1. Chapter 20 Electrochemistry

2. Oxidation and Reduction
Oxidation (loss of e-) Na Na+ + e-
Reduction (gain of e-) Cl + e Cl-
Oxidation-reduction (redox) reactions occur when electrons are transferred from an atom that is oxidized to an atom that is reduced.
Electron transfer can produce electrical energy spontaneously, but sometimes electrical energy is needed to make them occur (nonspontaneous).
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3. CHEM 1033
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4. Oxidation-Reduction (Redox) Reactions
BOTH reduction and oxidation must occur.
A substance that gives up electrons is oxidized and is called a reducing agent or reductant (causes another substance to be reduced).
A substance that accepts electrons is reduced and is therefore called an oxidizing agent or oxidant (causes another substance to be oxidized).
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5. Oxidation-reduction (Redox) Reactions
Is it a redox reaction??
Zn(s) + 2H+(aq) Zn2+(aq) + H2(g)
Quick hint:
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6. Oxidation Number Guidelines
Atoms in elemental form, oxidation number is zero.
(Cl2, H2, P4, Ne are all zero)
Monoatomic ion, the oxidation number is the charge on the ion. (Na+: +1; Al3+: +3; Cl-: -1)
O is usually -2. But in peroxides (like H2O2 and Na2O2) it has an oxidation number of -1.
H is +1 when bonded to nonmetals and -1 when bonded to metals.
(+1 in H2O, NH3 and CH4; -1 in NaH, CaH2 and AlH3)
The oxidation number of F is -1.
The sum of the oxidation numbers for the molecule is the charge on the molecule (zero for a neutral molecule).
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7. Example
Determine the oxidation state of all elements in ammonium thiosulfate (NH4)2(S2O3).
(NH4)2(S2O3)
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8. Determining Oxidation States
What is the oxidation state of Mn in MnO4-?
Answer: +7
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9. Balancing oxidation-reduction equations
We know: Balancing chemical equations follows law of conservation of mass.
AND, for redox reactions, gains and losses of electrons must also be balanced.
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10. Half-Reactions Separate oxidation and reduction processes in equation,
Sn2+(aq) + 2Fe3+(aq) Sn4+(aq) + 2Fe2+(aq)
Oxidation:
Reduction:
Overall, the number of electrons lost in the oxidation half reaction must equal the number gained in the reduction half reaction.
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11. Balancing Equations by the Method of Half-Reactions: Acidic
Consider :
MnO4-(aq) + C2O42-(aq) Mn2+(aq) + CO2(g)
Unbalanced half-reactions: MnO4-(aq) Mn2+(aq)
C2O42-(aq) CO2(g)
First, balance everything EXCEPT hydrogen and oxygen.
Deal with half-reactions SEPARATELY.
(acidic)
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12. Balancing Equations by the Method of Half-Reactions: Acidic
MnO4-(aq) + C2O42-(aq) Mn2+(aq) + CO2(g)
To balance O:
Add 4H2O to products to balance oxygen in reactants.
To balance H:
Add 8H+ to reactant side to balance the 8H in water.
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13. Balancing Equations by the Method of Half-Reactions: Acidic
Balance charge: Add up charges on both sides.
Add 5 electrons to reactant side.
5e H+(aq) + MnO4-(aq) Mn2+(aq) + 4H2O(l)
Mass balance of C in oxalate half-reaction.
C2O42-(aq) CO2(g)
Balance charge by adding two electrons to the products.
Last step: Cancel electrons and add reactions together.
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14. Balancing Equations by the Method of Half-Reactions: Acidic
5e H+(aq) + MnO4-(aq) Mn2+(aq) + 4H2O(l)
C2O42-(aq) CO2(g) + 2e-
Top reaction times 2. Bottom reaction times 5. ALL PARTS!!
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15. Balancing Equations by the Method of Half-Reactions: Acidic Summary
1. Divide equation into two incomplete half-reactions.
2. Balance each half-reaction
(a) balance elements other than H and O.
(b) balance O atoms by adding H2O.
(c) balance H atoms by adding H+ (basic conditions will require further work at this step).
