Electrochemistry
• “Marriage” of redox and thermo
• Spontaneous electron-transfer reactions
can result in spontaneous electric current
if the reactants are separated by a wire
– Voltaic (Galvanic) cells [Experiment 32!]
– We can use the “spontaneity” of the reaction
to do electrical work
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(Continued)
• We can “push” electrons through a cell in
order to make a nonspontaneous redox
reaction occur.
– Electrolysis cell [Experiment 32!]
– Doing work to “force” chemical reaction to
occur [opposite of voltaic cell]
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Balancing Redox Equations
• Deferred until later
• For now just know that:
– A half reaction has electrons written as a
reactant or a product
• Oxidation half reaction: A reactant gets oxidized
(loses electrons); electrons appear as a “product”
• Reduction half reaction: A reactant gets reduced
(gains electrons); electrons appear as a “reactant”
– A balanced redox equation does not show
electrons explicitly. #e-’s lost = #e-’s gained
(called “n”)
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Voltaic Cells
• Recall lab…
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5
• The spontaneous rxn occurs in
the cell
• e-’s flow from – to + (“get to go
where they want to go”)
• Anode = where ox occurs
• Cathode = where red occurs
• Salt bridge prevents charge
buildup (which would stop flow)
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Used when neither
redox species in a
half reaction (or
electrode) is a neutral
metal.
You used graphite in
place of Pt in lab for
Fe2+/Fe3+ and I2/Icells. A cheaper
“inert” electrode.
Neither is a
neutral metal
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Standard Reduction Potentials
(E°red)
• Recall lab
– Make a bunch of different cells, get different
Ecell values (Eºcell if at standard state).
– Clearly some reductions are more favorable
than others
• How do you know? [Which direction did e-’s flow?]
• By how much?
• Rank them? (Must pick a zero as reference.)
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Quick quiz
• NOTE: Every electrode compartment has one
oxidizing agent and one reducing agent (this pair
is called the redox “couple”)
• If an electrode has Ni(s) and Ni2+ ions in it, which
species is the oxidizing agent and which the
reducing agent (of the pair)?
Ni2+ (b/c it’s “more positive”; it has
Ox agent is ___
“room for an electron”
Ni (b/c it’s “more negative”; it
Red agent is ___
has an electron to give)
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Revisit Earlier Cell—Look at this as a “Competition
for the electrons”. Which oxidizing agent “wants
them more”?
Who is the (possible)
oxidizing agent on
Zn2+
the left? _____
Hint: The two
“players” are Zn
and Zn2+
Who wins? (Which
one “got” the
Cu2+
electrons?) ____
Who is the (possible)
oxidizing agent on
Cu2+
the right? _____
Hint: The two
“players” are Cu
and Cu2+.
 Cu2+ “pulled harder”
So…which of the
half reactions
shown at the right
is more favorable
(greater tendency to
happen)?
Cu2+ + 2 e-  Cu(s)
Zn2+ + 2 e-  Zn(s)
By how much?.....
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Reducing Cu2+ is more favorable than reducing
Zn2+ …by 1.10 V! (Measure it w/voltmeter!)
We define a “standard
reduction potential”,
E°red, for every reduction
half reaction such that:
E°cell = E°red(cathode) - E°red(anode)
Where reduction
takes place
The more positive the “E”
(Ecell, Ered, or Eox), the more
favorable the process
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Reducing Cu2+ is more favorable than reducing
Zn2+ …by 1.10 V! (Measure it w/voltmeter!)
E°cell = E°red(cathode) - E°red(anode)
1.10 V = E°red(Cu2+/Cu) - E°red(Zn2+/Zn)
NOTE:
If E°red(Zn2+/Zn) were 0 V,
E°red(Cu2+/Cu) would be +1.10 V
If E°red(Zn2+/Zn) were -1.0 V,
E°red(Cu2+/Cu) would be +0.10 V
If E°red(Zn2+/Zn) were +1.0 V,
E°red(Cu2+/Cu) would be +2.10 V
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 The “zero” is arbitrary, but must be chosen / agreed upon!
This electrode (SHE)
was ultimately the one
chosen by the scientific
community to be the
“zero” of potential.
2 H+ + 2 e-  H2 (g); E°red = 0.0 V
Upshot:
One can determine any
E°red experimentally by just
setting up a cell where one
of the half cells is SHE!
(next slide → )
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Determining a Standard Reduction Potential using the SHE
0.76 V = E°red(SHE) - E°red(Zn2+/Zn)
 0.76 V = 0 - E°red(Zn2+/Zn)
 E°red(Zn2+/Zn) = - 0.76 V
The reduction of H+ is more favorable
than the reduction of Zn2+…. by 0.76 V!
Ecell = Ecathode - Eanode
Both as
reductions
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H+ (not Zn2+) gets reduced
Could use this info to
predict that this direct
reaction would occur:
Use these values to:
predict which reactions
are spontaneous at
standard state
and to
find any E°cell!
E°cell = E°red(cat) - E°red(an)
2 H+(aq) + 2 e-
H2(g)
0
E°cell = 0 – (-0.76) = +0.76 V
Zn2+(aq) + 2 e-
Zn(s)
-0.76
H+ gets reduced; Zn2+ does
not (Zn gets oxidized):
2 H+ + Zn → H2(g) + Zn2+
is spontaneous: E°cell > 0
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Refers only to
species on the
left side of the
arrow. E.g., F2,
is a better ox
agent than H2O2
which is better
ox agent than
Au3+ (but all of
these species
are very good
oxidizing agents
relative to most!)
Refers only to
species on the
right side of the
arrow. E.g., F-,
is a poorer red
agent than H2O
which is poorer
red agent than
Au(s) (but all of
these are very
poor reducing
agents relative
to most!)
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Excerpt from Voltaic Cell lab reading
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Table 18.1 (continued)
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Recall earlier slide
We define a “standard
reduction potential”,
E°red, for every reduction
half reaction such that:
E°cell = E°red(cathode) - E°red(anode)
OR (could also write Ecell as…
Ecell = Ered + Eox
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Lab interlude
• See overhead / board
• The lab manual initially asks you to
pretend that the Ag+/Ag reduction potential
is 0.0 V just to show you the “arbitraryness” of this.
• Then it tells you that in reality, Ag+/Ag
reduction potential is 0.80 V if the H+/H2
potential (SHE) is 0.0 V
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Cu/Cu2+ &
Fe3+/Fe2+
0.32 V
Cu/Cu2+ Cu  Cu2+ + 2e-
Zn/Zn2+ &
Ag/Ag+
1.50 V
Zn/Zn2+
Zn  Zn2+ + 2e-
1.50 V
Ag+ + e- Ag
0.0 V
Zn/Zn2+
Zn  Zn2+ + 2e-
1.50 V
Cu2+ + 2e- Cu
-0.45 V
Cu/Cu2+ &
Zn/Zn2+
1.05 V
+0.45 V
Fe3+ + e- Fe2+
-0.13 V
**Circle the species that is the better oxidizing agent**
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From Text
Table
From Table A.1:
Fe3+ + e- Fe2+ -0.13 V
Ag+ + e- Ag
0.0 V
-0.13 V
-0.45 V
0.0 V
-1.50 V
(See next slide)
Ag+ + e- Ag
0.80 V
0.80 V
Fe3+ + e- Fe2+
0.67 V
0.77 V
Cu2+ + 2e- Cu
0.35 V
0.34 V
Zn2+ + 2e-  Zn
-0.70 V
-0.76 V
Cu2+ + 2e- Cu -0.45 V
Zn  Zn2+ + 2e- 1.50 V
Zn2+ + 2e-  Zn -1.50 V
Flip
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Table 18.1 (continued)
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Determine the cell reaction, calculate Ecell ,
identify the cathode and anode, label the (+) and
(-) electrodes, and show electron flow (see board)
Ni2+(aq) + 2 e-
Ni(s)
-0.23 V
Mn2+(aq) + 2 e-
Mn(s)
-1.18 V
The better oxidizing
its Ered is
Ni2+ (Because
agent is:___
more positive)
Ni2+ actually gets
So ___
reduced, and thus
electrons flow to the
right __ side, which
_____
must be the ____ode.
cath
So the Ni electrode
must be ______ive
posit
V
V - -1.18
E°cell = -0.23
_____
_____
+0.95 V
= _______
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Determine the cell reaction, calculate Ecell ,
identify the cathode and anode, label the (+) and
(-) electrodes, and show electron flow (see board)
Fe2+(aq) + 2 e-
Fe(s)
-0.45 V
Mg2+(aq) + 2 e-
Mg(s)
-2.37 V
The better oxidizing
Fe2+
agent is:___
Fe2+actually gets
So ___
reduced, and thus
electrons flow to the
right __ side, which
_____
must be the ____ode.
cath
So the Fe electrode
must be ______ive
posit
V
V - -2.37
E°cell = -0.45
_____
_____
+1.92 V
= _______
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Determine the cell reaction, calculate Ecell ,
identify the cathode and anode, label the (+) and
(-) electrodes, and show electron flow (see board)
e
Pb(s)
Fe2+(aq) + 2 e-
Fe(s)
Pb2+(aq) + 2 e-
Pb(s) -0.13 V
Fe(s)
Salt
bridge
1 M Fe2+
1 M Pb2+
-0.45 V
The better oxidizing
Pb2+
agent is:___
Pb2+actually gets
So ___
reduced, and thus
electrons flow to the
left __ side, which
_____
must be the ____ode.
cath
So the Pb electrode
must be ______ive
posit
V
V - -0.45
E°cell = -0.13
_____
_____
+0.32 V
= _______
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Nernst Equation
• See last week’s pink lab handout (Voltaic
Cells), board, and below
– Start with DG = DG + RTlnQ and substitute in
DG = -nFEcell and DG = -nFEcell
After some algebra (and substituting in values
for R, assuming T = 298 K, and converting to a
base 10 log):
Ecell = Ecell
o
o
0.0592 V

