1. Electrochemistry

2. homework
Page 863
1-13, 17, odd

3. Oxidation- Reduction Reactions
Redox or oxidation-reduction reactions are reactions that involve a transfer of electrons.
Oxidation is the loss of electrons.
Reduction is the gain of electrons.
(think of the charge, OIL RIG)
So in the reaction
4 K + O2 → 4 K O2-
Potassium get oxidized, oxygen get reduced

4. Oxidation States
Oxidation state is the theoretical charge on all atoms if all bonds were ionic.
The sum of the oxidation states must be equal to the charge of the ion or molecule.

5. Using oxidation states
In the reaction…
2 Na +2 H2O →2 NaOH + H2
Note the changes
Sodium went from 0 to 1
2 of the hydrogen atoms went from +1 to 0 (the other two were unchanged)

6. Identification of Redox Components.
Specify which of the following equations represents oxidation-reduction reactions, and indicate the oxidizing agent, the reducing agent, the species being oxidized, and the species being reduced.
CH4(g) + H2O(g)  CO(g) + 3H2(g)
2AgNO3(aq) + Cu(s)  Cu(NO3)2(aq) + 2Ag(s)
H+(aq) + 2CrO42-(aq)  Cr2O72-(aq) + H2O(l)

7. Half reactions Ce4+ + Sn2+ → Ce3+ + Sn4+ Half reactions
Ce4+ + e- → Ce3+
Sn2+ → 2e- + Sn4+
Electrons lost must equal electrons gained!
2 Ce4+ +2 e- →2 Ce3+
Merge the two half reactions
2 Ce4+ + Sn2+ → 2 Ce3+ + Sn4+

8. Redox reactions in acidic solutions
It will be noted in the problem
Balance all elements except hydrogen and oxygen.
Balance oxygen by adding H2O (which is always prevalent in an acidic solution)
Balance hydrogen by adding H+
Then balance the charge adding electrons and proceed normally.

9. Example In an acidic solution Cr2O7 2- + Cl- → Cr3+ + Cl2
Half reactions
Cr2O7 2- → Cr3+
Cl- → Cl2

10. Reduction side Cr2O7 2- → Cr3+ Cr2O7 2- → 2 Cr3+
Cr2O7 2- → 2 Cr H2O
Cr2O H+→ 2 Cr H2O
Cr2O H++ 6 e- →2Cr3++7 H2O

11. Oxidation side Cl- → Cl2 2 Cl- → Cl2 2 Cl- → Cl2 + 2 e-
I have to equal 6 e- so multiply by 3
6 Cl- → 3 Cl2 + 6 e-

12. Combine my half reactions
Cr2O H++ 6 e- → 2 Cr H2O
6 Cl- → 3 Cl2 + 6 e-
And you get
Cr2O H++6Cl-→2Cr3++3 Cl2+7H2O
The electrons cancel out .

13. Example In an acidic solution MnO4- + H2O2 → Mn2+ + O2 Half reactions
MnO4- → Mn2+
H2O2 → O2

14. Top Equation MnO4- → Mn2+ MnO4- → Mn2+ + 4 H2O
MnO H+→ Mn H2O
MnO H++ 5 e-→ Mn H2O

15. Bottom Equation H2O2 → O2 H2O2 → O2 + 2 H+ H2O2 → O2 + 2 H+ + 2 e-
I need to equal 5 e- so…
That won’t work…
2MnO H++ 10 e-→ 2 Mn H2O
5 H2O2 → 5 O H e-

16. Add them together 2MnO4- + 16 H++ 10 e-→ 2 Mn2+ + 8 H2O
5 H2O2 → 5 O H e-
And you get
2 MnO H++ 5 H2O2
→ 2 Mn O2 + 8 H2O
Notice the H+ canceled out as well.

17. Balancing Redox Equations in a basic solution
Look for the words basic or alkaline
Follow all rules for an acidic solution.
After you have completed the acidic reaction add OH- to each side to neutralize any H+.
Combine OH- and H+ to make H2O.
Cancel out any extra waters from both sides of the equation.

