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Electrolytic Cells & Electrolysis Reactions Outside electrical source provides electrons that force a non-spontaneous redox reaction. Electricity “splits”

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Presentation on theme: "Electrolytic Cells & Electrolysis Reactions Outside electrical source provides electrons that force a non-spontaneous redox reaction. Electricity “splits”"— Presentation transcript:

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: H 2 O(l) + electricity → H 2 (g) + O 2 (g)

2 Electrolysis Set Up Ex: NaCl(l) Single Cell filled with Single Cell filled with electrolyte with +/- ions Attach battery to two electrodes. Attach battery to two electrodes. Adds e- onto one electrode - Making it NEGATIVE Adds e- onto one electrode - Making it NEGATIVE Pulls e- off one electrode - Making it POSITIVE Pulls e- off one electrode - Making it POSITIVE Electrodes are made of an inert substance 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 At neg. electrode electrons are gained by ion (reduction at CATHODE) At positive electrode electrons are lost by ion (oxidation at ANODE) What is Oxidized/Reduced?

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 correctlyDIATOMIC!!!!!! 2Cl - Cl 2 + 2e- Rxn at Cathode: (Red) Na + + 1e- Na (Multiply by 2 to balance electrons) NET: 2Na + + 2Cl - 2Na + Cl 2

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 PbBr 2 (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 PbCl 2 (l) Oxidized: Cl - Reduced: Pb +2 Half Reactions Ox: Cl -1 Cl + 1e- 2Cl -1 Cl 2 + 2e- Red:Pb e-Pb Net:Pb Cl -1 Pb + Cl 2

11 Electrolysis of NaCl (aq)

12 Electrolysis of Water At Positive Electrode: Ox: O -2 O + 2e- but there is a diatomic! 2O -2 O 2 + 4e- At Negative Electrode Red:H e-H but there is a diatomic! 2H e-H 2 Net:2H 2 O 2H 2 + O 2 Electrolysis of Water (Animation) Electrolysis of Water (Simple)

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14 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.

15 Electroplating Negative Electrode Negative Electrode Is the OBJECT TO BE PLATED Is the OBJECT TO BE PLATED so the positive metal ions would go towards it and be REDUCED. so the positive metal ions would go towards it and be REDUCED. It is the CATHODE It is the CATHODE Red: Ag + + 1e- Ag 0

16 Electroplating Positive Electrode Positive Electrode Made of plating metal Made of plating metal It dissolves into solution as metal strip gets OXIDIZED. It dissolves into solution as metal strip gets OXIDIZED. It is the ANODE It is the ANODE This replenishes the ions for plating. This replenishes the ions for plating. Ox: Ag 0 Ag+ + 1e-

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

18 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-Ag 0

19 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. 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.

20 Electroplating Problems (Honors) If 10 amps are run through a CuSO 4 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 e-Cu 0 2 moles electrons make 1 mole of Cu 0 2 moles electrons make 1 mole of Cu 0 1 mole Cu = 63.5 grams

21 So….Let’s start here # coulombs = 10 amps x 300 seconds = 3000 coulombs = 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 96,500 coul 2 mole e- 1 mole Cu Mole ratio from Reduction half reaction Cu e-Cu 0

22 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

23 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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