Bromine reacts with sodium iodide to yield iodine and sodium bromide. Identify the oxidizing and reducing agents, and write the balanced equation.
The Correct Answer and Explanation is:
Balanced Chemical Equation:
Br2+2NaI→2NaBr+I2\text{Br}_2 + 2\text{NaI} \rightarrow 2\text{NaBr} + \text{I}_2Br2+2NaI→2NaBr+I2
Identifying the Oxidizing and Reducing Agents:
- Oxidizing Agent: Bromine (Br2\text{Br}_2Br2)
- Reducing Agent: Sodium iodide (NaI\text{NaI}NaI)
Explanation
This reaction involves halogen displacement, where a more reactive halogen (bromine) displaces a less reactive halogen (iodine) from its compound. To understand the redox process, we must examine the changes in oxidation states of the elements involved.
Bromine (Br2\text{Br}_2Br2) is in its elemental form, so its oxidation state is 0. In the product sodium bromide (NaBr\text{NaBr}NaBr), bromine exists as bromide ion (Br−\text{Br}^-Br−), which has an oxidation state of –1. This indicates that bromine has gained electrons, meaning it has been reduced. Therefore, bromine is the oxidizing agent, since oxidizing agents are substances that cause other species to be oxidized while they themselves are reduced.
On the other side, iodide ions in sodium iodide (NaI\text{NaI}NaI) start with an oxidation state of –1. In elemental iodine (I2\text{I}_2I2), each iodine atom has an oxidation state of 0. This change shows that iodine atoms lost electrons, which means they were oxidized. Substances that lose electrons in a redox reaction are reducing agents, so sodium iodide is the reducing agent.
This redox reaction highlights the ability of halogens to displace less reactive halogens from their ionic compounds. Bromine, being above iodine in the halogen group, is more reactive and thus capable of oxidizing iodide ions to iodine. The sodium ion (Na+\text{Na}^+Na+) is a spectator ion and does not participate in the redox change. It only helps to balance the charge in the compounds.
This reaction is often used in laboratories to demonstrate redox chemistry and the relative reactivity of halogens. It also provides a visible color change: the brownish-orange color of bromine disappears, and the violet or brown color of iodine appears, depending on the solvent.
