2Na  +  Cl$_2$ → 2NaCl In the reaction above, the specie that undergoes reduction is

Analysis

To determine which species undergoes reduction, we need to look at the change in the oxidation states of the elements involved in the reaction: 2Na + Cl$_2$ → 2NaCl

Concept: Oxidation and Reduction (Redox)

In chemistry, we use the acronym OIL RIG to remember redox reactions:

• Oxidation Is Loss (of electrons, increase in oxidation number).
• Reduction Is Gain (of electrons, decrease in oxidation number).

The oxidation state of an element in its free or uncombined state (like Na or Cl$_2$) is always 0.

Step-by-Step Solution

1. Assign oxidation numbers to the reactants: • Na (elemental form) = 0 • Cl$_2$ (elemental form) = 0

2. Assign oxidation numbers to the products: • In NaCl, Sodium (Na) is in Group 1, so it has an oxidation state of +1. • Chlorine (Cl) in NaCl has an oxidation state of -1.

3. Track the changes: • Na: Goes from 0 to +1 (Increase in oxidation number = Oxidation) • Cl: Goes from 0 to -1 (Decrease in oxidation number = Reduction)

4. Identify the species: The chlorine atoms started as part of the Cl$_2$ molecule. Since the oxidation state of the chlorine atoms decreased from 0 to -1, the species that underwent reduction is Cl$_2$.

Conclusion

The chlorine molecule (Cl$_2$) gained electrons to become chloride ions (Cl$^-$) in the NaCl compound. Therefore, Cl$_2$ is the species that is reduced.

Exam Tip

Always remember that in a reaction between a metal and a non-metal, the metal is almost always oxidized (loses electrons), and the non-metal is almost always reduced (gains electrons).