Teachers Notes:

Problem 1:
a. Anode                        Cathode


b. Anode reaction: CH₃OH + H₂O → CO₂ + 6 H⁺ + 6 e^-
Cathode reaction: O₂ + 4 H⁺ + 4 e^- → 2 H₂O
c. Since the electrons lost must equal the electrons gained
2(CH₃OH + H₂O → CO₂ + 6 H⁺ + 6 e^--) and
3(O₂ + 4 H⁺ + 4 e^- → 2 H₂O)
2CH₃OH + 2H₂O → 2CO₂ + 12 H⁺ + 12 e^-
3O₂ + 12 H⁺ + 12 e^- → 6 H₂O
Overall reaction: 2CH₃OH + 3O₂→ 2CO₂ + 4H₂O which can be simplified to:
CH₃OH + 3/2O₂→ CO₂2 + 2 H₂O

Problem 2:
Overall reaction: NiOOH + MH → Ni(OH)₂ + M
E°cell = 0.52 V - (-0.83 V) = 1.35 V.

Problem 3:
Overall reaction: Li_x/graphite + Li1_-xCoO₂ → Li_xCO₂ + Li_x-1/graphite
E°cell = 0.5 V - (-3.0 V) = 3.5 V.