Question

Balance the following net ionic equation:
Al(s) + Cu²⁺(aq) → Al³⁺(aq) + Cu(s)

Answer 1

The balanced net ionic equation for the reaction between solid aluminum and aqueous copper ions is 2Al(s) + 3Cu²⁺(aq) → 2Al³⁺(aq) + 3Cu(s), with aluminum being oxidized and copper ions being reduced.

Step-by-step explanation:

To balance the net ionic equation Al(s) + Cu²⁺(aq) → Al³⁺(aq) + Cu(s), one must ensure that the number of atoms and the charge are balanced on both sides of the equation. In this single replacement reaction, aluminum (Al) is oxidized to aluminum ions (Al³⁺) while copper ions (Cu²⁺) are reduced to copper (Cu). The balanced equation should reflect the transfer of electrons during the oxidation and reduction process.

We can represent this process by the following half-reactions:

1. Oxidation half-reaction: Al(s) → Al³⁺(aq) + 3e⁻
2. Reduction half-reaction: Cu²⁺(aq) + 2e⁻ → Cu(s)

To balance the net ionic equation, we need the same number of electrons lost and gained, which means we need to find the lowest common multiple of electrons transferred in the half-reactions. The lowest common multiple of 2 and 3 is 6, so we multiply the aluminum half-reaction by 2 and the copper half-reaction by 3, giving us:

1. Oxidation: 2Al(s) → 2Al³⁺(aq) + 6e⁻
2. Reduction: 3Cu²⁺(aq) + 6e⁻ → 3Cu(s)

Combining these balanced half-reactions, we get the balanced net ionic equation:

2Al(s) + 3Cu²⁺(aq) → 2Al³⁺(aq) + 3Cu(s)