Problem 96 - Chemistry: An Atoms-Focused Approach

Question

A method for determining the quantity of dissolved oxygen in natural waters requires a series of redox reactions. Balance the following chemical equations in that series under the conditions indicated: a. \(\mathrm{Mn}^{2+}(a q)+\mathrm{O}_{2}(g) \rightarrow \mathrm{MnO}_{2}(s)\) (basic solution) b. \(\mathrm{MnO}_{2}(s)+\mathrm{I}^{-}(a q) \rightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{I}_{2}(s)\) (acidic solution) c. \(\mathrm{I}_{2}(s)+\mathrm{S}_{2} \mathrm{O}_{3}^{2-}(a q) \rightarrow\) \(\mathrm{I}^{-}(a q)+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}(a q) \quad\) (neutral solution)

Step-by-Step Solution

Answer

Question: Balance the following chemical equations under different conditions: a. Mn²⁺(aq) + O₂(g) → MnO₂(s) in basic solution b. MnO₂(s) + I⁻(aq) → Mn²⁺(aq) + I₂(s) in acidic solution c. I₂(s) + S₂O₃²⁻(aq) → I⁻(aq) + S₄O₆²⁻(aq) in neutral solution Answer: a. In basic solution: Mn²⁺(aq) + 1/2 O₂(g) → MnO₂(s) b. In acidic solution: MnO₂(s) + 2I⁻(aq) + 4H⁺(aq) → Mn²⁺(aq) + I₂(s) + 2H₂O(l) c. In neutral solution: I₂(s) + 2S₂O₃²⁻(aq) → 2I⁻(aq) + S₄O₆²⁻(aq)

Step 1: Write the unbalanced equation

\(\mathrm{Mn}^{2+}(a q)+\mathrm{O}_{2}(g) \rightarrow \mathrm{MnO}_{2}(s)\)

Step 2: Identify the elements to balance

In this case, we need to balance the manganese (Mn) and oxygen (O) atoms.

Step 3: Balance the elements

Since there is already 1 Mn on each side, we need to balance the oxygen atoms by adding a coefficient to the oxygen gas: \(\mathrm{Mn}^{2+}(a q)+\frac{1}{2}\mathrm{O}_{2}(g) \rightarrow \mathrm{MnO}_{2}(s)\)

Step 4: Balance the charge (in this case, it's already balanced)

+2 charge on the left side and +2 charge on the right side.

Step 5: Double-check and simplify

The balanced equation in the basic solution is: \(\mathrm{Mn}^{2+}(a q)+\frac{1}{2}\mathrm{O}_{2}(g) \rightarrow \mathrm{MnO}_{2}(s)\) #b. Balancing equation in acidic solution#

Step 1: Write the unbalanced equation

$\mathrm{MnO}_{2}(s)+\mathrm{I}^{-}(a q) \rightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{I}_{2}(s)$

Step 2: Identify the elements to balance

In this case, we need to balance the manganese (Mn), oxygen (O), and iodine (I) atoms.

Step 3: Balance the elements

Mn: Already balanced. O: Add 4H+ ions on the right side to balance the oxygen atom: $\mathrm{MnO}_{2}(s)+\mathrm{I}^{-}(a q) \rightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{I}_{2}(s) + 2\mathrm{H}_2\mathrm{O}(l)$ I: Add the coefficient 2 in front of I- to balance the iodine atoms: $\mathrm{MnO}_{2}(s)+2\mathrm{I}^{-}(a q) \rightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{I}_{2}(s) + 2\mathrm{H}_2\mathrm{O}(l)$

Step 4: Balance the charge

Add 4H+ ions in the reactants side to balance the charge: $\mathrm{MnO}_{2}(s)+2\mathrm{I}^{-}(a q) + 4\mathrm{H}^{+}(a q) \rightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{I}_{2}(s) + 2\mathrm{H}_2\mathrm{O}(l)$

Step 5: Double-check and simplify

The balanced equation in the acidic solution is: $\mathrm{MnO}_{2}(s)+2\mathrm{I}^{-}(a q) + 4\mathrm{H}^{+}(a q) \rightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{I}_{2}(s) + 2\mathrm{H}_2\mathrm{O}(l)$ #c. Balancing equation in neutral solution#

Step 1: Write the unbalanced equation

$\mathrm{I}_{2}(s)+\mathrm{S}_{2} \mathrm{O}_{3}^{2-}(a q) \rightarrow \mathrm{I}^{-}(a q)+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}(a q)$

Step 2: Identify the elements to balance

In this case, we need to balance the iodine (I) and sulfur (S) atoms, as well as the oxygen (O) atoms as required.

