2. Khan Academy Galvanic Cell

3. Review of Terms
Electrochemistry – the study of the interchange of chemical and electrical energy
Oxidation–reduction (redox) reaction – involves a transfer of electrons from the reducing agent to the oxidizing agent
Oxidation – loss of electrons
Reduction – gain of electrons
Reducing agent – electron donor
Oxidizing agent – electron acceptor
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4. Electrochemistry
Electrochemistry is the study of the relationships between electricity and chemical reactions. It includes the study of both spontaneous and nonspontaneous processes.

5. Synopsis of Assigning Oxidation Numbers (as a Reminder)
Elements = 0
Monatomic ion = charge
F: –1
O: –2 (unless peroxide = –1)
H: +1 (unless a metal hydride = –1)
The sum of the oxidation numbers equals the overall charge (0 in a compound).

6. Oxidation Numbers
To keep track of what loses electrons and what gains them, we assign oxidation numbers.
If the oxidation number increases for an element, that element is oxidized.
If the oxidation number decreases for an element, that element is reduced.

7. Oxidation and Reduction
A species is oxidized when it loses electrons.
Zinc loses two electrons, forming the Zn2+ ion.
A species is reduced when it gains electrons.
H+ gains an electron, forming H2.
An oxidizing agent causes something else to be oxidized (H+); a reducing agent causes something else to be reduced (Zn).

8. Half-Reactions
The oxidation and reduction are written and balanced separately.
We will use them to balance a redox reaction.
For example, when Sn2+ and Fe3+ react,

9. For each half–reaction: Balance all the elements except H and O.
The Half–Reaction Method for Balancing Equations for Oxidation–Reduction Reactions Occurring in Acidic Solution
Write separate equations for the oxidation and reduction half–reactions.
For each half–reaction:
Balance all the elements except H and O.
Balance O using H2O.
Balance H using H+.
Balance the charge using electrons.
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10. Add the half–reactions, and cancel identical species.
The Half–Reaction Method for Balancing Equations for Oxidation–Reduction Reactions Occurring in Acidic Solution
If necessary, multiply one or both balanced half–reactions by an integer to equalize the number of electrons transferred in the two half–reactions.
Add the half–reactions, and cancel identical species.
Check that the elements and charges are balanced.
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11. The Half–Reaction Method for Balancing Equations for Oxidation–Reduction Reactions Occurring in Acidic Solution
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12. Cr2O72-(aq) + SO32-(aq) Cr3+(aq) + SO42-(aq)
How can we balance this equation?
First Steps:
Separate into half-reactions.
Balance elements except H and O.
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13. Method of Half Reactions
Cr2O72-(aq) 2Cr3+(aq)  
SO32-(aq) SO42-(aq)
How many electrons are involved in each half reaction?
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14. Method of Half Reactions (continued)
6e- + Cr2O72-(aq) 2Cr3+(aq) 
SO32-(aq) + SO42-(aq) + 2e-
How can we balance the oxygen atoms?
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15. Method of Half Reactions (continued)
6e- + Cr2O72-(aq) Cr3+(aq) + 7H2O 
H2O +SO32-(aq) SO42-(aq) + 2e-
How can we balance the hydrogen atoms?
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16. Method of Half Reactions (continued)
This reaction occurs in an acidic solution.
14H+ + 6e- + Cr2O Cr3+ + 7H2O 
H2O +SO SO e- + 2H+
How can we balance the electrons?
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17. Method of Half Reactions (continued)
14H+ + 6e- + Cr2O Cr3+ + 7H2O
3[H2O +SO SO e- + 2H+]
Final Balanced Equation:
Cr2O SO H Cr3+ + 3SO H2O
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18. Balancing Redox Equations: The Half-Reactions Method (a Synopsis)
Make two half-reactions (oxidation and reduction).
Balance atoms other than O and H. Then, balance O and H using H2O/H+.
Add electrons to balance charges.
Multiply by common factor to make electrons in half- reactions equal.
Add the half-reactions.
Simplify by dividing by common factor or converting H+ to OH– if basic.
Double-check atoms and charges balance!

19. The Half-Reaction Method
Consider the reaction between MnO4– and C2O42–:
MnO4–(aq) + C2O42–(aq)  Mn2+(aq) + CO2(aq)
Assigning oxidation numbers shows that Mn is reduced (+7  +2) and C is oxidized (+3  +4).

