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 given reactions are: (a) \(H_2S + Fe^{3+} + e^- \rightarrow S + Fe^{2+}\) (b) \(H_2S + Br_2 \rightarrow S + 2Br^- + 2e^-\) (c) \(2H_2S + MnO_4^- + 8H^+ \rightarrow 2S + Mn^{2+} + 3e^- + 4H_2O\) (d) \(H_2S + 2 HNO_3 \rightarrow S + 2 NO_2 + 2H_2O\)

Step-by-step solution

Reaction (a) - Reduction of \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+}\) by Hydrogen Sulfide

1. Write the unbalanced equation: First, write the unbalanced equation for the reduction. \(H_2S + Fe^{3+} \rightarrow S + Fe^{2+}\) 2. Identify the oxidation states: Identify oxidation states of elements involved to help balance the equation. In the given reaction, \(S\) in \(H_2S\) has an oxidation state of \(-2\), while elemental \(S\) has an oxidation state of \(0\). 3. Balance \(\mathrm{Fe}\) and \(\mathrm{S}\) atoms: Since there are equal number of \(\mathrm{Fe}\) and \(\mathrm{S}\) atoms on both sides, no balancing is needed for them. 4. Balance the charges: The charges on both sides of the equation should be balanced. The left side has overall \(+1\) charge, whereas the right side has overall zero charge. To balance the charges, add one electron on the left side: \(H_2S + Fe^{3+} + e^- \rightarrow S + Fe^{2+}\) 5. Balance Hydrogen and Oxygen: Since there are no Oxygen atoms present and Hydrogen is balanced, the balanced net ionic equation for this reaction is: \(H_2S + Fe^{3+} + e^- \rightarrow S + Fe^{2+}\)

Reaction (b) - Reduction of \(\mathrm{Br}_{2}\) to \(\mathrm{Br}^{-}\) by Hydrogen Sulfide

1. Write the unbalanced equation: First, write the unbalanced equation for this reduction. \(H_2S + Br_2 \rightarrow S + Br^-\) 2. Identify the oxidation states: Identify oxidation states of elements involved in the reaction. In the given reaction, \(S\) in \(H_2S\) has an oxidation state of \(-2\), while elemental \(S\) has an oxidation state of \(0\). For Bromine, the oxidation state of Br in \(Br_2\) is \(0\), and it changes to \(-1\) in \(Br^-\). 3. Balance \(\mathrm{Br}\) atoms: Balance Bromine atoms on both sides by adding a 2 in front of the \(Br^-\) ion on the right side. \(H_2S + Br_2 \rightarrow S + 2Br^-\) 4. Balance the charges: The charges on the left side have an overall \(0\), whereas the right side has an overall \(-2\) charge. To balance the charges, add two electrons on the right side. \(H_2S + Br_2 \rightarrow S + 2Br^- + 2e^-\) 5. Balance Hydrogen and Oxygen: Since there are no Oxygen atoms present and Hydrogen is balanced, the balanced net ionic equation for this reaction is: \(H_2S + Br_2 \rightarrow S + 2Br^- + 2e^-\)

Reaction (c) - Reduction of \(\mathrm{MnO}_{4}^{-}\) to \(\mathrm{Mn}^{2+}\) by Hydrogen Sulfide

1. Write the unbalanced equation: First, write the unbalanced equation for this reduction. \(H_2S + MnO_4^- \rightarrow S + Mn^{2+}\) 2. Identify the oxidation states: Identify oxidation states of elements involved in the reaction. In the given reaction, \(S\) in \(H_2S\) has an oxidation state of \(-2\), while elemental \(S\) has an oxidation state of \(0\). For Manganese, the oxidation state in \(MnO_4^-\) is \(+7\) and it changes to \(+2\) in \(Mn^{2+}\). 3. Balance the main atoms: Balance the number of Manganese, Sulfur, and Oxygen atoms on both sides of the equation. \(2H_2S + MnO_4^- \rightarrow 2S + Mn^{2+}\) 4. Balance the charges: The left side has an overall \(-1\) charge, whereas the right side has an overall \(+2\) charge. To balance the charges, add three electrons on the right side. \(2H_2S + MnO_4^- \rightarrow 2S + Mn^{2+} + 3e^-\) 5. Balance Hydrogen and Oxygen: Balance Oxygen atoms by adding 4 waters on the right side. Then balance the Hydrogen atoms adding 8 Hydrogen ions on the left side. \(2H_2S + MnO_4^- + 8H^+ \rightarrow 2S + Mn^{2+} + 3e^- + 4H_2O\) The balanced net ionic equation for this reaction is: \(2H_2S + MnO_4^- + 8H^+ \rightarrow 2S + Mn^{2+} + 3e^- + 4H_2O\)