(d) balance charge by adding e- to the side with greater overall positive charge.
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16. Balancing Equations by the Method of Half-Reactions: Summary
Multiply each half-reaction by an integer so that the number of electrons lost in one half-reaction equals the number gained in the other half-reaction.
4. Add the two half-reactions and cancel out all species appearing on both sides of the equation.
5. Check equation to make sure there are same number of atoms of each kind and the same total charge on both sides. Errors can be caught!!
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17. Balancing Equations by the Method of Half-Reactions: Basic
Balancing process is started using H+ and H2O, then adjusting with OH- to uphold reaction conditions.
(H+ does not exist in basic solutions.)
Balance the following reaction:
H2O2(aq) + ClO2(aq) ClO2-(aq) + O2(g)
(basic)
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18. Balancing Equations by the Method of Half-Reactions: Basic
Basic Redox Reactions
Split into two half-reactions.
H2O2(aq) O2(g)
ClO2(aq) ClO2-(aq)
Balance elements, then oxygen by adding H2O. Then, add H+ to balance H, just like an acidic redox reaction.
H2O2(aq) O2(g) + 2H+
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19. Balancing Equations by the Method of Half-Reactions: Basic
Basic Redox Reactions
Add OH- to both sides, enough to neutralize all H+ (basic reactions cannot support H+).
Combine H+ and OH- to form H2O.
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20. Balancing Equations by the Method of Half-Reactions: Basic
Basic Redox Reactions
Balance charge by adding e-.
2OH- + H2O2(aq) O2(g) + 2H2O +
+ ClO2(aq) ClO2-(aq)
Multiply each reaction so both have same e-. Then add them together, cancelling where possible.
2OH- + H2O2(aq) + 2ClO2(aq)
O2(g) + 2H2O + 2ClO2-(aq)
Double-check your answer!
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21. Balance this redox equation
Cu(s) + NO3-(aq) Cu2+(aq) + NO2(g) (acidic)
Ans: Cu(s) + 2NO3-(aq) + 4H+(aq) Cu2+(aq) + 2NO2(aq) + 2H2O(l)
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22. Balance this redox equation
NO2-(aq) + Al(s) NH3(aq) + Al(OH)4-(aq) (basic)
2Al(s) + NO2-(aq) +OH-(aq) + 5H2O(aq) Al(OH)4-(aq) + NH3(aq)
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23. Voltaic Cells
Voltaic (aka galvanic) cells: electrochemical reactions in which electron transfer occurs via an external circuit.
The energy released in a voltaic cell reaction can be used to perform electrical work.
Reactions are spontaneous.
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24. Voltaic Cells: Components
For example
Anode: Zn(s)  Zn2+(aq) + 2e ( half-reaction)
Cathode: Cu2+(aq) + 2e-  Cu(s) ( half reaction)
Salt bridge: cations move from anode to cathode, anions move from cathode to anode.
External circuit (wire): electrons move from anode to cathode (between two solid metal electrodes). Electron transfer can naturally occur in OTHER forms besides a wire (direct electron transport between solution and metal).
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25. Voltaic Cells: Ion Flow
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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26. Simplified Voltaic Cell
In any voltaic cell, the electrons flow from the anode through the external circuit to the cathode.
Anode labeled with negative sign (-) and cathode with positive sign (+).
Anions migrate toward anode. Cations toward the cathode.
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27. Voltaic Cells: Electron Flow
As oxidation occurs, Zn(s) is converted to Zn2+ and 2e-. And Cu2+ is converted to Cu(s).
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28. One-Pot Voltaic Cells
If a strip of Zn is placed in a solution of CuSO4, Cu is deposited on the Zn and the Zn dissolves, forming Zn2+.
Zn is spontaneously oxidized to Zn2+ by Cu2+ (Cu is the oxidizing agent).
The Cu2+ is spontaneously reduced to Cu0 by Zn (Zn is the reducing agent).
The entire process is spontaneous.
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29. The Atomic Level
A Cu2+(aq) ion comes into contact with a Zn(s) atom on the surface of the electrode. Two electrons are directly transferred from the Zn(s), forming Zn2+(aq), to the Cu2+(aq) forming Cu(s).
