Question from Section 20.6 - Nernst eqn

[Zellmer, Robert]

I got the following question from a fellow student today.  It deals with the Nernst eqn.  I got another question
a short while ago about this eqn.

"I was working on the mastering for 20.6 and I have a question. For some background info, I used the lnQ form of the Nernst equation for a question under nonstandard state conditions and got the question right. The next question was under standard state and I still used lnQ. I got it wrong because I had the wrong amount of e- transferred, but the hint said to use the log form which I understand is used for standard conditions. However, upon correcting my e- error I tried both lnQ and logQ and there was a 0.00001 difference between the two answers. So based on the sig figs for the problem they would come out to be the same. So that leads to my question... does it really matter whether you used lnQ or logQ?"


The Nernst eqn is used to relate cell potential under non-standard conditions, E, to
cell potential under standard conditions, E^o, the reaction quotient, Q, the temp
and moles of electrons transferred during the reaction.  It comes from the eqn
in Ch 19 that relates delta(G) to delta(G^o), temp and Q.  It can be written in
a couple of ways,

E = E^o  - (RT/nF)*ln(Q)

R = 8.314 J/mol*K
T = temp in kelvin
n = number of moles of electrons transferred
F = Faraday constant (96,485 Coulombs/mol e-, to 5 sig. fig.)

or

E = E^o  - 2.303*(RT/nF)*log(Q)  (base 10 log - not much reason to use this)

The 2.303 comes from the relationship between natural log and base 10 log.
The ln(10) = 2.302585093.
The log(10) = 1

To convert from ln(10) to base 10 log you need to multiply log(10) by
ln(10) which we generally round to 2.303.  So, that number is even a
rounded number.

At our normal REFERENCE temp of exactly 25 C (298.15 K) (this is NOT
part of the standard state definition), using base 10 log,

E = E^o  - (0.0592 V/n)*log(Q)

The 0.0592 Volts is to 3 sig. fig. enough in most cases for what we do.
The "n" is the same as above.

This last eqn is often how we use the Nernst eqn since we do most of our
work at the reference temp of 25 C.  We've discussed this very often in
class.  The tables at the back of the book for the thermodynamic quantities
we deal with, such as equil, constants, delta(H^o of formation), S^o,
delta(G^o of formation) and E^o values, are at 25 C (298.15 K).

If you see a question in the book, Mastering, quiz or exam that requires a
temp and it's not given assume 25 C.

Using the different equations should give the same results, with perhaps
slight differences in the last sig. fig. due to what you use for the constants,
(i.e. how many sig. fig.).  I showed how this can affect things in class.
I used an E^o value and then delta(G^o) = - nFE^o and delta(G^o) = - RT*ln(K)
to calculate K.  I showed what happens to the final answer when carrying
several extra digits, one extra digit or the correct sig. fig. from one step to
the next.  I showed that even using 298 K compared to 298.15 K can make
a difference (although this difference is much less than rounding in each
step rather than carrying extra digits).

It really shouldn't matter which eqn you use as long as you use the
constants given on the info sheet and carry extra digits from one step to
the next in a multi-step calculation.

Most people rarely use the 2nd eqn above anymore since we have
calculators that can do natural logs for us.  Back before calc. there
were easier ways to do base 10 logs w/o a calculator than natural
logs.  It's sort of still stuck around.  The last eqn is still extensively
used because we do most of our work at 298.15 K.

If you want to use the Nernst eqn at a different temp you can't use the
one with the 0.0592 V.  You need to use the first or second eqn given
above.  This has nothing to do with non-standard state conditions.

Standard state conditions are not related to a particular temp.  It
means what the standard physical state something is in at a
particular temp of interest.  For things in solution the standard conc.
is 1 M.  For gases, the std. st. is a partial pressure of 1 atm (technically
behaving ideally).  For solids or liquids, it's the pure substance.

That means we can do things that aren't under std. st. conditions at
25 C (298.15 K).  We can use any of the eqns above for this.

If you also want to do this at a temp other than 25 C you would need
to use the first or second eqn.

I hope this helps.

Dr. Zellmer
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