq)e.g. Pt immersed in a solution containing a redox couple4 + 3+Pt(s) Ce (aq),Ce (aq)Ce 4+ + e - Ce 3+o RT aE = E − lnF aGas electrode:Zn 2+ + 2e - Zn(s)Pt(s) | X 2 (g) | X +/− (aq)e.g. a standard hydrogen electrode+Pt(s)H2(g)H (aq)2H + + 2e - H 2 (g)Types of ElectrodesEEo= E − ln2+RTFo= E − ln a−Cl=RT2FRT−2F1ap3+Ce4+CeoH2E E ln2aZn+HSelf-test 6.8Write the half-reaction and the reaction quotient for a chlorine gas electrode.Standard Electrode PotentialsThe potential of a half-cell is measured relative to the standardhydrogen electrode (SHE) under standard conditions.Pt(s) | H 2 (g) | H + (aq)activities = 1 mol kg -1E o SHE = 0 V at all temperatures fugacities = 1 atmDetermining the standard potential of a silver-silver chloride electrode usinga Harned cell:Pt⏐H 2 (g)⏐HCl(aq), a H+ , a Cl- ⏐AgCl(s)⏐Ag(s)Half-reactions:H + + e − → ½H 2 (g)anode1e - E o SHE = 0 VAgCl(s) + e − → Ag(s) + Cl − (aq) cathodeRecall: E o is an intensive property!E o cell = E o indicator – E o referenceSHEOverall reaction:AgCl(s) + ½H 2 (g) → H + (aq) + Ag(s) + Cl − (aq)Nernst equation:o RT a + a −H ClEcell= Ecell− ln 1/2 F PH2o RTFor SHE: P H2 = 1 atm Ecell= Ecell− ln a + a −F H ClLet a H+ = γ m ± H+ and a Cl- = γ ± m o RT 2 2Cl- Ecell = Ecell− ln (mHClγ ± )Fγ ± = mean activity coefficientm = molality (concentration in mol kg -1 )4

o RT 2 2Ecell = Ecell− ln (mHClγ ± )Fo RT 2 RT 2Ecell = Ecell− ln (mHCl) − ln ( γ ± )FF2RTo 2RTEcell+ ln mHCl= Ecell− ln γ ±FFDebye-Hückel law: log γfor a 1:1 electrolyte:±= −log γ ±2 .303 log x = ln xA z z= −+−ln = −A′γ ±IA I = −AIf E o cell is known, then using E cellthe activity coefficient γ ± can becalculated for a solution ofmolality m.mHClm HCl1I =2 im z 2i i122I = [(m)( + 1) + (m )( −1)]H +Cl −21I = [ 2(m )]HCl2Evaluating a standard potential from two others:E o +Cu 2 + =/ Cu(s)E o + = +Cu / Cu(s)0.340 V0.522 VE o =Cu 2 + +/ CuCu 2+ (aq) + 2e - → Cu(s) E o = 0.340 V (A)Cu + (aq) + e - → Cu(s) E o = 0.522 V (B)(A) – (B): Cu 2+ (aq) + e - → Cu + (aq) E o (C) = ?NOTE:Thermodynamic Functions from the Cell PotentialThis is NOT a cell reaction!!→ the electrons do NOT cancel out?Redox couple Cu 2+ /Cu +Quote most positive 1streductionoxidation2RTo 2RTEcell+ ln mHCl= Ecell+ A′mFFHCly = c + m xMeasure the cell potential forvarious HCl concentrations.Plot y vs x and extrapolate to they-intercept to determine E o cellApplications of Standard Potentials1) The determination of activity coefficients2) The determination of equilibrium constants3) The electrochemical series4) The determination of thermodynamic functionsExample 2The mean activity coefficients of HBr in 5.0 and 20.0 mmol kg –1 are 0.930 and 0.879,respectively. Consider a hydrogen electrode in HBr(aq) solution at 25 °C operating at1.15 atm.Calculate the change in the electrode potential when the molality of the acid solutionis changed from 5.0 and 20.0 mmol kg –1 .Thermodynamics has the striking ability of to relate apparently unrelatedrelationships. ∆G = −nFEE o cell used to calculate ∆Go f of ions in solution.Using the relationship dE ∆S=dT nF ∂G= −S ∂T pThe temperature coefficient of E o cell gives the standard entropyof the cell reaction and hence entropies of ions in solution.Combining both relationships: dE∆H= ∆G+ T∆S= −nFE − T dT Non-calorimetric measurement of ∆H°and hence ∆H o f of ions in solution canbe determined.Example 3Devise a cell in which the cell reaction is:Mn(s) + Cl 2 (g) → MnCl 2 (aq)Give the half reactions at the electrodes and from the standard cell potential of 2.54 Vdeduce the standard potential for the Mn 2+ /Mn(s) redox couple.Given: E°(Cl 2 /Cl - ) = +1.36 VExample 4Estimate the cell potential at 25°C forAg(s)|AgBr(s)|KBr(aq, 0.050 mol kg –1 )||Cd(NO 3 ) 2 (aq,0.0034 mol kg –1 )|Cd(s)E°(R-H) = –0.40 V E°(L-H) = +0.07 V (assume non-ideal solutions)Write the spontaneous electrochemical reaction.5

