Question

$ClO_3^-$ in basic medium react with $N_2H_4$ and produce NO and $Cl^-$ as products. Which among the following statements regarding the reaction is correct?

• A. 4 mol electrons are provided by each mole $N_2H_4$ reacting
• B. 3 mol $N_2H_4$ is required to react with 1 mol$ClO_3^-$
• C. Electrons transferred per mol of $N_2H_4$ reduce 1.33 mol$ClO_3^-$
• D. 5 mol electrons are gained per mol of $ClO_3^-$ reacting

Answer 1

Answer: (C)

Electrons transferred per mol of $N_2H_4$ reduce 1.33 mol$ClO_3^-$

Answer 2

The reaction involves the reduction of $ClO_3^-$ to $Cl^-$ and the oxidation of $N_2H_4$ to $NO$. The reaction occurs in a basic medium.

1. Determine oxidation states:

• In $ClO_3^-$: Let the oxidation state of Cl be $x$. $x + 3(-2) = -1 \Rightarrow x = +5$.
• In $Cl^-$: The oxidation state of Cl is $-1$.
• In $N_2H_4$: Let the oxidation state of N be $y$. $2y + 4(+1) = 0 \Rightarrow 2y = -4 \Rightarrow y = -2$.
• In $NO$: Let the oxidation state of N be $z$. $z + (-2) = 0 \Rightarrow z = +2$.

2. Write and balance the half-reactions in basic medium:

Oxidation half-reaction ($N_2H_4 \rightarrow NO$):

• Balance N atoms: $N_2H_4 \rightarrow 2NO$

• Balance O atoms by adding $H_2O$: There are 2 O atoms on the right, so add $2H_2O$ to the left.

  $N_2H_4 + 2H_2O \rightarrow 2NO$

• Balance H atoms by adding $OH^-$ to the side deficient in H, and $H_2O$ to the other side (or simply add $OH^-$ to balance charge and $H_2O$ to balance H and O).

  There are $4H$ from $N_2H_4$ and $4H$ from $2H_2O$ (total $8H$) on the left. No H on the right. Add $8OH^-$ to the right.

  $N_2H_4 + 2H_2O \rightarrow 2NO + 8OH^-$

• Balance charge by adding electrons: The charge on the left is 0. The charge on the right is $-8$. Add $8e^-$ to the right.

  $N_2H_4 + 2H_2O \rightarrow 2NO + 8OH^- + 8e^-$

  Conclusion for Statement 1: Each mole of $N_2H_4$ provides 8 moles of electrons. So, statement 1 ("4 mol electrons are provided by each mole $N_2H_4$ reacting") is incorrect.

Reduction half-reaction ($ClO_3^- \rightarrow Cl^-$):

• Balance Cl atoms: Already balanced.

• Balance O atoms by adding $H_2O$: There are 3 O atoms on the left, so add $3H_2O$ to the right.

  $ClO_3^- \rightarrow Cl^- + 3H_2O$

• Balance H atoms by adding $OH^-$: There are $6H$ from $3H_2O$ on the right. No H on the left. Add $6OH^-$ to the left.

  $ClO_3^- + 6OH^- \rightarrow Cl^- + 3H_2O$

• Balance charge by adding electrons: The charge on the left is $-1 + (-6) = -7$. The charge on the right is $-1$. Add $6e^-$ to the left.

  $ClO_3^- + 6OH^- + 6e^- \rightarrow Cl^- + 3H_2O$

  Conclusion for Statement 4: Each mole of $ClO_3^-$ gains 6 moles of electrons. So, statement 4 ("5 mol electrons are gained per mol of $ClO_3^-$ reacting") is incorrect.

3. Combine the half-reactions to get the overall balanced equation:

To balance the electrons transferred, find the least common multiple of 8 (from oxidation) and 6 (from reduction), which is 24.

• Multiply the oxidation half-reaction by 3:

  $3(N_2H_4 + 2H_2O \rightarrow 2NO + 8OH^- + 8e^-)$

  $3N_2H_4 + 6H_2O \rightarrow 6NO + 24OH^- + 24e^-$

• Multiply the reduction half-reaction by 4:

  $4(ClO_3^- + 6OH^- + 6e^- \rightarrow Cl^- + 3H_2O)$

  $4ClO_3^- + 24OH^- + 24e^- \rightarrow 4Cl^- + 12H_2O$

• Add the two multiplied half-reactions and cancel common terms ($24OH^-$, $24e^-$, $H_2O$):

  $3N_2H_4 + 6H_2O + 4ClO_3^- + 24OH^- + 24e^- \rightarrow 6NO + 24OH^- + 24e^- + 4Cl^- + 12H_2O$

  $3N_2H_4 + 4ClO_3^- \rightarrow 6NO + 4Cl^- + 6H_2O$

4. Evaluate the remaining statements:

• Statement 2: 3 mol $N_2H_4$ is required to react with 1 mol $ClO_3^-$

  From the balanced equation, 3 moles of $N_2H_4$ react with 4 moles of $ClO_3^-$.

  Therefore, 1 mole of $ClO_3^-$ would require $3/4 = 0.75$ moles of $N_2H_4$.

  So, statement 2 is incorrect.

• Statement 3: Electrons transferred per mol of $N_2H_4$ reduce 1.33 mol $ClO_3^-$

  1 mole of $N_2H_4$ provides 8 moles of electrons (from the oxidation half-reaction).

  1 mole of $ClO_3^-$ gains 6 moles of electrons (from the reduction half-reaction).

  The number of moles of $ClO_3^-$ that can be reduced by 8 moles of electrons is:

  Moles of $ClO_3^- = \frac{\text{Moles of electrons provided}}{\text{Moles of electrons gained per mole of } ClO_3^-} = \frac{8}{6} = \frac{4}{3} \approx 1.33$ moles.

  So, statement 3 is correct.

The final answer is $\boxed{\text{Electrons transferred per mol of }N_2H_4\text{ reduce 1.33 mol}ClO_3^-}$.

Explanation of the solution:

1. Determine Oxidation States: Identify the change in oxidation states for N in $N_2H_4$ to $NO$ and Cl in $ClO_3^-$ to $Cl^-$.
   • $N_2H_4$: N is -2. In $NO$: N is +2. Change per N atom = +4. For $N_2H_4$ (2 N atoms), total electrons lost = $2 \times 4 = 8$.
   • $ClO_3^-$: Cl is +5. In $Cl^-$: Cl is -1. Total electrons gained = $5 - (-1) = 6$.
2. Analyze Electron Transfer:
   • 1 mole of $N_2H_4$ provides 8 moles of electrons.
   • 1 mole of $ClO_3^-$ gains 6 moles of electrons.
3. Evaluate Statements:
   • Statement 1: Incorrect, 8 mol electrons are provided by $N_2H_4$.
   • Statement 4: Incorrect, 6 mol electrons are gained by $ClO_3^-$.
   • Statement 2: To find the molar ratio, balance electrons (LCM of 8 and 6 is 24). $3N_2H_4$ (24e-) reacts with $4ClO_3^-$ (24e-). So, 3 mol $N_2H_4$ react with 4 mol $ClO_3^-$, not 1 mol. Incorrect.
   • Statement 3: 8 mol electrons (from 1 mol $N_2H_4$) can reduce $8/6 = 4/3 \approx 1.33$ mol $ClO_3^-$. Correct.

Answer: Electrons transferred per mol of $N_2H_4$ reduce 1.33 mol $ClO_3^-$.