Question

Balance each of the following half-reactions, which take place in acidic solution. a. \(\mathrm{SiO}_{2}(s) \rightarrow \mathrm{Si}(s)\) b. \(\mathrm{S}(s) \rightarrow \mathrm{H}_{2} \mathrm{S}(a q)\) c. \(\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{HNO}_{2}(a q)\) d. \(\mathrm{NO}_{3}(a q) \rightarrow \mathrm{NO}(g)\)

Short answer

The balanced half-reactions in acidic solutions are: a. \(4\mathrm{H^{+}}(aq) + 4\mathrm{e^{-}} + \mathrm{SiO}_{2}(s) \rightarrow \mathrm{Si}(s) + 2\mathrm{H_2O}(l)\) b. \(2\mathrm{H^{+}}(aq) + 2\mathrm{e^{-}} + \mathrm{S}(s) \rightarrow \mathrm{H_2S}(aq)\) c. \(3\mathrm{e^{-}} + \mathrm{NO_3^{-}}(aq) + 4\mathrm{H^{+}}(aq) \rightarrow \mathrm{HNO_2}(aq) +\mathrm{H_2O}(l)\) d. \(4\mathrm{e^{-}} + \mathrm{NO_3^{-}}(aq) + 4\mathrm{H^{+}}(aq) \rightarrow \mathrm{NO}(g) + 2\mathrm{H_2O}(l)\)

Step-by-step solution

Balance the atoms other than hydrogen and oxygen

In this case, the silicon atoms are already balanced with one Si atom on both sides.

Balance the number of oxygen atoms using water molecules (H2O)

We have 2 oxygen atoms on the left side, so we need to add 2 water molecules on the right side to balance the oxygen atoms: \(\mathrm{SiO}_{2}(s) \rightarrow \mathrm{Si}(s) + 2\mathrm{H_2O}(l)\)

Balance the number of hydrogen atoms using protons (H+)

Now, we have 4 hydrogen atoms on the right side, so we need to add 4 protons (H+) on the left side to balance the hydrogen atoms: \(4\mathrm{H^{+}}(aq) + \mathrm{SiO}_{2}(s) \rightarrow \mathrm{Si}(s) + 2\mathrm{H_2O}(l)\)

Balance the charges using electrons (e-)

The overall charge of the left side is +4, and the right side is 0. We need to add 4 electrons on the left side to balance the charges. \(4\mathrm{H^{+}}(aq) + 4\mathrm{e^{-}} + \mathrm{SiO}_{2}(s) \rightarrow \mathrm{Si}(s) + 2\mathrm{H_2O}(l)\) b. \(\mathrm{S}(s) \rightarrow \mathrm{H}_{2} \mathrm{S}(a q)\)

Balance the atoms other than hydrogen and oxygen

In this case, the sulfur atoms are already balanced with one S atom on both sides.

Balance the number of oxygen atoms using water molecules (H2O)

There are no oxygen atoms in this half-reaction, so we can move to the next step.

Balance the number of hydrogen atoms using protons (H+)

We have 2 hydrogen atoms on the right side, so we need to add 2 protons (H+) on the left side to balance the hydrogen atoms: \(2\mathrm{H^{+}}(aq) + \mathrm{S}(s) \rightarrow \mathrm{H_2S}(aq)\)

Balance the charges using electrons (e-)

The overall charge of the left side is +2, and the right side is 0. We need to add 2 electrons on the left side to balance the charges. \(2\mathrm{H^{+}}(aq) + 2\mathrm{e^{-}} + \mathrm{S}(s) \rightarrow \mathrm{H_2S}(aq)\) c. \(\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{HNO}_{2}(a q)\)

Balance the atoms other than hydrogen and oxygen

In this case, the nitrogen atoms are already balanced with one N atom on both sides.

