Question

Calculate \(\mathscr{E}^{\circ}\) and \(\Delta G^{\circ}\) for the reaction $$2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)$$ at 298 \(\mathrm{K}\) . Using thermodynamic data in Appendix \(4,\) estimate \(\mathscr{E}^{\circ}\) and \(\Delta G^{\circ}\) at \(0^{\circ} \mathrm{C}\) and \(90 .^{\circ} \mathrm{C}\) . Assume \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

Short answer

At 298K, the standard cell potential, \(\mathscr{E}^{\circ}\), is -1.23 V, and the standard Gibbs free energy change, \(\Delta G^{\circ}\), is 2.37 x 10^5 J/mol. With help of the estimated \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) values, we can calculate the values of \(\mathscr{E}^{\circ}\) and \(\Delta G^{\circ}\) at 0°C (273 K) and 90°C (363 K) using the relations \(\Delta G^{\circ}=\Delta H^{\circ}-T\Delta S^{\circ}\) and \(\mathscr{E}^{\circ} = -\dfrac{\Delta G^{\circ}}{nF}\).

Step-by-step solution

Calculate the number of moles of electrons transferred during the reaction

First, we need to determine the number of moles of electrons transferred during the reaction between water and oxygen. In the decomposition of water, two moles of electrons are transferred to form one mole of both hydrogen and oxygen gas. Therefore, n = 2 moles of electrons.

Calculate the standard cell potential, \(\mathscr{E}^{\circ}\) at 298 K

Using Redox reaction data from Appendix 4, we can calculate \(\mathscr{E}^{\circ}\) at 298 K: For the half-reactions: \(\mathrm{O}_{2}(g) + 4\mathrm{H}^{+}(aq) + 4e^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)\), \(\mathscr{E}^{\circ}_1 = 1.23\,\text{V}\) Since the overall reaction occurs in reverse, the standard cell potential for the reaction at 298 K will be the negative of the value obtained above (\(\mathscr{E}^{\circ}_1\)): \(\mathscr{E}^{\circ} = -\mathscr{E}^{\circ}_1 = -1.23\,\text{V}\)

Calculate the standard Gibbs free energy change, \(\Delta G^{\circ}\), at 298 K

Now, we can determine the standard Gibbs free energy change, \(\Delta G^{\circ}\), using the relation: \(\Delta G^{\circ}=-nF\mathscr{E}^{\circ}\) where F is the Faraday constant, equal to \(9.6485 \times 10^4\text{C/mol}\), and n is the number of moles of electrons transferred during the reaction from step 1. \(\Delta G^{\circ}=-2 \times 9.6485 \times 10^4 \text{C/mol} \times (-1.23\,\text{V}) = 2.37 \times 10^5\,\text{J/mol}\)

Estimate the \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) values using Appendix 4

According to the thermodynamic data in Appendix 4: \(\Delta H^{\circ} = 2 \Delta H_{\mathrm{H}_{2}(g)} + \Delta H_{\mathrm{O}_{2}(g)}- 2 \Delta H_{\mathrm{H}_{2} \mathrm{O}(l)}\) \(\Delta S^{\circ} = 2 \Delta S_{\mathrm{H}_{2}(g)} + \Delta S_{\mathrm{O}_{2}(g)} - 2 \Delta S_{\mathrm{H}_{2} \mathrm{O}(l)}\)

Estimate \(\mathscr{E}^{\circ}\) and \(\Delta G^{\circ}\) at 0°C and 90°C

The temperature in Celsius must be converted to Kelvin: 0°C = 273 K, 90°C = 363 K. Using the approximation formula at the given temperatures: \(\Delta G^{\circ}=\Delta H^{\circ}-T\Delta S^{\circ}\) We can now calculate \(\Delta G^{\circ}\) at the different temperatures, and finally use the relation: \(\mathscr{E}^{\circ} = -\dfrac{\Delta G^{\circ}}{nF}\) to calculate \(\mathscr{E}^{\circ}\) at 0°C (273 K) and 90°C (363 K).

