Problem 36 - Chemistry: An Atoms-Focused Approach

Question

Starting with the appropriate standard free encrigics of formation from Appendix 4, calculate the values of \(\Delta G^{*}\) and \(E_{\text {cell of the following reactions: }}\) a. \(\mathrm{FeO}(s)+\mathrm{H}_{2}(g) \rightarrow \mathrm{Fe}(s)+\mathrm{H}_{2} \mathrm{O}(\ell)\) b. \(2 \mathrm{Pb}(s)+\mathrm{O}_{2}(g)+2 \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow\) \(2 \mathrm{PbSO}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(\ell)\)

Step-by-Step Solution

Answer

Based on the given reactions and standard free energies of formation, the calculated values for Gibbs free energy change and cell potential are: Reaction (a): \(\Delta G^{*} = 7\, \text{kJ/mol}\) and \(E_{\text {cell}} = -0.036\, \text{V}\) Reaction (b): \(\Delta G^{*} = -1924 \, \text{kJ/mol}\) and \(E_{\text {cell}} = 5.00\, \text{V}\)

Step 1: Write down the standard free energies of formation

The standard free energies of formation for the species involved in this reaction can be found in Appendix 4. Based on the data provided, we have: \(\Delta G_{\text{f}}^\circ(\mathrm{FeO}) = -244 \, \text{kJ/mol}\) \(\Delta G_{\text{f}}^\circ(\mathrm{H_{2}O}) = -237 \, \text{kJ/mol}\)

Step 2: Calculate \(\Delta G^*\) for the reaction

For the given reaction, we can find the Gibbs free energy change as: \(\Delta G^{*} = \Delta G_{\text{f}}^\circ(\mathrm{Products}) - \Delta G_{\text{f}}^\circ(\mathrm{Reactants})\) \(\Delta G^{*} = (\Delta G_{\text{f}}^\circ(\mathrm{Fe}) + \Delta G_{\text{f}}^\circ(\mathrm{H_{2}O})) - (\Delta G_{\text{f}}^\circ(\mathrm{FeO}) + \Delta G_{\text{f}}^\circ(\mathrm{H_{2}}))\) Since the standard free energy of formation for the elements, \(\mathrm{Fe}\) and \(\mathrm{H_{2}}\), in their standard states are zero: \(\Delta G^{*} = (-237) - (-244) = 7\, \text{kJ/mol}\)

Step 3: Calculate \(E_\text{cell}\) for the reaction

Now, we can determine the cell potential, \(E_\text{cell}\), using the relationship between \(\Delta G^{*}\) and \(E_\text{cell}\): \(\Delta G^{*} = -nFE_\text{cell}\) Where \(n\) is the number of moles of electrons transferred in the reaction (in this case, it is 2 since the oxidation state of Fe changes from +2 to 0) and \(F\) is the Faraday's constant (96485 C/mol). Solving for \(E_\text{cell}\): \(E_\text{cell} = -\frac{\Delta G^{*}}{nF} = -\frac{7\times10^3}{2\times96485} = -0.036 \, \text{V}\). Reaction (b) $2 \mathrm{Pb}(s)+\mathrm{O}_{2}(g)+2 \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow$ \(2 \mathrm{PbSO}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(\ell)\) Similar to reaction (a), follow the steps to calculate \(\Delta G^{*}\) and \(E_{\text {cell }}\) for this reaction using the standard free energies of formation data from Appendix 4:

Step 1: Write down the standard free energies of formation

\(\Delta G_{\text{f}}^\circ(\mathrm{PbSO_4}) = -725 \, \text{kJ/mol}\)

Step 2: Calculate \(\Delta G^*\) for the reaction

\(\Delta G^{*} = (\Delta G_{\text{f}}^\circ(2 \mathrm{PbSO_4}) + \Delta G_{\text{f}}^\circ(2 \mathrm{H_{2}O})) - (\Delta G_{\text{f}}^\circ(2 \mathrm{Pb}) + \Delta G_{\text{f}}^\circ(\mathrm{O_{2}}) + \Delta G_{\text{f}}^\circ(2\mathrm{H_{2}SO_4}))\) Note that the standard free energy of formation for the elements, \(\mathrm{Pb}\) and \(\mathrm{O_{2}}\), in their standard states are zero: \(\Delta G^{*} = (2\times(-725) + 2\times(-237)) - (2\times0 + 0 + 2\times0) = -1924 \, \text{kJ/mol}\)

