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Ionic Solutions Structure of liquids Interactions in ionic solutions

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1 Ionic Solutions Structure of liquids Interactions in ionic solutions
Ion-ion interactions Debye-Huckel theory

2 Structure of liquids How do we describe structure of liquids?
Not random c.f. gas Not periodic c.f. solid Liquid are isotropic (same in all directions) Anisotropic  a property that is directionally dependent

3 Let’s compare distribution functions Ideal Gas
Radial distribution function g(r) completely featureless  equally likely to find a particle at any distance r from another particle V. weak molecular attraction % Probability Uncorrelated

4 Solid Strong correlation between atoms even at large separations
 lattice structure, fixed neighbours g(r) Nearest neighbour shell Next nearest neighbour shell

5 Liquid Well-defined 1st shell but broader than solid & less clear successive shells  short-range order Liquids form due to intermolecular potentials Broad Weak

6 Interactions in ionic molecules
Objective To find: Difficult to calculate from individual (microscopic) interactions using statistical mechanics techniques  Use macroscopic view of interactions average over 1023 molecules Interaction Potential H, G, S I,j

7 3 classes of interactions:
Solvent-solvent H (H2O) = +44 kJmol-1 H-bonding Solvent-ion Hsolv (K+Cl-) = -685 kJmol-1 Ion-ion Hsolid (Na+Cl-)= -719 kJmol-1 Ionic solutions dissolve in polar solvents  strong solute-solvent interactions compensate for high lattice enthalpies and lost solvent-solvent interactions Net G (KClaq) = -1.2 kJmol-1 Na+Cl- solid solute H2O solvent Na+aq Cl-aq solution strong interionic forces strong intermolecular strong solute-solvent interactions

8

9 Ion-ion interactions Energy between two ions carrying charge z1 and z2 separated by r is given by (from Coulomb’s Law): where 0 is the vacuum permittivity, r is the relative permittivity or dielectric constant of the solvent This is the energy needed to move the ions to a distance r apart, starting from an infinite separation

10 Like charge Opposite charge

11 Value of r correlates with the solvent polarity
r is extent to which a solvent reduces the energy of interaction between ions dissolved in it. Typical values at 25 C Value of r correlates with the solvent polarity  explains why H2O excellent solvent C6H6 2 (C2H5)2O 4 Pyridine 12 CH3OH 33 CH3CN 36 Me2SO 47 H2O 78

12 Polar molecules possess a permanent dipole moment, 
 = q . r Dipole moment: major source of solvation (hydration) energy reduces ion-ion interactions in solution Source of polarisation: molecular polarisability : dipole induced by a field E  =  . E -q +q r Universal property but small r (C6H6) ~ 2 - + E

13 Debye-Huckel theory Electrolyte solutions are non-ideal due to long-range ion interactions Electrostatic interactions decrease with distance r only as versus for interactions between neutral species. Distance between ions / Å Interaction strength % Ions Neutrals

14 Electroneutrality of system (solute+solvent) means net charge
around any ion is equal and opposite to its charge In ideal solutions this charge would be uniformly spread BUT Oppositely charged ions attract each other: arrangement of ions in solution is not random excess of counter-ions in vicinity (ion atmosphere) + - + - Time average - - + - - - + Ionic atmosphere

15 Chemical potential of an ideal solution is:
A = A + RT ln[A] For non-ideal (IONIC) solutions A = A + RT ln aA where aa = A . [A] and aA is the activity of A, A is the activity coefficient A < 1 for dilute ionic solutions due to electrostatics Electrostatic interactions between an ion and its ionic atmosphere lower its chemical potential

16 G = we =  - ideal = RTln 
The difference in  between the charged solution an ideal one without electrostatic interactions gives the energy of charging, we G = we =  - ideal = RTln  Neutral + Charges on - we G -ve if  < 1

17 The activity coefficient (deviation from ideality) depends upon the
ionic strength of the solution: where ci is the concentration of ion i, and zi its charge. A general salt has the form: z is charge on the anion/cation  gives stoichiometry sum includes all ions in solution Na+ Cl- Na2+ SO42-

18 Calculate the ionic strength of a 0.2 M CaCl2 solution
Calculate the ionic strength of a 0.2 M KCl solution cation anion I = Conc for 1:1 salt