18. Example We will use the same equation as before In a basic solution
MnO4- + H2O2 → Mn2+ + O2
2 MnO H++ 5 H2O2
→ 2 Mn O2 + 8 H2O

19. Basic solution Since this is a basic solution we can’t have excess H+.
We will add OH- to each side to neutralize all H+
2 MnO H++ 5 H2O2 + 6OH-
→2 Mn2+ +5 O2 +8 H2O + 6OH-
We added 6 OH- because there were 6H+

20. Cont.
H+ + OH- → H2O
Combine the hydroxide and hydrogen on the reactant side to make water
2 MnO H2O + 5 H2O2
→ 2 Mn2++ 5 O2+ 8 H2O + 6OH-
Cancel out waters on both sides
2 MnO H2O2
→2 Mn O2 +2 H2O +6OH-

21. Another example In a basic solution MnO4 − + SO32-→MnO4 2− + SO42-
Half reactions
MnO4 − → MnO4 2−
SO32-→ SO42-

22. Half reactions MnO4 − → MnO4 2− MnO4 - + e- → MnO4 2− SO32-→ SO42-
H2O + SO32-→ SO42-
H2O + SO32-→ SO H+
H2O + SO32-→ SO H+ +2e-
Double the top reaction

23. 2 MnO e- → 2 MnO4 2−
H2O + SO32-→ SO H+ +2e-
Combine them
2 MnO4 - + H2O + SO32-
→ 2 MnO4 2− +SO H+
Add OH-
2 MnO4 - + H2O + SO OH-
→ 2 MnO4 2−+SO H++2 OH-

24. 2 MnO4 - + H2O + SO OH-
→ 2 MnO4 2− +SO H2O
finishing
2 MnO4 - + SO OH-
→ 2 MnO4 2− +SO42- + H2O

25. Major Redox Points to Remember.
Any redox reaction can be treated as the sum of the reduction and oxidation half-reactions.
Mass (atoms) and charge are conserved in each half-reaction.
Electrons lost in one half-reaction are gained in the other.
Even though the half-reactions are treated separately, electron loss and electron gain occur simultaneously.

26. Galvanic Cell

27. Galvanic cell
A galvanic cell is a device in which chemical energy is changed to electrical energy.
A galvanic cell uses a spontaneous redox reaction to produce an electric current that can be used to do work. The system does work on the surroundings.
A redox reaction involves the transfer of electrons from the reducing agent to the oxidizing agent.

28. Oxidation v reduction Oxidation. * Involves a loss of electrons.
 * Increase in oxidation number.
 * “To get more positive.”
 * Occurs at the anode of a galvanic cell.
Reduction.
* Involves a gain of electrons.
* Decrease in oxidation number.
* “To get more negative.”
* Occurs at the cathode of a galvanic cell.

29. Production of Current
Oxidation Reactions involve a transfer of electrons.
Electric current is a movement of electrons.
In order to produce a usable current, the electrons must be forced across a set path (circuit).
In order to accomplish this, an oxidizing agent and something to oxidize must be separated from a reducing agent with something to reduce.

30. Pictures
Oxidation reduction reaction in the same container will have electrons transferring, but we can’t harness them.
Separating the oxidation reaction from the reduction reaction, but connecting them by a wire would allow only electrons to flow.
oxidation
reduction
Oxidation
Reduction

31. Closer look
X → X+ + e-
X+ + e- → X
Oxidation
Reduction
We now have excess electrons being formed in the oxidizing solution and a need for electrons in the reducing solution with a path for them to flow through.
However, if electrons did flow through the wire it would cause a negative and positive solution to form.

