Cell Potential and Gibbs Free Energy: Videos & Practice Problems

Understanding the relationship between Gibbs free energy (ΔG°), standard cell potential (E°cell), and the spontaneity of electrochemical reactions is crucial in chemistry. The equation:

$\mathrm{\Delta G}°=-nFE°\mathrm{cell}$

connects these concepts, where ΔG° is measured in kilojoules, 'n' is the moles of electrons transferred during the redox process, 'F' is Faraday's constant (96,485 coulombs per mole of electrons), and E°cell is the voltage. Faraday's constant, named after Michael Faraday, allows the conversion between coulombs and joules, using the factor that 1 coulomb equals 1 joule per volt. This conversion is essential for understanding the electrical work that can be produced in an electrochemical cell and for interchanging between different energy units in calculations.

Cell Potential and Gibbs Free Energy

In the context of electrochemical cells, the concept of Gibbs free energy, denoted as ΔG°, signifies the maximum electrical work that can be performed by the system under standard conditions. The relationship between spontaneity, Gibbs free energy, and standard cell potential is encapsulated in the equation:

$$\Delta G° = -n \cdot F \cdot E°_{cell}$$

In this equation, ΔG° is expressed in kilojoules, while the variables represent the following:

• n: This is the number of moles of electrons transferred during the redox reaction. In such reactions, one species undergoes oxidation (loses electrons) and another undergoes reduction (gains electrons).
• F: Known as Faraday's constant, it is valued at 96,485 coulombs per mole of electrons. This constant is crucial for converting the amount of charge to the number of moles of electrons.
• E°_cell: This represents the standard cell potential, measured in volts (V). It indicates the driving force behind the electrochemical reaction.

This equation effectively links the thermodynamic properties of the reaction (Gibbs free energy) with the electrochemical characteristics (cell potential), illustrating how the spontaneity of a reaction can be predicted based on these parameters. A negative ΔG° indicates a spontaneous reaction, while a positive ΔG° suggests non-spontaneity, thereby highlighting the importance of understanding these relationships in electrochemistry.

Cell Potential and Gibbs Free Energy Example

To calculate the maximum electrical work produced by a galvanic cell, we need to determine the Gibbs free energy change (ΔG^°). The reaction involves 3 moles of cobalt(II) ions reacting with 2 moles of chromium solid to produce 2 moles of chromium(III) ions and 3 moles of cobalt solid. The standard reduction potentials for the half-reactions are provided, which are essential for calculating the cell potential.

The formula for standard Gibbs free energy is given by:

$$\Delta G^{\circ} = -n \cdot F \cdot E^{\circ}_{cell}$$

where:

• n = number of moles of electrons transferred
• F = Faraday's constant (96,485 coulombs per mole of electrons)
• E^°_cell = standard cell potential

In this case, the half-reactions involve 3 electrons for cobalt and 2 electrons for chromium. To balance the number of electrons, we find the lowest common multiple, which is 6. Thus, we multiply the cobalt half-reaction by 2 and the chromium half-reaction by 3, resulting in a total of 6 moles of electrons transferred.

The standard cell potential (E^°_cell) is calculated as:

$$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$$

Identifying the cathode and anode, cobalt(II) is reduced (cathode) and chromium is oxidized (anode). The potentials are given as -0.28 V for cobalt and -0.74 V for chromium. Thus:

$$E^{\circ}_{cell} = -0.28\,V - (-0.74\,V) = 0.46\,V$$

Substituting the values into the Gibbs free energy equation:

$$\Delta G^{\circ} = -6 \cdot 96,485\,C/mol \cdot 0.46\,V$$

Since 1 volt is equivalent to 1 joule per coulomb, the units of coulombs will cancel out, leaving us with joules. To convert to kilojoules, we divide by 1,000:

$$\Delta G^{\circ} = -266.3\,kJ$$

This negative value indicates that the reaction is spontaneous and represents the maximum electrical work that can be obtained from the cell, which is -266.3 kJ.

Faraday's Constant in Electrochemistry

Faraday's constant, denoted as F, represents the charge in coulombs (C) associated with one mole of electrons, and it is approximately 96,485 C/mol. This constant is crucial in electrochemistry, particularly when analyzing electrochemical cells. The relationship between the charge (Q) that passes through an electrochemical cell and the moles of electrons (n) can be expressed with the formula:

Q = n × F

In this equation, Q is the total charge in coulombs, n is the number of moles of electrons, and F is Faraday's constant. This relationship allows for the calculation of charge based on the number of moles of electrons involved in the reaction.

