The Gibbs' energy for the decomposition of Al2O3 at 500° C is as follows: (2/3) Al2 O3 arrow (4/3) Al + O2 ; Δr G = + 960 kJ mol-1 The potential difference needed for the electrolytic reduction of aluminium oxide (Al2O3) at 500° C is at least

Question

The Gibbs' energy for the decomposition of Al2O3 at 500° C is as follows: (2/3) Al2 O3 arrow (4/3) Al + O2 ; Δr G = + 960 kJ mol-1 The potential difference needed for the electrolytic reduction of aluminium oxide (Al2O3) at 500° C is at least (A) 3.0 V (B) 2.5 V (C) 5.0 V (D) 4.5 V

Tardigrade Admin · Accepted Answer

Δ G° = - nFE° 960 × 1000 = - 4 × 96500 × E° E° = - 2.48 V So difference = 2.5 V

$3.0 V$

$2.5 V$

$5.0 V$

$4.5 V$

$Δ G^{\circ} = - n F E^{\circ}$
$960 \times 1000 = - 4 \times 96500 \times E^{\circ}$
$E^{\circ} = - 2.48 V$
So difference = $2.5 V$

Q. The Gibbs' energy for the decomposition of Al2​O3​ at 500∘C is as follows: 32​Al2​O3​→34​Al+O2​ ; Δr​G=+960kJmol−1 The potential difference needed for the electrolytic reduction of aluminium oxide (Al2​O3​) at 500∘C is at least

3.0V

2.5V

5.0V

4.5V

ΔG∘=−nFE∘ 960×1000=−4×96500×E∘ E∘=−2.48V So difference = 2.5V

$3.0 V$

$2.5 V$

$5.0 V$

$4.5 V$

$Δ G^{\circ} = - n F E^{\circ}$
$960 \times 1000 = - 4 \times 96500 \times E^{\circ}$
$E^{\circ} = - 2.48 V$
So difference = $2.5 V$