Identifying The Reduction Half-Reaction A Detailed Explanation

$2 I^{-}(aq) + Cl_2(g) \longrightarrow 2 Cl^{-}(aq) + I_2(aq)$

Introduction to Redox Reactions

At the heart of many chemical transformations lies the concept of redox reactions, short for reduction-oxidation reactions. These reactions involve the transfer of electrons between chemical species, leading to changes in their oxidation states. To fully grasp the intricacies of redox reactions, it's essential to dissect them into their constituent half-reactions: reduction and oxidation. Reduction refers to the gain of electrons by a species, causing its oxidation state to decrease, while oxidation signifies the loss of electrons, resulting in an increase in oxidation state. These two processes always occur in tandem, as one species loses electrons, another must gain them.

In the given reaction, we observe the interaction between iodide ions ($I^−$) and chlorine gas ($Cl_2$) in an aqueous solution. The reaction proceeds with the formation of chloride ions ($Cl^−$) and elemental iodine ($I_2$). To identify the reduction half-reaction, we must pinpoint the species that gains electrons during this transformation. By examining the changes in oxidation states, we can deduce the electron transfer process. Chlorine gas ($Cl_2$) is converted into chloride ions ($Cl^−$). The oxidation state of chlorine in $Cl_2$ is 0, while in $Cl^−$ it is -1. This decrease in oxidation state indicates that chlorine has gained electrons, thus undergoing reduction. Therefore, to accurately describe this reduction process, we need to formulate the half-reaction that illustrates the electron gain by chlorine.

Dissecting the Reaction: Identifying the Reduction Half-Reaction

The provided chemical equation, $2 I^{-}(aq) + Cl_2(g) \longrightarrow 2 Cl^{-}(aq) + I_2(aq)$, showcases a classic redox reaction where electrons are transferred between species, leading to changes in their oxidation states. To discern the reduction half-reaction, we need to meticulously analyze the changes in oxidation states of the reactants and products. In this scenario, we have iodide ions ($I^−$) reacting with chlorine gas ($Cl_2$) to produce chloride ions ($Cl^−$) and elemental iodine ($I_2$). Our primary focus is to identify the species that gains electrons, as this will define the reduction half-reaction.

Let's delve into the oxidation states of the involved elements. Chlorine gas ($Cl_2$), being an element in its standard state, has an oxidation state of 0. Upon reaction, it transforms into chloride ions ($Cl^−$), where chlorine exhibits an oxidation state of -1. This transition from 0 to -1 clearly indicates a gain of electrons by chlorine, thereby undergoing reduction. Conversely, iodide ions ($I^−$) are converted to elemental iodine ($I_2$). The oxidation state of iodine changes from -1 in $I^−$ to 0 in $I_2$, signifying a loss of electrons, which means oxidation.

With the reduction process pinpointed as the conversion of $Cl_2$ to $Cl^−$, we can now focus on formulating the correct half-reaction. The half-reaction must accurately depict the gain of electrons by chlorine. This involves balancing the equation in terms of both atoms and charge. By meticulously accounting for the number of electrons gained, we can arrive at the precise representation of the reduction half-reaction.

Formulating the Reduction Half-Reaction for Chlorine

To accurately represent the reduction half-reaction, we need to consider the transformation of chlorine gas ($Cl_2$) into chloride ions ($Cl^−$). This process involves the gain of electrons by chlorine atoms. The fundamental principle behind balancing half-reactions is to ensure that both the number of atoms and the charge are conserved on both sides of the equation. Let's systematically construct the half-reaction.

