Transcript Original powerpoint (~2.2 MB)

Chapter 20
Electrochemistry
Dr. Peter Warburton
[email protected]
http://www.chem.mun.ca/zcourses/1051.php
Galvanic cells
Electrochemical cells fall into
one of two basic types
Galvanic cells convert chemical
energy into electrical energy
(batteries)
Electrolytic cells convert
electrical energy into chemical
energy.
2
Here we put a
piece of zinc
metal into a Cu2+
ion solution. A
reaction occurs
where we get
Zn2+ ions and
solid copper
deposited on the
zinc surface.
3
This is an oxidationreduction (redox)
process where
electrons are
transferred from one
chemical to another.
One chemical loses
electrons in a
process called
oxidation, while the
other chemical gains
electrons in a
process called
reduction.
4
Zinc in Cu2+ solution is spontaneous
Since we actually see
this reaction
occurring, this
reaction must be
spontaneous!
The reverse reaction,
where we put copper
metal into a Zn2+ ion
solution is nonspontaneous!
5
Redox reaction
Zn (s) + Cu2+ (aq)  Zn2+ (aq) + Cu (s)
we can see that each zinc atom
gives away 2 electrons
to a copper (II) ion to give us a
copper atom and a zinc (II) ion
in the spontaneous reaction!
6
Half-reactions
To clarify the redox process, we often break a redox
reaction down into two separate steps (half-reactions).
In one half-reaction, a chemical loses electrons (is
oxidized)
Zn (s)  Zn2+ (aq) + 2 eWe call this the oxidation half-reaction.
Notice that we are effectively treating electrons as a
“product” of the half-reaction.
In the other half-reaction, we look at the reduction halfreaction, where a chemical gains electrons (is reduced)
Cu2+ (aq) + 2 e-  Cu (s)
7
Half-reactions
The sum of these half-reactions
must give us the overall reaction
of interest.
Zn (s)  Zn2+ (aq) + 2 eCu2+ (aq) + 2 e-  Cu (s)
Zn (s) + Cu2+ (aq) + 2 e-  Zn2+ (aq) + 2 e- + Cu (s)
Zn (s) + Cu2+ (aq)  Zn2+ (aq) + Cu (s)
8
Half-reactions
Why do we call them “half-reactions”?
Each half-reaction is written so we can see what is
happening to the electrons in the overall reaction.
In reality a half reaction CANNOT occur by itself to
any great extent.
The lost electrons in the oxidation half-reaction
MUST go somewhere.
The gained electrons in the reduction half-reaction
MUST come from somewhere.
Two half-reactions ALWAYS work together to give
an overall reaction that can occur to a great extent.
9
Half-reactions
M (s)
oxidation





reduction
M n  (aq)  ne-
The electrons stay
on the metal
electrode and are
NEVER found in
solution!
10
Since the chemicals
are in direct contact
with each other, the
electron transfer
occurs directly and
we can’t use the
electrons to do
anything useful.
How can we separate
the chemicals but
allow the electrons to
transfer indirectly so
we can use them?
11
Copper in Ag+ solution is spontaneous
12
The halfreactions
to take place in
separate
containers
(called half-cells).
Since a half-reaction cannot take place by itself we
need to connect the half-cells together. It turns out
that we must make a circuit (two connections!) for
the entire galvanic cell to work.
13
The left half-cell has a
solid copper electrode
in a Cu2+ ion solution,
while the right halfcell has a solid silver
electrode in a Ag+ ion
solution.
A wire can connect the two solid electrodes for the
electrons to move through. To connect the two
solutions so that ions can move between the halfcells requires us to use a salt bridge, which is just
another solution of ions.
14
Oxidation occurs
at the copper
electrode, which
we give the
special name
ANODE
Since the anode collects the electrons that are
lost, it has a negative charge and positive copper
ions leave the anode!
Cu (s) 
2+
Cu
(aq) + 2
e
15
Electrons move
from the
ANODE
to the silver
electrode through
the wire
We can get them to do
something useful, like light a
bulb!
16
Reduction occurs
at the silver
electrode, which
we give the special
name
CATHODE
Since the cathode collects the positive silver ions
so they can gain the electrons, the cathode has a
positive charge!
+
Ag
(aq) +
e
 Ag (s)
17
Positive ions leave
the anode while the
cathode collects
positive ions.
Alternatively,
negative ions
collect at the anode The ions are free to move
through the salt bridge
and move away
and are REQUIRED to
from the cathode
complete the circuit!
18
Overall, negative
charges (electrons and
negative ions) are
moving clockwise
Overall, positive
charges (positive ions
and electron “holes”)
are moving
counterclockwise
19
The overall reaction is
exactly the same as when
we place solid copper in a
Ag+ solution, but since we
have separated the halfcells, we can look at the
separate half-reactions
as they occur.
Cu (s)  Cu2+ (aq) + 2 e2 x [Ag+ (aq) + 1 e-  Ag (s)]
Note # of
e- must
balance!
Cu (s) + 2 Ag+ (aq) + 2 e-  Cu2+ (aq) + 2 e- + 2 Ag (s)
Cu (s) + 2 Ag+ (aq)  Cu2+ (aq) + 2 Ag (s)
20
Shorthand notation for galvanic cells
Drawing a diagram for a galvanic cell or describing
it as we did in the previous problem is too timeconsuming to do on a regular basis.