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30. Cell Electromotive Force (EMF)
Electromotive force (emf): the force required to push electrons through the external circuit.
Potential difference: difference per electrical charge between two electrodes. Units = Volts (V).
One volt is the potential required to impart one joule of energy to a charge of one coulomb:
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31. Ecell = Ered(cathode) - Ered(anode)
Cell EMF
Cell potential (Ecell): the emf of a cell. Aka cell voltage. Positive for spontaneous cell reactions.
Ecell strictly depends on reactions that occur at the cathode and anode, the concentration of reactants and products, and the temperature and is described by the equation:
Ecell = Ered(cathode) - Ered(anode)
For 1M solutions at 25˚C and 1 atm (standard conditions), the standard emf (standard cell potential), Ecell, is written as E˚cell.
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32. Standard Reduction Potentials
Potential associated with each half-reaction is chosen to be the potential for reduction to occur at that electrode. Hence, Standard Reduction Potentials.
Standard reduction potentials, E˚red, are measured relative to the standard hydrogen electrode (SHE).
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33. Standard Hydrogen Electrode (SHE)
SHE is assumed as the cathode (for consistency). Pt electrode in a tube containing 1M H+ solution. H2(g) is bubbled through the tube and equilibrium is established.
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34. Determining Standard Reduction Potentials
For the SHE:
2H+(aq, 1M) + 2e-  H2(g, 1 atm) E˚red =
Standard reduction potentials can then be calculated using the SHE (E˚= 0V) as E˚red(cathode).
Each calculated E˚ is rewritten as a reduction and tabulated.
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35. Finding the Standard Reduction Potential
E˚cell is measured, 0V (SHE) is used for E˚(cathode), and the reduction potential for Zn is found.
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36. Zn Standard Reduction Potential E˚cell = E˚red(cathode) - E˚red(anode)
We measure E˚cell relative to the SHE:
E˚cell = E˚red(cathode) - E˚red(anode)
0.76V = 0V - E˚red(anode).
Therefore, E˚red(anode) = -0.76V
And we find that -0.76V can be assigned to reduction of zinc.
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37. Zn Standard Reduction Potential
Since E˚red = V (negative!) we conclude that the reduction of Zn2+ in the presence of the SHE is not spontaneous.
However, the oxidation of Zn with the SHE is spontaneous.
Reactions with E˚red > 0 are spontaneous reductions relative to the SHE. E˚red < 0 are spontaneous oxidations.
Changing the stoichiometric coefficient does not affect E˚red.
2Zn2+(aq) + 4e-  2Zn(s) E˚red = V
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38. Full list in Appendix E
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39. EMF Trends
The larger the difference between E˚red values, the larger E˚cell.
Spontaneous voltaic cell: E˚red(cathode) is more positive than E˚red(anode), resulting in a positive E˚cell.
Recall
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40. Calculating E˚red from E˚cell
Zn(s) + Cu2+(aq, 1M) Zn2+(aq, 1M) + Cu(s) E˚cell = 1.10V
Given: Zn2+ + 2e Zn(s) E˚red = -0.76V
Calculate the E˚red for the reduction of Cu2+ to Cu.
What about E˚red for the oxidation of Cu(s) – reverse reaction?
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41. Oxidizing and Reducing Agents
We can use E˚red values to understand aqueous reaction chemistry.
e.g. Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
Since Cu2+ is responsible for the oxidation of Zn(s), Cu2+ is called the
Since Zn(s) is responsible for the reduction of Cu2+, Zn(s) is called the
E˚red for Cu2+ (0.34V) indicates that, compared to E˚red for Zn (-0.76V), it will be reduced.
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42. Example Problem: Cell emf
Calculate the standard emf for the following:
Ni(s) + 2Ce4+(aq) Ni2+(aq) + 2Ce3+(aq)
Ask yourself…What are the half-reactions?
Which one is oxidized? Reduced?
Ni2+(aq) + 2e Ni(s) E˚red = -0.28V
Ce4+(aq) + 1e Ce3+(aq) E˚red = 1.61V
is the oxidizing agent.
