Chapter 20 Electrochemistry

1
Chapter 20 Electrochemistry

• Dr. Peter Warburton
• peterw_at_mun.ca
• http//www.chem.mun.ca/zcourses/1051.php

2
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.

3
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.
4
This is an oxidation-reduction (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.
5
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 non-spontaneous!
6
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!

7
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 e-
• We 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 half-reaction, where a chemical gains
  electrons (is reduced)
• Cu2 (aq) 2 e- ? Cu (s)

8
Half-reactions

• The sum of these half-reactions must give us the
  overall reaction of interest.
• Zn (s) ? Zn2 (aq) 2 e-
• Cu2 (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)

9
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.

10
Half-reactions
The electrons stay on the metal electrode and are
NEVER found in solution!
11
Since the chemicals are in direct contact with
each other, the electron transfer occurs directly
and we cant 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?
12
Copper in Ag solution is spontaneous
13
The half-reactions 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.
14
The left half-cell has a solid copper electrode
in a Cu2 ion solution, while the right half-cell
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
half-cells requires us to use a salt bridge,
which is just another solution of ions.
15
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) ? Cu2 (aq) 2 e-
16
Electrons move from the ANODE to the silver
electrode through the wire
We can get them to do something useful, like
light a bulb!
17
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)
18
Positive ions leave the anode while the cathode
collects positive ions. Alternatively, negative
ions collect at the anode and move away from the
cathode
The ions are free to move through the salt bridge
and are REQUIRED to complete the circuit!
19
Overall, negative charges (electrons and negative
ions) are moving clockwise Overall, positive
charges (positive ions and electron holes) are
moving counterclockwise
20
The overall reaction is exactly the same as when
we place solid copper in a Ag solution, but
since we have separated the half-cells, we can
look at the separate half-reactions as they
occur.
Cu (s) ? Cu2 (aq) 2 e- 2 x Ag (aq) 1 e- ?
Ag (s) Cu (s) 2 Ag (aq) 2 e- ? Cu2 (aq)
2 e- 2 Ag (s) Cu (s) 2 Ag (aq) ? Cu2 (aq)
2 Ag (s)
21
Shorthand notation for galvanic cells

• Drawing a diagram for a galvanic cell or
  describing it as we did in the previous problem
  is too time-consuming 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)

22
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.

23
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.

24

• 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.

25
(through the wire connecting the electrodes)
26
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.
27
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.

28
Other shorthand notation considerations

• Consider this reaction
• Cu (s) Cl2 (g) ? Cu2 (aq) 2 Cl- (aq)
• Since we CANT 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)

29
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)

30
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)
31
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
32
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.
33
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.

34
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.

35
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!

36
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

37
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.

38
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 CV (one Coulomb-volt)

39
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!

40
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 Cmol-1

41
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!

42
Copper in Ag solution is spontaneous
43
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.

44
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 (_at_1 mol?L-1),
• all gas activities are 1 (_at_1 bar),
• and the temperature is specified
• (usually 25 C).

45
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.

46
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
47
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).
48
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
49
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!
Weve already seen that half-reactions cannot
occur without another half-reaction!
50
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.
51
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

52
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 e-
Reversing a reaction changes the sign of the
potential. For the S.H.E.
E?(red),anode - E?(red),cathode - E?H/H2
0 V
53
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!
54
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. Cu2 (aq) 2 e- ? Cu (s) E?(red)
0.340 V
55
If we reverse the half-reaction well get the
standard OXIDATION potential for the oxidation
of solid Cu to Cu2 ions Cu (s) ? Cu2 (aq) 2
e- E?(ox) -0.340 V
56
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!
57
We have found the standard potential of the
oxidation of solid zinc to zinc ions! This is a
standard oxidation potential. Zn (s) ? Zn2 (aq)
2 e- E?(ox) 0.763 V
58
If we reverse the half-reaction well get the
standard reduction potential for Zn2 ions to
solid zinc Zn2 (aq) 2 e- ? Zn (s) E?(red)
-0.763 V We report this value as the standard
electrode potential!
59
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!
60
Standard electrode potentials E?(red)
61
Using standard electrode potentials

• Using tabulated standard electrode potential data
  is accomplished much like a Hesss Law problem
  with one very important difference!
• Lets consider
• Zn (s) Zn2 (aq) Ag (aq) Ag (s)
• which has the balanced equation
• 2 Ag (aq) Zn (s) ? 2 Ag (s) Zn2 (aq)

62
Using standard electrode potentials

• Oxidation Zn (s) ? Zn2 (aq) 2 e-
• Reduction 2 Ag (aq) 1 e- ? Ag (s)
• 2 Ag (aq) Zn (s) ? 2 Ag (s) Zn2 (aq)
• E?cell 1.563 V

63

• Like Hesss Law, we look up the standard
  electrode potential reactions for both our sets
  of chemicals and then reverse the half-reaction
  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?

64
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!

65
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 electrode.
66
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)
67
Free energy and electrical work

• Weve 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.

68
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.

69
Free energy and electrical work

• Weve seen for a spontaneous process that DG lt 0.
  
• Weve also seen for a galvanic cell the overall
  reaction is spontaneous, and the cell potential
  is positive to indicate spontaneity.

70
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

71
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.