19 ONLY VALID FOR VERY DILUTE SOLUTIONS
Can now calculate activity (ion-ion interactions) where z is the charge on the anion/cation, A depends on T and solvent and I the ionic strength. Assuming: Only electrostatic interactions between hard-sphere ions (no ion-solvent) Electrolyte is fully dissociated and no ion-pairs exist Structureless solvent with uniform permittivity Debye-Huckel Limiting Law ONLY VALID FOR VERY DILUTE SOLUTIONS - + Ion-pair often formed in solvents of low polarity

20 Electrochemistry Electrodes Electrochemical cells Electrode potentials
Nernst Equation Electrode types

21 Nature dislikes charge separation
Electrodes Electroneutrality: Nature dislikes charge separation Consider immersing a Zn rod in water: small number of Zn atoms dissolve as Zn2+ ions electrons left behind on rod Zn(s)  Zn2+ + 2e-

22 rapidly inhibits this process  concentration of Zn2+ ions v. low
Build up of: –ve charge on rod +ve charge in solution rapidly inhibits this process  concentration of Zn2+ ions v. low This prohibition is the Electroneutrality Principle measure of work required to separate charges non-spontaneous (G > 0) Reaction can be enhanced by: draining excess e- from rod adding an e- acceptor to metal (e.g. H+) ~10-10 M  pure H2O acids attack metals

23 Later method is basis of electroplating:
If place Zn rod in aqueous CuSO4 instead of H2O Zinc metal quickly covered with black coating of finely-divided metallic copper. Simple oxidation-reduction process involving 2 e- from Zn to Cu Zn(s)  Zn2+ + 2e- Cu2+ + 2e-  Cu(s) Zn dissolution allowed as e-s removed from rod by copper ions and solution stays electrically neutral Net reaction: Zn(s) + Cu2+ Zn2+ + Cu(s) CuSO4 1 hr

24 Electrochemistry is the study of reactions in which charged
particles (ions/e-) cross the interface between 2 phases of matter: a metallic phase - electrode a conductive solution - electrolyte This kind of reaction is known generally as an electrode process. Electrode processes at the surface of electrodes produce a slight charge imbalance between the electrode and electrolyte. this produces an interfacial potential difference (can affect rate and extent of chemical reactions)

25 These interfacial potential differences () are small (a few volts)
BUT occur over extremely short distances. Electrodes immersed in solution have a thin, stable layer of water molecules and ions attached to their surface. Small voltages produce v. large potential gradient: 1 volt over 100 Å = 108 V cm-1 huge potential gradient ~ Å  +

26  = metal - solution = E (V)
ASIDE Interfacial potential difference  - arises from changes in chemical+electrical environments  = metal - solution = E (V) Zn Zn2+ SO42- Metal core Sea of e- Hydrated Free ion Anion interaction metal solution

27 Since electron-transfer reactions occur near electrode surfaces we cannot channel e-s through an instrument to measure this voltage BUT If have 2 metal-solution interfaces can easily measure potential difference between them - this arrangement is called an electrochemical (galvanic) cell. Typical cell consists of 2 metal electrodes e.g. Zn and Cu, each immersed in a salt solution of the corresponding metal 2 solutions are connected by a tube with a porous barrier (salt bridge) - this prevents rapid mixing but allows ions to diffuse through. Oxidation Reduction

28 Electrochemical cells
A special notation is used to describe electrochemical cells: Vertical bars indicate phase boundaries Double vertical bar in the middle denotes salt bridge Each electrode/solution pair is called a half-cell Cell reaction written so: Oxidation occurs at left-hand electrode Reduction occurs at right hand electrode Independent of actual location of 2 electrodes on bench!

