1. 1 © 2006 Brooks/Cole - Thomson Balancing Equations for Redox Reactions Some redox reactions have equations that must be balanced by special techniques. MnO 4 - + 5 Fe 2+ + 8 H + ---> Mn 2+ + 5 Fe 3+ + 4 H 2 O Mn = +7 Fe = +2 Fe = +3 Mn = +2

2. 2 © 2006 Brooks/Cole - Thomson Reduction of VO 2 + with Zn

3. 3 © 2006 Brooks/Cole - Thomson Balancing Equations Balance the following in acid solution— VO 2 + + Zn ---> VO 2+ + Zn 2+ VO 2 + + Zn ---> VO 2+ + Zn 2+ Step 1:Write the half-reactions OxZn ---> Zn 2+ RedVO 2 + ---> VO 2+ Step 2:Balance each half-reaction for mass. OxZn ---> Zn 2+ Red VO 2 + ---> VO 2+ + H 2 O 2 H + + Add H 2 O on O-deficient side and add H + on other side for H-balance.

4. 4 © 2006 Brooks/Cole - Thomson Balancing Equations Step 3:Balance half-reactions for charge. OxZn ---> Zn 2+ + 2e- Rede- + 2 H + + VO 2 + ---> VO 2+ + H 2 O Step 4:Multiply by an appropriate factor. OxZn ---> Zn 2+ + 2e- Red 2e- + 4 H + + 2 VO 2 + ---> 2 VO 2+ + 2 H 2 O Step 5:Add balanced half-reactions Zn + 4 H + + 2 VO 2 + ---> Zn 2+ + 2 VO 2+ + 2 H 2 O

5. 5 © 2006 Brooks/Cole - Thomson Tips on Balancing Equations Never add O 2, O atoms, or O 2- to balance oxygen ONLY add H 2 O or OH -.Never add O 2, O atoms, or O 2- to balance oxygen ONLY add H 2 O or OH -. Never add H 2 or H atoms to balance hydrogen ONLY add H + or H 2 O.Never add H 2 or H atoms to balance hydrogen ONLY add H + or H 2 O. Be sure to write the correct charges on all the ions.Be sure to write the correct charges on all the ions. Check your work at the end to make sure mass and charge are balanced.Check your work at the end to make sure mass and charge are balanced. PRACTICE!PRACTICE!

6. 6 © 2006 Brooks/Cole - Thomson Using Standard Potentials, E o In which direction do the following reactions go?In which direction do the following reactions go? Cu(s) + 2 Ag + (aq) ---> Cu 2+ (aq) + 2 Ag(s)Cu(s) + 2 Ag + (aq) ---> Cu 2+ (aq) + 2 Ag(s) –Goes right as written 2 Fe 2+ (aq) +2 Fe 2+ (aq) + Sn 2+ (aq) ---> 2 Fe 3+ (aq) + Sn(s) –Goes LEFT opposite to direction written What is E o net for the overall reaction?

7. 7 © 2006 Brooks/Cole - Thomson Cd --> Cd 2+ + 2e- or Cd 2+ + 2e- --> Cd Fe --> Fe 2+ + 2e- or Fe 2+ + 2e- --> Fe E o for a Voltaic Cell All ingredients are present. Which way does reaction proceed? Calculate E o for this cell.

8. 8 © 2006 Brooks/Cole - Thomson E at Nonstandard Conditions The NERNST EQUATIONThe NERNST EQUATION E = potential under nonstandard conditionsE = potential under nonstandard conditions n = no. of electrons exchangedn = no. of electrons exchanged F = Faraday’s constantF = Faraday’s constant R = gas constantR = gas constant T = temp in KelvinsT = temp in Kelvins ln = “natural log”ln = “natural log” Q = reaction quotientQ = reaction quotient

9. 9 © 2006 Brooks/Cole - Thomson E o and Thermodynamics E o is related to ∆G o, the free energy change for the reaction.E o is related to ∆G o, the free energy change for the reaction. ∆G˚ is proportional to –nE˚∆G˚ is proportional to –nE˚ ∆G o = -nFE o where F = Faraday constant = 9.6485 x 10 4 J/Vmol of e - (or 9.6485 x 10 4 coulombs/mol) and n is the number of moles of electrons transferred

10. 10 © 2006 Brooks/Cole - Thomson E o and ∆G o ∆G o = - n F E o For a product-favored reaction Reactants ----> Products Reactants ----> Products ∆G o 0 E o is positive For a reactant-favored reaction Reactants <---- Products Reactants <---- Products ∆G o > 0 and so E o 0 and so E o < 0 E o is negative

