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:
- Factor 1: More positive charge on electrode will make harder for Cu(+2) to get electrons from the electrode
- 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-),
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 :
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.
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.