Problem 36 - Chemistry The Molecular Science

Question

Hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4},\) can be used as the reducing agent in a fuel cell. $$ \mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{aq}) \longrightarrow \mathrm{N}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\ell) $$ (a) If \(\Delta_{t} G^{\circ}\) for the reaction is \(-598 \mathrm{~kJ},\) calculate the value of \(E^{\circ}\) expected for the reaction. (b) Suppose the equation is written with all coefficients doubled. Determine \(\Delta_{\mathrm{r}} G^{\circ}\) and \(E^{\circ}\) for this new reaction.

Step-by-Step Solution

Answer

(a) \(E^{\circ} = 1.55\, \text{V}\); (b) For doubled coefficients, \(Δ_{\mathrm{r}}G^{\circ} = -1196\, \text{kJ/mol}\), \(E^{\circ} = 1.55\, \text{V}\).

Step 1: Understand the relationship between ΔG° and E°

The relationship between standard Gibbs free energy change (ΔG°) and standard cell potential (E°) for a redox reaction is given by the equation:\[ΔG° = -nFE°\] where \(n\) is the number of moles of electrons transferred in the balanced equation, \(F\) is the Faraday constant \((96485 \ \text{C/mol})\), and \(E°\) is the standard cell potential.

Step 2: Solve for E° using ΔG°

The balanced equation for the reaction of hydrazine with oxygen shows the formation of nitrogen and water. To solve for \(E°\), rearrange the equation:\[E° = -\frac{ΔG°}{nF}\]We are given \(ΔG° = -598 \ \text{kJ/mol} = -598000 \ \text{J/mol}\). For this reaction, the number of electrons \(n\) transferred can be determined by balancing the half-reactions, which results in \(n = 4\).Substitute the known values:\[E° = -\frac{-598000 \ \text{J/mol}}{4 \times 96485 \ \text{C/mol}} \approx 1.55 \ \text{V}\]

Step 3: Double the Reaction Coefficients

Doubling all coefficients in the chemical equation results in:\[2\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{aq})+2\mathrm{O}_{2}(\mathrm{aq})\longrightarrow 2\mathrm{N}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O}(\ell)\] This change entails doubling the moles transferred \((n = 8)\), but doesn't affect \(E°\) as \(E°\) is an intensive property.

Step 4: Calculate ΔG° for the Doubled Equation

Because \(ΔG°\) is an extensive property and the reaction is doubled, \(ΔG°\) is also doubled:\[ΔG°_{doubled} = 2 \times (-598000 \ \text{J/mol}) = -1196000 \ \text{J/mol} = -1196 \ \text{kJ/mol}\]

Step 5: Determine E° for the Doubled Equation

Despite doubling the stoichiometry, \(E°\) remains the same because it is an intensive property. Therefore, even after doubling the coefficients, \(E°\) remains \(1.55 \ \text{V}\), as the driving force per electron remains unchanged.

Key Concepts

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are a type of chemical reaction involving the transfer of electrons between two species. In a redox reaction, one substance loses electrons (oxidation), while another substance gains electrons (reduction). This transfer of electrons is key to the energy conversions that occur in electrochemical cells, also known as batteries or fuel cells.
In the given reaction, hydrazine ( ₂H₄) acts as the reducing agent, meaning it loses electrons. Oxygen ( O₂) is reduced as it gains these electrons. This electron exchange is crucial because it creates a flow of electric current, which can be harnessed to do work, such as powering a device.
The balanced equation for hydrazine's reaction with oxygen shows how electrons are transferred. It is important to balance these reactions to ensure that the number of electrons lost by the reducing agent equals the number of electrons gained by the oxidizing agent. This balance helps determine the number of moles of electrons transferred, denoted as "n," which is crucial for calculating other properties of the reaction.

Gibbs Free Energy

Gibbs Free Energy ( Δ G) is a concept from thermodynamics that provides information about the spontaneity of a reaction. A negative Δ G indicates a spontaneous reaction, meaning the reaction will proceed without external input. In the specific exercise, Δ G^° for the reaction is -598 kJ, pointing to a spontaneous reaction.
The equation connecting Gibbs Free Energy to the electrochemical cell's potential is: Δ G^° = -nFE^°, where:

• Δ G^° is the standard Gibbs Free Energy change.
• n represents the number of moles of electrons transferred.
• F, the Faraday constant, is the charge of one mole of electrons (approximately 96485 C/mol).
• E^° is the standard cell potential.

By rearranging the equation to solve for E^°, students can explore how the energy change of a reaction is related to its electrochemical properties. Doubling the equation results in doubling Δ G as it is an extensive property, while E^° remains unchanged as it is intensive.

Standard Cell Potential

The standard cell potential (E^°) is a measure of the driving force behind an electrochemical reaction. It is expressed in volts and is determined under standard conditions (concentration of 1 M, pressure of 1 atm, and temperature of 298 K). E^° is crucial in predicting the capacity of a redox reaction to produce electrical energy.
In the hydrazine reaction, E^° was calculated as approximately 1.55 V. This value signifies the potential difference between the oxidation and reduction half-reactions. It's an intensive property, meaning it does not depend on the amount of material present; hence, doubling the entire reaction equation does not affect its value.
The calculation of E^° involves rearranging the Gibbs Free Energy equation: E^° = - Δ G^° / (nF). This equation indicates that a higher E^° value reflects a stronger tendency of the reaction to proceed spontaneously, aligning with the concept that reactions with a higher potential can supply more electrical energy under standard conditions.