ionic solutions structure of liquids interactions in ionic solutions ion-ion interactions...
TRANSCRIPT
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Ionic Solutions
• Structure of liquids
• Interactions in ionic solutions
• Ion-ion interactions
• Debye-Huckel theory
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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
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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
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g(r) Next nearest neighbour shell
Nearest neighbour shell
• Solid
Strong correlation between atoms even at large separations
lattice structure, fixed neighbours
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• Liquid
Well-defined 1st shell but broader than solid & less clear successive shells short-range order
Liquids form due to intermolecular potentials
Broad
Weak
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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
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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
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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
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Like charge
Opposite charge
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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
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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
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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
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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
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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
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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
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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-
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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
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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
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Electrochemistry
• Electrodes
• Electrochemical cells
• Electrode potentials
• Nernst Equation
• Electrode types
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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-
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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
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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
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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)
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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
+
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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
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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
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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
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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
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Problem
How do we find absolute values for individual half-cells?
(ER and EL)
Can only ever measure difference between 2 electrodes
?
2nd electrode
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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
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Standard Hydrogen Electrode (SHE)
Use under standard conditions:• 1 bar• 298 K• Unit activity (dilute solns)
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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)
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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
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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)
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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
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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
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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
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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
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In general,
Thermodynamics states:
KlnRTGG
since, = E (V)
Nernst Equation-nF
nFEG
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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
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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
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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
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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+
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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.
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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.
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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
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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
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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