1. Electrochemistry

3. Electrochemistry Electrochemistry involves either…
generating electricity by harnessing a spontaneous chemical reaction (one with K > 1) OR
using electricity to force a chemical reaction to occur (one that is non-spontaneous, K<1)
Understanding electrochemistry requires a basic understanding of the processes called “oxidation” and “reduction”.

4. Oxidation States
Oxidation states are numbers assigned to atoms that reflect the net charge an atom would have if the electrons in the chemical bonds involving that atom were assigned to the more electronegative atoms.
Oxidation states can be thought of as “imaginary” charges. They are assigned according to the following set of rules:

5. Assigning Oxidation States
Each atom in a pure element has an oxidation state of “0”.
e.g. F2 O3 K N Fe
For mono-atomic ions, the oxidation state is equal to the charge on the ion.
e.g. F- Al3+ Fe2+ O2-
The oxidation state of oxygen in a compound (e.g. CH3OH) is almost always “-2”.
The oxidation state of hydrogen in a compound (e.g. CH3OH) is almost always “+1”.
The sum of the oxidation states in a neutral compound must equal “0”.
The sum of the oxidation states in a complex ion must equal the charge on the complex ion.

6. Examples - assigning oxidation numbers
Assign oxidation states to all elements:

7. More Definitions Oxidizing Agent Reducing Agent
the substance in a chemical reaction which causes another species to be oxidized.
the oxidizing agent always gets reduced in the reaction.
Reducing Agent
the substance in a chemical reaction that causes another species to be reduced.
the reducing agent always gets oxidized!

8. Examples - labeling redox reactions
In each reaction, look for changes in oxidation state.
If changes occur, identify the substance being reduced, and the substance being oxidized.
Identify the oxidizing agent and the reducing agent.
5 Fe2+ + MnO H+ ® 5 Fe3+ + Mn H2O
H2 + CuO ® Cu + H2O
Zn HCl ® ZnCl2 + H2

9. Example - Balancing with Half-Reactions
Fe2+ + Cr2O72- ® Fe3+ + Cr3+
Fe2+ ® Fe3+
Fe2+ ® Fe3+ + e-
6 Fe2+ ® 6 Fe e-
Cr2O72- ® Cr3+
Cr2O72- ® 2 Cr3+
Cr2O72- ® 2 Cr H2O
Cr2O H+ ® 2 Cr H2O
Cr2O H+ + 6 e- ® 2 Cr H2O
Cr2O Fe H+ ® 2 Cr Fe H2O

10. Example - Balancing with Half-Reactions
Fe2+ + Cr2O72- ® Fe3+ + Cr3+
Fe2+ ® Fe3+
Fe2+ ® Fe3+ + e-
6 Fe2+ ® 6 Fe e-
Cr2O72- ® Cr3+
Cr2O72- ® 2 Cr3+
Cr2O72- ® 2 Cr H2O
Cr2O H+ ® 2 Cr H2O
Cr2O H+ + 6 e- ® 2 Cr H2O
Cr2O Fe H+ ® 2 Cr Fe H2O

11. Example - Balancing with Half-Reactions
Fe2+ + Cr2O72- ® Fe3+ + Cr3+
Fe2+ ® Fe3+
Fe2+ ® Fe3+ + e-
6 Fe2+ ® 6 Fe e-
Cr2O72- ® Cr3+
Cr2O72- ® 2 Cr3+
Cr2O72- ® 2 Cr H2O
Cr2O H+ ® 2 Cr H2O
Cr2O H+ + 6 e- ® 2 Cr H2O
Cr2O Fe H+ ® 2 Cr Fe H2O

12. Example - Balancing with Half-Reactions
Fe2+ + Cr2O72- ® Fe3+ + Cr3+
Fe2+ ® Fe3+
Fe2+ ® Fe3+ + e-
6 Fe2+ ® 6 Fe e-
Cr2O72- ® Cr3+
Cr2O72- ® 2 Cr3+
Cr2O72- ® 2 Cr H2O
Cr2O H+ ® 2 Cr H2O
Cr2O H+ + 6 e- ® 2 Cr H2O
Cr2O Fe H+ ® 2 Cr Fe H2O

13. What if the solution was basic?
Notice that the method has assumed the solution was acidic - we added H+ to balance the equation. The [H+] in a basic solution is very small. The [OH-] is much greater.
For this reason, we will add enough OH- ions to both sides of the equation to neutralize the H+ in the overall reaction.
The hydrogen and hydroxide ions will combine to make water, and you may have to do some canceling before you’re done.
Cr2O Fe H2O ® 2 Cr Fe OH-
Why is this reaction unlikely in basic solution?