72
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.
• Weve already seen that the faraday (F) 9.65 x
  104 Cmol-1

73
Free energy and electrical work

• DG welec -kelec Ecell
• DG welec -nFEcell
• and at standard conditions
• DG? welec -nFE?cell

74
Do the units match those for work?

• DG welec -nFEcell
• -(mol)(C?mol-1)(V)
• CV J
• Yes! The units match those for work.

75
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) E?Al3/Al -1.676 V
• Br2 (l) 2 e- ? 2 Br- (aq) 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.
76
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 .
77
Spontaneous change in redox reactions

• Weve already related the free energy change to
  the cell potential
• DG? -nFE?cell
• and we also know a spontaneous process
• MUST HAVE DG? lt 0
• WHICH MEANS E?cell gt 0
• for ALL spontaneous electrochemical
  (oxidation-reduction) processes.

78
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

79
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!

80
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) E?Sn4/ Sn2
  0.154 V
• Sn2 (aq) 2 e- ? Sn (s) E?Sn2/ Sn
  -0.137 V
• O2 (g) 4 H (aq) 4 e- ? 2 H2O (l)
  E?O2/H2O 1.229 V

81
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.
82
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.

83
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 !

84
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 gt 0
• (0 V) - E?(red),anode gt 0
• E?(red),anode lt 0

85
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)

86
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)

87
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

88
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 gt 0
• (0.956 V) - E?(red),anode gt 0
• E?(red),anode lt 0.956 V

89
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!

90
E?cell and Keq

• We have two equations relating free energy to Keq
  and E?cell
• DG? -RT ln Keq
• and
• DG? -nFE?cell

91
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

92
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 DONT HAVE TO!)

93
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!

94
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 youve seen!
95
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.
96
Problem

• Should the reaction of solid Sn with Pb2 ions go
  to completion at 25 ?C?
• Pb2 (aq) 2 e- ? Pb (s) E?Pb2/Pb -0.125 V
• Sn2 (aq) 2 e- ? Sn (s) 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.
97
Ecell as a function of concentration

• Zn (s) Zn2 (aq, 1M) Cu2 (aq, 1M) Cu (s)
• Weve 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.

98
Ecell as a function of concentration

• Since the actual reaction is
• Zn (s) Cu2 (aq) ? Zn2 (aq) Cu (s)
• then Le Chataliers 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.

99
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

100
Ecell as a function of concentration

• By substituting we see
• -nFEcell -nFE?cell RT ln Qeq
• or
• Ecell E?cell - (RT/nF) ln Qeq

101
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

102
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) logaZn2/aCu2

103
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

104
Ecell as a function of concentration
Zn (s) Cu2 (aq) ? Zn2 (aq) Cu (s)
105
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 Cmol-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)

106
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!

107
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)

108
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

109
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.

110
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) E?Pb2/Pb -0.125 V
• Sn2 (aq) 2 e- ? Sn (s) E?Sn2/Sn -0.137 V

111
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)

112
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!

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
• non-zero Ecell

114
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
115
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 e-
• 2 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
116
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 (RT/nF) log x2/12 Ecell 2.3026
(2RT/nF) (-log x)
117
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!
118
Concentration cells for finding Ksp

• Ag (s) Ag (satd AgI) Ag (0.100 M) Ag
  (s)
• Ag (s) ? 2 Ag (satd AgI) e-
• Ag (0.100 M) e- ? Ag (s)

Net reaction Ag (0.100 M) ? Ag (satd AgI)
E?cell will ALWAYS be zero for a concentration
cell
119
Concentration cells for finding Ksp
Net reaction Ag (0.100 M) ? Ag (satd 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
120
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 10-9 M

121
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 10-17
122
Problem

• If Ksp 1.8 x 10-10 for silver chloride then
  what would be Ecell for
• Ag (s) Ag (satd AgCl) Ag (0.100 M) Ag
  (s)

Answer Ecell 0.23 V
123
Problem

• Calculate the Ksp for lead iodide with the given
  concentration cell information
• Pb (s) Pb2 (satd PbI2) Pb2 (0.100 M) Pb
  (s)
• Ecell 0.0567 V

Answer Ksp 7.1 x 10-9
124
Electrolysis and electrolytic cells

• The reverse of every spontaneous chemical
  reaction is non-spontaneous.
• 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.

125
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 gt -Ecell

126
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 non-spontaneous!
127
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!
128
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 gt 0 so DG lt 0
• electrolysis reactants ? electrolysis products
• Ecell lt 0 so DG gt 0

129
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 gt 0
• so DGsum lt 0

130
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!

131
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.

132
Overpotentials

• Ebattery gt -(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

133
Competing reactions

• If we set up an electrolytic cell expecting
• Esum Ebattery Ecell gt 0
• will give us the reaction we want we may be
  surprised when we get a completely different
  reaction because
• Ebattery Eother gt Ebattery Ecell

134
Competing reactions

• Ebattery Eother gt Ebattery Ecell
• We saw in slides 78-79 that if we have competing
  reactions, then the one that is more
  spontaneous will preferentially occur.

135
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 e-
• at the anode.

136
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

137
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 non-standard
  conditions, which means
• Ecell ? E?cell

138
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!

139
Quantitative aspects of electrolysis
140
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 e-
• we see that one mole of electrons is enough to
  give us one-half a mole (35.5 g) of Cl2.

141
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)

142
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

143

• 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.

144
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
145
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