Problem 45 - Chemistry The Molecular Nature of Matter

Question

Balance the following equations for reactions occurring in an acidic solution. (a) \(\mathrm{Sn}+\mathrm{NO}_{3}^{-} \longrightarrow \mathrm{SnO}_{2}+\mathrm{NO}\) (b) \(\mathrm{PbO}_{2}+\mathrm{Cl}^{-} \longrightarrow \mathrm{PbCl}_{2}+\mathrm{Cl}_{2}\) (c) \(\mathrm{Ag}+\mathrm{NO}_{3}^{-} \longrightarrow \mathrm{NO}_{2}+\mathrm{Ag}^{+}\) (d) \(\mathrm{Fe}^{3+}+\mathrm{NH}_{3} \mathrm{OH}^{+} \longrightarrow \mathrm{Fe}^{2+}+\mathrm{N}_{2} \mathrm{O}\)

Step-by-Step Solution

Answer

The balanced equations are (a) \(\mathrm{Sn}^{2+} + 2\mathrm{NO}_{3}^{-} + 4\mathrm{H}^{+} \longrightarrow \mathrm{SnO}_{2} + 2\mathrm{NO} + 2\mathrm{H}_{2}\mathrm{O}\), (b) \(\mathrm{PbO}_{2} + 4\mathrm{Cl}^{-} + 4\mathrm{H}^{+} \longrightarrow \mathrm{PbCl}_{2} + \mathrm{Cl}_{2} + 2\mathrm{H}_{2}\mathrm{O}\), (c) \(3\mathrm{Ag} + 4\mathrm{H}^{+} + \mathrm{NO}_{3}^{-} \longrightarrow 3\mathrm{Ag}^{+} + 2\mathrm{H}_{2}\mathrm{O} + \mathrm{NO}_{2}\), and (d) \(2\mathrm{Fe}^{3+} + 2\mathrm{NH}_{3}\mathrm{OH}^{+} + 2\mathrm{H}_{2}\mathrm{O} \longrightarrow 2\mathrm{Fe}^{2+} + \mathrm{N}_{2}\mathrm{O} + 8\mathrm{H}^{+} + 6\mathrm{e}^{-}\).

Step 1: Balance atoms other than H and O

For reaction (a), balance tin (Sn) atoms by having same number of Sn atoms on both sides. The nitrogen (N) atoms are already balanced. For reaction (b), lead (Pb) is balanced; balance chlorine (Cl) by adding coefficients. In reaction (c), silver (Ag) is balanced but balance the nitrogen (N) atoms. For reaction (d), balance iron (Fe) and nitrogen (N).

Step 2: Balance oxygen atoms

Add water (H2O) molecules to balance the oxygen atoms. For reaction (a), add 1 H2O on the right side. For reaction (b), no water molecules needed as O is already balanced. For reaction (c), add 2 H2O to the right side. For reaction (d), add 1 H2O to the right side.

Step 3: Balance hydrogen atoms

Add protons (H+) to balance hydrogen atoms. For reaction (a), add 2 H+ on the left side. For reaction (b) no H+ needed as there's no O or H in products or reactants. For reaction (c), add 2 H+ to the left side. For reaction (d), add 4 H+ to the left side.

Step 4: Balance charges

Balance the overall charges by adding electrons (e-). For reaction (a), add 2 e- on the right side. For reaction (b), add 2 e- on the left side. For reaction (c), add 1 e- on the right side. For reaction (d), add 4 e- on the right side.

Step 5: Verify the balance

Check that all atoms and charges are balanced in each reaction. Ensure that the total number of each type of atom and total charges are equal on both sides of the equation for all reactions.

Key Concepts

Redox Reactions in Acidic Solution

Redox reactions are fundamental to understanding a wide variety of chemical processes, including those occurring in acidic solutions. In an acidic environment, a redox reaction involves the transfer of electrons between species and the subsequent balancing of hydrogen ions (H+) found in the acid.

To properly balance redox reactions in acidic solutions, we consider the oxidation states of the substances involved. We often have to add H+ ions when balancing hydrogen atoms and use water (H2O) molecules to balance oxygen atoms, due to the presence of the acidic medium. For example, when balancing the reaction \(\mathrm{Ag} + \mathrm{NO}_{3}^{-} \longrightarrow \mathrm{NO}_{2} + \mathrm{Ag}^{+}\), water and hydrogen ions are added to the less hydrogen-containing side to achieve balance.

It's essential to approach these problems systematically, beginning with atoms that are not hydrogen or oxygen, then balancing the latter using water and hydrogen ions, as required by the acidic environment. Finally, any excess charge is balanced by adding electrons (e-), thus ensuring both mass and charge conservation.

Balancing Chemical Equations

Balancing chemical equations is a foundational skill in chemistry that ensures the law of conservation of mass is obeyed. This law states that matter cannot be created or destroyed in a closed system. Therefore, a chemical equation must have the same number of each type of atom on both the reactant and product sides.

For instance, in the equation \(\mathrm{Sn} + \mathrm{NO}_{3}^{-} \longrightarrow \mathrm{SnO}_{2} + \mathrm{NO}\), we start by balancing the number of tin (Sn) atoms and proceed to nitrogen (N) and oxygen (O) atoms. The process involves adding coefficients before each substance in the reaction. Once all atoms are balanced, the last step checks for charge balance which, in the case of reactions in an acidic solution, may require the addition of H+ ions and electrons.

• Step 1: Balance atoms other than H and O
• Step 2: Balance oxygen atoms
• Step 3: Balance hydrogen atoms
• Step 4: Balance charges
• Step 5: Verify the balance

Careful attention to each step will ensure both sides of the equation have the same number of each atom and that the overall charge is equal.

Half-Reaction Method

The half-reaction method is a systematic way of balancing redox reactions, where the overall reaction is split into two separate equations: oxidation and reduction half-reactions. Each half-reaction is balanced independently, allowing for more straightforward management of the electron transfer process.

The steps typically involve the following:

• Separate the reaction into its oxidation and reduction components.
• Balance all elements in the half-reactions apart from oxygen and hydrogen.
• Balance oxygen by adding water molecules.
• Balance hydrogen by adding H+ ions, when in acidic solutions.
• Balance the charges by adding electrons.
• Equalize the number of electrons in both half-reactions if necessary and then combine them back into a single balanced equation.

For complex reactions, like \(\mathrm{Fe}^{3+} + \mathrm{NH}_{3}\mathrm{OH}^{+} \longrightarrow \mathrm{Fe}^{2+} + \mathrm{N}_{2} \mathrm{O}\), the half-reaction method is particularly helpful to ensure no detail is missed. Once all half-reactions are successfully balanced, they are added back together, yielding a fully balanced redox equation that honors the conservation of mass and charge.