In the reaction 2Al + 6NaCl → 2AlCl3 + 3Na, aluminum undergoes oxidation, losing electrons, and sodium undergoes reduction, gaining electrons.
In the reaction 2Al + 6NaCl → 2AlCl3 + 3Na, aluminum (Al) is oxidized, and sodium (Na) is reduced.
Oxidation involves the loss of electrons, and reduction involves the gain of electrons. In this reaction, aluminum starts with an oxidation state of 0 and ends up with an oxidation state of +3 in AlCl3. This change from 0 to +3 indicates that aluminum has lost electrons, and therefore, it has undergone oxidation.
On the other hand, sodium starts with an oxidation state of 0 and ends up with an oxidation state of +1 in NaCl. This change from 0 to +1 indicates that sodium has gained electrons, and therefore, it has undergone reduction.
The reaction can be split into two half-reactions to highlight the oxidation and reduction processes:
Oxidation half-reaction: 2Al → 2Al3+ + 6e-
Reduction half-reaction: 6Na+ + 6e- → 3Na
In summary, aluminum is oxidized as it loses electrons, and sodium is reduced as it gains electrons in the given chemical reaction.