Question

The minimum voltage required to electrolyse alumina in the Hall-Heroult process is [Given, $\Delta G^{o}_{f} (Al_{2} O_{3}) = - 1520$ kJ/mol and $\Delta G^{o}_{f} (CO_{2}) = -394$ kJ/mol]

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

Answer 1

Answer: (A)

In Hall-Heroult process the following reactions occur
$3C + 2Al_{2}O_{3} \to 4Al + 3CO_{2}$
$\Delta G^{o} = 3\Delta _{f}G^{o}(CO_{2}) - 2\Delta_{f} G^{o} (Al_{2}O_{3})$
$= 3 (-394) - 2 (-1520) = 1858 kJ$
$\Delta G^{o}= -nFE^o = or = -E^{o}$
$=\frac{\Delta G^{o}}{nF}=\frac{1858 \times1000}{12\times96500}=1.60 \,V$