If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Free energy and cell potential

Visit us (http://www.khanacademy.org/science/healthcare-and-medicine) for health and medicine content or (http://www.khanacademy.org/test-prep/mcat) for MCAT related content. These videos do not provide medical advice and are for informational purposes only. The videos are not intended to be a substitute for professional medical advice, diagnosis or treatment. Always seek the advice of a qualified health provider with any questions you may have regarding a medical condition. Never disregard professional medical advice or delay in seeking it because of something you have read or seen in any Khan Academy video.

Want to join the conversation?

  • blobby green style avatar for user falconbright303
    why is Gibbs free energy negative for spontaneous reactions?
    (4 votes)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user Darmon
      The underlying reason for this is merely that of convention; in thermodynamics, negative values represent energy RELEASED by a system, heat RELEASED by a system, or work DONE by a system. Conversely, positive values represent energy ABSORBED by a system, heat ABSORBED by a system, or work done UNTO a system.

      Taking just one step back by taking a look at the equation for Gibbs free energy, we see that G = H - TS. If H, or enthalpy, is negative, then we know that the reaction RELEASES heat energy, which contributes to spontaneity. Additionally, if the S, or entropy, term is positive, then we know that the reaction is associated with an increase in entropy, which contributes to spontaneity. This is why the entropy term is subtracted from the enthalpy term; a larger (more positive) entropy should make the Gibbs free energy more negative. :)
      (7 votes)
  • blobby green style avatar for user Jenna Moser
    At , I thought that standard temperature was zero degrees Celsius?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • piceratops seedling style avatar for user Jonita Stephen
    IS there a video for Kohlraush's law?
    (1 vote)
    Default Khan Academy avatar avatar for user

Video transcript

- [Voiceover] When you're dealing with a voltaic cell, it's important to relate the potential of the cell to the free energy of the redox reaction, and here's the equation that relates free energy to the cell potential. So, delta G, delta G is the change in free energy and we know, for a spontaneous reaction, delta G is negative, so let me go ahead and write that down here, so delta G is negative for our spontaneous redox reaction that we have in our voltaic cell, and we're going to calculate the value for delta G at the end of this video. E, over here, this represents our cell potential or our cell voltage, and this is easy to measure, just hook up a volt meter, and for this cell, the cell potential is positive 1.10 volts, so very easy to measure the voltage associated with a voltaic cell. Next, the n, the n represents the moles of electrons that are transferred in the redox reaction. In this example, we're talking about two moles of electrons are transferred in our redox reaction, and finally, let's talk about F, which represents Faraday's constant and Faraday's constant is the magnitude of charge that's carried by one mole of electrons, so we can calculate Faraday's constant, let's go ahead and do that up here, so one electron has a charge of 1.6 times 10 to the negative 19 coulombs, so coulombs is the unit for charge, so let's go ahead and write that. 1.6, actually it's 1.602 times 10 to the negative 19 coulombs for every one electron. So one electron has that charge. If we want to find the magnitutde of charge carried by one mole of electrons, we would need to multiply by Avogadro's number, so if we multiply by Avogadro's number, which we know is 6.022 times 10 to the 23rd. That's how many electrons there are in one mole of electrons. So that's how many electrons in one mole of electrons. So if you multiply these out, you're going to end up with charge per one mole of electron. This is 96,472. The units would be, this is coulombs per mole of electrons. So coulombs per mole. And if you do a more careful calculation, you'll get 96,485 coulombs per mole, and so sometimes you'll see textbooks use this number for different calculations. Most of the time, you can just round this to 96,500, so 96,500 coulombs per mole, and that's good enough for almost all the calculations that you will do. So that explains each term of our equation here. All right, so this cell potential that we talked about, 1.10 volts, this is actually the standard cell potential, which is the voltage measured when the cell operates under standard conditions, which is defined as all of your solids are in pure form, and that's the case for us here, because we're dealing with zinc metal, so that's a solid in its pure form, and copper metal, a solid in its pure form. Also defined as the solutions are at one molar concentration so we have a one molar concentration solution of zinc sulfate, which gives us one molar concentration of zinc two plus ions in solution, and a one molar concentration of copper sulfate, which gives us one molar concentration of copper two plus ions in solution, and our temperature is 25 degrees, so this is 25 degrees C, so if those are your conditions, and you hook up a volt meter, this is the voltage that you will get. So we can actually modify the equation that we've talked about, by adding in a superscript here. So we could write E zero instead of just E, and E zero is our standard cell potential. It just means that everything is under the conditions, under standard conditions, and then this would be delta G zero, so that's delta G zero now, the standard change in free energy. So let's go ahead and solve for delta G zero. We already know it's going to negative because this is a spontaneous redox reaction, but let's go ahead and plug in everything, so let's get a little bit more room down here. Let's plug in our numbers, so we're trying to solve for delta G zero, the standard change in free energy, so we have a negative sign here. I'll talk about what that means a little bit later. So, next, we have n. Remember what n represented here, so n is the moles of electrons that are transferred in our redox reaction which would be two, so this would be two moles of electrons and put these in parentheses here. Next, we have Faraday's constant, F, and so we've already seen that we can use 96,500 coulombs per mole for Faraday's constant, so this is 96,500 coulombs per mole. Next, we plug in the value for our standard cell potential, so E zero is 1.10 volts for this voltaic cell, and when we're thinking about units, so a volt is how many joules per coulomb? So, instead of plugging in volts here, we're going to plug in joules over coulomb, so our units will work out properly, so this is positive 1.10 joules per coulomb. All right, let's do this math. So let's look at our units first, and let's see what cancels out here, so we, our moles cancels out, our charge, coulombs, cancels out to give us joules for our answer. So let's go ahead and calculate how many joules, so let's do the calculation, so we have two times 96,500 and then we're going to multiply that by 1.10, so that would give us, this would give us 212,300 joules. So, remember this was in joules, and let's go ahead and convert that to kilojoules. So 212,000 joules would be 212 kilojoules, and remember we have a negative sign here, so don't forget about this negative sign. This would be negative 212 kilojoules, for our value for delta G zero. So, the standard change in free energy for this voltaic cell is negative 212 kilojoules, and we already know, right, that when delta G is negative, that's a spontaneous reaction, so this is a spontaneous redox reaction here. Notice that delta G zero and E zero have opposite signs, right? We had a positive sign for our voltage, and we ended up with a negative sign for delta G, and that's why we have this negative in our equation here, so a spontaneous reaction in a voltaic cell has a positive cell potential, so a positive value for the voltage, but a negative change in free energy because that indicates a spontaneous reaction.