What is the difference between #E# and #E^@# in electrochemistry?

Well, the mathematical difference is #E - E^@ = -(RT)/(nF)lnQ#... The conceptual difference is nonstandard vs. standard conditions.

We define standard conditions to be #25^@ "C"# and #"1 atm"# pressure, with #"1 M"# concentrations.

Many chemical functions, particularly in thermodynamics and electrochemistry, are temperature-dependent. Thus, we must be able to account for that...

We recall that for the Gibbs' free energy:

#DeltaG = DeltaG^@ + RTlnQ#

(if you do not recall this equation, look here for a derivation.)

and:

#DeltaG^@ = -nFE^@#

(which I will not derive as it is a simple unit conversion.)

The first equation uses #RTlnQ# as a correction factor for nonstandard conditions for the Gibbs' free energy. It turns out that the second equation also applies to the nonstandard #DeltaG#.

Hence:

#-nFE = -nFE^@ + RTlnQ#

Dividing by #-nF# gives:

#bb(E = E^@ - (RT)/(nF)lnQ)#

which is the purest version of the Nernst equation (before any simplifications), where:

• #n# is the number of electrons transferred in the redox reaction
• #F = "96485 C/mol e"^(-)# is the Faraday constant.
• #R# and #T# are known from the ideal gas law.
• #Q# is the reaction quotient, i.e. not-yet-equilibrium constant.
• #E# is the "electromotive force" for the cell process.
• #E^@# is, of course, #E# at standard conditions.

Likewise, #-(RT)/(nF)lnQ# is a correction factor to "convert" from standard to nonstandard conditions. If we are at standard conditions, then...

#E = E^@ - cancel((("8.314472 J/mol"cdot"K")("298.15 K"))/(nF)ln(1))^(0)#

#=> E = E^@#

at #25^@ "C"#, #"1 atm"# pressure, and #"1 M"# concentrations.