32. That’s not possible Or at least it would require a lot of energy.
A negative solution would theoretically be formed by adding electrons, and a positive one by removing electrons.
The negative solution would then repel the electrons and stop them from flowing in, and a positive solution would attract the electrons pulling them back where they came from.
Making it so the charged solutions wouldn’t form.
In order for this to work, I would need a way for ions to flow back and forth but keeping the solutions mostly separated.

33. The Salt Bridge or the Porous Disk.
These devices allow ion flow to occur (circuit completion) without mixing the solutions. They are typically made of sodium sulfate or potassium nitrate

34. e-
Closer look e- e-
Salt Bridge
X → X+ + e-
X+ + e- → X
Oxidation
Reduction
Now electrons can flow across the wire from the oxidation reaction to the reduction reaction.
As the oxidation reaction becomes positive, it removes negative ions and adds positive ions to the salt bridge.
The reduction reaction does the reverse.

35. Closer look Zooming in on the oxidizing side
Salt Bridge
- ion
+ ion
Oxidation side e-
+ ion
- ion
- ion
+ ion
+ ion
- ion
Zooming in on the oxidizing side
This would make the salt bridge positive…

36. Closer look (Zooming in on the reducing side)
Salt Bridge
- ion
+ ion
- ion e-
+ ion
- ion
+ ion
- ion
Reducing side
+ ion
(Zooming in on the reducing side)
if the reverse wasn’t happening on this side.

37. Close up of salt bridge
+ ion
+ ion
+ ion
- ion
- ion
- ion
- ion
- ion
+ ion
+ ion
+ ion
- ion
The ions keep flowing in the salt bridge to keep everything neutral.
Electrons do also travel across the salt bridge.
This decreases the cell’s effectiveness.

38. Electrochemical cell This is the basic unit of a battery.
It is also called a galvanic cell, most commercial batteries have several galvanic cells linked together.
Batteries always have two terminals.
The terminal where oxidation occurs is called the anode.
The terminal where reduction occurs is called the cathode.

39. Galvanic Cell

40. Cell Potential (Ecell)
Cell potential (electromotive force, emf) is the driving force in a galvanic cell that pulls electrons from the reducing agent in one compartment to the oxidizing agent in the other.
The volt (V) is the unit of electrical potential.
Electrical charge is measured in coulombs (C).
 A volt is 1 joule of work per coulomb of charge transferred: 1 V = 1 J/C.
A voltmeter is a device which measures cell potential.

41. Standard Reduction Potentials
The measured potential of a voltaic cell is affected by changed in concentration of the reactants as the reaction proceeds and by energy losses due to heating of the cell and external circuit.
 In order to compare the output of different cells, the standard cell potential (Eocell) is obtained at 298 K, 1 atm for gases, 1 M for solutions, and the pure solid for electrodes.

42. The Standard Hydrogen Electrode is considered the reference half-cell electrode, with a potential equal to 0.00 V.
 It is obtained when platinum is immersed in 1 M H+(aq), through which H2(g) is bubbled.

43. The Standard Electrode (Half-Cell) Potential (Ehalf-cell)
A standard electrode potential always refers to the half-reaction written as a reduction.
Oxidized form + n e-  reduced form Eohalf-cell
If you need the oxidation, you will have to reverse the reaction
Reversing a reaction changes the sign of the potential.
Eocell = Eoreduction + Eooxidation

44. Spontaneous reactions
As the potential increases in value (more positive), the reaction is more likely to occur (spontaneity occurs).
Eocell must be positive for the cell to produce electricity.
A substance will have a spontaneous reaction another substance with a lower Eocell.
Although some half-reactions must be manipulated with coefficients, NEVER MULTIPLY THE Eocell BY THE COEFFICIENT!!!

