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Electrochemistry
1
Oxidation-Reduction Rxns
•Oxidation-reduction rxns, called redox rxns,
are electron-transfer rxns.
•So the oxidation states of 1 or more
substances changes.
•Common redox rxns:
•iron rusting
•zinc reacting with acid
•combustion rxns
•batteries and electrolysis rxns
•photosynthesis
2
Oxidation States
• Remember how to determine oxidation numbers for atoms?
3
Oxidation States
• What are the oxidation states of all the elements in the
following rxn?
Zn(s) + HCl(aq) → ZnCl2(aq) + H2(g)
• What elements are changing oxidation states in the above?
• If an element loses electrons (oxidation state increases), then
it is being oxidized and is the reductant or reducing agent.
• What element is being oxidized in the above?
• If an element gains electrons (oxidation state decreases), then
it is being reduced and is the oxidant or the oxidizing agent.
• What element is being reduced in the above?
4
Galvanic or Voltaic Cells
•As electrons are being transferred, we
can use redox rxns to perform electrical
work.
•If we have a spontaneous redox rxn, we
can generate an electrical current.
•To do this, we have to separate the two
half-rxns instead of having them take
place in the same beaker.
5
• The two cells are connected two ways:
• An electrode in each cell are connected
together with a conducting wire.
• An electrode is simply a piece of metal
(or graphite), which may or may not be a
reactant or product.
• A salt bridge allows ions to flow through
from one cell to the other, so charge
balance is maintained (overall neutral
slns).
6
Galvanic Cell Connections
7
Galvanic Cell Connections
8
Galvanic Cell
9
Galvanic Cell
10
Shorthand Notation for Galvanic Cells
• We saw the galvanic cell for the following rxn (Zn
anode electrode, Pt cathode electrode):
Zn(s) + H+(aq) --> Zn2+(aq) + H2(g)
• There is a shorthand notation for this galvanic cell:
+
Zn(s) Zn (aq) H (aq) H 2 (g) Pt(s)
2+
11
Cell Potentials
• What drives the rxn in a galvanic cell?
• Or what forces the electrons to move from the
anode to the cathode?
• The driving force is an electrical potential
called the electromotive force or emf.
• The emf for a cell is also simply called the cell
potential, E, Ecell, or the cell voltage.
12
Cell Potentials
• The E cell potential is read with a voltmeter which is
connected to the cell circuit.
– The - terminal on the voltmeter must be connected
to the cell anode, while the + terminal on the
voltmeter must be connected to the cell cathode.
– In the lab, this is how the direction of a spontaneous
cell rxn is determined: when the voltmeter gives a +
reading, the connections are correct, and the anode
and cathode are identified. If the connections are
incorrect, a 0 or negative voltage reading is obtained
(as rxn occurs in reverse).
13
Units for Electrochemistry
• The unit for the cell voltage is volts, V
• Also:
• 1V = 1J/C (C is coulombs, electric charge)
• 1C = 1amp•s or 1C = 1A•s (amp or A is
current in amperes)
• 1watt = 1W = 1J/s (watt is the power)
• Since the voltage is related to energy as well as
electric charge, there MUST be a relationship
between a cell potential and energy!
14
G and E
• There is a mathematical relationship between
the 2:
G = -nFE
• where F = 96,500 C/mol e• and n = # mol e- transferred in cell
• Notice that if the cell rxn is spontaneous, the E
is + but the G is – as we would expect!
15
G and E
• If the cell is being run under standard state
conditions (1 atm for gases, 1M for solutions,
and at a specified temperature, usually given as
25°C), then:
G° = -nFE°
• Determine G° for the following cell rxn if
E° = 0.92V
Al(s) + Cr (aq) ® Al (aq) + Cr(s)
3+
3+
16
Standard Reduction Potentials for Half Cells,
E°red
• The E° for a cell rxn depends on both the E°
for the anode half cell rxn and the E° for the
cathode half cell rxn.
• If we know the E° for both half cells, we can
calculate the overall E° for the cell:
(Eq 1) E cell = E cathode + E anode OR E cell = E red + E ox
17
Standard Reduction Potentials for Half Cells,
E°red
• But we can’t measure an E° for a half cell as
there’s no rxn unless 2 half cells are coupled!
• So we have a reference half-cell rxn which has
been assigned an E° half-cell potential of 0V.
• This is the Standard Hydrogen Electrode or
SHE.
