1. Chapter 20
Electrochemistry

2. Voltaic Cells A chemical reaction can perform two types of work:
Produce a gas to perform PV work
Use movement of electrons from redox reactions to perform electrical work

3. Voltaic Cells
A voltaic (galvanic) cell is a device in which the transfer of electrons takes place through an external pathway rather than directly between reactants.
By physically separating the reduction half of a redox reaction from the oxidation half, we create a flow of electrons through an external circuit.
Used to accomplish electrical work

4. Voltaic Cells
Two solid metals that are connected by the external circuit are called electrodes.
Anode:
Cathode:
Electrodes may or may not participate in the reaction
Zn/Cu
Pt or other conducting material
-activity with conductivity testers
-Zn will participate and eventually dissolve and the copper electrode will have more mass, Pt just provides a reaction surface –will not gain or lose mass

5. Voltaic Cells
Each of the two components of a voltaic cell is called a half-cell
Oxidation half-cell
Reduction half-cell
For a voltaic cell to work, the solutions in the two half-cells must remain electrically neutral
Need migration of ions
Salt bridge or porous glass barrior
-define salt bridge

6. Voltaic Cells Anode  Acceptor of electrons: Oxidation
Cathode Source of electrons: Reduction
Anions always migrate toward the anode and cations toward the cathode.
-electrons migrate clockwise

7. Voltaic (Galvanic) Cell
--draw one on the board labeling the parts
Interactive:::

8. Describing a Voltaic Cell
The following redox reaction is spontaneous:
Cr2O72-(aq)+ 6I-(aq)  2Cr3+(aq) + 3I2(s)
A solution containing K2Cr2O7 and H2SO4 is poured into one beaker, and a solution of KI is poured into another. A salt bridge is used to join the beakers. A metallic conductor that will not react with either solution is suspended in each solution, and the two conductors are connected with wires through a voltmeter to detect an electric current. The resultant voltaic cell generates an electric current. Indicate the reaction occurring at the anode, the reaction at the cathode, the direction of electron migration, the direction of ion migration, and the signs of the electrodes.
-draw it?

9. Describing a Voltaic Cell
The two half-reactions in a voltaic cell are
Zn(s)  Zn2+(aq) + 2e-
ClO3-(aq) + 6H+(aq) + 6e-  Cl-(aq) + 3H2O(l)
(a) Indicate which reaction occurs at the anode and which at the cathode. (b) which electrode is consumed in the cell reaction? (c) Which electrode is positive?

10. Cell EMF Under Standard Conditions
Why do electrons transfer spontaneously during redox reactions?
Electrons flow from the anode of a voltaic cell to the cathode because of a difference in potential energy.
Potential energy higher at the anode
Electrons flow spontaneously toward the electrode with the more positive electrical potential.
-same reason as water flows down a water fall- difference in potential energy
-potential energy at the anode higher (like top of water fall) so electrons want to flow to the cathode
Emf animation:

11. Cell EMF Under Standard Conditions
The difference in potential energy per electrical charge between two electrodes is measured in units of volts.
1V = 1 (J/C)
Where V (volts), J (joule), and C (coulomb)

12. Cell EMF Under Standard Conditions
The potential difference between two electrodes provides a driving force that pushes electrons through the external circuit.
Electromotive Force (emf)
Emf of a cell is denoted as Ecell (the cell potential)
For spontaneous reactions, the cell potential will be positive
“electron motion” force
-measured in volts

13. Standard EMF Emf depends on
The particular cathode and anode half reactions
Concentrations of the reactants and products
Temperature
Tabulated values of standard reduction potentials denoted Eored to calculate Eocell
Eocell = Eored (cathode) – Eored (anode)
Standard: Eocell: 25C, 1atm, 1M
-only the potential associated with each electrode is chosen to be the potential for reduction to occur at that electrode.

14. Standard Emf
Indirectly measure the standard reduction potential of a half-reaction
Reference point:
2H+ (aq, 1 M) + 2e-  H2 (g, 1 atm)
Assigned a standard reduction potential of exactly zero volts
Called a standard hydrogen electrode (SHE)
SHE – uses inert Pt electrode
-example in book pg . 857

15. Standard Emf
When determining standard reduction potentials from other half-reactions, write the reaction as a reduction even though it is “running in reverse” as an oxidation reaction.
Whenever an electrical potential is assigned to a half-reaction, write the reaction as a reduction.
Eored are intensive properties
Intensive- because of ratio potential per electrical charge, therefore increasing moles will not affect value

16. Calculating Eored from Eocell
For the Zn-Cu2+ voltaic cell, we have
Zn + Cu2+  Zn2+ + Cu Eocell = 1.10V
Given that Eored of Zn2+ to Zn is V, calculate the Eored for the reduction of Cu2+ to Cu

17. Calculating Eored from Eocell
A voltaic cell is based on the half-reactions:
In+  In3+ + 2e-
Br2 + 2e-  2Br-
The standard emf for the cell is 1.46V and Eored for the reduction of bromine is +1.06V. Using this information, calculate Eored for the reduction of In3+ to In+.

