Question

Balance these redox reactions by the half reaction method. a) \(\mathrm{Br}^{-}+\mathrm{I}_{2} \rightarrow \mathrm{I}^{-}+\mathrm{Br}_{2}\) b) \(\mathrm{CrCl}_{3}+\mathrm{F}_{2} \rightarrow \mathrm{CrF}_{3}+\mathrm{Cl}_{2}\)

Short answer

(a) \(2\mathrm{Br}^- + \mathrm{I}_2 \rightarrow \mathrm{Br}_2 + 2\mathrm{I}^-\); (b) \(2\mathrm{CrCl}_3 + 3\mathrm{F}_2 \rightarrow 2\mathrm{CrF}_3 + 3\mathrm{Cl}_2\).

Step-by-step solution

Assign Oxidation Numbers

First, we determine the oxidation numbers of the elements involved in each reaction. For reaction (a), Br in \(\mathrm{Br}^-\) is -1, and I in \(\mathrm{I}_2\) is 0. In the products, I in \(\mathrm{I}^-\) is -1, and Br in \(\mathrm{Br}_2\) is 0. For reaction (b), Cr in \(\mathrm{CrCl}_3\) is +3, Cl in \(\mathrm{Cl}_2\) is 0, and F in \(\mathrm{F}_2\) is 0. In \(\mathrm{CrF}_3\), Cr remains +3 and Cl in \(\mathrm{Cl}_2\) is 0.

Write Half-Reactions

For reaction (a): \(\mathrm{Br}^- \rightarrow \mathrm{Br}_2\) is the oxidation half-reaction, while \(\mathrm{I}_2 \rightarrow \mathrm{I}^-\) is the reduction half-reaction. For reaction (b): \(\mathrm{CrCl}_3 \rightarrow \mathrm{CrF}_3\) and \(\mathrm{F}_2 \rightarrow \mathrm{Cl}_2\) since Cl in \(\mathrm{CrCl}_3\) precedes Cl in \(\mathrm{Cl}_2\) and F in \(\mathrm{F}_2\) changes to F in \(\mathrm{CrF}_3\). However, specific focus should be on F and Cl changes.

Balance Atoms Other than O and H

For reaction (a): Balance Br and I atoms. Each side needs 2 Br: \(2\mathrm{Br}^- \rightarrow \mathrm{Br}_2\). Each side needs 2 I: \(\mathrm{I}_2 \rightarrow 2\mathrm{I}^-\). For reaction (b): Balance Cr as they already appear balanced, focus on Cl and F first. Maintain the counts as the presence of Cl balances the half reactions upon balancing F.

Balance Charge with Electrons

For reaction (a): \(2\mathrm{Br}^- \rightarrow \mathrm{Br}_2 + 2e^-\) and \(\mathrm{I}_2 + 2e^- \rightarrow 2\mathrm{I}^-\). Charges are balanced when both gain and lose 2 electrons. For reaction (b): \(\mathrm{CrCl}_3 \rightarrow \mathrm{CrF}_3 + 3\mathrm{Cl}^-\) balances electron changes with \(\mathrm{F}_2 + 6e^- \rightarrow 2\mathrm{F}^-\) and thus \(3\mathrm{Cl}^-.\)

Equalize Electron Transfer

For reaction (a): Just combine half-reactions since electrons exchanged are already equal (2e each).For reaction (b): Multiply the entire \(\mathrm{CrCl}_3\) reduction by 2 to match \(\mathrm{F}_2\) changes, then combine with oxidations to ensure electrons canceled (6).

Write the Balanced Reaction

For reaction (a): Combine the half-reactions: \(2\mathrm{Br}^- + \mathrm{I}_2 \rightarrow \mathrm{Br}_2 + 2\mathrm{I}^-\).For reaction (b): Combine to form \(2\mathrm{CrCl}_3 + 3\mathrm{F}_2 \rightarrow 2\mathrm{CrF}_3 + 3\mathrm{Cl}_2\) properly balanced in charge and matter.

Key concepts

Half Reaction Method

To balance redox reactions effectively, the half reaction method is a crucial approach. This technique separates the redox equation into two half-reactions: one for oxidation and one for reduction. Each half-reaction is individually balanced for atoms and charge.
The advantage of using the half reaction method lies in its focus on electron transfer balancing. By separating into two parts, we can clearly see the number of electrons lost (oxidation) and gained (reduction). This enables precise balancing of these electrons.
In our original exercise, reaction (a) separates into: oxidation half-reaction \( \mathrm{Br}^- \rightarrow \mathrm{Br}_2 \) and reduction half-reaction \( \mathrm{I}_2 \rightarrow \mathrm{I}^- \). We balance each part separately then combine them. For reaction (b), though slightly complex, we identify the same divisions focused on species changes between \( \mathrm{Cl} \) and \( \mathrm{F} \).

• Identify substances involved in oxidation and reduction.
• Write out the separate half-reactions.
• Balance the atoms not involving oxygen or hydrogen.
• Balance charge with electrons specific to each half.

Oxidation Numbers

Understanding the changes in oxidation numbers is central to identifying redox reactions. Oxidation numbers are quasi-charges assigned based on the assumption that electrons in a molecule's covalent bonds belong to the more electronegative atom.
For example, in reaction (a), the \( \mathrm{Br}^- \) has an oxidation number of -1, contrasting sharply with its neutral state 0 in \( \mathrm{Br}_2 \). Similarly, \( \mathrm{I}_2 \) starts at an oxidation number of 0, going to -1 in \( \mathrm{I}^- \).
Assigning these numbers helps highlight the atoms that undergo oxidation or reduction. It's crucial because changes indicate electron transfer.
Here's how you can define them:

• Monatomic ions' oxidation number equals their charge.
• In compounds, elements have specific rules (e.g., oxygen in compounds is usually -2).
• Sum of oxidation numbers in a neutral compound equals zero while in ions equals the ion's charge.

Balancing Chemical Equations

Balancing chemical equations ensures that mass and charge are conserved in a reaction. For a redox reaction, this involves equalizing the quantity of each atom in reactants and products, as well as ensuring the charges are balanced.
There is an inherent structure to this process:

• First, balance the atoms that undergo changes (usually excluding oxygen and hydrogen initially).
• Determine the charge on each side of the reaction and add electrons as needed to balance the charge.
• Re-contextualize individual half-reactions into a complete equation that respects conservation of mass and charge.

In our exercise, balancing \( \mathrm{Br}^- \rightarrow \mathrm{Br}_2 \) by adding coefficients achieves atom balance, while electrons balance charge. Reaction (b) required ensuring the electron transfer quantity matched on both sides before finalizing the full equation.

Electron Transfer Balancing

Electron transfer balancing is a key detail in redox reactions, ensuring both oxidation and reduction half-reactions have equal numbers of electrons. This is vital because electrons act as the bridge for the conservation of charge across the reaction.
By explicitly writing out where electrons are gained or lost, you expose the core of the redox reaction. Techniques used include determining the least common multiple of electrons between reactions and adjusting coefficients accordingly.
For reaction (a), electron exchange is direct: both half-reactions directly balance electrons as 2 are involved in oxidation and reduction separately.
In reaction (b), initial discrepancies required multiplying half-reactions to equalize electron transfer before uniting them, indicating more complex electron balance is necessary. Follow these points:

• After determining electron loss and gain, adjust coefficients to achieve identical electron numbers in both half-reactions.
• Ensuring electron cancellation is crucial to completing redox reaction balancing.

This technique is necessary for consistent and accurately balanced redox equations.