Methanol is reacted with oxygen in a fuel cell to produce electrical energy. Write a half equation for the oxidation of methnol using an acidic electrolyte in a fuel cell.
what i got: 2CH3OH + 3O2 -> 2CO2+4H2O
what the answer says: CH3OH + H20 -> CO2 + 6H+ + 6e-
Also with questions like, Hydrogen can be produced by electrolysis of a dilute solution of potassium chloride. Give the overall equation for this reaction. what are your thought processes, and how would you answer it?
question 1:
what you wrote is the full reaction including that of the redution. methanol oxidation is going to be a redox half cell
starting with CH3OH as reactant and CO2 as product.
1. CH3OH -> CO2
2. CH3OH + H20 -> CO2 (balance oxygens by adding water)
3. CH3OH + H20 -> CO2 + 6H+ (balance hydrogens by adding H+ ions)
4. CH3OH + H20 -> CO2 + 6H+ + 6e- (balance charges by adding electrons)
question 2:
well when i think of electrolysis i think anode positive and cathode negative. i also quickly look at the ECS to see the reactant species, searching for strongest oxidant and strongest reductant.
Answer TIME!
so your reacting species are K+ ions, Cl- ions and H2O. looking at the ECS strongest reductant and oxidant is water. (NOTE: the chlorine exception doesn't apply as weak dilute solution).
thus the half reactions that occur are: (written as on ECS)
i) O2 + 4H+ + 4e- -> 2H2O
ii) 2H2O + 2e- -> H2 + 2OH-
now reverse the oxidation reaction (i), combine the two reactions and balance both sides.
6H2O -> 2H2 + 4OH- + 4H+ + O2