Question

The minimum voltage required to electrolyse alumina in the Hall-Heroult process is (Given, $\Delta G_{f}^{\circ}\left( Al _{2} O _{3}\right)=-1520\, kJ\, mol ^{-1}$;
$\Delta G_{f}^{\circ}\left( CO _{2}\right)=-394\, kJ\, mol ^{-1}$

• A. 1.575 V
• B. 1.60 V
• C. 1.312 V
• D. -2.62 V

Answer 1

Answer: (B)

In Hall-Heroult process, the following reactions occur
$3 C +2 Al _{2} O _{3} \longrightarrow 4 Al +3 CO _{2}$
$4 Al ^{3+}+12 e^{-} \longrightarrow 4 Al $
$\therefore \Delta G^{\circ} =3 \Delta G_{f}^{\circ}\left( CO _{2}\right)-2 \Delta G_{f}^{\circ}\left( Al _{2} O _{3}\right) $
$=3(-394)-2(-1520) $
$=1858\, kJ$
$\Delta G^{\circ} =-n F E_{\text {cell }}^{\circ}$
$-E_{\text {cell }}^{\circ} =\frac{1858 \times 1000}{12 \times 96500}=1.60\, V$