Question

For the Hall-Héroult process, how would you determine the two half-reactions (anode and cathode) which make up this process, complete with electrons? Also by balancing the half-cells and determining how the net-ionic equation is obtained?

Answer 1

Here's what I find.

The Hall–Héroult process involves the electrolysis of aluminium oxide in molten cryolite in a cell with graphite electrodes.

In the cell, the `"Al"_2"O"_3` exists as separate `"Al"^"3+"` and `"O"^"2-"` ions in the molten cryolite.

The process reduces the aluminium ions to molten aluminium and oxidizes the graphite anodes to carbon dioxide.

`"Cathode:"color(white)(m) "Al"^"3+"("in cryolite") + 3"e"^"-" → "Al(l)"`
`"Anode:"color(white)(mm)"C(s) +2O"^"2-"("in cryolite") → "CO"_2"(g)" +4"e"^"-"`

To get the net ionic equation, we equalize the number of electrons transferred and add the two equations.

`4 × ["Al"^"3+"("in cryolite") + 3"e"^"-" → "Al(l)"]`
`3 × ul(["C(s) +2O"^"2-"("in cryolite") → "CO"_2"(g)" +4"e"^"-"] color(white)(mmmmmmmmmmmmm))`
`color(white)(mmll)4"Al"^"3+"("in cryolite") + "6O"^"2-"("in cryolite") + "3C(s)" → "4Al(l)" + "3CO"_2"(g)"`