Chapter 19
Electrochemistry
Electrochemistry
Electrochemistry is the branch of chemistry that
examines the transformations between chemical and
electrical energy.
Electricity is the movement of electrons from one
point to another.
The chemical change responsible for producing
electrons is called oxidation.
Since matter cannot be created nor destroyed then
the electrons must go somewhere, thus causing a
reduction.
Electrochemistry
Electrochemistry is the application of spontaneous
and nonspontaneous reactions in our wonderful world.
Spontaneous reactions most often are found in
batteries, called galvanic cells while nonspontaneous
reactions are used to plate metals on top of metals
called electrolytic cells.
Oxidation-Reduction Reactions
Oxidation
Zn(s) ---> Zn2+(aq) + 2e-
Reduction
Cu2+(aq) + 2e- --> Cu(s)
Oxidizing Agent, the
substance reduced
Reducing Agent, the
substance oxidized
Voltaic Cell
(Galvanic Cell)
A spontaneous REDOX
reaction where the
electrons are forced to get
to the substance reduced
by flowing through an
external circuit.
Cells
An electrochemical cell is an apparatus that
converts chemical energy into electrical work
or electrical work into chemical energy.
In a voltaic cell, chemical energy is
transformed into electrical energy by a
spontaneous redox reaction.
This is a battery.
A cell diagram uses symbols to show how the
components of an electrochemical cell are
connected.
Parts of Cells
• An anode is an electrode at which an
oxidation half-reaction (loss of electrons)
takes place.
• A cathode is an electrode at which a
reduction half-reaction (gain of electrons)
takes place.
• A salt bridge connects the two solutions of
the cell to prevent a short-circuit.
Standard Potentials
• Standard Potentials (Eo) is the electromotive
force of a half-reaction in which all reactants
and products are their standard states.
• Standard Cell Potentials (Eocell) is the
electromotive force produced by an
electrochemical cell when all reactants and
products are in their standard states.
• Standard States - concentrations are 1M and
the partial pressure of gases are 1 bar.
• Eocell = Eocathode + Eoanode
Determining Standard Potentials
Standard reduction potentials are measured
relative to a hydrogen electrode with defined voltage
of 0.00 volts
The hydrogen electrode
Hydrogen gas is bubbled in at one atmosphere over a
platinum electrode
Since ions cannot serve as an electrode then platinum is
used (inert)
Connected via a salt bridge and wire with a volt
meter is another electrode
Determining Standard Potentials
Typically any metal, with the it’s metal ion in solution
to the extent of 1.0M
The volt meter measures the potential difference
between the two cells and is recorded as the
standard voltage
The voltage of the hydrogen cell is defined to be 0.0
at standard conditions
Standard conditions
Room temperature 25⁰C
1.0 M ion concentration
1.0 atm pressure
Standard Hydrogen Electrode
SHE is defined
as 0.000V
Determination of Eo with SHE
0.76 V = ESHE - EZn
0.76 V = 0.00 V - EZn
0.34 V = ECu - ESHE
0.34 V = Ecu - 0.00 V
Writing a Cell Diagram
1. Write the chemical symbol of the anode at the far
left and the symbol of the cathode at the far right.
2. Work from the electrodes toward the bridge using
vertical lines to indicate phase changes and the
symbols of ions or compounds that are changed
by the cell reaction.
3. Use a double vertical line to represent the bridge
connecting the anode and cathode half-reactions.
Cell Diagram Example
1. Zn(s) . . . . . . Cu(s)
2. Zn(s) | Zn2+(aq). . .Cu2+(aq) | Cu(s)
3. Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
Practice
Write the voltaic-cell reaction and sketch a cell
in which magnesium metal is oxidized to
magnesium ions and copper ions are reduced
to copper metal. Identify the cathode, anode,
and direction of the electron flow on the
sketch.
Practice
Ag+ + e-
Ag
Cu - 2e-
Cu2+ - 0.34v
0.80v
Practice
2(Ag+ + eCu - 2e2 Ag+ + Cu
Ag)
0.80v
Cu2+ - 0.34v
Ag + Cu2+
Practice
2(Ag+ + eCu - 2e2 Ag+ + Cu
Ag)
0.80v
Cu2+ - 0.34v
Ag + Cu2+
V = 0.46 volts
Practice
2(Ag+ + eCu - 2e2 Ag+ + Cu
Ag)
0.80v
Cu2+ - 0.34v
Ag + Cu2+
V = 0.46 volts
Where is the anode?
Practice
anode
2(Ag+ + eCu - 2e2 Ag+ + Cu
Ag)
0.80v
Cu2+ - 0.34v
Ag + Cu2+
V = 0.46 volts
Where is the anode?
Where is the cathode?
Practice
cathode
anode
2(Ag+ + eCu - 2e2 Ag+ + Cu
Ag)
0.80v
Cu2+ - 0.34v
Ag + Cu2+
V = 0.46 volts
Where is the anode?
