Oxidation and
Reduction
Lecture 9
Law of Mass Action
• Important to remember our equation
K = Õ ani i
i
describes the equilibrium condition. At non-equilibrium
conditions it is called the reaction quotient, Q.
• Written for the reaction H2CO3 = HCO3- + H+
K=
aHCO- aH +
3
aH 2CO3
• We can see that if the equilibrium constant is to remain
constant, addition of H+ must drive the reaction to the left (i.e,
activity of bicarbonate must decrease and that of carbonic
acid must increase).
• “Changing the concentration of one species in a reaction in
a system at equilibrium will cause a reaction in a direction that
minimizes that change”.
Le Chatelier’s Principle
• We can generalize this to pressure and
temperature:
dG = VdP - SdT
• An increase in pressure will drive a reaction in a
direction such as to decrease volume
• An increase in temperature will drive a reaction in a
direction such as to increase entropy.
• “When perturbed, a system reacts to minimize the
effects of perturbation.”
Temperature and Pressure
Dependence
• Since ∆Gr˚ = ∆Hr˚ - T∆Sr˚ and ∆Gr˚ = -RT ln K then
∆ H ro ∆ Sro
ln K = +
RT
R
• Temperature dependency can be found by taking
derivatives of this equation with respect to T and P:
æ ¶ln K ö DH ro
ç
÷ =
è ¶T øP RT 2
o (this is known as the van Hoff equation).
• For pressure, since
æ ¶DGr ö
ç
÷ = DVr
è ¶P øT
æ ¶ln K ö
DVro
ç
÷ =è ¶P øT
RT
Oxidation and Reduction
Oxidation refers to processes in which atoms gain or loss
electrons, e.g., Fe2+ Fe3+
Valence and Redox
• We define valence as the charge an atom acquires
when it is dissolved in solution.
• Conventions
o
o
o
o
Valence of all elements in pure form is 0.
Sum of valences much equal actual charge of species
Valence of hydrogen is +1 except in metal hydrides
it ispneumonic:
-1
Reallywhen
idiotic
Valence of O is -2 except in peroxides when it isLeo
-1. the Lion Says GRR:
Loss
Equals Oxidation
• Elements generally function as either
electron
donors or
Gain Refers to Reduction
acceptors.
o Metals in 0 valence state are electron donors (become positively charged)
o Oxygen is the most common electron acceptor (hence the term oxidation)
• Redox
o A reduced state can be thought of as one is which the availability of electrons
is high
o An oxidized state is one in which the availability of electrons is low.
Redox in
Aqueous
Solutions
Redox reactions occur over a wide range of
conditions: from groundwaters to magma.
They are approached differently. We begin
with aqueous solutions.
Electrochemical Cells
• A simple redox reaction
would be:
Feaq3+ + e-
2+
Feaq
o We want to know ∆G of the
reaction. Measuring energy of
it in electrochemical cell might
be good approach.
o However, such a cell can only
measure exchange of
electrons (e.g., between Zn
and Cu)
o We really want to know are
energies for individual redox
reactions such as:
Zns
2+
Znaq
+ 2e-
Hydrogen Scale Potential
• We assign a potential of 0 for the reaction:
½H2(g) = Haq+ + e-
o in practice one side has Pt electrode in H2 gas, the other acid with aH+ = 1.
• Then for the reaction
Zn2+
aq + H 2(g)
Zn s + 2H+
• The potential is assigned to
2+
Znaq
+ 2e-
Zns
• Potentials measured in this way are called hydrogen
scale potentials, written EH and have units of volts.
Table 3.3 EH˚ and pe˚ for half-cell reactions
• The sign convention for
EH is that the sign of the
potential is positive when
the reaction proceeds
from left to right.
o
Thus if a reaction has positive EH,
the metal ion will be reduced by
hydrogen gas to the metal. If a
reaction has negative EH, the
metal will be oxidized to the ion
and H+ reduced.
• The reactions are listed
so that a species will be
oxidized by (and
therefore will reduce) all
species listed below it.
o Thus Li will be oxidized in
preference to Ca.
EH and ∆G
• Electrochemical energy is a form of free energy. EH is
related to ∆Gr by:
∆Gr = -zFEH
• where F is the Faraday constant (96,485 coulombs) and
converts volts to joules.
