In the cathodic half-reaction of the Hall-Heroult process, how many moles of electrons are required — Metallurgy and Isolation of Elements Chemistry Question

💡 Solution & Explanation

The cathodic reduction reaction in the Hall-Heroult process is: $Al^{3+} + 3e^- \rightarrow Al_{(l)}$. To deposit 1 mole of Aluminium, 3 moles of electrons are consumed.