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Why are the ions in the salt bridge immune to the potential difference between Anode & Cathode ?

In the Galvanic or Voltaic cell example Only the electrons belonging to Anode (Zn metal) of neighbouring beaker are attracted to Cathode (Cu2+ ions) through the wire connecting them causing current but NOT the ions (Na+, Cl-) present in the salt bridge of it's own beaker... !

Why is this so ? The potential difference affecting the electrons of Anode should affect the ions of salt bridge also right? Instead the ions in the salt bridge seem to move in opposite direction to current opposing the potential difference ? How so ?

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  • $\begingroup$ Electrostatic field that pulls the electrons in the wire is omnipresent, it also acts on the ions in the salt bridge. The ions move against that force because there are other forces acting on them - short-range forces due to gradient of concentration of chemicals. $\endgroup$
    – Ján Lalinský
    Dec 11, 2020 at 13:14
  • $\begingroup$ What is the name of this other force? Where can i read more about them? $\endgroup$
    – Vignesh Sk
    Dec 12, 2020 at 9:07
  • $\begingroup$ To be more accurate, in this case, most of the energy needed to push the ions through the electrolyte and the salt bridge is probably due to electrochemical forces acting on the ions while they pass the electrode-electrolyte boundary. In the salt bridge, electrochemical forces are probably much weaker/negligible and electrostatic field actually is in the same direction as the positive ions motion. There may be some non-electric force in the salt bridge due to concentration gradient forces, but not much. $\endgroup$
    – Ján Lalinský
    Dec 12, 2020 at 13:11
  • $\begingroup$ You can find some info o this searching for "electrochemical forces", "electrochemical potential", "concentration cell". One case of forces due to gradients is explained in Landau, Lifshitz, Electrodynamics of Continuous Media, par. 28 Diffusion phenomena. $\endgroup$
    – Ján Lalinský
    Dec 12, 2020 at 13:11
  • $\begingroup$ Thank you for the resources. I will check them out 👍 $\endgroup$
    – Vignesh Sk
    Dec 14, 2020 at 6:07

1 Answer 1

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Struggled with the same doubt, so below is my understanding/resolution of the problem, comments are most welcome.

So, there are two types of ions, ions of half cell and external ions from the salt bridge.

Let's assume half cell rxn to be Cu+2 -> Cu in a CuSO4 medium,

1)Before Connection between two half cells, As the electrode will lose electrons,the positive charge will develop there, the tendency of Cu+ ions would be to be repelled from it and So4- ions to be attracted towards it, hence the double-layer formation and if the Net driving force for rxn becomes Zero => double layer formation gets completed/saturated,

There are two factors responsible for this saturation:

  1. Factor 1: More positive charge on electrode will make harder for Cu(+2) to get electrons from the electrode
  2. Factor 2:As there are more So4(-2) compared to Cu(+2) in the solution, it has developed a negative charge and it would be harder for Cu(+2) to leave the solution and go to the electrode. Both factors would be responsible for the opposing Electric field for Cu(+2).

2) After Connection between two half cells, but with no salt bridge, The first factor can be dealt away with, as we have an extra supply of electrons to neutralize it. But the second factor will be there and this again will lead to another condition of saturation after some more forward reaction.

2) After Connection between two half cells, but with salt bridge, We have introduced other ions, which does not interfere in our rxn but can neutralize the second factor (excess of So4 (2-),

What's happening:

One half cell solution has excess negative ions and the other has excess positive ions.

There are two ways to overcome the second factor, first, just make both chambers as one, but this will lead to direct reactions of "some" of Cu+ moving to Zn electrode. Reasons for such movement :

  1. Repulsion from their electrode and attraction towards Zn electrode, but this will not be very effective due to localized nature of double layer, so such ions would feel a zero effective charge from electrodes which are far away. So, mainly the movement would be due to diffusion.

  2. But there is another factor, excess of So4(2+) which will try to damp down such a migration but if we mix our electrolytes, we nullified this effect. This effect can be used positively in a salt bridge configuration. Lets call this oppostie ion effect.

The salt bridge.

By providing opposite ions to opposite chambers salt bridge can reduce the second factor, and while blocking the migration of Cu(+2) ions due to opposite ion effect. Ions of the salt bridge will selectively go to opposite half cells due to opposite ion effect. Such migration could be attained rapidly by higher conc. of salt in the salt bridge.A symmetrical movement ensure the electroneutrality principle and salt in the bridge will be replaced by neutral solvent molecules (effectively).

Will there still be chances of Cu(+2) migrating to the Zn electrode? yes, but the rate would be very slow and so it won't matter to us.

Will there be chances for So4(-2) ions migrating into the salt bridge? yes, when I say salt bridge ions gets replaced by solvent molecules, I mean the effective replacement of ions in the salt bridge. The Electroneutrality principle (we will not find any buildup of only one type of ions into parts of the salt bridge) will still hold in the salt bridge.

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