lnQ to logK

[Immi Lee - 1D]

For the Nernst Equation, I have in my notes: E = Eº - (RT/nF) ln Q, but then when plugging in values and T at standard conditions, I have E = Eº - 0.0592/n log Q, and when we are calculating K when Q=K at equilibrium, I have:

0 = Eº - 0.0592/n log K
K = 10 ^ (nEº / 0.0592)

Why did the equation change from ln to log?
When are we supposed to use each equation?

[Yichen Fan 3A]

Like you said in the question, we will use E = Eº - (RT/nF) lnQ in all the cases and E = Eº - 0.0592/n logQ at standard condition because basically in the second equation they calculated out all the constant for you. For problem of ln and log, you can always convert between them using:
log10(x) = ln(x) / ln(10)
ln(x) = log10(x) / log10(e)
I think they use log for the 2nd equation at standard condition because standard condition are very common and log base 10 is easier to use than natural log e. As far as I know, some older calculators cannot do natural log but they can do log.

[rhettfarmer-3H]

We can use the second equation with the log under standard conditions. This is just a faster route to your answer at 1 atm and 298k.

[Britney Tran IJ]

I'm pretty sure that the second one is for standard conditions

[Jarrett Sung 3B]

The log equation can only be used under standard conditions, so the units balance out to 0.0592. Under any other conditions, you have to use E = Enaught - (RT/nF)ln Q, but remember you can also use that equation under standard conditions.