Question

Complete and balance the following half-reactions. In each case indicate whether the half-reaction is an oxidation or a reduction. (a) \(\mathrm{Mo}^{3+}(a q) \longrightarrow \mathrm{Mo}(s)\) (acidic solution) (b) \(\mathrm{H}_{2} \mathrm{SO}_{3}(a q) \longrightarrow \mathrm{SO}_{4}^{2-}(a q)\) (acidic solution) (c) \(\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)\) (acidic solution) (d) \(\mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) (acidic solution) (e) \(\mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) (basic solution) (f) \(\mathrm{Mn}^{2+}(a q) \longrightarrow \mathrm{MnO}_{2}(s)\) (basic solution) (g) \(\mathrm{Cr}(\mathrm{OH})_{3}(s) \longrightarrow \mathrm{CrO}_{4}^{2-}(a q)\) (basic solution)

Short answer

(a) \(\mathrm{Mo}^{3+}(a q) + 3 \mathrm{e}^{-} \longrightarrow \mathrm{Mo}(s)\) (Reduction) (b) \(\mathrm{H}_{2} \mathrm{SO}_{3}(a q) + 2\mathrm{H}_{2}\mathrm{O}(l) + 2 \mathrm{e}^{-} \longrightarrow \mathrm{SO}_{4}^{2-}(a q)+ 4\mathrm{H}^{+}(a q)\) (Reduction) (c) \(4\mathrm{H}^{+}(a q)+ \mathrm{NO}_{3}^{-} (a q) + 3\mathrm{e}^{-} \longrightarrow \mathrm{NO}(g) + 2\mathrm{H}_{2}\mathrm{O}(l)\) (Reduction)

Step-by-step solution

Balance atoms other than H and O

In this case, the Mo atoms are already balanced.

Balance the O atoms

There are no O atoms in this half-reaction, so this step can be skipped.

Balance the H atoms

There are no H atoms in this half-reaction, so this step can be skipped.

Balance the charge

Add 3 electrons to the right side of the equation to balance the charge: \[\mathrm{Mo}^{3+}(a q) + 3 \mathrm{e}^{-} \longrightarrow \mathrm{Mo}(s)\]

Determine if it is an oxidation or reduction

Since the electrons are being gained on the right side, this is a reduction half-reaction. (b) \(\mathrm{H}_{2} \mathrm{SO}_{3}(a q) \longrightarrow \mathrm{SO}_{4}^{2-}(a q)\) (acidic solution)

Balance atoms other than H and O

The S atoms are already balanced.

Balance the O atoms

Add 2 \(\mathrm{H}_{2}\mathrm{O}\) molecules to the left side to balance the O atoms: \[\mathrm{H}_{2} \mathrm{SO}_{3}(a q) + 2\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{SO}_{4}^{2-}(a q)\]

Balance the H atoms

Add 4 \(\mathrm{H}^{+}\) ions to the right side to balance the H atoms: \[\mathrm{H}_{2} \mathrm{SO}_{3}(a q) + 2\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{SO}_{4}^{2-}(a q)+ 4\mathrm{H}^{+}(a q)\]

Balance the charge

Add 2 electrons to the right side of the equation to balance the charge: \[\mathrm{H}_{2} \mathrm{SO}_{3}(a q) + 2\mathrm{H}_{2}\mathrm{O}(l) + 2 \mathrm{e}^{-} \longrightarrow \mathrm{SO}_{4}^{2-}(a q)+ 4\mathrm{H}^{+}(a q)\]

Determine if it is an oxidation or reduction

Since the electrons are being gained on the right side, this is a reduction half-reaction. (c) \(\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)\) (acidic solution)

Balance atoms other than H and O

The N atoms are already balanced.

