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Electrochemistry & Solutions

1. Solutions and Mixtures

Year 1 – Module 38 Lectures

Dr Adam Leeafl2@york.ac.uk

Department of Chemistry

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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.

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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 ReadingR.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.

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Signifi cance of Areas, Lines and Points

Areas: single phase exists (specif y p and T state of matter)

Lines: 2 phases coexist in ium (specif y p or T need to know what 2 phases present)

P

T

S

L

G

P

T

S L

G

P

T

S

L

L

G

S

G

Phase Diagrams

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Signifi cance of Areas, Lines and Points

Areas: single phase exists (specif y p and T state of matter)

Lines: 2 phases coexist in ium (specif y p or T need to know what 2 phases present)

P

T

S

L

G

P

T

S L

G

P

T

S

L

L

G

S

G

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S-L line change in m.pt. with pressure S-G line " in sub.pt. with pressure L-G line " in b.pt. with pressure (isoteniscope explores how pressure aff ects boiling of liquid)

Triple Point: all 3 phases coexist in ium (unique value of p and T)

Critical Point: conditions (T and p) above which liquid state matter ceases to exist

VacuumPump

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Slope of Solid/ Liquid Line

I ce- water system: Slopes Backward: (rare) i.e. dT/ dP = - ve m.pt. pressure - ice-skating also note ice floats on water (ice< water)

CO2 system:

Slopes Forward: (usual) i.e. dT/ dP = + ve m.pt. pressure solid CO2 sinks in liquid CO2

L

S G

P

T

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These phenomena can be understood by simple (!) thermodynamics BUT FIRST

The Phase Rule obtained f rom experiment derived/ proven by TD allows rigorous discussion of phase behavior

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The Phase Rule

I t states: (only at ium)

F = c - p + 2

Components - smallest no. of substances needed to describe system e.g. H2O = 1 (ice, water and steam all H2O) EtOH + H2O = 2 H2O + NaCl = 2 (despite NaCl Na++Cl-)

No. of degrees of freedom

No. of components

No. of phases

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Phases - no. of unif orm, homogeneous regions within the phase diagram e.g. ice = 1 water + ice = 2 water + ice + steam = 3 Degrees of Freedom - smallest no. of independent State Variables needed to describe ium system

(e.g. p, V, T) A state function (or f unction of state) depends only on the current state of the system, and not on the route by which it was reached.

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Consider Water Diagram

At X P = 1 (solid) F = C - P + 2 = 2

At Y P = 2 (L + G) F = 1

At Z P = 3 (S + L + G) F = 0

S

L

G

P

T

X

Y

Z

must specif y 2 variables to locate position on diagram

position determined by only 1 variable

can only co-exist f or fi xed p and T

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Thermodynamics of Two-Phase Equilibria of a Single Component:

The Clapeyron Equation Thermodynamic eqns. needed 1. G = H - TS

2. G = 0 at ium (S=H/ T) 3. dG = Vdp - SdT (f undamental eqn. f or G)

concerned with equilibria along lines S

L

G

P

T

phase

phase

e.g. ice/water at m.pt.

1839 -1903

Josiah Willard Gibbs

Gibbs Free Energy

American mathematical physicist developed theory of chemical thermodynamics. First US engineering PhD…later Professor at Yale.

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Derivation of Clapeyron Eqn.:

convert 1 mol f rom state to state at ium (i.e. @ transition)

G = G - G = 0

G = G

dG = dG

Since, dG = Vdp - SdT

dG = Vdp - SdT = Vdp - SdT = dG

(V - V)dp = (S - S)dT

G.F.E = G G

1799 - 1864

Benoit Paul Emile Clapeyron

Parisian engineer and mathematician. Derived differential equation for determining heat of melting of a solid

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Application of Clapeyron Eqn. to Solid/ Liquid Phase

Process: f usion (melting)

T = melting point, Tm

H = enthalpy (fH)

V = volume (fV)

= slope of line =

I nverting: = change in Tm = with p

mdTdp

VTHfm

f

dpdTm

HVT

f

fm

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What is fV?

