Question

Volume of M/8 KMnO$_4$ solution required to react completely with 25.0 cm$^3$ of M/4 FeSO$_4$ in acidic medium is:

• A. 8.0 mL
• B. 5.0 mL
• C. 15.0 mL
• D. 10.0 mL

Hint:

In redox reactions involving KMnO$_4$ and FeSO$_4$, use the molarity equation to find the required volume of the reactants.

Answer 1

The Correct Option is D

Step 1: {Balance the Redox Reaction} 
The balanced ionic equation for the reaction is: \[ {MnO}_4^- + 5{Fe}^{2+} + 8{H}^+ \rightarrow {Mn}^{2+} + 5{Fe}^{3+} + 4{H}_2{O} \] Step 2: {Stoichiometric Relationship} 
From the equation, it is clear that 1 mole of KMnO$_4$ reacts with 5 moles of FeSO$_4$. 
Step 3: {Apply Molarity Equation} 
Use the molarity equation to balance the reaction: \[ \frac{M_1V_1}{n_1} \, ({KMnO}_4) = \frac{M_2V_2}{n_2} \, ({FeSO}_4) \] \[ \frac{1 \times V_1}{8 \times 1} = \frac{25 \times 5}{4 \times 5} \] Step 4: {Solve for \( V_1 \)} 
\[ V_1 = \frac{1 \times 25 \times 8}{4 \times 5} \] \[ V_1 = 10 { cm}^3 { or } V_1 = 10.0 { mL} \].

Answer 2

Step 1: Write the balanced redox reaction
In acidic medium, the redox reaction between potassium permanganate (KMnO₄) and iron(II) sulfate (FeSO₄) is:
\( 2MnO_4^- + 10Fe^{2+} + 16H^+ \rightarrow 2Mn^{2+} + 10Fe^{3+} + 8H_2O \)

Step 2: Understand the mole ratio
From the balanced equation:
2 moles of \( MnO_4^- \) react with 10 moles of \( Fe^{2+} \)
⇒ 1 mole of \( MnO_4^- \) reacts with 5 moles of \( Fe^{2+} \)

Step 3: Calculate moles of FeSO₄
Given volume of FeSO₄ = 25.0 cm³ = 25.0 mL = 0.025 L
Given concentration = M/4 = 0.25 M
Moles of FeSO₄ = Molarity × Volume = 0.25 × 0.025 = 0.00625 mol

Step 4: Use mole ratio to find moles of KMnO₄
From mole ratio (1 KMnO₄ : 5 FeSO₄):
Moles of KMnO₄ required = 0.00625 / 5 = 0.00125 mol

Step 5: Calculate volume of KMnO₄ solution required
Given concentration of KMnO₄ = M/8 = 0.125 M
Volume = Moles / Molarity = 0.00125 / 0.125 = 0.01 L = 10.0 mL

Final Answer: 10.0 mL