Problem 77 - General Chemistry: Principles and Modern Applications

Question

Balance these equations for reactions in acidic solution. (a) \(\mathrm{IBr}+\mathrm{BrO}_{3}^{-}+\mathrm{H}^{+} \longrightarrow \mathrm{IO}_{3}^{-}+\mathrm{Br}^{-}+\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NO}_{3}+\mathrm{Sn} \longrightarrow\) \(\mathrm{NH}_{2} \mathrm{OH}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{Sn}^{2+}\) (c) \(\mathrm{As}_{2} \mathrm{S}_{3}+\mathrm{NO}_{3}^{-} \longrightarrow \mathrm{H}_{3} \mathrm{AsO}_{4}+\mathrm{S}+\mathrm{NO}\) (d) \(\mathrm{H}_{5} \mathrm{IO}_{6}+\mathrm{I}_{2} \longrightarrow \mathrm{IO}_{3}^{-}+\mathrm{H}^{+}+\mathrm{H}_{2} \mathrm{O}\) (e) \(\mathrm{S}_{2} \mathrm{F}_{2}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{S}_{8}+\mathrm{H}_{2} \mathrm{S}_{4} \mathrm{O}_{6}+\mathrm{HF}\)

Step-by-Step Solution

Answer

Balanced equations: (a) \(2IBr + BrO_{3}^{-} + 6H^{+} \rightarrow 2IO_{3}^{-} + 5Br^{-} + 3H_{2}O\); (b) \(2CH_{3}CH_{2}NO_{3} + 2Sn \rightarrow 2NH_{2}OH + 2CH_{3}CH_{2}OH + 2Sn^{2+}\); (c) \(2As_{2}S_{3} + 13NO_{3}^{-} + 16H^{+} \rightarrow 2H_{3}AsO_{4} + 8S + 3NO + 10H_{2}O\); (d) \(H_{5}IO_{6} + 5I_{2} + 6H^{+} \rightarrow 6IO_{3}^{-} + 3H_{2}O\); (e) \(16S_{2}F_{2} + 48H_{2}O \rightarrow 8S_{8} + 16H_{2}S_{4}O_{6} + 32HF\).

Step 1: Determine oxidation states

Identify the oxidation states of the atoms in the reactants and products.

Step 2: Write down the half-reactions

The atoms that change the oxidation states are participating in the redox reaction. Write down the two half-reactions, one for oxidation and one for reduction.

Step 3: Balance the half-reactions

Using the mass balance, ensure that each atom appears the same number of times on both sides of the equation. To balance the charges, add protons \(H^{+}\) to the side that needs positive charges or add water molecules \(H_{2}O\) to provide the required amount of oxygen atoms, and if needed, add water molecules to the other side of the equation to compensate for the excess of hydrogen atoms generated.

Step 4: Combine the half-reactions

Combine the two half-reactions into one equation, ensuring that the electrons are balanced. Multiply the reactions if necessary until the number of electrons on both sides is the same. Make sure to cancel out the same substances on both sides of the equation whenever possible.

Step 5: Verify the balance

Check the complete balanced equation to ensure each atom and the overall charge are balanced.

Key Concepts

Oxidation States

Understanding oxidation states is crucial for balancing redox reactions. An oxidation state is a hypothetical charge that an atom would have if all bonds were ionic. Determining these states helps identify which atoms are oxidized and which are reduced. For example, knowing that iodine changes from +1 in IBr to +5 in IO₃⁻ shows it is oxidized. Similarly, recognizing that bromine shifts from +5 in BrO₃⁻ to -1 in Br⁻ reveals it is reduced.

Here are some tips to grasp oxidation states easily:

• Atoms in their elemental form have an oxidation state of 0.
• The sum of oxidation states in a neutral compound must be 0; in a polyatomic ion, it equals the charge of the ion.
• Group 1 metals are always +1, and Group 2 metals are always +2.

Practicing the determination of oxidation states will sharpen your skills in identifying redox processes.

Half-Reactions

Breaking down the reaction into half-reactions is a clear way to see what is being oxidized and reduced. A half-reaction focuses on either the oxidation or the reduction part of the redox process. For instance, in the reaction involving iodine and bromine, the oxidation could be expressed as:\[\text{IBr} \rightarrow \text{IO}_3^- + \text{e}^- \text{(Oxidation)}\]On the other hand, reduction could be shown as:\[\text{BrO}_3^- + \text{e}^- \rightarrow \text{Br}^- \text{(Reduction)}\]Balancing each half-reaction individually makes it easier to focus on the specific changes in oxidation states and the movement of electrons. This step-by-step approach ensures clarity and accuracy when dealing with more complex equations.

Mass Balance

Mass balance ensures that each atom is accounted for on both sides of the equation. This means if you start with a certain amount of each element in the reactants, you should have the same amount in the products. This principle is applied within independent half-reactions before they are combined.

Steps to achieve mass balance include:

• Balancing metals first, as they often participate directly in electron transfer.
• Balancing other elements next, leaving hydrogen and oxygen for last.
• Using water molecules to balance oxygen and hydrogen ions to balance hydrogen.

This structured approach guarantees that the redox reaction respects both the conservation of mass and charge.

Acidic Solution Redox Reactions

Balancing redox reactions in acidic solutions involves using protons (\(H^+\)) and water to balance the charges and atoms. These conditions require water to add oxygen atoms or protons to add hydrogen atoms where necessary. This helps maintain the delicate balance of the equation. For instance, if the half-reaction has more positive charge than the reactant side, you add \(H^+\) to balance it out, then use water to balance any unmatched oxygen. This sequence ensures a balanced redox reaction in an acidic medium.

Key points to remember:

• In acidic solutions, always add \(H^+\) and \(H_2O\) as needed.
• Double-check that both the atom count and the charge are balanced in the final equation.
• Verify electrolyte neutrality by making sure total positive and negative charges are equal on both sides.

Mastering these methods will allow you to handle redox reactions in various chemical environments efficiently.