Example 5The standard potential of the cell below at 25 °C is 0.95 V.Ag(s) |AgI(s) | AgI(aq) | Ag(s)Calculate: a) its solubility constant and b) the solubility of AgI .ELECTROCHEMICAL TECHNIQUESBulkInterfacialStatic (≡(m )DynamicConductivityPotentiometryControlledE I QTransport Thermodynamics KineticsThree main ways to transport ions to the surface of an electrode→ called mass transportConvectionMASS TRANSPORTDiffusionMigrationMOLECULAR MOTION IN LIQUIDSMass TransportMigration of ions - Conductivities of Electrolyte SolutionsMobilities of IonsDiffusion of IonsMovement due tomechanical intervention.e.g. stirring, pumping,shakingAffects molecules andions in solution.Movement due toconcentration gradient.e.g. movement across acell membraneAffects molecules andions in solution.Movement due to appliedelectric field.e.g. in an electrolysis cellAffects only ions insolution.Part of Chapter 20CONDUCTIVITIES OF ELECTROLYTE SOLUTIONSIons can be dragged through the solvent by applying a potential differencebetween two electrodes in solution.Measuring the conductivity of tap water:To measure conductivity:Incorporate conductivity cell intoone arm of resistance bridge andsearch for balance point.1 R =κ A1 κ =R ACκ =RR = resistance of solution (Ω)κ = conductivity (S cm -1 ) = distance between electrodesA = area of electrodesC = cell constantκ depends on the number of ions in a solution.R 3R 42R 1R R =1 2R R3R 4Resistance bridge R 1R= R R34 2Note: use ac current (ν~1kHz)to avoid electrolysis andpolarisation of the electrodeswhich would change thecomposition of the solution atthe electrode surfaces.Parallelplates293.3 µS / cm1 S = 1 Siemen= 1 Ω -16

κ=cΛΛm = molar conductivity (S cm 2 mol -1 )m c = molarityIf κ ∝ c, the molar conductivity should beindependent of the concentration of anelectrolyte. Is this the case?NO!Molar conductivity varies withconcentration because:- no. of ions in solution may not beproportional to concentration (especiallyfor weak electrolytes)- ions interact strongly with one anotherreducing effective chargeNB: Don’t confuse conductivity (κ) with conductance (G)Strong electrolyte – Λ mdecreases slightly asconc increases.Weak electrolyte –normal Λ m at low conc,but Λ m decreases sharplyas conc increases.1G =RKohlrausch’s law:At low concentrations, molar conductivities vary with the square root of conc.ΛmΛ o m= ν= Λ − K com+λ++ ν−λ−STRONG ELECTROLYTESΛ m o = limiting molar conductivity(ions are infinitely far apart no interaction)K → related to stoichiometry of electrolyteKohlrausch’s law of independent migration of ions:Λ m o → expressed as sum of contributions from its individual ions.λ = limiting molar conductivity of ionsν = no. of ions per formula unit of electrolyteH +34.96 OH -19.91Na + 5.01 Cl -7.63K + 7.35 Br -7.81Zn 2+ 10.56 SO 2-4 16.00e.g. for MgCl 2 : ν + = 1 ν − = 2Limiting ionic conductivies in water at 298 K, λ /mS m 2 mol -1WEAK ELECTROLYTESWeak electrolytes do NOT dissociate completely, conductivity of weak electrolytes depends on the degree of ionization (α)e.g. HA + H 2 O A - + H 3 O ++ −[H3O][A ]+K[ H3O] = αca =[HA]−[ A ] = αc2α cKa=[ HA] = ( 1−α)c1 − α(c = formal conc of HA)Example 6Calculate the degree of ionization and the acid dissociation constant at 298 K for a0.010 M acetic acid solution that has a resistance of 2220 Ω. The resistance of a 0.100M potassium chloride solution was also found to be 28.44 Ω.Λ m (0.1 M KCl) = 129 S cm 2 mol -1λ o (H + ) = 349.6 S cm 2 mol -1λ o (CH 3 COO - ) = 40.9 S cm 2 mol -1At infinite dilution the electrolyte is fullyionised :Λ o m= ν+λ++ ν−λ−At very low conc’s:Λ = αomΛ mOstwald’sdilutionlaw1 1 Λmc= +oΛmΛm K( Λ ) o 2am1 1 1= + Λmcoo 2ΛmΛm Ka( Λm)y c m xExample 7The molar conductivity of a strong electrolyte in water at 25 °C was found to be 109.9S cm 2 mol -1 for a concentration of 6.2 × 10 -3 mol L -1 and 106.1 S cm 2 mol -1 for aconcentration of 1.5 × 10 -3 mol L -1 .Estimate the limiting molar conductivity of the electrolyte.Example 8The limiting molar conductivities of KCl, KNO 3 , and AgNO 3 are 149,9, 145.0, and 133.4S cm 2 mol -1 , respectively (all at 25 °C).What is the limiting molar conductivity of AgCl at this temperature?7

