1. Electrolysis & Understanding Electrolytic Cells : When a non-spontaneous redox reaction is made to occur by putting electrical energy into the system. The battery (energy source) acts as a “pump” pushing electrons into the cathode and removing electrons from the anode. To maintain electrical neutrality, a redox reaction must occur within the cell consume electrons at the cathode - Reduction liberate electrons at the anode - Oxidation

2. Galvanic –vs-Electrolytic Cells:

3. Electrolysis Cell” Anode Cathode Ions present for current to flow DC voltage- with high enough voltage, chemical reactions will occur at the two electrodes.

4. Electrolysis of molten state Application: purification of metals Example: NaCl(l) -achieved only at 800°C. Na + attracted to cathode (-) and undergoes reductions. Cl - is attracted to the anode (+) and undergoes oxidation. 2Na + + 2e -  2Na(l) 2Cl -  2e - + Cl 2 (g)_____ 2Na + + 2Cl -  2Na(l) + Cl 2 (g)

5. Electrolysis of Aqueous Solutions Electrolysis in aqueous solutions also includes the presence of H 2 O which may undergo either oxidation or reduction (depending on energy requirements) Species present: [Na +, Cl -, H 2 O] Possible Reduction: Na + (aq) + e -  Na(s) 2H 2 O(l) + 2e -  H 2 (g) + 2OH - (aq) Possible Oxidation: 2Cl - (aq)  Cl 2 (g) + 2e - 2H 2 O(l)  2H 2 (g) + O 2 (g) + 4e - 2H 2 O(l) + 2Cl - (aq)  H 2 (g) + 2OH - (aq) +Cl 2 (g) Since the process is NOT spontaneous, E must have a net (-) value. Compare E(V) for each half reaction to determine what is occurring at each electrode. This cell is unique when we compare the oxidation of Cl- & H 2 O

6. Electroplating

7. Electrolysis and Electroplating Electric current is passed through a solution containing a salt of the metal to be plated. The object to be plated is the cathode and the metal ion is reduced on its surface.

8. Calculations & electroplating By knowing the # of moles e- that are required and the current flow/time one is able to calculate the mass of metal plated. Using a solution containing Ag + (aq) ions, metallic silver is deposited on the cathode. A current of 1.2A is applied for 2.4 hours. What is the mass of silver formed? (Useful conversions provided) Charge: 2.4hrs 3600s 1.12A = 9675.8C 1 hr Mass of Ag: 9676.8 C 1 mole e- 1 mole Ag(s) 107.9g 96,485C 1 mole e- 1mole Ag Answer: 10.8g

9. Useful Relationships: Used to relate electricity through an electrolytic cell and the amount of substances produced by the redox process. QuantityUnitRelationshipConversion Factor ChargeCoulomb (C)1C = 1A·S = 1J/V 1 mole e- = 96,480C CurrentAmpere (A)1A = 1C/s PotentialVolt (V)1V = 1J/C PowerWatt (W)1W = 1 J/s EnergyJoule (J)1J = 1V·C

10. Sample Problem: A current of 2.20A is passed through a solution containing Pb 2+ for 2.00 hours, with lead metal being deposited at the cathode. What mass of lead is deposited? 2.00 hr. 60 min. 60 sec. 2.20C 1mole e- 1mole Pb(s) 207.2 g Pb 1 hr. 1 min. S 96,500C2 mole e - 1mole Pb = 17.0g Pb

11. Sample Problem: Chromium metal can be electroplated from a water solution of potassium dichromate; the reduction half reaction is: Cr 2 O 7 2- (aq) + 14H + (aq) + 12 e -  2 Cr(s) + 7 H 2 O(l) How many grams of chromium will be plated by 1.00x10 4 C? ( Strategy: Coulombs  mole e-  mole Cr  mass Cr) Ans. = 0.898g Cr