The minimum voltage required to electrolyse alumina in the Hall-Heroult process is [Given, Δ Gof (Al2 O3) = - 1520 kJ/mol and Δ Gof (CO2) = -394 kJ/mol]

Question

The minimum voltage required to electrolyse alumina in the Hall-Heroult process is [Given, Δ Gof (Al2 O3) = - 1520 kJ/mol and Δ Gof (CO2) = -394 kJ/mol] (A) 1.60 V  (B) 1.575 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} \to 4 A l + 3 C O_{2}

Δ G^{o} = 3 Δ_{f} G^{o} (C O_{2}) - 2 Δ_{f} G^{o} (A l_{2} O_{3})

= 3 (- 394) - 2 (- 1520) = 1858 k J

Δ G^{o} = - n F E^{o} = or = - E^{o}

= \frac{Δ G ^{o}}{n F} = \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, $Δ G_{f}^{o} (A l_{2} O_{3}) = - 1520$ kJ/mol and $Δ G_{f}^{o} (C O_{2}) = - 394$ kJ/mol]

$1.60 V$

$1.575 V$

$1.312 V$

$- 2.62 V$

Q. The minimum voltage required to electrolyse alumina in the Hall-Heroult process is [Given, ΔGfo​(Al2​O3​)=−1520 kJ/mol and ΔGfo​(CO2​)=−394 kJ/mol]

1.60V

1.575V

1.312V

−2.62V

In Hall-Heroult process the following reactions occur 3C+2Al2​O3​→4Al+3CO2​ ΔGo=3Δf​Go(CO2​)−2Δf​Go(Al2​O3​) =3(−394)−2(−1520)=1858kJ ΔGo=−nFEo=or=−Eo =nFΔGo​=12×965001858×1000​=1.60V

Q. The minimum voltage required to electrolyse alumina in the Hall-Heroult process is [Given, $Δ G_{f}^{o} (A l_{2} O_{3}) = - 1520$ kJ/mol and $Δ G_{f}^{o} (C O_{2}) = - 394$ kJ/mol]

$1.60 V$

$1.575 V$

$1.312 V$

$- 2.62 V$