1. What’s in a solution? How far does a reaction “go”?
2. What factors influence how far a reaction “goes” and how fast it gets there?
3. How do atomic and molecular structure influence observed properties of substances?
Big Question #2
What factors influence how far a reaction “goes” and
how fast it gets there?
Sections 10.1-6
Phase Changes and Intermolecular Forces
A Macroscopic Comparison of Gases, Liquids, and Solids (Review from
State
Shape and Volume
Compressibility
Gas
Conforms to shape and volume of container
High
Liquid Conforms to shape of container;
Very low
volume limited by surface
Solid
Maintains its own shape and volume
Almost none
Ch. 1)
Ability to Flow
High
Moderate
Almost none
Energies of Phase Changes (Some review from Chapter 5, Some New)


heat of fusion (∆HFUS): enthalpy change for the melting of 1 mole of a
substance
e.g., H2O(s)  H2O(l)
∆H = ∆HFUS = +6.02 kJ
heat of vaporization (∆HVAP): enthalpy change for the vaporization of
1 mole of a substance
e.g., H2O(l)  H2O(g)
∆H = ∆HVAP = +44.0 kJ
PROTIP
Heat of fusion and heat of
vaporization are both
ALWAYS endothermic.
For a substance, the heat of
vaporization is always larger
than the heat of fusion.
At a phase change, all added/removed energy goes into the phase change and none goes into changing
the temperature. Thus, any measurement of the heat involved when passing through a phase change has
to be broken into steps. (Also recall from Chapter 5: Hess’s Law, Specific Heat, q=mcΔT)
e.g., Consider converting 1.5 mol of ice at –25 °C to liquid water at 35 °C.
What quantity of heat is required by this process in J?
The molar heat capacity of ice is 37.6 J·mol–1·°C –1. The molar heat capacity of liquid water is
75.4 J·mol–1·°C –1. The heat of fusion of water is 6.02 kJ·mol–1.


 37.6 J 
q =  (1.50 mol) 
  0 C  (  25 C)  
 mol C 



 6.02 kJ   1000 J  
+  (1.50 mol) 


 mol   kJ  



 75.4 J 
+  (1.50 mol) 
 (35 C  C) 
 mol C 


Big Question 2
1
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Equilibrium Nature of Phase Changes
~Figure 10.21 Liquid-vapor equilibrium
Effects of Temperature and Intermolecular Forces on Vapor Pressure
Figure 10.22 Vapor pressure vs.
temperature and intermolecular forces
Effect of temperature on molecular speeds
Big Question 2
2
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Phase Diagrams (below: ~Figures 10.25, 10.26)
Intermolecular Forces
Big Question 2
(Big Question #3 Alert!)
3
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Figure 10.7: Boiling points of Period 2, 3, 4, and 5 hydrides
Figure 10.8
H-bonds in DNA
Big Question 2
Table 10.3
Molar Mass & Boiling Point
4
Figure 10.15
Dispersion forces & molecular shape
3150:153-004/801
Sections 11.1–4
The Properties of Mixtures: Solutions and Colloids
Solution Process
Solution formation generally depends on relative strengths of intermolecular forces.
A solution forms if the driving forces can combine to give a favorable energy change.
We can model the solution process as a 3-step process and use Hess’s Law to determine the overall
enthalpy of the solution process (∆HSOLN):
1. Separation of solute particles (∆Hsolute endothermic)
2. Separation of solvent particles (∆Hsolvent endothermic)
3. Mixture of solute and solvent particles (∆Hmix exothermic)
∆HSOLN = ∆Hsolute + ∆Hsolvent + ∆Hmix
Case #1. Solution formation & the role of enthalpy (especially important for ionic solids in water)
Big Question 2
5
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Lattice Energy: energy associated with the coalescence of an ionic solid crystal from gaseous ions (p. 521)
e.g.,
Lattice energy is related to melting point, hardness, solubility of ionic compounds.
Lattice energy is proportional to the charge on the ions and inversely
proportional to the size of the ions (i.e., as the ions get smaller, the lattice
energy increases).
PROTIP
You would never be able to directly
measure lattice energy in a constantpressure calorimeter in 3150:152 lab.
Periodic Trends in Lattice Energy
Lattice Energies of Alkali Metal Halides
(kJ·mol–1)
F–
Cl– Br– I–
Li+ 1036 853 807 757
Na+
923 787 747 704
K+
821 715 682 649
Rb+
785 689 660 630
+
740 659 631 604
Cs
Lattice Energies of Compounds of
the OH– and O2– Ions (kJ·mol–1)
OH–
O2–
Na+
900
2,481
2+
3,006
3,791
Mg
Al3+ 5,627 15,916
Lattice energies of alkali metal halides
Case #2. Solution formation and the role of entropy
(important for mixtures of covalent cpds.)
Many mixtures of like nonpolar compounds form despite
∆HSOLN being nearly zero. Entropy drives the formation of
such solutions.
Big Question 2
6
solubility of ionic cpds. vs.
temperature
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Case #3. No solution formation


