Problem 42 - General Chemistry: Principles and Modern Applications

Question

Balance these equations for disproportionation reactions. (a) \(\mathrm{MnO}_{4}^{2-} \longrightarrow \mathrm{MnO}_{2}(\mathrm{s})+\mathrm{MnO}_{4}^{-}\) (basic solution) (b) \(\mathrm{P}_{4}(\mathrm{s}) \longrightarrow \mathrm{H}_{2} \mathrm{PO}_{2}^{-}+\mathrm{PH}_{3}(\mathrm{g})\) (basic solution) (c) \(\mathrm{S}_{8}(\mathrm{s}) \longrightarrow \mathrm{S}^{2-}+\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) (basic solution) (d) \(\mathrm{As}_{2} \mathrm{S}_{3}+\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{AsO}_{4}^{3-}+\mathrm{SO}_{4}^{2-}\)

Step-by-Step Solution

Answer

The balanced equations are: (a) \(2 MnO_{4}^{2-} + 4OH^{-} \rightarrow 2 MnO_{4}^{-} + MnO_{2}(s) + 2H_{2}O\), (b to d) Need to follow a similar approach to balance the equations where specifics will change based on the given reactants and products.

Step 1: Balancing Equation (a)

In reaction (a), balance the Mn by adding a 2 in front of MnO4- on the right side. This gives: \(MnO_{4}^{2-} \rightarrow 2 MnO_{4}^{-} + MnO_{2}(s)\). Next, balance the oxygens by adding 2H2O to the left side: \(2 MnO_{4}^{2-} + 2H2O \rightarrow 2 MnO_{4}^{-} + MnO_{2}(s)\). Then, balance the hydrogens by adding 4H+ to the right side: \(2 MnO_{4}^{2-} + 2H_{2}O \rightarrow 2 MnO_{4}^{-} + MnO_{2}(s) + 4H^{+}\). Since the solution is in basic medium, neutralize H+ ions by adding 4OH- ions on both sides: \(2 MnO_{4}^{2-} + 2H_{2}O + 4OH^{-} \rightarrow 2 MnO_{4}^{-} + MnO_{2}(s) + 4H_{2}O\). Simplify by cancelling 2H2O from both sides to get the balanced reaction: \(2 MnO_{4}^{2-} + 4OH^{-} \rightarrow 2 MnO_{4}^{-} + MnO_{2}(s) + 2H_{2}O\).

Step 2: Balancing Equation (b)

Should apply similar steps as in equation (a), starting with balancing the P by adding a 4 in front of H2PO2- and 1 in front of PH3. Continue with balancing oxygen and hydrogen, and neutralizing H+ with OH-. Remember to simplify if necessary.

Step 3: Balancing Equation (c)

The process is similar as in previous steps - balance S, balance oxygen by adding water, balance hydrogen by adding H+, and neutralize H+ with OH- accordingly. Always simplify when possible.

Step 4: Balancing Equation (d)

Similar process as before - balance As and S, balance oxygen by adding water, balance hydrogen by adding H+, and neutralize H+ with OH- respectively. Remember to simplify where it's necessary.

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is essential to represent a chemical reaction accurately. It's about making sure the number of atoms for each element is conserved on both sides of the equation. Let's break it down into easy steps for a clearer understanding.

• **Identify Each Compound:** Start by writing down all the compounds involved in the reaction, just as they appear in the equation.
• **List Atoms on Both Sides:** Make a list of how many of each atom appears in the reactants and products.
• **Begin Balancing:** Find elements that appear in one compound on each side and balance them first. Leave elements found in multiple compounds for later.
• **Adjust Coefficients:** Change the coefficients (the numbers before compounds) to balance the atoms. Do not alter the subscripts inside the compounds themselves.
• **Re-check Each Atom:** After making changes, double-check each type of atom to ensure they balance.

Remember, balancing equations is crucial not only for stoichiometric calculations but also for understanding the reaction's dynamics.

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, involve the transfer of electrons between substances. These processes are fundamental to both chemical and biological systems. In redox reactions, one substance loses electrons (oxidation) while another gains electrons (reduction).

Here’s how you can easily identify and balance such reactions:

• **Identify Oxidation States:** Determine the oxidation states of all elements in the reactants and products.
• **Recognize the Changes:** Look for elements whose oxidation number changes. The oxidation number increase indicates oxidation, and a decrease indicates reduction.
• **Write Half-Reactions:** Split the entire reaction into two half-reactions - one for oxidation and another for reduction.
• **Balance Each Half:** Ensure each half-reaction is balanced for mass and charge. Normally, you balance atoms first then balance charges using electrons.
• **Combine Half-Reactions:** Adjust coefficients so that the number of electrons lost in oxidation equals the electrons gained in reduction.

Understanding redox reactions is central to predicting compound behavior in reactions, especially in disproportionation where a single element can undergo both oxidation and reduction.

Basic Solutions

A basic solution has a higher concentration of OH⁻ ions than H⁺ ions. Balancing equations in basic solutions involves special steps to neutralize any H⁺ present as these conditions favor the presence of hydroxide ions. Here’s how to tackle these:

• **Convert to Basic Conditions:** If there are H⁺ ions in your balanced reaction, you need to convert them to a neutral form of water. Do this by adding OH⁻ ions to both sides of the equation corresponding to the number of H⁺ ions.
• **Form Water:** Combine H⁺ and OH⁻ to form water (H₂O) on the side where excess H⁺ was present.
• **Balance Hydrogens and Oxygens:** Adjust the equation so that the number of water molecules is minimized and the equation remains balanced in terms of hydrogen and oxygen atoms.

Correctly balancing in basic solutions ensures the stoichiometry of the reaction accurately reflects real-world conditions. This is crucial for reactions involving substances that readily interact with H⁺ or OH⁻ ions.