Chapter 20
Electrochemistry and
Oxidation-Reduction
Overview
Electrochemistry deals with
chemical changes produced by an
electric current and with the
production of electricity by
chemical reactions.
Overview
Energy changes are measured
electrically whenever possible
This allows for very accurate
measurements.
 DG and equilibrium constants can
be measured electrochemically.
Galvanic Cells and Cell
Potential
When a strip of magnesium is
placed in HCl, a reaction takes
place which we have represented
as a single replacement reaction:
Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)
Galvanic Cells and Cell
Potential
Looking at it as an Oxidation-
Reduction reaction it becomes:
Mg(s) + 2H3O+1(aq) + 2Cl-1(aq)
H2(g) + Mg+2(aq) + 2Cl-1(aq) + 2H2O(l)
Galvanic Cells and Cell
Potential
This reaction has two halves
– the magnesium oxidizing (gaining
charge by losing electrons)
Mg(s)
Mg+2(aq) + 2e-1
– the reduction of hydrogen (losing
charge by gaining electrons)
2H3O+1(aq) + 2e-1
H2(g) + 2H2O(l)
Galvanic Cells and Cell
Potential
The two half-reactions can be run
in two separate containers with
only a wire (e-1 pathway) and a salt
bridge (ion pathway) to connect
them.
Galvanic Cells and Cell
Potential
e-1
Salt Bridge
Mg
- ions
+ ions
Bubbles
of H2
MgCl2
HCl
Oxidation
Reduction
Galvanic Cells and Cell
Potential
This is an example of a galvanic
cell (sometimes called a voltage
cell).
A battery is also such an example.
– It converts chemical energy into
electrical energy.
Galvanic Cells and Cell
Potential
– The anode is where oxidation takes
place.
– The cathode is where reduction takes
place.
– Remember, Anode and oxidation
both start with a vowel.
Cell Potentials
Cell Potential (Ecell) is the “driving
force that pushes electrons through
the external circuit.
Ecell is measured in Volts (V)
This potential can be measured
with a Voltmeter
Cell Potentials
A galvanic cell has a positive
voltage when the cell reaction is
spontaneous.
The cell has a negative voltage
when the cell reaction is not
spontaneous.
Cell Potentials
Each half-cell also has a voltage
potential.
We can predict the potential of a
cell by using the half-cell
potentials provided in the
appendix.
Cell Potentials
The cell potential is the sum of the
half-cell potential of the anode,
Eox, and the half-cell potential of
the cathode, Ered.
Ecell = Eox + Ered
Cell Potentials
For the reaction:
Zn(s) + Cu+2(aq) Zn+2(aq) + Cu(s)
we can predict the overall potential
of the cell by finding the half-cell
potentials of each half-cell and
adding the two.
Cell Potentials
Anode Zn(s)
+2
Zn (aq)
Eox = 0.763V
Cath
+2
Cu (aq)
Sum
Zn(s) + Cu+2(aq) Zn+2(aq) + Cu(s) Ecell = 1.100 V
+ 2e
-1
+ 2e
-1
Cu(s)
Ered = 0.337 V
Standard Electrode
Potentials
The term electrode is used to
represent a complete half-cell.
– Zinc in a solution of zinc sulfate is
such an example.
– Platinum wire (inert conductor)in a
solution of bromine and the bromide
ion is another.
Standard Electrode
Potentials
There is no direct method of
determining the potential of such
an electrode.
To determine the potential, the
hydrogen electrode is assigned a
potential of Zero, and the other
electrodes are compared to this.
Standard Electrode
Potentials
The standard hydrogen electrode is
a gas electrode.
Such an electrode is set up by
bubbling the gas around an inert
conductor.
Standard Electrode
Potentials
The net ionic equation for the
electrode is:
2H3O+1 + 2e-1
H2 + H2O
where the hydrogen is being
reduced.
Standard Electrode
Potentials
By international standards, the
values of electrode potentials are
given for the reduction process.
Standard reduction potentials, Ered,
are measured with respect to
hydrogen at 25oC, 1 atm, and 1M
solution of ions.
Standard Electrode
Potentials
When a potential is measured for
an oxidation process, it can be
easily converted.
The reduction potential and the
oxidation potential of the same
electrode have the same absolute
value but of opposite sign.
Standard Electrode
Potentials
When a table of standard reduction
potentials is constructed (appendix
H) from most negative to most
positive, the activity series
correlates with many chemical
properties of the elements.
These correlations are:
Standard Electrode
Potentials
