Problem 70 - Chemistry Matter and Change

Question

Balance the following redox chemical equation. Rewrite the equation in full ionic form, then derive the net ionic equation and balance by the half- reaction method. Give the final answer as it is shown below but with the balancing coefficients. \(\mathrm{KMnO}_{4}(\mathrm{aq})+\mathrm{FeSO}_{4}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \rightarrow\) \(\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}(\mathrm{aq})+\mathrm{MnSO}_{4}(\mathrm{aq})+\) \(\quad \mathrm{K}_{2} \mathrm{SO}_{4}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1)\)

Step-by-Step Solution

Answer

The final balanced chemical equation is: \(\mathrm{KMnO}_{4}(\mathrm{aq}) + 5\mathrm{FeSO}_{4}(\mathrm{aq}) + 8\mathrm{H}_{2}\mathrm{SO}_{4}(\mathrm{aq}) \rightarrow 2\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}(\mathrm{aq}) + \mathrm{MnSO}_{4}(\mathrm{aq}) + 3\mathrm{K}_{2}\mathrm{SO}_{4}(\mathrm{aq}) + 4\mathrm{H}_{2}\mathrm{O}(\mathrm{l})\)

Step 1: Write the full ionic equation

Break down the given equation into its ionic components: \(\mathrm{K}^{+}(\mathrm{aq}) + \mathrm{MnO}_{4}^{-}(\mathrm{aq}) + \mathrm{Fe}^{2+} (\mathrm{aq}) + \mathrm{SO}_{4}^{2-} (\mathrm{aq}) + 2\mathrm{H}^{+} (\mathrm{aq}) + \mathrm{SO}_{4}^{2-} (\mathrm{aq})\) \(\rightarrow\) \(\mathrm{Fe}^{3+} (\mathrm{aq}) + 2\mathrm{SO}_{4}^{2-} (\mathrm{aq}) + \mathrm{Mn}^{2+} (\mathrm{aq}) + \mathrm{SO}_{4}^{2-} (\mathrm{aq}) + 2\mathrm{K}^{+} (\mathrm{aq}) + \mathrm{SO}_{4}^{2-} (\mathrm{aq}) + \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\)

Step 2: Identify oxidation and reduction half-reactions

We see that Fe undergoes oxidation from +2 to +3, and Mn undergoes reduction from +7 to +2. Hence, the half-reactions are: Oxidation: \(\mathrm{Fe}^{2+} (\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq}) + \mathrm{e}^{-}\) Reduction: \(\mathrm{MnO}_{4}^{-} (\mathrm{aq}) + 5\mathrm{e}^{-} + 8\mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{Mn}^{2+} (\mathrm{aq}) + 4\mathrm{H}_{2}\mathrm{O}(\mathrm{l})\)

Step 3: Balance half-reactions

For balancing the equations, we need the same number of electrons in both reactions. The least common multiple of 1 and 5 is 5. So, multiply the oxidation half-reaction by 5: \(5\mathrm{Fe}^{2+} (\mathrm{aq}) \rightarrow 5\mathrm{Fe}^{3+}(\mathrm{aq}) + 5\mathrm{e}^{-}\)

Step 4: Combine and simplify the balanced half-reactions

Add the two balanced half-reactions together and cancel like terms: \(\mathrm{MnO}_{4}^{-} (\mathrm{aq}) + 5\mathrm{Fe}^{2+} (\mathrm{aq}) + 8\mathrm{H}^{+} (\mathrm{aq}) \rightarrow \mathrm{Mn}^{2+} (\mathrm{aq}) + 5\mathrm{Fe}^{3+} (\mathrm{aq}) + 4\mathrm{H}_{2}\mathrm{O}(\mathrm{l})\) This is the balanced net ionic equation.

Step 5: Rewrite the balanced net ionic equation back to its overall chemical equation form

Replace the ions with their respective compounds: \(\mathrm{KMnO}_{4}(\mathrm{aq}) + 5\mathrm{FeSO}_{4}(\mathrm{aq}) + 8\mathrm{H}_{2}\mathrm{SO}_{4}(\mathrm{aq}) \rightarrow 2\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}(\mathrm{aq}) + \mathrm{MnSO}_{4}(\mathrm{aq}) + 3\mathrm{K}_{2}\mathrm{SO}_{4}(\mathrm{aq}) + 4\mathrm{H}_{2}\mathrm{O}(\mathrm{l})\) The final balanced chemical equation is: \(\mathrm{KMnO}_{4}(\mathrm{aq}) + 5\mathrm{FeSO}_{4}(\mathrm{aq}) + 8\mathrm{H}_{2}\mathrm{SO}_{4}(\mathrm{aq}) \rightarrow 2\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}(\mathrm{aq}) + \mathrm{MnSO}_{4}(\mathrm{aq}) + 3\mathrm{K}_{2}\mathrm{SO}_{4}(\mathrm{aq}) + 4\mathrm{H}_{2}\mathrm{O}(\mathrm{l})\)

