So there I was doing chapter 8 in the heinemann textbook (end of AOS1) when question 7 asks me to write the half-equations of the oxidation of ethanol and the reduction of oxygen. I have no idea how I was supposed to know that ethanol oxidised to CH3CHOOH and I also wasn't sure about the reduction of oxygen. Is this something I'm supposed to be able to work out or is it assumed knowledge?
Book's Question:
Blood alcohol content can also be determined by using a fuel cell. Write half equations describing the oxidation of ethanol and reduction of oxygen in an acidic electrolyte.
Book's answer:
CH3CH2OH + H2O --> CH3CHOOH + 4H+ + 4e–
O2 + 4H+ + 4e– --> 2 H2O
Hey!
It tells you write an oxidation reaction for ethanol and reduction for oxygen.
Remember: AN OIL RIG CAT
Oxidation Is Loss of electrons, thus, CH3CH2OH + H2O --> CH3CHOOH + 4H+ + 4e–
Reduction is gain of electrons, thus, O2 + 4H+ + 4e– --> 2 H2O
Pretty sure you can get these equations from the reduction potential table included in the textbook and data book.
Make sense?
