Cell Potential and Gibbs Free Energy - Online Tutor, Practice Problems & Exam Prep

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 ∆G Formula

Now in terms of electrochemical sounds, we're going to say that Gibbs free energy, which we're going to say here is a ΔG° so we're talking about standard Gibbs free energy represents the maximum or Max electrical work that can be created. Now here the connections between spontaneity, Gibbs free energy and standard cell potential are illustrated by the formula.

So here we're going to say that our standard Gibbs free energy equals -n * F * E°_cell. Here standard Gibbs free energy is usually in units of kilojoules. n represents the moles of electrons transferred. So remember within a redox reaction, one species is oxidized, meaning it donated to electrons to another species, which is reduced. So we're talking about the number of electrons transferred from one species to another.

Now we're going to say your F is known as Faraday's constant. Here it's in units of 96,485 coulombs per one mole of electrons. And then here we're going to say that E°_cell. Remember this represents our standard cell potential and here it's in units of volts which will represent by capital V. So here this equation is how we can connect Gibbs free energy, standard Gibbs free energy and our standard self potential.

Cell Potential and Gibbs Free Energy Example

Here it says to calculate the maximum electrical work that can be produced by this cell. So here we have 3 moles of cobalt 2 ion reacting with two moles of chromium solid to produce 2 moles of chromium 3 ion plus three moles of cobalt solid. With them we have their standard reduction potentials also given to us as half reactions.

All right. So remember, when they're asking us to determine the electrical work, they're really asking us to determine Gibbs free energy. So Gibbs free energy? Standard Gibbs free energy is

ΔG ∞ = − N × F × E cell

If we take a look here, N is the number of electrons that are transferred. The number of electrons in our half reactions are not equal. They have to be equal. We have 3 electrons in one and two in the other. What's their lowest common multiple? 6. To get to 6 we multiply this by two and we multiply this by three.

Remember, multiplying them by these numbers does nothing to their standard reduction potentials. These would say these same numbers. What it does is it gives us these coefficients to give us our balanced overall redox reaction. So now we know that there are 6 electrons that are transferred, so we have 6 moles of electrons. Paradise constant is 96,485 coulombs per one mole of electrons and we need our standard cell potential. Here our standard cell potential is equal to cathode minus anode.

Remember, the cathode is the site of reduction and the anode is inside of oxidation. If we look at our overall balance redox reaction, we see that cobalt 2 goes from +2 to neutral, so it's oxidation number decreased, therefore it was reduced. So cobalt is the cathode. Chromium goes from neutral, which is 0 to a +3 charge. Its oxidation number increased, so it was oxidized, meaning it is the anode. So we do

− 0.28   V − − 0.74   V

Remember, minus of a minus here really means you're adding them together, so you get a +.46 volts here. We plug in our .46 volts here, but just remember that a Volt is equal to a Joule per coulomb, so this is joules per coulomb. And in that way if we look what cancels out, moles cancel out, Coulomb's cancel out. We'll have initially our answer in joules, but remember we typically like to have Gibbs free energy and kilojoules. So one KJ is equal to 10 to the three joules here. This gives us

− 266.3   kJ

4 Gibbs free energy. So our standard Gibbs free energy, which helps us to determine the maximum electrical work, is

− 266.3   kJ .

Faraday's Constant

Now Faraday's constant represents the charge which we use capital C in Coulombs of 1 mole of electrons and is named after British scientist Michael Faraday. Now here we're going to say the charge that passes through the cell equals the moles of electrons times Faraday's constant. So if we have an electrochemical cell, we can determine that its charge is equal to moles of electrons N times Faraday's constant F.

Beyond this, we can save. The conversion factor between coulombs and joules is that one coulomb is equal to 1 Joule per Volt. So by incorporating Faraday's constant, it's possible for us to go between coulombs to joules or kilojoules, or vice versa. Now here we're going to say that Faraday's constant, remember, is 96,485 coulombs per mole of electrons.

And we just learned about this conversion factor. So here we want to get rid of coulomb, so we put on the bottom 1 coulomb, and on the top we're going to say that's equal to 1 Joule per Volt. Coulomb's cancel out, so it's still going to be the same 96,485 joules over volts times moles of electrons because the volts are denominated here. So they're staying a denominator here. Moles of electrons never got cancelled out, so they remain a denominator as well.

So just remember, by remembering this conversion factor, we can basically interchange between coulombs to joules or kilojoules, right? So this is all the things that we need to know in terms of Faraday's constant.

Cell Potential and Gibbs Free Energy Example

How many electrons were transferred between an anode and cathode that produced 482.425 kilojoules of energy? All right, so here they want us to find electrons transferred, or in essence, moles of electrons transferred. We've seen this type of terminology when it comes to Faraday's constant, which is 96,485 coulombs per one mole of electrons.

We also know the conversion factor, that one Coulomb is equal to 1 Joule per 1 Volt. So we're going to start out here with 482.425 kilojoules. And what I'm going to do here is I'm going to say that one KJ is equal to 10³ joules. Once I know joules now, we could cross multiply these two together so we can see that one Joule is equal to 1 coulomb times volts.

So that's conversion I can introduce here. We can say here that we have one Joule is equal to 1 coulomb times volts. And now, because we know this, we can say that Faraday's constant is 96,485 Coulombs per one mole of electrons. When I do that, that's going to give me 5 moles of electrons that are produced per Volt.

So here 5 moles of electrons would be involved in this transferring of this many kilojoules of energy.

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

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

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,