Question

In acidic medium one mole each of \( \text{MnO}_4^- \) and \( \text{Cr}_2\text{O}_7^{2-} \) is reduced by x and y moles of ferrous ions. The sum of x and y is

• A. 14
• B. 12
• C. 10
• D. 11

Hint:

Remember standard n-factors in acidic medium: \( \text{KMnO}_4 = 5 \) \( \text{K}_2\text{Cr}_2\text{O}_7 = 6 \)

Answer 1

The Correct Option is D

Step 1: Redox Stoichiometry Principle:

Equivalents of Oxidizing Agent = Equivalents of Reducing Agent. Moles of oxidant \( \times \) n-factor (oxidant) = Moles of reductant \( \times \) n-factor (reductant). The reducing agent is Ferrous ion (\( \text{Fe}^{2+} \)). Reaction: \( \text{Fe}^{2+} \to \text{Fe}^{3+} + e^- \). n-factor for \( \text{Fe}^{2+} = 1 \).
Step 2: Calculating x (for \( \text{MnO}_4^- \)):

In acidic medium: \( \text{MnO}_4^- \xrightarrow{+5e^-} \text{Mn}^{2+} \). Change in oxidation state of Mn: \( +7 \to +2 \). n-factor = 5. Equating equivalents: \( 1 \text{ mole} \times 5 = x \text{ moles} \times 1 \) \( x = 5 \).
Step 3: Calculating y (for \( \text{Cr}_2\text{O}_7^{2-} \)):

In acidic medium: \( \text{Cr}_2\text{O}_7^{2-} \xrightarrow{+6e^-} 2\text{Cr}^{3+} \). Change in oxidation state of Cr: \( +6 \to +3 \). Change per atom = 3. Since there are 2 Cr atoms, total n-factor = \( 2 \times 3 = 6 \). Equating equivalents: \( 1 \text{ mole} \times 6 = y \text{ moles} \times 1 \) \( y = 6 \).
Step 4: Summation:

Sum \( = x + y = 5 + 6 = 11 \).
Step 5: Final Answer:

The sum is 11.