We can use a shorthand
notation!
Cu (s) + 2 Ag+ (aq)  Cu2+ (aq) + 2 Ag (s)
Cu (s) | Cu2+ (aq) || Ag+ (aq) | Ag (s)
21
Cu (s) | Cu2+ (aq) || Ag+ (aq) | Ag (s)
A single vertical line indicates a change
in phase, like that between a solid
electrode and the solution its
immersed in.
A double vertical line indicates a salt
bridge.
What is not shown in the shorthand (but
is always implied) is the wire connecting
the two electrodes to complete the
circuit.
22
Cu (s) | Cu2+ (aq) || Ag+ (aq) | Ag (s)
If we read the shorthand notation from
left to right it says:
“A solid copper anode is in a solution
of copper (II) ions which is connected
by a salt bridge to a solution of silver
(I) ions into which a solid silver
cathode has been placed. The
electrodes are connected by a wire.”
23
We ALWAYS choose to write the cell notation
with the oxidation reaction first and then the
reduction reaction.
This means the leftmost chemical in the
notation is ALWAYS the anode, while the
rightmost chemical is ALWAYS the cathode.
Additionally, the electrons ALWAYS flow from
left to right through the wire, which is the way
we read the shorthand.
24
(through the wire
connecting the
electrodes)
25
Galvanic cells – Easy as ABC
Anode  Cathode
Negative  Positive
Oxidation  Reduction
Left  Right
“The anode is the negative electrode where
oxidation takes place. We put it on the left in
shorthand notation.”
“The cathode is the positive electrode where
reduction takes place. We put it on the right in
shorthand notation.”
26
Other shorthand notation considerations
Sometimes gases are involved in
galvanic cells.
Including them in the shorthand is
easy once we realize the gas is
just a separate phase and must
be separated from other phases
by a vertical line.
27
Other shorthand notation considerations
Consider this reaction
Cu (s) + Cl2 (g)  Cu2+ (aq) + 2 Cl- (aq)
Since we CAN’T use a gas as an electrode we
need some solid substance to do that job. In this
case we bubble the gas by a carbon rod
The cell notation with the carbon acting as the
cathode is
Cu (s) | Cu2+ (aq) || Cl2 (g) | Cl- (aq) | C (s)
28
Other shorthand notation considerations
Cu (s) | Cu2+ (aq) || Cl2 (g) | Cl- (aq) | C (s)
We can also be more specific by giving
concentration and pressure data for
any of the aqueous or gaseous
chemicals of the system.
e.g. Cu2+ (aq, 0.58 M)
and Cl- (aq, 0.34 M)
and Cl2 (g, 0.89 bar)
29
Problem
Write the shorthand notation for a
galvanic cell that uses the reaction
Fe (s) + Sn2+ (aq)  Fe2+ (aq) + Sn (s)
Fe (s) | Fe2+ (aq) || Sn2+ (aq) | Sn (s)
30
Problem
Write a balanced equation for the overall
cell reaction and give a brief description
of the galvanic cell represented by
Pb (s) | Pb2+ (aq) || Br2 (l) | Br- (aq) | Pt (s)
Pb (s) + Br2 (l)  Pb2+ (aq) + 2 Br- (aq)
The reduction of Br2 to Br- occurs on
the surface of a Pt cathode
31
Cell potentials for cell reactions
Electrons move from the
copper anode through the
wire to the silver cathode
because it is energetically
favourable for the
electrons to move!
An electron in a siver atom has less free energy
than the same electron in a copper atom.
Much like a ball wants to roll down a hill so it ends
up where it has the lower potential energy, an
electron wants to move to the atom where it has the
lower free energy.
32
Potential
The difference in the free energy for the
electrons in the anode and the cathode is
somewhat like the slope from the top to
the bottom of the hill.
If the hill is “steep”, the ball experiences
more of the force of gravity than it does on
a “gentle” hill.
33
Potential
The equivalent of the force of gravity to
the difference in the free energy of
electrons in different atoms is called the
electromotive force (emf) –
also known as the cell potential (E)
or the cell voltage (V).
Like a ball on a steep hill, electrons are
under a “greater” force to transfer from
the anode to the cathode when the cell
potential has a larger magnitude.
34
Potential
Because there is a free energy difference for
an electron in the anode as compared to the
same electron in the cathode, the electron
must lose free energy during the trip, just like
a ball loses potential energy (as motion!) as
it rolls down the hill.
The free energy change is negative and so
the movement of the electrons is a
spontaneous process!
35
Potential
The free energy change is negative
and so the movement of the electrons
is a spontaneous process!
This occurs when the
potential is positive so a
positive potential indicates a
spontaneous process
36
Potential
We can get energy out of a ball
(with its certain mass) rolling down
a slope (the experienced gravity),
We can get energy out of an
electron (with electrical charge)
that “rolls down the slope” that is
the potential difference of electron
free energy between the two
electrodes.
37
Potential
In terms of units, we can define
one Joule as the energy
we get from a
charge of one Coulomb
multiplied by the
potential of one volt.
