1. Chapter 21
Electrochemistry
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2. Oxidation-Reduction Reactions
oxidation: loss of electrons LEO
reduction: gain of electrons GER
Can be determined from change in oxidation numbers.
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3. Oxidation Number
Charge an atom has or would have if all the bonding electrons were assigned to the most electronegative element.
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4. Rules for assigning oxidation numbers
1). Oxidation number pure neutral element = 0 (e.g. Li, H2, C etc)
2). Oxidation number monatomic ion = ionic charge (Na+ = +1, Mg2+ = +2, etc)
3). Oxidation number of O = -2 in all its compounds except:
those with O-O bond (peroxides e.g. Na-O-O-Na then ox. number = -1) superoxides, (e.g. KO2 then ox. number = -1/2)
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5. Rules cont.
4). H is +1 in all its compounds (except metal hydrides then = -1)
5). F is - 1 in all its compounds (except, of course, F2)
6). Alkali metals are always +1; alkaline earths are always +2.
7). Sum of oxidation number is zero for neutral compounds, and equal to the overall charge for polyatomic ionic species.
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6. Oxidation and Reduction must occur simultaneously
Zn(s) + Cu2+(aq) > Zn2+(aq) + Cu(s)
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7. Electrochemical Cells
System consisting of electrodes that dip into an electrolyte and in which an oxidation-reduction reaction either uses or generates an electric current.
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8. Types of Electrochemical Cell
Voltaic or Galvanic
A spontaneous reaction generates an electric current
Electrolytic
An electric current is used to drive a nonspontaneous reaction
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10. Electrodes Cathode: electrode where reduction reaction occurs
Anode: electrode where oxidation reaction occurs.
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11. Balancing Redox Reactions
Neutral Solution
1. Split into two half-reactions
2. Balance each half-reaction
a. Balance the elements
b. Balance the charge
3. Combine half-reactions so that electrons cancel.
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12. Cl2(g) + Br-(aq) --> Br2(l) + Cl-(aq)
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13. Balancing Redox Reactions
Acidic Solution
1. Split into two half-reactions
2. Balance each half-reaction
a. Balance the elements
b. Balance oxygens by adding water
c. Balance hydrogens by adding hydrogen ions
d. Balance the charge
3. Combine half-reactions so that electrons cancel.
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14. Zn(s) + VO2+(aq) --> V2+(aq) + Zn2+(aq)
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15. Balancing Redox Reactions
Basic Solution
1. Split into two half-reactions
2. Balance each half-reaction
a. Balance the elements
b. Balance oxygens by adding water
c. Balance hydrogens by adding hydrogen ions
d. Neutralize the H+ ions by adding OH- ions
e. Balance the charge
3. Combine half-reactions so that electrons cancel.
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16. Zn(s) + Cl-(aq) --> Zn(OH)2(s) + Cl- (aq)
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17. Voltaic Cell Description anode: Zn ----------> Zn2+ + 2e-
cathode: Cu2+ + 2e > Cu
Consists of two half-cells that are electrically connected. The salt bridge consists of a tube of an electrolyte that is connected to two half-cells of the voltaic cell. Allows the flow of charge ions but prevents diffusional mixing of the different solutions.
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20. Volatic Cell Key Points
1. There are always two separate half-cells connected by a wire and a salt bridge
2. One of the compartments is the anode and the other is the cathode
3. If a metal participates in a cell reaction it is ordinarily chosen as the electrode. If no metal is involved in the half-reaction an electrically conduction solid like platinum or graphite is used
4. Electrons flow from the anode to the cathode.
5. Cations move to the cation and anions to the anode via the salt bridge.
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21. Cell Notation Zn/Zn2+ ll Cu2+/Cu anode reaction is shown at left
salt bridge is indicated by ll
cathode reaction is shown at right
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22. Fe3+(aq) + H2(g)  Fe2+(aq) + 2H+(aq)
Draw cell diagram and write the cell notation
Fe3+(aq) + H2(g)  Fe2+(aq) + 2H+(aq)
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23. Electrochemical Cells and Potentials
Emf
Electrons generated at the site of oxidation of a cell are thought to be "driven" or "pushed" toward the cathode by an electromotive force, or emf. This force is due to the difference in electric potential of an electron at the two electrodes.