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43. Example Problem: Cell emf
Which reaction will undergo reduction? What reaction occurs at the anode?
Sn4+(aq) + 2e Sn2+(aq) E˚red = 0.154V
MnO4-(aq) + 8H+(aq) + 5e Mn2+(aq) + 4H2O(l) E˚red = 1.51V
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44. Oxidizing and Reducing Agents: EMF
The more positive E˚red, the greater the tendency for the reactant in the half-reaction to be reduced. Therefore, the stronger the oxidizing agent.
The more negative E˚red, the greater the tendency for the product in the half-reaction to be oxidized. This means the product is tends to be a reducing agent.
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45. A species at higher left of the table of standard reduction potentials will spontaneously oxidize a species that is lower right in the table.
F2 will oxidize H2 or Li.
Ni2+ will oxidize Al(s).
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46. Example: Redox Agents
Which is the strongest reducing agent? Oxidizing agent?
Ce4+, Br2, H2O2, or Zn?
E˚red = 1.61V for Ce4+ (aq) V for Br2 (l)
1.776 for H2O2 (aq) V for Zn(s)
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47. Spontaneity of Redox Reactions
In a spontaneous voltaic cell, E˚red(cathode) must be more positive than E˚red(anode).
A positive E˚cell indicates a spontaneous voltaic cell process.
A negative E˚cell indicates a non-spontaneous process.
It all relates back to Gibbs Free Energy.
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48. Spontaneity of Redox Reactions
Cell emf and free-energy change are related by
G is the change in free-energy, n is the number of moles of electrons transferred, F is Faraday’s constant, and E is the emf of the cell. G units are J (assumed per mole).
1F = 96,500 C/mol = 96,500 J/Vmol
If standard conditions: G˚ = -nFE˚cell
If Ecell > 0 then , both of which indicate spontaneous processes.
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49. Example: Cell spontaneity
Calculate ΔG˚ for the following reaction. Is it spontaneous?
Cl2(g) + 2I-(aq) Cl-(aq) + I2(s) E˚cell = 0.823V
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50. Batteries
Defn: Self-contained electrochemical power source with one (or more) complete voltaic cell.
When multiple cells or multiple batteries are connected in series, greater emfs can be achieved.
Cathode labeled with a plus sign; the anode with a minus sign.
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51. How Batteries Work
Spontaneous redox reactions, as in voltaic cells, serve as the basis for battery operation.
Specific reactions at anode and cathode determine the voltage of the battery, and the usable life of the battery depends on the quantity of these substances.
cannot be recharged.
are rechargeable using an external power source after its emf has dropped below a usable level.
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52. Lead-Acid Batteries
A 12V car battery consists of 6 cathode/anode pairs, connected in series, each producing 2V.
Cathode: PbO2(s) on a metal grid in sulfuric acid
PbO2(s) + SO42-(aq) + 4H+(aq) + 2e-  PbSO4(s) + 2H2O(l)
Anode: Pb(s) in sulfuric acid
Pb(s) + SO42-(aq)  PbSO4(s) + 2e-
The overall electrochemical reaction is
PbO2(s) + Pb(s) + 2SO42-(aq) + 4H+(aq)  2PbSO4(s) + 2H2O(l)
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53. Lead-Acid Batteries Ecell = Ered(cathode) - Ered(anode)
= ( V) - ( V)
= V.
Wood or fiberglass spacers prevent electrode contact.
H2SO4 consumed during discharge, (voltage may vary with use).
Recharging reverses forward reaction:
PbO2(s) + Pb(s) + 2SO42-(aq) + 4H+(aq)  2PbSO4(s) + 2H2O(l)
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54. Alkaline Batteries
Most common nonrechargeable (primary) battery (100 billion produced annually).
Anode: Powdered Zn in gel in contact with cKOH solution
Zn(s) + 2OH-(aq)  Zn(OH)2(s) + 2e-
Cathode: MnO2 and C paste
2MnO2(s) + 2H2O(l) + 2e-  2MnO(OH)(s) + 2OH-(aq)
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55. Rechargeable Batteries
Lightweight batteries for use in cell phones, notebook computers, etc.