1 J = 1 C·V (one Coulomb-volt)
38
Potential
A Coulomb is a very large unit of
charge!
The charge on one electron is
1.60 x 10-19 C, so
one Coulomb is the charge of
about
6 billion billion electrons!
39
Potential
It is generally easier to talk about the charge
of one mole of electrons, which we give
the special name of
Faraday Constant or faraday (F)
1 faraday = 6.022 x 1023 mol-1 e- x 1.60 x 10-19 C
1 faraday = 9.65 x 104 C·mol-1
40
Potential
We can measure the potential between two
electrodes with a voltmeter, which should
give a positive reading when the positive
terminal of the voltmeter is connected to
the positive electrode (the cathode),
and
the negative terminal is connected to the
negative electrode (the anode).
When the voltmeter gives a positive
potential, we have identified the direction
of spontaneous change!
41
Copper in Ag+ solution is spontaneous
42
Standard cell potentials
Cell potentials depend on many factors
other than the chemicals in the system,
including the temperature, ion
concentrations, and pressure.
Like in the thermodynamics chapter, where
we defined a standard state of conditions for
enthalpy tables, we can do the same to define
standard cell potentials E°.
43
Standard cell potentials
We can only measure a standard cell
potential if we have
pure solids and liquids (activities of 1),
all solution activities are 1 (@1 molL-1),
all gas activities are 1 (@1 bar),
and the temperature is specified
(usually 25 °C).
44
Standard cell potentials
Zn (s) | Zn2+ (aq) || Cu2+ (aq) | Cu (s)
we can only measure the STANDARD cell
potential if the [Zn2+] and [Cu2+] are both
1 molL-1 , and the copper and zinc
electrodes are pure.
The E° for this cell is 1.10 V at 25 °C.
45
Standard electrode potentials
The standard cell potential E for
any galvanic cell can be expressed
as the difference of the standard
electrode potentials for the
cathode and the the anode

E
cell
=

E
(red),cathode
-

E
(red),anode
46
Standard electrode potentials
The standard electrode potential
depends on whether the electrode is
acting as the cathode or the anode.
However, the process at the cathode
(reduction) is the opposite process
that would occur if it were happening
at the anode (oxidation).
47
Standard electrode potentials
Reversing a process changes the
sign of the electrode potential
associated with the process.
Therefore we choose to report
ALL standard electrode potentials
as reduction processes
because for any specific electrode
E(red),cathode = - E(red),anode
48
Standard electrode potentials
It would be nice to create a table of
standard electrode potentials for all
possible electrodes, then we could find
standard cell potentials for any cell.
However, there is one problem!
We’ve already seen that
half-reactions cannot occur
without another half-reaction!
49
We got around a problem like this in
thermodynamics by defining the standard
enthalpy of formation of elements in
their standard states as ZERO.
We can do the same for electrode
potentials and set the potential for a
specific electrode as ZERO and measure
all other electrode potentials in
comparison to the standard.
50
Standard hydrogen electrode
The standard electrode for potentials is
the standard hydrogen electrode (S.H.E).
The electrode consists of hydrogen gas at
1 bar bubbling through a 1 molL-1 (actually
activity of 1) solution of H+ past a
platinum electrode. Therefore
2 H+ (aq, a = 1) + 2 e-  H2 (g, 1 bar)
E(red),cathode = EH+/H2 = 0 V
51
Standard hydrogen electrode
If the oxidation reaction
occurs instead in this
half-cell as an anode,
the overall reaction is
H2 (g, 1 bar) 
2 H+ (aq, 1 molL-1) + 2 eReversing a reaction
changes the sign of the
potential. For the S.H.E.
E(red),anode = - E(red),cathode
= - EH+/H = 0 V
2
52
Pt (s) | H2 (g) | H+ (aq) || Cu2+ (aq) | Cu (s)
The standard
potential for this cell
has been measured as
0.340 V at 25 C, and
our anode is the
standard hydrogen
electrode!
E cell  E (red),cathode  E (red),anode  E Cu 2 / Cu  E H  / H  0.340 V
2
53
E cell  E (red),cathode  E (red),anode  E Cu 2 / Cu  E H  / H  0.340 V
2
We have defined the standard
electrode potential of the
reduction of Cu2+ ions to solid Cu!
This is also known as a
standard reduction potential.
(aq) + 2  Cu (s)
E(red) = 0.340 V
2+
Cu
e
54
Ecell  E(red),cathode  E(red),anode  ECu 2 / Cu  0.000V  0.340V
If we reverse the half-reaction we’ll
get the
standard OXIDATION potential
for the oxidation of solid Cu to Cu2+
ions…
Cu (s) 
(aq) + 2
E(ox) = -0.340 V
2+
Cu
e
55
Zn (s) | Zn2+ (aq) || H+ (aq) | H2 (g) | Pt (s)
The standard
potential for this cell
has been measured as
0.763 V at 25 C, and
our anode is the zinc
electrode!
E cell  E (red),cathode  E (red),anode  E H  /H - E Zn2 /Zn  0.763 V
2
56
Ecell  E(red),cathode  E(red),anode  0.000V - EZn2 /Zn  0.763V
We have found the standard
potential of the oxidation of solid
zinc to zinc ions!
