A little more Thermodynamics,
Redox, Electrochemistry, and
Radioactivity
--Whew.
Entropy
• Entropy is a measure of disorder of a
system
• In general, in terms of entropy:
(s)<(l)<(aq)<<(g)
• Entropy (S), is measured in J/mol k
• The entropy of a pure substance, perfect
crystal, at absolute 0 = 0 J/ mol k
Is DS (+) or (-) ?
•
•
•
•
2NaHCO3 (s) Na2CO3 (s) + H2O (g)
CaCO3 (s) CaO (s) +CO2 (g)
H2 (g) +Cl2 (g) 2HCl (g)
N2 (g) + 3H2 (g) 2NH3 (g)
Changes in entropy
A: A system can lose entropy, becoming
more ordered.
B: As it does, the rest of the Universe
becomes less ordered
• B is always greater than A
Entropy always increases
Gibbs Free Energy
• For any reaction:
DG=DH-TDS
or
DGo=DHo-TDSo
(o=standard conditions)
and DG for a reaction = -DG for the
reverse reaction
Standard conditions (thermo)
o
25 C
(298k)
Solutes at 1.0M
Gasses at 1.0 atm.
Spontaneity
• DGo<0: the system will proceed forward
from standard conditions. This is called
spontaneous.
• DGo=0. Since concentrations are stable, a
system in equilibrium has no gain or loss
of enthalpy or entropy .
• DGo>0: Rxn is not spontaneous, the
reverse rxn is.
What is DSo , DHo , and DGo?
•
•
•
•
2NaHCO3 (s) Na2CO3 (s) + H2O (g)
CaCO3 (s) CaO (s) +CO2 (g)
H2 (g) + Cl2 (g) 2HCl (g)
N2 (g) + 3H2 (g) 2NH3 (g)
At what T does the spontaneity
change?
•
•
•
•
2NaHCO3 (s) Na2CO3 (s) + H2O (g)
CaCO3 (s) CaO (s) +CO2 (g)
H2 (g) + Cl2 (g) 2HCl (g)
N2 (g) + 3H2 (g) 2NH3 (g)
If DHo is & DSo is then DGo…
+
+
-
+
+
(or)
(or)
- spontaneous at any T
+Not spontaneous at any T
- spontaneous at high T
+Not spontaneous at low T
+Not spontaneous at high T
- spontaneous at low T
In (other) words…
Situation 1
• If an exothermic reaction leads to an
increase in entropy, then free energy is
released, and the reaction is
spontaneous at any temperature
In (other) words…
Situation 1
• If an exothermic reaction leads to an
increase in entropy, then free energy is
released, and the reaction is
spontaneous at any temperature
• (Can you state the other three situations?)
In (other) words…
Situation 2
• If an endothermic reaction leads to an
decrease in entropy, then free energy is
absorbed, and the reaction is
nonspontaneous at any temperature
In (other) words…
Situation 3
• If an endothermic reaction leads to an
increase in entropy, then free energy is
released at high temperatures, and the
reaction is spontaneous only when it is
hot enough
In (other) words…
Situation 4
• If an exothermic reaction leads to an
decrease in entropy, then free energy is
released only at low temperatures, and the
reaction is spontaneous only when it is
cool enough (It may be very slow at that
temperature)
So, how do you make products?
• Non-spontaneous does not mean that
reactants won’t make products. (It just
won’t make a whole lot before they start
decomposing just as fast.)
• Use Le Chatelier’s principle. Remove the
products (including heat), and the system
will keep making more.
Why must a non-spontaneous
reaction make some product?
Why must a non-spontaneous
reaction make some product?
• Entropy.
Why must a non-spontaneous
reaction make some product?
• Entropy.
• Pure reactants have less entropy than a
mixture of reactants and products
Come, let us reason together…
• Spontaneous=proceeding forward (more
products) from standard conditions
• Standard conditions=concentrations of 1 M
or 1 atm
• K=[products] / [reactants]
• (1)x/(1)y=1
therefore…
…if a reaction is spontaneous…
…K>1
DG<0 and K>1
for spontaneous reactions
DG>0 and K<1
for non-spontaneous reactions
Gibbs free energy and equilibrium
• If a system is spontaneous:
– It loses free energy if it proceeds forward from
standard conditions—so it will
– and K>1
• In any case:
DGo = -RT ln K
What is K (at 298k)?
