Question

Bromine is obtained from sea water by the following redox reaction: $$\mathrm{Cl}_{2}(\mathrm{g})+2 \mathrm{NaBr}(\mathrm{aq}) \rightarrow 2 \mathrm{NaCl}(\mathrm{aq})+\mathrm{Br}_{2}(\ell)$$ (a) What has been oxidized? What has been reduced? (b) Identify the oxidizing and reducing agents.

Short answer

Br- is oxidized; Cl2 is reduced. Oxidizing agent: Cl2. Reducing agent: NaBr.

Step-by-step solution

Write Down Oxidation States

Assign oxidation states to each element in the reactants and products:- In \( \mathrm{Cl}_2 \), each \( \mathrm{Cl} \) is in the 0 state.- In \( \mathrm{NaBr} \), \( \mathrm{Na} \) is +1 and \( \mathrm{Br} \) is -1.- In \( \mathrm{NaCl} \), \( \mathrm{Na} \) is +1 and \( \mathrm{Cl} \) is -1.- In \( \mathrm{Br}_2 \), each \( \mathrm{Br} \) is in the 0 state.

Determine Change in Oxidation States

Examine the changes in oxidation states:- \( \mathrm{Br} \) goes from -1 in \( \mathrm{NaBr} \) to 0 in \( \mathrm{Br}_2 \), indicating oxidation.- \( \mathrm{Cl} \) goes from 0 in \( \mathrm{Cl}_2 \) to -1 in \( \mathrm{NaCl} \), indicating reduction.

Identify Oxidized and Reduced Species

From Step 2:- \( \mathrm{Br}^- \) is oxidized to \( \mathrm{Br}_2 \).- \( \mathrm{Cl}_2 \) is reduced to \( \mathrm{Cl}^- \).

Identify Oxidizing and Reducing Agents

- The species oxidized is \( \mathrm{Br}^- \), making \( \mathrm{NaBr} \) the reducing agent.- The species reduced is \( \mathrm{Cl}_2 \), making \( \mathrm{Cl}_2 \) the oxidizing agent.

Key concepts

oxidation states

To understand redox reactions, it's essential to grasp what oxidation states are. Oxidation states, also known as oxidation numbers, represent the degree of oxidation of an atom in a compound. They help us track how many electrons an atom gains, loses, or shares during a reaction.
For elements in their standard state, like \( \mathrm{Cl}_2 \) and \( \mathrm{Br}_2 \), the oxidation state is zero. This is because they are not combined with any other elements. In contrast, in compounds like \( \mathrm{NaBr} \) and \( \mathrm{NaCl} \), each element has a specific oxidation state.
In \( \mathrm{NaBr} \), sodium (\( \mathrm{Na} \)) typically has an oxidation state of +1. Bromine (\( \mathrm{Br} \)) is at -1. In \( \mathrm{NaCl} \), the oxidation states are similar: sodium is +1, while chlorine (\( \mathrm{Cl} \)) is -1. Recognizing these values allows us to detect changes in oxidation states and understand the flow of electrons in reactions.

oxidizing agent

An oxidizing agent is a substance that brings about oxidation in another substance by gaining electrons. In doing so, the oxidizing agent is itself reduced. It's crucial in determining the direction of electron transfer.
For example, in the reaction \( \mathrm{Cl}_{2} + 2 \mathrm{NaBr} \rightarrow 2 \mathrm{NaCl} + \mathrm{Br}_{2} \), \( \mathrm{Cl}_2 \) acts as the oxidizing agent. Initially, it has an oxidation state of 0. During the reaction, it gains electrons and its oxidation state decreases to -1 in \( \mathrm{NaCl} \).
This electron gain by \( \mathrm{Cl}_2 \) is what defines it as the oxidizing agent. It's vital to remember that oxidizing agents are always reduced during redox reactions. They play an essential role in allowing the other substance – in this case, bromine – to be oxidized.

reducing agent

A reducing agent is the opposite of an oxidizing agent. It's a substance that donates electrons to another substance, reducing it. In this process, the reducing agent itself is oxidized.
In the reaction considered \( \mathrm{Cl}_{2}\) + \(2 \mathrm{NaBr}\) \(\rightarrow 2 \mathrm{NaCl}\) + \(\mathrm{Br}_{2}\), \(\mathrm{NaBr} \) acts as the reducing agent. Bromine in \( \mathrm{NaBr} \) starts with an oxidation state of -1 and changes to 0 when it becomes \( \mathrm{Br}_2 \).
As bromine loses electrons, it is oxidized, showing that \( \mathrm{NaBr} \) facilitated the electron donation process. Keep in mind that reducing agents are always oxidized in the reactions they are part of. Identifying them helps understand who donates electrons and who receives them.