Problem 110 - Chemistry The Science in Context

Question

When Soil Smells Bad In sediments and waterlogged soil, dissolved \(\mathrm{O}_{2}\) concentrations are so low that the microorganisms living there must rely on other sources of oxygen for respiration. Some bacteria can extract the oxygen from sulfate ions, reducing the sulfur in them to hydrogen sulfide gas and giving the sediments or soil a distinctive rotten-egg odor. a. What is the change in oxidation state of sulfur as a result of this reaction? b. Write the balanced net ionic equation for the reaction, under acidic conditions, that releases \(\mathrm{O}_{2}\) from sulfate and forms hydrogen sulfide gas.

Step-by-Step Solution

Answer

Answer: The change in the oxidation state of sulfur during this reaction is -8. The balanced net ionic equation under acidic conditions is given by: \(5\mathrm{SO}_{4}^{2-} + 6\mathrm{H}^{+} \rightarrow \mathrm{H}_{2}\mathrm{S} + 4\mathrm{O}_{2} + 4\mathrm{H}_{2}\mathrm{O}\).

Step 1: a. Determining the change in oxidation state of sulfur

To determine the change in oxygen state of sulfur, we have to find out its oxidation state before and after the reaction. In sulfate ions (\(\mathrm{SO}_{4}^{2-}\)), sulfur has an oxidation state of +6, whereas in hydrogen sulfide gas (\(\mathrm{H}_{2}\mathrm{S}\)), sulfur has an oxidation state of -2. The change in oxidation state of sulfur as a result of this reaction can be calculated by subtracting the final oxidation state from the initial oxidation state: Change in oxidation state = (-2) - (+6) = -8 Therefore, the change in the oxidation state of sulfur as a result of this reaction is -8.

Step 2: b. Writing the balanced net ionic equation for the reaction under acidic conditions

To write the balanced net ionic equation for the reaction under acidic conditions, we first need to identify the half-reactions involved in the process. We have sulfate ions (\(\mathrm{SO}_{4}^{2-}\)) being reduced to hydrogen sulfide gas (\(\mathrm{H}_{2}\mathrm{S}\)) along with the release of oxygen gas. So we have the following half-reactions: (1) Reduction of sulfate ions: \(\mathrm{SO}_{4}^{2-} \rightarrow \mathrm{H}_{2}\mathrm{S}\) (2) Release of oxygen gas: \(\mathrm{SO}_{4}^{2-} \rightarrow \mathrm{O}_{2}\) Now we need to balance the half-reactions: For the reduction of sulfate ions: \(\mathrm{SO}_{4}^{2-} + 6\mathrm{H}^{+} + 8\mathrm{e}^{-} \rightarrow \mathrm{H}_{2}\mathrm{S} + 4\mathrm{H}_{2}\mathrm{O}\) For the release of oxygen gas: \(\mathrm{SO}_{4}^{2-} \rightarrow \mathrm{O}_{2} + 2\mathrm{H}^{+} + 2\mathrm{e}^{-}\) Next, we need to multiply these half-reactions by appropriate factors so that the number of electrons lost in one reaction matches the number gained in the other reaction. In this case, we need to multiply the second half-reaction by 4 to balance the electrons: For the release of oxygen gas (multiplied by 4 to balance electrons): \(4\mathrm{SO}_{4}^{2-} \rightarrow 4\mathrm{O}_{2} + 8\mathrm{H}^{+} + 8\mathrm{e}^{-}\) Now we can add these two balanced half-reactions to obtain the balanced net ionic equation under acidic conditions: \(\mathrm{SO}_{4}^{2-} + 6\mathrm{H}^{+} + 8\mathrm{e}^{-} \rightarrow \mathrm{H}_{2}\mathrm{S} + 4\mathrm{H}_{2}\mathrm{O}\) \(+\) \(4\mathrm{SO}_{4}^{2-} \rightarrow 4\mathrm{O}_{2} + 8\mathrm{H}^{+} + 8\mathrm{e}^{-}\) After cancelling out electrons and protons, we get: \(5\mathrm{SO}_{4}^{2-} + 6\mathrm{H}^{+} \rightarrow \mathrm{H}_{2}\mathrm{S} + 4\mathrm{O}_{2} + 4\mathrm{H}_{2}\mathrm{O}\) Thus, the balanced net ionic equation under acidic conditions is given by: \(5\mathrm{SO}_{4}^{2-} + 6\mathrm{H}^{+} \rightarrow \mathrm{H}_{2}\mathrm{S} + 4\mathrm{O}_{2} + 4\mathrm{H}_{2}\mathrm{O}\).

Key Concepts

Understanding Sulfate Ions

Sulfate ions (\( \mathrm{SO}_{4}^{2-} \)) play a crucial role in various chemical reactions, especially in soil and sediment environments. These ions consist of one sulfur atom covalently bonded to four oxygen atoms, carrying a net charge of -2 due to the extra electrons. This negatively charged ion is quite stable and usually found dissolved in water. The presence of sulfate ions is significant in many natural processes, including respiration by certain bacteria.

In watery soils where oxygen is scarce, some microorganisms use sulfate ions as an alternative oxygen source for survival. Here, sulfate ions are reduced to form hydrogen sulfide gas (\( \mathrm{H}_{2}\mathrm{S} \)), which has a distinct rotten-egg smell. This process helps bacteria extract the energy they need in low-oxygen environments. Understanding this dynamic helps explain how sulfate ions can affect soil ecology, contributing to both nutrient cycling and unique geological formations.

Role of Hydrogen Sulfide

Hydrogen sulfide (\( \mathrm{H}_{2}\mathrm{S} \)) is a gas characterized by its recognizable odor of rotten eggs. This compound forms during the reduction of sulfate ions under anaerobic (oxygen-deficient) conditions, especially in sediments and stagnant waters. In these environments, sulfate-reducing bacteria use the sulfate ions as a source of oxygen, releasing hydrogen sulfide as a by-product.

Hydrogen sulfide is not only notable for its smell, but also plays a part in microbial metabolism. It allows bacteria to continue their respiration processes when other forms of oxygen are not available. The formation of hydrogen sulfide is a part of the sulfur cycle in nature, impacting the chemistry of water bodies and soil sediments. In high concentrations, it can be toxic to plant and animal life, but in small amounts, it is essential for certain ecological functions.

Balanced Net Ionic Equations Explained

A balanced net ionic equation is essential to describe the chemistry of the reaction accurately. This type of equation reflects only the chemical species that participate directly in the reaction, omitting spectator ions that do not undergo any change. It ensures that the atoms and charges are balanced on both sides of the equation, which is a fundamental principle in chemical reactions.

In the context of sulfate ions reducing to hydrogen sulfide under acidic conditions, the resulting balanced net ionic equation provides clarity on the stoichiometry of the process. Initially, we need to identify the half-reactions of the substances involved, specifically the reduction of sulfate ions and the release of oxygen gas. Balancing these reactions separately involves ensuring that electrons lost and gained are equal, which sometimes requires multiplying the equations by appropriate factors.

• This balanced form gives us a clear view of the chemical transformation.
• It shows the reduction of sulfate ions and the formation of hydrogen sulfide gas.
• It highlights the removal of protons and addition of water molecules as part of the balancing.

Knowing how to derive and balance a net ionic equation is crucial for chemistry students, helping them visualize and understand the underlying processes transforming one species into another.