Electrochemistry

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-

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

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

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

Table of Reduction Potentials Measured against the Standard Hydrogen Electrode

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

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

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 = + 1.10 V

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 | || |

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 = + 1.10 V

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-

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-

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):

Calculating an Equilibrium Constant from a Cell Potential Zn + Cu2+  Zn2+ + Cu E0 = + 1.10 V

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)

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).

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

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

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

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: 96 485 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- 96 485 C 1 s 1 mol e- 20.0 C 107.87 g 1 mol Ag = 2.2 x 102 s

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 = 96 485 C 1 mol e- 1 min 1 sec 0.187 mol 107.87 g = 20.1 g Ag 1 mol Ag

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: 96 485 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- 96 485 C 1 s 1 mol e- 20.0 C 107.87 g 1 mol Ag = 2.2 x 102 s