Ionic compounds do not dissolve in
nonpolar solvents.
Some ionic compounds are “insoluble”
in water.
Both situations result from ∆Hsolute being
much, much larger than ∆Hmix such that
∆HSOLN is far too endothermic to allow the
process to occur.
The moral of the story—a rule of thumb:
LIKE dissolves LIKE
Factors that Affect Solubility
1. Structure Effects
or
Vitamin C
2. Pressure Effects
Sgas = kH Pgas “Henry’s Law”
Vitamin A1
gases are more soluble in liquids at higher pressure
3. Temperature Effects
Generally, increasing temperature increases the rate of dissolving, but has various effects on the
amount of solute that will dissolve.
Gases: more soluble in liquids at lower temperature
Solids: depends on ∆HSOLN
Big Question 2
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Non-Liquid Solutions

Gas-gas
e.g., air
all gases are infinitely soluble in other gases.

Solid-solid
e.g., alloys, waxes
o substitutional alloy
vs.
interstitial alloy
Beyond Solutions: Colloids & Suspensions

Suspension
heterogeneous mixture

settles out eventually

suspended particles
visible to naked eye
Big Question 2

Colloid
Dispersed substance larger than
simple molecules, but small enough
to not settle out (sizes 1–1000 nm,
includes the range of visible
wavelengths) colloids scatter light
beams
8

Solution
homogeneous mixture

does not separate

visually uniform
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Chapter 14
Chemical Kinetics
the study of rates of chemical reactions. How fast is a reaction? The kinetics of a reaction must be
determined experimentally.
 reaction rate =
change in concentration
change in time
rate =
Δ(A)
d (A)
=
Δt
dt
The differential form of the rate shows that the rate = the slope of the tangent line to the curve.
Big Question 2
9
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Δ(H 2 O)
Δt
Δ(O 2 )
rate of appearance of O2 =
Δt
Δ(H 2 O 2 )
rate of disappearance of H2O2 = –
Δt
The rate of reaction changes during the course of the reaction. Looking at the H2O2 data:
So, for 2H2O2 → 2H2O + O2, rate of appearance of H2O =
Δ(H 2 O 2 )
Δt
Δ(H 2 O 2 )
rate2 = –
Δt
Δ(H 2 O 2 )
rate3 = –
Δt
Δ(H 2 O 2 )
rate4 = –
Δt
rate1 = –
=–
 0.500  mol·L1
= 2.31 × 10–5 mol·L−1·s−1
2.16  10 s
0.250  mol·L1

= 1.16 × 10–5 mol·L−1·s−1
=–
4
2.16  10 s
 0.125 mol·L1 = 0.58 × 10–5 mol·L−1·s−1
=–
2.16 104 s
 0.0625  mol·L1 = 0.29 × 10–5 mol·L−1·s−1
=–
2.16  104 s
4
In general, for aA + bB → cC + dD,
–
1 Δ(A)
1 Δ(B) 1 Δ(C)
1 Δ(D)
=–
=
=
a Δt
b Δt
c Δt
d Δt
◄ relationships between/among rates in terms
of different reactants and products
e.g., hydrogen peroxide disappears at twice the rate at which oxygen appears: –
Δ(H 2 O 2 )
 Δ(O 2 ) 
= 2