14. Balancing Redox Equations Practice
Balance in acidic solution:
H2C2O4 + MnO4- ® Mn2+ + CO2
5 H2C2O4 + 2 MnO H+ ® 2 Mn CO2 + 8 H2O
Balance in basic solution:
CN- + MnO4- ® CNO- + MnO2
3 CN- + 2 MnO4- + H2O ® 3 CNO- + 2 MnO2 + 2 OH-

15. Balancing Redox Equations Practice
Balance in acidic solution:
H2C2O4 + MnO4- ® Mn2+ + CO2
5 H2C2O4 + 2 MnO H+ ® 2 Mn CO2 + 8 H2O
Balance in basic solution:
CN- + MnO4- ® CNO- + MnO2
3 CN- + 2 MnO4- + H2O ® 3 CNO- + 2 MnO2 + 2 OH-

16. Balancing Redox Equations Practice
Balance in acidic solution:
H2C2O4 + MnO4- ® Mn2+ + CO2
5 H2C2O4 + 2 MnO H+ ® 2 Mn CO2 + 8 H2O
Balance in basic solution:
CN- + MnO4- ® CNO- + MnO2
3 CN- + 2 MnO4- + H2O ® 3 CNO- + 2 MnO2 + 2 OH-

17. Balancing Redox Equations Practice
Balance in acidic solution:
H2C2O4 + MnO4- ® Mn2+ + CO2
5 H2C2O4 + 2 MnO H+ ® 2 Mn CO2 + 8 H2O
Balance in basic solution:
CN- + MnO4- ® CNO- + MnO2
3 CN- + 2 MnO4- + H2O ® 3 CNO- + 2 MnO2 + 2 OH-

18. Balancing Redox Equations Practice
Balance in acidic solution:
H2C2O4 + MnO4- ® Mn2+ + CO2
5 H2C2O4 + 2 MnO H+ ® 2 Mn CO2 + 8 H2O
Balance in basic solution:
CN- + MnO4- ® CNO- + MnO2
3 CN- + 2 MnO4- + H2O ® 3 CNO- + 2 MnO2 + 2 OH-

19. Redox Reactions - What’s Happening?
Zinc is added to a blue solution of copper(II) sulfate
The blue colour disappears…the zinc metal “dissolves”, and solid copper metal precipitates on the zinc strip
The zinc is oxidized (loses electrons)
The copper ions are reduced (gain electrons)
Zn (s) + CuSO4 (aq)  ZnSO4 (aq) + Cu (s)
Zn (s) + Cu2+ (aq)  Zn2+ (aq) + Cu (s)

20. Copper ions (Cu2+)
collide with the zinc
metal surface
A zinc atom (Zn)
gives up two of its
electrons to the
copper ion
The result is a
neutral atom of Cu
deposited on the
zinc strip, and a
Zn2+ ion released
into the solution

21. Harnessing the Electricity
During the spontaneous redox reaction, electrons flow from the zinc atoms to the copper ions.
Zn ® Zn e- (oxidation)
Cu e- ® Cu (reduction)
Electricity can be thought of as the “flow of electric charge”.
How can we make use of the flow of electrons from the zinc atoms to the copper ions??
Answer: SEPARATE the copper ions from the zinc atoms. This will force the electrons from the zinc atoms to travel through an external path to reach the copper ions.

22. Electrochemical Cells Definitions
Voltaic/Galvanic Cell
An electorchemical cell/ device that uses a spontaneous redox reaction to produce electricity
Anode
the electrode where oxidation occurs
Cathode
the electrode where reduction occurs
Salt Bridge
connects two “half-cells” to complete the electric circuit.
for example, a U-tube filled with salt solution
Electrochemical Cells Definitions

23. Electrochemical Cells

25. Cell Potential
Water flows spontaneously over a waterfall because of a difference in potential energy between the top of the falls and the stream below.
In a similar fashion, electrons flow from the anode of a voltaic cell to the cathode because of a difference in potential energy.
The potential energy of electrons is higher in the anode than in the cathode, and they spontaneously flow through an external circuit from the anode to the cathode.
The potential difference between the two electrodes of an electrochemical cell provides the driving force that pushes electrons through the external circuit.
We call this potential difference, denoted Ecell, the cell potential.
Because Ecell is measured in volts, we often refer to it as the cell voltage. For any cell reaction that proceeds spontaneously, such as that in a voltaic cell, the cell voltage will be positive.