log Q
n
(T = 25C)
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Standard vs. Nonstandard Cell
Zn + Cu2+  Zn2+ + Cu; Q = ??
Recall lab—adding
NH3 to Cu2+ side!
Ecell = Ecell
o
o
0.0592 V

log Q
n
(T = 25C)
**Always write out the balanced redox equation before using the Nernst
Equation or predicting whether Ecell should increase or decrease.**
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Explain in detail an in a conceptual way why the cell potential goes up when the
NH3 is added. Is the driving force for the cell rxn greater or smaller after the NH3
is added?
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Relationship between variables
(at any conditions; Mines Fig)
DG = DG + RT ln Q
Q
Ecell = Ecell
o
o
0.0592 V

log Q
n
(T = 25C)
32
Relationship between variables
(at standard state conditions; From Tro)
(T = 25C)
33
EXAMPLE 18.8 Calculating Ecell Under Nonstandard Conditions
Determine the cell potential for an electrochemical cell based on the
following two half-reactions:
Oxidation:
Reduction:
E°cell = E°red(cat) - E°red(an)
OR
E°cell = E°red(cat) + E°ox(an)
**Need to write balanced equation before E o = E o  0.0592 V log Q
cell
cell
using Nernst! What will “n” be here?**
n
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Fig. 18.22
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47
What mass of gold is plated in 25 minutes if
the current is 5.5 A?
Au3+ (aq) + 3 e-  Au(s)
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