JEE Main & Advanced
Chemistry
Electrochemistry / विद्युत् रसायन
Question Bank
Faraday's law of electrolysis
question_answer
On the basis of the information available from the reaction \[\frac{4}{3}Al+{{O}_{2}}\to \frac{2}{3}A{{l}_{2}}{{O}_{3}},\Delta G=-827kJmo{{l}^{-1}}\] of \[{{O}_{2}}\], the minimum emf required to carry out an electrolysis of \[A{{l}_{2}}{{O}_{3}}\] is (F = 96500C \[mo{{l}^{-1}}\]) [CBSE PMT 2003]