Question

Write a balanced equation describing the reduction of \(\mathrm{H}_{2} \mathrm{SeO}_{4}\) by \(\mathrm{SO}_{2}\) to produce selenium.

Short answer

The balanced equation describing the reduction of H₂SeO₄ by SO₂ to produce selenium is: \[ 3 \mathrm{SO}_{2} + 2 \mathrm{H}_{2}\mathrm{SeO}_{4} \rightarrow 3\mathrm{S} + 3\mathrm{O}_{2} + 2\mathrm{Se} + 4 \mathrm{H}_{2}\mathrm{O} \]

Step-by-step solution

Write the unbalanced redox reaction

We start by writing the unbalanced redox reaction with the given reactants and products. \[ \mathrm{H}_{2}\mathrm{SeO}_{4} + \mathrm{SO}_{2} \rightarrow \mathrm{Se} + ? \]

Identify and balance the half-reactions

First, assign oxidation numbers to each element in the reactants and products: \(H_2SeO_4\): H = +1, Se = +6, and O = -2 \(SO_2\): S = +4 and O = -2 \(Se\): Se = 0 From the oxidation numbers, we can identify the half-reactions: Oxidation half-reaction: \(SO_2\) → S Reduction half-reaction: \(H_2SeO_4\) → Se Now, we balance the atoms in each half-reaction: Oxidation half-reaction: \[ \mathrm{SO}_{2} \rightarrow \mathrm{S} + \mathrm{O}_{2} \] Reduction half-reaction: \[ \mathrm{H}_{2}\mathrm{SeO}_{4} \rightarrow \mathrm{Se} + 2 \mathrm{H}_{2}\mathrm{O} \]

Balance the charges

To balance the charges, we add electrons to the appropriate side of each half-reaction. Oxidation half-reaction: Since S goes from +4 to 0, we need to add 4 electrons to the right side. \[ \mathrm{SO}_{2} \rightarrow \mathrm{S} + \mathrm{O}_{2} + 4 e^{-} \] Reduction half-reaction: Since Se goes from +6 to 0, we need to add 6 electrons to the left side. \[ 6 e^{-} + \mathrm{H}_{2}\mathrm{SeO}_{4} \rightarrow \mathrm{Se} + 2 \mathrm{H}_{2}\mathrm{O} \]

Combine the half-reactions

To combine the half-reactions, we need to multiply each by a factor that makes the number of electrons equal in both half-reactions. Since we have 4 electrons in the oxidation half-reaction and 6 in the reduction half-reaction, we can find the least common multiple, which is 12. Therefore, we need to multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2. Oxidation half-reaction x 3: \[ 3 \mathrm{SO}_{2} \rightarrow 3\mathrm{S} + 3\mathrm{O}_{2} + 12 e^{-} \] Reduction half-reaction x 2: \[ 12 e^{-} + 2 \mathrm{H}_{2}\mathrm{SeO}_{4} \rightarrow 2\mathrm{Se} + 4 \mathrm{H}_{2}\mathrm{O} \] Combining the balanced half-reactions: \[ 3 \mathrm{SO}_{2} + 2 \mathrm{H}_{2}\mathrm{SeO}_{4} \rightarrow 3\mathrm{S} + 3\mathrm{O}_{2} + 2\mathrm{Se} + 4 \mathrm{H}_{2}\mathrm{O} \]

Final Balanced Equation

Now, we have the balanced chemical equation for the reduction of H₂SeO₄ by SO₂ to produce selenium: \[ 3 \mathrm{SO}_{2} + 2 \mathrm{H}_{2}\mathrm{SeO}_{4} \rightarrow 3\mathrm{S} + 3\mathrm{O}_{2} + 2\mathrm{Se} + 4 \mathrm{H}_{2}\mathrm{O} \]

Key concepts

Oxidation Numbers

Understanding oxidation numbers is crucial when dealing with redox reactions. They help us determine how electrons are transferred during the reaction. In chemistry, oxidation numbers are theoretical charges assigned to atoms, assuming a complete transfer of electrons. For the molecule \( \mathrm{H}_{2}\mathrm{SeO}_{4} \), the oxidation number of hydrogen is +1, selenium is +6, and oxygen is -2. For \( \mathrm{SO}_{2} \), sulfur is +4, and oxygen remains -2. The change in oxidation numbers tells us which elements are oxidized and reduced.

