CHAP. 11: ELECTROCHEMISTRY [CONTENTS] 360
Example
During the electrolysis of HCl, the chargeQ= 1F passes through the system. The transport
number of the cation H+ist+= 0.82. Determine the concentration changes in the anode and
cathode compartments.
Solution
Hydrochloric acid dissociates into ions according to the reaction
HCl= H++ Cl−.
It thus holds thatz+=ν+=z−=ν−= 1. Both ions are discharged at the electrodes, i.e.
relations (11.18) and (11.20) apply. From the first of them we obtain
∆nan=−t+
Q/F
z+ν+
=− 0. 82 mol.
A total of 0.82 moles of hydrochloric acid were used up in the anode compartment.
Now we first calculate the anion transport number from relation (11.13)
t−= 1−t+= 1− 0 .82 = 0. 18.
and finally use relation (11.20) to obtain
∆ncat=−t−
Q/F
z−ν−
=− 0. 18 mol.
A total of 0.18 moles of hydrochloric acid were used up in the cathode compartment.
