The Gibb’s energy for the decomposition of Al2O3 at 500°C is as follows: 2/3 Al2O3arrow 4/3 Al+O2' Δr G=+966 KJ mol-1 The potential difference needed for electrolytic reduction of Al2O3 at 500°C is at least

Question

The Gibb’s energy for the decomposition of Al2O3 at 500°C is as follows: 2/3 Al2O3arrow 4/3 Al+O2' Δr G=+966 KJ mol-1 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 Al23++4e- arrow 4/3Al 2/3 O32- 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 Gibb’s energy for the decomposition of Al2​O3​ at 500∘C is as follows : 2/3Al2​O3​→4/3Al+O2′​Δr​G=+966KJmol−1 The potential difference needed for electrolytic reduction of Al2​O3​ at 500°C is at least

5.0 V

4.5 V

3.0 V

2.5 V

The ionic reactions are 2/3Al23+​+4e−→4/3Al 2/3O32−​→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 Gibb’s energy for the decomposition of $A l_{2} O_{3}$ at $50 0^{\circ} C$ is as follows :
$2/3 A l_{2} O_{3} \to 4/3 A l + O_{2^{'}} Δ_{r} G = + 966 K J m o l^{- 1}$
The potential difference needed for electrolytic reduction of $A l_{2} O_{3}$ at $500° C$ is at least