Question

Use the chemical equation to answer the question.Br2(l)+2NaI(aq)→I2(s)+2NaBr(aq) Which elements are oxidized and reduced in the reaction?(1 point)ResponsesBromine (Br) is oxidized, and iodine (I)is reduced.Bromine left parenthesis cap b r right parenthesis is oxidized, and iodine left parenthesis cap i right parenthesis is reduced.Sodium (Na) is oxidized, and bromine (Br) is reduced.Sodium left parenthesis cap n A right parenthesis is oxidized, and bromine left parenthesis cap b r right parenthesis is reduced.Iodine (I) is oxidized, and bromine (Br) is reduced.Iodine left parenthesis cap I right parenthesis is oxidized, and bromine left parenthesis cap b r right parenthesis is reduced.Bromine (Br) is oxidized, and sodium (Na) is reduced.

Answer 1

To identify which elements are oxidized and reduced in the reaction:
\[ \text{Br}_2(l) + 2 \text{NaI}(aq) \rightarrow \text{I}_2(s) + 2 \text{NaBr}(aq) \]

1. **Identify oxidation states:**
- In \(\text{Br}_2\), bromine has an oxidation state of 0 (elemental form).
- In \(\text{NaI}\), sodium (Na) is +1 and iodine (I) is -1.
- In \(\text{I}_2\), iodine has an oxidation state of 0 (elemental form).
- In \(\text{NaBr}\), sodium (Na) is +1 and bromine (Br) is -1.

2. **Analyze changes in oxidation states:**
- Bromine (\(\text{Br}_2\)) changes from 0 to -1 in \(\text{NaBr}\). This means bromine is **reduced**, as it gains electrons.
- Iodine (\(\text{I}\)) changes from -1 in \(\text{NaI}\) to 0 in \(\text{I}_2\). This means iodine is **oxidized**, as it loses electrons.

From this analysis:
- **Iodine (I) is oxidized**, and **bromine (Br) is reduced**.

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