Main content

## MCAT

### Unit 8: Lesson 17

Electrochemistry- Electrochemistry questions
- Electrochemistry
- Redox reaction from dissolving zinc in copper sulfate
- Introduction to galvanic/voltaic cells
- Electrodes and voltage of Galvanic cell
- Shorthand notation for galvanic/voltaic cells
- Free energy and cell potential
- Standard reduction potentials
- Voltage as an intensive property
- Using reduction potentials
- Spontaneity and redox reactions
- Standard cell potential and the equilibrium constant
- Calculating the equilibrium constant from the standard cell potential edited
- Nernst equation
- Using the Nernst equation
- Concentration cell
- Introduction to electrolysis
- Quantitative electrolysis
- Electrolysis of molten sodium chloride edited
- Lead storage battery
- Nickel-cadmium battery

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Standard cell potential and the equilibrium constant

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. Created by Jay.

## Want to join the conversation?

- how does logK appear after you multiply by the ln10 what is the rule?(7 votes)
- He did a very poor job explaining. He does nt show that he multiplies and divides the right side by ln10.

E0=((.0257ln10)/n)(lnk/ln10)

lnk/ln10= logk by change of base rule. substitute and you get what he got(14 votes)

- 3:48, did he make a mistake?

0.0257 X log 10 = 0.0257

Edit: Yeah, he did, just used calculator and log 10 is 1.

Edit 2: Nope, he didn't, so basically here is what happened. To convert Ln to Log 10, you have to times it by 2.303. Instead of showing the whole conversion, he made a mistake and did 0.0257 X log 10 = 0.0257.

So it should have been as follow:

0.0257 X 2.303 (ln to log 10) = 0.0592

then

E = (0.0592 / n) x Log K(0 votes)- There was no such mistake made in the video. Take a second look at Jay's equation at3:48: it reads 0.0257 x ln 10 = 0.0592, which is different from your interpretation of 0.0257 x log 10 = 0.0257. Note that he indeed used the natural log of 10, whereas you thought he wrote the log (base ten) of 10. Take the natural log of 10 and you will get approximately 2.303. :)(2 votes)

## 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. Let's solve for E zero. Let's solve for the standard cell potential. So to do that, we'll 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 going to solve for, we're going to solve for what this is equal to. So if we're at 25 degrees C, all right, 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 going to 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, right? 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 gives us RT over F, and so this gives 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, right, so we can rewrite our equation. So the one we had up here, so we're going to 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, right, the standard cell potential, E zero, to the equilibrium constant K. We can write this in a different way. So what we could do is we could take that .0257, .0257, and we could multiply that by the natural log of 10. We get .0592. And the reason why we could do this is to write our equation in log form. Up here we have natural log, so up here we 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 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, 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, another form of the same equation, right? Relating the standard cell potential, E zero, to the equilibrium constant K.