I found a term on this website that gives a unit of [A*m^2/mol]. The equation is:


where R is the gas constant [J/mol/K], K is ionic conductivity [S/m], T is the temperature [K], t+ is unitless, f is unitless and c is the concentration [mol/m^3], F is the Faraday constant [C/mol].

This equation gives the unit $\frac{A \cdot m^2}{mol}$. What kind of unit is this one? I can see that they define it as electrolyte diffusional conductivity, but I can't find this unit and better definition anywhere in the literature.

Can anyone help me with this?


1 Answer 1


First thing to say, I don't know much about electrochemistry. But concluding from thermodynamics, rewriting your equation a bit and making dimensional analysis there are some things to say:

  1. It seems that you interchange the words dimension and unit. So an example: Length is a dimension and meter, inches... are units that measure length. You are probably asking for the physical interpretation of the electrolyte diffusional conductivity and not for the specific unit it is measured in.

  2. The logarithm can be applied only to dimensionless numbers. In order to calculate it, you have to divide by some $c_0$ usually this is given in mol/l. In addition both terms in brackets have to be dimensionless, because there is a sum with a dimensionless number and in sums the dimensions have to match. So the only term relevant for the dimension, is the fraction.

  3. The resulting dimension of the fraction is in SI units [A / m] and not [A m$^2$ / mol]

  • $\begingroup$ Thank you for your answer. 1. Yes. I would like to know the interpretation of that conductivity. 2. I agree, but there is no $c_0$ in that equation. $c_e$ is solved using Fick's law, so it is in [mol/L]. If you look at Eq. 3.15, the term is multiplied with the logarithm of concentration again. 3. I agree. I get [A/m] if I ignore the term with the logarithm. But then the units in 3.15 don't match if that term is given in [A/m]. Or am I not getting something fundamental? $\endgroup$
    – Physther
    Nov 10, 2016 at 19:06
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    $\begingroup$ Units are not carried out from inside the logarithm or square radicals. Yes, this may lead to strange things like ln(m) unit that have no reasonable physical meaning. Furthermore, if an equation contains addition of values with different units, something is disastrously wrong. So I would cross-check the formula, as (1+k) member should by all common sense be fully unitless. I also see suspicious $d$, which may be a differentiation symbol, and than things become even more interesting. $\endgroup$
    – permeakra
    Nov 11, 2016 at 8:09
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    $\begingroup$ That said, not all units have names and finding a weir unit somewhere in a table is not surprising. Especially in borderline areas. $\endgroup$
    – permeakra
    Nov 11, 2016 at 8:10
  • $\begingroup$ I agree with everything you said. It is a weird notation, indeed. I have artificially assumed that 1 would have the unit [m3/mol], so that the units would be homogeneous. It is definitely incorrect, but I have the feeling that there is something weird about those equations, and it's not the first time I see it... And then there is the differential and the logarithm... So weird $\endgroup$
    – Physther
    Nov 11, 2016 at 9:22

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