Question

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

Answer 1

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

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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.