• The SHE half-cell may act as the anode or
cathode depending on what the other half-cell
is.
18
SHE Half-Cell
The rxn is and half-cell notations are:
2 H + (aq, 1 M) + 2 e - ® H 2 (g, 1 atm)
E red = 0 V
H + (aq, 1 M) H 2 (g, 1 atm) Pt(s) if it is the cathode and:
Pt(s) H 2 (g, 1 atm) H + (aq, 1 M)
if it is the anode
19
20
Standard Reduction Potentials for Half Cells,
E°red
• So half-cell standard potentials are measured
relative to the SHE half-cell.
• There are Tables of standard potentials for
half-cells, readily available (p 830 and
Appendix ii).
• Looking at these Tables you should notice that
all of the E° values are for the reduction halfcell rxn, E°red
21
22
Standard Reduction Potentials for Half Cells,
E°red
• Notice also that some of the values are -, while
others are +.
• This tells you the relative strengths of oxidizing
and reducing agents.
• The higher the E°red value, the stronger the
oxidizing agent in the half-cell.
23
Standard Reduction Potentials for Half Cells,
E°red
• For example, which is the stronger ox. agent,
Fe3+ or Fe2+?
Fe
3+
+ e ® Fe
Fe
2+
+ 2e ® Fe(s) E red = -0.45V
-
2+
E red = 0.77V
-
24
Standard Reduction Potentials for Half Cells,
E°red
• The E°red values also gives you the tendency of
2 coupled half-cell rxns to be spontaneous.
• They also tell you which half-cell will be the
anode (oxidation) and which will be the
cathode (reduction).
• The half-cell rxn with the higher E°red value
(more + value) WILL be the cathode.
• So of course, the more negative E°red value
will be the anode.
25
Standard Reduction Potentials for Half Cells,
E°red
• Let’s show this using Eq 1) and the fact that
E°red = - E°ox as they are just the reverse of
each other.
(Eq 1) E cell = E cathode + E anode OR E cell = E red + E ox
• Rearrange Eq 1) to reflect that Table values are
always given as E°red values.
26
Standard Reduction Potentials for Half Cells,
E°red
E cell = E red + E ox
but E ox = -E red (anode), and E red = E red (cathode) so this becomes:
(Eq 2) E cell = E red (cathode) - E red (anode)
We will use Eq 2) to find the overall E°cell
27
E°cell
• Before you start problems, there is a comment
to make about E°
• E° is an INTENSIVE properties, so that
means that E° DOES NOT depend on the
amount of a substance.
• This means that E° is the same whether you
have 1 mol or 100 mol!
• What does change is the TIME that the rxn
lasts; the more moles, the LONGER the rxn
lasts; but the standard voltage is constant.
28
E°cell
• Although this seems weird, look at the units of
a V: J/C
• So a V is a ratio of 2 extensive properties; but
both change by the SAME amount as the mol
changes.
• So there is NO net change for E°
• What does this mean for you?
• You ignore stoichiometry when calculating an
E°cell from 2 half-cell E°red values.
29
Non Standard State Conditions
• What if you aren’t at standard state?
• You already learned the relationship between
G and G°:
G = G° + RTlnQ
• But G = -nFE and G° = -nFE° so:
-nFE = -nFE° + RTlnQ
• Dividing both sides by -nF:
E = E° - (RT/nF)lnQ
• This is the Nernst Equation.
30
Nernst Equation
• The Nernst Equation is also written in terms of
log:
2.303RT
E = E° logQ
nF
• And if the temperature is 25°C, this simplifies
to:
0.0592V
E = E° logQ
n
• Obviously, you can only use the above at
25°C, otherwise use the complete Nernst Eq.
31
Nernst Equation
• How often are we at standard state?
• And if we start at standard state, do we stay at
standard state?
• What changes in the Nernst equation as a
reaction progresses?
• So how does E respond to this change?
• And when the rxn is complete (or at
equilibrium), what will E be?
2.303RT
E = E° logQ
nF
32
The Relationship between E° and K
2.303RT
E = E° logQ
nF
• At equilibrium, E = 0 and Q = K.
• So the Nernst Equation becomes:
33
The Relationship between E° and K
• Or at 25°C:
nE°
logK =
0.0592
• Solving for K gives:
K=e
nFE°
RT
• So, as you would expect, if E° > 0, then K > 1.
34