Cell Potential and Gibbs Free Energy - Video Tutorials & Practice Problems
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1
concept
Cell Potential and ∆G Formula
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Now in terms of electrochemical cells, we're gonna say that Gibbs free energy, which we're gonna say here is delta g naught, 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, negative n, times f times E naught sub 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 donates its electrons to another species, which is reduced. So we're talking about the number of electrons transferred from one species to another. Then we're gonna say your f is known as Faraday's constant. Here it's in units of 96,485 coulombs per 1 mole of electrons. And then here we're going to say that E naught sub cell, remember this represents our standard cell potential, and here it's in units of volts, which we'll represent by capital V. So here, this equation is how we can connect Gibbs free energy, standard Gibbs free energy, and our standard cell potential.
2
example
Cell Potential and Gibbs Free Energy Example
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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 2 moles of chromium solid to produce 2 moles of chromium 3 ion plus 3 moles of cobalt solid. With them, we have their standard reduction potentials also given to us as half reactions. Alright. 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 delta g naught equals negative n times Faraday's constant, times your standard cell potential. 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 1 and 2 in the other. What's their lowest common multiple? 6. To get to 6, we multiply this by 2 and we multiply this by 3. Remember multiplying them by these numbers does nothing to their standard reduction potentials. These would stay 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. Faraday's constant is 96,485 coulombs for 1 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 the site of oxidation. If we look at our overall balanced redox reaction, we see that cobalt 2 goes from plus 2 to neutral. So its oxidation number decreased, therefore it was reduced. So cobalt is the cathode. Chromium goes from neutral, which is 0, to a positive 3 charge. Its oxidation number increased, so it was oxidized, meaning it is the anode. So we do negative 0.28 volts minus a minus 0.74 volts. Remember, minus of a minus here really means you're adding them together so you get a positive 0.46 volts. Here we plug in our 0.46 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, coulombs cancel out. We'll have initially our answer in joules. But remember we typically like to have Gibbs free energy in kilojoules, so 1 kilojoule is equal to 10 to the 3 joules. Here this gives us negative 266 0.3 kilojoules for our Gibbs free energy. So our standard Gibbs free energy, which helps us to determine the maximum electrical work is negative 266.3 kilojoules.
3
concept
Faraday's Constant
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Now Faraday's constant represents the charge, which we use capital c, in columns 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 say the conversion factor between coulombs and joules is that 1 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 96,000 485 coulombs per mole of electrons. And we just learned about this conversion factor. So here we want to get rid of coulombs, so we put on the bottom 1 coulomb, and on the top we're gonna say that's equal to 1 joule per volt. Coulombs cancel out, so it's still gonna be the same number, 96,045 joules over volts times moles of electrons, because the volts are a denominator here, so they're staying a denominator here. Moles of electrons never got canceled out, so they remain a denominator as well. So just remember, by remembering this convergent 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.
4
example
Cell Potential and Gibbs Free Energy Example
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2m
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How many electrons were transferred between an anode and cathode that produced 482.425 kilojoules of energy? Alright, so here they want us to find electrons transferred, or in essence moles of electrons transferred. We've seen this type of term knowledge when it comes to Faraday's constant, which is 96,485 coulombs per 1 mole of electrons. We also know the conversion factor that 1 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 gonna do here is I'm gonna say that 1 kilojoule is equal to 10 to the 3 joules. Since I know joules now, we could cross multiply these 2 together so we can see that 1 joule is equal to 1 coulomb times volts. So that's conversion I can, introduce here. We can say here that we have 1 joule is equal to 1 coulomb times volts, And now because we know this, we can say that Faraday's constant is 96,480 5 coulombs per 1 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 quick kilojoules of energy.
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Problem
Problem
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
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Problem
Problem
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
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