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

##### 1 Answer
May 29, 2017

Well, the mathematical difference is $E - {E}^{\circ} = - \frac{R T}{n F} \ln Q$... The conceptual difference is nonstandard vs. standard conditions.

We define standard conditions to be ${25}^{\circ} \text{C}$ and $\text{1 atm}$ pressure, with $\text{1 M}$ 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:

$\Delta G = \Delta {G}^{\circ} + R T \ln Q$

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

and:

$\Delta {G}^{\circ} = - n F {E}^{\circ}$

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

The first equation uses $R T \ln Q$ as a correction factor for nonstandard conditions for the Gibbs' free energy. It turns out that the second equation also applies to the nonstandard $\Delta G$.

Hence:

$- n F E = - n F {E}^{\circ} + R T \ln Q$

Dividing by $- n F$ gives:

$\boldsymbol{E = {E}^{\circ} - \frac{R T}{n F} \ln Q}$

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

• $n$ is the number of electrons transferred in the redox reaction
• $F = {\text{96485 C/mol e}}^{-}$ 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.
• ${E}^{\circ}$ is, of course, $E$ at standard conditions.

Likewise, $- \frac{R T}{n F} \ln Q$ is a correction factor to "convert" from standard to nonstandard conditions. If we are at standard conditions, then...

$E = {E}^{\circ} - {\cancel{\frac{\left(\text{8.314472 J/mol"cdot"K")("298.15 K}\right)}{n F} \ln \left(1\right)}}^{0}$

$\implies E = {E}^{\circ}$

at ${25}^{\circ} \text{C}$, $\text{1 atm}$ pressure, and $\text{1 M}$ concentrations.