26. Standard Hydrogen Electrode (SHE)
It is impossible to measure the potential of a single electrode.
Using a voltmeter, we can measure the difference in potential between two electrodes.
Chemists arbitrarily assigned a potential of 0 V for the SHE.
By measuring the difference in potential between the SHE and other electrodes, potentials can then be assigned to other electrodes.
2 H+ + 2 e- ® H E° = 0 V

27. When a zinc electrode is connected to the SHE as shown above, the voltmeter reads: E°cell = + 0.76 V
The positive sign means electrons are flowing from zinc to the hydrogen cell. The zinc is being oxidized.
Since E°SHE= 0 V, we can conclude that E°Zn = V . This is the “oxidation potential” for the reaction: Zn ® Zn e-

28. Table of Standard Reduction Potentials

29. Calculating a Cell Potential
Step 1:
Which electrode is more likely to be reduced? This is the cathode.
Step 2:
Write half-reactions each half-cell - one reduction and one oxidation reaction. Include the values for E°red and E°ox from the table of standard reduction potentials.
Remember: E°red = -E°ox
Step 3:
Calculate the cell potential:
E°cell = E°red + E°ox

30. Step 1:
Since Cu has a higher reduction potential than Zn, copper forms the cathode. We could also argue that Zn has a higher oxidation potential…and forms the anode.
Step 2:
Cathode:
Cu2+ + 2e- ® Cu
E°red = V
Anode:
Zn ® Zn e-
E°ox = V
Step 3:
The overall reaction is:
Zn + Cu2+  Zn2+ + Cu
E°cell = E°red + E°ox = = 1.10 V Cu Zn

31. Zn | Zn2+ (1.0 M) || Cu2+ (1.0 M) | Cu
Line Notation
An electrochemical cell needs to be described in a more convenient way than drawing a diagram!
We use “line notation” to describe a cell. The zinc-copper standard cell is described:
Zn | Zn2+ (1.0 M) || Cu2+ (1.0 M) | Cu
The ANODE is described before the CATHODE.
Concentrations of ions are indicated in brackets.
A vertical line represents a phase boundary.
A double vertical line represents the salt bridge.

32. Line Notation - an example
Calculate the cell potential for the cell described below:
Mg | Mg2+ (1.0 M) || Ag+ (1.0 M) | Ag
Anode: Mg ® Mg e- E° = 2.37 V
(oxidation)
Cathode: Ag+ + e- ® Ag E° = 0.80 V
(reduction)
Overall: Mg + 2 Ag+  Mg Ag E° = 3.17 V
Review: What’s the oxidizing agent? the reducing agent?
What’s E° for the reverse reaction?

33. Link to animation
Click! And turn up the volume

34. The Nernst Equation
So far, the cells we have considered have operated under standard conditions M solutions and 1.0 atm pressures at 25°C.
Cell potential is dependent on concentration and on temperature, described by the Nernst Equation:
where “E°“ is the standard cell potential
“R” is the Ideal Gas Constant, 8.314
“T” is the temperature of the cell
“n” is the number of moles of electrons transferred
“F” is the Faraday, the charge on one mole of electrons, C/mol
“Q” is the reaction quotient (remember your equilibrium unit!)

35. Using the Nernst Equation
Calculate Ecell for the cell described below at 25°C
Mg | Mg2+ ( M) || Ag+ (2.0 M) | Ag
According to the balanced equation,
We have already seen that E°cell = 3.17 V
According to our balanced equation, n = 2 mol e-
Substituting into the Nernst Equation:
Thus, Ecell = 3.26 V

36. Calculating Equilibrium Constants
When an electrochemical cell operates, the concentrations of the ions change until Q = K .
When the cell reaches equilibrium, Ecell = 0.
Combining these facts with the Nernst equation gives:
Rearranging, we can derive an equation to calculate the equilibrium constant for a redox reaction:

37. Calculating K for a Redox Reaction
Calculate K for the reaction that occurs when Mg is added to a solution of AgNO3.
The net reaction is:
Mg + 2 Ag+  Mg Ag

38. Calculating K for a Redox Reaction
Calculate K for the reaction that occurs when Mg is added to a solution of AgNO3.
The net reaction is:
Mg + 2 Ag+  Mg Ag
247 = ln (K) so K = e247 = HUGE

39. Corrosion Oxidation of metals with oxygen to form a metal oxide
In some cases a thin layer of oxide coats the metal surface, preventing further oxidation (Al2O3)
When iron corrodes, the process is called “rusting”. The iron oxide (Fe2O3) flakes off, exposing more metal to corrosion.