20. Oxidation Half-Reaction
C2O42–  CO2
To balance the carbon, we add a coefficient of 2:
C2O42–  2 CO2

21. Oxidation Half-Reaction
C2O42–  2 CO2
The oxygen is now balanced as well. To balance the charge, we must add two electrons to the right side:
C2O42–  2 CO2 + 2e–

22. Reduction Half-Reaction
MnO4–  Mn2+
The manganese is balanced; to balance the oxygen, we must add four waters to the right side:
MnO4–  Mn H2O

23. Reduction Half-Reaction
MnO4–  Mn H2O
To balance the hydrogen, we add 8H+ to the left side:
8 H+ + MnO4–  Mn H2O

24. Reduction Half-Reaction
8 H+ + MnO4–  Mn H2O
To balance the charge, we add 5e– to the left side:
5e– + 8 H+ + MnO4–  Mn H2O

25. Combining the Half-Reactions
Now we combine the two half-reactions together:
C2O42–  2 CO2 + 2e–
5e– + 8 H+ + MnO4–  Mn H2O
To make the number of electrons equal on each side, we will multiply the first reaction by 5 and the second by 2:

26. Combining the Half-Reactions
5 C2O42–  10 CO2 + 10e–
10e– + 16 H+ + 2 MnO4–  2 Mn H2O
When we add these together, we get
10e– + 16 H+ + 2 MnO4– + 5 C2O42– 
2 Mn H2O + 10 CO2 +10e–

27. Combining the Half-Reactions
10e– + 16 H+ + 2 MnO4– + 5 C2O42– 
2 Mn H2O + 10 CO2 +10e–
The only thing that appears on both sides is the electrons. Subtracting them, we are left with
16 H+ + 2 MnO4– + 5 C2O42– 
2 Mn H2O + 10 CO2
(Verify that the equation is balanced by counting atoms and charges on each side of the equation.)

28. Br–(aq) + MnO4–(aq) Br2(l)+ Mn2+(aq)
EXERCISE!
Balance the following oxidation–reduction reaction that occurs in acidic solution.
Br–(aq) + MnO4–(aq) Br2(l)+ Mn2+(aq)
10Br–(aq) + 16H+(aq) + 2MnO4–(aq) 5Br2(l)+ 2Mn2+(aq) + 8H2O(l)
10Br-(aq) + 16H+(aq) + 2MnO4-(aq)  5Br2(l)+ 2Mn2+(aq) + 8H2O(l)
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29. The Half–Reaction Method for Balancing Equations for Oxidation–Reduction Reactions Occurring in Basic Solution
Use the half–reaction method as specified for acidic solutions to obtain the final balanced equation as if H+ ions were present.
To both sides of the equation obtained above, add a number of OH– ions that is equal to the number of H+ ions. (We want to eliminate H+ by forming H2O.)
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30. Check that elements and charges are balanced.
The Half–Reaction Method for Balancing Equations for Oxidation–Reduction Reactions Occurring in Basic Solution
Form H2O on the side containing both H+ and OH– ions, and eliminate the number of H2O molecules that appear on both sides of the equation.
Check that elements and charges are balanced.
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31. The Half–Reaction Method for Balancing Equations for Oxidation–Reduction Reactions Occurring in Basic Solution
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32. Galvanic Cell
Device in which chemical energy is changed to electrical energy.
Uses a spontaneous redox reaction to produce a current that can be used to do work.
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33. Voltaic Cells
In spontaneous redox reactions, electrons are transferred and energy is released.
That energy can do work if the electrons flow through an external device.
This is a voltaic cell.

34. A Galvanic Cell
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35. Galvanic Cell Oxidation occurs at the anode.
Reduction occurs at the cathode.
Salt bridge or porous disk – devices that allow ions to flow without extensive mixing of the solutions.
Salt bridge – contains a strong electrolyte held in a Jello–like matrix.
Porous disk – contains tiny passages that allow hindered flow of ions.
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36. Voltaic Cells
In the cell, electrons leave the anode and flow through the wire to the cathode.
Cations are formed in the anode compartment.
As the electrons reach the cathode, cations in solution are attracted to the now negative cathode.
The cations gain electrons and are deposited as metal on the cathode.

37. Electromotive Force (emf)
Water flows spontaneously one way in a waterfall.
Comparably, electrons flow spontaneously one way in a redox reaction, from high to low potential energy.

38. Electromotive Force (emf)
The potential difference between the anode and cathode in a cell is called the electromotive force (emf).
It is also called the cell potential and is designated Ecell.
It is measured in volts (V). One volt is one joule per coulomb (1 V = 1 J/C).