18. Calculating Eocell from Eored
Using the standard reduction potentials listed in Table 20.1, calculate the standard emf for the following voltaic cells:
Cr2O H+ + 6I-  2Cr3+ + 3I2 + 7H2O
2Al + 3I2  2Al3+ + 6I-

19. Standard EMF
For each of the half-cells in a voltaic cell, the standard reduction potential provides a measure of the driving force for reduction to occur.
The more positive the value of Eored, the greater the driving force for reduction under standard conditions.
The more positive Eored value identifies the cathode
-if both reactions are written as reduction and need to calc standard cell ptential- look at more positive number for the reaction assigned to the cathode

20. Determining Half-Reactions at Electrodes
A voltaic cell is based on the following two standard half-reactions:
Cd2+ + 2e-  Cd
Sn2+ + 2e-  Sn
By using your chart, determine (a) the half-reaction that occurs at the cathode and the anode, and (b) the standard cell potential
-see extra example on pg 860 if needed

21. Strengths of Oxidizing and Reducing Agents
Use Eored values to understand aqueous reaction chemistry
The more positive the Eored value for a half-reaction, the greater the tendency for the reactant of the half-reaction to be reduced and, therefore, to oxidize the other species.
Better oxidizing agent
Write reductions for F2 and H+, F2 is the best oxidizing agent
-best oxidizers are halogens and oxyanions on pg 860

22. Strengths of Oxidizing and Reducing Agents
The half-reaction with the smallest reduction potential is most easily reversed as an oxidation.
The more negative the Eored, the stronger the ability to act as the reducing agent
Show example of Li+, so small---oxidation here is highly favored , in water Li (the product) is the strongest reducing agent
(because equations are all listed as reductions) Only the reactant side can serve as oxidizers and the products on the right as reducers

23. Strengths of Oxidizing and Reducing Agents
-relative strength chart
-relate to strength of conjugate acids and bases
-as Ered becomes more positive, the species on the left becomes stronger and stronger oxidizing agents. And Ered becomes more negative, the species on the right become stronger and stronger reducing agents

24. Determining the Relative Strengths of Oxidizing Agents
Using standard reduction potentials:
Rank the following ions in order of increasing strength as oxidizing agents: NO3-, Ag+, Cr2O72-
Rank the following species from the strongest to the weakest reducing agent: I-, Fe, Al

25. Free Energy and Redox Reactions
Determining the spontaneity of redox reactions
Eo = Eored (reduction process) – Eored (oxidation process)
-any reaction that can occur in a voltaic cell to produce a positive emf must be spontaneous.
-value for spontaneity? (positive)
-modified equation

26. Spontaneous or Not?
Use standard reduction potentials to determine whether the following reactions are spontaneous under standard conditions.
Cu + 2H+  Cu2+ + H2
Cl2 + 2I-  2Cl- + I2

27. Spontaneous or Not?
Use standard reduction potentials to understand the activity series of metals
Activity series of metals: strongest reducing agent at the top
Calculate standard emf for
Ni + 2Ag+  Ni2+ + 2Ag
Value, positive, indicates that the displacement of silver by nickel is a spontaneous process

28. EMF and ΔG Relationship between G and EMF: ΔG = -nFE
Where n = number of electrons transferred n the reaction, G = Gibbs free energy, E = EMF, and F = Faraday’s constant
Faraday’s constant is the quantity of electrical charge on one mole of electrons (a faraday)
1 F = 96,485 C/mol = 96,485 J/V-mol
G is measure of spontaneity of a process that occurs at constant T and P
-positive E, n always positive, F always constant therefore G is negative (as suspected to have everything spontaneous)
-if under standard conditions: rewrite with degree signs
**relate to K

29. Determining ΔGº and K
(a) Use standard reduction potentials to calculate ΔGº and K at 298K for the reaction: 4Ag + O2 + 4H+  4Ag+ + 2H2O (b) Suppose the reaction in part (a) was written: 2Ag + ½ O2 + 2H+  2Ag+ + H2O What are values of Eº, ΔGº, and K when the reaction is written this way?

