The Gibbs energy for the decomposition of Al2O3 at 500 °C is as follows: 2/3Al2O3 arrow 4/3Al +O2 ; ΔrG=+966 kj/mol. The potential difference needed for electrolytic reduction of Al2O3 at 500 °C is at least

Question

The Gibbs energy for the decomposition of Al2O3 at 500 °C is as follows: 2/3Al2O3 arrow 4/3Al +O2 ; ΔrG=+966 kj/mol. The potential difference needed for electrolytic reduction of Al2O3 at 500 °C is at least (A) 5.0 V (B) 4.5 V (C) 3.0 V (D) 2.5 V

Tardigrade Admin · Accepted Answer

The ionic reactions are: 2/3 × (2Al3+)+4e-arrow 4/3Al 2/3 × (3O2-)arrow O2+4e- Thus, no. of electrons transferred, n = 4. Δ G=-nFE=-4×96500× E or 966×103 =-4×96500× E ⇒ E=-(966×103/4×96500) =-2.5 V

Q. The Gibbs energy for the decomposition of $A l_{2} O_{3}$ at $500^{\circ} C$ is as follows :
$2/3 A l_{2} O_{3} \to 4/3 A l + O_{2}$ ; $Δ_{r} G = + 966 kj / m o l$ .
The potential difference needed for electrolytic reduction of $A l_{2} O_{3}$ at $500^{\circ} C$ is at least

$5.0 V$

$4.5 V$

$3.0 V$

$2.5 V$

Q. The Gibbs energy for the decomposition of Al2​O3​ at 500∘C is as follows : 2/3Al2​O3​→4/3Al+O2​ ; Δr​G=+966kj/mol. The potential difference needed for electrolytic reduction of Al2​O3​ at 500∘C is at least

5.0V

4.5V

3.0V

2.5V

The ionic reactions are : 2/3×(2Al3+)+4e−→4/3Al 2/3×(3O2−)→O2​+4e− Thus, no. of electrons transferred, n=4. ΔG=−nFE=−4×96500×E or 966×103 =−4×96500×E ⇒E=−4×96500966×103​ =−2.5V

Q. The Gibbs energy for the decomposition of $A l_{2} O_{3}$ at $500^{\circ} C$ is as follows :
$2/3 A l_{2} O_{3} \to 4/3 A l + O_{2}$ ; $Δ_{r} G = + 966 kj / m o l$ .
The potential difference needed for electrolytic reduction of $A l_{2} O_{3}$ at $500^{\circ} C$ is at least

$5.0 V$

$4.5 V$

$3.0 V$

$2.5 V$