Δt
 Δt 
Differential Rate Laws
A general relationship between species concentration and reaction rate is called a rate law.
Assumptions: 1. only forward reactions (i.e., write rate laws in terms of reactants)
2. equilibrium avoided
We observed that the rate of reaction depended on (H2O2), but how exactly?
n = order of reaction
rate  (H2O2)n
rate = k (H2O2)n
k = rate constant
Δ(H 2 O 2 )
rate = –
= k (H2O2)n ◄a differential rate law: rate as a function of concentration
Δt
¡CAUTIONS!
1. n not necessarily equal to stoichiometric coefficient.
2. n and k are experimentally determined.
3. rates vary for each product/reactant. Must specify.
e.g., Consider this reaction:
2ClO2(aq) + 2OH–(aq) → ClO3–(aq) + ClO2–(aq) + H2O(l)
and the following initial rate data:
(ClO2)o/mol·L–1
(OH–)o/mol·L–1
initial rate/mol·L–1·s–1
0.0500
0.100
5.77 × 10–2
0.100
0.100
2.32 × 10–1
0.100
0.050
1.15 × 10–1
Determine the rate law for the reaction and the value of k (with units).
Big Question 2
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Integrated Rate Laws
Differential rate laws give information about rate as a function of concentration, but what about
concentrations as a function of time? We need to integrate the differential rate law.
The calculus involved for orders 0, 1, and 2 is below…it’s only necessary to use the final results and not
worry about the derivation.
Zero-Order
(A)
Rate = –
= k(A)0 = k
t
d (A)
or Rate = –
=k
dt
– d(A) = k dt
First-Order
(A)
= k(A)1 or
t
d (A)
Rate = –
= k(A)
dt
d (A)
= k dt
–
(A)
Rate = –
d(A) = – k dt
(A)

d (A) = –k
(A)o
t
Half-Life
0
let (A) =
d t
(A) – (A)o = –k(t – 0)
(A)o
at t = t½
2
(A)o
= –kt½
2
(A)o
= t½
2k
–
(A)o
= –kt½ + (A)o
(A) – (A)o = –kt
2
(A) = –kt + (A)o ◄an integrated rate law: concentration as a function of time
d (A)
= – k dt
(A)
(A)
d (A)
(A) (A) = –k
o
t
Half-Life
0
let (A) =
d t
ln(A) – ln(A)o = –k(t – 0)
ln(A) – ln(A)o = –kt
ln(A) = –kt + ln(A)o
(A)o
at t = t½
2
 (A)o 
ln 
 = –kt½ + ln(A)o
 2 
ln(A)o–ln 2 = –kt½+ln(A)o
– ln 2 = –kt½
ln 2 = kt½
ln 2
= t½
k
Second-Order
(A)
= k(A)2 or
t
d (A)
Rate = –
= k(A) 2
dt
d (A)
–
= k dt
(A)2
d (A)
= – k dt
(A)2
Rate = –
(A)
d (A)
(A) (A)2 = –k
o
t
d t
Big Question 2
0
 1 
1
– 
 = –k(t – 0)
(A)  (A) o 
1
1
+
= –kt
–
(A) (A) o
1
1
–
= kt
(A) (A) o
–
1
1
= kt +
(A)
(A) o
11
Half-Life
(A)o
at t = t½
2
1
1
= kt½ +
(A) o
 (A) o 


