Step 1 : Understanding the Question:
This question asks us to balance a classic redox reaction involving permanganate ions and iron(II) ions in an acidic aqueous medium. Once the complete equation is balanced, we need to identify the stoichiometric coefficient of water molecules ($\text{H}_2\text{O}$) on the product side. To achieve this, we can systematically apply the half-reaction method, which ensures that both mass and charge are conserved.
Step 2 : Key Formulas and Approach:
The ion-electron method is the most reliable approach to balance redox reactions in acidic conditions. The systematic steps are as follows:
1. Separate the overall reaction into oxidation and reduction half-reactions.
2. Balance all atoms in each half-reaction except for hydrogen and oxygen.
3. Balance oxygen atoms by adding water molecules ($\text{H}_2\text{O}$) to the oxygen-deficient side.
4. Balance hydrogen atoms by adding hydrogen ions ($\text{H}^+$) to the hydrogen-deficient side.
5. Balance the electrical charges by adding electrons ($e^-$) to the more positive side.
6. Multiply the half-reactions by appropriate integers so that the number of electrons lost equals the number of electrons gained, and then add them together.
Step 3 : Detailed Explanation:
We begin by identifying the oxidation states of the key elements. Manganese in $\text{MnO}_4^-$ has an oxidation state of $+7$ and is reduced to $+2$ in $\text{Mn}^{2+}$. Iron goes from $+2$ in $\text{Fe}^{2+}$ to $+3$ in $\text{Fe}^{3+}$, indicating oxidation.
Next, we write down the oxidation half-reaction: $\text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq)$. Since the iron atoms are already balanced, we balance the charge by adding one electron to the product side, yielding: $\text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq) + e^-$.
Now, we write down the reduction half-reaction skeleton: $\text{MnO}_4^-(aq) \rightarrow \text{Mn}^{2+}(aq)$.
The manganese atoms are already balanced with one on each side. To balance the four oxygen atoms on the reactant side, we add four water molecules to the product side: $\text{MnO}_4^-(aq) \rightarrow \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l)$.
Adding water introduces eight hydrogen atoms on the right. To balance these, we add eight hydrogen ions ($\text{H}^+$) to the reactant side: $\text{MnO}_4^-(aq) + 8\text{H}^+(aq) \rightarrow \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l)$.
To balance the net charge in the reduction half-reaction, we calculate the charges on both sides. The reactant side has a net charge of $(-1) + 8(+1) = +7$, while the product side has a net charge of $+2$. We add five electrons to the reactant side to balance the charge: $\text{MnO}_4^-(aq) + 8\text{H}^+(aq) + 5e^- \rightarrow \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l)$.
To combine the two half-reactions, we equalize the electron transfer. Since the oxidation half-reaction releases one electron and the reduction half-reaction consumes five, we multiply the oxidation half-reaction by five: $5\text{Fe}^{2+}(aq) \rightarrow 5\text{Fe}^{3+}(aq) + 5e^-$.
Finally, we sum the two adjusted half-reactions. The five electrons on each side cancel out, giving the final balanced equation: $\text{MnO}_4^-(aq) + 5\text{Fe}^{2+}(aq) + 8\text{H}^+(aq) \rightarrow \text{Mn}^{2+}(aq) + 5\text{Fe}^{3+}(aq) + 4\text{H}_2\text{O}(l)$. Looking at the water molecules, we see that the coefficient is 4.
Step 4 : Final Answer:
The stoichiometric coefficient of $\text{H}_2\text{O}$ in the final balanced equation is 4, which corresponds to Option (B).