Problem 91 - Chemistry and Chemical Reactivity

Question

A common analytical method for hydrazine involves its oxidation with iodate ion, \(\mathrm{IO}_{3}^{-},\) in acid solution. In the process, hydrazine acts as a four-electron reducing agent. \(\mathrm{N}_{2}(\mathrm{g})+5 \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+4 \mathrm{e}^{-} \rightarrow\) $$\mathrm{N}_{2} \mathrm{H}_{5}^{+}(\mathrm{aq})+5 \mathrm{H}_{2} \mathrm{O}(\ell) \quad E^{\circ}=-0.23 \mathrm{V}$$ Write the balanced equation for the reaction of hydrazine in acid solution \(\left(\mathrm{N}_{2} \mathrm{H}_{5}^{+}\right)\) with \(\mathrm{IO}_{3}^{-}(\mathrm{aq})\) to give \(\mathrm{N}_{2}\) and I. Calculate \(E^{\circ}\) for this reaction.

Step-by-Step Solution

Answer

The balanced equation is \( 3 \mathrm{N}_{2} \mathrm{H}_{5}^{+} + 2 \mathrm{IO}_{3}^{-} + 12 \mathrm{H}^{+} \rightarrow 3 \mathrm{N}_{2} + 2 \mathrm{I}^{-} + 9 \mathrm{H}_{2} \mathrm{O} \) and \( E^{\circ} = 1.43 \text{ V}. \)

Step 1: Write Half-Reaction for Iodate Ion

Start by writing the half-reaction for the reduction of iodate ion, \( \mathrm{IO}_{3}^{-} \). In acid solution, it reduces to iodide ion \( \mathrm{I}^{-} \). The balanced half-reaction is: \[ \mathrm{IO}_{3}^{-} + 6 \mathrm{H}^{+} + 6 \mathrm{e}^{-} \rightarrow \mathrm{I}^{-} + 3 \mathrm{H}_{2} \mathrm{O}. \] The standard electrode potential for this reaction is \( E^{\circ} = +1.20 \text{ V}. \) This potential may be determined from a standard electrode potential table.

Step 2: Write Half-Reaction for Hydrazine

The half-reaction for hydrazine is already provided: \( \mathrm{N}_{2}(\mathrm{g}) + 5 \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq}) + 4 \mathrm{e}^{-} \rightarrow \mathrm{N}_{2} \mathrm{H}_{5}^{+}(\mathrm{aq}) + 5 \mathrm{H}_{2} \mathrm{O}(\ell) \) with \( E^{\circ} = -0.23 \text{ V}. \) This indicates hydrazine acts as a reducing agent, losing electrons.

Step 3: Balance the Number of Electrons

To combine the half-reactions, balance the number of electrons transferred. The iodate reaction involves 6 electrons and the hydrazine reaction involves 4 electrons. The least common multiple of 4 and 6 is 12. Multiply the hydrazine reaction by 3 and the iodate reaction by 2 to balance the electrons exchanged: \[ 3(\mathrm{N}_{2}(\mathrm{g}) + 5 \mathrm{H}_{3} \mathrm{O}^{+} + 4\mathrm{e}^{-} \rightarrow \mathrm{N}_{2} \mathrm{H}_{5}^{+} + 5\mathrm{H}_{2} \mathrm{O}) \] \[ 2(\mathrm{IO}_{3}^{-} + 6 \mathrm{H}^{+} + 6 \mathrm{e}^{-} \rightarrow \mathrm{I}^{-} + 3 \mathrm{H}_{2} \mathrm{O}) \]