 2 
2
1
= kt½ +
(A) o
(A) o
let (A) =
1
= kt½
(A) o
1
= t½
k (A) o
3150:153-004/801
Summary of Rate Laws
Order
0
1
2
Differential Rate Law
d (A)
Rate = –
=k
dt
d (A)
Rate = –
= k(A)
dt
Rate = –
Integrated Rate Law (straight-line form)
(A) = –kt + (A)o
ln(A) = –kt + ln(A)o
1
1
= kt +
(A)
(A) o
d (A)
= k(A)2
dt
Using the Integrated Rate Laws
Example: What are the order and rate constant at 25 °C for the decomposition of N2O5?
2N2O5(g) → 4NO2(g) + O2(g)
Experimental method: place 0.1000 mol of N2O5 in a
1.000 L flask and measure (N2O5) as a function of time.
Experimental data:
time, s
(N2O5), mol·L–1
0
0.1000
50
0.0707
100
0.0500
200
0.0250
300
0.0125
400
0.00625
Approach: Make a guess at the reaction order, n. Plot the experimental data using the
corresponding integrated rate law. The integrated rate law that gives a straight line will give the
values of n and k.
Integrated rate law plots and reaction orders.
Half-Life, t½
 the amount of time required for the reactant concentration to decrease to half its initial value
1
i.e., the amount of time to reach (A) = 2(A)o
You can derive the formulas for half-life from the integrated rate laws; see above.
Only the first-order half-life formula is independent of the initial concentration.
Big Question 2
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Collision Model of Chemical Kinetics (CM)
Experimentally, reaction rate is affected by:
1. Concentration 2. Temperature
3. Orientation
Figure 14.17a
“Activated Complex” or “Transition State”:
 unstable species present as reactant bonds break and product bonds form.
 can’t generally be isolated.
 activation energy, Ea: energy required to overcome the barrier to reaction.
Big Question 2
13
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Evaluating Ea
1889, Svante Arrhenius
k = Zρ e
 Ea
RT
 Ea
RT
= Ae


 RTEa 
E
ln k = ln  Ae  = ln A + ln  e  = ln A – a
RT




E 1
ln k = – a   + ln A
R T
 Ea
RT
1
Determine k at various temperatures; plot ln k vs.   .
T
Or measure k at two temperatures (T1 and T2):
k  E  1 1 
ln  2  = a   
 k1  R  T1 T2 
¡CAUTIONS!
 need R = 8.3145 J·mol–1·K–1
 T must be in K
 Ea will be calculated in J, not kJ
Catalysts



increase rate of reaction
are not consumed overall by reaction (disappear, but later reappear)
provide alternative reaction pathways with lower activation energies
Reaction Mechanisms
 reaction mechanism: sequence of individual steps by which a reaction proceeds.
The slowest step governs the overall rate. Mechanistic steps MUST
o add to overall reaction.
o be physically reasonable. (For example, termolecular steps are unreasonable.)
o correlate with observed rate law.
e.g.,
NO2 + CO → CO2 + NO
observed rate law:
rate = k (NO2)2
observed rate law:
rate = k (NO)2(H2)
fast
2 NO º N2O2
slow
N2O2 + H2 → N2O + H2O
The slow second step causes some N2O2 to build up waiting to react. The first step reaches
equilibrium.
(N 2O2 )
K=
, so (N2O2) = K(NO)2
2
(NO)
The slow step governs the overall rate of the reaction: rate = k (N2O2)(H2)
But N2O2 can not be included in the rate law since it is not a reactant…..
e.g.,
2 NO + H2 → N2O + H2O
Proposed mechanism:
Big Question 2
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3150:153-004/801
3150:154 Experiment 4 (Chemical Kinetics)
6 I–(aq) + BrO3–(aq) + 6 H3O+ (aq) → 3 I2 (aq) + Br–(aq) + 9 H2O(l)
[BrO3 ]
–
rate = rate of loss of BrO3 = −
t
Δ[I 2 ]
rate = 13 (rate of formation of I2) = + 13
Δt
– n
– m
+ p
rate = k [I ] [BrO3 ] [H3O ]
The three pairings can be accomplished in a total of four experiments at room temperature, as covered
in the Procedure. Experiments 3 and 4 in Table 2 in the Procedure fulfill the requirements of case a:
[I–]3 ≠ [I–]4
[BrO3–]3 = [BrO3–]4
but
and
[H3O+]3 = [H3O+]4
Given these concentration values, we can use equation 4 to write the initial rates of reaction in the two
experiments:
rate3 = k ([I–]3)m ([BrO3–]4)n ([H3O+]4)p
rate4 = k ([I–]4)m ([BrO3–]4)n ([H3O+]4)p
The ratio of the rates is