This is a
standard oxidation potential.
Zn (s) 
(aq) + 2
E(ox) = 0.763 V
2+
Zn
e
57
Ecell  E(red),cathode  E(red),anode  0.000V - EZn2 /Zn  0.763V
If we reverse the half-reaction we’ll
get the standard reduction
potential for Zn2+ ions to solid zinc
(aq) + 2  Zn (s)
E(red) = -0.763 V
2+
Zn
e
We report this value as the
standard electrode potential!
58
Zn (s) | Zn2+ (aq) || Cu2+ (aq) | Cu (s)
The standard cell
potential for this cell
can be calculated if we
know the anode is the
zinc electrode and
the cathode is the
copper electrode!
E cell  E (red),cathode  E (red),anode  E Cu 2 /Cu - E Zn2 /Zn
 0.340V- (-0.763V)  1.103V
59
Standard electrode potentials E(red)
60
Using standard electrode potentials
Using tabulated standard electrode potential
data is accomplished much like a Hess’s Law
problem with one very important difference!
Let’s consider
Zn (s) | Zn2+ (aq) || Ag+ (aq) | Ag (s)
which has the balanced equation
2 Ag+ (aq) + Zn (s)  2 Ag (s) + Zn2+ (aq)
61
Using standard electrode potentials
Oxidation Zn (s)  Zn2+ (aq) + 2 eReduction 2 [Ag+ (aq) + 1 e-  Ag (s)]
2 Ag+ (aq) + Zn (s)  2 Ag (s) + Zn2+ (aq)
E cell  E (red),cathode  E (red),anode  E Ag  /Ag - E Zn2 /Zn
 0.800V- (-0.763V)  1.563V
Ecell = 1.563 V
62
Like Hess’s Law, we look up the standard
electrode potential reactions for both our sets
of chemicals and then reverse the halfreaction for the set undergoing oxidation
while changing the sign of the electrode
potential (- E(red),anode!) .
However, we DO NOT multiply the
potential for either half-reaction.
Why?
63
Potential
Recall the potential is
like the slope
of a hill.
A hill does not change its slope if we
have two (or more!) balls rolling
downhill instead of one ball.
Therefore the potential of an electrode
does not change if we multiply to get
the right number of electrons!
64
Problem
The standard cell potential for the following
galvanic cell is 0.78 V
Al (s) | Al3+ (aq) || Cr2+ (aq) | Cr (s)
The standard electrode potential for the Al
electrode is -1.676 V. Calculate the
standard electrode potential for the Cr
E
electrode. E  E

cell

(red),cathode

(red),anode
 E Cr 2 /Cr - E Al 3 /Al  E Cr 2 /Cr - (-1.676V)  0.78V
E Cr 2 /Cr  - 0.90V
65
Problem
Use the data from Table 20.1 to determine
the Ecell for the redox reaction in which
Fe2+ (aq) is oxidized to Fe3+ (aq) by MnO4(aq) in acidic solution. Also provide the
overall reaction.
Answer: Ecell = 0.74 V
5 Fe2+ (aq) + MnO4- (aq) + 8 H+ (aq) 
5 Fe3+ (aq) + Mn2+ (aq) + 4 H2O (l)
66
Free energy and electrical work
We’ve already seen for any system the
energy free to do work is given by
DG = DH - TDS
at standard conditions, or
DG = DH – TDS
at non-standard conditions.
67
Free energy and electrical work
Not all work has to be
expansion (PDV) work.
There are other types of work,
one of which is electrical work!
There must be a connection
between DG and electrical work
done by a galvanic cell.
68
Free energy and electrical work
We’ve seen for a spontaneous
process that DG < 0.
We’ve also seen for a galvanic
cell the overall reaction is
spontaneous, and the cell
potential is positive to indicate
spontaneity.
69
Free energy and electrical work
Therefore for a spontaneous process in a
galvanic cell, the change in free energy (the
electrical work done) must be directly
proportional to the negative of the potential.
DG  -Ecell
or
DG = welec = -kelec Ecell
70
Free energy and electrical work
The constant of proportionality kelec must
depend on two things. First, it depends on
how many electrons we have moved
through the wire. Twice as many electrons
should mean twice as much work…
We will usually measure
numbers of electrons in
moles and symbolize it by n.
71
Free energy and electrical work
The constant of proportionality kelec must
also depend on the charge of each
electron moving through the wire. Since we
are already talking about moles of
electrons, we should talk about the charge
of one mole of electrons.
We’ve already seen that the
faraday (F) = 9.65 x 104 C·mol-1
72
Free energy and electrical work
DG = welec = -kelec Ecell
DG = welec = -nFEcell
and at standard conditions
DG = welec = -nFEcell
73
Do the units match those for work?
DG = welec = -nFEcell
-1
= -(mol)(Cmol )(V)
= CV = J
Yes! The units match
those for work.
74
Problem
Use the given electrode potential data to
determine DG for the reaction
2 Al (s) + 3 Br2 (l) 
2 Al3+ (aq, 1 M) + 6 Br- (aq, 1 M)
Al3+ (aq) + 3 e-  Al (s)
Br2 (l) + 2 e-  2 Br- (aq)
EAl3+/Al = -1.676 V
EBr2/Br- = 1.065 V
Answer: Since Ecell is 2.741V and the rxn
involves 6 mol of e-, then DG is -1587 kJ.
