# What kind of unit is this [A*m^2/mol]?

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

$k_d^{eff}=\frac{2RT\kappa^{eff}}{F}(t^+-1)\left(1+\frac{\text{d}\ln(f)}{\text{d}\ln(c)}\right)$

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?

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]
• 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? Nov 10, 2016 at 19:06
• 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. Nov 11, 2016 at 8:09