On the basis of the information available from the reaction : <br> $\frac{4}{3} Al + O_2 \to \frac{2}{3} Al_2O_3, \Delta G = -827 \;KJ \; mol^{-1}$ of $O_2$, the minimum e.m.f required to carry out electrolysis of $Al_2O_3$ is $(F = 96500 \; C\; mol^{-1})$

Question

( A ) 6.42 V
( B ) 4.28 V
( C ) 8.56 V
( D ) 2.14 V