1. Electrochemical Cells (aka – Galvanic or Voltaic Cells) AP Chemistry Unit 10 Electrochemistry Chapter 17

2. Electrochemical/Galvanic Cells Chemical Energy is converted to Electrical Energy. Chemical Energy is converted to Electrical Energy. A spontaneous redox reaction that is used to generate a voltage (Electricity) A spontaneous redox reaction that is used to generate a voltage (Electricity) You separate the reactants (oxidizing and reducing agents) and force the electrons to travel through a wire. You separate the reactants (oxidizing and reducing agents) and force the electrons to travel through a wire.

3. Electrochemical Cell Salt Bridge Voltmeter

4. Cathode - Reduction Anode - Oxidation

5. Electrochemical Cells Anode – The electrode where oxidation occurs Anode – The electrode where oxidation occurs Cathode – The electrode where reduction occurs. Cathode – The electrode where reduction occurs. Salt Bridge – prevents build up of ions on one side of the cell and balances the charge. Salt Bridge – prevents build up of ions on one side of the cell and balances the charge. No more electricity can flow when the anode disappears! No more electricity can flow when the anode disappears!

6. Electrochemical Cell Animation Tutorial Link Electrochemical Cell Animation Tutorial Link Electrochemical Cell Animation Tutorial Link Electrochemical Cell Animation Tutorial Link

7. Cell Diagrams A voltaic/electrochemical cell may be represented by the following A voltaic/electrochemical cell may be represented by the following ANODECATHODE ANODECATHODE Zn(s) Zn 2+ (1M) Cu 2 (1M) Cu(s) Zn(s) Zn 2+ (1M) Cu 2 (1M) Cu(s) The single lines represent phase boundaries (e.g. solid anode to 1M liquid) and the double lines represent the salt bridge. The single lines represent phase boundaries (e.g. solid anode to 1M liquid) and the double lines represent the salt bridge. Standard conditions are 25 o C,1 atm and 1M solutions. Standard conditions are 25 o C,1 atm and 1M solutions.

8. The Cell Potential The potential is the ability of a cell to do electrical work. ξcell is measured in volts. 1 volt = 1 joule/ coulombA coulomb is the quantity of charge passing in 1 second when the current is 1 ampere 1 volt = 1 joule/ coulombA coulomb is the quantity of charge passing in 1 second when the current is 1 ampere 1C= 1Asec 1C= 1Asec

9. Standard Reduction Potentials Most reference tables for electrochemistry are written as reductions in a half-reaction format, with the most negative reduction on the top and the most positive on the bottom. By definition, all half reductions are compared to the hydrogen half reaction that has the standard value of 0.00 V under standard conditions. Most reference tables for electrochemistry are written as reductions in a half-reaction format, with the most negative reduction on the top and the most positive on the bottom. By definition, all half reductions are compared to the hydrogen half reaction that has the standard value of 0.00 V under standard conditions. Standard state: 25 o C, 1 atm Standard state: 25 o C, 1 atm 2H + + 2 e-  H 2 ξ 0 cell = 0.00 2H + + 2 e-  H 2 ξ 0 cell = 0.00 All other reduction potentials are based on this zero point. All other reduction potentials are based on this zero point.

10. Standard Reduction Potentials Strongest Oxidizing Agents/ Easily Reduced Strongest Reducing Agents/ Easily Oxidized

11. Rules for Using Standard Reduction Potentials 1)Read the half reactions as written 2) The more POSITIVE the reduction potential, the greater the tendency is for the substance to be reduced and therefore the better the oxidizing agent (keep the half-reaction with the more POSITIVE cell potential as written in the table—that is the reduction reaction) (keep the half-reaction with the more POSITIVE cell potential as written in the table—that is the reduction reaction) 3)The half-cell reactions ARE reversible. IF you need to reverse, you MUST change the sign of the ξ 0 cell. 4) If you change the stoichiometric coefficients, ξ 0 cell remains the same

12. Rules for Using Standard Reduction Potentials 5) Under standard state conditions: any species on the LEFT of a given half- reaction will react spontaneously with a species that is on the RIGHT and ABOVE it 6) The most positive values for ξcell mean that they are the strongest OXIDIZING AGENTS and therefore are themselves reduced. 7) The most negative values for ξcell mean that they are the strongest REDUCING AGENTS and therefore arethemselves oxidized.

13. The Total Cell Potential The total cell potential is the sum of the half-cell potentials The total cell potential is the sum of the half-cell potentials ξ 0 cell.=ξ 0 ox+ ξ 0 red A (+) ξ 0 cell means the reaction will happen spontaneously A (+) ξ 0 cell means the reaction will happen spontaneously A (-) ξ 0 cell means the reaction will not happen! A (-) ξ 0 cell means the reaction will not happen!

14. Determining ξ 0 cell Example: Example: What will be the overall reaction and the ξ o cell- total if Br 2 is added to a solution containing I 2 at 25 o C? Assume all species are in their standard states. What will be the overall reaction and the ξ o cell- total if Br 2 is added to a solution containing I 2 at 25 o C? Assume all species are in their standard states. Half reactions (as found in a standard table): Half reactions (as found in a standard table): I 2 (s)+ 2 e-  2I – (aq)ξ o cell = 0.53 V I 2 (s)+ 2 e-  2I – (aq)ξ o cell = 0.53 V Br 2 (l)+ 2e-  2Br ­ (aq)ξ o red = 1.07 V Br 2 (l)+ 2e-  2Br ­ (aq)ξ o red = 1.07 V

15. Using the standard reduction potentials, the fact that Br 2 has the more positive potential indicates that the Br 2 will be reduced. So, change the sign on the oxidation reaction and reverse it so that it is written as an oxidation. Using the standard reduction potentials, the fact that Br 2 has the more positive potential indicates that the Br 2 will be reduced. So, change the sign on the oxidation reaction and reverse it so that it is written as an oxidation. Br 2 (l)+ 2e-  2Br ­ (aq)ξ o red = 1.07 V Br 2 (l)+ 2e-  2Br ­ (aq)ξ o red = 1.07 V 2I – (aq)  I 2 (s) + 2 e-ξ o ox = -0.53 V 2I – (aq)  I 2 (s) + 2 e-ξ o ox = -0.53 V Add the reactions: Add the reactions: Br 2 (l) + 2e-  2Br ­ (aq)ξ o red = 1.07 V Br 2 (l) + 2e-  2Br ­ (aq)ξ o red = 1.07 V 2I – (aq)  I 2 (s)+2 e-ξ o ox = -0.53 V 2I – (aq)  I 2 (s)+2 e-ξ o ox = -0.53 V Br 2 (l)+2I – (aq)  2Br ­ (aq) + I 2 (s) Br 2 (l)+2I – (aq)  2Br ­ (aq) + I 2 (s) ξ o cell-total= +0.54 V

16. Sample Problem 2 What will be the overall reaction and the ξ o cell-total if Zn(s), a 1M solution of Zn 2+, Cu(s) and a 1M solutions of Cu 2+ are reacted? What will be the overall reaction and the ξ o cell-total if Zn(s), a 1M solution of Zn 2+, Cu(s) and a 1M solutions of Cu 2+ are reacted?