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CBSE Class 12-science Answered

salt bridge
Asked by | 31 May, 2008, 10:56: AM
Expert Answer

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

by using a piece of cotton wool soaked in one electrolyte), the ions of each solution

will diffuse into the other solution. The rate of diffusion per unit area of the

membrane (the flux, j

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

Answered by | 31 May, 2008, 09:27: PM
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