 [I ]3 
rate3 [I ]3 



rate4 [I  ] m  [I ]4 
4
m
m
In equation 5, rate3, rate4, [I–]3 and [I–]4 are numbers that you measure experimentally. Only the value
of the exponent m is unknown. Solve for m using the laws of logarithms:
 rate3 
 [I  ]3 
ln 
m
ln


  
 rate4 
 [I ]4 
m
ln  rate3 rate 4 
ln [I  ]3 [I  ]4 
Table 2. Components of Solutions A and B for Room Temperature Experiments
Components of Solution A (mL)
Exp.
Components of Solution B (mL)
H2O
Na2S2O3
KI
KBrO3
HCl
1
0
2.00
2.00
4.00
2.00
2
0
2.00
2.00
2.00
4.00
3
0
2.00
4.00
2.00
2.00
4
2.00
2.00
2.00
2.00
2.00
3H
0
2.00
4.00
2.00
2.00
3C
0
2.00
4.00
2.00
2.00
Big Question 2
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Chapter 18
Chemical Thermodynamics
First Law of Thermodynamics
 Energy can be converted from one form to another, but can not be created or destroyed.
ΔEUNIV = ΔESYS + ΔESURR = 0
Reaction Spontaneity
“spontaneous”: occurring without outside intervention
Entropy, S
Entropy is
 a measure of the tendency of energy to spread out, to
“diffuse”, to become “less concentrated”
 a measure of the number of ways energy can be
distributed among the motions of particles.
 a measure of a driving force.
 not necessarily conserved.
Entropy is NOT
 a driving force itself
 disorder or a measure of disorder
 applicable to macroscopic objects
Ludwig Boltzmann (late 1800s):
S = k ln W
Second Law of Thermodynamics
 In any spontaneous process, the entropy of the universe increases.
q
q
Experimentally, for a reversible process, ΔSSYS = SYS and ΔSSURR = SURR at constant T
T
T
Example: An iron skillet at 500 K is allowed to cool in the kitchen (300 K).
The skillet transfers 500 J of heat to the kitchen. Why does this occur spontaneously?
Big Question 2
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Alternate form of Second Law: Energy always flows as heat from hot objects to cold ones.
The Second Law allows predictions about what should occur, but says nothing about when.
Spontaneous processes can be very, very slow. Something holds back some spontaneous processes;
otherwise all spontaneous processes would be instant.
Obstructions to the Second Law make life possible!
∆S and Spontaneity Recap
ΔSUNIV + : spontaneous process
ΔSUNIV – : nonspontaneous process; reverse process spontaneous
ΔSUNIV = 0 : process at equilibrium (no net tendency in either direction)
Gibbs Free Energy, ∆G
We would like to be able to predict spontaneity based on a system variable rather than a universe one.
ΔSUNIV = ΔSSYS + ΔSSURR
(constant T)
and ΔH = qSYS
(constant P)
Heat is exchanged between the system and surroundings:
ΔHSYS = qSYS = –qSURR = –ΔHSURR
And ΔSSURR =
qSURR
ΔH SURR
ΔH SYS
=
=–
T
T
T
◄ΔSSURR in terms of system
ΔH SYS
T
◄ΔSUNIV in terms of system
Substituting:
ΔSUNIV = ΔSSYS –
Multiplying by –T:
–TΔSUNIV = –TΔSSYS + ΔHSYS = ΔHSYS –TΔSSYS
ΔGSYS = ΔHSYS –TΔSSYS
◄ TΔS: energy units
◄ ΔG ≡ Gibbs Free Energy
ΔG = ΔH –TΔS
∆G and Spontaneity
ΔSUNIV = –
ΔGSYS
T
ΔG – : spontaneous process
ΔG + : nonspontaneous process; reverse process spontaneous
ΔG = 0 : process at equilibrium (no net tendency in either direction)
Gibbs Free Energy is the amount of energy that is free to do useful work in a spontaneous process.
ΔG = wMAX (in theory—you never get that much work in real processes)

