Page:Elementary Text-book of Physics (Anthony, 1897).djvu/346

number of ions which pass through that cross-section in unit time in both directions is therefore $M(u+v)$ and the quantity of electricity carried through with them in both directions is $cM(u+v).$ But this quantity is equal to the current strength $I,$ and therefore $cM(u+v)=k(V_{1}-V_{2});$ or $u+v={\frac {k(V_{1}-V_{2})}{cM}}\cdot$ Now $cM$ is the quantity of current required to decompose the molecules in the cell, or the mass which is in solution in unit volume of the electrolyte; it may therefore be directly determined. Since equal volumes of the electrolytes contain the same number of univalent ions, this quantity of current is the same for all the cells, and since, with a known value of $I,$ we may determine the value of $k$ in each case by observations of $V_{1}-V_{2},$ the formula just obtained enables us to determine $u+v.$

This formula may be more conveniently used in another form. Let $n$ represent the weight of the hydrogen evolved by unit current in unit time, and $m$ the chemical equivalent of one of the products of electrolysis in the cell. Then $mn$ represents the weight of that product evolved by unit current in unit time, and ${\frac {1}{mn}}$ represents the current that will evolve unit weight in unit time. Now the electrolytes are prepared so that the weights of the constituents in the cells are given by $Nm,$ where $N$ is a number which is the same for all the cells. The current that will evolve these weights in the respective cells is therefore equal to ${\frac {N}{m}},$ and this current has been shown to be equal to $cM.$ Using this value of $cM$ in the equation for $u+v,$ we obtain $u+v={\frac {nk(V_{1}-V_{2})}{N}}\cdot$ In the experiments of Kohlrausch the difference of potential $V_{1}-V_{2}$ was the same for all the cells, and the value of ${\frac {k}{N}}$ determined for each cell. The values of $u+v$could then be calculated. The ratio ${\frac {k}{N}}$ is called the molecular conductivity.