Where is the cathode?
What is the purpose of
the salt bridge?
cathode
anode
Practice
2(Ag+ + eCu - 2e2 Ag+ + Cu
Ag)
0.80v
Cu2+ - 0.34v
Ag + Cu2+
V = 0.46 volts
Where is the anode?
Where is the cathode?
What is the purpose of
the salt bridge? To keep
the cells neutral
Cell Thermodynamics
Cell potential, Electrical Work, and Free Energy
In order to push electrons through a wire a force must
be present to push the electrons. It is called the
electromotive force EMF and is measured in volts.
A volt is j/C, one joule of force is required to push one
mole of electrons between two points of a circuit
having different potential energies.
When the system does work on the surroundings it is
considered to be negative, while work done on the
system is positive.
Cell Thermodynamics
Note that cell potential and work have opposite
signs
V=-w/q, where q is the charge in coulombs
Work here is negative, since we are referring to the system.
Since ΔE = q + w and if there is no loss of heat then V is
maximum work possible.
Since entropy of the universe is expanding, then maximum
work cannot occur, since heat is released to expand the
universe.
Maximum potential can be measured, without current flowing,
with a potentiometer or an efficient digital voltmeter. No
current flow implies no wasted energy
Cell Thermodynamics
The maximum potential difference is useful to
compare process in order to determine their
efficiency.
Faraday is the charge of one mole of electrons which
is 96485 C/mole e
Sample Problem
Calculate the work to push 5.97 moles of electrons
with a voltage of 2.10 volts.
2.10j
-W =
C
96485C 5.97 mole e kj
= 1210 Kj
mole e
103j
kj
96485C
5.97
mole
e
2.50j
-Wmax =
C mole e
103 j
= 1440 Kj
Cell Thermodynamics
The maximum potential difference is useful to
compare process in order to determine their
efficiency.
Faraday is the charge of one mole of electrons which
is 96485 C/mole e
Sample Problem % efficiency = 1210 X100 = 84.0%
1440
Calculate the work to push 5.97 moles of electrons
with a voltage of 2.10 volts.
2.10j
-W =
C
96485C 5.97 mole e kj
= 1210 Kj
mole e
103j
kj
96485C
5.97
mole
e
2.50j
-Wmax =
C mole e
103 j
= 1440 Kj
Voltage and Free Energy
G = -nFEcell
Cell Potential (Ecell) or Electromotive Force is
the voltage between the electrodes of a voltaic
cell (#V = #J/C; C - coulomb).
Faraday constant (F) is 9.65 x 104 C/(mol e-)
n - number of moles of electrons
Voltage and Electrical Work
Gcell = wmax = -qEmax = -nFEmax
welec = Gcell = -nFEcell
∆G=∆G°+RTlnQ
Voltage and Electrical Work
Gcell = wmax = -qEmax = -nFEmax
welec = Gcell = -nFEcell
∆G=∆G°+RTlnQ
-nFE = -nFE° + RTlnQ
Voltage and Electrical Work
Gcell = wmax = -qEmax = -nFEmax
welec = Gcell = -nFEcell
∆G=∆G°+RTlnQ
E = E° +RT/nF [lnQ]
-nFE = -nFE° + RTlnQ
E = E° - 0.0591 logQ
n
Cell Potential and Concentration
Consider the reaction:
Al(s) + 3 Mn3+ → 2 Al3+ + 3 Mn(s) E° = 0.48 v
What would happen if the concentration of Mn3+ were
increased greater than 1.0 M?
 The reaction would shift right
 Shifting right makes a smaller ∆G
 Should produce more volts
What would happen if the concentration of Mn3+ were
decreased less than 1.0 M?
Concentration Cells
What would happen if two silver half cells had
different concentrations of silver ions?