– The metals with large negative
reduction potentials at the top of the
series are good reducing agents in
the free state. They are the metals
most easily oxidized to their ions by
the removal of electrons.
Standard Electrode
Potentials
– The elements with large positive
reduction potentials at the bottom of
the series are good oxidizing agents
when in the oxidized form - that is,
when the metals are in the form of
ions and the nonmetals are in the
elemental state.
Standard Electrode
Potentials
– The reduced form of any element
reduces the oxidized form of any
element below it. For example,
metallic zinc reduces copper(II) ions
according to the equation:
Zn + Cu+2
Zn+2 + Cu
Calculation of Cell
Potentials
The values in Table 20.1 and
Appendix H can be sued to
determine standard state cell
potentials.
Two points are important to
remember when using these
values:
Calculation of Cell
Potentials
– Eo values are for reduction halfreactions, and the sign of a reduction
potential must be reversed when it is
used as a potential for an oxidation
half-reaction.
– Changing the stoichiometric
coefficients of a half-cell equation
does not change the value of Eo
Calculation of Cell
Potentials
Ex) Write the cell reaction and
determine the standard state
potential for the cell diagrammed
below:
Calculation of Cell
Potentials
e-1
Solution of
Fe+3 and Fe+2
Co
Solution
of Co+2
Anode
Cathode
Calculation of Cell
Potentials
The two half-reactions are:
Anode Co(s)
Co
+2
(aq)
+ 2e
-1
Cath
2Fe+3(aq) + 2e-1
2Fe+2(aq)
Sum
Co(s) + 2Fe+3(aq) Co+2(aq) + 2Fe+2(aq)
Eox =
0.277V
Ered =
0.771 V
Ecell =
1.048 V
Calculation of Cell
Potentials
For simplicity, a line notation can
be used to represent the electrodes
in a galvanic cell.
For the previous reaction we can
represent the cell as follows:
e-1
2
Co Co (1 M )
3
2
Fe (1 M ) Fe (1 M )
Calculation of Cell
Potentials
Ex) Determine the standard cell
potential, and write equations for
the half-reactions and the cell
reaction for the cell described by
the following line notation:
e-1
Fe Fe 2 (1 M )
MnO 4 1 (1 M ) Mn 2 (1 M ); H3O 1 (1 M ) Pt
Calculation of Cell
Potentials
The species to the left of the
double line are involved with the
anode half-reaction:
Fe(s)
Fe+2(aq) + 2e-1
Calculation of Cell
Potentials
The species to the right of the
double line are involved in the
cathode half-reaction:
MnO4-1(aq) + 8H3O+1(aq) + 5e-1
Mn+2(aq) + 12H2O(l)
Calculation of Cell
Potentials
The total equation would then be:
5Fe(s) + 2MnO4-1(aq) + 16H3O+1(aq) + 10e-1
5Fe+2(aq) + 2Mn+2(aq) + 24H2O(l)
Calculation of Cell
Potentials
The standard cell potential would
then be:
Anode
Eox = +0.440 V
Cathode
Ered = +1.51 V
Sum
Ecell = +1.95 V
Cell Potential, Electrical
Work, and Free Energy
Galvanic Cells are sources of
energy that can do work.
The amount of work depends on
the cell potential.
Since the work is done on the
surroundings, it will have a
negative value (-).
Cell Potential, Electrical
Work, and Free Energy
We can show the relationship
between the work available and the
cell potential by the following
equation:
w  nFEcell
Cell Potential, Electrical
Work, and Free Energy
– Where n = number of moles of
electrons
– and F = a constant called a faraday
(96.485 kJ/Vmole.
The maximum amount of work
available, wmax, is DG.
Cell Potential, Electrical
Work, and Free Energy
The equation can then be modified
to be:
DG   nFEcell
At standard state conditions we
have:
DG   nFE cell
Cell Potential, Electrical
Work, and Free Energy
Ex) Calculate the standard free
energy change at 25oC for the
reaction:
Cd(s) + Pb+2(aq)
Cd+2(aq) + Pb(s)
Cell Potential, Electrical
Work, and Free Energy
Anode Cd(s)
Cd+2(aq)
-1
-1
+ 2e
Cath
+2
Pb (aq)
Sum
Cd(s) + Pb+2(aq) Cd+2(aq) + Pb(s)
+ 2e
Pb(s)
Eox =
0.403V
Ered =
-0.126 V
Ecell =
0.227 V
Cell Potential, Electrical
Work, and Free Energy
Now we can use the formula:
DG   nFE cell
DG  2mol (96.485kJ / Vmol )(0.277V )
DG  535
. kJ
The negative value for DG means
that the reaction is spontaneous.
The Effect of Concentration
on Cell Potentials
We know that most reactions do
not occur at standard state.
We can derive a relationship
between the cell potential at
nonstandard states.
The Effect of Concentration
on Cell Potentials
Ecell  E

