Cell Potential and Equilibrium - Video Tutorials & Practice Problems
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concept
∆Eº, ∆G and K Formulas
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Here we're going to say that Gibbs free energy, which is delta g not, so we're talking about standard Gibbs free energy, is the bridge to the standard cell potential, which here that is E naught subcell, and the equilibrium constant k. Now this connection can be seen in the following way. So on the left side, we have the connection between our standard cell potential and Gibbs free energy. Here, we're talking about standard Gibbs free energy. It is equal to negative n times f times our standard cell potential. Here we'd say that our standard Gibbs free energy is in units of kilojoules, n is just the number of moles transferred within our redox reaction, f is Faraday's constant, which remember is 96,485 coulombs per moles of electrons. And then our standard cell potential here will be in units of volts, abbreviated v. On the right side, we have the connection between our equilibrium constant k and our standard Gibbs free energy. Here we'd say that standard Gibbs free energy, the change in it, is equal to negative rtlnk. So here r is our gas constant which is 8.314 joules over moles times k, t here equals temperature in Kelvin, k here is just our equilibrium constant. So we say that Gibbs free energy equals this equation, we say that Gibbs free energy equals this equation. Since Gibbs free energy equals both equations, they must be equal to one another. So in the middle here, we can say that the connection between our standard cell potential and our equilibrium constant k is negative n times f times standard cell potential equals negative rtlnk. Now here from this middle equation, we can isolate our standard cell potential. So here we want to isolate this variable here. To do that, you divide out negative n times f from both sides, So they cancel out here on the left side, the negative signs cancel out. So at the end we get here that our standard cell potential equals rt over n times f times ln of k. This represents the connection between our standard sub potential, your equilibrium constant, in its simplified form.
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example
Cell Potential and Equilibrium Example
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Here in this example question it says, a certain electrochemical reaction involves the transferring of 4 electrons. If the value of its equilibrium constant is 2.50 times 10 to the 12 at 25 degrees Celsius, calculate the standard cell potential. So we know the equation that connects our standard cell potential and our equilibrium constant is standard cell potential equals our gas constant, R, times our temperature in Kelvin, divided by the moles of electrons transferred, times Faraday's constant, and this will be multiplied by ln of k. So all we do now is we plug in the values that we have. R is 8.314 joules over moles times k. Our temperature here needs to be in Kelvin, so you're going to add 273.15 here to give us 298.15 Kelvin. So there goes our temperature. N is the moles of electrons transferred, which is 4 moles of electrons transferred. Faraday's constant is 96,0485 coulombs for moles of electrons, and then ln of k, which is 2.50 times 10 to the 12th. Here, when we look at the units that are canceling out, we see that what cancels out? Multiple electrons cancel out. Kelvins cancel out. K here has no units in itself. And then at the end what we're gonna see is we're gonna have joules, the per coulomb per moles of this question. So here when we do that, we're gonna get point 18335 joules per coulomb per mole. Joules per coulomb is the same thing as volts. Now here, we'll do it just a 3 sig fix, so that's gonna give me 0.183 volts. And here, it's important to realize that it's positive, 0.183 volts. We know that our cell potential here is positive since our equilibrium constant is greater than 1. Alright. So this will be our final answer.
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Problem
Problem
Calculate the equilibrium constant for the following reaction at 25ºC.