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Electrolysis.  Running a galvanic cell backwards.  Put a voltage bigger than the cell potential on the wire and reverse the direction of the redox reaction.

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Presentation on theme: "Electrolysis.  Running a galvanic cell backwards.  Put a voltage bigger than the cell potential on the wire and reverse the direction of the redox reaction."— Presentation transcript:

1 Electrolysis

2  Running a galvanic cell backwards.  Put a voltage bigger than the cell potential on the wire and reverse the direction of the redox reaction.  The Cathode & Anode swap as well as the flow direction of the ions.  Used for electroplating. Electrolysis

3 1.0 M Zn +2 e-e- e-e- Anode Cathode 1.10 Zn Cu 1.0 M Cu +2

4 1.0 M Zn +2 e-e- e-e- Anode Cathode A battery >1.10V Zn Cu 1.0 M Cu +2

5 Draw the following cell….  2I - (ag) + Cu 2+ (ag)  I 2(s) + Cu (s)  Which process takes place at the anode? Cathode?  What are the electrodes made of?  What is the direction of electron flow? Is the reaction spontaneous?  What Voltage is necessary to force electrolysis?

6 Electroplating  Place a current on the system which is greater then the potential at which the metal ion is reduced.  The metal will plate out.  The amount of metal reduced is directly related to the current in Amps. Amps = Coulombs/sec  Not a perfectly efficient process.

7 Calculating plating  Use Stoichiometry to solve.  G  moles  moles of e-  coulombs  Current  time  How long must 6.00 amp current be applied to produce 15.5 g of Au from Au 3+ ?

8 Another Approach  An aqueous solution of Lead II chloride contains 927g of lead chloride. What current (in Amps) is necessary to remove all the lead from the solution in 48 hours?

9 Other uses  Electrolysis of water.  Separating mixtures of ions.  More positive reduction potential means the reaction proceeds forward forming products.  These need less current to reduce.  Most positive reduction potential is easiest to plate out of solution.

10 Predict the order of reduction for the following ions…  Ag +, Zn 2+, IO 3 -  Ag + + e-  Ag  Zn 2+ + 2 e-  Zn  IO 3 - +6H + + 5e -  ½ I 2 + 3H 2 O


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