Balance the O atoms

Add 2 \(\mathrm{H}_{2}\mathrm{O}\) molecules to the right side to balance the O atoms: \[\mathrm{NO}_{3}^{-} (a q) \longrightarrow \mathrm{NO}(g) + 2\mathrm{H}_{2}\mathrm{O}(l)\]

Balance the H atoms

Add 4 \(\mathrm{H}^{+}\) ions to the left side to balance the H atoms: \[4\mathrm{H}^{+}(a q)+ \mathrm{NO}_{3}^{-} (a q) \longrightarrow \mathrm{NO}(g) + 2\mathrm{H}_{2}\mathrm{O}(l)\]

Balance the charge

Add 3 electrons to the left side of the equation to balance the charge: \[4\mathrm{H}^{+}(a q)+ \mathrm{NO}_{3}^{-} (a q) + 3\mathrm{e}^{-} \longrightarrow \mathrm{NO}(g) + 2\mathrm{H}_{2}\mathrm{O}(l)\]

Determine if it is an oxidation or reduction

Since the electrons are being gained on the right side, this is a reduction half-reaction. The solutions for the remaining half-reactions (d, e, f, and g) follow similar steps to the above examples. You can practice completing and balancing them to gain a better understanding. Keep in mind that basic solutions will require the addition of \(\mathrm{OH}^{-}\) ions to balance the H and O atoms.

Key concepts

Half-Reactions

Half-reactions are essential to understanding redox reactions in chemistry. They break down a redox reaction into two parts: oxidation and reduction. Each half-reaction shows either the gain or loss of electrons.
In the given exercise, we focus on balancing these half-reactions. This involves ensuring that both the number of atoms and the charge remain equal on both sides of the equation. By analyzing the movement of electrons in these half-reactions, we determine if each is an oxidation (loss of electrons) or a reduction (gain of electrons) process.
Understanding half-reactions is crucial because it allows chemists to study the electron transfer in a controlled manner. This breakdown helps in designing experiments and processes where specific redox reactions are needed.

• Oxidation results in electron loss.
• Reduction results in electron gain.
• Each half-reaction must have balanced charge and atoms.

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry. This involves ensuring that the number of each type of atom is equal on both sides of the equation. In addition to atoms, the charge must also be balanced.
For the half-reactions in the exercise, steps include balancing atoms other than hydrogen and oxygen first. Then, balance oxygen by adding water molecules, and hydrogen by adding hydrogen ions. Finally, balance the electrical charge by adding electrons. Balancing is important because it adheres to the law of conservation of mass. It ensures that no atoms are lost in the process, allowing chemists to predict the amounts of products and reactants accurately.

• Balance other atoms first.
• Balance oxygen using water.
• Balance hydrogen with hydrogen ions.
• Balance charge with electrons.

Oxidation and Reduction

Oxidation and reduction are two complementary processes in redox reactions. They describe the transfer of electrons between two substances. Oxidation involves the loss of electrons, leading to an increase in oxidation state. Reduction is the gain of electrons, decreasing the oxidation state. In every redox reaction, one species is oxidized while another is reduced. In the context of the exercise, determining whether the half-reaction is oxidation or reduction is done by observing the movement of electrons. If electrons are on the products side, it's reduction. If they're on the reactants side, it's oxidation. Both processes happening simultaneously keep the charge balanced.

• Redox reactions involve electron transfer.
• Oxidation: loss of electrons/increase in oxidation state.
• Reduction: gain of electrons/decrease in oxidation state.

Acidic and Basic Solutions

The medium of a chemical reaction, whether acidic or basic, can significantly influence the reaction process and products. In acidic solutions, hydrogen ions (\(\mathrm{H}^{+}\)) are plentiful, assisting in balancing hydrogen atoms in equations. Conversely, basic solutions require hydroxide ions (\(\mathrm{OH}^{-}\)) to maintain balance.For example, in the given exercise, reactions in acidic solutions often involve adding \(\mathrm{H}^{+}\) ions to the reaction. This changes when dealing with basic solutions, where \(\mathrm{OH}^{-}\) ions are used instead.This distinction is crucial in laboratory and industrial processes, as the pH of the solution can affect reaction rates and product stability.

• Acidic solutions: add \(\mathrm{H}^{+}\) ions.
• Basic solutions: add \(\mathrm{OH}^{-}\) ions.
• pH can influence reaction mechanisms and outcomes.