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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?

Some more info

The "wall" of the lambda sensor is as shown in the picture with the voltage $V_{ref}$ measured across:

enter image description here

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?

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  • $\begingroup$ As there is no chemical-potential tag, I really do not know which tags to add to this question. Please correct me on that. $\endgroup$
    – Steeven
    May 29, 2015 at 13:50
  • $\begingroup$ The tagging is correct. $\endgroup$
    – bon
    May 29, 2015 at 13:50
  • $\begingroup$ Perhaps this is Faraday's constant? $\endgroup$
    – LordStryker
    May 29, 2015 at 13:55

1 Answer 1

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

A quick dimensional analysis confirms that this makes sense.

$$\frac{\mathrm{J~K^{-1}~mol^{-1}~K}}{\mathrm{C~mol^{-1}}} = \frac{\mathrm{J}}{\mathrm{C}} = \mathrm{V}$$

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  • $\begingroup$ Oh dear! Thank you for the obvious, but hidden truth! $\endgroup$
    – Steeven
    May 29, 2015 at 14:05
  • $\begingroup$ I wish NIST or IUPAC or whoever would issue a ruling that Volts are an obsolete unit now that we know charge comes in integral units. It would be way easier if everything was in moles. A slightly less ideal (to me) solution, but still a massive improvement, would be to get rid of moles and measure everything in Coulombs. There is no reason to have both anymore. $\endgroup$
    – Curt F.
    May 29, 2015 at 14:07

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