Electrochemistry

1
Chapter 20

• Electrochemistry

2
I. Electrochemistry deals with
A) Voltaic cells - batteries - electricity
produced by _______________ .
B) Electrolysis - electrolytic cells - chemical
reactions produced by_________.
II. Importance
A) Application to the development of storage
batteries, fuel cells, dry cells.
B) Electrolysis - production and purification of
metals.
3
III. Conduction of electricity
A) Conduction through metals involves
1) the flow of electrons in a wire.
2) no change in the properties of the wire.
3) no appreciable amount of matter transferred.
4) an increase in resistance as the temperature
increases.
4
B) Electrolytic conduction involves
1) the movement of ions in _______________.
2) chemical reactions which take place at the
_________________.
3) the transfer of an ___________________
_________________.
4) the decrease in the ___________________ as the
temperature _________________.
5
IV. Important Units of Electricity
A) E electromotive force - EMF - measured in
_______________.
B) I current - measured in ____________.
C) R resistance - measured in ________
V. What is the relationship among the units?
A) E IR
B) I current - electrons are forced through a
circuit by an electrical potential difference
which is measured in _______________ .
6
C) Electrical energy - the unit is the Joule.
1) It takes 1 Joule of work to move 1 coulomb of
electricity from a lower to a higher potential
when the potential difference is 1 volt.
7
How is this related to unit of electrical energy
you buy from the electric company, the
_____________________.
8
VI. An electrochemical cell aka a voltaic cell or
a galvanic cell is
A) An arrangement of two electrodes in
appropriate electrolytes. Such a cell is capable
of producing an electric current as a result of a
chemical reaction within the cell.
Zn(s) Cu2 ? Zn2 Cu(s)
There is oxidation and reduction taking place
here.
9
Zn(s) ? Zn2 l-e-o
Cu2 ? Cu(s) g-e-r
2) To design a cell using this reaction as a
source of electrical energy, the electron
transfer must occur indirectly. The electrons
given off by the zinc atoms must be made to pass
through an external circuit before they
___________________________ .
10
(No Transcript)
11
3) A way to do this is to set up the following
12
What does the salt bridge do?
As the number of Cu2in the right compartment
decreases, it would become ______________.
The left compartment would become
__________________ because of the production of
__________.
This is an impossible situation. Why?
13
A salt solution is necessary to keep the
concentration of _____________________.
It completes the circuit with the movement of
ions. An inert substance is used KNO3 in agar in
a glass tube is the expensive way to do it. We
will use the cheap way of a piece of filter paper
dipped in a solution of KNO3 in the lab.
14
The K will migrate to the Cu2 solution. Two K
ions will be introduced for every Cu2 removed
from the solution and two NO3- ions will be
introduced for every _____________.
5) The concentration of the solution in the half
cell is important with respect to the measured
voltage.
15
For us the standard will be 1.0M. In advanced
courses the concept of activity is used instead
of in many cases. Activity and are not
usually equivalent, but for us the is close
enough to activity to be useful.
B) The electrode at which oxidation takes place
is called the anode (vowels a o).
16
1) It is always written (or drawn) on the left
side. It is always given a negative sign.
Zn(s) ? Zn2 anode - negative electrode
In Zn2solution the Zn2 concentration increases,
the mass as well as the size of the Zn electrode
________________.
A source of negative ions is required to balance
this increase in concentration of positive ions.
These ions must come from the ________________.
It is important that these ions do not react with
the zinc ions.
17
C) The electrode at which reduction occurs is
called the ___________ (consonants r and c).
1) It is always written on the ____________. It
is given a __________________, it takes electrons.
2) The Cu2 _______________, the Cu electrode
gets ________________, and the ________________
ions from the salt bridge migrate toward the
cathode.
18
3) Instead of drawing beakers, pieces of metal,
etc., we can abbreviate the cell this way
Zn(s) Zn2(aq) Cu2(aq) Cu(s)
is a phase boundary, i.e. separates different
states.
represents the salt bridge
oxidation is on the left and reduction is on the
right.
19
IN GENERAL FOR A CELL
ANODE MATERIAL ANODE SOLUTION CATHODE
SOLUTION CATHODE TERMINAL
D) Another example of a spontaneous reaction
which we can turn into a cell is the following
Zn(s) 2 H ? Zn2(aq) H2(g)
20
Zn(s) ? Zn2(aq) 2 e- _______________
2 H 2e- ? H2(g) __________________
Zn(s) Zn2(aq) H(aq) H2(g) Pt
21
(No Transcript)
22
Why do we need the Pt for the electrode?
In the lab we will not use Pt. Why not?
We will use graphite rods - we need something
which will _________________ and be __________ in
the chemical reaction studied.
23
VII. STANDARD ELECTRODE POTENTIALS
a) Electromotive force
1) For a pure metal in pure water, there is a
tendency for M(s) ? M(aq) 1 e-, with the
electron being left behind on the electrode.
After a period of time M(aq) 1 e- ? M(s).
Eventually an equilibrium is reached when the
rates of the 2 reactions are equal.
24
2) There is a tendency for different metals to
react this way to different extents, and they can
be ranked. For example
Mg(s) 34 Mg2(aq) 2 e- Zn(s) 34 Zn2(aq) 2 e
- H2(g) 34 2 H(aq) 2 e-
Cu(s) 34 Cu2(aq) 2 e-
25
3) For each half cell there exists at the surface
between the metal and the solution, a difference
in potential which is called the electrode
potential. The absolute electrode potential would
be the work required to bring an electron from
infinity to the electrode. We cannot measure this
absolute potential so what do we do?
______________
26
4) You can imagine that the cell emf is composed
of a contribution from the anode (whose value
depends on the ability of the oxidation
half-reaction to lose electrons) and a
contribution from the cathode (whose value
depends on the ability of the reduction half
reaction to gain electrons).
Ecell oxidation potential reduction potential
Ecell Eox Ered
27
5) the relative scale is based on the H2
electrode.
a) Under standard conditions, when we connect a
zinc half cell with an H2 half cell, a voltmeter
will read 0.76 volts, which is the sum of the 2
half reactions
Zn(s) ? Zn2(aq) 2 e- 2H 2 e- ? H2(g) 2H
Zn(s) ? Zn2(aq) H2(g)
28
(No Transcript)
29
0.76 volts the oxidation potential the
reduction potential.
E reduction for the H2 is given the value of
0.00V.
Eocell Eoox Eored
0.76V EoZn/Zn2 0
0.76V Eoox Zn/Zn2
30
The oxidation potential for a half-reaction
minus the reduction potential for the reverse
half-reaction.
Zn2(aq) 2 e- ? Zn(s) Eored Zn2/Zn -
0.76 V
By convention we tabulate the reduction
potentials. See page 856. Some appear on the
following slides.
31
(No Transcript)
32
If we put copper in with H2, and we set the
half-cells up in the same way as we did with Zn
and H2, the voltmeter will read a NEGATIVE value.
This means ____________________________________.
When we draw the cell we have to put oxidation on
the left, the H2 electrode is on the left side..

