Cell Potential: ∆G and K - Online Tutor, Practice Problems & Exam Prep

Understanding the spontaneity of chemical reactions is pivotal in chemistry. Spontaneous reactions are characterized by a standard cell potential (E_cell°) greater than zero, a negative Gibbs free energy change (ΔG°), and an equilibrium constant (K) greater than one. These variables are interconnected through fundamental equations. The Nernst equation relates E_cell° to K, indicating:

$E_{\mathrm{cell}°}=\left(\frac{RT}{NF}\right)ln\left(K\right)$

where R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, N is the number of moles of electrons transferred, and F is Faraday's constant (96,485 C/mol e-). The relationship between ΔG° and E_cell° is given by:

${\mathrm{\Delta G}}_{°}=-NF{E}_{\mathrm{cell}°}$

Lastly, the link between ΔG° and K is expressed by:

$K=e^{\frac{-{\mathrm{\Delta G}}_{°}}{RT}}$

These equations allow us to quantitatively assess reaction spontaneity and predict the direction of chemical processes.

Relationship between ∆Eº, ∆G and K

Now recall that when we talk about a spontaneous reaction, it will possess a standard cell potential that is greater than 0, a change in our standard Gibbs free energy that's less than 0, and an equilibrium constant k that's greater than 1. Now because we can talk about the conditions to make a reaction spontaneous with these variables, that means we can connect them to one another through equations. Here we're going to say the relationship between these three variables in spontaneity can be observed under the following diagram. So here we have our triangle diagram which connects our 3 variables together. So let's start on the left side, and we're looking at the equation that connects together our equilibrium constant k and our standard cell potential. So here we’d say that our standard cell potential equals rt/nfln(k). Here we're gonna say that Ecell0 equals our standard cell potential. We would say, ΔG (change in our standard Gibbs free energy) is Gibbs free energy. Our equilibrium constant is k, n here is just the moles of electrons transferred. We’d say f here is Faraday's constant in 96,485 coulombs per mole of electrons, so that's 485. And r here is our gas constant, which is 8.314 joules over moles times k. Now, let's move clockwise around this triangle, so now we're going over this one, and we're seeing the connection between our standard cell potential and Gibbs free energy. So here we'd say that change in standard Gibbs free energy equals negative n times f times our standard cell potential. So it's equal to the negative n, which is multiple electrons transferred, times Faraday's constant, times our standard cell potential. Let's keep going clockwise. Finally, we look at the last equation that connects our equilibrium constant k and Gibbs free energy. Here we would say that our equilibrium constant k equals e-ΔG₀/rt. So we realize that these three variables can help us determine if a reaction is spontaneous, because we can relate them to each other in terms of this qualitative analysis, we can look at their equations and how they connect, in terms of quantity to one another. So keep in mind the different formulas that connect together these three types of variables.

Cell Potential: G and K Example

Here it says calculate the standard cell potential for the following reaction if 10 moles of electrons are transferred. So here we have P₄ solid reacting with five moles of oxygen gas to produce one mole of tetraphosphorus decaoxide. So here we're going to say they're giving us Gibbs free energies of formations for each one of these compounds. Remember, if an element is found in its natural state, its Gibbs free energy of formation is equal to 0.

We're going to say here that change in standard Gibbs free energy equals negative N times Faraday's constant times our standard cell potential. We're told that 10 moles of electrons are transferred, so N is 10 moles of electrons. F is Faraday's constant, which is 96,485 Coulombs per mole of electrons. We're looking for our standard cell potential, so this is our unknown variable. Before we can find it, we also need to know the change in our standard Gibbs free energy.

How exactly do we find that? Well, since they're giving us Gibbs free energy of formation values, we can say that the change in the standard Gibbs free energy of my reaction equals products minus reactants. So we'd say that is 1 mole of P₄O₁₀, with each one being -2984 kilojoules per mole minus my reactants, both 0, so we don't even need to include them, so moles cancel out. So I'll have -2984 kilojoules.

What we need to remember here is that when it comes to our standard cell potential, it's in units of volts. But a Volt is equal to joules per coulomb. So that means that I need to convert these kilojoules into joules. One KJ is equal to 10³ joules, so that comes out to be -2,984,000 joules. What we do now, moles of electrons cancel out. Divide out -10 * 96,485 coulombs for both sides, so 96,485 coulombs.