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24. Standard Potential
Measure of the driving force of the cell reaction when all ions and molecules in the solution are at a concentration of 1 M and all gases are at a pressure of one atm.
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25. Calculating the Potential Eo of an Electrochemical Cell
Eotot = Eoox + Eored
Eoox: standard voltage for the oxidation half-reaction
Eored: standard voltage for the reduction half-reaction
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26. Standard Hydrogen Half-Cell
Cannot determine an individual half-reaction voltage therefore we arbitrarily take the standard voltage for the reduction of H+ ions to H2(g) to be zero (standard hydrogen half-cell):
2H+(aq,1M) + 2e > H2(g, 1atm) Eored = 0
With this assigned voltage others can be determined from measurement by hook-up with knowns.
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27. Standard Hydrogen Half-Cells
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28. Standard Potentials
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29. Using Standard Potentials
1. The Eo values are for reactions written in the form "oxidized form + electrons > reduced form." The species on the left side of the reaction is an oxidizing agent, and the species on the right is a reducing agent. All potentials are therefore for reduction reactions.
2. When writing the reaction "reduced form > oxidized form + electrons," the sign of Eo is reversed, but the value of Eo is unaffected.
3. All the half-reactions are reversible.
4. The more positive the value of Eo for the reactions the better the oxidizing ability of the ion or compound.
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30. 5. The more negative the value of the reduction potential Eo the less likely the reaction occurs as a reduction, and the more likely the reverse reaction occurs.
6. The reaction between any substance on the left in this table with any substance lower than it on the right is product-favored under standard conditions.
7. The algebraic sign of the half-reaction potential is the sign of the electron when it is attached to H2/H3O+ standard cell.
8. Electrochemical potential depends on the nature of the reactants and products and their concentrations, not on the quantities of material used.
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31. Large negative value means oxidation strongly favored; strong reducing agent.
Large positive value means reduction strongly favored; strong oxidizing agent.
Relative values in table give an indication that one half-reaction favored over other. Summing half-cell reactions allow determination of standard cell potential.
Half-cell potential intensive property  independent of amount of material  we don’t use stoichiometric coefficients for determining standard cell potentials.
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32. Br2(l) + 2I(aq)  I2(l) + 2Br(aq)
Determine the cell potential of
Br2(l) + 2I(aq)  I2(l) + 2Br(aq)
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33. 2Ag+(aq) + Cu(s)  2Ag(s) + Cu2+(aq)
Determine the cell potential of
2Ag+(aq) + Cu(s)  2Ag(s) + Cu2+(aq)
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34. MnO4-(aq) + Fe(s)  Fe2+(aq) + Mn2+(aq) (balanced?)
Determine cell potential:
MnO4-(aq) + Fe(s)  Fe2+(aq) + Mn2+(aq) (balanced?)
when it is operated galvanically. Which is the oxidizing agent? reducing agent?
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35. Determine if the reaction below is spontaneous in the direction written.
Fe3+(aq) + Ag(s)  ?
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36. Eotot and DGo 1. DG = free energy change
= maximum amount of useful work
= -wmax
2. Electric work
= charge x potential energy difference
= coulomb x volt
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37. Eotot and DGo
wmax = nFE = coulombs x volts = electrical energy in joules where F is the Faraday constant, 9.65 x 104 J/V mol.
DGo = -nFEo tot = nEo tot (in KJ)
Because spontaneous reactions have a negative free energy change, DG, the negative sign in the equation above confirms that all product-favored electron transfer reactions have a positive Eo.
