Teachers Notes: Problem 1: a. Anode Cathode
(38.0K)
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
|