Additionally, there is a conversion factor that relates coulombs to joules (J). Specifically, 1 C = 1 J/V, where V represents volts. This means that by incorporating Faraday's constant, one can convert between coulombs and joules or kilojoules. For instance, when converting Faraday's constant into joules, the expression becomes:

F = 96,485 J/V·mol

In this context, when performing conversions, if you want to eliminate coulombs, you can set up the conversion such that:

96,485 C × (1 J/V) = 96,485 J/V·mol

Here, the coulombs cancel out, leaving joules per volt multiplied by moles of electrons in the denominator. This demonstrates the interchangeability of units, allowing for seamless conversions between coulombs, joules, and kilojoules in electrochemical calculations. Understanding these relationships is essential for effectively working with electrochemical systems.

Faraday's Constant in Electrochemistry Example

To determine the number of electrons transferred between an anode and cathode that produced 482.425 kilojoules of energy, we can utilize Faraday's constant and the relationship between energy, charge, and voltage.First, convert kilojoules to joules:\[482.425 \text{ kJ} = 482.425 \times 10^3 \text{ J} = 482425 \text{ J}\]Next, we know that 1 joule is equivalent to 1 coulomb times volts. Therefore, we can express the energy in terms of coulombs:\[\text{Energy (J)} = \text{Charge (C)} \times \text{Voltage (V)}\]To find the total charge in coulombs, we need to know the voltage. However, since we are looking for moles of electrons, we can directly relate the total charge to moles of electrons using Faraday's constant, which states that 1 mole of electrons corresponds to 96,485 coulombs.Now, we can calculate the total moles of electrons transferred:\[\text{Moles of electrons} = \frac{\text{Total charge (C)}}{\text{Faraday's constant (C/mol)}}\]Substituting the values we have:\[\text{Moles of electrons} = \frac{482425 \text{ J}}{96,485 \text{ C/mol}} \approx 5 \text{ moles of electrons}\]Thus, approximately 5 moles of electrons were transferred during the process that produced 482.425 kilojoules of energy. This calculation highlights the relationship between energy, charge, and the fundamental unit of electric charge, allowing us to understand the electron transfer in electrochemical reactions.

What is the gibbs free energy change for the given reaction at 25ºC? 
 Au³⁺ (aq) + 3 Li (s)  →. Au (s) + 3 Li⁺ (aq)
Given the following reduction potentials: 
Au³⁺(aq) + 3 e^–  →. Au (s)                         E°_red = + 1.50 Volts
Li⁺ (aq) + e^–  →. Li (s)                                E°_red = – 3.04 Volts

• A. +1314 kJ
• B. -131.4 kJ
• C. -1314 kJ
• D. +109.5 kJ

Answer: (C)

The reduction of chlorate is given by the equation: 
ClO₃^– (aq) + 6 H₃O⁺ (aq) → Cl ^– (aq) + 9 H₂O (l)
If the standard cell potential is given as 1.373 V, how many electrons are transferred under standard conditions?

• A. 7
• B. 5
• C. 2
• D. 3

Answer: (A)

Here’s what students ask on this topic:

What is the relationship between Gibbs free energy and cell potential?

Gibbs free energy (ΔG) and cell potential (E_cell) are related through a fundamental equation in electrochemistry. The relationship is quantified by the equation:

ΔG = -nFE_cell

Here, ΔG represents the change in Gibbs free energy for a process, n is the number of moles of electrons transferred in the electrochemical reaction, F is the Faraday constant (approximately 96,485 coulombs per mole), and E_cell is the electromotive force (EMF) or cell potential of the electrochemical cell.

This equation tells us that when a reaction is spontaneous, the Gibbs free energy change is negative (ΔG < 0), and the cell potential is positive (E_cell > 0). Conversely, if the reaction is non-spontaneous, the Gibbs free energy change is positive (ΔG > 0), and the cell potential is negative (E_cell < 0). The cell potential is a measure of the driving force behind an electrochemical reaction, while the Gibbs free energy change provides information about the overall energy change and spontaneity of the reaction.

Based on the information in the table above, which of the following shows the cell potential and the Gibbs free energy change for the overall reaction that occurs in a standard galvanic cell?

To determine the cell potential (E^°_cell) and the Gibbs free energy change (ΔG^°) for the overall reaction in a standard galvanic cell, you would typically use the following relationships:

1. The cell potential (E^°_cell) is the difference between the reduction potentials of the cathode and the anode. For a standard galvanic cell, this is given by:
   $E^{°}{\mathrm{cell}}_{}=E^{°}{\mathrm{cathode}}_{}-E^{°}{\mathrm{anode}}_{}$
2. The Gibbs free energy change (ΔG^°) can be calculated from the cell potential using the formula:
   $\Delta G^{°}=-nF{E}^{°}{\mathrm{cell}}_{}$

Here, 'n' is the number of moles of electrons transferred in the reaction,