•
•
•
•
2 NaHCO3 (s) Na2CO3 (s) + H2O (g)
CaCO3 (s) CaO (s) +CO2 (g)
H2 (g) +Cl2 (g) 2 HCl (g)
N2 (g) + 3H2 (g) 2 NH3 (g)
Does it go forward from where it is?
If Q<K, yes! If you know DGo, use the
relationship:
DG = DGo + RT ln Q
If DG<0, it proceeds forwards
If DG=0, it is in equilibrium (Q=K)
If DG>0, it proceeds in reverse
Three Laws of Thermodynamics
1st Law: Energy is neither created nor
destroyed
2nd Law: Entropy always increases
3rd Law: The entropy of a pure substance,
perfect crystal, at absolute 0 = 0 kJ/ mol k
Redox
• A reduction is a gain of electrons, an
oxidation is a loss of electrons
• A reduction is always conjoined with an
oxidation (electrons are neither created
nor destroyed, charges must balance)
• Remember “OILRIG”
OILRIG
Oxidation is loss of electrons
Reduction is gain of electrons
LEO says GER
Loss of electrons is oxidation
Gain of electrons is reduction
Does a redox reaction occur?
• Look for an oxidizing agent and a reducing
agent.
• If there is one of each, then ask, “Can this
oxidizing agent oxidize this reducing
agent”
• Answer by comparing reduction potentials.
• (Don’t memorize a rule, compare the
values to a reaction you know will occur)
Does a redox reaction occur?
If you combine…
• Na+ and Fe+3?
• Cl- and Ag?
• Cu and K+ ?
• Pb+2 and I- ?
• Fe+2 and Mg?
Does a redox reaction occur?
If you combine…
• Na+ and Fe+3?—No. There is no reducer.
• Cl- and Ag?—No. There is no oxidizer.
• Cu and K+ ?—No. This oxidizer can’t do it
• Pb+2 and I- ?—No, but it will precipitate.
• Fe+2 and Mg?—Yes.
Fe+2 + Mg  Fe and Mg+2
Redox—half reactions
• Balance the atoms
• Rectify the electrons
• Add H2O and H+ to balance oxygen and
hydrogen
• Check that charges are balanced
• (Add OH- if the reaction is specified as in a
basic solution)
Try it.
• Sodium thiosulfate and nitric acid yield…
• Hydrogen peroxide and iron (II) sulfate
yield…
• Potassium dichromate and potassium
iodide yield…
• Potassium permanganate and ethanol
yield…
Try it.
• S2O3-2 + NO3- 
• H2O2 + Fe+2 
• Cr2O7-2 + I- 
• MnO4- + C2H5OH 
Try it.
• S2O3-2 + NO3-  SO4-2 + NO
• H2O2 + Fe+2  H2O + Fe+3
• Cr2O7-2 + I-  Cr+3 + I2
• MnO4- + C2H5OH  Mn+2 + CO2 + H2O
Try it.
3x(S2O3-2 + 5H2O 2 SO4-2 +10H + + 8 e-)
8x(NO3- +4H + + 3 e-  NO + 2 H2O)
3S2O3-2 + 15H2O 6 SO4-2 +30H + + 24 e-)
8NO3- +32H + + 24 e-  8NO + 16H2O)
3S2O3-2+8NO3-+2H+6SO4-2 +8NO+H2O
Reduction potentials
• Standard reduction potentials are
measured as compared to the reduction of
hydrogen ions to hydrogen gas. (0.00V)
• Half reactions that accomplish this have (-)
reduction potentials
• Half reactions that force the reverse have
(+) reduction potentials
Reduction potentials
• Specifically: Magnesium reduces H+
• 2H+ + Mg  Mg+2 + H2 which implies that
Mg+2 + 2 e-  Mg
has a Eo<0
• While bromine oxidizes hydrogen gas
• H2 + Br2  2H+ +2Br - which implies that;
Br2 + 2 e-  2Br –
has a Eo>0
Electrochemical Cells
• When half reactions are separated, and
the electrons are connected in a circuit.