29 Ecell = Eright – Eleft Electrochemical cell potential = Right - Left
Conventions Anode is where oxidation occurs Cathode is where reduction occurs If electrons flow from left  right electrodes spontaneously  potential of the RIGHT electrode > LEFT electrode Electrochemical cell potential = Right - Left Ecell = Eright – Eleft Ecell +ve if spontaneous -ve requires external drive Thermodynamics

30 Can only ever measure difference between 2 electrodes
Problem How do we find absolute values for individual half-cells? (ER and EL) Can only ever measure difference between 2 electrodes ? 2nd electrode

31 Standard Electrode Potentials
Can measure half-cell potentials relative to that of other half cells. Adopt a reference electrode (half-cell) whose potential is arbitrarily defined as zero. Measure potentials of various other electrodes against reference Also electrodes but:  (KCl) tiny Cancels anyway! E = (ER+ EKCl) – (EL+ EKCl) = ER - EL Salt bridge

32 Standard Hydrogen Electrode (SHE)
Use under standard conditions: 1 bar 298 K Unit activity (dilute solns)

33 (SEP) Carry out series of measurements for various Mx+/M systems
Construct table arranging these half-cell reactions in order of their potentials. Convention: Write half-cell reactions as reductions (so Mx+/M on RIGHT) e.g. Zn2+ + 2e  Zn(s)  Standard Electrode Potential E = EM2+/M - EH+/H2 (SEP) set = 0

34 SEP clearly related to spontaneity of process
Activity (Electromotive) Series Disfavoured Favoured SEP clearly related to spontaneity of process In practice H2 electrode rarely used in routine measurements: Difficult to prepare; Pt surface has to be specially treated Need H2 gas cumbersome and hazardous

35 Silver-Silver Chloride Electrode (SHE)
Want stable reference: Concentration of ionic species involved must be constant. Simplest method uses an electrode reaction involving saturated soln. of an insoluble salt of the ion e.g. silver-silver chloride electrode Ag | AgCl(s) | Cl- || AgCl(s) + e-  Ag(s) + Cl- Coating made using Ag rod in an electrolytic cell containing HCl - Ag+ ions combined with Cl- ions at the silver surface

36 Prediction of cell potentials
Given the E values for two half reactions, you can easily predict the voltage of the corresponding combined cell: SEP of the reduction half-cell minus SEP of the oxidation half-cell Example 1: Construct a Zn/Cu electrochemical cell & state cell reaction Zn(s) | Zn2+ || Cu2+ | Cu(s) or Cu(s) | Cu2+ || Zn2+ | Zn(s) Cell reaction Zn(s) + Cu2+ Zn2+ + Cu(s) or Zn2+ + Cu(s)  Zn + Cu2+ Ecell = ECu2+/Cu - EZn2+/Zn Ecell = EZn2+/Zn - ECu2+/Cu (-0.76) Ecell = Eright – Eleft

37 Ag(s) | AgCl(s) | Cl- || Cu2+ | Cu(s)
Example 2: Find the standard potential of the cell and the direction of e- flow Ag(s) | AgCl(s) | Cl- || Cu2+ | Cu(s) Half-cell reactions: AgCl(s) + e-  Ag(s) + Cl- Cu2+ + 2e-  Cu(s) Net cell reaction: 2Ag(s) + 2Cl- + Cu2+  AgCl(s) + Cu(s) Ecell = ECu2+/Cu - ECl-/AgCl = (+0.337) – (+0.222) = V Since Ecell is +ve reaction is spontaneous e- flow Ag  Cu Written as reductions

38 Thermodynamics: Nernst equation
Charge imbalance at electrode surfaces produces an interfacial potential difference  - reflects different electrochemical energies in electrode vs soln. Electrochemical potential of species A defined as: Recall: A= A zA FA F = Faradays constant amount of charge carried by 1 mole of electrons 1 F = Coulombs chemical electrical  =  + RT ln[A] A

39 Total reactants= Total products
Example: Fe3+ + e-(metal)  Fe2+ (Fe3+ + 3Fsolution) + (e- - Fmetal) = (Fe2+ + 2Fsolution) Recall for ANY reaction at equilibrium Fe2+ Fe3+ X- Fe3+ e- Fe2+ Rearrange F(metal - solution) = Fe3+ + e- - Fe2+ F  = Fe3+ + e- - Fe2+ Total reactants= Total products  =  RT ln[Fe3+] Fe3+  =  RT ln[Fe2+] Fe2+ where  = (Fe3+ + e- - Fe2+)