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38. Electrochemical Cells at Nonstandard Conditions
Voltage is a measure of reaction spontaneity.
Hence:
voltage is increased by increasing concentrations of reactants or decreasing concentrations of products.
voltage is decreased by decreasing concentrations of reactants or increasing concentrations of products.
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39. Quantitative: Nernst Equation
aA + bB > cC + dD
Etot = Eotot -RT/nF ln [products]/[reactants] raised to the power of their coefficients
where R = J/K mol, F = x 104 J/v mol and number of electrons transferred at 25 C
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40. Nernst Equation n = number of electrons transferred
Etot = Eotot /n ln[products]/[reactants] raised to the power of their coefficients
n = number of electrons transferred
[ ] use molarities for aqueous solutions, atm for gases
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41. K: Equilibrium Constant
Etot = Eotot /n ln [products]/[reactants] raised to the power of their coefficients
at equilibrium Etot = zero
and
K = [products]/[reactants] raised to the power of their coefficients
therefore
0 = Eotot /n ln K
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42. Eotot and K
ln K = nEotot/0.0257
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43. MnO4-(aq) + Fe(s)  Fe2+(aq) + Mn2+(aq)
Determine free energy and equilibrium constant for reaction below (unbalanced).
MnO4-(aq) + Fe(s)  Fe2+(aq) + Mn2+(aq)
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44. Determine cell potential and equilibrium constant of Cl2/Br2 cell.
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45. Ag(s)|AgCl(s)|Cl(1.0 M)||Ag+(1.0 M)|Ag(s).
The following cell has a potential of V at 25°C; determine Ksp.
Ag(s)|AgCl(s)|Cl(1.0 M)||Ag+(1.0 M)|Ag(s).
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46. Batteries
A battery is a package of one or more galvanic cells used for the production and storage of electric energy by chemical means. A galvanic cell consists of at least two half cells, a reduction cell and an oxidation cell. Chemical reactions in the two half cells provide the energy for the galvanic cell operations.
Each half cell consists of an electrode and an electrolyte solution. Usually the solution contains ions derived from the electrode by oxidation or reduction reaction.
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47. Types of Batteries
Batteries can be divided into two types: primary or disposable batteries and secondary or rechargeable batteries.
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48. Fuel cells are different from batteries in that they consume reactant, which must be replenished, whereas batteries store electrical energy chemically in a closed system.
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49. Corrosion of Metals/Cathodic Protection
Cathodic protection (CP) is a technique to control the corrosion of a metal surface by making that surface the cathode of an electrochemical cell.
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51. Cathodic Protection
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52. Electrolytic Cell
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53. Aqueous Electrolysis Cathode(reduction)
a. cation reduced to metal; usually occurs with transition metal cations
b. H+ ions reduced to H2; occurs with strong acids
c. H2O molecules reduced to H2; occurs with Group 1, Group 2 metals and Al
2H2O(l) + 2e > H2(g) + 2OH-(aq)
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54. Aqueous Electrolysis Anode(oxidation) a. anion oxidized to nonmetal
b. OH- ions oxidized to O2; occurs with strong bases
c. H2O molecules oxidized to O2; with NO3-, SO42- and F-
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55. Overall result NiCl2: Ni + Cl2 NaCl: H2, OH-, Cl2 CuSO4: Cu, O2, H+
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57. Stoichiometry of Electrolysis
Units
Faraday = charge = 9.65 x 104 C = 1 mole of e-
Coulomb = charge
Ampere = current = charge/time = C/s
Joule = electrical work = C x volts
watt = energy/time = 1 joule/second
volt = potential difference
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58. Electric Charge = electric current x time elapsed
Calculations
Electric Charge = electric current x time elapsed
Electric Energy = Coulombs x volts
watt = volts x amperes
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59. Determine amount of Cu2+ electrolyzed from solution at constant current of 6.00 A for period of 1.00 hour.
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