Question

Balance each of the following redox reactions: (a) \(\mathrm{Al}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{H}_{2}(g)\) (b) \(\mathrm{Fe}(s)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{FeCl}_{3}(s)\) (c) \(\mathrm{CO}(g)+\mathrm{I}_{2} \mathrm{O}_{5}(s) \longrightarrow \mathrm{I}_{2}(s)+\mathrm{CO}_{2}(g)\)

Short answer

(a) \(2\mathrm{Al} + 3\mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{Al}_2(\mathrm{SO}_4)_3 + 3\mathrm{H}_2\), (b) \(2\mathrm{Fe} + 3\mathrm{Cl}_2 \rightarrow 2\mathrm{FeCl}_3\), (c) \(5\mathrm{CO} + \mathrm{I}_2\mathrm{O}_5 \rightarrow 5\mathrm{CO}_2 + 5\mathrm{I}_2\)."

Step-by-step solution

Identify oxidation and reduction

For reaction (a), identify that aluminum (Al) is oxidized from 0 to +3, and hydrogen is reduced from +1 in \( \mathrm{H}_2\mathrm{SO}_4 \) to 0 in \( \mathrm{H}_2 \). In reaction (b), iron (Fe) is oxidized from 0 to +3, and chlorine (Cl) is reduced from 0 in \( \mathrm{Cl}_2 \) to -1 in \( \mathrm{FeCl}_3 \). In reaction (c), carbon in \( \mathrm{CO} \) is oxidized from +2 to +4, while iodine in \( \mathrm{I}_2\mathrm{O}_5 \) is reduced from +5 to 0.

Write oxidation and reduction half-reactions

For (a), oxidation half-reaction: \( 2\mathrm{Al} \rightarrow 2\mathrm{Al}^{3+} + 6e^- \), reduction half-reaction: \( 6H^+ + 6e^- \rightarrow 3\mathrm{H}_2 \). For (b), oxidation: \( 2\mathrm{Fe} \rightarrow 2\mathrm{Fe}^{3+} + 6e^- \), reduction: \( 3\mathrm{Cl}_2 + 6e^- \rightarrow 6\mathrm{Cl}^- \). For (c), oxidation: \( 5\mathrm{CO} \rightarrow 5\mathrm{CO}_2 + 10e^- \), reduction: \( \mathrm{I}_2\mathrm{O}_5 + 10e^- \rightarrow 5\mathrm{I}_2 \).

Balance charge and mass

For (a), multiply the reduction half-reaction by 1 to match the electrons from the oxidation half-reaction. For (b), simply add the oxidation and reduction reactions together since electrons already match. For (c), similarly, add the oxidation and reduction reactions after confirming the electron counts match.

Combine the half-reactions

Combine the half-reactions by adding them together for each case, ensuring all atoms and charges balance. Resulting equations: (a) \( 2\mathrm{Al} + 6\mathrm{H}^+ + 3\mathrm{SO}_4^{2-} \rightarrow \mathrm{Al}_2(\mathrm{SO}_4)_3 + 3\mathrm{H}_2 \).(b) \( 2\mathrm{Fe} + 3\mathrm{Cl}_2 \rightarrow 2\mathrm{FeCl}_3 \).(c) \( 5\mathrm{CO} + \mathrm{I}_2\mathrm{O}_5 \rightarrow 5\mathrm{CO}_2 + 5\mathrm{I}_2 \).

Check balance of atoms and charge

Ensure each element has the same number of atoms on both sides, and the total charge is the same on both sides for each reaction. Confirm these conditions for reactions (a), (b), and (c).

Key concepts

Oxidation-Reduction

In a redox reaction, one substance loses electrons, while another gains them. This process is essential in various chemical and biological reactions. The term "oxidation" refers to the loss of electrons, whereas "reduction" indicates a gain. You can easily remember this by using the mnemonic OIL RIG: Oxidation Is Loss, Reduction Is Gain.

Let's take a closer look at the reactions given in the exercise. For example, in (a), aluminum ( \( \text{Al} \)) loses electrons and gets oxidized. On the other hand, hydrogen, present in sulfuric acid ( \( \text{H}_2\text{SO}_4 \)), gains electrons and is reduced. This swapping creates a balance that drives the reaction.

• Oxidation involves an increase in oxidation state, as observed with aluminum turning from 0 to +3.
• Reduction involves a decrease in oxidation state, like hydrogen moving from +1 to 0.

Half-Reactions

Half-reactions are a way to break down redox reactions into simpler parts. By separating oxidation and reduction processes, you can understand each component's role in the overall reaction.

Let's consider reaction (a) again. In the oxidation half-reaction, aluminum transforms into aluminum ions, releasing electrons:\[ 2\text{Al} \rightarrow 2\text{Al}^{3+} + 6e^- \]
Here, two aluminum atoms become ions, losing a total of six electrons. In the reduction half-reaction, hydrogen ions gain electrons:\[ 6H^+ + 6e^- \rightarrow 3\text{H}_2 \]
By adding these half-reactions, you can ensure the electrons lost and gained match, leading to a balanced overall equation. This step is key for understanding how electron transfer balances the reaction.

Balancing Chemical Equations

Balancing chemical equations ensures that both mass and charge are conserved across a reaction. This principle is foundational in chemistry and is crucial for accurately representing chemical changes.

Balancing involves adjusting the coefficients of reactants and products to make sure both sides of the equation have equal numbers of each type of atom. For example, in reaction (b), the equation becomes effectively balanced by ensuring that both the number of iron and chlorine atoms match:\[ 2\text{Fe} + 3\text{Cl}_2 \rightarrow 2\text{FeCl}_3 \]

• Double-check by counting the atoms: 2 Fe before and after, and 6 Cl on both sides.
• Make sure the charges balance, even though this reaction is neutral.
• Balancing is a trial-and-error process but always aim to find the simplest whole number ratios.

Electrons

Electrons are the tiny particles responsible for electricity and chemical bonding. In redox reactions, they play a pivotal role as they move from one atom to another, causing oxidation and reduction.

In the given reactions, electron movement is what facilitates chemical change. Take reaction (c), for instance. Carbon monoxide ( \( \text{CO} \)) donates electrons to iodine pentoxide ( \( \text{I}_2\text{O}_5 \)), enabling the formation of carbon dioxide and solid iodine:\[ 5\text{CO} \rightarrow 5\text{CO}_2 + 10e^- \]\[ \text{I}_2\text{O}_5 + 10e^- \rightarrow 5\text{I}_2 \].

• Electrons are carriers of energy in chemical reactions.
• The balance of electron transfer ensures that all redox reactions stay neutral in charge, maintaining equilibrium.
• Tracking electron flow gives insights into how and why chemical transformations occur.