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

Integrating from one point to another in the electrolyte,

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

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,

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

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

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

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

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 (${\displaystyle '}$ and ${\displaystyle ''}$).

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

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

I think this is identical with your equation (${\displaystyle V}$) 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., ${\displaystyle \textstyle \sum _{0}\displaystyle c_{0}\gamma _{0}'=\textstyle \sum _{0}\displaystyle c_{0}\gamma _{0}''}$. I use (₀) for an y ion, (₁) for any cation, (₂) for any anion. [The accents (${\displaystyle '}$) and (${\displaystyle ''}$) 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 ${\displaystyle v}$.

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

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