45. Galvanic Cell Problems
Consider a galvanic cell based on the following reactions. For each give the balanced half cell reactions and calculate Eocell.
Cu2+(aq) + Zn(s)  Zn2+(aq) + Cu(s)
Fe3+(aq) + Cu(s)  Fe2+(aq) + Cu2+(aq)
Al (s) + Cd2+ (aq)  Cd(s) + Al3+(aq)

46. Calculating an Unknown Eohalf-cell from Eocell.
A voltaic cell based on the reaction between aqueous Br2 and vanadium (III) ions has
Eocell = 1.39 V:
Br2(aq) +2V3+(aq) + 2H2O(l)
 2VO2+(aq) + 4H+(aq) +2Br-(aq)
What is the standard electrode potential for the reduction of VO2+ to V3+?

47. Line Notations.
The components of the anode compartment are written to the left of the cathode compartment.
Double vertical lines separate the half-cells and represents the wire and salt bridge.
 Within each half-cell, a single vertical line represents a phase boundary. A comma separates half-cell components in the same phase.
Half-cell components appear in the same order as in the half-reaction, while electrodes appear at the extreme left and right of the notation.
Mg(s) Mg2+(aq) Al3+(aq) Al(s)
Fe(s) Fe2+(aq) H+, MnO4-(aq), Mn2+(aq) Pt(s)

48. Describing a Galvanic Cell.
A cell will always run spontaneously in the direction that produces a positive cell potential.
A complete description of a galvanic cell always includes four items:
1. The cell potential (always positive for a galvanic cell) and the balanced cell reaction.
2. The direction of electron flow, obtained by inspecting the half-reactions and using the direction that gives a positive Eocell.

49. 3. Designation of the anode and cathode.
4. 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.

50. Diagramming Voltaic Cells
In one compartment of a voltaic cell, a graphite rod dips into an acidic solution of K2Cr2O7 and Cr(NO3)3; in the other, a tin bar dips into a Sn(NO3)2 solution. A KNO3 salt bridge joins the half-cells. The tin electrode is negative relative to the graphite.
Diagram the cell, show balanced equations, and write the cell notation.

51. Description of a Galvanic Cell.
Describe completely the galvanic cell based on the following half-reactions under standard conditions:
Ag+ + e-  Ag Eocell = 0.80 V (1)
Fe e-  Fe2+ Eocell = 0.77 V (2)
In addition, draw the cell and write the line notation.

52. Cell Potential, Electrical Work
The work that can be accomplished when electrons are transferred (emf) is defined in terms of a potential difference (in volts) between two circuits.
emf = potential difference (V) = work (J)
charge (C)
or 1 V = 1 J/C
Work is viewed from the point of view of the system.
Therefore, E = -w/q or -w = q E

53. Maximum work The maximum work is defined as wmax = -q Emax
*Achieving maximum work is impossible.
In any real, spontaneous process, some energy is always wasted. The actual work realized is always less than the calculated maximum.

54. Free Energy
The Faraday (F) is defined as the charge of 1 mole of electrons.
F = 96,485 C/mol e-
The purpose of a voltaic cell is to convert the free energy change of a spontaneous reaction into the KE of electrons moving through an external circuit.
wmax = DG

55. For a galvanic cell DG = -n FEmax or DGo = -n FEomax
Where n is the number of moles of electrons transferred in the redox reaction.
Note the units of F is C/mol, and the units of E, volts, is J/C
-nFE = (mol)(C/mol )(J/C) = J
This equation is in Joules, not Kilojoules!!

56. Spontaneity
If Ecell > 0, then DG < 0 and the process is spontaneous.
If Ecell < 0, then DG > 0 and the process is nonspontaneous.
If Ecell = 0, then DG = 0 and the process is at equilibrium.

57. Calculating DGo for a Cell Reaction.
Using the Standard Reduction Potential Chart, calculate DGo for the reaction
Cu2+(aq) + Fe(s)  Cu(s) + Fe2+(aq)
Is the reaction spontaneous?
Predict whether 1 M HNO3 will dissolve gold metal to form a 1 M Au3+ solution.
Below is the half reaction of nitric acid acting as an oxidizing agent
4 H+ + NO e-  NO + 2 H2O Eo = .96 V

58. The Relationship to the Equilibrium Constant
DGo = -R T ln K and DGo = -n FEocell
 therefore -R T ln K = -n FEocell
 Eocell = RT ln K /n F
Calculating K and DGo from Eocell
When cadmium metal reduces Cu2+ in solution, Cd2+ forms in addition to copper metal. If DGo = -143 kJ, calculate K at 25o C. What would be Eocell in a voltaic cell that used this reaction?

