Question

In aqueous solution, hydrogen sulfide reduces (a) \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+}\), (b) \(\mathrm{Br}_{2}\) to \(\mathrm{Br}^{-}\), (c) \(\mathrm{MnO}_{4}^{-}\) to \(\mathrm{Mn}^{2+}\), (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.

Short answer

The balanced net ionic equations for the four reactions are: (a) \(2H_{2}S + 2Fe^{3+} + 4e^{-} \rightarrow 2Fe^{2+} + 2S + 4H^{+}\) (b) \(H_{2}S + Br_{2} \rightarrow 2Br^{-} + S + 2H^{+}\) (c) \(5H_{2}S + 2MnO_{4}^{-} + 16H^{+} \rightarrow 2Mn^{2+} + 5S + 8H_{2}O + 10e^{-}\) (d) \(3H_{2}S + 4HNO_{3} + 2e^{-} \rightarrow 3NO_{2} + 3S + 4H_{2}O + 6H^{+}\)

Step-by-step solution

Write the unbalanced equation

Begin by writing down the unbalanced equation for this reaction: \(H_{2}S + Fe^{3+} \rightarrow Fe^{2+} + S\)

Balance the atoms

Balance the atoms in the equation, starting with elements that appear only once on each side, excluding hydrogen and oxygen: \(2H_{2}S + 2Fe^{3+} \rightarrow 2Fe^{2+} + 2S\)

Balance the charge

Make sure the charge is balanced on both sides of the equation: \(2H_{2}S + 2Fe^{3+} + 4e^{-} \rightarrow 2Fe^{2+} + 2S + 4H^{+}\) Now, the net ionic equation for the first reaction is: \(2H_{2}S + 2Fe^{3+} + 4e^{-} \rightarrow 2Fe^{2+} + 2S + 4H^{+}\) (b) Hydrogen sulfide reduces \(\mathrm{Br}_{2}\) to \(\mathrm{Br}^{-}\):

Write the unbalanced equation

Begin with the unbalanced equation for this reaction: \(H_{2}S + Br_{2} \rightarrow Br^{-} + S\)

Balance the atoms

Balance the atoms, again starting with elements that appear only once on each side: \(H_{2}S + Br_{2} \rightarrow 2Br^{-} + S + 2H^{+}\)

Balance the charge

Check that the charge is balanced on both sides of the equation: Now, the net ionic equation for the second reaction is: \(H_{2}S + Br_{2} \rightarrow 2Br^{-} + S + 2H^{+}\) (c) Hydrogen sulfide reduces \(\mathrm{MnO}_{4}^{-}\) to \(\mathrm{Mn}^{2+}\):

Write the unbalanced equation

Start with the unbalanced equation for this reaction: \(H_{2}S + MnO_{4}^{-} \rightarrow Mn^{2+} + S\)

Balance the atoms

Balance the atoms, starting with elements that appear only once on each side, excluding hydrogen and oxygen: \(5H_{2}S + 2MnO_{4}^{-} \rightarrow 2Mn^{2+} + 5S\)

Balance the oxygen atoms and charge

Add water molecules to balance the oxygen atoms, and add hydrogen ions and electrons to balance the charge on both sides of the equation: \(5H_{2}S + 2MnO_{4}^{-} + 16H^{+} \rightarrow 2Mn^{2+} + 5S + 8H_{2}O + 10e^{-}\) Now, the net ionic equation for the third reaction is: \(5H_{2}S + 2MnO_{4}^{-} + 16H^{+} \rightarrow 2Mn^{2+} + 5S + 8H_{2}O + 10e^{-}\) (d) Hydrogen sulfide reduces \(\mathrm{HNO}_{3}\) to \(\mathrm{NO}_{2}\):

Write the unbalanced equation

Write down the unbalanced equation for this reaction: \(H_{2}S + HNO_{3} \rightarrow NO_{2} + S\)

Balance the atoms

Balance the atoms, starting with elements that appear only once on each side, excluding hydrogen and oxygen: \(3H_{2}S + 4HNO_{3} \rightarrow 3NO_{2} + 3S\)

Balance the oxygen atoms and charge

Add water molecules to balance the oxygen atoms, and add hydrogen ions and electrons to balance the charge on both sides of the equation: \(3H_{2}S + 4HNO_{3} + 2e^{-} \rightarrow 3NO_{2} + 3S + 4H_{2}O + 6H^{+}\) Now, the net ionic equation for the fourth reaction is: \(3H_{2}S + 4HNO_{3} + 2e^{-} \rightarrow 3NO_{2} + 3S + 4H_{2}O + 6H^{+}\)

Key concepts

Balancing Chemical Equations

Balancing chemical equations is an essential skill for understanding chemical reactions. When balancing equations, the goal is to have the same number of atoms for each element on both sides of the equation. This ensures that matter is neither created nor destroyed. Here's a simple approach to balancing them:

• Start by writing the unbalanced equation.
• Focus on one element at a time, avoiding hydrogen and oxygen initially.
• Adjust the coefficients (the numbers before molecules) to ensure each type of atom is equal on both sides.
• Lastly, balance hydrogen and oxygen, which often appear in multiple compounds.

Remember, coefficients should be whole numbers. Balancing equations not only helps in portraying the reaction accurately but also in predicting the amount of products formed in reactions.

Net Ionic Equations

Net ionic equations simplify the representation of a reaction by focusing only on the ions and molecules directly involved. They are particularly useful in redox reactions, where electron transfer is central. Here's how to write them:

• Start with a balanced molecular equation.
• Identify the states of the compounds, noting which are aqueous or solid.
• Break down aqueous solutions into their ions.
• Cancel out spectator ions, which do not change during the reaction.
• Write the resulting net ionic equation focused on the reacting species.

This method is crucial in aqueous solutions, as it allows students to see which substances are actually participating in the reaction process and helps in understanding redox processes at a deeper level.

Aqueous Solutions

Aqueous solutions are those where the solvent is water. They play a critical role in chemistry since many reactions occur in this medium. In these solutions, ionic compounds dissolve into individual ions, facilitating various reactions, especially redox ones. Key points to understand include:

• Ionic compounds tend to dissociate in water, forming electrolyte solutions.
• Reactions in aqueous solutions often involve the exchange or transformation of ions.
• Insoluble substances will precipitate instead of dissolving, which helps in predicting reaction outcomes.
• Water's polar nature plays a crucial role in dissolving ionic substances.

Understanding how substances behave in aqueous solutions is vital for writing net ionic equations and predicting the course of reactions. This knowledge is foundational for studying chemical processes in biological and environmental systems.