V = 1/ density = 1/ = 1/ gcm-3 = cm3g-1

fV = 1/ l - 1/ s

Sign of slope: Water dp/ dT = - ve = and fH = positive fV = - ve

s < l hence ice floats

VTHfm

f

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Modified eqn. for Liquid/Gas and Solid/Gas Lines:

Clausius-Clapeyron Eqn. Maths dx/x = dlnx dp/p = dlnp

= = =

=

=

dxxn

1nx 1n

2x

dx

12x 12

x1

2T

dT

T1

dTdp

VTH Clapeyron

Equation

Rudolf Julius Emmanuel Clausius

1822 -1888

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Derivation

e.g. liquid/ gas system

V = Vg - Vl

Vg >> Vl

V Vg

For an I deal Gas,

pVg = RT (per mol)

Vg = RT/ p = V

substituting into Clapeyron eqn

= =

dTdp

RTTpH

pdp

2RT

dTH

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Hence,

dlnp = =

I ntegrating,

lnp = - + constant

I soteniscope expt.

NB - normal boiling point is T at 1 atmosphere - T is always in K not C

2TR

dTHdT

plnd2RT

H

T1

RH

y = m x + c

lnp

1/ T

slope = -H/ R X

X X

X

X

Clausius-ClapeyronEquation

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Example Problems The densities of a substance in solid & liquid states at the melting point (23 C) are 0.875 & 0.901 gcm3 respectively. Heat of f usion is 276 J g-1. At what temperature will the substance melt at an applied pressure of 100 atmos.?

Use, =

fV = 1/ l - 1/ s = 1/ 0.901 - 1/ 0.875 = -0.033 cm3g-1 = (23+273) x (-0.033)

convert cm-3m-3..... x 10-6

= -296 x 0.033x10-6 KPa-1

= -0.35 K

mdTdp

VTHfm

f

dpdTm T V

(cm3g-1)

H 276 (J g-1)

dpdTm

276

99

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The vapour pressure of liquid napthelene is 15.5 mmHg at 95 C and 505.7 mmHg at 200.5 C. Calculate the heat of vapourisation and the normal boiling point of napthalene.

Use, ln =

ln =

H = 47,900 J mol = 47.9 kJ mol-1

Normal b.pt. (i.e. Tb f or p = 1atm = 760 mmHg)

ln =

Tb = 490 K (217C)

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v

T1

T1

RH

2

1

p

p

5.157.505

338

15.473

1314.8

Hv

5.15760

338

1T1

314.8900,47

b

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23

24

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Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp

Since pV = nRT,

2

1

2

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2lnp

pnRT

p

nrTdpVdpG

22 lnpnRTGG If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = Go

B + nBRT ln pB

since GA = nAA

A = Ao + RT ln pA

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Phase Diagrams in Binary

Systems

Composition:

mol f raction of A = nA / nA+nB

f or liquids - xA, xB

xA =

f or vapour phase - yA, yB

nA pA, nB pB (pA,pB - partial pressures)

yA = pA / (pA + pB) = pA / p

BBAA

AA

M/wM/wM/w

xA + xB = 1

yA + yB = 1

p = pA + pB

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yA yA yB yB

nA nB nA+ nBGinitial = nAA + nBB

= nA[ + RTlnp] + nB[B + RTlnp]

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp

+ nBlnyBp - + nBlnp]

ya

yb

ya ya yb yb

Initial Final

A B

Gmixing

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yA yA yB yB

ya

yb

yA yA yB yB

Gmixing Smixing

ya

yb

001

1

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Chemical Potential (in English!)

GibbsFreeEnergy

Pressure

Molecules acquire more spare energy

Greater “chemical potential”

Low Pressure

Constant Temperature

High Pressure

G ln(pressure)

Gmolar = G molar + RT lnp

Energy free for moleculesto “do stuff”at STP

Effect of environmenton this free energy

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For single component e.g. pure H2O

Why do we use Chemical Potential?

