Problem 104 - Chemistry The Molecular Science

Question

Hydrogen sulfide gas is bubbled into an acidified solution of potassium permanganate; elemental sulfur precipitates. The unbalanced equation is \(\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})+\mathrm{MnO}_{4}^{-}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{S}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\ell)\) (a) Potassium ions are not given in the equation. Explain why. (b) Explain why this is a redox equation. (c) Identify the oxidizing and reducing agents.

Step-by-Step Solution

Answer

(a) Potassium ions are spectators. (b) There is electron transfer. (c) Oxidizing agent: MnO4⁻; Reducing agent: H2S.

Step 1: Understanding why Potassium Ions are Excluded

The equation provided is written as a net ionic equation, meaning it only includes the species that participate in the chemical reaction. Potassium ions (\( \mathrm{K}^+ \)) do not participate in the redox reaction itself; they are spectators and thus are not shown.

Step 2: Recognizing Redox Reactions

In a redox reaction, oxidation and reduction occur simultaneously. The presence of \( \mathrm{MnO}_4^- \) is a clue because permanganate ions are commonly involved in redox chemistry, where manganese changes its oxidation state.

Step 3: Assigning Oxidation States

To determine the oxidizing and reducing agents, assign oxidation states: Sulfur in \( \mathrm{H}_2\mathrm{S} \) is \(-2\), and after the reaction, as solid sulfur, it is 0 (indicating oxidation). Manganese in \( \mathrm{MnO}_4^- \) is +7, and in \( \mathrm{Mn}^{2+} \), it is +2 (indicating reduction).

Step 4: Identifying Oxidizing and Reducing Agents

The species that is reduced (gains electrons) is the oxidizing agent. Here, \( \mathrm{MnO}_4^- \) is reduced, acting as the oxidizing agent. The species that is oxidized (loses electrons) is the reducing agent. Here, \( \mathrm{H}_2\mathrm{S} \) is oxidized, acting as the reducing agent.

Key Concepts

Oxidation States

Oxidation states, also known as oxidation numbers, are vital in understanding redox reactions. They help us keep track of electron transfer between atoms in different substances. Think of oxidation states as a bookkeeping tool that allows us to determine which atoms gain or lose electrons during a reaction. In redox reactions, one substance always becomes more negative (reduction), and another becomes more positive (oxidation).

Let's look at the unbalanced reaction: \[ \mathrm{H}_{2}\mathrm{~S}(\mathrm{~g})+\mathrm{MnO}_{4}^{-}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{S}(\mathrm{s})+\mathrm{H}_{2}\mathrm{O}(\ell) \]

• Sulfur in \( \mathrm{H}_2\mathrm{S} \) starts with an oxidation state of \(-2\) because hydrogen is generally +1 and there are two hydrogen atoms.
• In the resulting solid sulfur, the oxidation state changes to 0. This shift indicates that sulfur was oxidized (lost electrons).
• Manganese in \( \mathrm{MnO}_4^- \) has an oxidation state of +7, but after the reaction, it is reduced to +2 in \( \mathrm{Mn}^{2+} \). Here, manganese is reduced (gains electrons).

Understanding oxidation states is crucial because it is the basis for identifying which reactant is oxidized and which is reduced.

Oxidizing Agents

An oxidizing agent is a substance that causes another substance to oxidize, meaning it allows another substance to lose electrons. During the redox reaction, the oxidizing agent itself is reduced. This might sound a bit backwards: it gains electrons (is reduced) while the other substance loses electrons (is oxidized).

In our specific reaction,

• \( \mathrm{MnO}_4^- \) serves as the oxidizing agent. This is because it facilitates the oxidation of \( \mathrm{H}_2\mathrm{S} \), accepting the electrons that sulfur loses.
• The permanganate ion \( \mathrm{MnO}_4^- \) starts with a high oxidation state of +7. Thus, it has a strong tendency to gain electrons and reduce to \( \mathrm{Mn}^{2+} \), signifying its role as an oxidizing agent.
• The reduction of \( \mathrm{MnO}_4^- \) is confirmed through its change in oxidation state from +7 in \( \mathrm{MnO}_4^- \) to +2 in \( \mathrm{Mn}^{2+} \).

Recognizing the oxidizing agent is pivotal in redox chemistry as it highlights the electron transfer dynamics driving the reaction.

Net Ionic Equation

Net ionic equations provide a streamlined view of the core chemical changes in reactions, especially in aqueous solutions. By showing only the species that actively participate in the reaction, these equations exclude spectator ions that remain unchanged.

In the given reaction:

• Only the species undergoing a change, like \( \mathrm{H}_2\mathrm{S} \) and \( \mathrm{MnO}_4^- \), are represented.
• Potassium ions, \( \mathrm{K}^+ \), which might initially seem part of the process due to being part of potassium permanganate, are left out as they don't change; they're just there to balance charge in the complete molecular equation.
• This approach focuses on the essence of the reaction, removing irrelevant components and allowing students to focus on the process of redox transformation.

Learning to write net ionic equations is a skill that provides clarity in complex chemical reactions, simplifying analysis and interpretation.