To understand conductivity measurements, we need to understand whyions move at different rates?Motion of ions is largely random, but in the presence of an electric field themotion is biased → migrationDC, uniform electricfield is applied+V source-∆φ+Mobilities of IonsElectric field:∆φε =∆φ = potential difference = distance between electrodesε = electric fieldForce acting on the ion due to electric field:ze∆φze = charge of ionF e= zeε=e = 1.602×10 -19 Ccauses ion to accelerate towards oppositelycharge electrode.But ion undergoes frictional retardationwhen moving through the solvent:F Fricviscosity drag= fs = 6πηass = speed of ionη = viscosity of solventa = radius of ionThe two forces (one due to the electric field force and the other due to the viscositydrag, interactions between ions, etc.) act in the opposite directions.Eventually, these forces are balancedF e = F Fricand the ions reach a terminal speed called the drift speed.zeε= 6πηaszeεDrift speed: s =6πηastronger electric field ε faster ions drifts = uεu = ionic mobility =ze6πηa∴ expect the conductivity to decrease with increasing solution viscosity andincreasing ion size.Ionic mobilities in water at 298 K,u /10 -8 m 2 s -1 V -1H + 36.23 OH - 20.64Na + 5.19 Cl - 7.91K + 7.62 Br - 8.09Zn 2+ 5.47 SO 2-4 8.29The drift speedgoverns the rateat which chargeis transported.However, we see the larger ion (with thesame charge) has a larger mobility.Need to consider:a = hydrodynamic radiusi.e. includes hydration sphereSmall ions are more extensivelyhydrated that larger ions (due to thestronger electric field)Self-study:What is the Grotthaus mechanism which is used to explain the very high mobility ofH + in water?Ion-ion interactions affecting mobilities:In ionic solutions the ion-atmosphere also needs to beconsidered when ions migrate.When an electric field is introduced, two effects occur:theoretical measurableLINK between mobility and molar conductivityFor cations and anions:λ = zuFAnd for very diluted solutions:oΛ m = ( ν+ z+u++ ν−z−u− )FFaraday’s constant = F = N A eSince Λ o m = ν+ λ++ν−λ− Relaxation effect:When an electric field is introduced the ionic atmosphere isdistorted as it drags behind moving charge.Central ion and atmosphere opposite in charge retardationof moving ion. Electrophoretic Effect:The ion-atmosphere moves in the opposite direction to thecentral ion which lead to the reduction in ion mobility. Both effects results in lower mobilities!Fick’s first law of diffusion:The flux (J) of particles is proportional to the concentration gradient.dcJ = −DdxD = diffusion coefficientc = molar concentrationx = distanceDIFFUSION OF IONSc 1 > c 2Applies to ions and uncharged particles in a solution.c 1c 2JcDo Self-test 20.4dcdxxSelf-test 20.4Use the experimental value of the mobility to evaluate the diffusion coefficient, thelimiting molar conductivity and the hydrodynamic radius of the NH 4+ ion inaqueous solution. (u = 7.63 × 10 -8 m 2 s -1 V -1 )Einstein equation:Stokes-Einstein equation:Nernst-Einstein equation:RTD = uzFkTD =6πηasubstitute expression for uk = Boltzmann constant = 1.381×10 -23 J K -1Λ =F 2o 2( ν+ z+D++ ν−z2 Or for am −D−)λ+ =F 2o( ν + z2+ D+)RTsingle ion: RT8

Fick’s second law of diffusion or thediffusion equation:2∂c∂ c= D2∂t∂x→ describes timedependence of thediffusion processDiffusion coefficients at 298 K, 10 -9 m 2 s -1 Example 9H + in water9.31The mobility of the NO - 3 ion in aqueous solution at 25 °C is 7.40×10 -8 m 2 s -1 V -1 .Na + in water1.33(Viscosity of water is 0.89110 -3 kg m -1 s -1 ).sucrose in water 0.522Calculate its diffusion coefficient and the effective radius at this temperature.It shows that the rate of change of conc is proportional to thecurvature (the second derivative) of the concentration withrespect to the distance.This equation shows that:1) If the concentration changes sharply from point topoint, then the concentration changes rapidly in time.cvscxx2) If the curvature = 0, then the concentration does notchange in time.3) If the concentration decreases linearly with distance,then the concentration is constant at any point,because the inflow is exactly balanced by the outflowof particles or ions.cxRemember: in the calculations you have to show the work on units.Without that, the work might be considered as not done at all.9