Problem 43 - Chemistry The Central Science In Si Units

Question

In aqueous solution, hydrogen sulfide reduces (a) \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+},(\mathbf{b}) \mathrm{Br}_{2}\) to \(\mathrm{Br}^{-},(\mathbf{c}) \mathrm{MnO}_{4}^{-}\) to \(\mathrm{Mn}^{2+},(\mathbf{d}) \mathrm{HNO}_{3}\) to \(\mathrm{NO}_{2}\) In all cases, under appropriate conditions, the product is elemental sulfur. Write a balanced net ionic equation for each reaction.

Step-by-Step Solution

Answer

Net ionic equations: (a) \( \text{H}_2\text{S} + 2\text{Fe}^{3+} \rightarrow 2\text{Fe}^{2+} + \text{S} + 2\text{H}^+ \) (b) \( \text{H}_2\text{S} + \text{Br}_2 \rightarrow 2\text{Br}^- + \text{S} + 2\text{H}^+ \) (c) \( 5\text{H}_2\text{S} + 2\text{MnO}_4^- + 16\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 5\text{S} + 8\text{H}_2\text{O} \) (d) \( 2\text{H}_2\text{S} + 2\text{HNO}_3 \rightarrow 2\text{NO}_2 + 2\text{S} + 2\text{H}_2\text{O} \).

Step 1: Understand the Redox Reaction Components

In each part, we identify the oxidizing agent (species being reduced) and the reducing agent (hydrogen sulfide, which is being oxidized to sulfur). Identify the oxidation states engaged in each part: \( \text{H}_2\text{S} \rightarrow \text{S}^0 \) and the given changes for (a) \( \text{Fe}^{3+} \rightarrow \text{Fe}^{2+} \), (b) \( \text{Br}_2 \rightarrow \text{Br}^- \), (c) \( \text{MnO}_4^- \rightarrow \text{Mn}^{2+} \), and (d) \( \text{HNO}_3 \rightarrow \text{NO}_2 \).

Step 2: Write Oxidation Half-Reactions

For all parts, the oxidation half-reaction of \( \text{H}_2\text{S} \) in an aqueous solution is: \[ \text{H}_2\text{S} \rightarrow \text{S} + 2\text{H}^+ + 2\text{e}^- \]

Step 3: Write Reduction Half-Reactions

For each reduction reaction: (a) \( \text{Fe}^{3+} + \text{e}^- \rightarrow \text{Fe}^{2+} \) (b) \( \text{Br}_2 + 2\text{e}^- \rightarrow 2\text{Br}^- \) (c) \( \text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \) (d) \( \text{HNO}_3 + \text{e}^- \rightarrow \text{NO}_2 + \text{OH}^- \). Ensure each is mass and charge balanced.

Step 4: Balance Electron Transfer and Combine

Match electrons in oxidation and each reduction half-reaction by multiplying to obtain equivalent electrons transferred:(a) Multiply oxidation by 1, combine to: \[ \text{H}_2\text{S} + 2\text{Fe}^{3+} \rightarrow 2\text{Fe}^{2+} + \text{S} + 2\text{H}^+ \](b) Multiply oxidation by 2, combine to: \[ \text{H}_2\text{S} + \text{Br}_2 \rightarrow 2\text{Br}^- + \text{S} + 2\text{H}^+ \](c) Multiply oxidation by 5, combine to: \[ 5\text{H}_2\text{S} + 2\text{MnO}_4^- + 16\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 5\text{S} + 8\text{H}_2\text{O} \](d) Multiply oxidation by 2, combine to:\[ 2\text{H}_2\text{S} + 2\text{HNO}_3 \rightarrow \text{2NO}_2 + \text{2S} + \text{2H}_2\text{O} \].{

Step 5: Validate Charge and Atoms Balance

Ensure total atoms and charges are balanced in each overall reaction. Each compound's oxidation state change should be verified within the net equations. Check electron continuity and verify each equation depicts a feasible redox process.

Key Concepts

Oxidation And Reduction

In the world of chemistry, oxidation and reduction reactions are central to understanding how substances interact at the elemental level. These types of reactions are collectively known as redox reactions. Redox reactions involve the transfer of electrons from one substance to another.
For every redox reaction, there are two parts: oxidation, where electrons are lost, and reduction, where electrons are gained. In the original exercise, hydrogen sulfide (\(\text{H}_2\text{S}\)) loses electrons as it gets oxidized to form sulfur (\(\text{S}^0\)), while other chemical species like \(\text{Fe}^{3+}\) gain these electrons and undergo reduction.

• Oxidizing agent is a substance that gains electrons and is reduced.
• Reducing agent is a substance that loses electrons and is oxidized.

To identify the oxidizing and reducing agents for each component:
(a) \(\text{Fe}^{3+}\) acts as the oxidizing agent, while hydrogen sulfide is the reducing agent.
(b) \(\text{Br}_2\) is the oxidizing agent here, reduced to \(\text{Br}^-\).
(c) \(\text{MnO}_4^-\) reduces to \(\text{Mn}^{2+}\).
(d) \(\text{HNO}_3\) reduces to \(\text{NO}_2\).
After determining these, the next step involves converting these findings into balanced equations.

Balancing Chemical Equations

Balancing chemical equations is crucial in ensuring that the law of conservation of mass is observed. This means every chemical equation must have the same number of each type of atom on both sides of the reaction arrow.
Balancing involves several steps that can make this process straightforward, especially for redox reactions.

• First, write down the unbalanced equation with all reactants and products.
• Next, focus on balancing the atoms involved in the oxidation and reduction half-reactions.
• Then, balance the number of electrons transferred in the half-reactions by multiplying them adequately.
• Finally, add the balanced half-reactions together to get the overall balanced equation.

An example from the exercise is:
For the reaction where hydrogen sulfide reduces iron: \(\text{H}_2\text{S} + 2\text{Fe}^{3+} \rightarrow 2\text{Fe}^{2+} + \text{S} + 2\text{H}^+\)
It's essential to ensure molecular and electron balance in each equation formed after combining the half-reactions.

Net Ionic Equations

Net ionic equations are a simplified form of chemical equations that show only the reacting ions or molecules. This stripped-down version makes it easier to focus on the crux of the redox process without getting sidetracked by non-reactive spectator ions.
To write a net ionic equation:

• Start by writing the balanced full ionic equation.
• Identify and remove spectator ions, which are ions that appear unchanged on both sides of the equation.
• What remains is the net ionic equation that highlights the actual chemical change occurring in the reaction.

Considering the reduction of bromine example:
Full ionic for the bromine example:\(\text{H}_2\text{S} + \text{Br}_2 \rightarrow 2\text{Br}^- + \text{S} + 2\text{H}^+\)
In this particular case, no spectator ions are present, as all components are actively participating. The net ionic equation remains unaffected and provides a clear view of the chemistry taking place!