59. The Effects of Concentration on E
For the cell reaction
2Al(s) + 3Mn2+(aq) ⇌ 2Al3+(aq) + 3Mn(s) Eocell = 0.48 V
predict whether Ecell is larger or smaller than Eocell for the following cases.
a) [Al3+] = 2.0 M, [Mn2+] = 1.0 M
b) [Al3+] = 1.0 M, [Mn2+] = 3.0 M

60. Nernst Equation
The Nernst equation gives the relationship between the cell potential and the concentrations of cell components.
E = Eo – RT ln Q nF
Calculations with this equation have been removed from the AP test, however, qualitative understanding of how concentration affects a cell are still on the test.

61. The Effect of Q
A cell in which the concentrations are not in their standard states will continue to discharge until equilibrium is reached.
Q values are never negative.
For Q values less than 1, the concentration of reactant are higher than product initially, and the reaction will shift to the right to reach equilibrium, increasing the Ecell compared to Eocell.
When Q < 1, [reactant] > [product], and therefore Ecell > Eocell.

62. Continued effect of Q
For Q values higher than 1, the concentration of product are higher than reactant, and the reaction will shift to the left, decreasing the Ecell.
When Q > 1, [reactant] < [product], and therefore Ecell < Eocell.
When Q = 1, [reactant] = [product], Ecell = Eocell.

63. K and Q At equilibrium, K = Q and Ecell = 0.
At equilibrium, the components in the two cell compartments have the same free energy, and DG = 0.
At this point, the cell can no longer do work!

64. Problems Consider a cell based on the reaction
Fe(s) + Cu2+(aq)  Fe2+(aq) + Cu (s)
If [Cu2+] = .30 M, what [Fe2+] is needed to increase Ecell by 0.25 V above Eocell at 25o C?
[Fe2+] would have to be 4.3x10^-9 M, which shows to make large voltage changes you need exceptionally large changes in concentration. Nearst equations problems generally have small changes in electric potential.

65. Describe the cell based on the following half-reactions:
VO H+ + e-  VO H2O
Eo = V
Zn e-  Zn Eo = V
where T = 25o C
[VO2+ ] = 2.0 M
[H+ ] = M
[VO2+ ] = M
[Zn2+ ] = 0.1 M

66. Concentration Cells
Concentration cells are constructed with the exact same half-reactions, with the exception of a difference in concentrations.
Voltages are typically small as electrons are transferred from the cell of higher concentration to the cell of lower concentration.
Eocell = 0.00 V, but these never deal with standard conditions because concentrations are not standard (not 1 M).

67. Calculating the Potential of a Concentration Cell.
A concentration cell is built using two Au/Au3+ half-cells. In half-cell A, [Au3+] = 7.0 x 10-4 M, and in half-cell B, [Au3+] = 2.5 x10-2 M. What is Ecell, and which electrode is the anode?

68. Corrosion

69. Corrosion
Corrosion-An oxidization of a metal, and the oxide flaking off.
Oxidized metal is commonly called rust
Most commonly oxygen will oxidize a metal.
Either by
[Metal] + O2 → [Metal]O Or
[Metal] + H2O → [Metal]O + H2

70. The Electrochemical Corrosion of Iron
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71. Resisting corrosion
Most metals resist corrosion by an oxide layer forming on the outside that protects the metal inside.
It protects the inside metal by preventing the oxygen (or other oxidizing agent) from being able to reach it.

