Question

In acidic medium $\text{MnO}_4^-$ oxidises $\text{NO}_2^-$ to $\text{NO}_3^-$. How many moles of $\text{NO}_2^-$ are oxidised by 10 moles of $\text{MnO}_4^-$?

• A. 20
• B. 10
• C. 25
• D. 15

Hint:

In redox reactions, balance the half-reactions by ensuring the number of electrons lost equals the number gained. Use the mole ratio from the balanced equation to find the number of moles of reactants or products.

Answer 1

The Correct Option is C

Step 1: Write the Half-Reactions for the Redox Process
In acidic medium, \( \text{MnO}_4^- \) is reduced to \( \text{Mn}^{2+} \), and \( \text{NO}_2^- \) is oxidized to \( \text{NO}_3^- \). Let’s write the balanced half-reactions.
Reduction half-reaction (\( \text{MnO}_4^- \)):
\[ \text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \] Each \( \text{MnO}_4^- \) gains 5 electrons.
Oxidation half-reaction (\( \text{NO}_2^- \)):
\[ \text{NO}_2^- + \text{H}_2\text{O} \rightarrow \text{NO}_3^- + 2\text{H}^+ + 2\text{e}^- \] Each \( \text{NO}_2^- \) loses 2 electrons. Step 2: Balance the Number of Electrons Transferred
To balance the electrons, find the least common multiple (LCM) of the electrons involved: 5 (from the reduction) and 2 (from the oxidation). The LCM of 5 and 2 is 10.
Multiply the reduction half-reaction by 2 (to involve 10 electrons):
\[ 2\text{MnO}_4^- + 16\text{H}^+ + 10\text{e}^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O} \] Multiply the oxidation half-reaction by 5 (to involve 10 electrons):
\[ 5\text{NO}_2^- + 5\text{H}_2\text{O} \rightarrow 5\text{NO}_3^- + 10\text{H}^+ + 10\text{e}^- \] Step 3: Combine the Half-Reactions
Add the two half-reactions, canceling the electrons:
\[ 2\text{MnO}_4^- + 16\text{H}^+ + 5\text{NO}_2^- + 5\text{H}_2\text{O} \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O} + 5\text{NO}_3^- + 10\text{H}^+ \] Simplify by canceling \( 10\text{H}^+ \) on both sides and \( 5\text{H}_2\text{O} \):
\[ 2\text{MnO}_4^- + 6\text{H}^+ + 5\text{NO}_2^- \rightarrow 2\text{Mn}^{2+} + 3\text{H}_2\text{O} + 5\text{NO}_3^- \] The balanced equation shows that 2 moles of \( \text{MnO}_4^- \) oxidize 5 moles of \( \text{NO}_2^- \). Step 4: Calculate Moles of \( \text{NO}_2^- \) Oxidized by 10 Moles of \( \text{MnO}_4^- \)
From the balanced equation, the mole ratio of \( \text{MnO}_4^- \) to \( \text{NO}_2^- \) is 2:5.
\[ \text{Moles of } \text{NO}_2^- = 10 \text{ moles of } \text{MnO}_4^- \times \frac{5 \text{ moles of } \text{NO}_2^-}{2 \text{ moles of } \text{MnO}_4^-} \] \[ = 10 \times \frac{5}{2} = 10 \times 2.5 = 25 \, \text{moles} \] So, 10 moles of \( \text{MnO}_4^- \) oxidize 25 moles of \( \text{NO}_2^- \). Step 5: Analyze Options
Option (1): 20. Incorrect, as the calculated moles of \( \text{NO}_2^- \) is 25, not 20.
Option (2): 10. Incorrect, as the calculated moles of \( \text{NO}_2^- \) is 25, not 10.
Option (3): 25. Correct, as it matches our calculated number of moles.
Option (4): 15. Incorrect, as the calculated moles of \( \text{NO}_2^- \) is 25, not 15.