1. Galvanic Cells
Electrochemistry: The area of chemistry concerned with the interconversion of chemical and electrical energy
Galvanic (Voltaic) Cell: A spontaneous chemical reaction that generates an electric current
Electrolytic Cell: An electric current that drives a nonspontaneous reaction
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2. Galvanic Cells Zn2+(aq) + Cu(s) Zn(s) + Cu2+(aq)
Oxidation half-reaction:
Zn2+(aq) + 2 e
Zn(s)
Reduction half-reaction:
Cu(s)
Cu2+(aq) + 2 e
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3. Zn-Cu2+ Spontaneous Redox Reaction
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4. Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq)
Voltaic Cells
Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq)
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5. Zn-Cu2+ Voltaic Cell
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6. Voltaic Cells Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq)
Anode
Cathode
salt bridge
Electrodes (anode & cathode)
Allow e- to pass in/out of solution.
A salt bridge (or porous barrier) is required...
e- move across an external conductor.
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7. Galvanic Cells Anode: The electrode where oxidation occurs
The electrode where electrons are produced
Is what anions migrate toward
Has a negative sign
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8. Galvanic Cells Cathode: The electrode where reduction occurs
The electrode where electrons are consumed
Is what cations migrate toward
Has a positive sign
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9. Voltaic Cells Zinc is removed: Zn(s) → Zn2+(aq) + 2 e-
Oxidation at anode (vowels!)
The anode loses mass
Zn supplies e-.
Anode has “-” charge.
Copper is deposited: Cu2+(aq) + 2 e- → Cu(s)
Reduction at cathode (consonants!)
The cathode gains mass
Cu2+ accepts e-.
Cathode has “+” charge.
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10. Salt Bridge
It is an inverted glass U-tube with glass wool plugged at each end
Contains a salt paste (e.g. K2SO4,KNO3 ).
A porous plug restricts bulk flow
Ions pass into/out of the cells
Stops charge buildup.
K2SO4
SO42- K+ Zn Cu
Zn2+
Cu2+
porous plug
SO42-
Zn2+
SO42- released
Zn2+ removed
as Zn → Zn2+
2 K+ released
SO42- removed
as Cu2+ → Cu
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11. Voltaic Cells In all voltaic cells:
The redox reaction must be product-favored.
Oxidation occurs at the anode.
Reduction occurs at the cathode.
Electrons move from anode to cathode through an external ‘wire’.
The electrical circuit is completed by movement of ions into and out of a salt bridge.
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12. Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
Shorthand Notation for Galvanic Cells
Anode half-reaction:
Zn(s)
Zn2+(aq) + 2 e
Cathode half-reaction:
Cu2+(aq) + 2 e
Cu(s)
Overall cell reaction:
Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)
Salt bridge
Anode half-cell
Cathode half-cell
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
Electron flow
Phase boundary
Phase boundary
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13. Show the Cell Notation for the Following Voltaic Cell
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14. Electrochemical Cell Notations
Pt|H2 (P=1 bar)|H+ (aq,1.0M) ||Fe+3 (aq,1.0M) ,Fe2+(aq,1.0M) |Pt
Electrodes
Typically
The anode is made of the metal that is oxidized.
The cathode is made of the same metal as is produced by the reduction.
If the redox reaction involves the oxidation or reduction of an ion to a different oxidation state, or the oxidation or reduction of a gas, we may use an inert electrode.
An inert electrode is one that not does participate in the reaction but just provides a surface on which the transfer of electrons can take place.
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15. Cell Potentials and Free-Energy Changes for Cell Reactions
Electromotive Force (emf): The force or electrical potential that pushes the negatively charged electrons away from the anode ( electrode) and pulls them toward the cathode (+ electrode)
It is also called the cell potential (E) or the cell voltage.
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16. SI unit of electric potential
Cell Potentials and Free-Energy Changes for Cell Reactions
1 J = 1 C × 1 V
joule (J)
SI unit of energy
volt (V)
SI unit of electric potential
coulomb (C)
Electric charge
1 coulomb is the amount of charge transferred when a current of 1 ampere (A) flows for 1 second.