No real need to use

Gibbs Free Energy (G) is total energy in entire system available to “do stuff”

- includes all molecules, of all substances, in all phases

G = nAA + nBB

For mixturese.g. H2O/C2H5OH

G = nH2OH2O

Free energy only comes from H2O

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH

tells us how much fromH2O versus C2H5OH

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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

Why do we different molecules have different Chemical Potentials?

Ethanol can soak up much more energy in extra vibrational modes and chemical bonds

- will respond differently to pressure/temperature increases

FreeEnergy(G)

&

Gas-phase molecule

Liquid-phaseSolid-state

Pressure

VolatileInvolatile

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Raoult's law

I deal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with

the mol f ractions xA and xB in a liquid.

i.e.

pA = poA xA

pB = poB xB

p = pA + pB = poA xA + po

B xB

y = m . x + c (=0)

Raoult's laweqns. of straight linespassing thru origin

Total pressure above boiling liquid

Volatility

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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

Pure A

0

lnxA

Pure B

xA

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Dilution

Mixing alwayslowers

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Vapour-Pressure Diagrams

Plots, on the same diagram, of vapour

pressure against composition of (i) the liquid

phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition

(the liquid line)

p = pA + pB = poA xA + po

B xB

= poA xA + po

B (1 - xA)

= poA xA + po

B - poB xA

= poB + (po

A - poB) xA

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BUT

p depends on liquid composition (x)

A + B

P I f applied pressure > p

vapour liquid

Liquid poA

p

1 0 xoA

0 1 xoB

Pure A

Pure B

Volatile Involatile

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however

(b) Vapour pressure vs. vapour composition

p is not a linear f unction of y

poA

poB p

1 0 yoA

0 1 yoB

Pure A

Pure B

Vapour

Volatile Involatile

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The Vapour-Pressure Diagram

po

A

poB p

mol f raction A

liquid line

liquid linevapour line

L + V

B

liquid only

vapour only

liquid+ vapour

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Deviations f rom I deality

Associated with intermolecular attractions

(e.g. Van der Waals, H-bonding)

Strength of interaction heat of mixing

i.e. mH f or A-B system

3 cases

A B

A

B Case 1: Ideal

A equally attracted by B

as by itself + vice-versa

(e.g. alkanes) mixH = 0 Ideal mix

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A B

A

BCase 2: -ve deviation

A more attracted by B(e.g. CHCl3 + acetone)

mixH = < 0

A B

A

BCase 3: +ve deviation

A less attracted by B(e.g. EtOH + water)

mixH = > 0

b.pt. > ideal

b.pt. < ideal

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Summary: Raoult’s Law for Solvents

pA = xA . pAΘ

Liquid po

A

1 0 x o

A

p Partial pressure of A

Total pressure

0 1 xoB

Partial pressure of B

pB = xB . pBΘ

pBΘ

Proportionality constant

Involatilelow vapour pressure

Volatilehigh vapour pressure

A B

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p0ө = vapour pressure

= tendancy of system to increase S

High order: low S Less order: higher S

Strong desire to S

Boiling of Afavoured

Less need to S

A happier in liquid

A

A

AA

High p0ө(A) Low p0

ө(A)

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Proportionality constant

Amount in solution

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

Dissolution isEXOTHERMIC

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Consider O2 dissolution in water:

Important in Green Chemistry for selective oxidation

Cinnamyl AlcoholCinnamic Acid

Cinnamaldehyde

Solvent: H2O Solute: O2

pH2O

pO2

Aspects of Allylic Alcohol OxidationAdam F. Lee et al, Green Chemistry 2000, 6, 279

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

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.