40. Galvanized Iron - Preventing Corrosion
Iron can be protected from corrosion by coating it with zinc metal - a metal that will oxidize first, instead of the iron
Note that E°oxidation of zinc is greater than that of iron.

41. Cathodic Protection - Preventing Corrosion
To protect underground
pipelines, a sacrificial
anode is added.
The water pipe is turned into the cathode and an active metal is used as the sacrificial anode.
Magnesium is used as the sacrificial anode, since it is more easily
oxidized than the
iron water pipe.

42. Dry-Cell Batteries
Zn + 2 MnO2 + 2 NH4+ ® Zn+2 + Mn2O3 + 2 NH3 + H2O

43. The Alkaline Battery Zn(s) + 2 MnO2(s) ---> ZnO(s) + Mn2O3(s)
(KOH is the electrolyte)
Zn(s) MnO2(s) ---> ZnO(s) + Mn2O3(s)

44. Lead Storage Battery (Car Battery)
Anode:
Pb (s) + HSO4- (aq) 
PbSO4 (s) + H+ (aq) + 2 e- Eo = V
Cathode:
PbO2 (s) + 3 H+ (aq) + HSO4- (aq) + 2 e- 
PbSO4(s) + 2 H2O () Eo = V
Overall:
Pb (s) + PbO2 (s) + 3 H+ (aq) + 2HSO4 
2 PbSO4(s) + 2 H2O () Eo = V

45. A Picture of a Car Battery
A 12-V battery has six cells connected in series.
Each cell generates 2 V.

46. Fuel Cells Anode: 2 H2 (g) + 4 OH- (aq) 2 H2O () + 4 e-
Cathode: O2 (g) + 2 H2O () + 4 e-  4 OH- (aq)
Overall: 2 H2 (g) + O2 (g)  2 H2O ()

47. Electrolysis of Molten Sodium Chloride
Anode: 2 Cl- ()  Cl2 (g) + 2 e-
Cathode: 2 Na+ () + 2 e-  Na()
Overall: 2 Cl- () 2 Na+ ()  Cl2 (g) + Na()
Molten sodium
metal is formed
at the cathode.
Chlorine gas
is formed at the
anode.

48. Electrolysis of Water Anode: 2 H2O  O2 + 4 H+ + 2 e-
Cathode: 4 H2O + 4 e-  2 H2 + 4OH-
Overall:
6 H2O  O2 + 2 H2 + 4 OH- + 4 H+
Since 4 OH- + 4 H+ ® 4 H2O
Net:
2 H2O  O2 + 2 H2 H2 O2
H2O
cathode
anode
Na2SO4

49. Electrolysis of Aqueous Sodium Bromide
Two possible reactions at the cathode:
2 H2O + 2 e-  H2 +2OH- E° = V
Na+ + e- ® Na E° = V
Two possible reactions at the anode:
2 Br-  Br2 + 2 e- E° = V
2 H2O ® O2 + 4 H+ + 4 e- E° = V
Overall reaction: 2 Br- + 2 H2O  Br2 + H2 + 2 OH-
Eocell = Eored + Eoox = (-1.09)
Eocell = V

51. Quantitative Electrolysis
Q = I t
Charge = Current x time
(Coulombs) (Amperes) (seconds)
1 F = C/mol e-

52. What mass of aluminum can be produced in 8
What mass of aluminum can be produced in 8.00 min by passing a constant current of 100 A through a molten mixture of aluminum oxide, Al2O3?

53. What mass of aluminum can be produced in 8
What mass of aluminum can be produced in 8.00 min by passing a constant current of 100 A through a molten mixture of aluminum oxide, Al2O3?
Al e-  Al (s)

54. What mass of aluminum can be produced in 8
What mass of aluminum can be produced in 8.00 min by passing a constant current of 100 A through a molten mixture of aluminum oxide, Al2O3?
Al e-  Al (s)

55. = 4.47 g of Aluminum will be produced
What mass of aluminum can be produced in 8.00 min by passing a constant current of 100 A through a molten mixture of aluminum oxide, Al2O3?
Al e-  Al (s)
= 4.47 g of Aluminum will be produced