Problem 83 - Chemistry and Chemical Reactivity

Question

Bromine is obtained from sea water by the following reaction: $$\mathrm{Cl}_{2}(\mathrm{g})+2 \mathrm{NaBr}(\mathrm{aq}) \longrightarrow 2 \mathrm{NaCl}(\mathrm{aq})+\mathrm{Br}_{2}(\ell)$$ (a) What has been oxidized? What has been reduced? (b) Identify the oxidizing and reducing agents. (c) What mass of \(\mathrm{Cl}_{2}\) is required to react completely with \(125 \mathrm{mL}\) of \(0.153 \mathrm{M} \mathrm{NaBr} ?\)

Step-by-Step Solution

Answer

(a) Br is oxidized, Cl is reduced. (b) Cl2 is the oxidizing agent, NaBr is the reducing agent. (c) 0.678 grams of Cl2 is required.

Step 1: Determine Oxidation States

First, we need to find the oxidation states of all elements in the reaction. In \( \mathrm{Cl}_{2} \), chlorine has an oxidation state of 0 because it is in its elemental form. In \( \mathrm{NaBr} \), sodium (Na) has an oxidation state of +1 and bromine (Br) -1. In \( \mathrm{NaCl} \), sodium (Na) is +1, while chlorine (Cl) is -1. Finally, in \( \mathrm{Br}_{2} \), bromine is 0 since it is in its elemental form.

Step 2: Identify Oxidized and Reduced Elements

Oxidation involves an increase in oxidation state, while reduction involves a decrease. In the given reaction, chlorine goes from an oxidation state of 0 in \( \mathrm{Cl}_{2} \) to -1 in \( \mathrm{NaCl} \), indicating it is reduced. Bromine goes from -1 in \( \mathrm{NaBr} \) to 0 in \( \mathrm{Br}_{2} \), showing it is oxidized.

Step 3: Determine Oxidizing and Reducing Agents

The substance that gets reduced is the oxidizing agent, and the one that gets oxidized is the reducing agent. Hence, \( \mathrm{Cl}_{2} \) is the oxidizing agent because it is reduced, while \( \mathrm{NaBr} \) is the reducing agent because it is oxidized.

Step 4: Calculate Moles of NaBr

We need to find the moles of \( \mathrm{NaBr} \) using its concentration and volume.\[ \text{Moles of } \mathrm{NaBr} = 0.153 \mathrm{M} \times 0.125 \mathrm{L} = 0.019125 \text{ moles.} \]

Step 5: Use Stoichiometry to Find Moles of Cl2

According to the balanced reaction equation, 2 moles of \( \mathrm{NaBr} \) react with 1 mole of \( \mathrm{Cl}_{2} \). Therefore, the moles of \( \mathrm{Cl}_{2} \) needed are half the moles of \( \mathrm{NaBr} \): \[ \text{Moles of } \mathrm{Cl}_{2} = \frac{0.019125}{2} = 0.0095625. \]

Step 6: Calculate Mass of Cl2

To find the mass of \( \mathrm{Cl}_{2} \), we multiply the moles by the molar mass of \( \mathrm{Cl}_{2} \) (70.906 g/mol):\[ \text{Mass of } \mathrm{Cl}_{2} = 0.0095625 \times 70.906 = 0.678 \text{ grams.} \]

Key Concepts

Oxidation States

Oxidation states are numerical values assigned to each element in a chemical compound, indicating the number of electrons an atom gains, loses, or appears to use when forming compounds. These numbers help track the transfer of electrons during chemical reactions, especially in redox (reduction-oxidation) reactions.

• In pure elements like \( \mathrm{Cl}_{2} \) and \( \mathrm{Br}_{2} \), the oxidation state is 0 since there is no electron gain or loss.
• The oxidation state of \( \mathrm{Na} \) in \( \mathrm{NaBr} \) and \( \mathrm{NaCl} \) is +1. Sodium typically loses one electron in compounds.
• For \( \mathrm{Br} \) in \( \mathrm{NaBr} \), the oxidation state is -1, since bromine gains one electron. In \( \mathrm{Br}_{2} \), it returns to 0.
• Similarly, \( \mathrm{Cl} \) in \( \mathrm{NaCl} \) has an oxidation state of -1, indicating gain of an electron compared to its elemental state of 0 in \( \mathrm{Cl}_{2} \).

Understanding oxidation states helps identify which elements undergo oxidation or reduction in a reaction. This is pivotal in figuring out how electrons are transferred in redox reactions.

Stoichiometry

Stoichiometry is the part of chemistry that quantitatively analyzes the relationships between reactants and products in a chemical reaction. By understanding stoichiometry, you can predict the amount of substances consumed and produced in a reaction.
In our example, the balanced equation is:\[\mathrm{Cl}_{2}(\mathrm{g}) + 2 \mathrm{NaBr}(\mathrm{aq}) \longrightarrow 2 \mathrm{NaCl}(\mathrm{aq}) + \mathrm{Br}_{2}(\ell)\] Here, stoichiometry guides us in knowing that:

• Every 1 mole of \( \mathrm{Cl}_{2} \) reacts with 2 moles of \( \mathrm{NaBr} \).
• If you start with a specific quantity of \( \mathrm{NaBr} \), the stoichiometric ratio tells us how much \( \mathrm{Cl}_{2} \) is needed and how much \( \mathrm{Br}_{2} \) and \( \mathrm{NaCl} \) will form.

In the exercise, we used stoichiometry to determine the moles of \( \mathrm{Cl}_{2} \) required from a given amount of \( \mathrm{NaBr} \), eventually finding the mass of \( \mathrm{Cl}_{2} \) needed for the reaction to proceed.

Oxidizing Agent

In a redox reaction, the oxidizing agent is the species that accepts electrons during the reaction. It gains electrons and is thereby reduced. This is a crucial role as it helps driven the entire reaction process by taking electrons from the reducing agent.
In the given reaction:

• \( \mathrm{Cl}_{2} \) acts as the oxidizing agent. Initially, it is in the elemental form with an oxidation state of 0.
• As the reaction proceeds, it gains electrons from \( \mathrm{NaBr} \) and its oxidation state decreases to -1 in \( \mathrm{NaCl} \).

Through this electron acceptance, \( \mathrm{Cl}_{2} \) facilitates the oxidation of bromine in \( \mathrm{NaBr} \) to \( \mathrm{Br}_{2} \), making the reaction possible.

Reducing Agent

A reducing agent donates electrons in a redox reaction, becoming oxidized in the process. It drives the reduction of another chemical by giving up electrons. Understanding the role of a reducing agent is essential, as it highlights the source of electrons for the reaction.
In our chemical reaction:

• \( \mathrm{NaBr} \) operates as the reducing agent.
• In \( \mathrm{NaBr} \), bromine, which has an oxidation state of -1, loses an electron and its oxidation state goes to 0 in \( \mathrm{Br}_{2} \).
• This electron loss enables the reduction of \( \mathrm{Cl}_{2} \) to form \( \mathrm{NaCl} \).

By transferring electrons to \( \mathrm{Cl}_{2} \), \( \mathrm{NaBr} \) allows the reaction to occur, making it an integral player in achieving the desired chemical transformation.