So this all cancels on the right side. And at the end what are we going to have? We're going to have joules per coulomb which is equal to volts, so our standard cell potential here will come out to be 3.09 volts. So this would be our final answer.

Given the following standard reduction potentials, determine Ksp for Hg₂Cl₂₍s) at 25 °C.

Hg₂²⁺ (aq) + 2 e^– → 2 Hg (l)                           E°red = + 0.789 V

Hg₂Cl₂ (s) + 2 e^– → 2 Hg (l) + 2 Cl ^– (aq)                E°red = + 0.271 V

• A. 4.93 x 10¹¹
• B. 2.18 x 10¹⁶
• C. 3.27 x 10¹⁷
• D. 6.74 x 10⁴

What is the value of the cell potential for the 4 electron transfer reaction below if the equilibrium mixture contains 0.255 M of CH₄, 1.10 M CO₂, 0.388 M CO and 0.250 M H₂ at 25ºC?

                                     CH₄ (g) + CO₂ (g)  ⇌. 2 CO (g) + 2 H₂ (g)

• A. +1.2720 V
• B. -0.2810 V
• C. +0.3021 V
• D. -0.0218 V

Given the reaction: 2 Cl2 (g) + 2 H2O (g) ⇌ 4 HCl (g) + O2 (g) Kp = 7.5x10-2, calculate the Gibbs Free Energy change for the reaction below at 30ºC.

8 HCl (g) + 2 O2 (g) ⇌ 4 Cl2 (g) + 4 H2O (g)

• A. -1.3 x 104 J/mol
• B. -2.9 x 103 J/mol
• C. +4.3 x 105 J/mol
• D. +7.7 x 107 J/mol

Here’s what students ask on this topic:

What is the relationship between standard cell potential (E°cell) and the equilibrium constant (K)?

The relationship between the standard cell potential (E°cell) and the equilibrium constant (K) is given by the Nernst equation:

$E_{\mathrm{cell}°}=\frac{RT}{NF}ln\left(K\right)$

Here, R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, N is the number of moles of electrons transferred, and F is Faraday's constant (96,485 C/mol e-). This equation shows that a higher equilibrium constant (K) corresponds to a higher standard cell potential (E°cell), indicating a more spontaneous reaction.

How do you calculate the change in Gibbs free energy (ΔG°) from the standard cell potential (E°cell)?

The change in Gibbs free energy (ΔG°) can be calculated from the standard cell potential (E°cell) using the following equation:

${\mathrm{\Delta G}}_{°}=-NF{E}_{\mathrm{cell}°}$

In this equation, N is the number of moles of electrons transferred, and F is Faraday's constant (96,485 C/mol e-). This relationship indicates that a positive standard cell potential (E°cell) results in a negative ΔG°, signifying a spontaneous reaction.

What is the significance of a negative ΔG° in a chemical reaction?

A negative ΔG° (change in Gibbs free energy) in a chemical reaction signifies that the reaction is spontaneous under standard conditions. This means that the reaction can proceed without any external input of energy. A negative ΔG° is associated with a positive standard cell potential (E°cell) and an equilibrium constant (K) greater than one, indicating that the products are favored over the reactants at equilibrium.

How are the equilibrium constant (K) and Gibbs free energy (ΔG°) related?

The equilibrium constant (K) and Gibbs free energy (ΔG°) are related by the following equation:

$K=e^{\frac{-{\mathrm{\Delta G}}_{°}}{RT}}$

Here, R is the gas constant (8.314 J/mol·K) and T is the temperature in Kelvin. This equation shows that a negative ΔG° results in a large equilibrium constant (K), indicating that the reaction favors the formation of products at equilibrium.

What are the conditions for a reaction to be spontaneous?

For a reaction to be spontaneous, it must meet the following conditions:

• The standard cell potential (E°cell) must be greater than zero.
• The change in Gibbs free energy (ΔG°) must be less than zero.
• The equilibrium constant (K) must be greater than one.

These conditions are interconnected through fundamental equations, allowing us to predict the spontaneity of a reaction based on these variables.