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Electrochemistry. Electrochemistry Terminology #1  Oxidation  Oxidation – A process in which an element attains a more positive oxidation state Na(s)

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Presentation on theme: "Electrochemistry. Electrochemistry Terminology #1  Oxidation  Oxidation – A process in which an element attains a more positive oxidation state Na(s)"— Presentation transcript:

1 Electrochemistry

2 Electrochemistry Terminology #1  Oxidation  Oxidation – A process in which an element attains a more positive oxidation state Na(s)  Na + + e -  Reduction  Reduction – A process in which an element attains a more negative oxidation state Cl 2 + 2e -  2Cl -

3 Electrochemistry Terminology #2 G ain E lectrons = R eduction An old memory device for oxidation and reduction goes like this… LEO says GER L ose E lectrons = O xidation

4 Electrochemistry Terminology #3 Oxidizing agent  Oxidizing agent The substance that is reduced is the oxidizing agent Reducing agent  Reducing agent The substance that is oxidized is the reducing agent

5 Electrochemistry Terminology #4 Anode  Anode The electrode where oxidation occurs Cathode  Cathode The electrode where reduction occurs Memory device: Reduction at the Cathode

6 Table of Reduction Potentials Measured against the Standard Hydrogen Electrode

7 Measuring Standard Electrode Potential Potentials are measured against a hydrogen ion reduction reaction, which is arbitrarily assigned a potential of zero volts.

8 Galvanic (Electrochemical) Cells Spontaneous redox processes have: A positive cell potential, E 0 A negative free energy change, (-  G)

9 Zn - Cu Galvanic Cell Zn 2+ + 2e -  Zn E = -0.76V Cu 2+ + 2e -  Cu E = +0.34V From a table of reduction potentials:

10 Zn - Cu Galvanic Cell Cu 2+ + 2e -  Cu E = +0.34V The less positive, or more negative reduction potential becomes the oxidation… Zn  Zn 2+ + 2e - E = +0.76V Zn + Cu 2+  Zn 2+ + Cu E 0 = + 1.10 V

11 Line Notation Zn(s) | Zn 2+ (aq) || Cu 2+ (aq) | Cu(s) An abbreviated representation of an electrochemical cell Anodesolution Anodematerial Cathodesolution Cathodematerial ||||

12 Calculating  G 0 for a Cell  G 0 = -nFE 0 n = moles of electrons in balanced redox equation F = Faraday constant = 96,485 coulombs/mol e - E 0 Zn + Cu 2+  Zn 2+ + Cu E 0 = + 1.10 V

13 The Nernst Equation Standard potentials assume a concentration of 1 M. The Nernst equation allows us to calculate potential when the two cells are not 1.0 M. R = 8.31 J/(mol  K) T = Temperature in K n = moles of electrons in balanced redox equation F = Faraday constant = 96,485 coulombs/mol e -

14 Nernst Equation Simplified At 25  C (298 K) the Nernst Equation is simplified this way:

15 Equilibrium Constants and Cell Potential equilibrium At equilibrium, forward and reverse reactions occur at equal rates, therefore: 1. 1. The battery is “dead” 2. The cell potential, E, is zero volts Modifying the Nernst Equation (at 25  C):

16 E 0 Zn + Cu 2+  Zn 2+ + Cu E 0 = + 1.10 V Calculating an Equilibrium Constant from a Cell Potential

17 Concentration Cell Step 1: Determine which side undergoes oxidation, and which side undergoes reduction. Both sides have the same components but at different concentrations. ???

18 Concentration Cell Both sides have the same components but at different concentrations. The 1.0 M Zn 2+ must decrease in concentration, and the 0.10 M Zn 2+ must increase in concentration Zn 2+ (1.0M) + 2e -  Zn (reduction) Zn  Zn 2+ (0.10M) + 2e - (oxidation) ??? Cathode Anode Zn 2+ (1.0M)  Zn 2+ (0.10M)

19 Concentration Cell Step 2: Calculate cell potential using the Nernst Equation (assuming 25  C). Both sides have the same components but at different concentrations. ??? Cathode Anode Zn 2+ (1.0M)  Zn 2+ (0.10M) Concentration Cell

20 Nernst Calculations Zn 2+ (1.0M)  Zn 2+ (0.10M)

21 Electrolytic Processes A negative -E 0 cell potential, ( -E 0 ) +  G A positive free energy change, ( +  G ) NOT Electrolytic processes are NOT spontaneous. They have:

22 Electrolysis of Water In acidic solution Anode rxn: Cathode rxn: -1.23 V -0.83 V -2.06 V

23 Electroplating of Silver Anode reaction: Ag  Ag + + e - Electroplating requirements: 1. Solution of the plating metal 3. Cathode with the object to be plated 2. Anode made of the plating metal 4. Source of current Cathode reaction: Ag + + e -  Ag

24 Solving an Electroplating Problem Q: How many seconds will it take to plate out 5.0 grams of silver from a solution of AgNO 3 using a 20.0 Ampere current? 5.0 g Ag + + e -  Ag 1 mol Ag 107.87 g 1 mol e - 1 mol Ag 96 485 C 1 mol e - 1 s 20.0 C = 2.2 x 10 2 s


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