If ΔG –, work that can be done on surroundings by system

If ΔG +, work that must be done on system by surroundings to force the nonspontaneous process to
occur.
Big Question 2
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3150:153-004/801
Many bodily processes are kept from reaching equilibrium by biochemical coupling. A system at
equilibrium can not do any useful work.
Equilibrium: point of minimum free energy and maximum entropy
Third Law of Thermodynamics
 The entropy of a perfect, crystalline substance at absolute zero is 0.
Standard State conditions
compounds
 gas: pressure of 1 bar (100 kPa, ~1 atm)
 pure liquid/solid
 solution: 1 M concentration
elements
 form in which element exists at 1 bar and T specified
ΔHF° and ΔGF° of an element in its standard state ≡ 0/
∆H, ∆S, T, and Spontaneity
1. ΔH –, ΔS +: always spontaneous (ΔG always –) because both ΔH and ΔS are favorable
2. ΔH +, ΔS –: always nonspontaneous (ΔG always +) because both ΔH and ΔS are unfavorable
3. ΔH –, ΔS –: sign of ΔG depends on T and magnitudes of ΔH and ΔS
4. ΔH +, ΔS +: sign of ΔG depends on T and magnitudes of ΔH and ΔS
Big Question 2
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3150:153-004/801
For reactions that change in spontaneity depending on T, we can calculate the “crossover temperature”:
Example:
Hg(l) º Hg(g)
ΔH° = 60.83 kJ
ΔS° = 97.49 J·K−1
∆G, ∆G°, and K
ΔG° = free energy change when reactants in standard states are converted to products in standard states
Why bother with ΔG°?
1. Can compare relative tendencies of reactions to occur
2. Need it to calculate ΔG under nonstandard conditions
ΔG = ΔG° + RT ln Q
3. Provides information about equilibrium position
At equilibrium, ΔG = 0 and Q = K
Example: What is ΔG° for the auto-ionization of water?
Big Question 2
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Chapter 19
Electrochemistry
Review: Reduction-Oxidation Reactions (See Chapter 4)
Driving force: transfer/shift of electrons
“LEO the lion says GER”
Loss of Electrons is Oxidation; Gain of Electrons is Reduction
“OIL RIG”
Oxidation Is Loss; Reduction Is Gain
“ELMO”
Electron Loss Means Oxidation
The substance that is oxidized ≡ the reducing agent.
The substance that is reduced ≡ the oxidizing agent.
A reaction is a “redox reaction” if the oxidation numbers of one or
more atoms changes.
e.g.,
2Zn(s) + O2(g)  2ZnO(s)
ON of Zn changes from 0 to +2
ON of O changes from 0 to –2
 This is a redox reaction,
and
Zn was oxidized and was the reducing agent,
and
O2 was reduced and was the oxidizing agent.
Electrochemistry


the study of interchange of chemical and electrical energy
involves redox reactions
Goal: to be able to do useful things (work) by generating an electric current.
Much of the experimental evidence for thermodynamic concepts arose from electrochemical
experiments!
Experimentally, this reaction is spontaneous (i.e., ΔG° is negative): Zn(s) + Cu2+(aq)  Cu(s) + Zn2+(aq)
The problem is that the electron transfer occurs at the interface between the substances. To do useful
electrical work, we need to force the electrons to travel through a wire; i.e., the reduction and oxidation
processes must be separated!
Big Question 2
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We need to add an apparatus that will allow electrical neutrality to be maintained in both beakers
without allowing the contents of the beakers to mix.


