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# Calculating the equilibrium constant from the standard cell potential

Example problem for calculating the equilibrium constant K using the standard cell potential. Created by Jay.

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• Why does he use log (K), haven't we been using natural log (K)? I did it using natural log and got 6.85*10^96.
• Unfortunately, you will see both bases used in this chemistry. Which is used depends on who did the math, but it is more common to use log base 10.

There is no good reason for using the two bases of the logs, but either will work, you just have to use slightly different constants that incorporate the change in base of a log from the natural log.
• Why didn't you use G=-RTlnk to solve for K? why is the value for K different when using this equation? :(
• He actually did use it to arrive to this "new" equation : E°= (0.0592)* log(K)
He explains it in the previous video of this series but I will try to explain :
He uses deltaG°=-RTlnK and deltaG°=-nFE°. With these two equations you have RTlnK = nFE°. You then leave E° on the right and bring the rest on the left : E° = (RT/nF) lnK. He then calculates RT/F, which you can because you know the value of R and F but also T since here its E° which means we are in standard conditions (25°C). This gives him a certain value we'll call b. For certain reasons in chemistry, log is more used than natural log (ln) so he gave an equation using the log. To do that, you need to divide ln(K) by ln(10) so we divide the equation by ln(10) which gives us : E°/ln(10) = (b/n)log(K). You then multiply by ln(10) to have E° instead of E°/ln(10) and you find : E° = (b * ln10/n)log(K). He calculated b*ln10 and found 0.0592V. In the end, your equation is this : E° = (0.0592/n)log(K)

When you use the equation G=-RTlnK, you might find something slightly different. That is because he calculated constants in advance, so maybe you have more precise numbers, but if you round it up, you should get the same thing.

Hope this helps, even two years later :)
• What happened to the units for n?
• n is simply a number (of electrons) and has no units
• Very beginning of video..You can't balance this equation with the bridge method? Why is that? This is the first time I've ever had this problem. I think it has something to do with the fact you picked an equation which has cations and anions both on product side.
• I am confuse with what equilibrium constant is telling us, like if it is large it only goes on a certain way (arrow part?) and if it is small the arrow to the left is larger than arrow to the right... what is equilibrium constant?
• you can think of k (equilibrium constant) as a kind of pool to fill in

1. if it's huge like the one in the video, your reaction should keep going for a long time to make that huge amount of products per 1 set of reactant to fill that pool (as k is proportional to "mols of product/mols of reactants")

2. on the other hand, it k is small or even less than 1 you can fill that pool quite easily and quickly

3. then the spillover (reverse reaction) starts

• I don't understand... Where did you determine that you originally had 2 electrons?
• I2(s) (2 iodine atoms in 1 solid molecule) reduces to form 2 I- ions in solution by gaining 2 e- per molecule. This is set up as a half reaction for the problem. Is this what you are asking about?
• at isn't E*cell = Ecathode - Eanode ? why did he add those two?
(1 vote)
• He changed the sign of the standard reduction potential for the oxidation half-reaction - I suggest watching the rest of Jay's videos on the subject of voltaic cells.
• How would you calculate the reaction quotient for this reaction?
• I thought Ecell was cathode (reduction) - anode (oxidation), is that only for galvanic cells? Why do you add the Ecell of the two half reactions?
(1 vote)
• So Ecell is cathode - anode, but only if you're using reduction potentials since most potential tables give the potentials of the reduction reactions. So if you did this using the reduction potentials given at the start of the problem then Ecell would be: 0.54 - (-1.66), which is the same thing as 0.54 + 1.66 because of the double negative.

Jay just turned the reduction potential of the aluminum (-1.66) into an oxidation potential (+1.66) by changing the sign of the potential. And it's this oxidation potential of aluminum we can add to the reduction potential of iodine to get Ecell. But it's obviously the same math as before using only the reduction potentials.

So you can find Ecell either way, you just have to conscious of the signs. Using the reduction potentials it is: Ecell = cathode - anode. Using the reduction AND oxidation potentials it is: Ecell = cathode + anode.

Hope that helps.