Electrons would spontaneous flow in a direction
so as to make the concentrations equal to each
other V=0 or ∆G= 0
Electrons should flow from the less
concentrated to the more concentrated half cell
Examples of concentration cells
•
•
•
•
In animals
Potato batteries
Different areas of the
ocean
Voltage and Electrical Work
Gcell = wmax = -qEmax = -nFEmax
welec = Gcell = -nFEcell
∆G=∆G°+RTlnQ
-nFE = -nFE° + RTlnQ
E = E° +RT/nF [lnQ]
E = E° - 0. 591 logQ
n
Nernst Equation
Remember K=Q at equilibrium
Note: RTlnQ = 0.0591logQ
Concentration Cells
Calculate the initial voltage of the following cell:
Zn(s) | 0.1M Zn2+(aq) || 0.2 Zn2+(aq) | Zn(s)
0.0591
2+
log
Zn
2
1
0.0591
log
Ecathode = E° 2
Zn2+
Eanode = E° -
Concentration Cells
Calculate the initial voltage of the following cell:
Zn(s) | 0.1M Zn2+(aq) || 0.2 Zn2+(aq) | Zn(s)
0.0591
2+
log
Zn
2
1
0.0591
log
Ecathode = E° 2
Zn2+
Eanode = E° -
Subtract bottom equation from top equation
2+
Zn
0.0591
log Zn2+
Eanode – Ecathode = - 2
Concentration Cells
Calculate the initial voltage of the following cell:
Zn(s) | 0.1M Zn2+(aq) || 0.2 Zn2+(aq) | Zn(s)
0.0591
2+
log
Zn
2
1
0.0591
log
Ecathode = E° 2
Zn2+
Eanode = E° -
Subtract bottom equation from top equation
0.1
0.0591
log
Eanode – Ecathode =
0.2
2
Concentration Cells
Calculate the initial voltage of the following cell:
Zn(s) | 0.1M Zn2+(aq) || 0.2 Zn2+(aq) | Zn(s)
0.0591
2+
log
Zn
2
1
0.0591
log
Ecathode = E° 2
Zn2+
Eanode = E° -
Subtract bottom equation from top equation
0.1
0.0591
log
= 0.9 V
Eanode – Ecathode =
0.2
2
Ion-Selective Electrodes
Because cell potential is sensitive to the concentrations
of the reactants and products involved in the cell
reaction, measured potentials can be used to
determine concentration of an ion.
pH meters are an example of this.
A glass electrode contains a reference cell, plus a thin
glass membrane at the end that measure ion
concentrations.
The membrane can be made sensitive to ions other
than H, just by varying the type of glass.
pH Meter
Pt | H2(g, 1atm)| H+(xM) || H+ (1M) | H2 (g 1atm) | Pt
pH Meter
Eanode – Ecathode = - 0.0591
2
Eanode – Ecathode = - 0.0591
2
[H+]2
log 2
1
2 log
[H+]
12
[H+]
Eanode – Ecathode = - 0.0591 log 12
+
Eanode – Ecathode = - 0.0591 log [H ]
Ecell = 0.0591 pH
Ecell
pH =
0.0591
Calculating Equilibrium constants
•
•
•
•
•
∆G=∆G°+RTlnQ
-nFE°=-RTlnEeq
∆G°=-RTlnKeq
RTlnKeq
E°=
nF
Use the measured voltage for E
Use the E° voltage
Calculate K
The Dry Cell
Anode:
Zn (s) → Zn2+ (aq) + 2 eCathaode:
2MnO2(s) + 2 H+ → Mn2O3 + HOH
2MnO2(s) + 2 NH4+ + → Mn2O3 +2 NH3+ HOH
Cell Reaction:
Zn (s) + 2MnO2(s) + 2 NH4+ → Zn + Mn2O3 + 2NH3 + HOH
Lead Storage Battery
Pb2+
PbO2
Pb
Pb - 2ePbO2 + 4H+ + 2 ePb + PbO2 + 4H+
Pb2+
Pb2+ + 2H2O
2Pb2+ + 2H2O
(Anode Rx)
(Cathode Rx)
(Cell Rx)
Lead Storage Battery
Charging
Electron flow
A
Pb2+
Pb
PbO2
charging
Pb - 2ePb2+
PbO2 + 4H+ + 2 ePb2+ + 2H2O
Pb + PbO2 + 4H+
2 Pb2+ + 2H2O
(Anode Rx)
(Cathode Rx)
(Cell Rx)
Fuel Cell
Cathode:
Anode:
H2 → 2H+ + 2
e-
½ O2 + 2H+ +2e- → HOH
A cell that converts
chemical energy directly
into electrical energy
Similar to a battery,
except the reactants are
continually supplied from
an external source.
PEM
Fuel Cell
The anode is separated from the cathode by a proton
exchange membrane (PEM)
Only protons can pass through, while the electrons are
forced through an external circuit.
On the other side of the PEM protons, electrons, and
oxygen combine to form water
Only about 0.7 volts per cell, thus many cells are linked
in series.
Much current research into PEM membranes.
Corrosion: Unwanted Voltaic Cells
Corrosion can be viewed as returning metals to their
natural stable state
Corrosion involves oxidation of a metal
Since one fifth of the steel produced annually
replaces rusted metal
Metals oxidize easily, since their outer shell
electrons are loosely held
Oxidation of metals by oxygen is a
spontaneous process giving a positive voltage,
or a negative free energy; gold and platinum
are the exceptions.
Corrosion: Unwanted Voltaic Cells
Most metals form a thin oxide coating protecting the
inner metal atoms from oxidation, such as aluminum,
magnesium, nickel, and chromium
Copper will produce an exterior green carbonate (called
patina), while silver produces Ag2S, where does the
sulfur come from?
Corrosion: Unwanted Voltaic Cells
Consider a water droplet on the surface of steel.
Some of the iron dissolves into the water to produce
Fe2+ ions
The lost electrons flow along the steel where they
come into contact with moist oxygen to form
hydroxide ions.
