Problem 53 - Chemistry and Chemical Reactivity

Question

A common analytical method for hydrazine involves its oxidation with iodate ion, \(10_{3}^{-}\), in acid solution. In the process, hydrazine acts as a four- electron reducing agent. $$ \begin{array}{c} \mathrm{N}_{2}(\mathrm{g})+5 \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+4 \mathrm{e}^{-} \longrightarrow \mathrm{N}_{2} \mathrm{H}_{5}^{+}(\mathrm{aq})+5 \mathrm{H}_{2} \mathrm{O}(\ell) \\ E^{\circ}=-0.23 \mathrm{V} \end{array} $$ 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 \(\mathrm{I}_{2}\) Calculate \(E^{\circ}\) for this reaction.

Step-by-Step Solution

Answer

Balanced equation: \( 5\text{N}_2\text{H}_5^+ + 4\text{IO}_3^- + 24\text{H}^+ \rightarrow 5\text{N}_2 + 4\text{I}_2 + 24\text{H}_2\text{O} \), \(E^\circ = 1.43 \, \text{V}\).

Step 1: Determine Oxidation Half-Reaction

First, identify the half-reaction where the iodate ion, \( \text{IO}_3^- \), is reduced. In acidic conditions, \( \text{IO}_3^- \) converts to \( \text{I}_2 \). The balanced half-reaction is: \[ \text{IO}_3^- + 6\text{H}^+ + 5\text{e}^- \rightarrow \text{I}_2 + 3\text{H}_2\text{O} \] The standard reduction potential for this reaction is \( E^\circ = 1.20 \, \text{V} \).

Step 2: Write the Hydrazine Oxidation Half-Reaction

The given oxidation half-reaction for hydrazine \( \text{N}_2\text{H}_5^+ \) is as follows: \[ \text{N}_2\text{H}_5^+ \rightarrow \text{N}_2 + 5\text{H}_3\text{O}^+ + 4\text{e}^- \] with \( E^\circ = -0.23 \, \text{V} \).

Step 3: Balance Electrons Between Half-Reactions

To balance the transfer of electrons between the two half-reactions, multiply the oxidation half-reaction by 5 and the reduction half-reaction by 4. This ensures the electron transfer cancels out when the two half-reactions are combined.

Step 4: Multiply and Add Half-Reactions

After multiplying, combine the reactions. For hydrazine: \[ 5(\text{N}_2\text{H}_5^+ \rightarrow \text{N}_2 + 5\text{H}_3\text{O}^+ + 4\text{e}^-) \] For iodate ion: \[ 4(\text{IO}_3^- + 6\text{H}^+ + 5\text{e}^- \rightarrow \text{I}_2 + 3\text{H}_2\text{O}) \] The combined reaction is: \[ 5\text{N}_2\text{H}_5^+ + 4\text{IO}_3^- + 24\text{H}^+ \rightarrow 5\text{N}_2 + 4\text{I}_2 + 24\text{H}_2\text{O} \]

Step 5: Calculate the Standard Cell Potential \(E^\circ\)

Finally, calculate the standard cell potential using the formula: \( E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \). Here, \( E^\circ_{\text{cathode}} = 1.20 \, \text{V} \) and \( E^\circ_{\text{anode}} = -0.23 \, \text{V} \). Thus, \( E^\circ_{\text{cell}} = 1.20 - (-0.23) = 1.43 \, \text{V} \).

Key Concepts

Hydrazine Oxidation

Hydrazine, which is chemically represented as \( \mathrm{N}_2\mathrm{H}_4 \), is a strong reducing agent. When it reacts in an acidic solution, it undergoes oxidation. During this process, it loses electrons to form \( \mathrm{N}_2 \), nitrogen gas, along with \( \mathrm{H}_3\mathrm{O}^+ \) ions and water as byproducts. In redox reactions, oxidation refers to the loss of electrons and since we are dealing with hydrazine as the reducing agent, it loses a total of four electrons. This change occurs in the presence of an oxidizing agent, such as the iodate ion. The oxidation of hydrazine is not just a simple chemical transformation; it is key in many industrial applications and analytical chemistry methods. An understanding of this oxidation process is crucial for formulating balanced redox reactions, such as those involving hydrazine and iodate ions, which allows scientists and chemists to predict the outcomes in various chemical environments.

Iodate Reduction

The iodate ion, represented as \( \mathrm{IO}_3^- \), is a potent oxidizing agent. In acidic solutions, the iodate ion undergoes reduction to form iodine (\( \mathrm{I}_2 \)) and water as side products. This reduction involves gaining electrons. Specifically, \( \mathrm{IO}_3^- \) gains a total of five electrons during this reaction. This electron transfer is vital in defining the characteristics of a redox reaction involving iodate, where the reduction potential is positive, indicating a favorable energy release when it gains electrons.In a balanced chemical equation, iodate reduction plays a significant part in ensuring that the electrons lost during hydrazine oxidation are equally gained in a stoichiometric manner, leading to the formation of stable products. This process is crucial in many laboratory methodologies where iodine is needed — particularly in titrations and analytical measurements where the presence of iodine is detected as an indication of reaction progress.

Electrochemistry

Electrochemistry is the study of chemical changes that produce electrical currents and the changes caused by the passage of electricity. In the context of the reaction between hydrazine and iodate, electrochemistry is foundation to understanding how ions interact within the solution. This study involves analyzing how the chemical components rearrange, which is crucial when calculating standard cell potentials. The redox reactions, like those involving hydrazine oxidation and iodate reduction, play out as a system that can be examined using electrochemical cells. Electrochemistry allows for the measurement of these potential differences and gives insights into how energy is transferred during the reactions. By focusing on the role of electrons, this domain brings clarity to how reactions can be harnessed in batteries and sensors, where the flow of charge correlates to the work being done by the system.

Standard Cell Potential

The standard cell potential \( E^\circ \) provides a measure of the cell's ability to drive an electrical current through an external circuit. It is defined as the difference in potential between the cathode and anode under standard conditions. During the reaction between hydrazine and iodate, this potential represents the voltage produced by the complete redox reaction.To calculate the \( E^\circ \) for any given reaction, one would use the formula \( E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \). Here, \( E^\circ_{\text{cathode}} \) is the potential where reduction takes place (iodate to iodine with 1.20 V), while \( E^\circ_{\text{anode}} \) is where oxidation occurs (hydrazine to \( N_2 \) with -0.23 V).The resulting value for \( E^\circ_{\text{cell}} \) gives a direct indication of the reaction's ability to proceed spontaneously. For the hydrazine-iodate reaction, a cell potential of 1.43 V suggests a strongly favorable process, highlighting its practical applications in electrochemical cells and analytical chemistry methods.