CBSE Class 10 Science 2017 · Paper 531/1 · Question 7

Ethanol (${\text{C}}_2{\text{H}}_6\text{O}$) changes into ethanoic acid (${\text{C}}_2{\text{H}}_4{\text{O}}_2$) by gaining one oxygen atom and losing two hydrogen atoms, the oxygen being added by an oxidising agent (alkaline potassium permanganate or acidified potassium dichromate). Gain of oxygen and loss of hydrogen are oxidation, so the conversion is an oxidation reaction: ${\text{CH}}_3{\text{CH}}_2\text{OH}+2[\text{O}]\to {\text{CH}}_3\text{COOH}+{\text{H}}_2\text{O}$ (with alkaline ${\text{KMnO}}_4$ or acidified ${\text{K}}_2{\text{Cr}}_2{\text{O}}_7$, on heating) Difference: this is a partial, controlled oxidation. Only the $-{\text{CH}}_2\text{OH}$ end changes to $-\text{COOH}$, the two carbon atoms stay joined, and the product is another carbon compound. In combustion, ethanol burns in oxygen and the whole molecule is oxidised completely to carbon dioxide and water, with the release of a large amount of heat and light: ${\text{C}}_2{\text{H}}_5\text{OH}+3{\text{O}}_2\to 2{\text{CO}}_2+3{\text{H}}_2\text{O}+\text{heat and light}$