Cell Potential and Equilibrium - Online Tutor, Practice Problems & Exam Prep

Gibbs free energy (ΔG∘) serves as a pivotal link between standard cell potential (E∘cell) and the equilibrium constant (K). The relationship is encapsulated by two fundamental equations:

ΔG°=-nFE°_cell

and

ΔG°=-RTlnK

Here, n represents the number of moles of electrons transferred in the redox reaction, F is Faraday's constant (96,485 C/mol e^-), R is the gas constant (8.314 J/mol K), and T is the temperature in Kelvin. By equating the two expressions for ΔG°, we derive the equation that directly relates the standard cell potential to the equilibrium constant:

E°_cell=RT/nF lnK

This equation highlights the interdependence of electrochemical cell potential and chemical equilibrium, crucial for understanding redox reactions and their spontaneity.

∆Eº, ∆G and K Formulas

Here we're going to say that Gibbs free energy, which is ΔG0 so. We're talking about standard Gibbs free energy is the bridge to the standard cell potential, which here that is Ecell and the equilibrium constant K. Now this connection can be seen in the following way.

So on the left side we have the connection between our standard cell potential and Gibbs free energy. Here we're talking about standard Gibbs free energy. It is equal to -N⋅F⋅Ecell. Here we'd say that our standard gives free energy is in units of kilojoules, and it's just the number of moles transferred within our redox reaction. F is Faraday's constant which remember is 96,485 coulombs per moles of electrons and then our standard cell potential here will be in units of volts abbreviated V.

On the right side we have the connection between our equilibrium constant K and our standard gift free energy. Here we say that standard Gibbs free energy the change in it is equal to -RT⋅lnK. So here R is our gas constant which is 8.314 joules over moles times T here equals temperature in Kelvin. K here is just R equilibrium constant. So we say that Gibbs free energy equals this equation.

We say that gives free energy equals this equation. Since Gibbs free energy equals both equations, they must be equal to one another. So in the middle here we can say that the connection between our standard cell potential and our equilibrium constant K is -N⋅F⋅Ecell=-RT⋅lnK. Now here from this middle equation we can isolate our standard cell potential.

So here we want to isolate this variable here. To do that, you divide out -N⋅F from both sides so they can slot here on the left side. The negative signs cancel out. So at the end we'd get here that our standard cell potential equals RTNF⋅lnK. This represents the connection between our standard sub potential, your equilibrium constant in its simplified form.

Cell Potential and Equilibrium Example

Here in this example question, it says a certain electrochemical reaction involves a transferring of four electrons if the value of its equilibrium constant is 2.50 * 10¹² and 25¬∞C. Calculate the standard cell potential. So we know the equation that connects our standard cell potential and our equilibrium constant is standard subpotential equals our gas constant R times our temperature in Kelvin divided by the moles of the moles of electrons transferred times Faraday's constant and this will be multiplied by lane of K.

So what we do now is we plug in the values that we have. R is 8.314 joules over moles times K. Our temperature here needs to be in Kelvin, so you're going to add 273.15 here to give us 298.5 Kelvin. So there goes our temperature and is the moles of electrons transferred which is 4 moles of electrons transferred. Faraday's constant is 96,045 coulombs per moles of electrons and then lane of K which is 2.5 * 10¹² here.

When we look at the units that are canceling out we see that what cancels out multiple electrons cancel out Kelvin's cancel out. K here has no units in itself. And then at the end what we're going to see is we're going to have Joules, the per coulomb, per moles of this question. So here when we do that, we're going to get point 18335 joules per coulomb, per mole. Joules per coulomb is the same thing as volts.

Now here we'll do it just a three six fix. So that's going to give me .183 volts. And here it's important to realize that it's positive 0.183 volts. We know that our cell potential here is positive since our equilibrium constant is greater than one. All right. So this will be our final answer.

Calculate the equilibrium constant for the following reaction at 25ºC. 

Fe (s) + I₂ (s)  →  Fe²⁺ (aq) + 2 I ^– (aq)

Given the following reduction potentials: 

Fe²⁺(aq) + 2 e^–  →. Fe (s)                          E°_red = – 0.45 V

I₂ (s) + 2 e^–  →. 2 I ^– (aq)                           E°_red = + 0.54 V

• A. 1.15 x 1011
• B. 2.96 x 1033
• C. 3.91 x 105
• D. 8.17 x 10-3

Here’s what students ask on this topic:

When considering the relationship among standard free energy change, equilibrium constants, and standard cell potential, what is the equation for ${\mathrm{\Delta G}}_{°}?$

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The equation that relates the standard free energy change (${\mathrm{\Delta G}}_{°}$), the equilibrium constant (K), and the standard cell potential ($E_{°}$) is given by the following thermodynamic equation:

${\mathrm{\Delta G}}_{°}=\; -RT\; ln(K)$

Here, ${\mathrm{\Delta G}}_{°}$ is the standard free energy change, R is the universal gas constant (8.314 J/(mol*K)), T is the temperature in Kelvin, and ln(K) is the natural logarithm of the equilibrium constant, K.

Additionally, there's a direct relationship between ${\mathrm{\Delta G}}_{°}$ and the standard cell potential ($E_{°}$) for an electrochemical cell, which is described by the equation:

${\mathrm{\Delta G}}_{°}=\; -nFE{°}_{}$

In this equation, n is the number of moles of electrons transferred in the reaction, F is the Faraday constant (approximately 96485 C/mol), and

What would the equilibrium potential be for K<sup>+</sup> if there were equal concentration inside and outside the cell?

If the concentration of potassium ions (K<sup>+</sup>) is equal on both sides of the cell membrane, the equilibrium potential for K<sup>+</sup> would be 0 millivolts (mV). This is because the equilibrium potential, also known as the Nernst potential, is determined by the ratio of the external and internal concentrations of the ion. When these concentrations are equal, the logarithmic term in the Nernst equation becomes zero, because log(1) is 0.

The Nernst equation for calculating the equilibrium potential (E<sub>ion</sub>) is given by:

$E_{\mathrm{ion}}=\frac{RT}{zF}\cdot ln\left(\frac{[\mathrm{ion}outside]}{[\mathrm{ion}inside]}\right)$

Where:

• R is the universal gas constant
• T is the temperature in Kelvin
• z is the charge of the ion
• F is Faraday's constant
• [ion outside] and [ion inside] are the external and internal ion concentrations, respectively

Since [ion outside]/[ion inside] would be 1 (because the concentrations are equal), ln(1) is 0, and thus E<sub>ion</sub> = 0