39. Standard Reduction Potentials
Reduction potentials for many electrodes have been measured and tabulated.
The values are compared to the reduction of hydrogen as a standard.

40. CONCEPT CHECK!
Order the following from strongest to weakest oxidizing agent and justify. Of those you cannot order, explain why. Fe Na F- Na+ Cl2
Cl2 is a better oxidizing agent than Na+ (Cl2 has the larger reduction potential). The others cannot be ordered because we do not know their reduction potentials (although we can predict that F- will not easily be reduced, we do not have knowledge of quantitative proof from Table 18.1).

41. Standard Hydrogen Electrode
Their reference is called the standard hydrogen electrode (SHE).
By definition as the standard, the reduction potential for hydrogen is 0 V:
2 H+(aq, 1M) + 2e–  H2(g, 1 atm)

42. Standard Cell Potentials
The cell potential at standard conditions can be found through this equation:
Ecell
= Ered (cathode) – Ered (anode)
Because cell potential is based on the potential energy per unit of charge, it is an intensive property.

43. Voltaic Cell: Cathode Reaction
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44. Voltaic Cell: Anode Reaction
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45. Cell Potentials For the anode in this cell, E°red = –0.76 V
For the cathode, E°red = V
So, for the cell, E°cell = E°red (anode) – E°red (cathode) = V – (–0.76 V) = V

46. Oxidizing and Reducing Agents
The more positive the value of E°red, the greater the tendency for reduction under standard conditions.
The strongest oxidizers have the most positive reduction potentials.
The strongest reducers have the most negative reduction potentials.

47. Cell Potential
A galvanic cell consists of an oxidizing agent in one compartment that pulls electrons through a wire from a reducing agent in the other compartment.
The “pull”, or driving force, on the electrons is called the cell potential ( ), or the electromotive force (emf) of the cell.
Unit of electrical potential is the volt (V).
1 joule of work per coulomb of charge transferred.
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48. Galvanic Cell
All half-reactions are given as reduction processes in standard tables.
Table 18.1
1 M, 1atm, 25°C
When a half-reaction is reversed, the sign of E° is reversed.
When a half-reaction is multiplied by an integer, E° remains the same.
A galvanic cell runs spontaneously in the direction that gives a positive value for E°cell.
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49. Example: Fe3+(aq) + Cu(s) → Cu2+(aq) + Fe2+(aq)
Half-Reactions:
Fe3+ + e– → Fe2+ E° = 0.77 V
Cu2+ + 2e– → Cu E° = 0.34 V
To balance the cell reaction and calculate the cell potential, we must reverse reaction 2.
Cu → Cu2+ + 2e– – E° = – 0.34 V
Each Cu atom produces two electrons but each Fe3+ ion accepts only one electron, therefore reaction 1 must be multiplied by 2.
2Fe3+ + 2e– → 2Fe E° = 0.77 V
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50. Overall Balanced Cell Reaction
2Fe3+ + 2e– → 2Fe E° = 0.77 V (cathode)
Cu → Cu2+ + 2e– – E° = – 0.34 V (anode)
Balanced Cell Reaction:
Cu + 2Fe3+ → Cu2+ + 2Fe2+
Cell Potential:
E°cell = E°(cathode) – E°(anode)
E°cell = 0.77 V – 0.34 V = 0.43 V
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51. Line Notation Used to describe electrochemical cells.
Anode components are listed on the left.
Cathode components are listed on the right.
Separated by double vertical lines which indicated salt bridge or porous disk.
The concentration of aqueous solutions should be specified in the notation when known.
Example: Mg(s)|Mg2+(aq)||Al3+(aq)|Al(s)
Mg → Mg2+ + 2e– (anode)
Al3+ + 3e– → Al (cathode)
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52. Description of a Galvanic Cell
The cell potential (always positive for a galvanic cell where E°cell = E°(cathode) – E°(anode)) and the balanced cell reaction.
The direction of electron flow, obtained by inspecting the half–reactions and using the direction that gives a positive E°cell.
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53. Description of a Galvanic Cell
Designation of the anode and cathode.
The nature of each electrode and the ions present in each compartment. A chemically inert conductor is required if none of the substances participating in the half–reaction is a conducting solid.
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54. Sketch a cell using the following solutions and electrodes. Include:
CONCEPT CHECK!
Sketch a cell using the following solutions and electrodes. Include:
The potential of the cell
The direction of electron flow
Labels on the anode and the cathode
Ag electrode in 1.0 M Ag+(aq) and Cu electrode in 1.0 M Cu2+(aq)
Copper is the anode, silver is the cathode (electrons flow from copper to silver). The cell potential is 0.46 V.
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55. Sketch a cell using the following solutions and electrodes. Include:
CONCEPT CHECK!
Sketch a cell using the following solutions and electrodes. Include:
The potential of the cell
The direction of electron flow
Labels on the anode and the cathode
Zn electrode in 1.0 M Zn2+(aq) and Cu electrode in 1.0 M Cu2+(aq)
Zinc is the anode, copper is the cathode (electrons flow from zinc to copper). The cell potential is 1.10 V.
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56. CONCEPT CHECK!
Consider the cell from part b. What would happen to the potential if you increase the [Cu2+]? Explain. The cell potential should increase.
Since the copper(II) ion is the reactant in the overall equation of the cell, the cell potential should increase (LeChâtelier's principle applies here).
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57. Work Work is never the maximum possible if any current is flowing.
In any real, spontaneous process some energy is always wasted – the actual work realized is always less than the calculated maximum.
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58. Maximum Cell Potential
Directly related to the free energy difference between the reactants and the products in the cell.
ΔG° = –nFE°
F = 96,485 C/mol e–
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59. Free Energy and Redox
Spontaneous redox reactions produce a positive cell potential, or emf.
E° = E°red (reduction) – E°red (oxidation)
Note that this is true for ALL redox reactions, not only for voltaic cells.
Since Gibbs free energy is the measure of spontaneity, positive emf corresponds to negative ΔG.
How do they relate? ΔG = –nFE (F is the Faraday constant, 96,485 C/mol.)