Balance the number of oxygen atoms using water molecules (H2O)

We have 3 oxygen atoms on the left side and 2 on the right side. So, we need to add 1 water molecule (H2O) to the right side to balance the oxygen atoms: \(\mathrm{NO_3^{-}}(aq) \rightarrow \mathrm{HNO_2}(aq) + \mathrm{H_2O}(l)\)

Balance the number of hydrogen atoms using protons (H+)

We have 4 hydrogen atoms on the right side, so we need to add 4 protons (H+) on the left side to balance the hydrogen atoms: \(\mathrm{NO_3^{-}}(aq) + 4\mathrm{H^{+}}(aq) \rightarrow \mathrm{HNO_2}(aq) + \mathrm{H_2O}(l)\)

Balance the charges using electrons (e-)

The overall charge of the left side is +3, and the right side is 0. We need to add 3 electrons on the left side to balance the charges. \(3\mathrm{e^{-}} + \mathrm{NO_3^{-}}(aq) + 4\mathrm{H^{+}}(aq) \rightarrow \mathrm{HNO_2}(aq) +\mathrm{H_2O}(l)\) d. \(\mathrm{NO}_{3}(a q) \rightarrow \mathrm{NO}(g)\)

Balance the atoms other than hydrogen and oxygen

In this case, the nitrogen atoms are already balanced with one N atom on both sides.

Balance the number of oxygen atoms using water molecules (H2O)

We have 3 oxygen atoms on the left side and 1 on the right side. So, we need to add 2 water molecules (H2O) to the right side to balance the oxygen atoms: \(\mathrm{NO_3^{-}}(aq) \rightarrow \mathrm{NO}(g) + 2\mathrm{H_2O}(l)\)

Balance the number of hydrogen atoms using protons (H+)

We have 4 hydrogen atoms on the right side, so we need to add 4 protons (H+) on the left side to balance the hydrogen atoms: \(\mathrm{NO_3^{-}}(aq) + 4\mathrm{H^{+}}(aq) \rightarrow \mathrm{NO}(g) + 2\mathrm{H_2O}(l)\)

Balance the charges using electrons (e-)

The overall charge of the left side is +4, and the right side is 0. We need to add 4 electrons on the left side to balance the charges. \(4\mathrm{e^{-}} + \mathrm{NO_3^{-}}(aq) + 4\mathrm{H^{+}}(aq) \rightarrow \mathrm{NO}(g) + 2\mathrm{H_2O}(l)\)

Key concepts

Half-Reactions

Half-reactions are an essential part of balancing redox reactions. They provide a way to focus on the oxidation or reduction component separately. This means that in one part of the balanced equation, you deal with the oxidation process (the loss of electrons), and in another part, you handle the reduction process (the gain of electrons). By separating the overall reaction into two half-reactions, it becomes easier to manage complex reactions and make sure that all elements, electrons, and charges are balanced correctly.

• Start by identifying the substances that are oxidized and reduced.
• Split the main reaction into two reactions: one for oxidation and one for reduction.
• Balance each half-reaction independently before combining them.

By using half-reactions, you lessen the difficulty of observing multiple changes simultaneously, like the number of atoms, addition of protons, and movement of electrons all at once.

Acidic Solution

When balancing redox reactions in acidic solutions, it’s important to consider the role of hydrogen ions ( H^+ ). Acidic conditions imply an abundance of H^+ ions, which helps in balancing oxidation and reduction equations. In such solutions:

• Protons are added to help balance the hydrogen atoms in the equation.
• Water molecules ( H_2O ) are used to balance oxygen atoms.
• The excess protons ensure that hydrogen atoms are properly balanced on both sides of the equation.

These components are crucial because they reflect the actual environment in which the reaction occurs, which is often in an acidic medium. Incorporating protons and water molecules allows flexibility in achieving a properly balanced equation, reflecting realistic chemical reactions in nature.

Electrons

Electrons play a central role in redox reactions as they are both gained and lost during the process. When balancing reaction equations, keeping track of electrons ensures that mass and charge are balanced across both sides of the equation. Here are a few steps to keep in mind:

• Determine how many electrons need to be added or removed to balance the charge of each half-reaction.
• Add electrons to the side of the half-reaction that needs them to equalize the charges.
• If electrons still don’t balance, check if more adjustments are needed in the other components of the equation.

Electrons provide the key to understanding the transfer of charge and energy between substances. Tracking them carefully facilitates the understanding of how matter undergoes chemical transformations and maintains the equilibrium during the reaction process. This is essential for precisely predicting how substances react chemically.