Corrosion: Unwanted Voltaic Cells
Finally the Fe2+ reaches some oxygen, being further
oxidized and converted into Fe2O3
Moisture must be present to help transport ions much
like a salt bridge.
Corrosion: Unwanted Voltaic Cells
The iron oxide then combines with water to make
various hydrates, thus different colors of rust
Corrosion Prevention
Prevention of corrosion is an important way of
conserving our natural resources of energy and metals.
•
Primary means is paint
•
Chromium and tin are often used to plate steel
because they oxidize fo form a durable, effective
oxide coating
•
Zinc, also used to coat steel a process called
galvanizing (mixed oxide carbonate coating)
Since zinc is a more active metal, losses it
electrons more easily, then any oxidation
dissolves zinc rather than iron.
•
Corrosion Prevention
•
Alloying is also used to prevent corrosion
 Stainless steel contains chromium and nickel,
both of which form oxide coatings that change
steel’s reduction potential to one characteristic
of the noble metals
 Ion bombardment is being used to produce a
thin layer of stainless steel on the surface,
thus cheaper
Cathodic Protection
When another metal more easily oxidized is
electrically connected to the metal that is being
protected
Cathodic Protection
For example the Alaska pipeline is protected by wire
wrapped around the pipe and connected to magnesium
pounded into the ground
Cathodic Protection
Reduction (water)
happens at the
cathode and is
called cathodic
protection;
making steel act
as a cathode and
therefore does not
oxidize
Alaska Pipeline
O2 + 2H2O + 4e → 4OHEº=0.40
2(Mg - 2e → Mg 2+)
Eº= 2.38 v
O2 + 2 H2O + 2 Mg
Mg2+ + 4 OH- Ecell = 2.88 v
Electrolytic Cells
A nonspontaneous reaction
When REDOX reactions are forced to go in the
opposite direction
Electrolysis is used to plate metals, or to purify
metals, or produce chemicals
Salt bridge is not necessary, but a battery is required
Platinum electrodes are used
Voltage must add to a negative value
Electrolytic Cells
Electrollysis of salt water
Possible Anode Reactions
H2O - 4eO2 + 4H+ v = -1.23
2Cl -
- 2e-
Cl2
v = -1.36
Possible Cathode Reactions
HOH + 2eNa+ + e-
H2 + OH - v= -0.83
Na+
v=-2.71
Electrolytic Cells
Electrollysis of salt water
Possible Anode Reactions
H2O - 4eO2 + 4H+ v = -1.23
2Cl -
- 2e-
Cl2
v = -1.36
Possible Cathode Reactions
HOH + 2eNa+ + e-
H2 + OH - v= -0.83
Na+
v=-2.71
Which combinations do we choose?
Electrolytic Cells
Electrolysis of salt water
Possible Anode Reactions
H2O - 4eO2 + 4H+ v = -1.23
2Cl -
- 2e-
Cl2
v = -1.36
Possible Cathode Reactions
HOH + 2eNa+ + e-
H2 + OH - v= -0.83
Na+
v=-2.71
Which combinations do we choose? One of each having
the smallest voltage, except overvoltage (0.4-0.6v) must be
supplied to over come high energy of activation forming gas
on a metal surface. In light of this, chlorine is oxidized and
water is reduced to produce chlorine gas at the anode and
hydrogen gas at the cathode.
Electrolytic Cells
Electrolysis of salt water
2Cl -
- 2e-
HOH + 2e2Cl - + HOH
Cl2
v = -1.36
(anode Rx_
H2 + OH- v= -0.83 (cathode Rx)
H2 + Cl2 V=-2.19 (cell Rx)
Electrolytic Cells
Electrolysis of salt water
2Cl -
- 2e-
HOH + 2e2Cl - + HOH
Cl2
v = -1.36
(anode Rx_
H2 + OH- v= -0.83 (cathode Rx)
H2 + Cl2 + OH- V=-2.19 (cell Rx)
Note: Use inert electrodes (Pt) and a battery larger
than 2.19 volts.
What do you think might happen if
phenolphthalein were added to the electrolyte?
Yes, a pink color would develop around the
cathode and eventually spread through the
solution
Electrolytic Cells
Electrolysis of salt water
2Cl -
- 2e-
HOH + 2e2Cl - + HOH
Cl2
v = -1.36
(anode Rx_
H2 + OH- v= -0.83 (cathode Rx)
H2 + Cl2 + OH- V=-2.19 (cell Rx)
Note: Use inert electrodes (Pt) and a battery larger
than 2.19 volts.
What do you think might happen if
phenolphthalein were added to the electrolyte?
Electroplating
How long must a current of 5.00 A
be applied to a solution of Ag+ to
produce 12.0 g of silver on the
spoon in the figure?
Amp=C/s
Ag+ + e-
Ag (s)
Electroplating
How long must a current of 5.00 A
be applied to a solution of Ag+ to
produce 12.0 g of silver on the
spoon in the figure?