Key Concepts

Ionic Equations

Ionic equations are a convenient way to represent chemical reactions that occur in aqueous solutions. They focus on the species that are actually involved in the reaction, leaving out the spectator ions that do not participate in the actual chemical change. In any ionic equation, we split the soluble ionic compounds into their individual ions. For instance, in the full ionic equation of our exercise above, we break down compounds like

• \(\text{K}^{+}(\text{aq}) + \text{MnO}_{4}^{\-}(\text{aq}) + \text{Fe}^{2+} (\text{aq}) + \text{SO}_{4}^{2-} (\text{aq}) + 2\text{H}^{+} (\text{aq}) + \text{SO}_{4}^{2-} (\text{aq})\)

into their ionic forms. This gives a clearer view of the actual chemical events, especially when it comes to tracking which atoms or ions are participating in the reaction.After writing out the full ionic equation, the next step is to derive the net ionic equation, which summarizes the raw chemical process by removing the spectator ions—those ions that don't undergo a change throughout the reaction. By focusing on the changes, we can easily observe how the reactants transform into the products, sharpening our understanding of the interaction at a molecular level.

Half-Reaction Method

The half-reaction method is a technique used to balance redox (reduction-oxidation) equations. This process involves separating the oxidation and reduction processes into two half-reactions and balancing them individually. By breaking down the redox reaction, it becomes easier to balance both mass and charge.

• The oxidation half-reaction shows the species that loses electrons (gets oxidized), such as \(\text{Fe}^{2+} (\text{aq}) \rightarrow \text{Fe}^{3+}(\text{aq}) + \text{e}^{-}\).
• The reduction half-reaction depicts the species gaining electrons (getting reduced), for example, \(\text{MnO}_{4}^{-} (\text{aq}) + 5\text{e}^{-} + 8\text{H}^{+}(\text{aq}) \rightarrow \text{Mn}^{2+} (\text{aq}) + 4\text{H}_{2}\text{O}(\text{l})\).

After writing both half-reactions, each needs to be balanced individually with respect to mass and charge. To balance the charge, electrons are added where needed. Next, you adjust the numbers so that the electrons lost in oxidation equal those gained in reduction, often necessitating multiplying one or both half-reactions by appropriate coefficients.

Balancing Chemical Equations

Balancing chemical equations is a crucial skill in chemistry that ensures the law of conservation of mass is obeyed. This law states that mass cannot be created or destroyed in a closed system through ordinary chemical reactions. Thus, the number of atoms of each element must be the same on both sides of the chemical equation. To balance the chemical equation using the half-reaction method, follow these steps:

• Write each half-reaction and balance the atoms that undergo redox changes.
• Balance the charges by adding electrons as needed.
• Make sure that the numbers of electrons in the oxidation and reduction half-reactions are the same by finding the least common multiple and adjusting the coefficients.

Once both half-reactions are balanced in terms of atoms and charge, they are combined to give a complete balanced equation. This process reflects how balanced chemical equations account for both mass and charge, providing a correct depiction of the chemical reaction.

Oxidation and Reduction

Oxidation and reduction reactions, collectively known as redox reactions, involve the transfer of electrons between species. These reactions are fundamental in understanding chemical processes such as metabolism, combustion, and corrosion.In any given redox reaction:

• Oxidation is the process where an element or compound loses electrons. This is observable with iron in the exercise, where \(\text{Fe}^{2+}\) becomes \(\text{Fe}^{3+}\).
• Reduction involves the gain of electrons. As seen with the permanganate ion transforming to manganese(II), it gains electrons and hydrogen ions.

Redox reactions are characterized by changes in oxidation state. The oxidizing agent (species reduced) gains electrons, and the reducing agent (species oxidized) loses electrons. Understanding these electron systems not only helps in balancing chemical reactions but also explains the underlying principles driving a myriad of chemical transformations.