11. 11 © 2006 Brooks/Cole - Thomson E o and Equilibrium Constant  G o = -RT ln K  G o = -n F E o

12. 12 © 2006 Brooks/Cole - Thomson Electrolysis Using electrical energy to produce chemical change. Sn 2+ (aq) + 2 Cl - (aq) ---> Sn(s) + Cl 2 (g)

13. 13 © 2006 Brooks/Cole - Thomson Electrolysis Electric Energy ---> Chemical Change Electrolysis of molten NaCl. Electrolysis of molten NaCl. Here a battery “pumps” electrons from Cl - to Na +. Here a battery “pumps” electrons from Cl - to Na +. NOTE: Polarity of electrodes is reversed from batteries. NOTE: Polarity of electrodes is reversed from batteries.

14. 14 © 2006 Brooks/Cole - Thomson Electrolysis of Molten NaCl Figure 20.18

15. 15 © 2006 Brooks/Cole - Thomson Electrolysis of Molten NaCl Anode (+) 2 Cl - ---> Cl 2 (g) + 2e- Cathode (-) Na + + e- ---> Na E o for cell (in water) = - 4.07 V (in water) External energy needed because E o is (-).

16. 16 © 2006 Brooks/Cole - Thomson Electrolysis of Aqueous NaI Anode (+): 2 I - ---> I 2 (g) + 2e- Cathode (-): 2 H 2 O + 2e- ---> H 2 + 2 OH - E o for cell = -1.36 V

17. 17 © 2006 Brooks/Cole - Thomson Electrolysis of Aqueous CuCl 2 Anode (+) 2 Cl - ---> Cl 2 (g) + 2e- Cathode (-) Cu 2+ + 2e- ---> Cu E o for cell = -1.02 V Note that Cu 2+ is more easily reduced than either H 2 O or Na +.

18. 18 © 2006 Brooks/Cole - Thomson Michael Faraday 1791-1867 Originated the terms anode, cathode, anion, cation, electrode. Discoverer of electrolysiselectrolysis magnetic props. of mattermagnetic props. of matter electromagnetic inductionelectromagnetic induction benzene and other organic chemicalsbenzene and other organic chemicals Was a popular lecturer.

19. 19 © 2006 Brooks/Cole - Thomson Quantitative Aspects of Electrochemistry Consider electrolysis of aqueous silver ion. Ag + (aq) + e- ---> Ag(s) 1 mol e----> 1 mol Ag If we could measure the moles of e-, we could know the quantity of Ag formed. But how to measure moles of e-?

20. 20 © 2006 Brooks/Cole - Thomson But how is charge related to moles of electrons? Quantitative Aspects of Electrochemistry = 96,500 C/mol e- = 1 Faraday Michael Faraday 1791-1867

21. 21 © 2006 Brooks/Cole - Thomson Quantitative Aspects of Electrochemistry 1.50 amps flow through a Ag + (aq) solution for 15.0 min. What mass of Ag metal is deposited? Solution (a)Calc. charge Charge (C) = current (A) x time (t) = (1.5 amps)(15.0 min)(60 s/min) = 1350 C

22. 22 © 2006 Brooks/Cole - Thomson Quantitative Aspects of Electrochemistry Solution (a)Charge = 1350 C (b)Calculate moles of e- used 1.50 amps flow through a Ag + (aq) solution for 15.0 min. What mass of Ag metal is deposited? (c)Calc. quantity of Ag

23. 23 © 2006 Brooks/Cole - Thomson Quantitative Aspects of Electrochemistry The anode reaction in a lead storage battery is Pb(s) + HSO 4 - (aq) ---> PbSO 4 (s) + H + (aq) + 2e- If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last? Solution a)454 g Pb = 2.19 mol Pb b)Calculate moles of e- c)Calculate charge 4.38 mol e- 96,500 C/mol e- = 423,000 C 4.38 mol e- 96,500 C/mol e- = 423,000 C

24. 24 © 2006 Brooks/Cole - Thomson Quantitative Aspects of Electrochemistry The anode reaction in a lead storage battery is Pb(s) + HSO 4 - (aq) ---> PbSO 4 (s) + H + (aq) + 2e- If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last? Solution a)454 g Pb = 2.19 mol Pb b)Mol of e- = 4.38 mol c)Charge = 423,000 C About 78 hours d)Calculate time