Amp=C/s
Ag+ + e-
Ag (s)
Electroplating
How long must a current of 5.00 A
be applied to a solution of Ag+ to
produce 12.0 g of silver on the
spoon in the figure?
Amp=C/s
Ag+ + e-
amp-s
C
Ag (s)
Electroplating
How long must a current of 5.00 A
be applied to a solution of Ag+ to
produce 12.0 g of silver on the
spoon in the figure?
Amp=C/s
Ag+ + e-
Ag (s)
amp-s
C
5.00 amp
Electroplating
How long must a current of 5.00 A
be applied to a solution of Ag+ to
produce 12.0 g of silver on the
spoon in the figure?
Amp=C/s
Ag+ + e-
amp-s
C
Ag (s)
96500C
5.00 amp mole e-
Electroplating
How long must a current of 5.00 A
be applied to a solution of Ag+ to
produce 12.0 g of silver on the
spoon in the figure?
Amp=C/s
Ag+ + e-
amp-s
C
Ag (s)
96500C mole e5.00 amp mole e- mole Ag
Electroplating
How long must a current of 5.00 A
be applied to a solution of Ag+ to
produce 12.0 g of silver on the
spoon in the figure?
Amp=C/s
Ag+ + e-
amp-s
C
Ag (s)
96500C mole e5.00 amp mole e- mole Ag
Electroplating
How long must a current of 5.00 A
be applied to a solution of Ag+ to
produce 12.0 g of silver on the
spoon in the figure?
Amp=C/s
Ag+ + e-
amp-s
C
Ag (s)
96500C mole e5.00 amp mole e- mole Ag
mole Ag
107.9 g Ag
Electroplating
How long must a current of 5.00 A
be applied to a solution of Ag+ to
produce 12.0 g of silver on the
spoon in the figure?
Amp=C/s
Ag+ + e-
amp-s
C
Ag (s)
96500C mole e5.00 amp mole e- mole Ag
mole Ag
12.0 g Ag
107.9 g Ag
Electroplating
How long must a current of 5.00 A
be applied to a solution of Ag+ to
produce 12.0 g of silver on the
spoon in the figure?
Amp=C/s
Ag+ + e-
amp-s
C
Ag (s)
96500C mole e5.00 amp mole e- mole Ag
mole Ag
12.0 g Ag hr
107.9 g Ag
3600 s
Electroplating
How long must a current of 5.00 A
be applied to a solution of Ag+ to
produce 12.0 g of silver on the
spoon in the figure?
Amp=C/s
Ag+ + e-
amp-s
C
Ag (s)
96500C mole e5.00 amp mole e- mole Ag
mole Ag
12.0 g Ag hr
107.9 g Ag
3600 s
= 0.596 hrs
Commercial Applications
Charles Martin Hall (1863-1914)
was a student at Oberlin College
in Ohio when he first became
interested in aluminum. One of
his professors commented that
anyone who could manufacture
aluminum cheaply would make a
fortune.
Aluminum Production
Aluminum Production
Aluminum is the third most abundant element on the
earth
Since aluminum is a very active metal, it is found in
nature as its oxide in and ore called bauxite
Production of aluminum metal from its ore proved to be
more difficult than of most other metals
Reduction of ions requires dissolving the ions so that
they can migrate to electrodes
Water is not suitable, since it is more easily reduced
compared to aluminum
Aluminum Production
Cryolite (Na3AlF6) was the solvent that finally worked to
dissolve the aluminum oxide
Aluminum oxide melts at 2000°C, but a mixture of
cryolite and aluminum oxide melts at 1000°C
The anode, made of graphite, oxidizes the oxide to
oxygen gas
The cathode the inner shell of the cell also made of
graphite reduces the aluminum to molten metal, which is
drained off from the bottom
Aluminum Production
Thus aluminum can be reduced electrolytically
A mixture of aluminum oxide and crolite melts at
about 1000°C
Bauxite is reacted with OH- to remove other
metal oxides since they will not dissolve
Production of Copper
An interesting application is the refining, or purification of
copper metal
Impure copper is used as the anode in an electrolytic
cell containing aqueous copper sulfate as the electrolyte.
The cathode of the cell is made of pure copper.
When electrolysis is carried out the voltage across the
cell is adjusted so that only copper and other more
active metals, such as iron or zinc are able to dissolve at
the anode.
Production of Copper
Since copper is the ion most easily reduced, then it is
the only ion plated on the pure copper electrode
The other ions stay in solution
The sludge that does not dissolve, called anode sludge,
contains gold, platinum and silver
Down’s cell
Down’s cell
Made of brick, since high temperature to melt NaCl
The anode pokes up from the floor, with a hood
over it to capture the chlorine gas
The cathode, doughnut shaped, around the anode,
with a pipe in the top to drain off the sodium metal,
which is less dense than the molten sodium
chloride
ChemTour: Zinc-Copper Cell
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This ChemTour illustrates the reactions that occur at the
electrodes of a typical zinc–copper battery and explores
how the energy released by a voltaic cell is used to do work
on the surroundings.
