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Electrochemistry: Balancing Oxidation-Reduction Reactions and EMF Calculation

Learn how to balance oxidation-reduction equations using the half-reaction method and calculate EMF under nonstandard conditions using the Nernst equation.

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Electrochemistry: Balancing Oxidation-Reduction Reactions and EMF Calculation

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  1. Electrochemistry

  2. Oxidation Reduction Reactions • (1) Oxidation: Loss e- Increase in Oxidation Number Zn(s)Zn2+ + 2e- • (2) Reduction: Acceptance of e- Decrease in Oxidation Number Cl2(g) + 2e- 2Cl-

  3. Balancing Oxidation-Reduction Equations: Use Half-Reaction Method The half-reactions for Sn2+(aq) + 2Fe3+(aq)  Sn4+(aq) + 2Fe3+(aq) are Sn2+(aq)  Sn4+(aq) +2e- 2Fe3+(aq) + 2e- 2Fe2+(aq) • Oxidation Half-Reaction: electrons are products. • Reduction Half-Reaction: electrons are reactants.

  4. Half-Reaction Method for Balancing Oxidation-Reduction Equations 1. Separate the equation into the two half-reactions. Write down the two half reactions. 2. Balance each half reaction: a. First balance all elements other than H and O. b. Then balance O by adding water. c. Then balance H by adding H+ if have acidic solution. d. Finish by balancing charge by adding electrons.

  5. 3. Multiply each half reaction to make the number of electrons equal. 4. Add the two half-reactions and simplify. To simplify, remove components common to both reactant and product sides. 5. Check!

  6. Example Balance: (acidic) MnO4- (aq) + Fe2+ (aq)  Mn2+ (aq) + Fe3+ (aq) • The two incomplete half reactions are MnO4-(aq)  Mn2+(aq) Fe2+ (aq)  Fe3+ (aq)

  7. 2. Balance each half reaction: MnO4- Mn2+ + 4 H2O Fe2+ (aq)  Fe3+ (aq)

  8. MnO4- + 5e- Mn2+ + 4 H2O Fe2+ (aq)  Fe3+ (aq) + e-

  9. 8H+ + MnO4- + 5e- Mn2+ + 4 H2O Fe2+ (aq)  Fe3+ (aq) + e-

  10. 8H+ + MnO4- + 5e- Mn2+ + 4 H2O 5 (Fe2+ (aq)  Fe3+ (aq) + e-)

  11. 8H+ + MnO4- + 5e- Mn2+ + 4 H2O 5 (Fe2+ (aq)  Fe3+ (aq) + e-) 8H+ + MnO4- + 5 Fe2+ (aq) Mn2+ + 4 H2O + 5Fe3+ (aq)

  12. Example Balance: (basic) MnO4- (aq) + I- (aq)  MnO2+ I2 • The two incomplete half reactions are MnO4-(aq)  MnO2 I- (aq)  I2

  13. 2. Balance each half reaction: MnO4-(aq) + 3e- MnO2 2I- (aq)  I2 + 2e-

  14. MnO4-(aq) + 3e- MnO2 + 4OH- 2I- (aq)  I2 + 2e-

  15. 2H2O + MnO4-(aq) + 3e- MnO2 + 4OH- 2I- (aq)  I2 + 2e-

  16. 2 (2H2O + MnO4-(aq) + 3e- MnO2 + 4OH-) 3 (2I- (aq)  I2 + 2e-) 4H2O + 6I- (aq) +MnO4-(aq)  2MnO2 + 8OH- + 3I2

  17. Problem • Complete and balance the following equations, and identify the oxidizing and reducing agents • Cr2O72- (aq) + I-(aq)  Cr3+(aq) + IO3-(aq) • (acidic solution) • Pb(OH)42- (aq) + ClO- (aq)  PbO2(s) + Cl-(aq) • (basic solution)

  18. Zn Cu

  19. salt bridge (K+) Cation flow → (Cl-) flow ←

  20. Brief Activity Series

  21. Strong Reducing Agent.

  22. Strong Oxidizing Agent

  23. How many moles of Cu can be plated out of a solution containing Cu2+ ions if a 1.50 amp current is passed through for 300 s? Cu2+ + 2e-→ Cu(s) Notes: 1 amp = C/s F = 9.65 x 104 c/ mole e- Find: c → moles e-s → moles Cu 300 s 1.50 c s mole e- 9.65 x 104 C 1 mole Cu 2 mole e-s =moles Cu 2.33 x 10-3

  24. Effect of Concentration on EMF A Battery going dead. ∆G = ∆Go + RT lnQ -nFE = -nFEo + RT ln Q Dividing by –nF gives Rise to the Nerst Equation.

  25. The Nerst Equation: Q: aA + bB ↔ cC + dD n = # of e- s transferred

  26. The Nerst Equation allows us to find voltage under nonstandard conditions Example: Determine E (voltage) for: Fe(s) + Cd2+(aq)→ Fe2+(aq) + Cd(s) When [Fe2+] = 0.10 M and [Cd2+] = 1.0 M @ 298K Half RXNs: Fe(s) → Fe2+ + 2e- Eo = +0.44 V Cd2+ + 2e-→ Cd(s) Eo = -0.40 V Eo = +0.04 V + Value means spontaneous

  27. Now try: [Fe2+] = 1.0 M and [Cd2+] = 0.01 M @ 298K - Value means non spontaneous and will go in opposite direction

  28. What does an E = 0 value mean? For: Fe(s) + Cd2+(aq) → Fe2+(aq) + Cd(s) Eo = +0.04V K = 22

  29. The pray graet Dariush in Takht-e-jamshid My god preservation this country ( Iran ) from ENEMY , FALSE and ARID YEAR

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