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My book mentions sodium sulfate ($\ce{Na2SO4}$) as an electrolyte that can be used in a salt bridge. I'm asking this question because ChemLibreTexts says that

It is also important that the charge balance in each of the half cells facilitated by the ions in the salt bridge occurs at the same rates. That means that the halide anions moving from the salt bridge into the anode to balance out the excess $\ce{Zn^{2+}}$ ions do so at the same rate as the alkali cations moving from the salt bridge into the cathode to balance out the depletion of $\ce{Cu^{2+}}$ ions. Ions have a property known as mobility and the mobility of an ion depends on its size. Smaller ions have a higher mobility than larger ions. That means that the ideal species for a salt bridge should have a cation and anion of the same size and charge.

In $\ce{Na2SO4}$, the sizes of $\ce{Na+}$ and $\ce{SO_4^{2-}}$ are approximately equal (0.242nm and 0.227nm). The problem is that the magnitudes of the charges on them are not equal ($+1$ and $-2$). Also, there are twice as many $\ce{Na+}$ ions than $\ce{SO_4^{2-}}$ ions. So, will $\ce{Na+}$ ions and $\ce{SO_4^{2-}}$ ions enter the solution at the same rates?

My thoughts:

Probably the positive and negative charges are flowing in the two half-cells at the same rate. It's true that the charge on $\ce{SO_4^{2-}}$ is $-2$ (twice that of $\ce{Na+}$'s charge) and $\ce{Na+}$ has a $+1$ charge, but the number of $\ce{Na+}$ ions is twice that of $\ce{SO_4^{2-}}$ ions. So, probably the rate at which the ions are entering the cells is the same. Am I correct?

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It is of course better to use cations and anions of the same size and charge. But it is not a requirement. The cell and the bridge are also working if these parameters are different. The only effect to be observed at the end of the life of the cell is a depletion of ions in the copper zone. This will increase the inner resistance of the cell, and decrease the available voltage observed between the electrodes.

On the other hand, $\ce{Na2SO4}$ can be used to fill a bridge in the Daniell cell ($\ce{Zn/Cu}$). It must absolutely be avoided if one of the electrode is made of lead, because $\ce{Na2SO4}$ will react with $\ce{Pb^{2+}}$ and produce a white insoluble precipitate of $\ce{PbSO4}$, which will destroy the cell.

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