ChemTour: Free Energy
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Students learn how the potential of an electrochemical cell
can be used to determine the free energy available to do
work, and explore the relationships between free energy,
cell potential, and the equilibrium constant.
ChemTour: Cell Potential
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This ChemTour explores the concept of cell potential (Ecell)
as a measure of how much electrical energy is stored in an
electrochemical cell.
ChemTour: Alkaline Battery
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This ChemTour explores the oxidation–reduction reactions
that power common alkaline batteries and describes the
changes in reaction quotient as a battery loses its charge.
ChemTour: Fuel Cell
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Students learn how fuel cells use a redox reaction between
hydrogen and oxygen to produce electrical energy.
Given that the standard reduction
potential of Ag+ (aq) is 0.80 V, will a
silver rod dissolve in an acidic solution
of pH 0.0 under standard conditions?
A) Yes
Immersion of Ag Rod in Acid Solution
B) No
C) Depends
Given that the standard reduction
potential of Ag+ (aq) is 0.80 V, will a
silver rod dissolve in an acidic solution
of pH 0.0 under standard conditions?
Ag+
+ e-
A) Yes
Immersion of Ag Rod in Acid Solution
Ag (s)
B) No
0.80 V
C) Depends
Given that the standard reduction
potential of Ag+ (aq) is 0.80 V, will a
silver rod dissolve in an acidic solution
of pH 0.0 under standard conditions?
Ag+
+ e-
Ag (s)
0.80 V
Oxidation or reduction?
A) Yes
Immersion of Ag Rod in Acid Solution
B) No
C) Depends
Given that the standard reduction
potential of Ag+ (aq) is 0.80 V, will a
silver rod dissolve in an acidic solution
of pH 0.0 under standard conditions?
Ag+
+ e-
Ag (s)
0.80 V
Oxidation or reduction? Reduction!
Dissolving?
A) Yes
Immersion of Ag Rod in Acid Solution
B) No
C) Depends
Given that the standard reduction
potential of Ag+ (aq) is 0.80 V, will a
silver rod dissolve in an acidic solution
of pH 0.0 under standard conditions?
Ag+
+ e-
Ag (s)
0.80 V
Oxidation or reduction? Reduction!
Dissolving? No, Dissolving is oxidation
A) Yes
Immersion of Ag Rod in Acid Solution
B) No
C) Depends
Given that the standard reduction
potential of Ag+ (aq) is 0.80 V, will a
silver rod dissolve in an acidic solution
of pH 0.0 under standard conditions?
Ag (s) + e-
Ag+ (aq)
- 0.80 V
Dissolving is oxidation and is non spontaneous.
A) Yes
Immersion of Ag Rod in Acid Solution
B) No
C) Depends
Given that the standard reduction
potential of Ag+ (aq) is 0.80 V, will a
silver rod dissolve in an acidic solution
of pH 0.0 under standard conditions?
Ag (s) + e-
Ag+ (aq)
- 0.80 V
Dissolving is oxidation and is non spontaneous.
A) Yes
Immersion of Ag Rod in Acid Solution
B) No
C) Depends
Under standard conditions, copper will
plate out onto a nickel rod immersed in a
solution containing Cu2+ ions, but
aluminum will not plate onto a nickel rod
immersed in a solution containing Al3+
ions. Which of the following is the
strongest oxidizing agent?
A) Ni(s)
B) Al3+(aq)
Oxidizing Strength of Al3+, Ni, and Cu2+
C) Cu2+(aq)
Under standard conditions, copper will
plate out onto a nickel rod immersed in a
solution containing Cu2+ ions, but
aluminum will not plate onto a nickel rod
immersed in a solution containing Al3+
ions. Which of the following is the
strongest oxidizing agent?
Cu2+ + Ni
Ni2+
+ Cu E°>0
Ni2+
+ Al E°<0
o.a.
Al3+ + Ni
o.a
A) Ni(s)
B) Al3+(aq)
Oxidizing Strength of Al3+, Ni, and Cu2+
C) Cu2+(aq)
Under standard conditions, copper will
plate out onto a nickel rod immersed in a
solution containing Cu2+ ions, but
aluminum will not plate onto a nickel rod
immersed in a solution containing Al3+
ions. Which of the following is the
strongest oxidizing agent?
Cu2+ + Ni
Ni2+
+ Cu E°>0
Ni2+
+ Al E°<0
o.a.
Al3+ + Ni
o.a
A) Ni(s)
B) Al3+(aq)
Oxidizing Strength of Al3+, Ni, and Cu2+
C) Cu2+(aq)
Under standard conditions, copper will
plate out onto a nickel rod immersed in a
solution containing Cu2+ ions, but
aluminum will not plate onto a nickel rod
immersed in a solution containing Al3+
ions. Which of the following is the
strongest oxidizing agent?
