Question

Write balanced equations for the following halfreactions. Specify whether each is an oxidation or reduction. (a) \(\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) \rightarrow \mathrm{O}_{2}(\mathrm{g})\) (in acid) (b) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{aq}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})\) (in acid) (c) \(\mathrm{NO}_{3}^{-}(\mathrm{aq}) \rightarrow \mathrm{NO}(\mathrm{g})\) (in acid) (d) \(\mathrm{MnO}_{4}^{-}(\mathrm{aq}) \rightarrow \mathrm{MnO}_{2}(\mathrm{s})\) (in base)

Short answer

(a) Oxidation: \(\text{H}_2\text{O}_2 \rightarrow \text{O}_2 + 2\text{H}^+ + 2\text{e}^-\); (b) Oxidation: \(\text{H}_2\text{C}_2\text{O}_4 \rightarrow 2 \text{CO}_2 + 2 \text{H}^+ + 2 \text{e}^-\); (c) Reduction: \(\text{NO}_3^- + 4\text{H}^+ + 3\text{e}^- \rightarrow \text{NO} + 2\text{H}_2\text{O}\); (d) Reduction: \(\text{MnO}_4^- + 2\text{H}_2\text{O} + 3\text{e}^- \rightarrow \text{MnO}_2 + 4 \text{OH}^-\).

Step-by-step solution

Identify Oxidation States for (a)

The oxidation state of oxygen in \(\text{H}_2\text{O}_2\) is \(-1\), and in \(\text{O}_2\) it is \(0\). Since the oxidation state of oxygen increases, this is an oxidation reaction.

Balance Atoms and Charge for (a)

Start with the unbalanced equation: \(\text{H}_2\text{O}_2 \rightarrow \text{O}_2\). To balance it in acidic solution, we can add \(2\text{H}^+\) and \(2\text{e}^-\) on the right side to balance charges and hydrogens: \(\text{H}_2\text{O}_2 \rightarrow \text{O}_2 + 2\text{H}^+ + 2\text{e}^-\).

Identify Oxidation States for (b)

The oxidation state of carbon in \(\text{H}_2\text{C}_2\text{O}_4\) is \(+3\), and in \(\text{CO}_2\) it is \(+4\). Since the oxidation state increases, this is an oxidation reaction.

Balance Atoms and Charge for (b)

Starting from \(\text{H}_2\text{C}_2\text{O}_4 \rightarrow 2 \text{CO}_2\), in an acidic solution, add \(2 \text{H}^+\) and \(2 \text{e}^-\) to balance charges: \(\text{H}_2\text{C}_2\text{O}_4 \rightarrow 2 \text{CO}_2 + 2 \text{H}^+ + 2 \text{e}^-\).

Identify Oxidation States for (c)

The oxidation state of nitrogen in \(\text{NO}_3^-\) is \(+5\), and in \(\text{NO}\) it is \(+2\). This decrease in oxidation state indicates a reduction reaction.

Balance Atoms and Charge for (c)

Start with \(\text{NO}_3^- \rightarrow \text{NO}\). In acidic solution, add \(3 \text{H}_2\text{O}\), \(4\text{H}^+\), and \(3\text{e}^-\) to fully balance: \(\text{NO}_3^- + 4\text{H}^+ + 3\text{e}^- \rightarrow \text{NO} + 2\text{H}_2\text{O}\).

Identify Oxidation States for (d)

The oxidation state of manganese in \(\text{MnO}_4^-\) is \(+7\), and in \(\text{MnO}_2\) it is \(+4\). The decrease in oxidation state indicates a reduction reaction.

Balance Atoms and Charges for (d)

Since the reaction is in a basic environment, start with \(\text{MnO}_4^- \rightarrow \text{MnO}_2\). Add \(2 \text{H}_2\text{O}\), \(3\text{e}^-\), and \(4 \text{OH}^-\) to balance: \(\text{MnO}_4^- + 2\text{H}_2\text{O} + 3\text{e}^- \rightarrow \text{MnO}_2 + 4 \text{OH}^-\).

Key concepts

Oxidation

When talking about oxidation in chemistry, it refers to the process where an atom, ion, or molecule loses electrons. This can be identified through a change in oxidation state or number. Remember that oxidation does not always involve oxygen, despite the name suggesting that. It's all about losing electrons!

Take the example from the exercise, where - In - \(H_2O_2 \)- The oxidation state of - oxygen increases from -from - -1 to - 0 in - \(O_2 \). This increase is a clear sign that oxidation has occurred.

Here are key indicators of oxidation:

• An increase in oxidation state
• Loss of electrons
• Possibly forming more oxygens molecules by losing hydrogen or other attached groups

Reduction

Reduction is the opposite of oxidation. During a reduction process, an atom, ion, or molecule gains electrons. This leads to a decrease in oxidation state, which is the main indicator that reduction has taken place.

Using the exercise, in the transformation of \(NO_3^- \ ightarrow NO\), the oxidation state of nitrogen reduces from \( +5 \) to \( +2 \). This decrease highlights a reduction.

To spot reduction, look for:

• A decrease in oxidation state
• Gain of electrons
• Reduction in the number of oxygen atoms \( if they are present\)

Balancing Chemical Equations

Balancing chemical equations ensures that the Law of Conservation of Mass is satisfied. This means that the number and type of atoms on reactants side must equal those on the products side. This is crucial for accurately describing a chemical reaction.

In redox reactions, like those in the exercise, sometimes balance can be tricky due to electron transfer. Balance involves:

• Balancing elements other than - \(O \)
• Adding - \(H^+ \)ions to balance hydrogen
• Using - electrons to account for oxidation or reduction activity

For instance, the conversion of \(H_2C_2O_4 \ ightarrow 2 CO_2\) is balanced by adding - \(2H^+ \) and - \(2e^- \) as shown in the solution steps. This ensures that the atoms and charges are both balanced on both sides.

Here's what to consider:

• Balance metals first
• Hydrogens are balanced last, often using - \(H^+ \)
• Ensure electrons lost equal those gained

Oxidation States

Oxidation states (or numbers) help chemists keep track of electrons in atoms, ions, or molecules. They are assigned based on a set of rules and can reveal the electron flow in a reaction, helping to find oxidation and reduction processes accurately.

In simple terms, oxidation states show the degree of oxidation of an atom, indicating whether it has lost or gained electrons. For instance, in \(MnO_4^- \ ightarrow MnO_2\), manganese changes from an oxidation state of \( +7 \) to \( +4 \). This decrease illustrates a reduction.

Rules to determine oxidation states:

• The oxidation state of an element in its standard state is zero.
• For a simple monatomic ion, the oxidation state is the charge of the ion.
• In compounds, hydrogen is \( +1 \) and oxygen is usually \( -2 \).

Understanding oxidation states is key to mastering redox reactions, helping predict how substances will interact in chemical processes.