cell
RT

ln Q
nF
The equation is called the Nernst
Equation.
The Effect of Concentration
on Cell Potentials
Where:
– R = the gas constant (8.314 J/K)
– T = the Kelvin temperature
– F = the faraday constant (96.485kJ/Vmol)
– n = the number of moles of electrons
– Q = the reaction quotient.
The Effect of Concentration
on Cell Potentials
For reactions at 25oC, we can
rearrange the equation to be:
Ecell  E

cell
0.05916 V

log Q
n
– the constant, 0.05916 V, contains the
value of RT/F and has been
converted to the logarithmic scale.
The Effect of Concentration
on Cell Potentials
Ex) Calculate the potential at 25oC
for the cell
+2
+2
Cd Cd (2.00M) Pb (0.0010M) Pb
The cell reaction and cell potential
at standard states is:E   0.277 V
cell
Cd(s) + Pb+2(aq)
Cd+2(aq) + Pb(s)
The Effect of Concentration
on Cell Potentials
Since the cell reaction is at 25oC,
we can use the simplified equation:
Ecell  E

cell
0.05916 V

log Q
n
– in this case, n = 2
The Effect of Concentration
on Cell Potentials
2
[Cd ]
2.00
Q
2 
[ Pb ] 0.0010
0.05916 V

Ecell  Ecell 
log Q
n
0.05916 V
2.00
 0.277V 
log
2
0.0010
The Effect of Concentration
on Cell Potentials
0.05916 V
 0.277V 
(3.301)
2
 0.277V  0.0976V  0179
. V
The cell potential decreases from
0.277V at standard state to 0.179V
at nonstandard concentrations.
The Effect of Concentration
on Cell Potentials
The magnitude of the cell potential
is measure of spontaneity of a
reaction.
– A decrease in potential indicates a
decrease in spontaneity.
– Also, an increase in Q indicates a
decrease in spontaneity.
Cell Potential and the
Equilibrium Constant
Since DG is related to the
equilibrium constant and the cell
potential, we can see that the cell
potential of an Redox reaction is
related to the equilibrium constant,
K.
Cell Potential and the
Equilibrium Constant
The generalized formula for this
is:
E

cell
RT

ln K
nF
At 25oC, 0
we
can
simplify
this:
.05916V
E

cell

n
log K
Cell Potential and the
Equilibrium Constant
Ex) Determine the equilibrium
constant at 25oC for the reaction:
Br2(l) + 2Cl-1(aq)
2Br-1(aq) + Cl2(g)
Cell Potential and the
Equilibrium Constant
The standard cell potential can be
determined as follows:
Anode 2Cl
Cath
Sum
-1
-1
Cl2(g) + 2e
(aq)
Br2(l) + 2e-1
Br2(l) + 2Cl
-1
2Br-1(aq)
(aq)
-1
2Br
(aq)
+ Cl2(g)
Eox =
-1.3595
Ered =
+1.0652
Ecell =
-0.2943
Cell Potential and the
Equilibrium Constant
We see that two electrons are
involved in the reaction so n=2
0.05916V
E 
log K
n
0.05916V
 0.2943 
log K
2

