Question

*Which elements are oxidized and reduced in this reaction? * Br2 (l) + 2NaI(aq) → I2 (s) + 2NaBr(aq) * 1 point Iodine (I) is oxidized, and bromine (Br) is reduced. Bromine (Br) is oxidized, and iodine (I) is reduced. Sodium (Na) is oxidized, and bromine (Br) is reduced. Bromine (Br) is oxidized, and sodium (Na) is reduced

Answer 1

To determine which elements are oxidized and reduced in the reaction \( \text{Br}_2 (l) + 2\text{NaI}(aq) \rightarrow \text{I}_2 (s) + 2\text{NaBr}(aq) \), we need to look at the oxidation states of the elements involved.

1. In sodium iodide (\(\text{NaI}\)), sodium (Na) has an oxidation state of +1, while iodine (I) has an oxidation state of -1.
2. In bromine (\(\text{Br}_2\)), bromine has an oxidation state of 0 (as it is in its elemental form).
3. In iodine (\(\text{I}_2\)), iodine in the product also has an oxidation state of 0 (as it is in its elemental form).
4. In sodium bromide (\(\text{NaBr}\)), sodium (Na) still has an oxidation state of +1, while bromine (Br) has an oxidation state of -1 in this compound.

Now let's analyze the changes:
- **Iodine (I)** goes from -1 in \(\text{NaI}\) to 0 in \(\text{I}_2\) - it is oxidized (loss of electrons).
- **Bromine (Br)** goes from 0 in \(\text{Br}_2\) to -1 in \(\text{NaBr}\) - it is reduced (gain of electrons).

So, the correct answer is:
**Iodine (I) is oxidized, and bromine (Br) is reduced.**