• A salt bridge is needed to allow charges
to migrate to offset the motion of electrons
• Cathode=reduction
• An electrode carries electrons to or from a
half reaction
Shorthand notation
• The Danielle Cell, using copper and zinc,
Zn|Zn+2||Cu+2|Cu
…makes 1.1 V
Zn|Zn+2||Cu+2|Cu
(or, in general)
product reactant
Anode of
of
cathode
oxidation reduction
If non-metals are used…
Pt|H2|H2O||O2|H2O|Pt
• The (non-reactive) metal electrode is
noted outside the bars
Standard cell potentials
• Eo=Ered-Eox
Be able to:
• Sketch a cell (include salt bridge and
circuit)
• Label anode and cathode
• Write the half reactions, complete reaction
• Calculate Eo, show direction of electron
flow
• Describe the oxidation and reduction—
with mass changes, observations.
• Read and write the shorthand notation
The Nernst equation
• Among all of the ways to make nonstandard conditions, concentrations not
1.00 M is the most likely. In this case:
• Ecell
=Eo-(RT/nF) ln Q
(n=the number of electrons transferred)
?Emf
• What is the emf of a Cu/Ag cell if
[Cu+2]=.4M and [Ag+]=.1M?
?Emf
• What is the emf of a Cu/Ag cell if
[Cu+2]=.4M and [Ag+]=.1M?
Cu + 2Ag+ Cu+2 +2Ag
n=2 and Q=[Cu+2]/[Ag+]2=40
?Emf
• What is the emf of a Cu/Ag cell if
[Cu+2]=.4M and [Ag+]=.1M?
Cu + 2Ag+ Cu+2 +2Ag
n=2 and Q=[Cu+2]/[Ag+]2=40
You did
remember to
square it didn’t
you?
?Emf
• What is the emf of a Cu/Ag cell if
[Cu+2]=.4M and [Ag+]=.1M?
Cu + 2Ag+ Cu+2 +2Ag
n=2 and Q=[Cu+2]/[Ag+]2=40
• Ecell
=Eo-(RT/nF) ln Q
?Emf
• What is the emf of a Cu/Ag cell if
[Cu+2]=.4M and [Ag+]=.1M?
Cu + 2Ag+ Cu+2 +2Ag
n=2 and Q=[Cu+2]/[Ag+]2=40
• Ecell
=Eo-(RT/nF) ln Q
=.33V-(8.31x298/2/96500)ln40
=.28V
?Emf
• What is the emf of a Cu/Ag cell if
[Cu+2]=.4M and [Ag+]=.1M?
Cu + 2Ag+ Cu+2 +2Ag
n=2 and Q=[Cu+2]/[Ag+]2=40
• Ecell
=Eo-(RT/nF) ln Q
=.33V-(8.31x298/2/96500)ln40
=.28V
It’s running down. It
will get worse as the
battery is used
?Emf
• What is the emf of a Cu/Ag cell if
[Cu+2]=.02M and [Ag+]=.4M?
?Emf
• What is the emf of a Cu/Ag cell if
[Cu+2]=.02M and [Ag+]=.4M?
Cu + 2Ag+ Cu+2 +2Ag
n=2 and Q=[Cu+2]/[Ag+]2=.125
• Ecell
=Eo-(RT/nF) ln Q
=.33V-(8.31x298/2/96500)ln.125
=.36V
The Nernst equation
• You may find these equivalencies helpful
• Ecell
=Eo-(RT/nF) ln Q
=Eo-2.303RT/nF log Q
=Eo-.0592/n log Q
Equilibrium and cell potential
Let us reason together.
• Q=K at equilibrium and
• Ecell=Eo-(RT/nF) ln Q and
• your batteries go dead (Ecell=0)
eventually.
and these ideas imply that…
Equilibrium and cell potential
Let us reason together.
• Q=K at equilibrium and
• Ecell=Eo-(RT/nF) ln Q and
• your batteries go dead (Ecell=0)
eventually.
and these ideas imply that…
Eo=(RT/nF) ln K
Gibb’s free energy and cell
potential
• As a battery operates, it gives off free
energy in electrical work. There are two
relationships of note:
DGo = -RT ln K and Eo=(RT/nF) ln K
Therefore…
Gibb’s free energy and cell
potential
• As a battery operates, it gives off free
energy in electrical work. There are two
relationships of note:
DGo = -RT ln K and Eo=(RT/nF) ln K
Therefore…
DGo = -nFEo
K
Use…
Use…
DGo = -RT ln K
o
DG
Eo=(RT/nF) ln K
DGo = -nFEo
Use…
Eo
Counting electrons—Faraday’s
constant
• The unit of electrical charge, 1 coulomb, is
a small fraction of a mole.
