Answer:
Al(s) + 3CuCl₂(aq) → 3Cu(s) + 2AlCl₃(aq)
Explanation:
We define the reactants:
Al(s) and CuCl₂ (aq)
Our products are: Cu(s) and AlCl₃
This is a redox reaction where the Al and the Cu, change the oxidation state
Al(s) → Al³⁺ + 3e⁻ ⇒ Oxidation
Cu²⁺(aq) + 2e⁻ → Cu(s) ⇒ Reduction
We multiply both reactions, in order to have the electrons balanced
(Al(s) → Al³⁺ + 3e⁻ ) .2 ⇒ 2Al(s) → 2Al³⁺ + 6e⁻
(Cu²⁺(aq) + 2e⁻ → Cu(s) ) .3 ⇒ 3Cu²⁺(aq) + 6e⁻ → 3Cu(s)
We sum both half reactions, to cancel the electrons:
2Al(s) + 3Cu²⁺(aq) + 6e⁻ → 3Cu(s) + 2Al³⁺ + 6e⁻
We add the chlorides, to get the balanced reaction
Balanced reaction is: Al(s) + 3CuCl₂(aq) → 3Cu(s) + 2AlCl₃(aq)
