Ionic Solutions

• Structure of liquids

• Interactions in ionic solutions

• Ion-ion interactions

• Debye-Huckel theory

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

Structure of liquids

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

Uncorrelated

V. weak molecularattraction

% P

roba

bilit

y

g(r) Next nearest neighbour shell

Nearest neighbour shell

• Solid

Strong correlation between atoms even at large separations

lattice structure, fixed neighbours

• Liquid

Well-defined 1st shell but broader than solid & less clear successive shells short-range order

Liquids form due to intermolecular potentials

Broad

Weak

Objective• To find:

Difficult to calculate from individual (microscopic) interactions using statistical mechanics techniques

Use macroscopic view of interactions

Interactions in ionic molecules

Interaction PotentialI,j

H, G, Saverage over 1023 molecules

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

Na+Cl- solid solute

H2Osolvent

Na+aq

+ Cl-aq

solution

strong interionicforces

strong intermolecular forces

strong solute-solvent interactions

Net G (KClaq) = -1.2 kJmol-1

Energy between two ions carrying charge z1 and z2 separated by ris 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

r1

.4

zzEnergy

r0

21

Ion-ion interactions

Like charge

Opposite charge

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

Polar molecules possess a permanent dipole moment,

= q . r

Dipole moment:

1. major source of solvation (hydration) energy

2. reduces ion-ion interactions in solution

Source of polarisation:

1. molecular polarisability : dipole induced by a field E

= . E

r

-q +q

- ++

++-- -

E

Universal property but small r (C6H6) ~ 2

Electrolyte solutions are non-ideal due to long-range ioninteractions

Electrostatic interactions decrease with distance r only as versus for interactions between neutral species.

Debye-Huckel theory

r1

6r

1

0

20

40

60

80

100

0 2 4 6 8 10

1/rPower (1/r6)

Distance between ions / Å

Inte

ract

ion

stre

ngth

%

IonsNeutrals

Electroneutrality of system (solute+solvent) means net chargearound 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

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

The difference in between

1. the charged solution 2. an ideal one without electrostatic interactions

gives the energy of charging, we

G = we = - ideal = RTln

Neutral

+

Charges on

-

--+

+

+

-

we

G -ve

if < 1

The activity coefficient (deviation from ideality) depends upon theionic 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

XM zz

sum includes all ions in solutioni

2

iizc21

I

Na+ Cl-

Na2+ SO4

2-

1. Calculate the ionic strength of a 0.2 M CaCl2 solution

2. Calculate the ionic strength of a 0.2 M KCl solution

M6.0

])1(x4.0)2(x2.0[21

I 22

cation anion

M2.0

])1(x2.0)1(x2.0[21

I 22

I = Conc for 1:1 salt

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:

1. Only electrostatic interactions between hard-sphere ions (no ion-solvent)

2. Electrolyte is fully dissociated and no ion-pairs exist

3. Structureless solvent with uniform permittivity

often formed in solvents of low polarity

- +Ion-pair

Debye-Huckel Limiting Law

IzzAlog10

ONLY VALID FOR VERY DILUTE SOLUTIONS

Electrochemistry

• Electrodes

• Electrochemical cells

• Electrode potentials

• Nernst Equation

• Electrode types

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-

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:1. draining excess e- from rod2. adding an e- acceptor to metal (e.g. H+)

~10-10 M pure H2O

acids attack metals

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

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)

These interfacial potential differences () are small (a few volts)

BUT

occur over extremely short distances.

Electrodes immersed in solution havea 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

~10-100 Å

+

ASIDE

Interfacial potential difference

- arises from changes in chemical+electrical environments

= metal - solution = E (V)

Zn

Zn

Zn

Metal coreSea of e-

metal solution

Zn2+

SO42-

HydratedFree ionAnion interaction

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 electrodese.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

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!

Electrochemical cells

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

Problem

How do we find absolute values for individual half-cells?

(ER and EL)

Can only ever measure difference between 2 electrodes

?

2nd electrode

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

Standard Electrode Potentials

Salt bridge

Also electrodes but:• (KCl) tiny• Cancels anyway!