• We use Faraday’s constant
1F=96,500 C/mole
• as the conversion. Also:
• Potential energy(V):1 Volt = 1J/C, and
• Current (I)
1 Amp = 1C/s
Electrolytic Cells
• Applying an external voltage will allow a
non-spontaneous reaction to occur.
2H2O2H2+O2 is not spontaneous (Right?)
If you apply a voltage to water (with some
electrolyte added to carry a charge), it will
decompose (or electrolyse)
Electrolytic Cells
• Applying an external voltage will allow a
non-spontaneous reaction to occur.
2H2O2H2+O2 is not spontaneous (Right?)
If you apply a voltage to water (with some
electrolyte added to carry a charge), it will
decompose (or electrolyse)
Anode reaction: 2H2O4H+ + O2 + 4eCathode reaction: 4 H+ + 4 e-2 H2
Electrolytic Cells
• Stoichiometric relationships:
Current x time#electronsmass product
If a copper (II) sulfate solution is
electroplated for 12 minutes at 15 amps,
Electrolytic Cells
• Stoichiometric relationships:
Current x time#electronsmass product
If a copper (II) sulfate solution is
electroplated for 12 minutes at 15 amps,
15 C/s x (12 x 60)s = 10800 C
Electrolytic Cells
• Stoichiometric relationships:
Current x time#electronsmass product
If a copper (II) sulfate solution is
electroplated for 12 minutes at 15 amps,
15 C/s x (12 x 60)s = 10800 C
10800 C x 1 mole / 96,500 C = .112 mole e-
Electrolytic Cells
• Stoichiometric relationships:
Current x time#electronsmass product
If a copper (II) sulfate solution is
electroplated for 12 minutes at 15 amps,
15 C/s x (12 x 60)s = 10800 C
10800 C x 1 mole / 96,500 C = .112 mole e.112 mole e- x 1 Cu+2/ 2 e- = .0560 mole Cu
.
Electrolytic Cells
• Stoichiometric relationships:
Current x time#electronsmass product
If a copper (II) sulfate solution is
electroplated for 12 minutes at 15 amps,
15 C/s x (12 x 60)s = 10800 C
10800 C x 1 mole / 96,500 C = .112 mole e.112 mole e- x 1 Cu+2/ 2 e- = .0560 mole Cu
.0560 mole Cu x 63.55 g/ mole= 3.6 g Cu
Electrolytic Cells
• Voltage concerns:
The easiest ox. & red. reaction will occur
If you run an electrolytic cell with platinum
electrodes, in a sodium nitrate solution:
Cathode reaction:
Na+ + e-Na or 4 H+ + 4 e-2 H2
Electrolytic Cells
• Voltage concerns:
The easiest ox. & red. reaction will occur
Choose
If you run an electrolytic cell with platinum
thisin aone!
electrodes,
sodium nitrate solution:
Cathode reaction:
Na+ + e-Na or 4 H+ + 4 e-2 H2
Electrolytic Cells
• Voltage concerns:
The easiest ox. & red. reaction will occur
If you run an electrolytic cell with platinum
electrodes, in a sodium nitrate solution:
Anode reaction:
Pt  e- + Pt+2 or 2H2O4H+ + O2 + 4e?
Electrolytic Choose
Cells
• Voltage concerns:
The easiest ox. & red. reaction will occur
this one!
If you run an electrolytic cell with platinum
electrodes, in a sodium nitrate solution:
Anode reaction:
Pt  e- + Pt+2 or 2H2O4H+ + O2 + 4e?
Nuclear Chemistry
• --breaks the rules that one atom cannot be
converted to another.
Chemistry is the dance of the
electrons—nuclear reactions
change the nuclei of atoms
• --charge and mass are still conserved.
Nuclide Notation
• A nuclide is a nucleus or atom of a specific
isotope of an element
39
19
K
• This is potassium-39. It has 19 protons
(atomic number = 19), making it
potassium, and 20 neutrons, making a
mass number of 39
How many p, n, e- in each?
