11678_Lecture15 2-14-07 Voltaic Cells
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Transcript of 11678_Lecture15 2-14-07 Voltaic Cells
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A Hydrogen ElectrodeCell potentials can bemeasured
Half-cell potential cannot be
measured
An arbitrary choice sets
the half cell potential of astandard hydrogen
electrode = 0
Eo1/2(H+/H2) = 0 V
Half Reaction is always
written as a reduction
2 H+ (aq) + 2 e- H2(g)
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Questions
Anode?
Cathode?
Ox An
Zn (s) Zn2+ (aq) + 2 e-
Red Cat
2H+ (aq) + 2e- H2 (g)
Spontanteous Cell Reaction?
2H+ (aq) + Zn (s) H2 (g) + Zn2+ (aq)
Zinc metal dissolves in 1 M H+
Zn/H2 Voltaic Cell
Cell Potential = 0.763 Vunder standard conditions
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Voltage of Voltaic Cells
All voltaic cells are based on a
spontaneous chemical reaction
Therefore, G for the reaction must be
negative This means that the voltage of a voltaic
cell is always positive.
G = -nFe
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Standard Potential of Zn/H2 Cell2H+ (aq) + Zn (s) H2 (g) + Zn2+ (aq)
Cathode: 2 H+ (aq) + 2e- H2 (g)Anode: Zn (s) Zn2+ (aq) + 2 e-
Eo
(cell) = E1/2o
(H+
/H2) E1/2o
(Zn2+
/Zn)=+0.763V
E0(cell) = 0 E1/2o (Zn2+/Zn) = 0.763 V
E1/2o (Zn2+/Zn) = -0.763 V
Zn2+ (s) + 2e- Zn(s) E1/2o = -0.763 V
Cathode Anode
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Determination of Half-Cell PotentialsAnode and cathode?
H+/H2 is anode and Cu2+/Cu is
cathode
H2 (g) + Cu2+ 2 H+ (aq) + Cu (s)
E0(cell) = E01/2(Cu2+/Cu) E01/2(H
+/H2)
E0(cell) = E01/2(Cu2+/Cu) 0 = 0.340 V
E01/2(Cu2+/Cu) = +0.340 V
Cu/H2 Voltaic CellCell Potential = 0.340 V
under standard conditions
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Cell EMFCell EMF
Standard Reduction PotentialsStandard Reduction Potentials
The larger the difference
between Ered values, thelargerEcell.
In a voltaic cell (spontaneous)
Ered(cathode) is more positivethan Ered(anode).
RecallEcell = Ered(cathode) -
Ered(anode)
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Half-reaction
cells
Eo(cell) = 1.10 V (measured)
E1/2o (Cu2+/Cu) E1/2
o (Zn2+/Zn)
Eo (cell) = 0.340 V (-0.763 V)
Eo (cell) = 1.10 V (Calculated from E01/2)
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Selected StandardElectrode Potentials at 25 oC
EOS
StrongerO
xidizingAge
nts
StrongerReducing
Agents
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Cell EMFCell EMF
Oxidizing and Reducing AgentsOxidizing and Reducing Agents
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Cell Potential Determine the cell potential for a galvanic cellbased on the redox reaction.
Cu(s) + Fe+3(aq) Cu+2(aq) + Fe+2(aq)
Fe+3(aq) + e- Fe+2(aq) E1/20= 0.77 V
Cu+2(aq)+2e- Cu(s) E1/20= 0.34 V
Cu(s) Cu+2(aq)+2e- -E1/20= -0.34 V
2Fe+3(aq) + 2e- 2Fe+2(aq) E1/20= 0.77 V
Cu(s) + 2Fe3+ Cu2+ + 2Fe2+ Ecell0= 0.43V
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Shorthand Cell Notation
eg: Cu(s) + Fe+3(aq) Cu+2(aq) + Fe+2(aq)
solidAqueousAqueoussolid
Anode on the leftCathode on the right
Single line different phases.
Double line porous disk or salt bridge. If all the substances on one side are
aqueous, a platinum electrode is indicated.
Cu(s)Cu+2(aq)Fe+2(aq),Fe+3(aq)Pt(s)Anode CathodeSolution Bridge Solution
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Shorthand Notation Voltaic Cell
Zn Zn2+
(anode)
porous
plate
Cu2+ Zn2+
Cu2+ Cu(cathode)
e-
1.10 V
Start HereEnd Here
Tell the story. What do you see during your trip from the anode to the cathode?
Zn(s) Zn2+ (aq, 1M) Cu2+ (aq, 1M) Cu(s)
1.00 M1.00 M
Trip
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Practice Completely describe the galvanic cell based on
the following half-reactions under standardconditions.
MnO4- + 8 H+ +5e- Mn+2 + 4H2O E=1.51
Fe+3
+3e-
Fe(s) E=0.036V Anode? Cathode?
Cell Potential? Oxidizing agent and reducing agent?
Shorthand cell notation?
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Fe3+ (aq)
MnO4- + 8 H+ +5e- Mn+2 + 4H2O E=1/2 1.51
E0cell = 1.51V 0.036V = 1.47V
Fe+3 +3e- Fe(s)E1/2=0.036V
Fe(s) Fe3+ + 3 e- -E1/2 o = -0.036V
Questions
1. Whats missing in Diagram?
2. Anode?
3. Cathode?
4. Cell Reaction?
5. Direction electron flow?
6. Cell Potential?
7. Shorthand Representation?Fe(s)Fe3+MnO4- (aq H+), Mn2+Pt
Practice3 MnO
4
- + 24 H+ + 5 Fe(s)
3 Mn2+ + 12 H2O + 5 Fe3+
e-e-
e-
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Spontaneity of Redox ReactionsSpontaneity of Redox Reactions
Consider the displacement of silver by nickel:
Ni(s) + 2Ag+(aq) Ni2+(aq) + 2Ag(s)
E0 = E0red(Ag+/Ag) E0red(Ni2+/Ni)= (0.80 V) - (-0.28 V)
= 1.08 VThe positive E0 indicates a
spontaneous process.
EMF and FreeEMF and Free--Energy ChangeEnergy Change
G= -nFE
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Spontaneity of Redox ReactionsSpontaneity of Redox Reactions
EMF and FreeEMF and Free--Energy ChangeEnergy Change
G is the change in free-energy, n is the
number of moles of electrons transferred, FisFaradays constant, and Eis the emf of the
cell.
Since n and Fare positive, ifG > 0 then E