The Gibbs Free Energy and Cell Voltage - Chemistry LibreTexts
Given the relationship between the criteria for spontaneity (ΔG° < 0). In chemistry, a spontaneous processes is one that occurs without the addition of external energy. More rigorous Gibbs free energy / spontaneity relationship. A positive voltage that forms across the electrodes of a voltaic cell indicates that the oxidation-reduction What is the relationship between these two values?.
To use cell potentials to calculate solution concentrations. Changes in reaction conditions can have a tremendous effect on the course of a redox reaction. The relationship between voltage and concentration is one of the factors that must be understood to predict whether a reaction will be spontaneous. The total amount of energy produced by an electrochemical cell, and thus the amount of energy available to do electrical work, depends on both the cell potential and the total number of electrons that are transferred from the reductant to the oxidant during the course of a reaction.
The resulting electric current is measured in coulombs Can SI unit that measures the number of electrons passing a given point in 1 s. A coulomb relates energy in joules to electrical potential in volts. Michael Faraday — Faraday was a British physicist and chemist who was arguably one of the greatest experimental scientists in history.
The son of a blacksmith, Faraday was self-educated and became an apprentice bookbinder at age 14 before turning to science. His experiments in electricity and magnetism made electricity a routine tool in science and led to both the electric motor and the electric generator. He discovered the phenomenon of electrolysis and laid the foundations of electrochemistry.
Next, we're going to write the oxidation half-reaction for aluminum, so here is aluminum, so we're going to write an oxidation half-reaction, so we need to reverse this reduction half-reaction. So we write solid aluminum, right, is going to aluminum three plus.
To do that, it needs to lose three electrons, so loss of electrons is oxidation. So we need to find the standard oxidation potential for this half-reaction now. And we've done this several times. When you reverse a reduction half-reaction and turn it into an oxidation half-reaction which is what we've done down here, you just change the sign on the standard reduction potential.
So the standard reduction potential is negative 1. So the standard oxidation potential is positive 1. If we wanted to write the overall reaction, we need to balance everything. So we need to get our number of electrons equal.
So we could do that by multiplying our first half-reaction here by three, because that gives us six electrons.
- Voltage and Spontaneity
- 17.2: The Gibbs Free Energy and Cell Voltage
- Spontaneity and redox reactions
So three times two gives us six electrons. The number of electrons needs to be the same, so we need to multiply our second half-reaction by two, because two times three also gives us six electrons.
So let's do that calculation here.
Gibbs free energy and spontaneity
So we have three Pb two plus, so for reduction, we would have three Pb two plus, plus three times two electrons, gives us six electrons, and this would give us three solid Pb. So three Pb solid. So we're multiplying our half-reaction, but remember, we do not multiply our standard reduction potential by three because voltage is an intensive property.
So we leave our standard reduction potential as negative. All right, let's do our next half-reaction, so our oxidation half-reaction. So we would have two aluminum, so two aluminum, and then two, this is going to give us two Al three plus, and six electrons, two times three gives us six electrons.
And once again, we do not multiply our oxidation potential by two because voltage is an intensive property. So our standard oxidation potential remains positive 1.
To get our overall reaction, we just add together our two half reactions, right, so let's add these together, so we're going to add these together to get our overall.
Gibbs free energy and spontaneity (article) | Khan Academy
All right, so the electrons would cancel out, so six electrons, so let's write our reactants, which are right here, so we would have three Pb two plus Plus two Al, so plus two Al, giving us, so for our products, we would have three Pb, so three Pb, plus two Al three plus. So we have two Al three plus. So is this reaction spontaneous? So is our overall reaction spontaneous? We could figure that out by calculating the standard cell potential, which is what our problem asked us to do.
So what is the standard cell potential? We know we just have to add up our standard reduction potential and our standard oxidation potential. So just like we add our half-reactions, we add our potentials together.
So we're going to add our standard reduction potential for that half-reaction, and we're going to add our standard oxidation potential for that half reaction, to get the standard cell potential for the cell. So that's equal to positive 1. And remember, when your cell potential is positive, that means a spontaneous reaction.
So this is a spontaneous reaction, which is what we predicted. We predicted that lead two plus could oxidize aluminum, all right, so let's go back up here to our problem again, so remember, our problem asked us, can lead two plus oxidize aluminum? And we predicted yes, by using the diagonal rule here, by drawing this arrow down here, and then we just calculated the standard cell potential, and we saw that it confirmed our prediction.
All right, next, let's do copper. So we said that lead two plus will not be able to oxidize copper. Let's find the standard cell potential to confirm our prediction. So we're going to keep this half-reaction, right?
So we're going to keep this half-reaction for lead, and then for copper, we need to reverse this half reaction as it's written. So we need to write the reverse of this reaction, and when you're writing the reverse, remember, you need to change the sign, this is positive. So let's go back down here, where we have some more room, and let's figure out the standard cell potential.