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Chapter 20
Electrochemistry
Prentice Hall © 2003
Chapter 20
• Oxidation-reduction reactions
• Oxidation numbers
• Oxidation of metals by acids and salts
• The activity series
ALL these to be done in class
Prentice Hall © 2003
Chapter 20
Oxidation-Reduction
Reactions
• In the reaction
Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g).
• Which element is oxidised and which one is reduced?
• Oxidation – loss of e• Reduction – gain of e-
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Chapter 20
Balancing OxidationReduction Reactions
• Law of conservation of mass: the amount of each element
present at the beginning of the reaction must be present at
the end.
• Conservation of charge: electrons are not lost in a
chemical reaction.
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Chapter 20
Half Reactions
Half-reactions are a convenient way of separating oxidation
and reduction reactions.
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Chapter 20
• The half-reactions for
Sn2+(aq) + 2Fe3+(aq)  Sn4+(aq) + 2Fe3+(aq)
are………
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Chapter 20
Balancing Equations by the Method of
Half Reactions
The two incomplete half reactions are
MnO4-(aq)  Mn2+(aq)
C2O42-(aq)  2CO2(g)
Balance the overall reaction equation in an acidic solution
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Chapter 20
Balancing Equations for Reactions
Occurring in Basic Solution
• We use OH- and H2O rather than H+ and H2O.
• The same method as for acidic solution is used, but OHis added to “neutralize” the H+ used.
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Chapter 20
Voltaic Cells
• If a strip of Zn is placed in a solution of CuSO4, Cu is
deposited on the Zn and the Zn dissolves by forming
Zn2+.
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Chapter 20
• Zn is spontaneously oxidized to Zn2+ by Cu2+.
• The Cu2+ is spontaneously reduced to Cu0 by Zn.
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Chapter 20
“Rules” of voltaic cells:
1. At the anode electrons are products. (Oxidation)
2. At the cathode electrons are reagents. (Reduction)
3. Electrons can’t swim!
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Chapter 20
• Anions and cations move through a porous barrier or salt
bridge.
• Cations move into the cathodic compartment to
neutralize the excess negatively charged ions
• Anions move into the anodic compartment to neutralize
the excess Zn2+ ions formed by oxidation
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Chapter 20
A Molecular View of Electrode
Processes
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Chapter 20
Cell EMF
• e- flow from anode to cathode because the cathode has a
lower electrical potential energy than the anode.
• 1 V is the potential difference required to impart 1 J of
energy to a charge of one coulomb:
1J
1V 
1C
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Chapter 20
• 1 V is the potential difference required to impart 1 J of
energy to a charge of one coulomb:
1J
1V 
1C
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Chapter 20
• Electromotive force (emf) is the force required to push
electrons through the external circuit.
• Cell potential: Ecell is the emf of a cell.
• For 1M solutions at 25 C (standard conditions), the
standard emf (standard cell potential) is called Ecell.
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Chapter 20
Standard Reduction (Half-Cell)
Potentials
• Standard reduction potentials, Ered are measured relative
to the standard hydrogen electrode (SHE).
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Chapter 20
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Chapter 20
• For the SHE, we assign
2H+(aq, 1M) + 2e-  H2(g, 1 atm)
• Ered = 0.
Ecell  Ered cathode  Ered anode
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Chapter 20
• For Zn:
Ecell = Ered(cathode) - Ered(anode)
0.76 V = 0 V - Ered(anode).
• Therefore, Ered(anode) = -0.76 V.
• Standard reduction potentials must be written as
reduction reactions:
Zn2+(aq) + 2e-  Zn(s), Ered = -0.76 V.
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Chapter 20
• Changing the stoichiometric coefficient does not affect
Ered.
• Therefore,
2Zn2+(aq) + 4e-  2Zn(s), Ered = -0.76 V.
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Chapter 20
• Reactions with Ered < 0 are spontaneous oxidations
relative to the SHE.
• The larger the difference between Ered values, the larger
Ecell.
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Chapter 20
Oxidizing and Reducing Agents
• The more positive Ered the stronger the oxidizing agent
on the left.
• The more negative Ered the stronger the reducing agent
on the right.
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Chapter 20
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Chapter 20
• More generally, for any electrochemical process
Ecell  Ered reductionprocess  Ered oxidationprocess
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Chapter 20
Example: For the following cell, what is the cell
reaction and Eocell?
Al(s)|Al3+(aq)||Fe2+(aq)|Fe(s)
Al3+(aq) + 3e- → Al(s); EoAl = -1.66 V
Fe2+(aq) + 2e- → Fe(s); EoFe = -0.41 V
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Chapter 20
Example: When an aqueous solution of CuSO4 is
electrolysed, Cu metal is deposited:
Cu2+(aq) + 2e- → Cu(s)
If a constant current was passed for 5.00 h and 404 mg
of Cu metal was deposited, what was the current?
Ans: 6.81 x 10-2 A
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Chapter 20