75
Problem
The hydrogen-oxygen fuel cell is a galvanic cell
with a reaction
2 H2 (g) + O2 (g)  2 H2O (l)
Using the given data calculate Ecell for this
reaction:
DGf (H2O) = -237.1 kJmol-1
DGf (H2) = 0.00 kJmol-1
DGf (O2) = 0.00 kJmol-1
Answer: Since DG is -474.2 kJ and the rxn
involves 4 mol of e-, then Ecell is 1.229 V .
76
Spontaneous change in redox reactions
We’ve already related the free energy
change to the cell potential
DG = -nFEcell
and we also know a spontaneous process
DG < 0
WHICH MEANS Ecell > 0
MUST HAVE
for ALL spontaneous electrochemical
(oxidation-reduction) processes.
77
Problem
When sodium metal is added to seawater,
which has [Mg2+] = 0.0512 M, no
magnesium metal is obtained. According
to the data below, should this reaction
occur? What reaction does occur?
Na+ (aq) + 1 e-  Na (s)
ENa+/Na = -2.713 V
Mg2+ (aq) + 2 e-  Mg (s)
EMg2+/Mg = -2.356 V
2 H2O (l) + 2 e-  H2 (g) + 2 OH- (aq) EH2O/H2 = -0.828 V
78
Problem answer
For the reaction
2 Na (s) + Mg2+ (aq)  2 Na+ (aq) + Mg (s)
Ecell = 0.357 V and the reaction should be
spontaneous. However, the reaction of sodium with
water is
2 Na (s) + 2 H2O (l)  2 Na+ (aq) + H2 (g) + 2 OH- (aq)
and has Ecell = 1.885 V and this reaction should also
be spontaneous.
Since this reaction is “more spontaneous” (higher
Ecell) sodium preferentially reacts with water and not
magnesium ions!
79
Problem
Without using the data for a detailed
calculation, explain why Sn2+ solutions
must be protected from oxygen. One way
to protect them is to add metallic (solid) tin.
Sn4+ (aq) + 2 e-  Sn2+ (aq)
Sn2+ (aq) + 2 e-  Sn (s)
O2 (g) + 4 H+ (aq) + 4 e-  2 H2O (l)
ESn4+/ Sn2+ = 0.154 V
ESn2+/ Sn = -0.137 V
EO2/H2O = 1.229 V
80
Problem answer
For both possible reactions the
reduction of oxygen is the cathode
half-cell reaction.
Since Ecell = E(red),cathode - E(red),anode,
then the anode half-cell reaction that is
more negative will give the higher (“more
spontaneous”) Ecell reaction that will
preferentially occur.
81
Metals and acids
Some metals will react with acidic
solutions to form H2 gas and metal ions
in solution while others will not.
We now know that those metals that do
react with acid do so because the
reaction is spontaneous while those that
do not react do not because the reaction
is non-spontaneous.
82
Metals and acids
In MOST cases the reduction reaction of
metals in acidic solutions is
2 H+ (aq) + 2 e-  H2 (g)
E(red),cathode = EH+/H2 = 0 V
This is the standard hydrogen
electrode half-reaction !
83
Metals and acids
IF this is the preferred reduction (cathode)
reaction, and since a spontaneous process
must have a positive potential then for a metal
to react with the H+ of an acid means
Ecell = E(red),cathode - E(red),anode > 0
(0 V) - E(red),anode > 0
E(red),anode < 0
84
Metals and acids
The metals that CAN REACT
with H+ are the ones with a
negative Ered value like
Na (ENa+/Na = -2.713 V)
or Al (EAl3+/Al = -1.676 V)
or Pb (EPb2+/Pb = -0.125 V)
85
Metals and acids
The metals that CAN NOT
REACT with H+ have a
positive Ered value like
Ag (EAg+/Ag = +0.800 V)
or Au (EAu3+/Au = +1.52 V)
or Cu (ECu2+/Cu = +0.340 V)
86
Metals and acids
Some acids, like HNO3 have a
different preferred reduction
(cathode) reaction. For example
NO3- (aq) + 4 H+ (aq) + 3 e- 
NO (g) + 2 H2O (l)
E(red),cathode = ENO3-/NO
= +0.956 V
87
Metals and acids
IF this is the preferred reduction (cathode)
reaction, and since a spontaneous process
must have a positive potential then for a metal
to react with the NO3- of nitric acid means
Ecell = E(red),cathode - E(red),anode > 0
(0.956 V) - E(red),anode > 0
E(red),anode < 0.956 V
88
Metals and acids
In nitric acid all the metals
that usually react with acids
will still react, but now
Ag (EAg+/Ag = +0.800 V)
WILL ALSO REACT!
89
Ecell and Keq
We have two equations relating
free energy to Keq and Ecell
DG = -RT ln Keq
and
DG = -nFEcell
90
Ecell and Keq
By setting the equations equal to each
other we find the relationship between
cell potential and the thermodynamic
equilibrium constant
-RT ln Keq = -nFEcell
Ecell = (RT/nF) ln Keq
91
Ecell and Keq
Ecell = (RT/nF) ln Keq
We have two constants (R and F)
in this equation and we often
perform reactions at 298.15 K.
