Electrochemistry & Solutions 1. Solutions and Mixtures Year 1 – Module 3 8 Lectures Dr Adam Lee

1. Department of Chemistry Electrochemistry & Solutions 1. Solutions and Mixtures Year 1 – Module 3 8 Lectures Dr Adam Lee afl2@york.ac.uk

2. Aims To: • Understand physical chemistry of solutions and their thermodynamic properties  predict/control physical behaviour  improve chemical reactions • Link electrochemical properties to chemical thermodynamics  rationalise reactivity.

3. Synopsis • Phase rule • Clapeyron & Clausius-Clapeyron Equations • Chemical potential • Phase diagrams • Raoults law (Henry’s law) • Lever rule • Distillation and Azeotropes • Osmosis • Structure of liquids • Interactions in ionic solutions • Ion-ion interactions • Debye-Huckel theory • Electrodes • Electrochemical cells • Electrode potentials • Nernst Equation • Electrode types Recommended Reading R.G. Compton and G.H.W. Sanders, Electrode Potentials Oxford Chemistry Primers No 41. P. W. Atkins, The Elements of Physical Chemistry, OUP, 3rd Edition, Chapters 5, 6 & 9. P. W. Atkins, Physical Chemistry, OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

4. Phase Diagrams

6. Vacuum Pump

9. No. of degrees of freedom No. of phases No. of components

12. Josiah Willard Gibbs 1839 -1903 Gibbs Free Energy American mathematical physicist developed theory of chemical thermodynamics. First US engineering PhD…later Professor at Yale.

13. Benoit Paul Emile Clapeyron 1799 - 1864 Parisian engineer and mathematician. Derived differential equation for determining heat of melting of a solid

17. Modified eqn. for Liquid/Gas and Solid/Gas Lines: Clausius - Clapeyron Eqn. Maths dx/x = dlnx dp/p = dlnp dx 2 1 - + n 1 1 + x - x n x dx = = = ò ò 2 x x 2 1 n 1 - + + dT 1 - = ò 2 T T Rudolf Julius Emmanuel Clausius H dp D Clapeyron = T V dT D Equation 1822 -1888

19. Clausius-Clapeyron Equation

20. 99

25. In general for a mixture AB: GA = GoA + nART ln pA, GB = GoB + nBRT ln pB since GA = nAA A = Ao + RT ln pA Consider ideal gas at constant temperature: dG = Vdp – SdT = Vdp Since pV = nRT, If initial state 1 = STP (1 atm)

26. p = pA + pB

27. nA+ nB ya ya yb yb yA yA yB yB Ginitial = nAA + nBB = nA[ + RTlnp] + nB[B + RTlnp] Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp] G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp + nBlnyBp - + nBlnp] nA nB A B Initial Final Gmixing ya yb

28. yA yA yB yB yA yA yB yB Gmixing Smixing ya ya 0 1 yb yb 0 1

29. Greater “chemical potential” Effect of environment on this free energy Energy free for molecules to “do stuff”at STP Chemical Potential (in English!) Molecules acquire more spare energy Gibbs Free Energy  G  ln(pressure) Pressure Low Pressure High Pressure Constant Temperature Gmolar = Gmolar + RT lnp

30. For mixtures e.g. H2O/C2H5OH For single component e.g. pure H2O G = nH2OH2O G = nH2OH2O+nEtOHEtOH Free energy only comes from H2O  tells us how much from H2O versus C2H5OH Why do we use Chemical Potential? Gibbs Free Energy (G) is total energy in entire system available to “do stuff” - includes all molecules, of all substances, in all phases G = nAA + nBB No real need to use  Free energy from 2 sources

31. Gas-phase molecule Free Energy (G) &  Liquid-phase Solid-state Pressure Why do we different molecules have different Chemical Potentials? Involatile Volatile Ethanol can soak up much more energy in extra vibrational modes and chemical bonds - will respond differently to pressure/temperature increases Chemical Potential : 1. A measure of "escaping tendency" of components in a solution 2. A measure of the reactivity of a component in a solution

32. Volatility Total pressure above boiling liquid Raoult's law eqns. of straight lines passing thru origin

33. 0 lnxA Pure B Pure A 0 1 xA xA runs between 0 (none present) to 1 (pure solution) Mixing (diluting a substance) always lowers xA  this means mixing ALWAYS gives a negative lnxA Mixing always lowers  Dilution

35. Volatile Involatile

36. Volatile Involatile

39. Case 2: -ve deviation A more attracted by B (e.g. CHCl3 + acetone) mixH = < 0 Case 3: +ve deviation A less attracted by B (e.g. EtOH + water) A A A A B B B B mixH = > 0 b.pt. > ideal b.pt. < ideal

40. Proportionality constant Total pressure o · p A Liquid p · Partial pressure of A Partial pressure of B 1 0 o x A o x 0 1 B A B Summary: Raoult’s Law for Solvents pA = xA . pAΘ Volatile high vapour pressure Involatile low vapour pressure pBΘ pB = xB . pBΘ

41. High p0ө(A) Low p0ө(A) High order: low S Less order: higher S A A A A Strong desire to  S Less need to  S Boiling of A favoured A happier in liquid p0ө = vapour pressure = tendancy of system to increase S

42. Proportionality constant Amount in solution

43. Dissolution is EXOTHERMIC For dissolution of oxygen in water, O2(g) O2(aq), enthalpy change under standard conditions is -11.7 kJ/mole.

44. pO2 pH2O Cinnamaldehyde Cinnamic Acid Solvent: H2O Solute: O2 O2 H2O Henry's law accurate for gases dissolving in liquids when concentrations and partial pressures are low. As conc. and partial pressures increase, deviations from Henry's law become noticeable Consider O2 dissolution in water: Important in Green Chemistry for selective oxidation Cinnamyl Alcohol Similar to behavior of gases - deviate from the ideal gas law at high P and low T. Solutions obeying Henry's law are therefore often called ideal dilute solutions. Aspects of Allylic Alcohol Oxidation Adam F. Lee et al, Green Chemistry 2000, 6, 279

46. A yA xA 

48. Lever Rule B GAS GAS B B A A LIQUID B B B A A A B B A LIQUID A A Volatile Involatile B A Tie-line A-B Composition Liquid-Gas Distribution

50. Boiling/ Condensation Temperature Example Problem The following temperature/composition data were obtained for a mixture of octane (O) and toluene (T) at 760 Torr, where x is the mol fraction in the liquid and y the mol fraction in the vapour at equilibrium The boiling points are 110.6 C for toluene and 125.6 C for octane. Plot the temperature/composition diagram of the mixture. What is the composition of vapour in equilibrium with the liquid of composition: 1. x(T) = 0.25 2. x(O) = 0.25 Liquid Vapour x(T) = 1, T = 110.6 C x(O) = 1, x(T) = 0, T = 125.6 C P.W. Atkins, Elements of Phys.Chem. page 141