Step 4: Combine the Half-Reactions

Add the two balanced half-reactions, ensuring electrons cancel out. This gives the overall balanced chemical equation: \[ 3 \mathrm{N}_{2} + 15 \mathrm{H}_{3} \mathrm{O}^{+} + 2 \mathrm{IO}_{3}^{-} \rightarrow 3 \mathrm{N}_{2} \mathrm{H}_{5}^{+} + 2 \mathrm{I}^{-} + 5 \mathrm{H}_{2} \mathrm{O} \] + \[6 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{H}_2 \mathrm{O}. \] After canceling out terms that appear on both sides, simplify to obtain \[ 3 \mathrm{N}_{2} \mathrm{H}_{5}^{+} + 2 \mathrm{IO}_{3}^{-} + 12 \mathrm{H}^{+} \rightleftharpoons 3 \mathrm{N}_2 + 2 \mathrm{I}^{-} + 9 \mathrm{H}_2 \mathrm{O}. \]

Step 5: Calculate Standard Cell Potential

Calculate the overall cell potential \( E^{\circ} \) by adding the standard potentials of the two half-reactions: \( E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = 1.20 \text{ V} - (-0.23 \text{ V}) = 1.43 \text{ V}. \)

Key Concepts

Half-Reactions

To understand redox reactions, we start by exploring half-reactions. These are simplified equations that show either the oxidation or reduction process separately. Each half-reaction involves the transfer of electrons, essential for balancing redox reactions.

**Why Half-Reactions Matter**
Half-reactions help clarify what's being oxidized and what's reduced. This separation makes it easier to apply the principle of electron balance. For the iodate ion (\( \mathrm{IO}_{3}^{-} \)), its reduction involves electrons turning it into iodide (\( \mathrm{I}^{-} \)).

**Example: Half-Reaction for Iodate Ion**
The half-reaction for reducing iodate ion is:

• \( \mathrm{IO}_{3}^{-} + 6 \mathrm{H}^{+} + 6 \mathrm{e}^{-} \rightarrow \mathrm{I}^{-} + 3 \mathrm{H}_{2} \mathrm{O} \)
  This shows the iodate ion gaining electrons (reduction). The balanced electron count is crucial.

For hydrazine, its half-reaction showcases it as a reducing agent, losing electrons to form \( \mathrm{N}_{2}\mathrm{H}_5^{+} \). By dealing with each part separately, we can better manage the overall reaction.

Electrode Potential

Electrode potential, also referred to as standard electrode potential (\( E^{\circ} \)), is a measure of the tendency of a chemical species to be reduced. It's like a scorecard for how likely a reaction is to gain electrons, measured in volts.

**Understanding Electrode Potential**

• A positive \( E^{\circ} \) value signals a strong drive to gain electrons, suggesting it's a good oxidizing agent.
• A negative \( E^{\circ} \) suggests the species prefers to lose electrons, thus acting as a reducing agent.

In our exercise, the iodate reaction has \( E^{\circ} = +1.20 \text{ V} \), hinting at a strong reduction tendency.

**Calculating Overall Cell Potential**
The overall cell potential (\( E^{\circ}_{\text{cell}} \)) is derived by subtracting the anode potential from the cathode potential:

• \( E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \)

In the exercise, this calculation yields: \( 1.20 \text{ V} - (-0.23 \text{ V}) = 1.43 \text{ V} \). This positive result reinforces the feasibility and spontaneity of the reaction.

Balancing Chemical Equations

Balancing chemical equations is central in redox reactions; it ensures the law of conservation of mass and charge. We need to balance elements and charges, especially when dealing with electrons in the half-reactions.

**Steps to Balance Equations**

• Write the separate half-reactions for oxidation and reduction.
• Balance electrons between the half-reactions. This might include multiplying each reaction to equalize the electron transfer.
• Combine them to create a balanced overall reaction.

In our case, the hydrazine half-reaction was multiplied by 3 and the iodate half-reaction by 2 to balance the electrons (12 each), before combining them.

**Final Balanced Equation**
After combining and simplifying, the overall equation from the exercise was:

• \(3 \mathrm{N}_{2}\mathrm{H}_5^{+} + 2\mathrm{IO}_{3}^{-} + 12\mathrm{H}^{+} \rightarrow 3 \mathrm{N}_2 + 2 \mathrm{I}^{-} + 9 \mathrm{H}_{2} \mathrm{O} \)

This process of balancing ensures a clear, accurate depiction of the chemical reaction, maintaining consistency with the initial and final states.