Nickel-cadmium (NiCad) battery still common.
Cadmium oxidized at anode, nickel oxyhydroxide [NiO(OH)(s)] reduced at cathode.
Cathode:
2NiO(OH)(s) + 2H2O(l) + 2e-  2Ni(OH)2(s) + 2OH-(aq)
Anode:
Cd(s) + 2OH-(aq)  Cd(OH)2(s) + 2e-
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56. Rechargeable Batteries
Solid reaction products adhere to the electrodes, permitting electrode reactions to be reversed during recharging.
Drawbacks of NiCad battery:
Toxicity of Cd (disposal problems), relatively heavy.
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57. NiMH and Li-ion Batteries
Cathode reaction in nickel metal hydride (NiMH) battery is same as for Ni-Cd battery.
2NiO(OH)(s) + 2H2O(l) + 2e-  2Ni(OH)2(s) + 2OH-(aq)
Anode consists of a metal alloy, e.g. ZrNi2, capable of absorbing hydrogen atoms. During oxidation, the hydrogen atoms lose electrons (used by cathode), and the resulting H+ ions react with OH- to form water (drives cathode reaction to the right).
Another type:
Li-ion battery - lightweight with very high energy density.
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58. Fuel Cells Direct, efficient production of electricity.
NOT batteries as they are not self-contained (some products must be released).
On Apollo moon flights, the H2-O2 fuel cell was the primary source of electricity.
Cathode: 2H2O(l) + O2(g) + 4e-  4OH-(aq)
Anode: 2H2(g) + 4OH-(aq)  4H2O(l) + 4e-
What’s the overall reaction?
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59. Corrosion Generally the result of an undesirable redox reaction.
Spontaneous! A metal is attacked by some substance in its environment to form an unwanted compound.
For many metals oxidation is thermodynamically favored at RT.
Corrosion can form an insulating protective coating, preventing further oxidation. (e.g., Al forms a hydrated form of Al2O3)
~20% of iron produced annually in USA is used to replace rusted items!!
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60. Corrosion of Iron Rusting of iron requires both oxygen and water.
Other factors that can accelerate rusting of iron: -
Rusting of iron is electrochemical and iron itself conducts electricity.
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61. Corrosion of Iron
Recall: E˚red Fe2+ (-0.44V) < E˚red O2 (1.23V), so iron can be oxidized by oxygen.
Cathode: O2(g) + 4H+(aq) + 4e-  2H2O(l) E˚red= 1.23V
Anode: Fe(s)  Fe2+(aq) + 2e- E˚red= -0.44V
Dissolved oxygen in water usually causes Fe oxidation.
Fe2+ initially formed can be further oxidized to Fe3+ which forms rust, Fe2O3 . xH2O(s).
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62. Corrosion of Iron
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63. Corrosion of Iron
Oxidation occurs more readily at the site with the greatest concentration of O2 (water-air junction).
Salts produce the necessary electrolyte required to complete the electrical circuit.
Preventing Corrosion of Iron
Corrosion can be prevented by coating the iron with paint or another metal.
Galvanized iron (commonly used for nails) is coated with a thin layer of zinc.
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64. Corrosion Prevention: Galvanization
Zinc protects iron since Zn is more easily oxidized (E˚redZn is more negative than E˚redFe).
Anode: Zn2+(aq) +2e-  Zn(s) E˚red = V
Cathode: Fe2+(aq) + 2e-  Fe(s) E˚red = V
Zn is oxidized INSTEAD OF IRON, preventing corrosion.
Protects even if the surface coat is broken.
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65. Corrosion Prevention: Galvanization Preventing Corrosion of Iron
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66. Corrosion Prevention: Sacrificial Anodes
To protect underground pipelines.
The water pipe is the cathode and a more active metal is used as the anode (cathodic protection).
Often, Mg is used as the sacrificial anode:
Mg2+(aq) +2e-  Mg(s) E˚red = V
Fe2+(aq) + 2e-  Fe(s) E˚red = V
Since E˚red of Mg is more negative, it will oxidize first.