Step 3: Balance the elements

I: Add the coefficient 2 in front of I- to balance the iodine atoms: $\mathrm{I}_{2}(s)+\mathrm{S}_{2} \mathrm{O}_{3}^{2-}(a q) \rightarrow 2\mathrm{I}^{-}(a q)+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}(a q)$ S: Add the coefficient 2 in front of the S2O3 ions to balance: $ \mathrm{I}_{2}(s)+2\mathrm{S}_{2} \mathrm{O}_{3}^{2-}(a q) \rightarrow 2\mathrm{I}^{-}(a q)+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}(a q)$ O: oxygen atoms are balanced.

Step 4: Balance the charge

In this case, the charges are already balanced with -2 on both sides.

Step 5: Double-check and simplify

The balanced equation in the neutral solution is: $\mathrm{I}_{2}(s)+2\mathrm{S}_{2} \mathrm{O}_{3}^{2-}(a q) \rightarrow 2\mathrm{I}^{-}(a q)+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}(a q)$

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is a crucial skill in understanding any type of chemical reaction, including redox reactions. It involves making sure that the number of atoms for each element is the same on both sides of the equation. This ensures that mass is conserved, as required by the law of conservation of mass. When balancing, it's helpful to follow a systematic approach:

• First, write the unbalanced equation with all reactants and products clearly listed.
• Next, identify the elements that need balancing—in redox reactions, this usually includes metals and oxygen or hydrogen atoms.
• Balance the elements one at a time by adjusting coefficients; start with elements that appear in only one reactant and one product.
• Remember that coefficients affect all atoms in a molecule.
• Finally, double-check your work to make sure all elements are balanced and the smallest whole number of coefficients is used.

Breaking down complicated reactions into these steps can make the process much more manageable. With practice, balancing equations becomes an intuitive part of understanding chemistry.

Acidic and Basic Solutions

Understanding whether a reaction takes place in an acidic or basic solution is necessary for correctly balancing some chemical equations. The conditions of the solution affect the species present and how charges are balanced.

• Acidic solutions contain an excess of hydrogen ions ( H^{+} ), which can be used to balance equations that involve oxygen atoms. Often, water ( H_2O ) is also formed.
• In basic solutions, hydroxide ions ( OH^{-} ) are present, which lead to the formation of water when needed for balancing.

It's important to incorporate sodium hydroxide ( NaOH ) in basic reactions when water formation is indicated, and hydrochloric acid ( HCl ) might be seen in acidic solutions. The medium in which a chemical reaction is balanced often determines the appropriate balancing method and highlights the importance of understanding solution chemistry.

Charge Balancing

Charge balancing is a critical aspect of balancing chemical equations, especially in redox reactions. In any chemical equation, the total charge on the reactant side must equal the total charge on the product side to maintain electrostatic neutrality. This means that when you balance equations, you not only care about the number of atoms but also their charges. When determining how to balance charges:

• Identify the oxidation state of the species involved.
• Add or remove electrons as needed to ensure that the net charge is the same on both sides of the equation.
• In reactions in acidic or basic media, use H^{+} ions or OH^{-} ions to adjust charge discrepancies.

Double-checking that the number of electrons lost in oxidation matches the number gained in reduction ensures that your equation correctly represents the redox process.

Oxidation States

Oxidation states, or oxidation numbers, are a fundamental concept for understanding redox reactions. They help determine how electrons are transferred between species during the reaction. Steps to assign oxidation states:

• Elements in their natural form, like O_2 or H_2 , have an oxidation state of zero.
• For ions, the oxidation state is equal to the charge of the ion.
• In molecules or compounds, the sum of the oxidation states of all atoms must equal the overall charge of the molecule.
• Common rules include assigning -1 for fluorine in compounds, +1 for hydrogen, and -2 for oxygen unless bonded to fluorine.

By identifying changes in oxidation states, you can see where oxidation (loss of electrons) and reduction (gain of electrons) occur. This process helps in the identification and balancing of redox equations, showcasing the flow and conservation of electrons throughout the reaction.