For the reaction: NH3+OCl−→N2H4+Cl−in basic medium, the coefficients of NH3,OCl−and N2H4 for the balanced equation are respectively2,2,22,2,12,1,14,4,2

To balance the redox reaction $\text{NH}_3 + \text{OCl}^- \rightarrow \text{N}_2\text{H}_4 + \text{Cl}^-$ in a basic medium, we will follow a systematic approach.

1. Break Down the Problem

• Identify the oxidation states of the elements involved.
• Determine the half-reactions for oxidation and reduction.
• Balance each half-reaction.

2. Relevant Concepts

In basic conditions:

• Oxidation involves the loss of electrons.
• Reduction involves the gain of electrons.
• Balance all atoms except for O and H first, then balance O using water (H₂O) and H using OH⁻.

3. Analysis and Detail

1. Identify Oxidation States:

   • In $\text{NH}_3$, N is -3.
   • In $\text{N}_2\text{H}_4$, N is -2 (each N atom).
   • In $\text{OCl}^-$, Cl is +1.
   • In $\text{Cl}^-$, Cl is -1.

2. Half-Reactions:

   • For oxidation (NH₃ to N₂H₄): $2 \text{NH}_3 \rightarrow \text{N}_2\text{H}_4 + 2 \text{e}^-$
   • For reduction (OCl⁻ to Cl⁻): $\text{OCl}^- + 2 \text{e}^- \rightarrow \text{Cl}^- + \text{OH}^-$

3. Combine the Half-Reactions:

   • Multiplying the first half-reaction by 2: $2 \text{NH}_3 \rightarrow \text{N}_2\text{H}_4 + 2 \text{e}^-$
   • Combine with the reduction half-reaction to get: $2 \text{NH}_3 + 2 \text{OCl}^- \rightarrow \text{N}_2\text{H}_4 + 2 \text{Cl}^- + 2 \text{OH}^-$

4. Verify and Summarize

• Count atoms:
  • Nitrogen: 2 from $\text{NH}_3$, 2 in $\text{N}_2\text{H}_4$
  • Chlorine: 2 from $\text{OCl}^-$, 2 in $\text{Cl}^-$
  • Oxygen matches balance using OH⁻ added.

Final Answer

The balanced coefficients for the equation are:

• $\text{NH}_3 = 2$
• $\text{OCl}^- = 2$
• $\text{N}_2\text{H}_4 = 1$ Thus the coefficients are $2, 2, 1$.