Salt bridge
o allows ions to flow, maintaining electrical neutrality
o contains a strong electrolyte in a gelatinous matrix or a porous/absorbent substance
saturated with a strong electrolytes solution, with the ends closed with a glass frit.
o is an “external source” of ions.
Porous disk
o prevents solutions from mixing significantly but allows ions to cross, maintaining charge
balance.
o is an “internal source” of ions.
voltaic (galvanic) cell: device in which chemical energy is converted to electrical energy
electrodes: metallic conductors that make electrical contact with contents of half-cells.
anode: electrode at which oxidation occurs
cathode: electrode at which reduction occurs
In the salt bridge:
Big Question 2
anions flow toward/into anode compartment
cations flow toward/into cathode compartment
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Cell Potential, E (E°)
measured in volts
1 volt = 1 joule energy per coulomb charge transferred [1 V = 1 J·C−1]
E°CELL = E°OVERALL = E°OXIDATION + E°REDUCTION = E°CATHODE – E°ANODE
E°OXIDATION = – E°ANODE
E°REDUCTION = E°CATHODE
e.g., the two tabulated reductions Zn2+ + 2 e–  Zn
E°RED = –0.76 V
Cu2+ + 2 e–  Cu
E°RED = +0.34 V
and
Reverse Zn half-reaction: Zn  Zn2+ + 2 e–
E°OX = +0.76 V (= – E°ANODE = – (–0.76 V))
Add to Cu half-reaction: Cu2+ + 2 e–  Cu
E°RED = +0.34 V (= E°CATHODE)
For Zn + Cu2+  Cu + Zn2+:
E°CELL = +1.10 V
We need a reference to measure half-cell E°s…
“Standard Hydrogen Electrode” (SHE) (really a half-cell)
 Assigned exactly zero volts
H2(g)  2H+(aq) + 2e–
OR
2H+(aq) + 2e–  H2(g)
Cell Shorthand Notation
anode
solid
species oxidized,
if different
oxidation
product
species
reduced
reduction
product
cathode solid,
if different
e.g.,
For Zn(s) + Cu2+(aq)  Cu(s) + Zn2+(aq):
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
e.g.,
For 2H+(aq) + Fe(s)  H2(g) + Fe2+(aq):
Fe(s) | Fe2+(aq) || H+(aq) | H2(g) | Pt(s)
E° and Reaction Spontaneity—or—Predicting Redox Reactions
 Redox reactions that have a positive overall E° are spontaneous as written.
∆G = –nFE
∆G° = –nFE°
–
n, F always positive
n = moles e transferred
F = Faraday constant
∆G, ∆G° negative
if E, E° positive
96,485 coul/mol e–
Big Question 2
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Example: Can acid oxidize iron?
Nernst Equation Cell potentials at other-than-standard conditions
ECELL = E°CELL !
R = 8.314 J·mol−1·K−1
T in Kelvin
n = moles e– transferred
F = 96,485 coul·mol−1
Q = reaction quotient
RT
ln Q
nF
Application: Concentration Cells
Same half-reactions, but
different concentrations.
Cell will run so as to
equalize concentrations.
Big Question 2
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Application: Measurement of Equilibrium Constants
RT
ECELL = E°CELL !
ln Q
At equilibrium, E = 0 and Q = K, so
nF
ln K =
nFE 
RT
A Couple Selected Redox/Electrochemistry Applications

Corrosion (ingredients: anode + cathode + electrolyte + conductor)
Some metals form durable oxide coatings that prevent further oxidation.
Many do not, however, like iron…..
Corrosion Prevention
1. Galvanizing—coat steel with a layer of zinc
Fe  Fe2+ + 2 e–
E°OX = +0.44 V
2+
–
E°OX = +0.76 V
Zn  Zn + 2 e
2. Alloying—add metals that form a durable oxide coating, such as Cr or Ni
3. Cathodic Protection—connect iron to a metal with a more positive E°OX with a wire.
o used for buried tanks/pipes, piers, lock gates, ship hulls, bridge decks
o some cathodic protection uses external current to force the protected steel to be the cathode.
Fe  Fe2+ + 2 e–
Mg  Mg2+ + 2 e–
Big Question 2
E°OX = +0.44 V
E°OX = +2.37 V
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I-70 Indianapolis
“The advantage of cathodic protection is that it can halt the progress of corrosion without the removal
of chloride-contaminated concrete. Corrosion requires an anode, a point on the reinforcing steel
where ions are released. Cathodic protection is the application of direct current such that the steel
becomes cathodic to artificial anodes located on the deck. These anodes usually consist of sheets of
thin wire mesh. A relatively small DC rectifier operating on “AC” line voltage and a control panel are
normally located beneath the bridge.
“Cathodic protection systems do not need to operate 24 hours per day to be beneficial. Therefore, they
can be powered by solar panels or in line with highway lighting systems. Cathodic protection systems
should be considered for locations where traffic maintenance costs are very high and where a few
years of additional service between repairs would be advantageous.”
Source: Indiana Department of Transportation
 Electrolytic Cells
Using electric current to drive nonspontaneous reaction
e.g., plating Cr(s) on iron
e.g.,
Fe  Fe2+ + 2 e–
E°OX = +0.44 V
Cr3+ + 3 e–  Cr
E°RED = –0.74 V
How many grams of solid chromium could be deposited by running 10 amps of current through
a Cr3+ solution for 2 hours?
60 min 60 s 10 coul
mol e
1 mol Cr 52.00 g Cr





 13 g Cr
2 hr ×
hr
min
s
96485 coul 3 mol e
mol Cr
Big Question 2
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