60. Free Energy, Redox, and K How is everything related?
ΔG° = –nFE° = –RT ln K

61. A Concentration Cell
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62. Nernst Equation Remember, ΔG = ΔG° + RT ln Q So, –nFE = nFE° + RT ln Q
Dividing both sides by –nF, we get the Nernst equation:
E = E° – (RT/nF) ln Q
OR E = E° – (2.303 RT/nF) log Q
Using standard thermodynamic temperature and the constants R and F,
E = E° – (0.0592/n) log Q

63. Nernst Equation
The relationship between cell potential and concentrations of cell components
At 25°C: or
(at equilibrium)
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64. CONCEPT CHECK!
Explain the difference between E and E°. When is E equal to zero? When the cell is in equilibrium ("dead" battery). When is E° equal to zero? E  is equal to zero for a concentration cell.
ε is the cell potential at any condition, and ε is the cell potential under standard conditions (1.0 M or 1 atm, and 25C).
ε equals zero when the cell is in equilibrium ("dead" battery). ε is equal to zero for a concentration cell.
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65. EXERCISE!
A concentration cell is constructed using two nickel electrodes with Ni2+ concentrations of 1.0 M and × 10-4 M in the two half-cells.
Calculate the potential of this cell at 25°C.
0.118 V
The cell potential equals V. ε = ε° - (0.0591/n)logQ. For a concentration cell, ε° = 0.
ε = 0 - (0.0591/2)log(1.0×10-4 / 1.0) = V
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66. You make a galvanic cell at 25°C containing:
CONCEPT CHECK!
You make a galvanic cell at 25°C containing:
A nickel electrode in 1.0 M Ni2+(aq)
A silver electrode in 1.0 M Ag+(aq)
Sketch this cell, labeling the anode and cathode, showing the direction of the electron flow, and calculate the cell potential.
1.03 V
Nickel is the anode, silver is the cathode (electrons flow from nickel to silver).
To find the cell potential, use the Nernst equation: ε = ε° - (0.0591/n)logQ.
The overall equation is 2Ag+ + Ni → 2Ag + Ni2+. Therefore , Q = [Ni2+] / [Ag+]2 = (1.0 M) / (1.0 M)2.
ε = 1.03 V – (0.0591/2)log(1) = 1.03 V.
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67. Concentration Cells
Notice that the Nernst equation implies that a cell could be created that has the same substance at both electrodes, called a concentration cell.
For such a cell,
would be 0, but Q would not.
Ecell
Therefore, as long as the concentrations are different, E will not be 0.

68. Some Applications of Cells
Electrochemistry can be applied as follows:
Batteries: a portable, self-contained electrochemical power source that consists of one or more voltaic cells.
Batteries can be primary cells (cannot be recharged when “dead”—the reaction is complete) or secondary cells (can be recharged).
Prevention of corrosion (“rust-proofing”)
Electrolysis

69. One of the Six Cells in a 12–V Lead Storage Battery
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70. A Common Dry Cell Battery

71. A Mercury Battery
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72. Schematic of the Hydrogen-Oxygen Fuel Cell

73. Involves oxidation of the metal.
Process of returning metals to their natural state – the ores from which they were originally obtained.
Involves oxidation of the metal.
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74. The Electrochemical Corrosion of Iron
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75. Corrosion
Corrosion is oxidation.
Its common name is rusting.