Cu2+ + Ni
Ni2+
+ Cu E°>0
Ni2+
+ Al E°<0
o.a.
Al3+ + Ni
o.a
A) Ni(s)
B) Al3+(aq)
Can Ni(s) ever be a reducing agent?
Oxidizing Strength of Al3+, Ni, and Cu2+
C) Cu2+(aq)
Under standard conditions, copper will
plate out onto a nickel rod immersed in a
solution containing Cu2+ ions, but
aluminum will not plate onto a nickel rod
immersed in a solution containing Al3+
ions. Which of the following is the
strongest oxidizing agent?
Cu2+ + Ni
Ni2+
+ Cu E°>0
Ni2+
+ Al E°<0
o.a.
Al3+ + Ni
o.a
A) Ni(s)
B) Al3+(aq)
C) Cu2+(aq)
Can Ni(s) ever be a reducing agent? Only if it
can reduce. Ever hear of Ni2-?
Oxidizing Strength of Al3+, Ni, and Cu2+
For the reaction
Cu2+(aq) + 2 Ag(s)
Cu(s) + 2 Ag+(aq),
G° = 88.3 kJ/mole. Under standard
conditions, will the Cu/Ag cell pictured to
the left produce a current?
A) Yes
Ag/Cu Electrochemical Cell
B) No
C) Depends
For the reaction
Cu2+(aq) + 2 Ag(s)
Cu(s) + 2 Ag+(aq),
G° = 88.3 kJ/mole. Under standard
conditions, will the Cu/Ag cell pictured to
the left produce a current?
Is the forward reaction spontaneous?
A) Yes
Ag/Cu Electrochemical Cell
B) No
C) Depends
For the reaction
Cu2+(aq) + 2 Ag(s)
Cu(s) + 2 Ag+(aq),
G° = 88.3 kJ/mole. Under standard
conditions, will the Cu/Ag cell pictured to
the left produce a current?
Is the forward reaction spontaneous?
No!
A) Yes
Ag/Cu Electrochemical Cell
B) No
C) Depends
For the reaction
Cu2+(aq) + 2 Ag(s)
Cu(s) + 2 Ag+(aq),
G° = 88.3 kJ/mole. Under standard
conditions, will the Cu/Ag cell pictured to
the left produce a current?
How about the reverse reaction?
A) Yes
Ag/Cu Electrochemical Cell
B) No
C) Depends
For the reaction
Cu2+(aq) + 2 Ag(s)
Cu(s) + 2 Ag+(aq),
G° = 88.3 kJ/mole. Under standard
conditions, will the Cu/Ag cell pictured to
the left produce a current?
How about the reverse reaction? Yes
A) Yes
Ag/Cu Electrochemical Cell
B) No
C) Depends
For the reaction
Cu2+(aq) + 2 Ag(s)
Cu(s) + 2 Ag+(aq),
G° = 88.3 kJ/mole. Under standard
conditions, will the Cu/Ag cell pictured to
the left produce a current?
How about the reverse reaction? Yes
Then copper will oxidize producing e’s
A) Yes
Ag/Cu Electrochemical Cell
B) No
C) Depends
For the reaction
Cu2+(aq) + 2 Ag(s)
Cu(s) + 2 Ag+(aq),
G° = 88.3 kJ/mole. Under standard
conditions, will the Cu/Ag cell pictured to
the left produce a current?
How about the reverse reaction? Yes
Then copper will oxidize producing e’s
A) Yes
Ag/Cu Electrochemical Cell
B) No
C) Depends
For the oxidation of Fe2+ by Ag+,
Fe2+(aq) + Ag+(aq)
Fe3+(aq) + Ag(s),
H° and S° are both negative. Which of the following
plots shows the correct relationship between the
electromotive force, E°, (y-axis) and the temperature (xaxis)?
A)
Oxidation of Fe + by Ag+
B)
C)
For the oxidation of Fe2+ by Ag+,
Fe2+(aq) + Ag+(aq)
Fe3+(aq) + Ag(s),
H° and S° are both negative. Which of the following
plots shows the correct relationship between the
electromotive force, E°, (y-axis) and the temperature (xaxis)? ΔG = ΔH - TΔS
A)
Oxidation of Fe + by Ag+
B)
C)
For the oxidation of Fe2+ by Ag+,
Fe2+(aq) + Ag+(aq)
Fe3+(aq) + Ag(s),
H° and S° are both negative. Which of the following
plots shows the correct relationship between the
electromotive force, E°, (y-axis) and the temperature (xaxis)? ΔG = ΔH - TΔS Spontaneous or nonspontaneous?