72. Examples Aluminum very readily loses electrons.
You would expect it to “rust” easily.
However, aluminum is a very useful metal because it doesn’t corrode like other metals can.
An aluminum oxide layer forms on the outside, stopping further oxidation from occurring.
This oxide gives aluminum a dull color.

73. Steel
Steel corrodes very readily because iron oxide doesn’t stick to the surface.
It instead falls off exposing new metal to be oxidized.
This makes iron less useful and explains why ancient people would prefer other metals.
However, the abundance and other properties of iron have made it useful.

74. Preventing oxidation
Iron can be protected by painting the surface or coating it with a different material to prevent the corrosion.
Galvanized steel is steel coated with zinc to prevent oxidation.
Zinc actually oxidizes more readily than iron.

75. Galvanic corrosion
Two different metals placed next to each other with an electrolytic solution connecting will cause an oxidation reduction reaction to occur.
Just like the galvanic cell.
Electrons will flow from a more active metal to a less active metal.
One metal will end up oxidizing the other, but in the process will itself become reduced.
This rapidly oxidized or rusts the one metal but prevents the less active metal from oxidizing (rusting)

76. Galvanic corrosion

77. Galvanic corrosion You can also see galvanic corrosion on a battery.
Batteries that are hooked up to a circuit for an extended period of time tend to become rusted.

78. Cathodic Protection
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79. High temperature corrosion
An oxidation reaction like any other reaction occurs faster when heated.
Metals that are constantly heated tend to rust more quickly.

80. Noble metals There are certain metals that don’t form an oxide.
Gold and silver are noble metals.
Silver will oxidize with sulfur, but not with oxygen.
Gold does not readily oxidize in nature.

81. Electrolysis
Electrolysis-Forcing a current through to produce a chemical reaction.
Water can be electrolysized
H2O → H2 + O2
This reaction is very important for fuel cell cars.
It uses electricity to create a combustible fuel for an internal combustion engine.

82. Refining metals
Metals are found as metal oxides (ores) in nature commonly.
An electrolysis reaction is commonly used to produce metals from these ores.
Sodium metal can be produced by melting sodium chloride and passing an electric current through the melt.

83. Hall-Heroult Process Before 1886 aluminum was a very expensive metal.
Even though it is very abundant on the Earth’s surface, it is only found as bauxite, an oxide.
Since aluminum is so reactive no reducing agent could easily turn the ore into a metal.
It was so valuable the Napoleon served his honored guests aluminum silverware and gave the others gold or silver.

84. Charles Hall
A student in a chemistry course at Oberlin College in Ohio was told by his professor, that if anyone could a cheap method to manufacture aluminum from bauxite they could make a fortune.
Using crude galvanic cells Charles Hall was able to achieve this using an electrolysis reaction.
Yes, he did make a fortune with it.

85. Electrolysis
An electrolytic cell uses electrical energy to drive a nonspontaneous process.
 The process is called electrolysis, which involves forcing a current through a cell to produce a chemical reaction for which the cell potential is negative.
Everything is the same as a galvanic cell except the signs of the anode and cathode.

86. Stoichiometry of Electrolysis
Faraday’s Law of Electrolysis: the amount of a substance produced at each electrode is directly proportional to the amount of electric charge flowing through the cell.
 The SI Unit of current is the ampere (A).
1 ampere = 1 coulomb/second or 1 A = 1 C/s
Applying the Relationship Among Current, Time, and Amount of a Substance.

87. Problem
Using a current of 4.75 A, how many minutes does it take to plate 1.50 g Cu onto a sculpture from a CuSO4 solution?

88. Batteries history
Battery- combination on 2 or more electrochemical cells that convert chemical energy into electrical energy.
Luigi Galvini and Allesandro Volta are credited with the invention of the first batteries.
Galvini came up with the galvanic cell. Volta connected them together in a series.
The name battery was coined by Benjamin Franklin, because the batteries at the time were a series of connected jars which reminded him of a battery of cannons.