What is the mass number and
atomic number?
3
1
59
26
H
+3
Fe
36
17
Cl
90
38
+2
Sr
131
53
228
90
I
Th
Natural decays
•
a—the loss of a particle from a nuclide
--The a particle is composed of 2p and 2n,
the 4He nucleus
--decreases the mass by 4 and the atomic
number by 2
• b—emission of an electron (b particle) from the
nucleus by the conversion of a n  p + e--the electron is the b particle
--increases the atomic number by 1, does
not affect mass
Write the reaction
•
•
•
•
•
•
Argon-39 undergoes a b decay
Thorium-228 undergoes an a decay
An a decay forms lead-204
A b decay forms nitrogen-14
A natural decay forms Sc-45 from Ca-45
A natural decay forms Ac-227 from Pa-231
Notice what they do
• A b decay lowers the n/p ratio in small
nuclei, or when the ratio is too large.
• An a decay lowers the total size, and
raises the n/p in large nuclides, or when
the ratio is too small.
What is “too large” or “too small”?
What is “too large” or “too small”?
b decay
a decay
What is “too large” or “too small”?
What is “too large” or “too small”?
n:p = 2:1
n:p = 1:1
This just in.
• Researchers report the first creation of the
long-lived nucleus hassium-270, a "doubly
magic" combination of 108 protons and
162 neutrons. Its long lifetime of 22
seconds supports the theory of an "island
of stability" for the heaviest elements.
(J Dvorak et al. 2006 Phys. Rev. Lett. 97,
242501)
Nuclear reactions
• Many nuclear reactions involve colliding
nuclei or smaller particles at some
significant fraction of the speed of light,
• --find the missing particle by balancing
mass and charge.
Particles might include…
•
•
•
•
•
•
p
n
e- (AKA b)
d
a
g (OK, it’s not a particle, but it’s often
written in)
Nuclear reactions
Condensed notation
Nuclide1(gets hit by…,produces…and)Nuclide2
• For example:
238U(n,2n)237U
means
238U+1n21n+237U
(or so they say)
Condensed notation
• Write the reaction
• 7Li(1p,4He)4He
• Write the condensed notation
• 14N +4a1p +17O
In comparison
• Physical changes:
joules/mole range
• Chemical changes:
kilojoules/mole range
• Nuclear changes:
megajoules/mole range
For example:
• If a nuclear reaction releases 45 MJ/mol,
what is the wavelength of the photons
produced?
Here is a problem
• The mass of a He-4 atom =4.00260 amu
Do you see the problem?
• 1p + 1e- has a mass of 1.007825 amu
• 1n has a mass 1.008665 amu
• The mass of a He-4 atom =4.00260 amu
Binding energy
• Very careful mass measurements lead to
the conclusion that matter and energy can
be inter-converted.
• The mass lost in a nuclear reaction is
converted to energy.
• Nuclei have a mass defect representing
the binding energy in the nucleus
What is the binding energy?
• The mass of a He-4 atom =4.00260 amu
• 2 x p+e=
• 2 x n=
2.01565 amu
2.01733 amu
4.03298 amu
What is the binding energy?
• The mass of a He-4 atom =4.00260 amu
• 2 x p+e=
• 2 x n=
-
2.01565 amu
2.01733 amu
4.03298 amu
4.00260 amu
0.03038 amu!
What is the binding energy?
0.03038 amu! Where did it go?
This is the binding energy. To convert to energy,
use E=mc2
.03038 amu x1g/6.022 x 1023amu= 5.045 x 10-26g
= 5.045 x 10-29kg
E=mc2=(5.045x10-29kg)(3.00x108m/s)2=4.54x10-12J
(or 1.14x10-12J/nucleon since there are 4 particles
in the nucleus)
What is the binding energy?
(in J/nucleon)
1) The mass of I-127 atom=126.9004 amu
2) The mass of Bi-209 atom=208.9804 amu
3) The mass of a 0-16 atom =15.995 amu
What is the binding energy?
(in J/nucleon)
1) The mass of I-127 atom=126.9004 amu
(1.36 x 10-12J/nucleon)
1) The mass of Bi-209 atom=208.9804 amu
(1.26 x 10-12J/nucleon)
1) The mass of a 0-16 atom =15.995 amu
(1.28 x 10-12J/nucleon)
What is the mass defect?