With these three fixed values we
can simplify this equation (but we
DON’T HAVE TO!)
92
Ecell and Keq
Ecell = (0.025693 V/n) ln Keq
using R = 8.3145 JK-1mol-1
BE CAREFUL!
This form ONLY applies
at 298.15 K!
93
Everything is connected!
We COULD also put kinetics and rate constants
and how they relate both to thermochemistry
and equilibrium in this diagram to show ALL the
possible connections in the chemistry you’ve seen!
94
Problem
Should the reaction of solid Al with Cu2+
ions go to completion at 25 C if Ecell for
the reaction is 2.016 V?
2 Al (s) + 3 Cu2+ (1 M)  3 Cu (s) + 2 Al3+ (1 M)
Answer: The reaction involves 6
moles of electrons, so Keq = e471 =
very large, so the reaction goes to
completion.
95
Problem
Should the reaction of solid Sn with Pb2+
ions go to completion at 25 C?
Pb2+ (aq) + 2 e-  Pb (s)
Sn2+ (aq) + 2 e-  Sn (s)
EPb2+/Pb = -0.125 V
ESn2+/Sn = -0.137 V
Answer: Since Ecell = 0.012 V and 2
moles of electrons are involved in the
process Keq = 2.5 and the reaction
does not go to completion.
96
Ecell as a function of concentration
Zn (s) | Zn2+ (aq, 1M) || Cu2+ (aq, 1M) | Cu (s)
We’ve seen that Ecell is 1.103 V for this
reaction at standard conditions.
However, what happens to the cell at
non-standard conditions?
Zn (s) | Zn2+ (aq, 0.10 M) || Cu2+ (aq, 2.0 M) | Cu (s)
If we set up this cell and measure the
potential, then Ecell is 1.142 V.
97
Ecell as a function of concentration
Since the actual reaction is
Zn (s) + Cu2+ (aq)  Zn2+ (aq) + Cu (s)
then Le Chatalier’s Principle tells us
that decreasing [Zn2+] from 1 M to 0.10
M should shift the reaction towards
products and increasing [Cu2+] from 1
M to 2.0 M should shift the reaction
towards products as well.
98
Ecell as a function of concentration
Both new concentrations serve to make the
reaction more complete (or more
spontaneous), and so we expect a more
positive potential!
Recall that
DG = DG + RT ln Qeq
and
DG = -nFEcell
99
Ecell as a function of concentration
By substituting we see
-nFEcell = -nFEcell + RT ln Qeq
or
Ecell = Ecell - (RT/nF) ln Qeq
100
Ecell as a function of concentration
Sometimes we prefer log to ln!
Since ln x = 2.3026 log x
we can change
Ecell = Ecell - (RT/nF) ln Qeq
into the Nernst Equation
Ecell = Ecell - 2.3026
(RT/nF) log Qeq
101
Ecell as a function of concentration
For this reaction
Zn (s) + Cu2+ (aq)  Zn2+ (aq) + Cu (s)
Qeq = aZn2+ / aCu2+ and so
Ecell = Ecell - 2.3026
(RT/nF) log{aZn2+/aCu2+}
102
Ecell as a function of concentration
If we plot Ecell versus log Qeq
we should get a
straight line with a
slope of [- (2.3026 RT/nF)]
and y-intercept of Ecell
103
Ecell as a function of concentration
Zn (s) + Cu2+ (aq)  Zn2+ (aq) + Cu (s)
Qeq = aZn2+ / aCu2+
104
Ecell as a function of concentration
Ecell = Ecell - (2.3026 RT/nF) log Qeq
Since R (8.3145 JK-1mol-1)
and F (9.65 x 104 C·mol-1)
are constants, and if we choose the
temperature to be 298.15 K, then we can
replace these three fixed values
as we have done before (slide 93)
105
Nernst equation at 298.15 K
Ecell = Ecell
– (0.0592 V/n) log Qeq
It makes the most sense to
memorize the Nernst
Equation and substitute
rather than remembering this
form for one temperature!
106
Problem
Calculate Ecell for the for the following
galvanic cells at 298.15 K. Will the
reactions be spontaneous?
Al (s) | Al3+ (0.36 M)
|| Sn4+ (0.086 M), Sn2+ (0.54 M) | Pt(s)
Pt(s) | Cl2 (1 atm) | Cl- (1.0 M)
|| Pb2+ (0.050 M), H+ (0.10 M) | PbO2(s)
107
Problem data
Sn4+ (aq) + 2 e-  Sn2+ (aq)
ESn4+/Sn2+ = 0.154 V
Al3+ (aq) + 3 e-  Al (s)
EAl3+/Al = -1.676 V
PbO2 (s) + 4 H+ (aq) + 2 e-  Pb2+ (aq) + 2 H2O (l)
EPbO2/Pb2+ = 1.455 V
Cl2 (g) + 2 e-  2 Cl- (aq)
ECl2/Cl- = 1.358 V
108
Problem answer
3 Sn4+ (0.086 M) + 2 Al (s) 
3 Sn2+ (0.54 M) + 2 Al3+ (0.36 M)
Ecell = 1.830 V and Ecell = 1.815 V
PbO2 (s) + 4 H+ (0.10 M) + Cl2 (1 atm) 
Pb2+ (0.050 M) + 2 H2O (l) + 2 Cl- (1.0 M)
Ecell = 0.097 V and Ecell = 0.017 V
Since both Ecell values are positive, both
reactions will be spontaneous at the given
conditions.
