Answer:
Oxidation half-reaction : Na(s) → Na⁺ + 1 e-
Reduction half-reaction: Cl₂ + 2 e- → 2 Cl⁻
Explanation:
Oxidation half-reaction: solid sodium (Na(s)) has an oxidation number of 0. It loses 1 electron and forms the cation Na⁺. So, the half-reaction is the following:
Na(s) → Na⁺ + 1 e-
Reduction half-reaction: chlorine gas (Cl₂) has an oxidation number of 0. Each atom of Cl gains 1 electron to form two Cl⁻ ions, according to the following half-reaction:
Cl₂ + 2 e- → 2 Cl⁻
The total oxidation-reduction reaction is obtained by adding the oxidation half-reaction multiplied by 2 (to balance the electrons) and the reduction half-reaction, as follows:
2 x (Na(s) → Na⁺ + 1 e-)
Cl₂ + 2 e- → 2 Cl⁻
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2Na(s) + Cl₂ → 2NaCl
