• # question_answer On the basis of the information available from the reaction$\frac{4}{3}Al+{{O}_{2}}\xrightarrow{{}}\frac{2}{3}A{{l}_{2}}{{O}_{3}},\,\Delta G=-827\,kJ\,mo{{l}^{-1}}$ of ${{O}_{2}}$, the minimum emf required to carry out an electrolysis of $A{{l}_{2}}{{O}_{3}}$is $(F=96500C\,\,mo{{l}^{-1}})$ A)  $2.14V$            B)  $4.28V$ C)  6.42V                                    D)  $8.56V$

$\Delta G=-827\,kJ\,mo{{l}^{-1}}$ We know that  $\Delta G=-nEF\{n=4\}$ $\therefore$  $-827\times {{10}^{3}}J=-4\times E\times 96500$ $\therefore$  $E=\frac{827\times {{10}^{3}}}{4\times 96500}$ $\therefore$  $E=2.14V$