Problem 69 - Chemistry The Molecular Science

Question

Given the reaction $$\mathrm{Cl}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{HOCl}(\mathrm{aq})$$ (a) identify the oxidizing agent and the reducing agent. (b) write the equilibrium constant expression for the reaction. (c) calculate the concentration of \(\mathrm{HOCl}\) in equilibrium with $$\mathrm{Cl}_{2}(\mathrm{~g}) \text { at } 1.0 \mathrm{~atm} . K_{\mathrm{c}}=2.7 \times 10^{-5}$$.

Step-by-Step Solution

Answer

The oxidizing and reducing agent is \(\mathrm{Cl}_2\). \(K_c = \frac{[\mathrm{H}^{+}][\mathrm{Cl}^{-}][\mathrm{HOCl}]}{[\mathrm{Cl}_2]}\). \([\mathrm{HOCl}] = 3.0 \times 10^{-2} \text{ M}\).

Step 1: Identify Oxidation States

To determine the oxidizing and reducing agents, start by identifying the oxidation states of the elements involved. In the reactants, Cl in \(\mathrm{Cl}_2\) has an oxidation state of 0, and elements in their standard state like \(\mathrm{H}_2\) in water also have an oxidation state of 0. In the products, the oxidation state of \(\mathrm{Cl}^{-}\) is -1, \(\mathrm{H}^{+}\) is +1, and in \(\mathrm{HOCl}\), O is usually -2, Cl is +1, and H is +1. Thus, \(\mathrm{Cl}_2\) goes from 0 to -1 and +1, suggesting it undergoes both reduction and oxidation.

Step 2: Identify the Oxidizing and Reducing Agents

The species that is reduced gains electrons; \(\mathrm{Cl}_2\) gains electrons to become \(\mathrm{Cl}^{-}\). The species that is oxidized loses electrons; \(\mathrm{Cl}_2\) loses electrons to form \(\mathrm{HOCl}\). Thus, \(\mathrm{Cl}_2\) serves as both the oxidizing and reducing agent in this reaction.

Step 3: Write Equilibrium Constant Expression

The general formula for the equilibrium constant \(K_c\) in terms of concentrations is given by the ratio of the products to the reactants, each raised to the power of their coefficients in the balanced equation. For this reaction, \[ K_c = \frac{[\mathrm{H}^{+}][\mathrm{Cl}^{-}][\mathrm{HOCl}]}{[\mathrm{Cl}_2]} \] since \([\mathrm{H}_2O]\) is a liquid, its concentration is essentially constant and is omitted from the expression.

Step 4: Calculate HOCl Concentration

To find the concentration of \(\mathrm{HOCl}\) at equilibrium, use the given \(K_c\) value and solve the equilibrium expression for \([\mathrm{HOCl}]\). Given \([\mathrm{Cl}_2] = 1.0\) atm, initially assumed, the system achieves equilibrium as: \[ 2.7 \times 10^{-5} = \frac{[x][x][x]}{[1.0 - x]} \] Solving for \(x\), assuming \(x\) (change in concentration) is very small compared to 1, simplifying gives \([\mathrm{HOCl}] \approx \sqrt[3]{2.7 \times 10^{-5}}\). You can further approximate: \[ [\mathrm{HOCl}] \approx 3.0 \times 10^{-2} \text{ M} \].

Key Concepts

Oxidizing Agent

In the context of chemical reactions, an oxidizing agent plays a crucial role by accepting electrons from another substance. It essentially causes the oxidation of the species from which it takes electrons. In the given reaction involving chlorine gas (\(\mathrm{Cl}_2\)) and water, \(\mathrm{Cl}_2\) acts as the oxidizing agent. Typically, the oxidizing agent is the species being reduced, which means it gains electrons during the reaction. Here’s how it works: - Chlorine starts with an oxidation state of 0.- Through the reaction process, some chlorine atoms are reduced to chloride ions (\(\mathrm{Cl}^{-}\)), changing their oxidation state to -1. - This reduction process confirms that \(\mathrm{Cl}_2\) is the oxidizing agent in this scenario.

Reducing Agent

The reducing agent is equally important in a redox reaction. It donates electrons to another substance, effectively reducing that substance while being oxidized itself. In our example with chlorine gas and water, again, \(\mathrm{Cl}_2\) is responsible for being not only the oxidizing but also the reducing agent.- Initially, chlorine has an oxidation state of 0.- Part of it is oxidized to form hypochlorous acid (\(\mathrm{HOCl}\)), where the chlorine attains an oxidation state of +1.- Hence, \(\mathrm{Cl}_2\) oxidizes itself by losing electrons in this part of the reaction, making it the reducing agent as well. This intriguing aspect of the molecule being both an oxidizing and a reducing agent in the same reaction is known as a disproportionation reaction.

Equilibrium Constant Expression

An equilibrium constant expression is a mathematical representation that describes the ratio of the concentrations of products to the reactants for a reversible reaction at equilibrium. In the reaction we consider here, the equilibrium constant expression, represented as \(K_c\), can be derived using the concentrations of the reactants and products.- For the reaction \(\mathrm{Cl}_{2} + \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}^{+} + \mathrm{Cl}^{-} + \mathrm{HOCl}\), the expression for \(K_c\) becomes: \[ K_c = \frac{[\mathrm{H}^{+}][\mathrm{Cl}^{-}][\mathrm{HOCl}]}{[\mathrm{Cl}_2]} \]- Note that the concentration of water \([\mathrm{H}_2O]\), being a liquid, can be considered constant and is not included in the expression.This expression helps quantify the position of equilibrium and predict how the system behaves under various conditions.

Oxidation States

Understanding oxidation states is fundamental to identifying oxidizing and reducing agents in a reaction. Oxidation states help describe the charge of an atom if it were to lose or gain electrons entirely to bonds. Here is a quick overview of oxidation states relevant to our reaction:- \(\mathrm{Cl}_2\) begins with an oxidation state of 0. - When reduced, chlorine turns into \(\mathrm{Cl}^{-}\) with an oxidation state of -1.- In \(\mathrm{HOCl}\), chlorine achieves an oxidation state of +1.These changes highlight the redox processes occurring:- A decrease in oxidation state (from 0 to -1) represents reduction.- An increase in oxidation state (from 0 to +1) implies oxidation.By following the alteration in oxidation states, it becomes clear which species are undergoing reduction or oxidation in any given chemical reaction.