The goal of a salt bridge is to maintain electroneutrality in both of the half-cells.

At the anode half-cell, the anode is oxidized. The resulting electrons travel up the wire to the load while the electrolyte gains a positive charge from the extra solvated cations. At the cathode half-cell, the cathode is reduced. The incoming electrons allow the solvated cations in the electrolyte to solidify and deposit onto the cathode. These two redox reactions lead to the formation of a positive charge in the anode half-cell and a negative charge in the cathode half-cell.

Of course, to keep the current flowing through the load, electroneutrality must be maintained: the two half-cell electrolytes must remain neutral. To do so, a crude solution would be to just use a second wire, or a water bridge. Several answers already point out that doing so just means that the current flowing through the load will be severely reduced as the half-cell electrolyte ions would just flow through the water bridge instead of the load.

Using a salt bridge avoids this, as the chosen salt is usually inert and does not react with the electrolytes. However, the salt bridge eventually runs out of the initial salt. The anion of the salt keeps flowing into the anode while the cation of the salt keeps flowing into the cathode and they eventually end up completely depleted.

If I understand correctly (according to this discussion), the salt bridge does not, in fact, run out, since the ions of the electrolytes actually take over as charge carriers. However, in this case, are we not just back to square one, which is to say, just a water bridge? Or is there some benefit to the reaction being already underway?

  • $\begingroup$ A water bridge works poorly until the ions from both solutions flow into it. A salt bridge works well from the start. $\endgroup$ Commented Dec 20, 2021 at 21:38
  • $\begingroup$ Ions flow to the bridge too, not only from it. // The second wire would create 2 cells from 1. $\endgroup$
    – Poutnik
    Commented Dec 20, 2021 at 21:43
  • $\begingroup$ @Mithoron it does confirm my understanding that the original charge carriers of the salt bridge can get depleted, in which case the electrolyte ions of the half-cells take over as charge carriers in the bridge. However, it doesn't confirm whether the performance of the salt bridge drops once the initial charge carriers are depleted. $\endgroup$
    – David Cian
    Commented Dec 20, 2021 at 21:50
  • $\begingroup$ As a toy example, just imagine a salt bridge with NaCl, except it has a single molecule of NaCl. That single molecule would get very quickly depleted, as the Na+ would go to the cathode and the Cl- would go to the anode. Was there even any point in using a salt bridge with so little salt? $\endgroup$
    – David Cian
    Commented Dec 20, 2021 at 21:52
  • $\begingroup$ Negative electrode at the anode and a positive electrode at the cathode ?. $\endgroup$ Commented Dec 20, 2021 at 23:22

1 Answer 1


Apparently your problem is not chemical. It is related to electric charges being created and/or destroyed in solution.

Let's examine the "fate" of electric charges with a simple example : Zinc metal as anode and a copper plate dipped into $\ce{Cu^{2+}}$ as cathode (plus the same amount of corresponding negative ions like sulfate $\ce{SO4^{2-}}$ ions).

When the cell works, new $\ce{Zn^{2+}}$ ions are created at the anode, and $\ce{Cu^{2+}}$ ions are disappearing at the cathode. So positive charges are created around the anode, and the same charges are disappearing from the cathode, also creating an excess of negative ($\ce{SO4^{2-}}$) ions around the cathode. Without a bridge, the anode will become more and more positively charged, and the cathode more and more negatively charged. After a while, no more positive ions could be created in such a highly positive anodic region.

The cell can only work if these charge excesses disappear. This is done by allowing the new anodic cations $\ce{Zn^{2+}}$ to join the excess of negative ions from the cathodic region. Of course both ions $\ce{Zn^{2+}}$ and $\ce{SO4^{2-}}$ are attracting each other and will move to join in the salt bridge. But the bridge is filled with a salt solution containing positive and negative charges like $\ce{K+}$ and $\ce{NO3^-}$. At one entrance of the bridge, the positive $\ce{Zn^{2+}}$ ions attract the negative $\ce{NO3^-}$ ions and repell the positive $\ce{K+}$ ions. On the other side of the bridge, the negative $\ce{SO4^{2-}}$ ions produce exactly the same chemical migration. And the cell may work on. If the migration rate are forgotten, the local charge concentrations do not change, even though the chemical concentrations are changing all the time. The solutions are always electrically neutral.

In the long run, $\ce{Zn^{2+}}$ will slowly replace $\ce{K^+}$ ions in the bridge, and $\ce{SO4^{2-}}$ will slowly replace $\ce{NO3^-}$ in the same bridge.

So from the beginning to the end of this long "cell story", the electrical neutrality of the solutions is maintained, even though the composition of the solutions do change a lot.


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