Electrochemistry

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Electrochemistry
A galvanic or voltaic cell produces electricity
by means of a spontaneous redox reaction
Anode The electrode at which oxidation occurs
Cathode The electrode at which reduction occurs
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The Daniell Cell
Zn is the anode
Cu is the cathode
Zn dissolves
.
Cu plates onto the copper electrode
Electrons flow from anode to cathode
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The Standard Hydrogen Electrode
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Zn (s)Zn2H(aq)H2 (g) Pt (s)
Experiment shows that Zn (s) dissolves.
Therefore, we know that zinc is oxidized. This
means that hydrogen ions are more readily
reduced than zinc ions
.
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H2(g)Pt(s)H(aq)Cu2(aq)Cu (s)
Experiment shows that Cu (s) plates out.
Therefore, we know that copper is reduced. This
means that Copper ions are more readily reduced
than hydrogen ions
.
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Determining Standard Reduction Potential
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• Half-Reaction Eored (Volts)
• K e- -----gt K -2.924
• Ca2 2e- -----gt Ca -2.87
• Na e- -----gt Na -2.71
• Mg2 2e- -----gt Mg -2.375
• Al3 3e- -----gt Al -1.706
• Zn2 2e- -----gt Zn -0.7628
• Fe2 2e- -----gt Fe -0.409
• Ni2 2e- -----gt Ni -0.23
• Fe3 3e- -----gt Fe -0.036
• 2 H 2e- -----gt H2 0.0000...
• Cu2 2e- -----gt Cu 0.3402
• Ag e- -----gt Ag 0.7996
• Br2(aq) 2e- -----gt 2 Br- 1.087
• O2 4H 4e- -----gt 2 H2O 1.229
• Cl2(g) 2e- -----gt 2 Cl - 1.3583
• Au e- -----gt Au 1.68
• O3 2H 2e- ---gt O2 H2O 2.07
• F2(g) 2e- ---gt 2 F-(aq) 2.87

Strongest reducing agent
Least readily reduced
Strongest Oxidzing agent
Most readily reduced
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Determining the Eocell
The standard emf of a cell, Eocell is the sum of
the standard oxidation potential and the standard
reduction potential
Eocell Eoox Eored
For the Daniell cell
Zn (s) ---gt Zn2 (aq) 2e- Eo 0.76 V
Eoox
Eored
Cu2 (aq) 2e- ---gt Cu (s) Eo 0.34 V
Eo 1.10 V
Zn(s) Cu2(aq) ---gt Zn2 (aq) Cu(s)
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Problem 20.9 on Page 801
Calculate the standard emf of a cell that uses
Ag/Ag and Al/Al3 half-cell reactions. Write the
overall cell reaction that occurs under
standard-state conditions.
From the table we find that Eored for Al3 -
1.66 V and Eored for Ag 0.80 V. Therefore
Al3 is oxidized.
Al (s) ---gt Al3 (aq) 3e- Eo 1.66 V
Eoox
Eored
3Ag (aq) 3e - ---gt 3Ag (s) Eo 0.80 V
Eo 2.46 V
Al(s) 3Ag (aq) ---gt Al3 (aq) 3 Ag(s)
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EMF is an Intensive Property
EMF does not depend on the amount of material in
an electrochemical cell.
Changing stoichiometric coeffiecients of a
half-cell reaction does not effect the value of Eo
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Basic Electrical Units
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Relationship of Charge and Current

• one coulomb is the quantity of electric charge
  that flows through a circuit in one second if the
  current is one ampere.
• FORMULA Q It
• Q charge in coulombs
• I current in amperes
• t time in seconds
• 1 A 1 C/sec

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Electrolytic Cells
An electrolytic cell converts electrical energy
to chemical energy The process that takes place
in an electrolytic cell is called
electrolysis These cells are used to make
non-spontaneous reactions occur
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Electrolytic Cells
An electrolytic cell has the voltmeter replaced
by an external source of electricity such as a
battery The battery forces a non-spontaneous
reaction to occur
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An Electrolytic Cell
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Characteristics of an Electrolytic Cell
(connected to positive)
(connected to negative)
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The Daniell Cell A Galvanic Cell
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The Daniell Cell Characteristics of a Galvanic
Cell
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Electroplating
Electrolytic cells may be used to electroplate a
thin layer of a more desirable metal over a less
desirable metal object to be plated (or covered)
is the cathode metal to be plated (form a thin
layer) is the anode
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Faradays Law of Electrolysis
Faradays Law The amount (mass) of a substance
produced or consumed in an electrolytic cell is
directly proportional to the quantity of
electricity that flows through the cell and the
molar mass of the substance. The charge (Q) on a
mole of electrons is known as one faraday (1
F) 1 F 96,485 C/mol 96,500 C/mol Q neF
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Relationship of Charge and Current

• one coulomb is the quantity of electric charge
  that flows through a circuit in one second if the
  current is one ampere.
• FORMULA Q It
• Q charge in coulombs
• I current in amperes
• t time in seconds
• 1 A 1 C/sec

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Example 20.10 on Page 796
A current of 0.452 A is passed through an
electrolytic cell containing molten CaCl2 for
1.50 hours. Write the electrode reactions and
calculate the mass of chlorine gas released.
2 Cl- ---gt Cl2 (g) 2e-
anode
Ca2 2e- ---gt Ca (l)
cathode
Ca2 2 Cl- ---gt Ca (l) Cl (g)
The amount of charge flowing in 1.50 hrs
0.452 C
3600 s
1.50 hr
2 440 C
1 hr
1 s
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Example 20.10 on Page 796 Continued
A current of 0.452 A is passed through an
electrolytic cell containing molten CaCl2 for
1.50 hours. Write the electrode reactions and
calculate the mass of chlorine gas released.
The amount of charge flowing in 1.50 hrs
0.452 C
3600 s
1.50 hr
2 440 C
1 hr
1 s
The coefficients here stand for moles. 1 F 1
mole electrons.
2 Cl- ---gt Cl2 (g) 2e-
1 mole Cl2
70.90 g Cl2
1 F
2 440 C
0.896 g Cl2
1 mole Cl2
2 F
96 500 C