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17. Cell Potentials and Free-Energy Changes for Cell Reactions
faraday or Faraday constant
The electric charge on 1 mol of electrons and is equal to 96,500 C/mol e
DG = nFE or
DG° = nFE°
Free-energy change
Cell potential
Number of moles of electrons transferred in the reaction
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18. Cell Potentials and Free-Energy Changes for Cell Reactions
The standard cell potential at 25 °C is 1.10 V for the reaction:
Zn2+(aq) + Cu(s)
Zn(s) + Cu2+(aq)
Calculate the standard free-energy change for this reaction at 25 °C.
DG° = nFE°
mol e
96,500 C
1 C V
1 J
1000 J
1 kJ
= (2 mol e)
(1.10 V)
DG° = 212 kJ
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19. Standard Reduction Potentials
Anode half-reaction:
H2(g)
2 H+(aq) + 2 e
Cathode half-reaction:
Cu2+(aq) + 2 e
Cu(s)
Overall cell reaction:
H2(g) + Cu2+(aq)
2 H1+(aq) + Cu(s)
The standard potential of a cell is the sum of the standard half-cell potentials for oxidation at the anode and reduction at the cathode:
E°cell = E°ox + E°red
The measured potential for this cell: E°cell = 0.34 V
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20. Standard Reduction Potentials
The standard hydrogen electrode (S.H.E.) has been chosen to be the reference electrode.
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21. Standard Reduction Potentials
The standard hydrogen electrode (S.H.E.) has been chosen to be the reference electrode.
H2(g, 1 atm)
2H+(aq, 1 M) + 2e
E°ox = 0 V
E°red = 0 V
2H+(aq, 1 M) + 2e
H2(g, 1 atm)
Gas + soln.
in contact with Pt
Compact notation:
H+(aq), 1M| H2(g, 1bar) | Pt
The Standard Hydrogen Electrode (SHE) is assigned a value of E° = 0 V:
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22. Standard Reduction Potentials
Anode half-reaction:
H2(g)
2 H+(aq) + 2 e
Cathode half-reaction:
Cu2+(aq) + 2 e
Cu(s)
Overall cell reaction:
H2(g) + Cu2+(aq)
2 H+(aq) + Cu(s)
E°cell = E°ox + E°red
0.34 V = 0 V + E°red
A standard reduction potential can be defined:
Cu(s)
Cu2+(aq) + 2 e
E° = 0.34 V
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23. Standard Reduction Potentials
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24. As a standard reduction potential:
Standard Reduction Potentials
Anode half-reaction:
Zn(s)
Zn2+(aq) + 2 e
Cathode half-reaction:
2 H+(aq) + 2 e
H2(g)
Overall cell reaction:
Zn(s) + 2 H+(aq)
Zn2+(aq) + H2(g)
E°cell = E°ox + E°red
0.76 V = E°ox + 0 V
Zn2+(aq) + 2 e
Zn(s)
E° = 0.76 V
As a standard reduction potential:
Zn(s)
Zn2+(aq) + 2 e
E° = 0.76 V
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25. Standard Reduction Potentials
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26. (If E°cell < 0 the reaction is reactant favored).
Cell Potentials & Half Cells
A standard potential (E°) occurs when:
All [solute] = 1 M (or saturated if solubility < 1 M).
All gases have P = 1 bar.
All solids are pure.
If E°cell is positive the reaction is product favored
(If E°cell < 0 the reaction is reactant favored).
E° does not depend on nreactant or nproduct
(i.e not on the stoichiometric coefficients).
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27. Using Standard Half-Cell Potentials
Potential differences between SHE many other half-cells have been made.
Standard reductions are listed.
Each half-reaction can occur in either direction
More positive E° = easier reduction.
Less positive E° = easier oxidation for the reverse reaction.
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28. Cell Potentials Consider the following table: Identify:
Half Reaction
Half Cell
E° (V)
Au3+(aq) + 3 e- →
Au(s)
Au3+(aq)|Au(s)
+1.52
Cl2(g) + 2 e-
2 Cl-(aq)
Cl2(g)|Cl−(aq)|Pt
+1.358
Al3+(aq) + 3 e-
Al(s)
Al3+(aq)|Al(s)
-1.676
K+(aq) + e-
K(s)
K+(aq)|K(s)
-2.925
Identify:
the strongest oxidizing agent.
the weakest oxidizing agent.
the strongest reducing agent.
the weakest reducing agent.