109
Problem
For what ratio of [Sn2+] / [Pb2+] will
the given cell reaction NOT be
spontaneous in either direction?
Sn (s) | Sn2+ (aq) || Pb2+ (aq) | Pb (s)
Pb2+ (aq) + 2 e-  Pb (s)
Sn2+ (aq) + 2 e-  Sn (s)
EPb2+/Pb = -0.125 V
ESn2+/Sn = -0.137 V
110
Problem answer
The reaction is NOT spontaneous in
either direction ONLY at equilibrium,
where Ecell = 0.
Since for this cell Ecell = 0.012 V the
equilibrium occurs when
Qeq = [Sn2+] / [Pb2+] = Keq = 2.5
(see slide 96)
111
Concentration cells
We know if we mix two solutions of the
same chemical but with different
concentrations, then the final solution
will have a single uniform
concentration.
The mixing is a
spontaneous process!
112
Concentration cells
We can set up the mixing process
as an electrochemical cell!
The different concentrations in the
two half-cells
will lead to a
non-zero Ecell
113
Concentration cells
We can set up the mixing process
as an electrochemical cell!
The different concentrations in the
two half-cells
will lead to a
Ecell
different from
Ecell
114
Concentration cells for H+
Pt (s) | H2 (1 atm) | H+ (x M) || H+ (1 M) | H2 (1 atm) | Pt (s)
H2 (g, 1 atm)  2 H+ (x M) + 2 e2 H+ (1 M) + 2 e-  H2 (g, 1 atm)
Net reaction: 2 H+ (1 M)  2 H+ (x M)
Ecell
will ALWAYS
be zero for a
concentration
cell…
115
Concentration cells for H+
Net reaction: 2 H+ (1 M)  2 H+ (x M)
Ecell = Ecell - 2.3026
(RT/nF) log Qeq
Ecell = 0 - 2.3026
2
2
(RT/nF) log x /1
Ecell = 2.3026 (2RT/nF) (-log x)
116
Concentration cells for H+
Ecell = 2.3026 (2RT/nF) (-log x)
if x = [H+] then (-log x) = pH
At 298.15 K we can replace our three
fixed values R, F, and T to give
Ecell = 0.0592 V (pH)
This is the basis for electronic
pH meters!
117
Concentration cells for finding Ksp
Ag (s) | Ag+ (sat’d AgI) || Ag+ (0.100 M) | Ag (s)
Ag (s)  2 Ag+ (sat’d AgI) + eAg+ (0.100 M) + e-  Ag (s)
Net reaction: Ag+ (0.100 M)  Ag+ (sat’d AgI)
Ecell
will ALWAYS
be zero for a
concentration
cell…
118
Concentration cells for finding Ksp
Net reaction: Ag+ (0.100 M)  Ag+ (sat’d AgI)
We measure Ecell for this concentration cell
and we find it to be 0.417 V
Since
Q = [Ag+] / [Ag+]
and
Ecell = Ecell
– (0.0592 V/n) log Qeq
and n = 1
119
Concentration cells for finding Ksp
Net reaction: Ag+ (0.100 M)  Ag+ (satd AgI)
Ecell = Ecell - (0.0592 V) log Qeq
0.417 V = 0 - (0.0592 V) log Qeq
log Qeq = 0.417 V / (-0.0592 V)
log Qeq = -7.044
Qeq = [Ag+]/(0.100 M) = 9.04 x 10-8
So
+
[Ag ]
= 9.04 x
-9
10 M
120
Concentration cells for finding Ksp
+
Since the [Ag ]
came from a
saturated AgI solution, then
[Ag+] = [I-] = 9.04 x 10-9 M
+
and Ksp = [Ag ] [I ]
Ksp = (9.04 x 10-9) (9.04 x 10-9)
Ksp = 8.3 x
-17
10
121
Problem
If Ksp = 1.8 x 10-10 for silver chloride then
what would be Ecell for
Ag (s) | Ag+ (sat’d AgCl) || Ag+ (0.100 M) | Ag (s)
Answer: Ecell = 0.23 V
122
Problem
Calculate the Ksp for lead iodide with the
given concentration cell information
Pb (s) | Pb2+ (sat’d PbI2)||Pb2+ (0.100 M) | Pb (s)
Ecell = 0.0567 V
Answer: Ksp = 7.1 x 10-9
123
Electrolysis and electrolytic cells
The reverse of every spontaneous
chemical reaction is nonspontaneous.
If we apply electric current to a
chemical system, it is possible to force
non-spontaneous chemical reactions
occur in a process called electrolysis,
in what we call electrolytic cells.
124
Electrolysis and electrolytic cells
The potential we apply to the
electrolytic cell must be greater
than that for the spontaneous
reaction, and must be applied
in the opposite direction.