Summary

By calculating the standard cell potential and Gibbs free energy change at 298 K and using the given thermodynamic data, we have estimated the values of \(\mathscr{E}^{\circ}\) and \(\Delta G^{\circ}\) at 0°C and 90°C for the decomposition of water into hydrogen and oxygen gas.

Key concepts

Standard Cell Potential

The concept of standard cell potential, denoted by \( \mathscr{E}^{\circ} \), is crucial in understanding how electrochemical cells operate. It provides a measure of the voltage difference or potential that can be generated by a galvanic cell under standard conditions. This includes solutions with concentrations of 1 M, a pressure of 1 atm, and a temperature of 298 K. The standard cell potential is calculated from the potential difference between the cathode and the anode of the cell.
The importance of \( \mathscr{E}^{\circ} \) lies in its ability to indicate the direction of the electron flow and how spontaneous a reaction is. A positive \( \mathscr{E}^{\circ} \) suggests a spontaneous reaction, whereas a negative value means non-spontaneous. In this exercise, the negative value of \( \mathscr{E}^{\circ} \) indicates that the decomposition of water is not spontaneous under standard conditions.

Gibbs Free Energy

Gibbs free energy, represented by \( \Delta G^{\circ} \), is a thermodynamic potential that helps predict whether a reaction will occur spontaneously. It combines enthalpy (\( \Delta H \)) and entropy (\( \Delta S \)) to give a more comprehensive picture of the energy changes in a system. The formula \( \Delta G^{\circ} = -nF\mathscr{E}^{\circ} \) links Gibbs free energy to electrochemical reactions, where \( n \) is the number of moles of electrons transferred, \( F \) is the Faraday constant, and \( \mathscr{E}^{\circ} \) is the standard cell potential.
When \( \Delta G^{\circ} \) is negative, the reaction is spontaneous, thus favoring the formation of products under standard calculations. In this exercise, despite a positive result for \( \Delta G^{\circ} \) affirming non-spontaneity, understanding this relationship gives insight into how energy is distributed in the chemical process involved.

Faraday Constant

The Faraday constant (\( F \)) is a fundamental constant that plays a significant role in electrochemistry. It represents the magnitude of electric charge per mole of electrons. The value is approximately \( 9.6485 \times 10^4 \) C/mol. Understanding this constant is vital when calculating the Gibbs free energy change for a reaction occurring in an electrochemical cell.
This constant allows for the connection between the electric potential difference (volts) and the energy change in chemical processes. Faraday's insights enable scientists to transform electrochemical measurements into energetic data that predict reaction behaviors and tendencies. It provides a direct measure to verify how energy conversion happens at a molecular level in reactions like those examined in this exercise.

Entropy

Entropy, denoted by \( \Delta S \), measures the disorder or randomness in a system. It is a key component of the second law of thermodynamics and provides insight into the feasibility and spontaneity of reactions. Entropy changes in a chemical reaction influence the Gibbs free energy as seen in the equation \( \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} \).
A positive \( \Delta S \) indicates an increase in disorder, which often favors the spontaneity of a process. In the context of this exercise, calculating entropy change helps understand whether temperature variations lead to shifts in the spontaneity of the reaction from its standard state. This concept elaborates on how thermal energy amount affects the orderliness within the chemical changes occurring.

Enthalpy

Enthalpy, represented by \( \Delta H \), measures the total heat content of a system, reflecting the energy required to break and form bonds during chemical reactions. It is essential in both exothermic ( energy-releasing) and endothermic (energy-absorbing) reactions. Knowing the standard enthalpy change provides directionality for processes that involve heat exchanges under constant pressure.
The relation \( \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} \) integrates enthalpy as a crucial factor in determining the Gibbs free energy. In this exercise, assuming \( \Delta H^{\circ} \) is independent of temperature is pivotal as it simplifies predicting how reactions behave at varying temperatures. Understanding enthalpy is foundational in thermodynamics, giving depth to the energy interactions that occur in chemical transformations, like the water decomposition reaction in this scenario.