1. CE Chemistry Module 8

2. A. Involves electron changes (can tell by change in charge) 0 +1 -1 +1 -1 0 Cl 2 + 2 NaBr 2NaCl + Br 2 B. Oxidation 1. First used to apply to reactions with oxygen (rusting, burning) 2. Now means loss of e- (charge becomes more +) 3. oxidation: 2Br - Br 2 + 2e- C. Reduction 1. First used to mean metal oxide ores “reduced” to the metals. 2. Now means gain of e- (charge becomes more -) 3. reduction: 2e- + Cl 2 2 Cl -

3. You can remember what happens to electrons by: 1. LEO the Lion says GER Losing Electrons is Oxidation; Gaining Electrons is Reduction 2. OIL RIG – Oxidation Is Loss of e-; Reduction Is Gain of e- D. Agents (O.A., R.A.) 1. Oxidizing agent (OA) * Causes the oxidation * Takes e-, is reduced 2. Reducing agent (RA) * Causes the reduction * Gives e-, is oxidized LEO OIL RIG

4. 1. Assign charges. 2. Find the element whose charge went up – this is the element oxidized. Connect the two with a line bracket going up. 3. Find the element whose charge went down – this is the element reduced. Connect the two with a line bracket going down. 4. Example (on board)

5. 1. Free element = 0 (Na, Cu, Cl 2, H 2 ) 2. Monatomic ion is equal to charge on ion (Na + = +1) 3. H in compounds = +1 4. O in compounds = -2 5. Sum of charges equals the charge of the whole particle if not neutral (for NO 3 -, the charges add to -1, so the O = -2 as usual for a total of -6 on the 3 atoms of O. The N must be +5 to add to the ion’s charge of -1) 6. Group 1 = +1, Group 2 = +2, Group 13 = +3 in compounds Examples: (on board)

6. 1. Split the equation into reduction and oxidation equations. Choose one to start with (ox or red) 2. Balance red/ox atoms 3. Show the electrons gained (reactant) or lost (product) 4. Balance charge: if acidic - use hydrogen ions, if basic - use hydroxide ions 5. Balance hydrogen atoms with water, check oxygen (should be balanced) 6. Balance the other half equation 7. Make electrons equal for the two half-equations 8. Combine the two equations (add), eliminating electrons from both sides and anything else that can cancel 9. Rewrite as balanced redox equation

7. A. Electrolytic conduction is the migration of e- (induced or forced by the current) B. The negative electrode is the cathode, the positive electrode is the anode. C. Positive ions are attracted to the cathode – called cations D. Negative ions are attracted to the anode – called anions E. Reduction occurs at the cathode, oxidation occurs at the anode (consonants/vowels) F. Electrolysis is a chemical change produced by an electrical current.

8. A. Energy (electricity) produced by a chemical process (spontaneous) 1. Dry cell – flashlight batteries (electrolyte paste inside - can be acid or base (alkaline)) 2. Lead storage cell – car battery (electrolyte solution inside – sulfuric acid)

9. 2. Car Battery a. 2 electrodes, Pb and PbO 2 as solid plates b. Concentrated H 2 SO 4 sulfuric acid acts as an electrolyte c. Reaction: 0 +4 -2 +1 +6 -2 +2 +6 -2 +1 -2 Pb (s) + PbO 2(s) + 2 H 2 SO 4(aq) → 2PbSO 4(s) + 2H 2 O (l) Oxidation: Pb → Pb 2+ + 2e - Reduction: Pb 4+ + 2e - → Pb 2+ C. Electrons flow from anode (+) to cathode (-)

10. C. Cell shorthand notation – ABC (anode/bridge/cathode) 1. Anode written first: Oxidation shown with line between Zn|Zn 2+ 2. Bridge: Double line represents salt bridge ║ 3. Cathode written last: Reduction shown with line between Cu 2+ |Cu 4. Overall: Zn|Zn 2+ ║ Cu 2+ |Cu

11. Reducing Agent Oxidizing Agent e-e- e-e- e-e- e-e- e-e- e-e- AnodeCathode

12. * Electrons are produced (lost) at the anode * They travel through the wire to where you are powering or charging something (like your cell phone) * They arrive at the cathode where they reduce the cations * The anions flow through the salt bridge to the anode and the cations flow to the cathode to neutralize charge buildup.

14. * We start at the anode * Electrons are lost there * And go through the wire * And through the load on fire * They enter the cathode * And reduce the cations * And the anions flow through the salt bridge back to where? * * Last time at end: THE ANODE!!!!

15. A. Electrode potential – potential difference between two half-cells in a voltaic cell (measured in volts) B. The hydrogen half-cell is used as a reference point – given the value of zero C. Tables are set up as reduction tables. Substances higher in the table are better oxidizing agents (have more positive E o ), lower ones are better reducing agents (have more negative E o ). Find the reduction potential. D. To get the oxidation potential, the sign is changed and the equation is reversed. Oxidation half-reaction: H 2 S (g)  2H + (g) + S (s) + 2 e - Find this equation on the table written: 2H + (g) + S (s) + 2 e -  H 2 S (g) E o red = 0.144 V Reverse the reaction to get E o ox and change the sign of E. H 2 S (g)  2H + (g) + S (s) + 2 e - E o ox = -0.144 V E. To determine voltage, the potentials of the two reactions are added together – it is only spontaneous if the result is positive

16. Solution: First determine the oxidation numbers on both sides of the reaction. Cu: +2 to 0 (reduced) H: +1 both S: -2 to 0 (oxidized) Reduction half-reaction: Cu 2+ (aq) + 2 e -  Cu (s) E o red = 0.339 V Oxidation half-reaction: H 2 S (g)  2H + (g) + S (s) + 2 e - Find this equation on the table written: 2H + (g) + S (s) + 2 e -  H 2 S (g) E o red = 0.144 V Reverse the reaction to get E o ox and change the sign of E. H 2 S (g)  2H + (g) + S (s) + 2 e - E o ox = -0.144 V Then: E o cell = E o red + E o ox = 0.339 V + (-0.144 V) = 0.195 V Spontaneous because it is positive

17. * For all voltaic (galvanic) cells, E o is a positive quantity, so choose the oxidation and reduction that will give you a positive value. * The quantities E o, E o ox, and E o red are independent of the coefficients in the equation for the cell reaction. NEVER MULTIPLY by the coefficients – use just as they are written in the table.

18. 1. Calculate the cell voltage for the following cell: Zn|Zn 2+ ║ Cu 2+ |Cu (change sign for ox. potential) Oxidation: Zn → Zn 2+ + 2e- E o ox = +.7618 V Reduction: Cu 2+ + 2e- → CuE o red = +.3419 V Overall: Zn + Cu 2+ → Zn 2+ + Cu E o cell = +1.1037 V 2. Is this reaction spontaneous as written? Al 3+ + Ni → Ni 2+ + Al Oxidation: 3(Ni → Ni 2+ + 2e-) E o ox = +.257 V Reduction: 2(Al 3+ + 3e- → Al)E o red = -1.662 V Overall: 3Ni + 2Al 3+ → 3Ni 2+ + 2Al E o cell = -1.405 V Not spontaneous (must be positive volts to be spont.)

19. 3. Putting these half-cells together, what will be spontaneous reaction? Ag │ Ag + E o red = +.7996 V Fe 3+ │ Fe E o red = -.037 V (reverse and change sign) Fe │ Fe 3+ E o ox = +.037 V Have to decide which sign can be changed (oxidation) so that the sum of the two (overall cell voltage) will be positive. The iron should be the oxidation so that when added together gives positive voltage.7996 V +.037 V =.8366 V