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67. Corrosion Prevention: Sacrificial Anodes
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68. Electrolysis
Defn: Use of electrical energy to drive a non-spontaneous redox reaction.
In voltaic and electrolytic cells
reduction occurs at the cathode, and
oxidation occurs at the anode.
However, in electrolytic cells, electrons are forced to flow from the anode to cathode.
Anode is now designated as POSITIVE(+).
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69. Electrolysis of Molten Salts
Used industrially to produce metals such as Na and Al.
Requires high temperatures.
Example: Electrolysis of molten NaCl: NaCl(s)  Na+(l) + Cl-(l)
Two reactants: Na+ and Cl-
Cathode: 2Na+(l) + 2e-  2Na(l) E˚red = -2.71V
Anode: 2Cl-(l)  Cl2(g) + 2e- E˚red = V
This is clearly a non-spontaneous reaction.
E˚cell =
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70. Electrolysis of Molten Solutions
Electrons still flow from anode to cathode, but are forced.
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71. Electrolysis of Aqueous Solutions
Aqueous solution electrolysis is complicated by water.
Ask: Is water oxidized (to form O2) or reduced (to form H2) in reference to the other components?
e.g. For an aqueous solution of NaF in an electrolytic cell, possible reactants are Na+, F- and H2O.
Both Na+ and H2O can be reduced but not F-.
Thus possible reactions at cathode are:
Na+(aq) + e Na(s) E˚red = V
2H2O(l) + 2e H2(g) + 2OH-(aq) E˚red = V
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72. Electrolysis of Aqueous Solutions
More positive the E˚red favors reduction reactions.
Reduction of H2O (E˚red = -0.83V) occurs at cathode with H2 gas produced.
At anode, either F- or H2O must be oxidized since Na+ cannot lose additional electrons.
2F-(aq) F2(g) + 2e E˚red = V
4OH-(aq) O2(g) + 2H2O(l) + 4e E˚red = V
Thus oxidation of H2O occurs at anode (more negative E˚red favors oxidation reactions).
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73. Electrolysis of Aqueous Solutions
Cathode 4H2O(l) + 4e H2(g) + 4OH-(aq) E˚red = V
Anode OH-(aq) O2(g) + 2H2O(l) + 4e E˚red = V
Process is non-spontaneous (as expected).
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74. Electroplating Active electrodes - take part in electrolysis.
Example: electrolytic plating.
Defn: Electrolysis used to deposit a thin layer of one metal on another in order to improve beauty or resistance to corrosion. e.g. electroplating nickel on a piece of steel.
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75. Electroplating
Consider an active Ni electrode and another metallic electrode placed in an aqueous solution of NiSO4.
Anode: Ni(s)  Ni2+(aq) + 2e-
Cathode: Ni2+(aq) + 2e-  Ni(s)
With voltage supplied, Ni plates on the steel electrode.
Electroplating is important in protecting objects from corrosion (chrome).
E˚cell = 0V, so outside source of voltage needed!
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76. Quantitative Aspects of Electrolysis
How much material can we obtain with electrolysis?
Consider the reduction of Cu2+ to Cu(s).
Cu2+(aq) + 2e-  Cu(s).
2 mol of electrons will plate 1 mol of Cu
The charge of 1 mol of electrons is 96,500 C (=1F).
Using Q = It
The amount of Cu can be calculated from the current used (I) and time (t) allowed to plate.
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77. Calculate the number of grams of aluminum produced in 1
Calculate the number of grams of aluminum produced in 1.00h by the electrolysis of molten AlCl3 if the electrical current is 10.0 A.
Example
Ans. 3.36g
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78. Electrical Work
Free-energy is a measure of the maximum amount of useful work that can be obtained from a system.
We know
If work is negative, then work is performed by the system and E is positive (which means it’s spontaneous)
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79. Electrical Work
The emf can be thought of as a measure of the driving force for a redox reaction to proceed.
In order to drive non-spontaneous, electrolytic reactions, the external, applied emf must be greater than Ecell.
If Ecell = V, then external emf must be at least V.
Note: work can be expressed in Watts (W)
1 W = 1 J/s.
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