Reaction (d) - Reduction of \(\mathrm{HNO}_{3}\) to \(\mathrm{NO}_{2}\) by Hydrogen Sulfide

1. Write the unbalanced equation: First, write the unbalanced equation for this reduction. \(H_2S + HNO_3 \rightarrow S + NO_2\) 2. Identify the oxidation states: Identify oxidation states of elements involved in the reaction. In the given reaction, \(S\) in \(H_2S\) has an oxidation state of \(-2\), while elemental \(S\) has an oxidation state of \(0\). For Nitrogen, the oxidation state in \(HNO_3\) is \(+5\) and it changes to \(+4\) in \(NO_2\). 3. Balance the main atoms: Balance Sulfur and Nitrogen atoms on both sides of the equation. \(H_2S + 2 HNO_3 \rightarrow S + 2 NO_2\) 4. Balance the charges: The left and right side both have overall \(+2\) charges, so no need to add any electrons. 5. Balance Hydrogen and Oxygen: Balance Oxygen atoms by adding two waters on the left side. Then balance the Hydrogen atoms as they are already balanced. \(H_2S + 2 HNO_3 \rightarrow S + 2 NO_2 + 2H_2O\) The balanced net ionic equation for this reaction is: \(H_2S + 2 HNO_3 \rightarrow S + 2 NO_2 + 2H_2O\)

Key concepts

Balancing Redox Equations

Mastering redox reactions requires an understanding of how to balance redox equations. This involves ensuring that the number of atoms and the charge are equal on both sides of the reaction. To start, separate the equation into two halves: the oxidation half and the reduction half. Then identify the number of electrons lost in the oxidation and gained in the reduction and use these to balance the reaction. In complex scenarios such as the reduction of \(MnO_4^-\) to \(Mn^{2+}\), we balance the oxygens by adding water molecules on one side, then balance hydrogens by adding \(H^+\) ions. By methodically balancing the atoms and charge, you can arrive at an accurate depiction of the chemical process.

Pro tip: Double-check that electrons lost and gained are equal and that the charges balance out in both the half-reactions and the final equation to ensure accuracy.

Oxidation State Determination

To properly balance a redox equation, one must first determine the oxidation states of the elements involved. Oxidation state, or oxidation number, represents the degree of oxidation of an atom in a compound. It is essentially a bookkeeping system to help balance redox reactions. The rules to determine oxidation states include assigning an oxidation state of zero to pure elements and respecting the typical charges for ions when they are part of a compound. For example, in the reaction \(H_2S + Fe^{3+} \rightarrow S + Fe^{2+}\), \(S\) goes from \( -2 \) to \( 0 \) where its oxidation state increases, indicating it has been oxidized. Learning these rules will allow you to effectively balance a redox equation by knowing when and where electrons are being transferred.

Remember: Always assign oxidation states before attempting to balance a redox equation to clearly identify the species being oxidized and reduced.

Net Ionic Equations

Net ionic equations show the substances that are actively participating in a chemical reaction, removed from their spectator ions. This simplifies the equation and focuses on the essence of the change. In the reduction of \(Br_2\) to \(Br^-\), the net ionic equation helps visualize the actual redox process without unnecessary components. Start by writing down all ionic substances as dissociated ions. Then cancel out all species that appear on both the reactant and product sides, as they do not participate in the reaction. What is left is the net ionic equation, which showcases the heart of the redox process.

Attention: Not all reactions have net ionic equations. If all components are spectator ions, no net reaction occurs; there won't be a net ionic equation to write.

Chemical Reduction Processes

Chemical reduction is a half of the redox reaction where a substance gains electrons, reducing its oxidation state. In our examples, substances like \(Fe^{3+}\), \(Br_2\), and \(MnO_{4}^-\) are reduced to \(Fe^{2+}\), \(Br^-\), and \(Mn^{2+}\), respectively. Hydrogen sulfide, \(H_2S\), acts as a reducing agent and donates electrons to these species, which is why sulfur ends up in its elemental form, \(S\). Understanding the role of each component in a reaction helps to predict products and apply this knowledge to real-world processes, like corrosion prevention or the manufacturing of materials.

Insight: Keep in mind that in a reduction process, the reducing agent gets oxidized, as it loses electrons. Identifying both agents accurately is crucial for a thorough understanding of redox reactions.