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I am having trouble with the concept of a junction potential as it relates to the salt bridge of an electrochemical cell. I understand that a junction potential occurs between two solutions of differing concentrations and that it depends on the mobilities of the ions in solution.

For example, between a solution of sodium chloride and deionized water, a junction potential will develop when the ions of the sodium chloride solution are allowed to diffuse into the deionized water. The junction potential occurs because the chloride anions will diffuse more quickly into the deionized water (due to its greater mobility) than the sodium cations. This leaves the sodium chloride side of the junction between the two solutions more positive and the deionized water side more negative, creating a junction potential.

However, what I do not understand is how a junction potential develops in an electrochemical cell. A junction potential develops at both ends of the salt bridge. Are the junction potentials at either end different? If so, why? If not, how would a junction potential occur between two solutions that don't directly interact (the two solutions in this case being the solutions of the respective half cells)?

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    $\begingroup$ First you do realize that the junction potential would depend somewhat on the amount of current flowing in the cell. With modern high impedance preamps there wouldn't be much current. // Since the ions and ionic strength in the two half cells doesn't have to be the same there is reason to believe that the junction potentials at opposite ends of the salt bridge could be different. The gist is that the salt bridge would develop a small charge. Over both half cells and the salt bridge then the net charge would be zero of course. $\endgroup$ – MaxW Apr 6 '16 at 21:24
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    $\begingroup$ Having a charge in the salt bridge just means that the system isn't at a real equilibrium. So long as the solutions in the two half cells and the salt bridge are different, then the system really isn't at equilibrium, but rather at a steady state. $\endgroup$ – MaxW Apr 6 '16 at 21:26

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