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I understand the basics of electrochemsitry and galvanic cell, where there are half cells where either an oxidation or reduction reaction occur, and electrons moving between the wires provide the current.

What confuses me is the salt bridge. In this diagram here, nitrate ions, which are from the solution of a half cell, go into the salt bridge.

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However, other diagrams (and answers like those to Do the salt bridge ions have to match the ions in the half-cells solutions?) do not state that the ions have to be the same (you don't need nitrate ions in the salt bridge if you have a silver nitrate solution for example).

enter image description here

So do ions like nitrate from the solution in a half cell enter the salt bridge?

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    $\begingroup$ Of course they do enter. How else would the charge be transferred? $\endgroup$ – Ivan Neretin Feb 3 '20 at 9:14
  • $\begingroup$ But in the second picture the salt bridge has KCl. So would sulfate ions be moved into the salt bridge? $\endgroup$ – Mason R Feb 3 '20 at 9:35
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    $\begingroup$ Of course they would. Again, how else would the charge be transferred? $\endgroup$ – Ivan Neretin Feb 3 '20 at 9:46
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    $\begingroup$ @IvanNeretin I disagree. If the concentration in the salt bridge is higher than that of the half cells, you could have anions from the salt bridge diffuse into the oxidation half cell (making up the missing negative charges lost by electrons leaving through the wire) and cations from the salt bridge diffuse into the reduction half cell (neutralizing the negative charge electrons from the wire bring in). $\endgroup$ – Karsten Theis Feb 3 '20 at 19:54
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You must think electrically, without thinking first to the chemistry. In the anodic solution first, the reaction produces positive ions $\ce{Mg^2+}$ or $\ce{Zn^2+}$. It does not really matter here. The problem is that this cannot stay like this. Either the positive ions have to move where negative ions are available. Or new negative ions must arrive in the solution. Or both.

The simplest way for solving this difficulty is to mix the two solutions. Because a similar difficulty appears on the cathodic region : positive ions ($\ce{Ag+}$ or $\ce{Cu^2+}$) are here discharged, leaving an excess of positive ions around the cathode. By mixing the solutions the excess negative ions produced around the cathode move to the anode where they are needed. But mixing the solutions is the least thing to do, because now $\ce{Ag+}$ or $\ce{Cu^2+}$ ions can touch the $\ce{Zn}$ or $\ce{Mg}$ anode. And the chemical reaction will go on in the contact between positive ion and metal. The same electrons of the half-reaction are produced but they will not go through the long external circuit form the anode to the cathode. They prefer the shortest circuit direct from metal to ion in solution, and metallic $\ce{Ag}$ will soon be formed on the $\ce{Mg}$ anode (or $\ce{Cu}$ on the $\ce{Zn}$ plate). No current will be emitted in the outer circuit. It is the end of the cell. So it is absolutely necessary to prevent the $\ce{Cu^2+}$ or $\ce{Ag+}$ ions to touch the anode, so the two compartments must be separated.

That is why we introduce the salt bridge. It must contain ions that do not react with any ion contained in the two compartments. Their nature is not important. Usually it is $\ce{K+}$ and $\ce{NO3-}$.

So let's go back to what's happening electrically in solution when the cell is working. There is an excess of positive ions around the anode, and an excess of positive ions around the cathode. To compensate this situation, the negative ions of the bridge are attracted by the positive ions in excess at the anode, and the positive ions of the bridge are attracted by the cathode. Both quit the bridge and get mixed in the solutions. This allow the cell to work.

After some time, the bridge is loosing all its stuff, so some positive ions of the anode ($\ce{Zn^2+}$or $\ce{Mg^2+}$) may enter the bridge and replace the initial positive ions $\ce{K+}$that have moved to the cathode. And symmetrically some negative ions ($\ce{NO3-}$ or $\ce{SO4^2-}$) from the anode will enter the bridge to replace the initial ions $\ce{NO3-}$.

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    $\begingroup$ Sorry if this is a basic question, but don't ions like potassium react with things like sulfate? Or is that not a reaction? $\endgroup$ – Mason R Feb 3 '20 at 9:54

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