1. Unit 5: Electrochemistry An AWESOME presentation by Dana and Brendan

2. Part 1: Oxidation and Reduction  Determining oxidation numbers  Oxidizing and Reducing agents  Oxidation and Reduction half-reactions  Balance redox reactions in acidic and basic solutions  Determining oxidation numbers  Oxidizing and Reducing agents  Oxidation and Reduction half-reactions  Balance redox reactions in acidic and basic solutions

3. Determining Oxidation Numbers  Used to determine whether any elements are changing oxidation states.  Look at the charge of the element, use your knowledge about the charge of the anion to determine what the charge of the cation must be to balance the charge of the anion and produce the final, overall charge.  Used to determine whether any elements are changing oxidation states.  Look at the charge of the element, use your knowledge about the charge of the anion to determine what the charge of the cation must be to balance the charge of the anion and produce the final, overall charge. Ex. 1 MnO 4 - We know O has a charge of 2- X+4(-2)=-1 X=+7

4. Ex. 2 Zn(s) + 2H + (aq) Zn 2+ (aq) + H 2 (g) 01+ 2+0 The oxidation number of Zn changes from 0 to 2+ and that of H changes from 1+ to 0 so this is an oxidation-reduction reaction.

5. Oxidizing and Reducing Agents  LEO the lion goes GER  OIL RIG  The substance that loses its electrons is oxidized and the one that gains electrons is reduced.  LEO the lion goes GER  OIL RIG  The substance that loses its electrons is oxidized and the one that gains electrons is reduced. From Ex. 1, Zn went from 0 to 2+ so it loses electrons and is oxidized. H goes from 1+ to 0 so it gains electrons and is reduced.

6. Oxidation and Reduction Half- Reactions  When there is an equation, separate the half-reactions.  Then balance the elements other than H and O.  Balance Oxygen by adding H 2 0  Balance the H atoms by adding H +  Last, balance the charge by adding e -  When there is an equation, separate the half-reactions.  Then balance the elements other than H and O.  Balance Oxygen by adding H 2 0  Balance the H atoms by adding H +  Last, balance the charge by adding e -

7. Ex. 3 MnO 4 – (aq)+ C 2 O 4 2– (aq) Mn 2+ (aq) + CO 2 (g) 1. Separate the reactions MnO 4 - (aq) Mn 2+ (aq) C 2 O 4 2- (aq) CO 2 (g) 2. Balance the half-reactions 5e - +8H + +MnO 4 - (aq) Mn 2+ (aq)+4H 2 O(l) C 2 O 4 2- (aq) 2CO 2 (g)+2e -

8. Ex. 3 continued 3. Multiply the equations to make the amount of electrons of each equal. [5e - +8H + +MnO 4 - (aq) Mn 2+ (aq)+4H 2 O(l)]2 [C 2 O 4 2- (aq) 2CO 2 (g)+2e - ]5 4. Add the equations 10e - +16H + +2MnO 4 - (aq) 2Mn 2+ (aq)+8H 2 O(l) 5C 2 O 4 2- (aq) 10CO 2 (g)+10e - 16H + +2MnO 4 - (aq)+5C 2 O 4 2- (aq) 2Mn 2+ (aq)+10CO 2 (g)+8H 2 0(l)

9. In Acidic and Basic Solutions  The previous example is how you would balance reactions in acidic solutions, but there are extra steps to balance them in basic solutions.  You must balance the half-reactions as you would for acidic half-reactions then add OH - to both sides to cancel out the H +  The previous example is how you would balance reactions in acidic solutions, but there are extra steps to balance them in basic solutions.  You must balance the half-reactions as you would for acidic half-reactions then add OH - to both sides to cancel out the H +

10. Ex. 4 NO 2 - (aq) + Al(s)NH 3 (aq) + Al(OH) 4 - (aq) 1. Solve as you would in acidic solution 6e - +7H + +NO 2 - (aq) NH 3 (aq)+2H 2 O 4H 2 O+Al(s) Al(OH) 4 - (aq)+4H + +3e - 2. Add OH - to both sides 7OH - +6e - +7H + +NO 2 - (aq) NH 3 (aq)+2H 2 O+7OH - 4OH - +4H 2 O+Al(s) Al(OH) 4 - (aq)+4H + +3e - +4OH- H + and OH - combine to form H 2 O

