Page:Scientific Papers of Josiah Willard Gibbs.djvu/467

Integrating from one point to another in the electrolyte,

$\phi \int {\frac {dx}{\textstyle \sum \displaystyle c_{1}k_{1}\gamma _{1}}}=At\int {\frac {\textstyle \sum \displaystyle \mp {\frac {k_{1}}{M_{1}}}d\gamma _{1}}{\textstyle \sum \displaystyle c_{1}k_{1}\gamma _{1}}}+V'-V''.$

The evaluation of these integrals which denote the resistance and electromotive force for a finite part of the electrolyte depends on the distribution of the ions in the cell. For one salt with varying concentration,

$\phi {\frac {dx}{c_{1}k_{1}\gamma _{1}+c_{2}k_{2}\gamma _{2}}}=At{\frac {-{\frac {k_{1}}{M_{1}}}d\gamma _{1}+{\frac {k_{2}}{M_{2}}}d\gamma _{2}}{c_{1}k_{1}\gamma _{1}+c_{2}k_{2}\gamma _{2}}}-dV,$

or, since $c_{1}\gamma _{1}=c_{2}\gamma _{2}$ and $c_{1}d\gamma _{1}=c_{2}d\gamma _{2}$,

$\phi {\frac {dx}{c_{1}k_{1}\gamma _{1}+c_{2}k_{2}\gamma _{2}}}=At{\frac {-{\frac {k_{1}}{c_{1}M_{1}}}+{\frac {k_{2}}{c_{2}M_{2}}}}{k_{1}+k_{2}}}{\frac {d\gamma _{1}}{\gamma _{1}}}-dV,$

$\phi {\frac {dx}{c_{1}k_{1}\gamma _{1}+c_{2}k_{2}\gamma _{2}}}=At{\frac {-{\frac {k_{1}}{c_{1}M_{1}}}+{\frac {k_{2}}{c_{2}M_{2}}}}{k_{1}+k_{2}}}\log {\frac {\gamma _{1}''}{\gamma _{1}'}}+V'-V''.$

The resistance depends on the concentration throughout the part of the cell considered, but the electromotive force depends only on the concentration at the terminal points ($'$ and $''$).

For $c_{1}M_{1}$ and $c_{2}M_{2}$ we may write ${\frac {v_{1}}{a_{\text{H}}}}$ and ${\frac {v_{2}}{a_{\text{H}}}}$, where $v_{1}$ and $v_{2}$ are the "valencies" of the molecules. This gives

$V''-V'=a_{\text{H}}At{\frac {{\frac {k_{1}}{v_{1}}}-{\frac {k_{1}}{v_{1}}}}{k_{1}+k_{2}}}\log {\frac {\gamma _{1}'}{\gamma _{1}''}},$for$\phi =0$(circuit open).

I think this is identical with your equation ($V$) when your ions have the same valency.

Planck's problem is less simple.^[1] We may regard it as relating to a tube connecting the two great reservoirs filled with different electrolytes of same concentration, i.e., $\textstyle \sum _{0}\displaystyle c_{0}\gamma _{0}'=\textstyle \sum _{0}\displaystyle c_{0}\gamma _{0}''$. I use (₀) for an y ion, (₁) for any cation, (₂) for any anion. [The accents ($'$) and ($''$) refer to the two reservoirs.]

The tube is supposed to have reached a stationary state and dissociation is complete. The number of ions is immaterial, but they all must have the same valency $v$.

Now by equations (3) and (4), since $c_{0}M_{0}={\frac {v}{a_{\text{H}}}}$,

$\phi _{0}=-{\frac {aAt}{v}}k_{0}{\frac {d\gamma _{0}}{dx}}\mp k_{0}{\frac {dV}{dx}}\gamma _{0},$

1. [Planck, Wied. Ann., vol. xl (1890), p. 561.]