The minimum voltage required to electrolyse alumina in the Hall-Heroult process is (Given, Δ Gf°( Al 2 O 3)=-1520 kJ mol -1; Δ Gf°( CO 2)=-394 kJ mol -1

Question

The minimum voltage required to electrolyse alumina in the Hall-Heroult process is (Given, Δ Gf°( Al 2 O 3)=-1520 kJ mol -1; Δ Gf°( CO 2)=-394 kJ mol -1 (A) 1.575 V (B) 1.60 V (C) 1.312 V (D) -2.62 V

Tardigrade Admin · Accepted Answer

In Hall-Heroult process, the following reactions occur

3 C + 2 A l_{2} O_{3} ⟶ 4 A l + 3 C O_{2}

4 A l^{3 +} + 12 e^{-} ⟶ 4 A l

∴ Δ G^{\circ} = 3Δ G_{f}^{\circ} (C O_{2}) - 2Δ G_{f}^{\circ} (A l_{2} O_{3})

= 3 (- 394) - 2 (- 1520)

= 1858 k J

Δ G^{\circ} = - n F E_{cell}^{\circ}

- E_{cell}^{\circ} = \frac{1858 \times 1000}{12 \times 96500} = 1.60 V

Q. The minimum voltage required to electrolyse alumina in the Hall-Heroult process is (Given, ΔGf∘​(Al2​O3​)=−1520kJmol−1; ΔGf∘​(CO2​)=−394kJmol−1

1.575 V

1.60 V

1.312 V

-2.62 V

In Hall-Heroult process, the following reactions occur 3C+2Al2​O3​⟶4Al+3CO2​ 4Al3++12e−⟶4Al ∴ΔG∘=3ΔGf∘​(CO2​)−2ΔGf∘​(Al2​O3​) =3(−394)−2(−1520) =1858kJ ΔG∘=−nFEcell ∘​ −Ecell ∘​=12×965001858×1000​=1.60V

Q. The minimum voltage required to electrolyse alumina in the Hall-Heroult process is (Given, $Δ G_{f}^{\circ} (A l_{2} O_{3}) = - 1520 k J m o l^{- 1}$ ;
$Δ G_{f}^{\circ} (C O_{2}) = - 394 k J m o l^{- 1}$