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.

Short answer

The balanced net ionic equations for the given reactions are: Case (a): \(2 \, H_2S + 4Fe^{3+} \rightarrow S_2 + 4H^+ + 4Fe^{2+}\) Case (b): \(2 \, H_2S + Br_2 \rightarrow S_2 + 4H^+ + 4Br^-\) Case (c): \(5 \, H_2S + 4MnO_4^- + 16H^+ \rightarrow 2S_2 + 20H^+ + 4Mn^{2+} + 8H_2O\) Case (d): \(2 \, H_2S + 2HNO_3 \rightarrow S_2 + 4H^+ + 2NO_2 + 2H_2O\)

Step-by-step solution

Case (a): Fe³⁺ to Fe²⁺

For this reaction, we will write the oxidation half-reaction of hydrogen sulfide and the reduction half-reaction of Fe³⁺. Oxidation half-reaction: \(2 \, H_2S \rightarrow S_2 + 4H^+ + 4e^-\) Reduction half-reaction: \(Fe^{3+} + e^- \rightarrow Fe^{2+}\) To balance the charges and electrons, we need to multiply the reduction half-reaction by 4. This allows us to combine the two half-reactions to create the balanced net ionic equation: Final balanced net ionic equation: \(2 \, H_2S + 4Fe^{3+} \rightarrow S_2 + 4H^+ + 4Fe^{2+}\)

Case (b): Br₂ to Br⁻

For this case, we will write the oxidation half-reaction of hydrogen sulfide and the reduction half-reaction of Br₂. Oxidation half-reaction: \(2 \, H_2S \rightarrow S_2 + 4H^+ + 4e^-\) Reduction half-reaction: \(Br_2 + 2e^- \rightarrow 2Br^-\) To balance the charges and electrons, we need to multiply the reduction half-reaction by 2. This allows us to combine the two half-reactions to create the balanced net ionic equation: Final balanced net ionic equation: \(2 \, H_2S + Br_2 \rightarrow S_2 + 4H^+ + 4Br^-\)

Case (c): MnO₄⁻ to Mn²⁺

For this case, we will write the oxidation half-reaction of hydrogen sulfide and the reduction half-reaction of MnO₄⁻. Oxidation half-reaction: \(2 \, H_2S \rightarrow S_2 + 4H^+ + 4e^-\) Reduction half-reaction: \(MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O\) To balance the charges and electrons, we need to multiply the oxidation half-reaction by 5/2 and the reduction half-reaction by 2. This allows us to combine the two half-reactions to create the balanced net ionic equation: Final balanced net ionic equation: \(5 \, H_2S + 4MnO_4^- + 16H^+ \rightarrow 2S_2 + 20H^+ + 4Mn^{2+} + 8H_2O\)

Case (d): HNO₃ to NO₂

For this case, we will write the oxidation half-reaction of hydrogen sulfide and the reduction half-reaction of HNO₃. Oxidation half-reaction: \(2 \, H_2S \rightarrow S_2 + 4H^+ + 4e^-\) Reduction half-reaction: \(2HNO_3 + 2e^- \rightarrow 2NO_2 + 2H_2O\) We can simply combine the two half-reactions to create the balanced net ionic equation: Final balanced net ionic equation: \(2 \, H_2S + 2HNO_3 \rightarrow S_2 + 4H^+ + 2NO_2 + 2H_2O\)

Key concepts

Oxidation Half-reaction

In a redox reaction, the oxidation half-reaction involves the loss of electrons. This part of the reaction showcases the substance donating electrons. By understanding this, you can see why it's called an 'oxidation' process. The electrons are given away during this process—which can be a bit scary to imagine, but it’s quite a normal procedure in chemistry!

For example, in our exercise, the oxidation half-reaction for hydrogen sulfide (H₂S) is given by the equation:

• \(2 \, H_2S \rightarrow S_2 + 4H^+ + 4e^-\)

Here, hydrogen sulfide (H₂S) is oxidized to form elemental sulfur (S₂). During this process, four electrons (4e^-) are released.

Understanding this concept helps in identifying how the sulfide ions lose electrons and turn into elemental sulfur. This helps demonstrate the fundamental role of electron transfer in redox reactions.

Reduction Half-reaction

The reduction half-reaction in a redox reaction involves the gain of electrons. This part of the reaction involves a substance that accepts electrons. Think of it as the substance getting an electron 'birthday gift' — receiving electrons!

Using our examples from the solutions:

• In case (a), the reduction half-reaction of ferric ion (Fe^{3+}) is:
  \(Fe^{3+} + e^- \rightarrow Fe^{2+}\)
• In case (b), bromine (Br₂) is reduced to bromide ions (Br^−) with the equation:
  \(Br_2 + 2e^- \rightarrow 2Br^-\)
• In case (c), permanganate ion (MnO₄^-) reduces to manganese ion (Mn^{2+}):
  \(MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O\)
• In case (d), nitric acid (HNO₃) is reduced to nitrogen dioxide (NO₂):
  \(2HNO_3 + 2e^- \rightarrow 2NO_2 + 2H_2O\)

Excitingly, through these reactions, the substances achieve a more stable electronic state. The gain of electrons means the substance is reduced, which sounds backward, but remember—in chemistry, gaining electrons reduces the oxidation state.

Net Ionic Equation

A net ionic equation shows only the ions and molecules directly involved in the chemical reaction. The goal is to focus on the chemical species actually participating in the transformation. It helps to highlight the essential changes without extra clutter.

In our exercise, we balanced each case by finding the oxidation and reduction half-reactions individually and then combining them. Here’s how we did it:

• For case (a) that involves Fe^{3+} and H_2S, the balanced net ionic equation is:
  \(2 \; H_2S + 4Fe^{3+} \rightarrow S_2 + 4H^+ + 4Fe^{2+}\)
• In case (b), with Br_2 and H_2S, the net ionic equation is:
  \(2 \; H_2S + Br_2 \rightarrow S_2 + 4H^+ + 4Br^-\)
• For case (c), involving MnO_4^-, the net ionic equation results in:
  \(5 \; H_2S + 4MnO_4^- + 16H^+ \rightarrow 2S_2 + 20H^+ + 4Mn^{2+} + 8H_2O\)
• In case (d), the reaction of HNO_3 and H_2S gives:
  \(2 \; H_2S + 2HNO_3 \rightarrow S_2 + 4H^+ + 2NO_2 + 2H_2O\)

These equations help us comprehend the actual changes taking place, leaving spectator ions out of the picture.