33
H2(g) ? 2 H(aq) 2 e- (ox)
(anode) Cu2(aq) 2 e- ? Cu(s) (red) (ca
thode)
The voltmeter reads 0.34 volts.
0.34 volts 0 EoCu2/Cu
Eoreduction for Cu2/Cu 0.34 volts.
34
For the zinc and copper cell, the half-reactions
are
Zn(s) ? Zn2 2e- l-e-o Cu2 2e- ? Cu(s)
g-e-r
Zn(s) Cu2 ? Zn2 Cu(s)
Eocell Eoox Eored
In terms of reduction potentials
Eocell - EoZn EoCu
35
E0cell ______ V
36
We can use the sign of the emf of the cell to
predict the spontaneity of redox reactions.
1) Under standard state conditions, for reactants
and products ( 1 M, 1 atm, 25oC), the redox
reaction is spontaneous in the forward direction
if the standard emf of the cell is positive. If
it is negative it is spontaneous in the opposite
direction.
37
2) That means the sum of Eoox Eored or (-
Eored Eored) must be positive for the reaction
as written to be spontaneous.
3) The more positive Eo for the reduction is, the
greater the tendency for the substance to be
reduced.
F2 2 e- ? 2F- Eo 2.87 V
Thus F2 is the strongest oxidizing agent because
it has the greatest tendency to be reduced, i.e.
gain electrons.
38
4) At the other extreme Li e- ? Li Eo - 3
.04 V
Li is the weakest oxidizing agent, the most
difficult species to reduce.
It has the greatest tendency to go to the left.
The strongest reducing agents in a table of
standard electrode potentials are the reduced
species corresponding to the half-reactions with
the smallest (most negative) Eo values.
39
(No Transcript)
40
On the table, the species on the left, the
oxidizing agent, increases in strength from top
to bottom, and the reducing agent, the species on
the right increases in strength from bottom to
top.
F2 is the best oxidizing agent - the best at
taking electrons.
F- is the worst reducing agent - the worst at
giving electrons.
41
Li is the best reducing agent - the best at
giving electrons.
5) Half-cell reactions are reversible - depending
on the conditions any electrode can act as either
the anode or the cathode.
6) Under standard conditions, any species on the
left of a given half-cell reaction will react
spontaneously with a species that appears on the
right of any half-cell located above it on the
table.
42
7) Is the spontaneous reaction
Cu2(aq) 2 Ag(s) ? 2 Ag Cu(s)
OR
Cu(s) 2 Ag(aq) ? Cu2(aq) 2 Ag(s)
From the table we see Ag is on the left and Cu
is on the right and above it, so we predict the
spontaneous reaction is
Cu(s) 2 Ag(aq) ? Cu2(aq) 2 Ag(s)
43
To make the cell, the Cu half cell will be on the
left, and the Ag half-cell will be on the right.
Cu(s) Cu2(aq) Ag(aq) Ag(s)
Cu(s)? Cu 2(aq) 2 e- -0.34V
2 Ag(aq) 2e- ? 2 Ag(s) 0.80V
Cu 2Ag ? Cu2 2 Ag(s) 0.46 V
44
Look carefully at the value 0.80 volts. While I
have multiplied the half reaction by 2, I do not
multiply the Eo by 2.
The half-cell potentials are independent of the
amount of substance they are intensive
properties of the substances.
45
VIII. SPONTANEITY OF REDOX REACTIONS
A) We have used DGo to determine the spontaneity
of a reaction. If DGo is (-), the reaction is
spontaneous as written. If DGo is () the
reaction is not spontaneous as written, but the
reverse reaction is spontaneous.
We have used Eocell as the measure of the
tendency of a net cell reaction to occur as
written.
46
Since both DGo and Eo are criteria for
spontaneity, they must be related to each other.
Recall that DGo is the maximum useful work that
can be exchanged between the system and
surroundings in passing from an initial to a
final state.
Electrical work (energy quantity of electrons X
potential difference)

