Electrons Transferred In Redox Reaction: A Chemistry Guide

Hey everyone! Today, we're diving deep into the fascinating world of redox reactions, focusing specifically on how to determine the number of electrons transferred in a balanced reaction. We'll be using the given half-reactions for the reaction $Au(OH)_3 + HI ightarrow Au + I_2 + H_2O$ as our example. So, buckle up, and let's get started!

Understanding Redox Reactions and Half-Reactions

First things first, what exactly are redox reactions? Redox reactions, short for reduction-oxidation reactions, are chemical reactions where electrons are transferred between reactants. This electron transfer results in changes in the oxidation states of the atoms involved. One substance loses electrons (oxidation), while another gains electrons (reduction). Think of it like a see-saw; one side goes up (oxidation), and the other goes down (reduction).

Now, let's break down half-reactions. To better understand the electron transfer, we split the overall redox reaction into two half-reactions: the oxidation half-reaction and the reduction half-reaction. The oxidation half-reaction shows the loss of electrons, while the reduction half-reaction shows the gain of electrons. These half-reactions provide a clear picture of what's happening at the atomic level. In our example, we have:

• Reduction half-reaction: $Au^{+3} + 3e^- ightarrow Au$
• Oxidation half-reaction: $2I^- ightarrow I_2 + 2e^-$

Deep Dive into the Reduction Half-Reaction

In this reduction half-reaction, the gold ion ($Au^{+3}$) gains three electrons ($3e^-$) to become neutral gold ($Au$). This gain of electrons is the defining characteristic of reduction. The oxidation state of gold changes from +3 to 0. So, essentially, gold is being reduced in this process. Think of it like this: the positive charge is being reduced by the addition of negatively charged electrons.

Exploring the Oxidation Half-Reaction

Conversely, in the oxidation half-reaction, two iodide ions ($2I^-$) lose two electrons to form iodine ($I_2$). This loss of electrons is oxidation. The oxidation state of iodine changes from -1 to 0. Iodine is being oxidized, meaning it's losing electrons. Remember, oxidation is the loss of electrons, and it often involves an increase in oxidation state.

Determining the Number of Electrons Transferred

Okay, now for the million-dollar question: How many electrons are transferred in the overall reaction? This is where balancing the half-reactions becomes crucial. The key principle here is that the number of electrons lost in the oxidation half-reaction must equal the number of electrons gained in the reduction half-reaction. This is the golden rule of balancing redox reactions!

Looking at our half-reactions, we see that:

• The reduction half-reaction involves a gain of 3 electrons.
• The oxidation half-reaction involves a loss of 2 electrons.

These numbers aren't equal, so we need to find a common multiple to balance them. The least common multiple of 3 and 2 is 6. Therefore, we need to manipulate the half-reactions so that both involve 6 electrons.

Balancing the Electrons: A Step-by-Step Approach

To get 6 electrons in the reduction half-reaction, we multiply the entire reaction by 2:

$2 imes (Au^{+3} + 3e^- ightarrow Au)$ becomes $2Au^{+3} + 6e^- ightarrow 2Au$

Now, the reduction half-reaction involves 6 electrons being gained.

To get 6 electrons in the oxidation half-reaction, we multiply the entire reaction by 3:

$3 imes (2I^- ightarrow I_2 + 2e^-)$ becomes $6I^- ightarrow 3I_2 + 6e^-$

Now, the oxidation half-reaction involves 6 electrons being lost.

The Grand Finale: Combining the Balanced Half-Reactions

Now that both half-reactions involve the same number of electrons (6), we can combine them to get the balanced overall redox reaction. We simply add the two balanced half-reactions together:

$(2Au^{+3} + 6e^- ightarrow 2Au) + (6I^- ightarrow 3I_2 + 6e^-)$

This gives us:

$2Au^{+3} + 6e^- + 6I^- ightarrow 2Au + 3I_2 + 6e^-$

Notice that we have 6 electrons on both sides of the equation. These electrons cancel out, because they are transferred but not consumed in the reaction. This is the magic of redox reactions: the electrons lost by one species are gained by another.

The Balanced Redox Reaction

After canceling the electrons, we get the balanced ionic equation:

$2Au^{+3} + 6I^- ightarrow 2Au + 3I_2$

This balanced equation tells us the stoichiometry of the reaction, meaning the ratio in which the reactants and products react. We can see that 2 gold ions react with 6 iodide ions to produce 2 gold atoms and 3 iodine molecules.

Answering the Question: The Number of Electrons Transferred

So, finally, we have our answer! Based on the balanced half-reactions and the overall balanced redox reaction, the number of electrons transferred in this reaction is 6. This is the number of electrons that are effectively