1. Electrochemistry

2. Electrochemistry Terminology #1
Oxidation – A process in which an element attains a more positive oxidation state
Na(s)  Na+ + e-
Reduction – A process in which an element attains a more negative oxidation state
Cl2 + 2e-  2Cl-

3. Electrochemistry Terminology #2
An old memory device for oxidation and reduction goes like this…
LEO says GER
Lose Electrons = Oxidation
Gain Electrons = Reduction

4. Electrochemistry Terminology #3
Oxidizing agent
The substance that is reduced is the oxidizing agent
Reducing agent
The substance that is oxidized is the reducing agent

5. Electrochemistry Terminology #4
Anode
The electrode where oxidation occurs
Cathode
The electrode where reduction occurs
Memory device:
Reduction
at the
Cathode

6. Table of Reduction Potentials
Measured against the Standard Hydrogen Electrode

7. Measuring Standard Electrode Potential
Potentials are measured against a hydrogen ion reduction reaction, which is arbitrarily assigned a potential of zero volts.

8. Galvanic (Electrochemical) Cells
Spontaneous redox processes have:
A positive
cell potential, E0
A negative free energy change, (-G)

9. Zn - Cu Galvanic Cell Zn2+ + 2e-  Zn E = -0.76V
The less positive, or more negative reduction potential becomes the oxidation…
Zn2+ + 2e-  Zn E = -0.76V
Cu2+ + 2e-  Cu E = +0.34V
Zn  Zn2+ + 2e E = +0.76V
Zn + Cu2+  Zn2+ + Cu E0 = V

10. Line Notation Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) | || |
An abbreviated representation of an electrochemical cell
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
Anode
material
Anode
solution
Cathode
solution
Cathode
material | || |

11. Calculating G0 for a Cell
G0 = -nFE0
n = moles of electrons in balanced redox equation
F = Faraday constant = 96,485 coulombs/mol e-
Zn + Cu2+  Zn2+ + Cu E0 = V

12. The Nernst Equation R = 8.31 J/(molK) T = Temperature in K
Standard potentials assume a concentration of 1 M. The Nernst equation allows us to calculate potential when the two cells are not 1.0 M.
R = 8.31 J/(molK)
T = Temperature in K
n = moles of electrons in balanced redox equation
F = Faraday constant = 96,485 coulombs/mol e-

13. The Nernst Equation R = 8.31 J/(molK) T = Temperature in K
Standard potentials assume a concentration of 1 M. The Nernst equation allows us to calculate potential when the two cells are not 1.0 M.
R = 8.31 J/(molK)
T = Temperature in K
n = moles of electrons in balanced redox equation
F = Faraday constant = 96,485 coulombs/mol e-

14. Equilibrium Constants and Cell Potential
At equilibrium, forward and reverse reactions occur at equal rates, therefore:
The battery is “dead”
The cell potential, E, is zero volts
Modifying the Nernst Equation (at 25 C):

15. Calculating an Equilibrium Constant from a Cell Potential
Zn + Cu2+  Zn2+ + Cu E0 = V

16. Both sides have the same components but at different concentrations.
???
Concentration Cell
Both sides have the same components but at different concentrations.
Anode
Cathode
Step 1: Determine which side undergoes oxidation, and which side undergoes reduction.
The 1.0 M Zn2+ must decrease in concentration, and the 0.10 M Zn2+ must increase in concentration
Zn  Zn2+ (0.10M) + 2e- (oxidation)
Zn2+ (1.0M) + 2e-  Zn (reduction)
Zn2+ (1.0M)  Zn2+ (0.10M)

17. Both sides have the same components but at different concentrations.
Concentration Cell
0.030
???
Concentration Cell
Both sides have the same components but at different concentrations.
Anode
Cathode
Zn2+ (1.0M)  Zn2+ (0.10M)
Step 2: Calculate cell potential using the Nernst
Equation (assuming 25 C).

18. Electrolytic Processes
Electrolytic processes are NOT spontaneous. They have:
A negative
cell potential, (-E0)
A positive free energy change, (+G)

19. Electrolysis of Water In acidic solution Anode rxn: Cathode rxn:
-1.23 V
Cathode rxn:
-0.83 V
-2.06 V

20. Electroplating of Silver
Anode reaction:
Ag  Ag+ + e-
Cathode reaction:
Ag+ + e-  Ag
Electroplating requirements:
1. Solution of the plating metal
2. Anode made of the plating metal
3. Cathode with the object to be plated
4. Source of current

21. Solving an Electroplating Problem
Q: How many seconds will it take to plate out 5.0 grams of silver from a solution of AgNO3 using a 20.0 Ampere current?
Ag+ + e-  Ag
Step 2: Compare moles of silver to moles of electrons
Step 3: coulombs ( C ) are required to move each mole of electrons
Step 1: Convert grams of silver to moles of silver
Step 4: 20.0 amperes is 20.0 coulombs per second
5.0 g
1 mol Ag
1 mol e- C
1 s
1 mol e-
20.0 C g
1 mol Ag
= 2.2 x 102 s

22. How many grams of silver can be plated by applying 10. 0 A for 30
How many grams of silver can be plated by applying 10.0 A for 30.0 min?
30.0 min
60 sec
10.0 C
1 mol e-
1 mol Ag = C
1 mol e-
1 min
1 sec
0.187 mol g
= 20.1 g Ag
1 mol Ag

23. Solving an Electroplating Problem
Q: How many seconds will it take to plate out 5.0 grams of silver from a solution of AgNO3 using a 20.0 Ampere current?
Ag+ + e-  Ag
Step 2: Compare moles of silver to moles of electrons
Step 3: coulombs ( C ) are required to move each mole of electrons
Step 1: Convert grams of silver to moles of silver
Step 4: 20.0 amperes is 20.0 coulombs per second
5.0 g
1 mol Ag
1 mol e- C
1 s
1 mol e-
20.0 C g
1 mol Ag
= 2.2 x 102 s