How many Faradays of charge are required to reduce of ions to metal? — Mole Concept and Some Basic Concepts of Chemistry Chemistry Question

💡 Solution & Explanation

Reduction half-reaction: $Al^{3+} + 3e^- \rightarrow Al$. <br>To reduce $1 \text{ mole}$ of $Al^{3+}$, $3 \text{ moles}$ of electrons are required. <br>Since $1 \text{ mole}$ of electrons carries $1 \text{ Faraday}$ of charge, the answer is 3 F.