Step 3: Calculate \(E_\text{cell}\) for the reaction

In this reaction, the number of moles of electrons transferred is 4 (since the oxidation state of Pb changes from 0 to +2 and 2 Pb atoms are involved): \(E_\text{cell} = -\frac{\Delta G^{*}}{nF} = -\frac{-1924\times10^3}{4\times96485} = 5.00 \, \text{V}\). So, for reaction (a) we have \(\Delta G^{*} = 7\, \text{kJ/mol}\) and \(E_{\text {cell}} = -0.036\, \text{V}\), and for reaction (b) we have \(\Delta G^{*} = -1924 \, \text{kJ/mol}\) and \(E_{\text {cell}} = 5.00\, \text{V}\).

Key Concepts

Electrochemistry

Electrochemistry is the study that combines electricity and chemistry. It examines how chemical reactions involve the flow of electricity. This field looks at reactions in which electrons are transferred between molecules, known as redox reactions. These reactions occur in electrochemical cells, which can either produce electrical energy from chemical reactions, like in batteries, or drive chemical reactions using electrical energy, like in electrolysis.

In electrochemistry, we frequently work with two main types of cells: galvanic cells and electrolytic cells. **Galvanic cells** produce electricity. A classic example is a zinc and copper cell, where spontaneous reactions generate an electric current. **Electrolytic cells** require electricity input to induce chemical reactions. An example is the electrolysis of water to produce hydrogen and oxygen gases.

The driving force behind the movement of electrons in these cells is known as the cell potential or electromotive force (EMF). This potential difference between the anode and cathode is vital for determining how much work a cell can do, such as in biological systems or technological applications.

Electrode Potential

Electrode potential, often called reduction potential, is a measure of the tendency of a chemical species to acquire electrons and be reduced. Each element or compound in an electrochemical system can have multiple potentials, depending on its state of oxidation and the conditions it's under.

The **standard electrode potential (E^0e)** is measured under standard conditions: 1 M concentration, 25°C temperature, and 1 atm pressure. These potentials are fundamental in determining the **cell potential**, which is found by combining the standard reduction potentials of the two half-cells involved in a reaction. The equation for cell potential is expressed as: \[ E_{ ext{cell}} = E_{ ext{cathode}} - E_{ ext{anode}} \]

A positive cell potential indicates a spontaneous reaction, which means it can proceed without external energy. Conversely, a negative cell potential suggests that energy input is needed to drive the reaction, as is often the case in electrolytic cells. Understanding electrode potential helps in predicting the direction of electron flow and the feasibility of a redox reaction occurring spontaneously. This concept is crucial in the study of batteries, corrosion, and electroplating.

Thermodynamics

Thermodynamics is the branch of physics that deals with heat and temperature, and their relation to energy and work. In the context of electrochemistry, thermodynamics helps understand how electrical energy relates to changes in chemical potential energy.

One of the key equations in this area relates the Gibbs free energy change (\Delta G^{*}) to the cell potential (E_{ ext{cell}}) in an electrochemical system: \[ \Delta G^{*} = -nFE_{ ext{cell}} \] where e is the number of moles of electrons transferred in the reaction, and Fe is the Faraday constant. This equation is a bridge between energy and electrochemistry, showing how cell potential relates to the spontaneity of a reaction.

• A negative \Delta G^{*} indicates a spontaneous reaction, much like a positive E_{ ext{cell}} does.
• A positive \Delta G^{*} indicates a non-spontaneous process, requiring external energy input.

Potential energy changes involved in chemical processes are crucial for developing batteries, fuel cells, and understanding biological energy exchanges. Thermodynamics thus provides a powerful tool to predict and explain these changes, helping students link theory with practical chemical reactions.