= Au3+(aq)
= K+(aq)
= K(s)
= Au(s)
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29. Cell Potentials Consider the following table: Will:
Half Reaction
Half Cell
E° (V)
Au3+(aq) + 3 e- →
Au(s)
Au3+(aq)|Au(s)
+1.52
Cl2(g) + 2 e-
2 Cl-(aq)
Cl2(g)|Cl−(aq)|Pt
+1.358
Al3+(aq) + 3 e-
Al(s)
Al3+(aq)|Al(s)
-1.676
K+(aq) + e-
K(s)
K+(aq)|K(s)
-2.925
Will:
Cl2(g) oxidize Au(s) to Au3+(aq)?
K(s) reduce Cl2(g) to Cl-(aq)?
and
c) What can be reduced by Al(s)?
= No
= Yes
= Cl2 and Au3+
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30. Using Standard Reduction Potentials
Zn2+(aq) + 2 e
Zn(s)
E° = (0.76 V)
Cu(s)
Cu2+(aq) + 2 e
E° = 0.34 V
E° = 1.10 V
Zn2+(aq) + Cu(s)
Zn(s) + Cu2+(aq)
Ag(s)]
2 × [Ag+(aq) + e
E° = 0.80 V
Cu2+(aq) + 2 e
Cu(s)
E° = 0.34 V
2 Ag(s) + Cu2+(aq)
2 Ag+(aq) + Cu(s)
E° = 0.46 V
Half-cell potentials are intensive properties.
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31. The Nernst Equation DG = DG° + RT ln Q DG = nFE and DG° = nFE°
Using:
DG = nFE and DG° = nFE° nF RT
Nernst Equation:
E = E° 
ln Q or nF
2.303RT
E = E°
log Q V
E = E° 
log Q
in volts, at 25 °C n
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32. The Nernst Equation Consider a galvanic cell that uses the reaction:
Cu(s) + 2 Fe3+(aq)
Cu2+(aq) + 2 Fe2+(aq)
What is the potential of a cell at 25 °C that has the following ion concentrations?
[Fe3+] = 1.0 × 104 M
[Cu2+] = 0.25 M
[Fe2+] = 0.20 M
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33. The Nernst Equation 0.0592 V E = E°  log Q n Cu(s) + 2 Fe3+(aq)
Cu2+(aq) + 2 Fe2+(aq)
Calculate E°:
Cu(s)
Cu2+(aq) + 2 e
E° = 0.34 V
Fe3+(aq) + e
Fe2+(aq)
E° = 0.77 V
E°cell = 0.34 V V = 0.43 V
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34. The Nernst Equation 0.0592 V E = E°  log Q n Cu(s) + 2 Fe3+(aq)
Cu2+(aq) + 2 Fe2+(aq)
Calculate E: V
[Cu2+][Fe2+]2
E = E° –
log n
[Fe3+]2 V
(0.25)(0.20)2
= 0.43 V 
log 2
(1.0 x 104)2
E = 0.25 V
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35. Standard Cell Potentials and Equilibrium Constants
Using
DG° = nFE°
and
DG° = RT ln K
nFE° = RT ln K nF RT
2.303 RT
E° =
ln K =
log K nF V
E° =
log K
in volts, at 25 °C n
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36. Standard Cell Potentials and Equilibrium Constants
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37. Standard Cell Potentials and Equilibrium Constants
Three methods to determine equilibrium constants:
K =
[A]a[B]b
[C]c[D]d
K from concentration data: RT
DG°
ln K =
K from thermochemical data: nF RT
nFE°
K from electrochemical data:
E° =
ln K or
ln K = RT V
E° =
log K
in volts, at 25 °C n
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38. Example
Consider the following reaction at 25°C:
Find K
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39. Example Solution: Oxidation numbers Half-reactions: n:
Multiply the oxidation half-reaction by 3 to cancel out the three electrons in the reduction half-reaction
The electrons cancel out: n = 3
+e-
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40. Example 𝐾=10 𝑛 𝐸 ° 0.0592 =10 3𝑋0.165 0.0592 = 2.3 X108 0.0592 V E° =
log K
in volts, at 25 °C, n=3,E = 0.165V n
𝐾=10 𝑛 𝐸 ° =10 3𝑋 = 2.3 X108
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41. Concentration Cells Identical cells? E may not be zero…
Example Zn|Zn2+(aq,0.01M)||Zn2+(aq,1M)|Zn
Left-to-right…
Zn(s) → Zn2+ (0.01M) + 2 e-
Zn2+ (1M) + 2 e- → Zn(s)
Zn2+(1M) → Zn2+(0.01M) (net reaction)
E = E° − log
E = 0 − log
[Zn2+]dil
[Zn2+]conc
0.010 1
0.0592 2
Same cell
E = V
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42. Determination of pH
Cell potential measurements can allow one to determine the pH of a solution by determining [H+]. According to the Nernst equation, cell potential depends on the H+ concentration of the test solution.
