Electrolysis & Understanding Electrolytic Cells
Electrolytic cells involve non-spontaneous redox reactions driven by electrical energy. Electroplating utilizes electric current to deposit metal ions onto a surface. Learn about electrolysis, galvanic vs. electrolytic cells, and applications like purification of metals. Understand the calculations involved in electroplating processes.
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Presentation Transcript
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
Electrolysis Cell DC voltage- with high enough voltage, chemical reactions will occur at the two electrodes. Ions present for current to flow Cathode Anode
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- + Cl2(g)_____ 2Na+ + 2Cl- 2Na(l) + Cl2(g)
Electrolysis of Aqueous Solutions Electrolysis in aqueous solutions also includes the presence of H2O which may undergo either oxidation or reduction (depending on energy requirements) Species present: [Na+, Cl-, H2O] Possible Reduction: Na+(aq) + e- Na(s) 2H2O(l) + 2e- H2(g) + 2OH-(aq) Possible Oxidation: 2Cl-(aq) Cl2(g) + 2e- 2H2O(l) 2H2(g) + O2(g) + 4e- 2H2O(l) + 2Cl-(aq) H2(g) + 2OH-(aq) +Cl2(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- & H2O
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.
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
Useful Relationships: Used to relate electricity through an electrolytic cell and the amount of substances produced by the redox process. Quantity Unit Relationship Conversion Factor Charge Coulomb (C) 1C = 1A S = 1J/V 1 mole e- = 96,480C Current Ampere (A) 1A = 1C/s Potential Volt (V) 1V = 1J/C Power Watt (W) 1W = 1 J/s Energy Joule (J) 1J = 1V C
Sample Problem: A current of 2.20A is passed through a solution containing Pb2+ 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 96,500C 2 mole e- 1mole Pb = 17.0g Pb
Sample Problem: Chromium metal can be electroplated from a water solution of potassium dichromate; the reduction half reaction is: Cr2O72-(aq) + 14H+(aq) + 12 e- 2 Cr(s) + 7 H2O(l) How many grams of chromium will be plated by 1.00x104C? ( Strategy: Coulombs mole e- mole Cr mass Cr) Ans. = 0.898g Cr
