On the basis of the information available from the reaction. (4/3) Al + O 2 arrow (2/3) Al 2 O 3, Δ G =-827 KJ mol -1 of O2, the minimum EMF required to carry out the electrolysis of Al2O3 is (F = 96500 C mol-1)

Question

On the basis of the information available from the reaction. (4/3) Al + O 2 arrow (2/3) Al 2 O 3, Δ G =-827 KJ mol -1 of O2, the minimum EMF required to carry out the electrolysis of Al2O3 is (F = 96500 C mol-1) (A) 2.14 V (B) 4.28 V (C) 6.42 V (D) 8.56 V

Tardigrade Admin · Accepted Answer

(4/3)Al+O2 arrow (2/3)Al2O3, Δ G=-827 kJ mol-1 Δ G=-nEF0 (n=4) -827 × 103 J=-4 × E0 × 96500 E= (827 × 103/4 × 96500 ) E0=2.14 V

Q. On the basis of the information available from the reaction.
$\frac{4}{3} A l + O_{2} \to \frac{2}{3} A l_{2} O_{3}, Δ G = - 827 K J m o l^{- 1}$
of $O_{2},$ the minimum EMF required to carry out the electrolysis of $A l_{2} O_{3}$ is
( $F = 96500 C m o l^{- 1})$

2.14 V

4.28 V

6.42 V

8.56 V

$\frac{4}{3} A l + O_{2} \to \frac{2}{3} A l_{2} O_{3},$
$Δ G = - 827 k J m o l^{- 1}$
$Δ G = - n E F^{0} (n = 4)$
$- 827 \times 1 0^{3} J = - 4 \times E^{0} \times 96500$
$E = \frac{827 \times 1 0 ^{3}}{4 \times 96500}$
$E^{0} = 2.14 V$

Q. On the basis of the information available from the reaction. 34​Al+O2​→32​Al2​O3​,ΔG=−827KJmol−1 of O2​, the minimum EMF required to carry out the electrolysis of Al2​O3​ is (F=96500Cmol−1)