To obtain the relationship between Ecell and pH, substitute the following into the Nernst equation:
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43. Determination of pH
On the left, a small, commercial electrode is pictured.
On the right is a sketch showing the construction of a glass electrode for measuring hydrogen concentrations.
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44. Batteries Lead Storage Battery Anode: Pb(s) + HSO4(aq)
PbSO4(s) + H+(aq) + 2 e
Cathode:
PbO2(s) + 3 H+(aq) + HSO4(aq) + 2e
PbSO4(s) + 2 H2O(l)
Overall:
Pb(s) + PbO2(s) + 2 H+(aq) + 2 HSO41(aq)
2 PbSO4(s) + 2 H2O(l)
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45. Batteries Dry-Cell Batteries Leclanché cell Anode: Zn(s)
Zn2+(aq) + 2 e
Cathode:
2 MnO2(s) + 2 NH4+(aq) + 2 e
Mn2O3(s) + 2 NH3(aq)+ H2O(l)
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46. Corrosion Corrosion: The oxidative deterioration of a metal
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47. Corrosion
For some metals, oxidation protects the metal (aluminum, chromium, magnesium, titanium, zinc, and others). For other metals, there are two main techniques.
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48. Prevention of Corrosion
Galvanization: The coating of iron with zinc
When some of the iron is oxidized (rust), the process is reversed since zinc will reduce Fe2+ to Fe:
Fe(s)
Fe2+(aq) + 2 e
E° = 0.45 V
Zn(s)
Zn2+(aq) + 2 e
E° = 0.76 V
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49. Prevention of Corrosion
Cathodic Protection: Instead of coating the entire surface of the first metal with a second metal, the second metal is placed in electrical contact with the first metal:
Anode:
Mg(s)
Mg2+(aq) + 2 e
E° = 2.37 V
Cathode:
O2(g) + 4 H+(aq) + 4 e
2 H2O(l)
E° = 1.23 V
Attaching a magnesium stake to iron will corrode the magnesium instead of the iron. Magnesium acts as a sacrificial anode.
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50. Sacrificial Anode
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51. Electrolysis and Electrolytic Cells
Electrolysis: The process of using an electric current to bring about chemical change
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52. Electrolysis of Molten Sodium Chloride
Electrolysis and Electrolytic Cells
Electrolysis of Molten Sodium Chloride
Anode:
2 Cl(l)
Cl2(g) + 2 e
Cathode:
2 Na+(l) + 2 e
2 Na(l)
Overall:
2 Na+(l) + 2 Cl(l)
2 Na(l) + Cl2(g)
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53. Electroplating
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54. Quantitative Aspects of Electrolysis
Charge(C) = Current(A) × Time(s)
1 mol e
Moles of e = Charge(C) ×
96,500 C
Faraday constant
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55. Counting Electrons
What mass of gold will be electroplated from aqueous gold(III) chloride in 5.0 min by a 10.0 A current?
Au3+(aq) + 3 e- → Au(s)
Charge = (10.0 A)(5.0 min) = 3.0 x 103 C
60 s
1 min
1 mol e-
96,500 C
197.0 g
1 mol Au
3 mol e-
3.0 x 103 C = 2.0 g
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56. Counting Electrons
How long will it take to produce 14 g of Al from Al3+(melt)? Assume a current of A.
Al e- → Al
96,500 C
1 mol e-
1 mol Al
26.98 g
3 mol e-
14. g = 1.50 x 105 C
charge
current
Time = = = 3.0 x 102 s
1.50 x 105 C
500.0 A
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