• Iron-56 is the most stable nuclide, having
1.40 x 10-12 J/ nucleon binding energy.
What is the mass defect of the atom?
What is the mass defect?
• Iron-56 is the most stable nuclide, having
1.40 x 10-12 J/ nucleon binding energy.
What is the mass defect of the atom?
(.52 amu)
Fission vs Fusion
• Fission=breaking up large nuclei—
--natural radioactive decay of large atoms
--used for nuclear power
• Fusion=combining small nuclei
--occurs naturally in stars
--prospects for nuclear energy—no
radioactive byproducts
Both are transmutations—one nuclide is
converted into another
Rates of Radioactive decay
• Natural radioactive decays are first order
reactions, so use the first order reaction
rate laws
Rate= k[A]
and
ln [A]t – ln [A]o= - k t
Consider the relationships
•
•
•
•
•
•
Half life
Original amount
Final amount
Time elapsed
Rate constant
Rate of decay
Try it.
• If you start with 1.38 mg of U-234 and
k=2.84 x 10-6/yr for its decay
--how much is left after 20,000 years?
--how long will it take to reach .010 mg?
--what is the initial rate of decay?
--what is the final rate of decay?
--what is the half-life?
Try it.
• Br-82 has a half life of 35.3 hours. If
you start with a 6.5 mg sample of Br-82
--how much is left after 2 days?
--what is the initial rate of decay?
--what is the rate constant?
--how long will it take to reach 1.5 mg?
Try it.
• A .350 mg sample of K-42 decays to
only .066 mg in 29.7 hours.
--what is the rate constant?
--how much was left after 20.0 hours?
--what was the initial rate of decay?
--what is the half life?
--how long will it take to reach .010 mg?
Amount of radioactivity
• 1 rad=.00001 J/g (often converted to rems
in medicine)
Officially:
• 1 Curie= 3.70 x 10 10 decays/s
…where decays/s = Rate x Avogadro’s #
Try it.
• A .350 mg sample of K-42 decays to
only .066 mg in 29.7 hours.
--what is the rate constant?
--how much was left after 20.0 hours?
--what was the initial rate of decay?
--what is the half life?
--how longHow
will radioactive
it take to reach .010 mg?
is the sample?
Th-th-that’s all, folks.
Atomic theory
• All matter is composed of atoms.
--atoms of one element are identical
--atoms of different elements are
different
--reactions form different combinations
of atoms, not different atoms
• Atoms are composed of protons,
neutrons, and electrons.
(Are all of the little kids in bed?—Now we
can tell you the real story…)
Modern Theory
In this Universe,
you will find:
Fermions
and
Bosons
(force mediating
particles)
Modern Theory
Bosons (force mediating particles)
include:
g, (electromagnetic force)
Zo, (weak nuclear force)
Gluons (Strong nuclear force)
Graviton (gravity)
W±, (weak nuclear force)
Higgs (mediates mass)
Modern Theory
Fermions are the
fundamental
particles, including:
Quarks
and
Leptons
Modern Theory
up
down
Types of
quarks
include:
top (aka
“truth”)
charm
bottom (aka
“beauty”)
strange
(and their antiparticles)
Modern Theory
Leptons include:
Electrons, e-,
Muons, m,
Tauons, t,
and three types
of neutrinos
e, m, t
(and their
antiparticles)
Modern Theory
Combinations
of quarks make
hadrons, either:
Mesons
(2 quarks each)
Including:
p+, K-, r+, B0, c
Baryons
(3 quarks each)
including
p, n, p-, L, -
Modern Theory
In this Universe, you will find
fermions which
include the
fundamental
particles
Quarks
u,d,c,s,t,b
(and antiparticles)
Quarks make
up the
hadrons,
either
and
Bosons (force mediating particles)
g, (electromagnetic force
Zo, (weak nuclear force)
Gluons (Str. nuclear force)
Graviton (gravity)
W±,(weak nuclear force)
Higgs (mediates mass)
Leptons e-, m, t, e, m, t
(and antiparticles)
Mesons (2 quarks
each) p+, K-, r+, B0, c
baryons (3 quarks each)
including p, n, p-,L,-