47
Electrical work Joules n X F X Eo
48
Electrical work is the maximum useful work at
maximum cell potential. That is, when the current
flow is equal to zero. This is never realized
since it takes work to drive a current through
the cell itself, there is some frictional
heating, some energy is always wasted.
49
DGo maximum useful work
DGo - nFEo
(- sign is necessary since the reaction is
spontaneous when DGo is (-) and Eo is ().
1F 96,500 C/mole of electrons 96,500 J/volt
mole electrons
DGo 96.5 kJ/v mol e X n x Eo kJ
50
Example
DGo -n FEo -96.5 kJ/mole X 2 mol e X 1.10V
DGo - 212kJ (the reaction is spontaneous.)
51
IX. THE NERNST EQUATION
A) LeChatelier's Principle (again)
If a system in chemical equilibrium is disturbed
by a change in conditions, the system shifts in
equilibrium composition to minimize the effect of
the imposed changes.
52
B) we apply this to the half-reaction
Zn 34 Zn2 (aq 1M) 2e-
As the Zn2 becomes larger than 1 M, the
equilibrium shifts to reduce the Zn2. Fewer
electrons are made available, and a smaller
negative charge is associated with the zinc
electrode. There is a lower value for the zinc
electrode potential.
53
Conversely, if we decrease the Zn2 below 1 M,
the relative tendency for the Zn metal to give
electrons to increase the Zn2 is increased and
the measured electrode potential is greater than
0.76 V.
C) The value of the cell voltages at 's other
than 1 M maybe quantitatively calculated from the
Nernst equation.
54
The dependence of cell potential on concentration
results directly from the dependence of free
energy on concentration.

Recall from the last chapter
DG DGo RT ln Q
And in this chapter
DGo -nFEo (maximum useful work)
DG - nFE
55
- nFE -nFEo RTln Q
change ln to log (base e to base 10)
- nFE -nFEo 2.303 RTlog Q
divide by - nF
And we obtain the Nernst equation
56
(No Transcript)
57
2) An example
Ni(s) Ni2(0.600M) Sn2(0.300M) Sn(s)
According to the reduction potentials
2 e- Ni2 34 Ni(s) -0.230 V
2 e- Sn2 34 Sn(s) -0.140V
One of these needs to be reversed to get a
positive voltage, and a spontaneous reaction.
58
Ni(s) 34 2 e- Ni2 0.230 V
2 e- Sn2 34 Sn(s) -0.140V
Ni(s) Sn234 Ni2 Sn(s) 0.090 V Eo
59
E) Calculation of the equilibrium constant
1) at equilibrium E 0 Q ____
From the Nernst Equation
For the cell Ni(s) Ni2(0.600M) Sn2(0.300M)
Sn(s)

