Write a half-equation for the overall oxidation of ethanol into ethanoic acid.
This was the answer: CH3CH2OH + H2O = CH3COOH + 4H+ + 4e–
But, i dont know how they got it. Can anyone explain?
The oxygen of water (boldt print) reacts with ethanol.
CH3CH2OH + H2O = CH3COOH + 4H+ + 4e–
At the same time, the hydrogen atoms of water and the hydrogen atoms of the methylene group (CH2) in ethanol are ionized (in bold print above). That is why 4 H+ ions and 4e- come into being.
The oxygen of water (boldt print) reacts with ethanol.
CH3CH2OH + H2O = CH3COOH + 4H+ + 4e–
At the same time, the hydrogen atoms of water and the hydrogen atoms of the methylene group (CH2) in ethanol are ionized (in bold print above). That is why 4 H+ ions and 4e- come into being.
CH3CH2OH + H2O = CH3COOH + 4H+ + 4e–
That does help quite a lot! I've always thought of it in terms of oxidation number, and i couldnt work out these oxidation numbers for them!