40 Nernst Equation since,  = E (V) In general, Thermodynamics states:
-nF

41 Maximum amount of work, w, is given by
Proof: Maximum amount of work, w, is given by w = -VQ where V is emf, Q charge flowing. Q = nF for 1 mole of cell reaction Since dG = dw(T, Pconstant) and when E +ve Spontaneous

42 Often need to describe non-standard conditions M2+(aq) + 2e-  M(s)
What does it all mean?! Often need to describe non-standard conditions M2+(aq) + 2e-  M(s) If [M2+(aq)] is increased, the equilibrium shifts to the right (i.e. more e- are absorbed from the electrode) Actual value of E M2+/M is more positive than E M2+/M Nernst Equation: quantifies this relationship where = RT/F

43 The effect of n: Consider the reaction
Mn(s) | Mn2+(aq) || Fe3+(aq) | Fe2+(aq) First write down the two half-cell reactions. Mn2+ + 2e-  Mn(s) E = V Fe3+ + e-  Fe2+ E = V Cell reaction: 2Fe3+ + Mn  Fe2+ + Mn2+  E cell = (-1.182) = 1.95 V Even though Mn reaction involves 2 e- and the iron only 1 e-!! Hence ½ Mn2+ + e-  ½Mn (s) E = V Mn2+ + 2e- Mn(s) E = V

44 G doubles as expected  twice as much work to transfer 2 e-
How can this be true?! Consider, So if n doubles: G doubles as expected  twice as much work to transfer 2 e- but E remains the same Care must be taken when making comparisons G terms are truly additive convert to G if unsure 2+

45 Calculating how far reactions go?
Since And We can combine the two equations such that We can therefore determine EΘ values from K and vice versa Consider our first reaction. Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) EΘcell = V EΘcell = = ln K Hence, ln K = or K = 1.62 x 1037 n The reaction therefore goes to completion.

46 5 HMnO42- (aq) + 3 H+(aq)  4 MnO4-(aq) + Mn2+(aq) + 4 H2O(l)
More importantly, electrochemistry allows the measurement of very large or very small values of K. Manganate (VI) ions are unstable with respect to disproportionation into Mn(VII) and Mn(II) in acidic solution. 5 HMnO42- (aq) + 3 H+(aq)  4 MnO4-(aq) + Mn2+(aq) + 4 H2O(l) This reaction can be expressed as: HMnO42-(aq) + 7 H+(aq) + 4 e-  Mn2+(aq) + 4 H2O(l) EΘ = V 4 MnO42- (aq) + 4 H+(aq) + 4 e-  4 HMnO4-(aq) EΘ = V Overall E = 0.73 V, hence K = since n = 4. High concentrations of MnO42- cannot be achieved in acidic solution.

47 Calculation of an unknown electrode potential from two
others using G. Given: (a) E (Pu3+/Pu) = V (b) E (Pu4+/Pu3+) = 0.98 V find E (Pu4+/Pu) = ? First write out the half reactions and express in terms of G (a) Pu e-  Pu (s) E = -2.03; G = +6.09F (b) Pu4+ + e-  Pu3+ E = +0.98; G = -0.98F (c) Pu4+ + 4e-  Pu (s) G = -4FE Construct a suitable thermodynamic cycle Gc= Ga + Gb Gc= +6.09F – 0.98F = 5.11F = - 4FE  -4FE = E (Pu4+/Pu) = V

48 Summary Ionic solids require polar solvents to overcome high lattice energies Even in ionic solutions electrostatic interactions between ions cause strong deviations from ideal neutral solutions Deviations increases with ion concentrations/charge Electrochemical potential of ions differs from neutral (solid) phase Interfacial potential difference drives electrochemical reactions (ion reactivity described by electromotive series) Interfacial potential related to Gibbs free energy for e- transfer Electrochemical reactions subject to G>0

49 Electrochemistry Key Equations
Chemical Potentials A = A + RT ln[A] IDEAL A = A + RT ln aA NON-IDEAL aa = A . [A] activity A< 1 for dilute solutions activity coefficient Ionic Strength zi = charge on each ion ci = conc. of each ion Electrochemical Cell Potential Ecell = Eright – Eleft Ecell +ve spontaneous ve non-spontaneous Gibbs Free Energy Nernst Equation Q = reaction quotient  equilibrium constant


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