• and
∆Gr˚ = -zFE˚
o Values of E˚ available in compilations (e.g., Table 3.3)
• Since
∆ Gr = ∆ Gro + RT ln Õ aini
• then
• This is known as the Nernst Equation.
pe
• Consider again the reaction:
Feaq3+ + e-
2+
Feaq
• The equilibrium constant expression for this reaction
is
aq
aFe
K =? aq
2+
aFe3+ ae-
• In log form:
logK = logaFe2+ - logaFe3+ - logae-
• We define pe as:
• So
pe = - log ae-
log K = log
aFe2+
aFe3+
+ pe
Standard State pe and
Relation to EH
• Continuing with the reaction
log K = log
aFe2+
aFe3+
Feaq3+ + e-
2+
Feaq
+ pe
• In an aqueous solution, the standard state activities
are?
• Therefore
pe˚ = log K
1
• More generally,
pe˚= log K
z
o So for this reaction:
pe = pe˚- log
• pe is related to EH as:
pe =
aFe2+
aFe3+
FEH
5039EH
=
2.303RT
T
What pe is really telling us
• We have defined pe as the negative log of the
activity of the electron. So a high pe means a low
activity and concentration of electrons in our
system. A low concentration of electrons implies an
oxidized system; a high concentration (and low pe)
implies a reduced system.
• Same is true of EH.
• So these are parameters that tell us about the redox
state of our system (just as pH tells us about acidity).
Speaking of pe and pH…
•
•
A commonly used diagram to
illustrate chemical variation in
aqueous solutions is the pepH diagram (or EH-pH)
Water only stable over limited
range, so we start by setting
boundaries.
½O2(g) + 2e- + 2H+ = H2O
1
log K = log aH 2O - log PO2 + 2 pe + 2 pH = 41.56
2
o
o
•
•
In the standard state:
pe = 20.78-pH
The is a line with intercept of 20.78 and
slope of -1.
Similarly:
H+ + e- = ½H2(g)
and pe = -pH
pe-pH Diagrams
• To construct the
diagrams
1. Write a reaction relating
species of interest.
2. Redox reactions should contain
e3. pH dependent reactions should
contain H+
4. Write the equilibrium constant
expression.
5. Get in log form, solve for pe
with equation of the form
pe = a + bpH
5. Find or calculate value of log K.
Drawing stability
boundaries
• Now consider:
Feaq3+ + elog K = log
2+
Feaq
aFe2+
aFe3+
+ pe
• For equal activities of the two
species,
pe = log K
o (horizontal line with intercept = K)
• Next Fe(OH)2+ ⇋ Fe(OH)2+:
Fe(OH)2+ + OH– ⇋ Fe(OH)2+
Fe(OH)2+ + H2O ⇋ Fe(OH)2+ + H+
log K = log
aFe(OH )2+
aFe(OH )+
2
- pH = -3.6
Use H+
rather
than OH-!
Drawing stability
boundaries
• Now consider
Fe3+ ⇋ Fe(OH)2+
Fe3+ + H2O = Fe(OH)2+ + H+
log K = log
aFe(OH )2+
aFe3+
- pH = -2.2
• Next:
Fe(OH)2+ ⇋ Fe2+
Fe(OH)2++e- + H2O ⇋ Fe2++ H+
2+
2+
Fe(OH) –Fe
Fe(OH )aq + e + H
Feaq + H 2O
• Our reaction is:
• Equilibrium constant expression is:
2+
log K = log
• In the form we want:
aFe2+
aFe(OH )2+
-
+
2+
- pe - pH
pe = log K - pH
• We can write it as as the sum of two reactions,
o we sum
Feaq3+ + e+
Fe(OH )2+
aq + H
o to yield
+
Fe(OH)2+
aq + e + H
2+
Feaq
Feaq3+ + H 2O
2+
Feaq
+ H 2O
• The log equilibrium constant of the net reaction is
the sum of the log equilibrium constants of the two.
Line 5 has a slope of -1 and an
intercept of log K.
We can also use pe-pH diagrams to
illustrate stability of solid phases in
presence of solution. In this case, we
must choose concentration.
More about pe-pH
diagrams
• pe-pH diagrams are a
kind of stability or
predominance diagram.
• They differ from phase
diagrams because lines
indicate not phase
boundaries, but equal
concentrations.
o There is only 1 phase in this this
diagram – an aqueous solution.
• Regions are regions of
predominance.
o The aqueous species continue to
exist beyond their fields, but their
concentrations drop off
exponentially.
Environmental
Interpretation of pe-pH