Firstly, we begin with the species involved in the reduction: $Cl_2$ and $Cl^−$. We can write the initial form of the half-reaction as:

$Cl_2(g) \longrightarrow Cl^−(aq)$

Next, we need to balance the number of chlorine atoms. On the left side, we have two chlorine atoms in $Cl_2$, while on the right side, we have only one in $Cl^−$. To balance the atoms, we add a coefficient of 2 in front of $Cl^−$:

$Cl_2(g) \longrightarrow 2 Cl^−(aq)$

Now, we turn our attention to balancing the charge. On the left side, the overall charge is 0, as $Cl_2$ is a neutral molecule. On the right side, we have two chloride ions, each carrying a charge of -1, resulting in a total charge of -2. To balance the charge, we need to add electrons ($e^−$) to the side with the greater positive charge (or lesser negative charge). In this case, we add electrons to the left side:

$Cl_2(g) + 2 e^− \longrightarrow 2 Cl^−(aq)$

This equation is now balanced in terms of both atoms and charge. It accurately represents the reduction half-reaction, where chlorine gas gains two electrons to form two chloride ions. This balanced half-reaction is crucial for understanding the electron transfer process in the overall redox reaction.

Analyzing the Options: Identifying the Correct Half-Reaction

Given the reaction $2 I^{-}(aq) + Cl_2(g) \longrightarrow 2 Cl^{-}(aq) + I_2(aq)$, we've established that the reduction half-reaction involves the conversion of chlorine gas ($Cl_2$) to chloride ions ($Cl^−$). Through careful balancing, we derived the correct half-reaction: $Cl_2(g) + 2 e^− \longrightarrow 2 Cl^−(aq)$. Now, let's examine the provided options to pinpoint the one that accurately matches our derived half-reaction.

The options typically present various formulations of half-reactions, some of which may be incorrect due to errors in balancing atoms, charges, or both. The key to identifying the correct option lies in meticulously comparing each option against the balanced half-reaction we obtained. We need to ensure that the option correctly represents the gain of electrons by chlorine and maintains the balance of atoms and charges.

Let's consider a hypothetical set of options:

A. $Cl_2(g) \longrightarrow 2 Cl^−(aq) + 2 e^−$ B. $Cl_2(g) + e^− \longrightarrow Cl^−(aq)$ C. $Cl_2(g) + 2 e^− \longrightarrow 2 Cl^−(aq)$ D. $2 Cl^−(aq) \longrightarrow Cl_2(g) + 2 e^−$

Option A is incorrect because it shows electrons as products, indicating oxidation rather than reduction. Option B is incorrect as it does not balance the number of chlorine atoms. Option D is also incorrect because it represents the reverse process, the oxidation of chloride ions to chlorine gas. Option C, $Cl_2(g) + 2 e^− \longrightarrow 2 Cl^−(aq)$, perfectly matches our derived balanced half-reaction, making it the correct choice. This underscores the importance of systematically balancing half-reactions to accurately represent redox processes.

Conclusion: The Correct Reduction Half-Reaction

In summary, for the reaction $2 I^{-}(aq) + Cl_2(g) \longrightarrow 2 Cl^{-}(aq) + I_2(aq)$, the reduction half-reaction correctly describes the gain of electrons by chlorine gas ($Cl_2$) to form chloride ions ($Cl^−$). Through a detailed analysis of oxidation states and a meticulous balancing process, we identified the correct half-reaction as:

$Cl_2(g) + 2 e^− \longrightarrow 2 Cl^−(aq)$

This half-reaction accurately portrays the electron transfer process, where chlorine gas gains two electrons, resulting in the formation of two chloride ions. Understanding and formulating half-reactions is crucial for comprehending the fundamental principles of redox chemistry. By correctly identifying and balancing half-reactions, we can unravel the complexities of electron transfer processes in chemical reactions.

The correct option is A. $Cl_2(g) + 2 e^− \longrightarrow 2 Cl^−(aq)$

This option accurately represents the reduction half-reaction, where chlorine gas gains two electrons to form chloride ions. The other options either misrepresent the direction of electron transfer or fail to balance the atoms and charges correctly.

Repair Input Keyword

Which half-reaction correctly describes the reduction process taking place in the following reaction: $2 I^{-}(aq) + Cl_2(g) \longrightarrow 2 Cl^{-}(aq) + I_2(aq)$?