76. Corrosion Prevention
Application of a coating (like paint or metal plating)
Galvanizing
Alloying
Cathodic Protection
Protects steel in buried fuel tanks and pipelines.
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77. Preventing Corrosion
Corrosion is prevented by coating iron with a metal that is more readily oxidized.
Cathodic protection occurs when zinc is more easily oxidized, so that metal is sacrificed to keep the iron from rusting.

78. Preventing Corrosion
Another method to prevent corrosion is used for underground pipes.
A sacrificial anode is attached to the pipe. The anode is oxidized before the pipe.

79. Cathodic Protection
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80. Forcing a current through a cell to produce a chemical change for which the cell potential is negative.
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81. Electrolysis
Nonspontaneous reactions can occur in electrochemistry IF outside electricity is used to drive the reaction.
Use of electrical energy to create chemical reactions is called electrolysis.

82. Stoichiometry of Electrolysis
How much chemical change occurs with the flow of a given current for a specified time?
current and time  quantity of charge 
moles of electrons  moles of analyte 
grams of analyte
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83. Electrolysis and “Stoichiometry”
1 coulomb = 1 ampere × 1 second
Q = It = nF
Q = charge (C)
I = current (A)
t = time (s)
n = moles of electrons that travel through the wire in the given time
F = Faraday’s constant
NOTE: n is different than that for the Nernst equation!

84. Stoichiometry of Electrolysis
current and time  quantity of charge
Coulombs of charge = amps (C/s) × seconds (s)
quantity of charge  moles of electrons
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85. CONCEPT CHECK!
An unknown metal (M) is electrolyzed. It took 52.8 sec for a current of 2.00 amp to plate g of the metal from a solution containing M(NO3)3. What is the metal? gold (Au)
The metal has a molar mass of 197 g/mol. The metal is gold (Au).
To find the molar mass, divide the number of grams of the metal by the number of moles of metal.
To find the moles of metal:
52.8 sec × (2.00 C/sec) × (1 mol e–/96485 C) × (1 mol M/3 mole e–) = 3.65 × 10–4 mol M
To find the molar mass:
g / 3.65 × 10–4 mol M = 197 g/mol
Note: Use the red box animation to assist in explaining how to solve the problem.
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86. CONCEPT CHECK!
Consider a solution containing 0.10 M of each of the following: Pb2+, Cu2+, Sn2+, Ni2+, and Zn2+. Predict the order in which the metals plate out as the voltage is turned up from zero. Cu2+, Pb2+, Sn2+, Ni2+, Zn2+ Do the metals form on the cathode or the anode? Explain.
Use the reduction potentials from Table Once that voltage has been surpassed, the metal will plate out. They plate out on the cathode (since they are reduced).
Order: Cu2+, Pb2+, Sn2+, Ni2+, Zn2+
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87. Production of aluminum Purification of metals Metal plating
Electrolysis of sodium chloride
Production of chlorine and sodium hydroxide
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88. Producing Aluminum by the Hall-Heroult Process
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89. Electroplatinging a Spoon
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90. The Downs Cell for the Electrolysis of Molten Sodium Chloride
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91. Some Examples of Batteries
Lead–acid battery: reactants and products are solids, so Q is 1 and the potential is independent of concentrations; however, made with lead and sulfuric acid (hazards).
Alkaline battery: most common primary battery.
Ni–Cd and Ni–metal hydride batteries: lightweight, rechargeable; Cd is toxic and heavy, so hydrides are replacing it.
Lithium-ion batteries: rechargeable, light; produce more voltage than Ni-based batteries.

92. Some Batteries
Lead–Acid Battery
Alkaline Battery

93. Lithium-Ion Battery

94. Fuel Cells
When a fuel is burned, the energy created can be converted to electrical energy.
Usually, this conversion is only 40% efficient, with the remainder lost as heat.
The direct conversion of chemical to electrical energy is expected to be more efficient and is the basis for fuel cells.
Fuel cells are NOT batteries; the source of energy must be continuously provided.

95. Hydrogen Fuel Cells In this cell, hydrogen and oxygen form water.
The cells are twice as efficient as combustion.
The cells use hydrogen gas as the fuel and oxygen from the air.