# The property F in an equation for chemical potential

While studying the lamba sensor that works on yttria-strabilised-zirconium (YSZ) I came across the following equation:

$$p_{O_2}^{gas}=p_{O_2}^{ref} \cdot \exp\left(\frac{4FV_{ref}}{RT}\right)$$

What is $F$ in this equation?

The "wall" of the lambda sensor is as shown in the picture with the voltage $V_{ref}$ measured across: The equation apparently is derived directly from the chemical equilibrium (chemical potentials equals to zero):

$$\mu_{O_2}+4\mu_e=2\mu_{O^{--}}$$

They come to the following where they end at the final formula:

$$V_{ref}=\frac{RT}{4F}\ln\left(\frac{p_{O_2}^{gas}}{p_{O_2}^{ref}}\right) \implies p_{O_2}^{gas}=p_{O_2}^{ref} \cdot \exp\left(\frac{4FV_{ref}}{RT}\right)$$

I understand the partial pressures. The gas constant $R$ and temperature $T$ are also clear. If I am correct, $V_{ref}$ is the voltage across the material of the sensor wall, as in the picture, so this is also clear.

The equation is the final result from which the oxygen amount in the gas can be found, since the partial pressure of the oxygen in the gas can be calculated. My question is: What is $F$ in the equation?

• As there is no chemical-potential tag, I really do not know which tags to add to this question. Please correct me on that. – Steeven May 29 '15 at 13:50
• The tagging is correct. – bon May 29 '15 at 13:50
• Perhaps this is Faraday's constant? – LordStryker May 29 '15 at 13:55

$F$ is the Faraday constant: $9.64853399(24)\times10^4~\mathrm{C~mol^{-1}}$.
$$\frac{\mathrm{J~K^{-1}~mol^{-1}~K}}{\mathrm{C~mol^{-1}}} = \frac{\mathrm{J}}{\mathrm{C}} = \mathrm{V}$$