11. Ex. 4 continued 7H 2 O+6e - +NO 2 - (aq) NH 3 (aq)+2H 2 O+7OH - [4OH - +4H 2 O+Al(s) Al(OH) 4 - (aq)+3e - +4H 2 O]2 3. Add the equations 7H 2 O+6e - +NO 2 - (aq) NH 3 (aq)+2H 2 O+7OH - 8OH - +8H 2 O+2Al(s) 2Al(OH) 4 - (aq)+6e - +8H 2 O OH - +5H 2 O+2Al(s)+NO 2 - (aq)NH 3 (aq)+2Al(OH) 4 - (aq)

12. Part 2: Electrochemical cells

13. 1. Half Cells (Voltaic Cell or Galvanic cell)

14.  Anions flow toward the anode, cations flow toward the cathode.  Electrons flow from the anode to the cathode.  The oxidation (losing electrons) reaction takes place at the anode.  The reduction (gaining electrons) reaction takes place at the cathode.  Anions flow toward the anode, cations flow toward the cathode.  Electrons flow from the anode to the cathode.  The oxidation (losing electrons) reaction takes place at the anode.  The reduction (gaining electrons) reaction takes place at the cathode. Points to Remember:

15. Anode (oxidation half-reaction): Zn(s) Zn 2+ (aq)+2e - Cathode (reduction half-reaction): Cu 2+ (aq)+2e - Cu(s)

16. 2. Using Standard Reduction Potentials to predict Spontaneous Reactions The larger the Potential (in V), the more readily reduced the substance is. The smaller the Potential, the more readily oxidized it is. So, to choose which is oxidized and which is reduced, look to the table and the one with the higher value is reduced. E o cell = E o red (cathode) - E o red (anode) Find this experimentally One will be given If E cell is positive, the reaction is spontaneous

17. 3. More about standard reduction potential and Cell Potential  Cell potential, also called electromotive force, or emf, is measured in volts and sometimes referred to as the cell voltage.  For E o cell, the circle denotes that this is the cell potential at standard conditions.  Standard conditions are 1 atm, 25 degrees Celsius, and one mole.  This measures the difference in potential energy between the cathode and the anode. If there is a greater amount of potential energy in the anode then the cathode, then the electrons will flow from the anode to the cathode spontaneously.  Cell potential, also called electromotive force, or emf, is measured in volts and sometimes referred to as the cell voltage.  For E o cell, the circle denotes that this is the cell potential at standard conditions.  Standard conditions are 1 atm, 25 degrees Celsius, and one mole.  This measures the difference in potential energy between the cathode and the anode. If there is a greater amount of potential energy in the anode then the cathode, then the electrons will flow from the anode to the cathode spontaneously. E o cell = E o red (cathode) - E o red (anode)

18. 4. Cell potential at non-standard conditions  Nernst Equation: E = E o - (RT/nF) ln Q  Nernst Equation: E = E o - (RT/nF) ln Q Already Solved for 8.314 J/molK Temperature(K) Number of electrons transferred 96485 C/mol Reaction Quotient: [Reactants]/[Products]

19. 5. Gibbs Free Energy  A measure of the spontaneity of the process. If ∆G is negative, the reaction is spontaneous. ∆G o = -nFE o n: number of electrons transferred F: Faraday’s constant E o : Cell potential  A measure of the spontaneity of the process. If ∆G is negative, the reaction is spontaneous. ∆G o = -nFE o n: number of electrons transferred F: Faraday’s constant E o : Cell potential

20. 6. Equilibrium Constant  If the reaction is spontaneous, K will be a large, positive number. ∆G o = -RT ln K R: 8.314 J/molK T: Temperature (K)  If the reaction is spontaneous, K will be a large, positive number. ∆G o = -RT ln K R: 8.314 J/molK T: Temperature (K)

21. 7. Relationship between electrode mass, current, and time Given type of metal, current, amount of days, and cell potential, solve for the amount of grams of element lost. Ex. 5 Metal: Zinc, Current: 50 mA, Days: 12, Cell Potential:1.30 V Note: Normally this information will be provided in a word problem Zn Zn 2+ +2e - 12 days x 86400 s x.050 C x 1 mole e - x 1 mol Zn x 65.4 g = 17.57 grams 1 1 day 1 s 96485 C 2 mol e - 1 mol Zn