Ebattery > -Ecell
125
Zinc in Cu2+ solution is spontaneous
Since we actually see
this reaction
occurring, this
reaction must be
spontaneous!
The reverse reaction,
where we put copper
metal into a Zn2+ ion
solution is nonspontaneous!
126
Zinc in Cu2+ solution is spontaneous
If we want to get
zinc from this cell,
we must force the
non-spontaneous
reaction to occur by
applying a potential
in the direction
opposite that for the
spontaneous
process!
Reduction ALWAYS
occurs at the
cathode!
Cathode is –ve!
Reduction!
Anode is +ve!
Oxidation!
127
Electrolysis as coupled reactions
When we are doing electrolysis we
are using the spontaneous battery
reaction to drive the non-spontaneous
electrolysis reaction.
battery reactants  battery products
Ebattery > 0 so DG < 0
electrolysis reactants  electrolysis products
Ecell < 0 so DG > 0
128
Electrolysis as coupled reactions
In our setup we are coupling
(adding) the reactions which means
we add their potentials
(or free energies).
battery reactants + electrolysis reactants 
electrolysis products + battery products
Esum = Ebattery + Ecell > 0
so DGsum < 0
129
Complicating factors in electrolysis
While adding the potentials to
get the potential for the
coupled reaction is
straightforward in theory,
in practice there are
complicating factors!
130
Overpotentials
The electrolytic cell has electron transfers
occurring at the surface of the electrodes. If
solutions are involved then there is generally a
good contact to the electrode.
However, if gases are contacting the electrode
the contact is problematic.
As the contact to the electrode gets worse we
often need to apply an overpotential (extra
Eoverpotential) to make up for this problem.
131
Overpotentials
Ebattery > -(Ecell + Eoverpotential)
For example, a solid platinum
electrode generally has a
near-zero volt Eoverpotential
while the formation of H2 gas on
the surface of a mercury cathode
has Eoverpotential  1.5 V
132
Competing reactions
If we set up an electrolytic cell expecting
Esum = Ebattery + Ecell > 0
will give us the reaction we want we may
be surprised when we get a completely
different reaction because
Ebattery + Eother > Ebattery + Ecell
133
Competing reactions
Ebattery + Eother > Ebattery + Ecell
We saw in slides 78-79 that if we
have competing reactions, then
the one that is “more
spontaneous” will preferentially
occur.
134
Competing reactions
Often, but not always, when we do
electrolysis in aqueous solutions we
get the competing reactions
2 H2O (l) + 2e-  H2 (g) + 2 OH- (aq)
at the cathode and
2 H2O (l) O2 (g) + 4 H+ (aq) + 4 eat the anode.
135
Competing reactions
When we do electrolysis in aqueous
solution
we must identify which of the two possible
reduction reactions is “more
spontaneous” when forced and
we must identify which of the two possible
oxidation reactions is “more
spontaneous” when forced.
See pages 850-851 and Example Problem
20-11 in the text for more info on this
very important topic
136
Non-standard conditions
Industrially we try to maximize
product with minimum energy and
money input. This often means that
we do electrolysis on cells at nonstandard conditions, which means
Ecell  Ecell
137
Electrodes
Platinum is an inert electrode that only
provides a surface for the true reactants to
transfer electrons.
An active electrode is an actual reactant
in the half-cell reaction.
Using a different electrode on one side
of the electrolytic cell might change the
half-cell reaction on that side!
138
Quantitative aspects of electrolysis
139
Quantitative aspects of electrolysis
If we pass 1 mole of electrons through
the cell, from the balanced equation
Na+ (l) + e-  Na (s)
we see we will get 1 mole (23.0 g) of
solid sodium out. At the other
electrode, where
2 Cl- (l)  Cl2 (g) + 2 ewe see that one mole of electrons is
enough to give us one-half a mole
(35.5 g) of Cl2.
140
Quantitative aspects of electrolysis
How many electrons pass through
the cell depends on the current,
which is charge per unit time, (the
ampere A, which is a C/s) and the
time the current was allowed to pass
though the cell…
Charge (C) = Current (C/s) x time (s)
Charge (C) = Current (A) x time (s)
141
Quantitative aspects of electrolysis
We saw earlier that one mole of
electrons has a charge equal to one
Faraday
1 F = 9.65 x 104 Cmol-1
moles of e- = Charge (C) / Faraday
moles of e- =
(Current x time)
9.65 x 104 Cmol-1
142
The flowchart shows how to find
the amount of substance that
comes from electrolysis based on
a known current and time.
If we want to know the current or
time we used to get a certain
amount of substance, we
reverse the order of the
flowchart.
143
Problem
How many kilograms of aluminum can be
produced in 8.00 h by passing a constant
current of 1.00 x 105 A for an electrolytic cell
with the following half reaction at the
cathode?
Al3+ + 3 e-  Al
Molar mass of Al is 26.9815 gmol-1
Answer: 268 kg
144
Problem
A layer of silver is electroplated (an
electrolytic process) on a coffee server using
a constant current of 0.100 A. How much
time is required to deposit 3.00 g of silver?
Molar mass of silver is 107.868 gmol-1
Answer: 7.45 hours
145