60
(No Transcript)
61
F)The pH meter is a special case of the Nernst
Equation
1/2 H2 ? H 1 e-
Q H
62
pH - log H
63
E Eo 0.0592 pH
The scale on the pH meter is marked off so that a
change of 1 pH unit equals 0.0592 volts or 59.2
millivolts.
64
X. Electrolysis - Electrolytic cells
A) Two electrodes, generally inert metals, are
placed in a solution of an electrolyte, or a
molten metal salt and a current is passed through
the solution or the molten salt.
B) Cations, () ions, are attracted to the
cathode, the negative electrode in this case.
Anions, (-) ions, are attracted to the anode, the
positive electrode in this case.
65
C) Oxidation takes place at the anode (always).
Anions lose e's to the anode and are oxidized.
D) At the cathode, cations pick up electrons and
are reduced. Reduction takes place at the cathode.
E) Examples
66
(No Transcript)
67
1) The dotted line represents a membrane to keep
the components separated.
2) At the anode a yellow-green gas is formed, and
at the cathode a silvery metal is formed.
3) At the anode 2 Cl- ? Cl2 2 e- (l-e-o)
4) At the cathode 2 Na 2e- ? 2 Na (g-e-r)
5) The overall reaction is 2 Na(l) 2 Cl-(l) ?
2 Na(l) Cl2(g)
68
The industrial version looks like this
69
F) In the electrolysis of a molten salt, the
possiblehalf-reactions are usually limited to
those involving ions from the salt. When you
electrolyze an aqueous solution of an ionic
compound, however, you MUST consider the
possibility that _____________ is involved at one
or both electrodes.
70
G) What happens when a water solution of NaCl is
electrolyzed? This is a less expensive, more
convenient procedure. Why?________________________
___________________
Are the products the same as those obtained in
the electrolysis of the molten salt? _____
71
Here we have 2 Cl- ? Cl2 2 e- (ox-anode)
2 H2O 2 e- ? H2 2 OH- (red-cathode)
72
5) The overall reaction is 2 H2O 2 Cl- ? Cl2
H2 2 OH-
Why isn't sodium produced at the cathode as in
the molten salt version?
Na 1 e- ? Na Eo -2.71 V
2 H2O 2e- ? H2 2 OH- Eo -0.83 V
Water is more easily reduced than Na or any
other Group I,II, or Al3 ion.
73
H) IN THE ABSENCE OF COMPLICATING FACTORS, the
reaction we predict is the one requiring the
lower applied voltage. (A complicating factor is
overvoltage, the voltage required for a reaction
may be considerably higher than the electrode
potential indicates.)
74
At the other electrode, there is also a
complication. For
2 Cl- ? Cl2(g) 2 e- Eo -1.36 V
And for
H2O ? O2(g) 4 H(aq)
4 e- Eo -1.23 V What about these values?
What do you think would be the relationship
between the concentration of NaCl and what is
produced at the anode?
75
I)What happens in a water solution of CuSO4? What
substance would you predict would be predict to
be produced at the anode? at the cathode?
Water is easier to oxidize than sulfate ion.
76
XI. QUANTITATIVE aspects of ELECTROLYSIS
A) These are the questions I will want you to
answer
How much material will be generated or deposited?
In how much time? with how many amps?
77
B) Faraday discovered that the quantity of a
substance undergoing chemical change at each
electrode is proportional to the quantity of
electricity which passes through the cell.
C) The quantity of electricity can be expressed
in the number of electrons, Faradays, or Coulombs.
78
D) Example Na 1 e- ? Na(s) What does this me
an?
E) Important Relationships
1) 1 coulomb 1 ampere flowing for 1 second,
which is 1 amp x 1 sec.
79
(No Transcript)
80
F) Sample Problems 1) How many grams of copper ca
n be deposited from a solution of Cu(NO3)2 by
5.00 Faradays of electricity?
First write a balanced equation for the reduction
reaction of Cu2.
Then determine the number of Faradays per mole of
Cu.
Then multiply by ________.
81
Then look up the molar mass of Cu.
Then multiply by ________ .
The answer is ____________ .
2) If 5.00 amps of current flow through a
solution of AgNO3 for 193 seconds, how many grams
of silver can be deposited at the cathode?
82
3) How much time in hours is required to produce
1000.0 kg of Mg by the passage of 1.50 X 105 amps
through molten MgCl2?
XII. Balancing oxidation-reduction equations
(Chapter 4 pp 156-158 and handout sheet)
83
(No Transcript)