The analysis of the Hittorf method for the determination of transference numbers was greatly abridged in Section 12-6, and the more detailed presentation appears here.
We consider first the system shown in Fig. 12-6, for which a qualitative analysis was given. The more exact bookkeeping is as follows, based on 1 & of electricity passing through the cell.
Cell I. A n o d e compartment
A n o d e reaction: A g = A g+ + e-: gain of 1 equiv of A g+ migration of A g+: loss of /+ equiv of A g+ migration of N 03 -: gain of t_ equiv of N Os~ . Net change: Gain of 1 — t+ or f_ equiv of Ag+, gain of f_ equiv of N03~~;
or gain of f_ equiv of A g N 03. Cell HI. Cathode compartment
Cathode reaction: A g+ + e~ = A g : loss o f 1 equiv o f A g+ migration of A g+: gain of /+ equiv of A g+ migration of N 03 -: loss of i_ equiv of N03~~.
Net change: Loss of 1 — t+ or f_ equiv of A g+, loss of i_ equiv of N 03 _; or loss of r_ equiv of A g N 03.
The overall change per faraday is that the anode compartment gains t_ equiv of A g N 03 and the cathode compartment loses f_ equiv. The middle compartment should not change in content since the gains and losses due to the migration of ions past the dividing lines 1—II and 11—III exactly compensate.
This analysis applies to any cell containing a single electrolyte and having electrodes that generate or consume the cation of the electrolyte. The cell of Fig. 12-6 might have been filled with C u S 04 solution, for example, and have copper electrodes. We would conclude that, per faraday, the anode compartment would gain t_ equiv of C u S 04 and the cathode compartment would lose t_ equiv.
Alternatively, the electrodes might be reversible to the anion, that is, the electrode reaction might produce and consume anion. This would be the case if, for example, the cell were filled with NaCl solution, and the electrodes were silver coated with silver chloride. The bookkeeping is now as follows:
Cell I. A n o d e compartment
The opposite working of the electrode reaction thus has the effect of making the anode compartment undergo a net loss of t+ equiv of NaCl and the cathode compartment undergo a net gain of t+ equiv of NaCl, in contrast to the preceding example.
It is not necessary, of course, that the electrode reaction involve either of the ions of the electrolyte. If platinum electrodes were used in the example just given, the analysis would be as follows:
This last example illustrates some further points. First, the net change can be expressed either in terms of gains and losses of individual types of ions or in terms
of gains and losses of complete electrolytes; the alternative statements are seen on examination to be entirely equivalent. The purpose of the middle compartment now becomes apparent. The analysis assumes that only Na+ and Cl~ ions carry current past the I—II and II—III dividing lines. The electrode reactions are producing H+ and OH~ ions, however, and if mixing occurs in compartments I and III
SPECIAL TOPICS, SECTION 2 485 during the electrolysis, then these electrode products will carry part of the current between compartments. The gains and losses of N a+ and Cl~ ions would then be less than expected, and the overall analysis would be in error. The test of whether electrolysis products reached compartment II would, in this case, be whether the pH of II changed or not.
Notice that all of these analyses are couched in terms of amounts gained or lost during electrolysis. The concentrations present do not enter directly. The transference numbers are themselves concentration-dependent, however, since λ+ and λ_ will in general change differently with concentration. It is desirable for this reason that no great change in composition occur during the electrolysis.
Ordinarily, then, the amount of electricity passed through the cell will be some small fraction of a faraday.
It is, of course, necessary to measure the quantity of electricity involved in a Hittorf experiment so that the results can be put on a per faraday basis. It is not necessary, however, to know the applied potential or the actual current—only the total quantity of electricity. One usually obtains this by determining the amount of an electrode reaction that has occurred. Thus in the first example, the loss in weight of the silver anode gives the number of faradays used. It would be possible in the third example to titrate the solution in the anode compartment to find the amount of H+ ion produced. However, it is generally more convenient t o have a second cell in series with the Hittorf cell, which functions as a coulometer.
This might consist of a silver anode dipping into silver nitrate solution and a platinum cathode. Either the loss in weight of the anode or the amount of silver deposited on the cathode would show the number of faradays that had passed through the cell. A numerical example is as follows.
Example. W e c a n construct a n illustrative problem using the cell in which a n N a C l solution analysis, the loss should be /+ equiv of N a C l per faraday. The transference number of Na+ is therefore 0.385 and that of CI" is 0.615. The equivalent conductivity of 0.1 m N a C l is 106.74, so that AN a+ = (0.385)(106.74) = 41.09 and XCi- = (0.615)(106.74) = 65.65.
The use of a rinse of original solution is a characteristic procedure in a transference experiment.
The purpose of the rinsing, of course, is to displace all of the electrolyzed solution, and the use of original solution for the rinse simplifies the ensuing calculation. W e are not interested in con
centrations and need only to compare the amount of N a C l present in electrolyzed solution plus rinse with the amount originally associated with the quantity of water in the combined solutions.
It is thus not necessary to k n o w exactly h o w m u c h solution was originally present in the anode compartment.
A n alternative but less accurate procedure would have been to fill the anode compartment with a known volume of the 0.1 m solution and to determine the change in concentration resulting from the electrolysis. For example, had the anode compartment originally contained 200 c m3 of solution, or approximately 0.02 equiv of N a C l , then at the end of the electrolysis the amount remaining w o u l d have b e e n 0.02 — 0.00685 = 0.01315 equiv, corresponding to a 0.0657 m
N a C l solution. T h e solution w o u l d also have contained 0.01780 equiv of HC1, or 0.0890 m HC1.
When we calculate transference numbers by the Hittorf method we implicitly assume that the solvent water does not migrate during the electrolysis. The water of hydration of the ions should move with them, however, and Washburn suggested the ingenious experiment of adding a sugar to the solutions in a Hittorf cell and referring the calculations to the sugar as the n o n -migrating species. The result showed that several molecules of water d o in fact migrate with each ion.