30. Determining n and K For the reaction
3 Ni2+ + 2Cr(OH)3 + 10OH-  Ni + 2CrO H2O
(a) What is the value of n? (b) Given that ΔGº equals +87 kJ/mol, calculate K at a temperature of 298K

31. Cell EMF Under Nonstandard Conditions
As a voltaic cell is discharged , the reactants of the reaction are consumed and the products are generated
The concentrations of these substances changes
EMF drops until E = 0, and the concentration of reactants and products are at equilibrium
How does cell emf depend on the concentration of reactants and products?

32. The Nernst Equation Dependence of cell emf on concentration
At 298K with units of volts, the equation simplifies to:

33. The Nernst Equation
The Nernst equation helps us understand why the emf of a voltaic cell drops as the cell discharges
Increasing the concentration of reactants or decreasing the concentration of products increases the driving force (higher emf)
Decreasing the concentration of reactants or increasing the concentration of the products decreases the driving force (lower emf)
Discuss mathmatical equation of changing Q
-what happens as reactants converted to products? Increase products, increase Q, lower E towards equilibrium

34. Voltaic Cell EMF Under Nonstandard Conditions
Calculate the emf at 298K generated by:
Cr2O72-(aq)+ 14H+(aq) + 6I-(aq)  Cr3+(aq) + 3I2(s) + 7H2O(l)
When [Cr2O72-] = 2.0M, [H+] = 1.oM, [I-] = 1.0M, [Cr3+] = 1.0x10-5M

35. Voltaic Cell EMF Under Nonstandard Conditions
Calculate the emf at 298K generated by:
2Al (s)+ 3I2(s) 2Al3+ (aq) + 6I- (aq)
When [Al3+ ]= 4.0x10-3M and [I- ]0.010M

36. Calculating Concentrations in a Voltaic cell
If the voltage of the following cell is +0.45V at 298K when [Zn2+] = 1.0M and PH2= 1atm, what is the concentration of H+?
Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g)

37. Batteries and Fuel Cells
A battery is a portable, self-contained electrochemical power source that consists of one or more voltaic cells.
When cells are connected in series, the battery produces a voltage that is the sum of the emfs of the individual cells.
Multiple cells in series
Multiple batteries in series
-productvie use of spontaneous redox reactions
Greater voltages can be achieved by using multiple voltaic cells in a single battery.
Series- cathode of one attached to the anode of another

38. Batteries and Fuel Cells
The substances that are oxidized at the anode and reduced at the cathode determine the emf of the battery.
The usable life of the battery depends on the quantities of these substances.
Need a porous barrier between anode and cathode compartments
Primary and Secondary batteries
Primary batteries- cannot be recharged –must be thrown away when emf=0
Secondary batteries may be recharged from an external power source

39. Batteries and Fuel Cells
Lead-Acid Battery (12-v car battery, 6 voltaic cells in series that each produce 2V)
Alkaline Battery (most common primary battery)
Nickel-Cadmium, Nickel-Metal-Hydride, and Lithium-Ion Batteries (secondary batteries)
Cadmium- heavy metal not good for the environment (toxic), increases the weight of the battery
Li- lightweight, greater energy density

40. Lead-Acid Batteries

41. Corrosion
Corrosion reactions are spontaneous redox reactions in which a metal is attacked by some substance in its environment and converted to an unwanted compound.
Oxidation is a thermodynamically favored process in air at room temperature
Undesirable redox reactions
-if not inhibited, very destructive

42. Corrosion
Prevent corrosion by forming a protective oxide layer that is impermeable to O2 and H2O
Examples:
Al3+ forms protective Al2O3 layer Mg

43. Corrosion of Iron Rusting of iron requires both oxygen and water
pH of solution, presence of salts, contact with metals more difficult to oxidize than iron, and stress on the iron can accelerate rusting
-20% iron produced in US is made to replace iron objects subjected to rust damage
-one section of metal acts as anode, another as cathode
-rust forms at cathode where oxygen is more readily available
-anode is corroded (pitted)
-adding salt is like having a salt bridge
-galvanized iron– have a Zn coating, Zn is easier to oxidize and becomes anode- if barrier is broken, zinc is corroded, not iron
-can paint or coat iron with another metal to help protect it

44. Corrosion of Iron
Cathodic protection: protecting a metal from corrosion by making it the cathode in an electrochemical cell.
The metal that is oxidized while protecting the cathode is called the sacrificial anode.

45. Corrosion of Iron
Cathodic protection