cell
Cell Potential and the
Equilibrium Constant
 9.949  log K
10
 9 .949
K
112
.  10
10
K
Balancing OxidationReduction Reactions
To balance Redox Reactions, we
can categorize them into three
groups:
– Molecular Redox Reactions
– Acidic Solutions
– Basic Solutions
Balancing OxidationReduction Reactions
A Molecular Redox Reaction is
when two ions in a reaction have a
change in their oxidation reactions.
One will increase (oxidize) while
the other will decrease (reduce) in
Oxidation.
Balancing OxidationReduction Reactions
– I will use the following example and
show step-by-step how to balance
the equation:
Sn + HNO3
SnO2 + NO2 + H2O
Balancing OxidationReduction Reactions
Step 1: Assign oxidation numbers to
each element to identify the elements
being oxidized and those being
reduced:
Sn + HNO3
SnO2 + NO2 + H2O
0
+1 +5 -2
+4
-2
+4
-2
+1
-2
Balancing OxidationReduction Reactions
Step 2: Now write two equations
using only the elements that change
in ON. Then add electrons to bring
the equation into electrical balance:
Sn0
Sn+4 + 4 e-1
oxidation
N+5 + 1 e-1
N+4
reduction
Balancing OxidationReduction Reactions
Step 3: Multiply the two equations by
the smallest whole numbers to cancel
out the electrons:
1(Sn0
Sn+4 + 4 e-1)
oxidation
(N+5 + 1 e-1
N+4)4
reduction
Balancing OxidationReduction Reactions
Step 4: Transfer the coefficient in
front of each substance in the halfreactions to the corresponding
substance in the original equation:
Sn + 4HNO3
SnO2 + 4NO2 + H2O
Balancing OxidationReduction Reactions
Step 5: Balance the remaining
elements which were not oxidized or
reduced in the usual manner:
Sn + 4HNO3
SnO2 + 4NO2 + 2H2O
(balanced
Balancing OxidationReduction Reactions
An Ionic Redox Reaction involves
ions.
Ionic Redox Reactions are usually
found in acidic or basic solutions,
so the addition of H2O, H3O+1
and/or OH-1 is done to balance the
overall oxidation numbers.
Balancing OxidationReduction Reactions
– Starting with an example of an ionic
Redox reaction in an acidic solution,
I will use the following example and
show you the steps as before:
MnO4-1 + S-2
Mn+2 + S0
(acidic
solution)
Balancing OxidationReduction Reactions
Step 1: Assign oxidation numbers to
each element to identify the elements
being oxidized and those being
reduced:
MnO4-1 + S-2
Mn+2 + S0
+7
-2
-2
+2
0
Balancing OxidationReduction Reactions
Step 2: Now write two equations
using only the elements that change
in ON. Then add electrons to bring
the equation into electrical balance:
Mn+7 + 5e-1
Mn+2 (reduction)
S-2
S0 + 2e-1
(oxidation)
Balancing OxidationReduction Reactions
Step 3: Multiply the two equations by
the smallest whole numbers to cancel
out the electrons:
2(Mn+7 + 5e-1
Mn+2) (reduction)
(S-2
S0 + 2e-1)5 (oxidation)
Balancing OxidationReduction Reactions
Step 4: Transfer the coefficient in
front of each substance in the halfreactions to the corresponding
substance in the original equation:
2MnO4-1 + 5S-2
2Mn+2 + 5S0
Balancing OxidationReduction Reactions
Step 5: Since the reaction takes place
in an acidic environment, we now
add hydronium ions to the more
negative side to balance the charges:
16 H3O+1 + 2MnO4-1 + 5S-2
2Mn+2 + 5S0
Balancing OxidationReduction Reactions
Step 6: As an aqueous solution, water
is also available as part of the
reaction. Balance the Hydrogens
and Oxygens by adding water to
either side of the equation:
16 H3O+1 + 2MnO4-1 + 5S-2
2Mn+2 + 5S0 + 24 H2O
Balancing OxidationReduction Reactions
– Steps 1 through 4 and step 6 are the
same for a basic solution. In step 5,
instead of adding a hydronium ion,
we add an hydroxide ion. As an
example:
CrO4-2 + Fe(OH)2
Cr(OH)3 + Fe(OH)3
Balancing OxidationReduction Reactions
– After step 4 we get:
CrO4-2 + 3Fe(OH)2
Cr(OH)3 + 3Fe(OH)3
Balancing OxidationReduction Reactions
Step 5: Since the reaction takes place
in a basic environment, we now add
hydroxide ions to the more positive
side to balance the charges:
CrO4-2 + 3Fe(OH)2
Cr(OH)3 + 3Fe(OH)3 + 2OH-1
Balancing OxidationReduction Reactions
Step 6: As an aqueous solution, water
is also available as part of the
reaction. Balance the Hydrogens
and Oxygens by adding water to
either side of the equation:
4H2O + CrO4-2 + 3Fe(OH)2
Cr(OH)3 + 3Fe(OH)3 + 2OH-1