By comparing the initial and final states, selenium's oxidation number dropping from +6 to 0 indicates reduction, as it gains electrons. Conversely, sulfur's change from +4 to 0 signifies oxidation, as it loses electrons. Assigning these numbers correctly is the first step in balancing redox reactions.

Half-Reactions

In redox reactions, it's helpful to split the process into half-reactions. This separation isolates the oxidation and reduction processes. For our reaction, the reduction half-reaction involves \( \mathrm{H}_{2}\mathrm{SeO}_{4} \) converting to \( \mathrm{Se} \), while the oxidation half-reaction involves \( \mathrm{SO}_{2} \) converting to \( \mathrm{S} \).

Reduction indicates a gain of electrons. In this example, \( \mathrm{H}_{2}\mathrm{SeO}_{4} \) gains electrons to form \( \mathrm{Se} \). The reaction can be written as:

• \( \mathrm{H}_{2}\mathrm{SeO}_{4} \rightarrow \mathrm{Se} + 2 \mathrm{H}_{2}\mathrm{O} \)
• Adding 6 electrons: \( 6 e^{-} + \mathrm{H}_{2}\mathrm{SeO}_{4} \rightarrow \mathrm{Se} + 2 \mathrm{H}_{2}\mathrm{O} \)

Oxidation, on the other hand, involves the loss of electrons, as shown in \( \mathrm{SO}_{2} \) converting to \( \mathrm{S} \). This process is represented as:

• \( \mathrm{SO}_{2} \rightarrow \mathrm{S} + \mathrm{O}_{2} \)
• Adding electrons: \( \mathrm{SO}_{2} \rightarrow \mathrm{S} + \mathrm{O}_{2} + 4 e^{-} \)

By isolating these processes, we ensure a streamlined approach to balancing the redox equation.

Electron Transfer

Electron transfer is at the heart of redox reactions, marking the transition of electrons from one element to another. In our redox equation, electrons are moving between sulfur and selenium. The concept of electron transfer explains how oxidation and reduction occur simultaneously in a redox reaction.

Upon separating the reaction into half-reactions, each half-reaction involves a balance of electron transfer. In the oxidation half-reaction, \( \mathrm{SO}_{2} \) loses 4 electrons, while in the reduction half-reaction, \( \mathrm{H}_{2}\mathrm{SeO}_{4} \) gains 6 electrons. To synchronize these processes, both half-reactions must exchange an equal quantity of electrons.

Thus, the least common multiple of 4 and 6 (which is 12) is used, ensuring that each half-reaction is adjusted to involve 12 electrons. By multiplying the oxidation half-reaction by 3 and the reduction half-reaction by 2, we equalize the electron transfer, achieving a balanced redox equation.

Balancing Chemical Equations

Balancing chemical equations ensures conservation of mass and charge, important fundamentals of chemistry. With redox reactions, achieving a balanced equation means matching the number of atoms and the overall charge on both sides of the equation. This is where our understanding of oxidation numbers and electron transfers integrates.

• The oxidation half-reaction is balanced by multiplying by 3: \( 3 \mathrm{SO}_{2} \rightarrow 3 \mathrm{S} + 3 \mathrm{O}_{2} + 12 e^{-} \).
• For the reduction half-reaction, we multiply by 2: \( 12 e^{-} + 2 \mathrm{H}_{2}\mathrm{SeO}_{4} \rightarrow 2 \mathrm{Se} + 4 \mathrm{H}_{2}\mathrm{O} \).

Combining these adjusted half-reactions results in the final balanced equation:

\( 3 \mathrm{SO}_{2} + 2 \mathrm{H}_{2}\mathrm{SeO}_{4} \rightarrow 3 \mathrm{S} + 3 \mathrm{O}_{2} + 2 \mathrm{Se} + 4 \mathrm{H}_{2}\mathrm{O} \).
By balancing, we verify that the number of electrons in oxidation matches reduction, confirming the transaction's symmetry and adherence to chemical laws. Understanding this step brings clarity to the larger picture of converting raw reactants into final products in any reaction.