Question

Balance the following redox reaction in acidic medium and determine the stoichiometric coefficient of $\text{H}_2\text{O}$ in the final balanced equation.

\[\text{MnO}_4^-(aq) + \text{Fe}^{2+}(aq) \rightarrow \text{Mn}^{2+}(aq) + \text{Fe}^{3+}(aq)\]

• A. 2
• B. 4
• C. 6
• D. 8

Hint:

When an exam problem strictly asks for the coefficient of \(\text{H}_2\text{O}\) in a permanganate (\(\text{MnO}_4^-\)) reduction, you can bypass balancing the entire equation! Since all oxygens in the reaction come solely from \(\text{MnO}_4^-\) and end up entirely inside \(\text{H}_2\text{O}\), the 4 oxygen atoms in \(\text{MnO}_4^-\) will *always* demand exactly 4 \(\text{H}_2\text{O}\) molecules to balance out, regardless of what species is being oxidized.

Answer 1

The Correct Option is B

Concept: To balance a redox reaction in an acidic medium, we use the Ion-Electron Method (also known as the Half-Reaction Method). This involves breaking the overall chemical skeleton into an oxidation half-reaction and a reduction half-reaction, balancing atoms and charges individually for each half, and then recombining them so that net electrons cancel out completely. The procedural rules for balancing an acidic medium half-reaction are:
• Balance all atoms except hydrogen (\(\text{H}\)) and oxygen (\(\text{O}\)).
• Balance oxygen atoms by adding water molecules (\(\text{H}_2\text{O}\)) to the deficient side.
• Balance hydrogen atoms by adding hydrogen ions (\(\text{H}^+\)) to the deficient side.
• Balance the net ionic charge by adding electrons (\(e^-\)).

Step 1: Separating the reaction into half-reactions and balancing the oxidation part.
Let's look at the change in oxidation states to identify the tracks:
• Oxidation half-reaction: Iron changes from a \(+2\) charge to a \(+3\) charge. \[ \text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq) \] Since the iron atoms are already balanced, we only need to balance the charge by adding 1 electron to the product side: \[ \text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq) + e^- \quad \cdots \text{(Equation 1)} \]

Step 2: Balancing the reduction half-reaction ($\text{MnO}_4^- \rightarrow \text{Mn}^{2+}$).

• Reduction half-reaction skeleton: Manganese goes from a \(+7\) state to a \(+2\) state. \[ \text{MnO}_4^-(aq) \rightarrow \text{Mn}^{2+}(aq) \] Now let's apply the acidic balancing guidelines step-by-step:
• Manganese atoms: Already balanced (1 on each side).
• Oxygen atoms: There are 4 oxygen atoms on the reactant side. To balance them, add 4 water molecules (\(\text{H}_2\text{O}\)) to the product side: \[ \text{MnO}_4^-(aq) \rightarrow \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l) \]
• Hydrogen atoms: Adding water introduced 8 hydrogen atoms to the product side. Balance this by adding 8 hydrogen ions (\(\text{H}^+\)) to the reactant side: \[ \text{MnO}_4^-(aq) + 8\text{H}^+(aq) \rightarrow \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l) \]
• Charge balancing: Let's calculate the net charge on both sides: Reactant side charge &= (-1) + 8(+1) = +7
Product side charge &= (+2) + 0 = +2 To bridge the gap from \(+7\) to \(+2\), add 5 electrons (\(e^-\)) to the reactant side: \[ \text{MnO}_4^-(aq) + 8\text{H}^+(aq) + 5e^- \rightarrow \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l) \quad \cdots \text{(Equation 2)} \end{align*} \end{itemize} Step 3: {\color{red}Equalizing electrons and combining both halves.} To ensure that no free electrons remain in our final chemical equation, we multiply Equation 1 by **5** so it matches the 5 electrons consumed in Equation 2: \[ 5\text{Fe}^{2+}(aq) \rightarrow 5\text{Fe}^{3+}(aq) + 5e^- \] Now, add this adjusted oxidation half directly to Equation 2. The 5 electrons on both sides cancel out perfectly: \[ \text{MnO}_4^-(aq) + 5\text{Fe}^{2+}(aq) + 8\text{H}^+(aq) \rightarrow \text{Mn}^{2+}(aq) + 5\text{Fe}^{3+}(aq) + 4\text{H}_2\text{O}(l) \] Looking closely at the coefficient attached to the water molecules on the product side, we find it is exactly 4.