Chapter 20 Electrochemistry
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Transcript of Chapter 20 Electrochemistry
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Prentice Hall © 2003 Chapter 20
Chapter 20Chapter 20ElectrochemistryElectrochemistry
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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
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• 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-
Oxidation-Reduction Oxidation-Reduction ReactionsReactions
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Prentice Hall © 2003 Chapter 20
• 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.
Balancing Oxidation-Balancing Oxidation-Reduction ReactionsReduction Reactions
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Half ReactionsHalf-reactions are a convenient way of separating oxidation and reduction reactions.
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• The half-reactions for
Sn2+(aq) + 2Fe3+(aq) Sn4+(aq) + 2Fe3+(aq)
are………
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Prentice Hall © 2003 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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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 OH- is added to “neutralize” the H+ used.
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• 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+.
Voltaic CellsVoltaic Cells
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Prentice Hall © 2003 Chapter 20
• Zn is spontaneously oxidized to Zn2+ by Cu2+.• The Cu2+ is spontaneously reduced to Cu0 by Zn.
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“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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Prentice Hall © 2003 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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A Molecular View of Electrode Processes
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• 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:
Cell EMFCell EMF
C 1J 1
V 1
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Prentice Hall © 2003 Chapter 20
• 1 V is the potential difference required to impart 1 J of energy to a charge of one coulomb:
C 1J 1
V 1
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• 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 Ecell.
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Standard Reduction (Half-Cell) Potentials
• Standard reduction potentials, Ered are measured relative to the standard hydrogen electrode (SHE).
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• For the SHE, we assign
2H+(aq, 1M) + 2e- H2(g, 1 atm)
• Ered = 0.
anodecathode redredcell EEE
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• For Zn:
Ecell = Ered(cathode) - Ered(anode)
0.76 V = 0 V - Ered(anode).
• Therefore, Ered(anode) = -0.76 V.
• Standard reduction potentials must be written as reduction reactions:
Zn2+(aq) + 2e- Zn(s), Ered = -0.76 V.
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• Changing the stoichiometric coefficient does not affect Ered.
• Therefore,
2Zn2+(aq) + 4e- 2Zn(s), Ered = -0.76 V.
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• Reactions with Ered < 0 are spontaneous oxidations relative to the SHE.
• The larger the difference between Ered values, the larger Ecell.
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Oxidizing and Reducing Agents
• The more positive Ered the stronger the oxidizing agent on the left.
• The more negative Ered the stronger the reducing agent on the right.
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• More generally, for any electrochemical process
processoxidation processreduction redredcell EEE
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Example: For the following cell, what is the cellreaction and Eo
cell?
Al3+(aq) + 3e- → Al(s); EoAl = -1.66 V
Fe2+(aq) + 2e- → Fe(s); EoFe = -0.41 V
Al(s)|Al3+(aq)||Fe2+(aq)|Fe(s)
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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