E = (ER+ EKCl) – (EL+ EKCl)

= ER - EL

Standard Hydrogen Electrode (SHE)

Use under standard conditions:• 1 bar• 298 K• Unit activity (dilute solns)

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 = EM2+/M - E

H+/H2

set = 0(SEP)

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

Activity(Electromotive)

Series

Disfavoured

Favoured

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

Silver-Silver Chloride Electrode (SHE)

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 = ECu2+/Cu - E

Zn2+/Zn Ecell = EZn2+/Zn - E

Cu2+/Cu

0.337 - (-0.76) -0.76 - 0.337

Prediction of cell potentials

Ecell = Eright – Eleft

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 = ECu2+/Cu - E

Cl-/AgCl = (+0.337) – (+0.222) = + 0.115 V

Since Ecell is +ve reaction is spontaneous e- flow Ag Cu

Written as reductions

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 FA

chemical electrical

F = Faradays constant

• amount of charge carried by 1 mole of electrons• 1 F = 96467 Coulombs

= + RT ln[A]A A

Example:

Fe3+ + e-(metal) Fe2+

(Fe3+ + 3Fsolution) + (e- - Fmetal) = (Fe2+ + 2Fsolution)

Recall for ANY reaction at equilibrium

Fe2+

Fe3+X-

Total reactants= Total products

Rearrange

F(metal - solution) = Fe3+ + e- - Fe2+

F = Fe3+ + e- - Fe2+

= + RT ln[Fe3+]Fe3+ Fe3+

= + RT ln[Fe2+]Fe2+ Fe2+

Fe3+ e- Fe2+

where

= (Fe3+ + e- - Fe2+)

F

1

In general,

Thermodynamics states:

KlnRTGG

since, = E (V)

Nernst Equation-nF

nFEG

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

nFEG nFEG

0nFEG

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

)]([)]([

ln2

0256.0

/ /22

sMaqM

EE nMM MM

where 0.0256 = RT/F

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 = -1.182 VFe3+ + e- Fe2+ E = +0.769 V

Cell reaction:2Fe3+ + Mn Fe2+ + Mn2+ E cell = +0.769 - (-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 = -1.182 V Mn2+ + 2e- Mn(s) E = -1.182 V

Care must be taken when making comparisons- G terms are truly additive- convert to G if unsure

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

nFEG

12 kJmol228182.1965002nFE)Mn/Mn(G

13 kJmol74769.0965001nFE)Fe/Fe(G

1)Mn/

2Mn()Fe/

3Fe( kJmol302228)2.74(GGcellG

2+

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) EΘ

cell = 1.101 V

EΘcell = 1.101 = 0.0257 ln K Hence, ln K = 85.6 or K = 1.62 x 1037

n

nFEG

KlnRTG

KlnnF

RTE

The reaction therefore goes to completion.

More importantly, electrochemistry allows the measurement of very large or very small values of K.

)]aq(Cu)][[s(Zn[

)]s(Cu)][aq(Zn[K

2

2

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: HMnO4

2-(aq) + 7 H+

(aq) + 4 e- Mn2+(aq) + 4 H2O(l) EΘ = +1.63 V

4 MnO42- (aq) + 4 H+

(aq) + 4 e- 4 HMnO4-(aq) EΘ = +0.90 V

Overall E = 0.73 V, hence K = 1050 since n = 4.

High concentrations of MnO42- cannot be achieved in acidic solution.

Calculation of an unknown electrode potential from two others using G.

Given:(a) E (Pu3+/Pu) = -2.03 V(b) E (Pu4+/Pu3+) = 0.98 Vfind E (Pu4+/Pu) = ?

First write out the half reactions and express in terms of G (a) Pu3+ + 3 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 cycleG

c= Ga + G

b

Gc= +6.09F – 0.98F = 5.11F = - 4FE

-4FE = 5.11 E (Pu4+/Pu) = -1.28 V

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

Electrochemistry Key EquationsChemical PotentialsA =

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

i

2iizc

21

I

nFEGQ = reaction quotient equilibrium constant