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 sounds, we're gonna say that Gibbs free energy, which we're gonna say here is a delta G not. 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 gives free energy and standard cell potential are illustrated by the formula. So here we're going to say that our standard gives free energy equals N negative N times F times E not sub CLL. Here standard gives 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. Now, we're gonna say your f is known as far 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 knot sub 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 attention.
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 three moles of Cobalt two ion reacting with two moles of chromium solid to produce two moles of chromium, three 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 gives free energy. So gibbs free energy standard gives free energy is delta G knot equals negative N times fay is constant times your standard self potential. If we take a look here and 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 three electrons in one and two in the other. What's their lowest common multiple? Six to get to six? 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 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 six electrons that are transferred. So we have six moles of electrons, Faraday 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 a node. Remember that cathode is the site of reduction and the A node is the site of oxidation. If we look at our overall balance redox reaction, we see that Cobalt two goes from plus two to neutral. So its oxidation number decreased, therefore, it was reduced. So Cobalt is the cathode chromium goes from neutral, which is zero to a positive three charge, its oxidation number increased. So it was oxidized, meaning it is the ao 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 vault is equal to a jewel per cooler. So this is jules perula. And in that way, if we look what cancels out moles, cancel out Poulos cancels out, we'll have initially our answer in jewels. But remember we typically like to have gives free energy in kilojoules. So one kilojoule is equal to 10 to the three joules here. This gives us negative 266.3 kilojoules four gives free energy. So our standard gives free energy which helps us to determine the maximum electrical work is negative 266.3 kg.
3
concept
Faraday's Constant
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Now, Fay's constant represents the charge which we use capital C in columns of one 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 constant. So if we have an electrochemical cell, we can determine that its charge is equal to moles of electrons N times fays constant f Beyond this, we can say that the conversion factor between coulombs and joules is that one Coulomb is equal to one joule per volt. So by incorporating Ferris constant where it's possible for us to go between Coombs to Js or kilo tools or vice versa. Now, here we're going to say that Ferris constant remember is 96,485 Coolum per mole of electrons. And we just learned about this conversion factor. So here we want to get rid of Coolum. So we put on the bottom one Coolum and on the top, we're gonna say that's equal to one joule per volt who lets cancel out. So it's still gonna be the same number. 96,085 Joes over volts times moles of electrons because the volts are 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 conversion factor, we can basically interchange between cools to jewels or killer jewels, right? So this is all the things that we need to know in terms of far is constant.
4
example
Cell Potential and Gibbs Free Energy Example
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How many electrons were transferred between an ade 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 fair days constant, which is 96,485 coulombs per one mole of electrons. We also know the conversion factor that one poem is equal to one joule per one volt. So we're gonna start out here with 482.425 kilojoules. And what I'm gonna do here is I'm gonna say that one kilojoule is equal to 10 to the three jolts. Since I know jewel now we could cross multiply these two together. So we can see that one joule is equal to one poum times vaults. So that's conversion I can um introduce here, we can say here that we have one jewel is equal to one colon times faults. And now because we know this, we can say that F constant is 96,004, 85 cools per one mole of electrons. When I do that, it's gonna give me five moles of electrons that are produced per volt. So here five 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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