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
Answers
GPT-4o mini
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 analyze the oxidation states of the relevant elements before and after the reaction.
1. **Reactants:**
- In \(\text{Br}_2\), bromine has an oxidation state of 0 (since it's in elemental form).
- In \(\text{NaI}\), sodium (\(\text{Na}\)) has an oxidation state of +1, and iodine (\(\text{I}\)) has an oxidation state of -1.
2. **Products:**
- In \(\text{I}_2\), iodine has an oxidation state of 0 (again, as it is in elemental form).
- In \(\text{NaBr}\), sodium (\(\text{Na}\)) has an oxidation state of +1, and bromine (\(\text{Br}\)) has an oxidation state of -1.
Now let's track the changes in oxidation states:
- **Bromine (\(Br\))** changes from 0 in \(Br_2\) to -1 in \(NaBr\) (it gains electrons, so it is reduced).
- **Iodine (\(I\))** changes from -1 in \(NaI\) to 0 in \(I_2\) (it loses electrons, so it is oxidized).
Thus, we find that:
- Iodine (I) is oxidized (oxidation state goes from -1 to 0).
- Bromine (Br) is reduced (oxidation state goes from 0 to -1).
The correct answer is: **Iodine (I) is oxidized, and bromine (Br) is reduced.**
\[ \text{Br}_2 (l) + 2\text{NaI} (aq) \rightarrow \text{I}_2 (s) + 2\text{NaBr} (aq) \]
we need to analyze the oxidation states of the relevant elements before and after the reaction.
1. **Reactants:**
- In \(\text{Br}_2\), bromine has an oxidation state of 0 (since it's in elemental form).
- In \(\text{NaI}\), sodium (\(\text{Na}\)) has an oxidation state of +1, and iodine (\(\text{I}\)) has an oxidation state of -1.
2. **Products:**
- In \(\text{I}_2\), iodine has an oxidation state of 0 (again, as it is in elemental form).
- In \(\text{NaBr}\), sodium (\(\text{Na}\)) has an oxidation state of +1, and bromine (\(\text{Br}\)) has an oxidation state of -1.
Now let's track the changes in oxidation states:
- **Bromine (\(Br\))** changes from 0 in \(Br_2\) to -1 in \(NaBr\) (it gains electrons, so it is reduced).
- **Iodine (\(I\))** changes from -1 in \(NaI\) to 0 in \(I_2\) (it loses electrons, so it is oxidized).
Thus, we find that:
- Iodine (I) is oxidized (oxidation state goes from -1 to 0).
- Bromine (Br) is reduced (oxidation state goes from 0 to -1).
The correct answer is: **Iodine (I) is oxidized, and bromine (Br) is reduced.**