H2O O2

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The Tie-Line

Horizontal line thru a point on V-P diagram

defi ning p and composition of the system

join compositions of co-existing

phases at equilibrium

poA

poB p

mol f raction (liq, vap)

L

V

A B

A

xAyA

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The Lever Rule

tie-lines composition of phases But how much total material is in each phase Defi ne no. of mols in liquid & vapour:

nl = (nA)l + (nB)l

nv = (nA)v + (nB)v

Lever Rule:

mol f raction

poA

poB p

A B

l v a

allengthavlength

n

n

v

l

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A A

AB

A

A

BA

B B

BA

B

B

AB

Tie-line

LIQUID

GAS

LIQUID

GAS

Lever Rule

A-B Composition Liquid-Gas Distribution

B A

Volatile Involatile

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Boiling Point Diagrams

1. liquids boil when p = prevailing pressure

(f or normal b.p. , p = 1atm.)

2. b.p. is 1/ vapour pressure (po)

B.P. diagrams are of opposite slope to V-P

Thus

Fixed T Fixed p

0 xA 1

liquid

vapour

0 xA 1

vapour

liquid TB,A

TB,B L + V L + V

T

poB

poA

p

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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

x(Toluene) y(Toluene) Temperature / C0.908 0.923 110.90.795 0.836 1120.615 0.698 1140.527 0.624 115.80.408 0.527 117.3

0.3 0.41 1190.203 0.297 1200.097 0.164 123

P.W. Atkins, Elements of Phys.Chem. page 141

Boiling/Condensation Temperature

Liquid Vapour

x(T) = 1, T = 110.6 C

x(O) = 1, x(T) = 0, T = 125.6 C

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Principle of Distillation Take liquid x1 & heat to bubble line

vapour composition y

Cool vapour liquid composition x2 REPEAT!!

T

mol fraction 1 0

x1

y2

y1

boiling

x2

cool

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Extreme Deviations f rom I deality

Result in:

maxima/ minima in B.Pt + V.P. diagrams

AZEOTROPES

a - without zeo - boil

tropos - change

means composition does not change during boiling

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A B

A

BCase 2: -ve deviation

A more attracted by B(e.g. CHCl3 + acetone)

mixH = < 0

A B

A

BCase 3: +ve deviation

A less attracted by B(e.g. EtOH + water)

mixH = > 0

b.pt. > ideal

b.pt. < ideal

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Vapour-Pressure diagrams

Boiling-Point diagrams

P

A composition B

Vapour

Liquid

L+V L+V

L+V L+V

A composition B

Vapour

Liquid

+- ve - ve

T

A composition B

Vapour

Liquid

L+V L+V

L+V L+V

A composition B

Vapour

Liquid

azeotropic composition

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Distillation of Azeotropes

T

A composition B

V

L

L+V

A B

P

T

A composition B

V

L

L+V

A

B

Distillate always ends up at azeotropic composition Residue is pure A or B

Distillate always ends up as pure A or B Residue is mixture with azeotropic composition

-EtOH cannot be dried by distillation P = 96% EtOH, 4% H2O

A

B

A B

Residue Distlllate

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Topics Covered (lectures 2-4)

Chemical Potential - A(l) =

A(l) + RTlnxA

- A(g) = A(g) + RTlnpA

Mol fractions - A = nA / nA+nB

Raoult’s Law - pA = po

A xA pB = poB xB

- ideal solutions

- +ve/-ve deviations

Vapour-pressure diagrams - Tie-lines

- Lever Rule

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Solutions EquationsPhase Rule

F = c - p + 2 c = componentsdegrees of freedom p = no. of phases

Clapeyron EquationH = enthalpy of phase change

V = volume change associated with phase change

Clausius-Clapeyron Equation

or

Mol fractionsxA = nA / nA+nB ni = mols of i

yA = pA / pA + pB pi = partial pressure of i

Raoults LawpA = po

A xA and pB = poB xB

Lever Rule (for tie-line joining phases via point a)

nl =no. moles in liquid phasenv =no. moles in liquid phase

dTdp

VTH

dTplnd

2RTH

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v

2

1

T1

T1

RH

pp

ln

allengthavlength

nn

v

l

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Contains solute (e.g. NaCl, glucose)

WHY?!!!

Solvent AA(l) =

A(l) + RTlnxA

A

A(solution) < A(solvent)

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