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### Course: Class 12 Chemistry (India) > Unit 2

Lesson 2: Standard cell potentials- Standard reduction potentials
- Using reduction potentials
- Voltage as an intensive property
- Worked example: Identifying the best oxidising agent using Eo(red.) values.
- Worked example: Constructing a galvanic cell using reduction potential values
- Spontaneity and redox reactions
- Free energy and cell potential
- Standard cell potential and the equilibrium constant
- Calculating the equilibrium constant from the standard cell potential

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# Standard cell potential and the equilibrium constant

Finding the relationship between standard cell potential and equilibrium constant K. Created by Jay.

## Want to join the conversation?

- At5:07, can you explain the math behind multiplying by ln10? How does multiplying by ln10 changed lnK into logK? Can you work that out?(11 votes)
- He is using the change of base formula to go from base e to base 10. Using the change of base formula, logK=lnK/ln10, which gives you lnK=logK*ln10. By replacing "lnK" by "logK*ln10", he is able to get the formula in terms of logK.(21 votes)

- What exactly is the definition of a standard cell potential?(4 votes)
- The standard cell potential is the electromotive force (EMF) that drives electrons from the anode to the cathode, where the concentrations at each electrode are 1 M, the temperature is 298 K, and the pressure is 1 atm.(8 votes)

- In this video, at0:03two formulas for Gibbs free energy are used. None of these are shown to have any relationship to the definition deltaG = H-TS which was introduced in an earlier video in the Chemistry playlist. Is it obvious how they are related or have I simply missed some key videos?(5 votes)
- The derivation of those formulas is pretty heavy chemistry, requiring a knowledge of calculus and beyond the scope of the introductory chemistry courses this video is designed to help with. If you want to do some research yourself, you can always google it, but, to save you some time, here is a link to the Wikipedia article on Gibbs Free Energy. https://en.m.wikipedia.org/wiki/Gibbs_free_energy#Derivation(2 votes)

- Could you please explain how or from where did you get the value of R 8.314J/mol.K?(2 votes)
- The value of R (8.314 J/mol.K) is a universal gas constant. It is a bit like the number pi (3.141) that simply exists for practical uses(4 votes)

- why did you choose 10 for to plug in K in natural log?(2 votes)
- He's converting a natural log to a log with base 10. You do this by multiplying by 2.303, which happens to be ln(10).(2 votes)

- I don't understand how InK turned into ln10 at4:12?(1 vote)
- ln(k) does not turn into ln(10). By manipulating the constant, 0.0257 V by multiplying it by ln(10), it allows us to use a different formula using log instead of ln.

I hope this helps!(2 votes)

- "n is the number of moles transferred." Moles of electrons?(1 vote)
- Is there any way to get the standard electrode potential from electronegativity?(1 vote)
- in a previous video, he mentioned that the potential is an intensive property which doesn't depend on the moles of electrons transferred but in this equation, the potential does depend on the moles transferred so where am I wrong?(1 vote)
- At4:15, would explain why you substitute the k in log with 10 ?(1 vote)
- So, when I first saw this I had to write it down to check what he did, he should have explained it. He wanted to write the equation using log(x) instead of ln(x)

To do this, you use the change of base property: loga(b)=logc(b)/logc(a)

or, in this case, log(K)=ln(K)/ln(10) If you multiply both sides by ln(10) you get ln(K)=log(K)*ln(10), which is what he substituded.

Because ln(10) = 2.302... you multiply it by .0257 to get .0592(1 vote)

## Video transcript

- [Voiceover] We've already
seen the equation on the left, which relates the standard
change of free energy, delta G zero, to the standard
cell potential, E zero. The equation on the right
is from thermodynamics and it relates the standard
change in free energy, delta G zero, to the
equilibrium constant, K. So, we can set these equal to each other to relate the standard cell potential to the equilibrium constant. Since both of these are
equal to delta G zero, we can say that this is equal to this. So, now we have negative NFE zero is equal to negative RT natural log of the
equilibrium constant, K. Now let's solve for E zero. Let's solve for the
standard cell potential. So, to do that, we need
to divide both sides by negative NF. So, we get E zero is equal to positive RT over NF natural log of our equilibrium constant. Next, we're gonna solve
for what this is equal to. So, if we're at 25 degrees C, alright, so our temperature
under standard conditions for our cells that we've
been talking about, T is in Kelvin, so we need to convert degrees Celsius into Kelvin, and to do that, you need to add 273.15. So, that gives us 298.15 Kelvin. So, that's what this T is, it's our absolute temperature in Kelvin. R is the gas constant. So, R is equal to 8.314 joules over mole times Kelvin, here. So, we're gonna multiply that
by our absolute temperature. And so, our absolute
temperature was 298.15, so, this is 298.15 Kelvin. This is all over Faraday's constant. Alright, so remember, F
is Faraday's constant. So, this F right here,
Faraday's constant, which is 96,500 coulombs per mole, so, the charge of one mole of electrons. So, this give us RT over F. And so, let's get out the calculator and find what this is equal to, here. So, we have 8.314 times 298.1, lemme go back here, .15, and then we're going to divide that by Faraday's constant, 96,500. And so, we get .0257. So, let's round that. .0257, so this give us .0257. What will the units be? Well, Kelvin would cancel out, here, and, let's see, what else cancels out, the moles would cancel out, and that gives us joules over coulombs. Which, of course, is equal to volts. Alright, so we can rewrite our equation, so, the one we had up here. So, we're gonna plug
in for RT over F, now. So, now we would have the
standard cell potential, E zero, is equal to, well this would be .0257 and that was volts over N. Remember, N is the number of moles that are transferred
in your redox reaction. And this is times the natural log of K, our equilibrium constant, here. So, this is one form of the equation that relates the standard cell potential, alright, the standard
cell potential E zero, to the equilibrium constant, K. We can write this in a different way. Alright, so what we could do is we could take that .0257, .0257, and we can multiply that
by the natural log of ten. So, let's do that. So, we have .0257 times the natural log of ten. And that gives us .0592. So, we get .0592. And the reason why we could do this is to write our equation in log form. Alright, up here, we have natural log, so up here, we have a natural log, but now, we can write it in log form. So, now we have the
standard cell potential, E zero, is equal to, well now, we'd have .0592 volts. So, we have .0592 volts, once again, over N, the number of moles of electrons transferred
in our redox reaction. And this time, it would
be times the log of K. So, not the natural log, the log of K. So, we've taken care of
that in our calculation. So, this is just another
form of the same equation. Alright, relating the
standard cell potential, E zero, to the equilibrium constant, K.