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Electrolytic Cells & Electrolysis Reactions

Electrolytic Cells & Electrolysis Reactions. Outside electrical source provides electrons that force a non-spontaneous redox reaction . Electricity “splits” a compound into it’s neutral elements Ex: H 2 O(l) + electricity → H 2 (g) + O 2 (g). Electrolysis Set Up Ex: NaCl(l).

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Electrolytic Cells & Electrolysis Reactions

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  1. Electrolytic Cells & Electrolysis Reactions Outside electrical source provides electrons that force a non-spontaneous redox reaction. Electricity “splits” a compound into it’s neutral elements Ex: H2O(l) + electricity → H2(g) + O2(g)

  2. Electrolysis Set Up Ex: NaCl(l) • Single Cell filled with electrolyte with +/- ions • Attach battery to two electrodes. • Adds e- onto one electrode - Making it NEGATIVE • Pulls e- off one electrode - Making it POSITIVE • Electrodes are made of an inert substance (like platinum or graphite) that conducts.

  3. Which Way do the Ions Move? To electrode of opposite charge

  4. What is Oxidized/Reduced? At neg. electrode electrons are gained by ion (reduction at CATHODE) At positive electrode electrons are lost by ion (oxidation at ANODE)

  5. Remember AN OX RED CAT Anode is where oxidation happens Cathode is where reduction happens

  6. Half Reactions & Net Equation Rxn at Anode: (Ox) Cl- Cl + 1e- Or more correctly DIATOMIC!!!!!! 2Cl- Cl2 + 2e- Rxn at Cathode: (Red) Na+ + 1e- Na (Multiply by 2 to balance electrons) NET: 2Na+ + 2Cl- 2Na + Cl2

  7. Electrolysis of Molten NaCl (l)

  8. Determining Voltage Needed (Honors) Use the Voltage Table to determine the total voltage “needed” to run the Electrolytic cell. Total voltage should be a NEGATIVE number

  9. Electrolysis of PbBr2(l) What is oxidized? What is reduced? What are the ox/red half reactions? What is the net equation? Negative Electrode Positive Electrode

  10. Electrolysis of PbCl2(l) Oxidized: Cl- Reduced: Pb+2 Half Reactions Ox: Cl-1 Cl + 1e- 2Cl-1 Cl2 + 2e- Red: Pb+2 + 2e- Pb Net: Pb+2 + 2Cl-1 Pb + Cl2

  11. Electrolysis of NaCl (aq)

  12. Electrolysis of Water At Positive Electrode: Ox: O-2 O + 2e- but there is a diatomic! 2O-2 O2 + 4e- At Negative Electrode Red: H+1 + 1e- H but there is a diatomic! 2H+1 + 2e- H2 Net: 2H2O 2H2 + O2 Electrolysis of Water (Animation) http://youtu.be/2t13S-KpGeE Electrolysis of Water (Simple) http://youtu.be/HQ9Fhd7P_HA

  13. Electroplating Electrolysis reaction used to coat a substance with a thin layer of metal. Often coating is a less reactive metal that is not easily oxidized or corroded.

  14. Electroplating • Negative Electrode • Is the OBJECT TO BE PLATED • so the positive metal ions would go towards it and be REDUCED. • It is the CATHODE Red: Ag+ + 1e- Ag0

  15. Electroplating • Positive Electrode • Made of plating metal • It dissolves into solution as metal strip gets OXIDIZED. • It is the ANODE • This replenishes the ions for plating. Ox: Ag0 Ag+ + 1e-

  16. Electroplating Problems (Honors) Coulomb = measure of electrical charge 1 mole e- = 96,500 coulombs # coulombs = # amps x seconds

  17. Electroplating Problems (Honors) Reduction: Happens on object to be plated Look at Reduction half reaction Look at mole relationships between electrons and metal atoms. Ex: Ag+ + 1e- Ag0

  18. Electroplating Problems (Honors) • You can now answer questions regarding the amount of a substance in moles or grams that can be electroplated over a certain amount of time.

  19. Electroplating Problems (Honors) If 10 amps are run through a CuSO4 solution for 5 minutes, calculate the grams of Cu that will plate onto the spoon. We Know: 1 mole e- = 96,500 coulombs # coulombs = # amps x seconds Red: Cu+2 + 2e- Cu0 2 moles electrons make 1 mole of Cu0 1 mole Cu = 63.5 grams

  20. So….Let’s start here # coulombs = 10 amps x 300 seconds = 3000 coulombs 3000 coul. x 1 mole e- x 1 mole Cu x 63.5g Cu = .987 grams 96,500 coul 2 mole e- 1 mole Cu Mole ratio from Reduction half reaction Cu+2 + 2e- Cu0

  21. You Try One How long will it take to deposit 20 grams of silver from a solution of AgCl onto a copper tray if a current of 5 amps is used? Answer = 3, 574 sec or 59.5 minutes or about 1 hour

  22. You Try One How many amps are needed to deposit .504g. of Iron in 40 minutes by passing a current through a solution of Iron II Sulfate? Answer: .72 amps

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