A)
Oxidation of Fe + by Ag+
B)
C)
For the oxidation of Fe2+ by Ag+,
Fe2+(aq) + Ag+(aq)
Fe3+(aq) + Ag(s),
H° and S° are both negative. Which of the following
plots shows the correct relationship between the
electromotive force, E°, (y-axis) and the temperature (xaxis)? ΔG = ΔH - TΔS Yes spontaneous at low T
A)
Oxidation of Fe + by Ag+
B)
C)
For the oxidation of Fe2+ by Ag+,
Fe2+(aq) + Ag+(aq)
Fe3+(aq) + Ag(s),
H° and S° are both negative. Which of the following
plots shows the correct relationship between the
electromotive force, E°, (y-axis) and the temperature (xaxis)? ΔG = ΔH - TΔS Yes spontaneous at low T
A)
Oxidation of Fe + by Ag+
B)
C)
For the cell pictured to the left, the voltage,
E, is measured to be 0.80 V. Given the
following standard reduction potentials,
Ag+ + e→ Ag
2 H+ + 2 e- → H2

E red= 0.80 V

E red = 0.00 V
what can be said of the pH in the right half
of the cell if the partial pressure of H2(g) is
1.0 atm?
A) pH > 0
pH of an Electrochemical Cell
B) pH = 0 C) pH < 0
For the cell pictured to the left, the voltage,
E, is measured to be 0.80 V. Given the
following standard reduction potentials,
Ag+ + e→ Ag
2 H+ + 2 e- → H2

E red= 0.80 V

E red = 0.00 V
what can be said of the pH in the right half
of the cell if the partial pressure of H2(g) is
1.0 atm?
Oxidation occurs in which half cell?
A) pH > 0
pH of an Electrochemical Cell
B) pH = 0 C) pH < 0
For the cell pictured to the left, the voltage,
E, is measured to be 0.80 V. Given the
following standard reduction potentials,
Ag+ + e→ Ag
2 H+ + 2 e- → H2

E red= 0.80 V

E red = 0.00 V
what can be said of the pH in the right half
of the cell if the partial pressure of H2(g) is
1.0 atm?
Oxidation occurs in which half cell? Yes, the hydrogen cell
A) pH > 0
pH of an Electrochemical Cell
B) pH = 0 C) pH < 0
For the cell pictured to the left, the voltage,
E, is measured to be 0.80 V. Given the
following standard reduction potentials,
Ag+ + e→ Ag
2 H+ + 2 e- → H2

E red= 0.80 V

E red = 0.00 V
what can be said of the pH in the right half
of the cell if the partial pressure of H2(g) is
1.0 atm?
In order to have 0.00 V in the hydrogen half cell, what must
the proton concentration be?
A) pH > 0
pH of an Electrochemical Cell
B) pH = 0 C) pH < 0
For the cell pictured to the left, the voltage,
E, is measured to be 0.80 V. Given the
following standard reduction potentials,
Ag+ + e→ Ag
2 H+ + 2 e- → H2

E red= 0.80 V

E red = 0.00 V
what can be said of the pH in the right half
of the cell if the partial pressure of H2(g) is
1.0 atm?
In order to have 0.00 V in the hydrogen half cell, what must
the proton concentration be? Yes, one molar
A) pH > 0
pH of an Electrochemical Cell
B) pH = 0 C) pH < 0
For the cell pictured to the left, the voltage,
E, is measured to be 0.80 V. Given the
following standard reduction potentials,
Ag+ + e→ Ag
2 H+ + 2 e- → H2

E red= 0.80 V

E red = 0.00 V
what can be said of the pH in the right half
of the cell if the partial pressure of H2(g) is
1.0 atm?
In order to have 0.00 V in the hydrogen half cell, what must
the proton concentration be? Yes, one molar. And the pH?
A) pH > 0
pH of an Electrochemical Cell
B) pH = 0 C) pH < 0
For the cell pictured to the left, the voltage,
E, is measured to be 0.80 V. Given the
following standard reduction potentials,
Ag+ + e→ Ag
2 H+ + 2 e- → H2

E red= 0.80 V

E red = 0.00 V
what can be said of the pH in the right half
of the cell if the partial pressure of H2(g) is
1.0 atm?
In order to have 0.00 V in the hydrogen half cell, what must
the proton concentration be? Yes, one molar. And the pH?
A) pH > 0
pH of an Electrochemical Cell
B) pH = 0 C) pH < 0
For the electrochemical cell
Cu | Cu2+(1.0 M) || Cu2+(0.1 M) | Cu,
what will happen to the color in the
darker solution (1.0 M Cu2+) once the
circuit is completed?
This is a concentration cell and electrons flow from one half
cell to the other until the solutions have equal concentrations.
A) It gets darker.
B) It gets lighter. C) It stays the same.
Cu/Cu + Concentration Cell
For the electrochemical cell
Cu | Cu2+(1.0 M) || Cu2+(0.1 M) | Cu,
what will happen to the color in the
darker solution (1.0 M Cu2+) once the
circuit is completed?
This is a concentration cell and electrons flow from one half
cell to the other until the solutions have equal concentrations.
A) It gets darker.
B) It gets lighter. C) It stays the same.
Cu/Cu + Concentration Cell
The End