CBSE Class 12-science Answered

When two electrolyte solutions are in contact through a porous membrane (or even

i /mol cm2 s-1 ) is given by Fick's First Law, which for onei = Di Ø [I] / Ø xΛi .≠ Di. Hence, different ions will diffuse at different1 and c2 in

dimension is:

j

The flux of species i will depend on its molar conductivity, since, by the

Nernst-Einstein Equation,

rates. Consider placing two solutions of HCl, with concentrations c

contact.Ø[I] / Øx), and will thus cause diffusion+ and Cl- from high concentration to low concentration. At the moment+ ions will equal that- ions. Initially DH+ p DCl-, since H+ ions can diffuse by the Grotthuss- have to diffuse past water molecules. Thus a potential

There will be a large concentration gradient (

of both H

that the interface is formed, the concentration gradient of H

of Cl

Mechanism, whereas Cl

difference is set up across the interface, and the solution of lower concentration

will become positively chargedLJP) is given byLJP = (t+ - t-) {RT/F} ln {c2/c1}1 and c2 are the different concentrations of the (same) electrolyte.3, NH4NO3. This is possible since in these electrolytes, t+ l t- l 0.5, andLJP l 0.

Ultimately, a steady state will be attained in which a potential difference will exist

across the thin boundary region between the two solutions. This is known as a liquid

junction potential. For 1:1 electrolyte solutions in concentration cells, the size of

this potential difference (E

E

where c

To overcome the problem of LJPs in electrochemical cells, we use a salt bridge that

allows both solutions to be in contact with a third solution thatgives rise to no LJP,

e.g. KCl, KNO

hence E