89. Types of batteries Two major types are:
Wet Cell batteries- use a liquid electrolyte to allow the ions to freely exchange during the redox reaction.
Car batteries or batteries with a liquid inside.
Dry Cell battery- use a paste that immobilizes the electrolyte.
AA, AAA, C, D, 9V etc.

90. The electrolyte This is the salt bridge discussed earlier.
It allows ions to flow freely while the electrons travel across our load, the thing you are trying to power.
The electrolyte normally needs to be acidic or basic to make the redox reaction occur.
Sulfuric acid is commonly used, it is commonly called battery acid.

91. Why not HCl
HCl would be a very poor choice because of the redox reaction
2 HCl → H2 + Cl2
Hydrogen typically gets reduced
2 H+ +2e- → H2
But chlorine getting oxidized is very dangerous
2 Cl- → 2e- + Cl2
Because of the poisonous gas produced.

92. Wet Cell Batteries Car batteries are wet cell batteries.
The obvious problem with these batteries is the need to be keep them upright or the electrolyte, sulfuric acid, will leak out.
However the power they produce is quite substantial.

93. Lead-Acid
The standard battery used in a car was invented in 1859 by Gaston Planté.
It uses a Lead plate and a Lead Dioxide plate in a sulfuric acid solution.
Here is the unbalanced redox reaction
Pb + PbO2 + H2SO4 ⇌ PbSO4
Reduction half
PbO2 + H2SO4 ⇌ PbSO4
Oxidation half
Pb +H2SO4 ⇌ PbSO4

94. One of the Six Cells in a 12–V Lead Storage Battery
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95. Rechargeable
The nice thing about this battery is it is easily rechargeable.
PbSO4 will readily form Pb and PbO2 if electric current is added back to the cell.
This happened completely by chance since there was no practical way to recharge the battery when it was invented.
Later the generator would be invented and from that a car’s alternator and easily recharge the battery while you drive.

96. Alkaline Batteries Normal AA AAA C and D batteries are alkaline.
These are dry cell batteries
The reaction is
Zn + MnO2 →ZnO + Mn2O3
This occurs in a paste of KOH.
This reaction is not reversible!

97. A Common Dry Cell Battery
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98. These may leak if you try to recharge them.

99. Strangely enough
A single AA, AAA, C or D “battery” is not a battery by definition.
They are all single cells.
They are not a battery until you connect them together, like you have to in most devices.
A 9 V battery is a battery because it has 6 cells linked together in the rectangular case.
Car batteries also have 6 cells linked together.

100. Lithium Ion Batteries
Commonly used in cell phones, laptops and other portable electronic devices.
Not to be confused with Lithium single use batteries (like energizer e2).
These batteries are rechargeable.
There use a lithium compound as the cathode and variety of possibilities for the anode material.

101. Li-Ion

102. Lithium Ion Batteries
These batteries are very light for the power the produce
They can be built to a variety of shapes to fit their device.
Over time, the battery will not be able to hold as much of a charge so it will need to be recharged more often.
It will take less time to recharge when this occurs.

103. Other batteries
Zinc-carbon battery - Also known as a standard carbon battery, zinc-carbon chemistry is used in all inexpensive AA, C and D dry-cell batteries. The electrodes are zinc and carbon, with an acidic paste between them that serves as the electrolyte.
Nickel-cadmium battery (NiCd)- The electrodes are nickel-hydroxide and cadmium, with potassium-hydroxide as the electrolyte (rechargeable).
Nickel-metal hydride battery (NiMh)- This battery is rapidly replacing nickel-cadmium because it does not suffer from the memory effect that nickel-cadmiums do (rechargeable).

104. Other batteries
Lithium-iodide battery - Lithium-iodide chemistry is used in pacemakers and hearing aides because of their long life.
Zinc-air battery - This battery is lightweight and rechargeable.
Zinc-mercury oxide battery - This is often used in hearing-aids.
Silver-zinc battery - This is used in aeronautical applications because the power-to-weight ratio is good.

105. A Mercury Battery
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