# Electrochemical cells. Why is a salt bridge needed at all when the system is open? [closed]

I get that the purpose of salt bridges is maintaining the amount of negative and positive charges in the solutions of two connected half-cells when electrons flow from one to the other. But when the system is open, as it is when using a high impedance voltmeter to meassure the potential of the cell. There is no flow of electrons to counter. Does a salt bridge have any purpose then?

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• When using a voltmeter there is a very very small current flow. There must be some current to get a voltage, remember that $V=iR$. – MaxW Sep 22 '18 at 16:24
• But isn't that current flow negligible? Is it enough to need a salt bridge? – G. Ayala Sep 22 '18 at 16:36
• With no salt bridge there is no current flow at all. $V = iR$ now let $i=0$, so $V = 0\times R = 0$ – MaxW Sep 22 '18 at 16:39
• Now why would there be no current flow without the salt bridge? Electric current travels through the wire connecting both half-cells (in this case through a voltmeter that allows a very tiny amount of current through). The salt bridge only exhanges ions with the solutions to maintain the amount of charges in both solutions. – G. Ayala Sep 22 '18 at 16:51

Below is an image of the Daniel cell from Wikipedia which I modified a bit.

Again from Wikipedia...

In the Daniell cell, copper and zinc electrodes are immersed in a solution of copper(II) sulfate and zinc sulfate respectively. At the anode (on the left), zinc is oxidized per the following half reaction:

$$\ce{Zn(s) → Zn^2+(aq) + 2e−} \tag{EMF -0.7618 V}$$

At the cathode (on the right), copper is reduced per the following reaction:

$$\ce{Cu^2+(aq) + 2e^− → Cu(s)}\tag{EMF +0.340 V}$$

The total reaction being:

$$\quad\quad\quad\quad\ce{Zn(s) + Cu^{2+}(aq) → Zn^{2+}(aq) + Cu(s)}\quad\quad\quad\text{(Open Circuit voltage 1.1018 V)}$$

Now from the zinc half cell electrons to the copper half cell via the wire. Considering what would happen if the flow of electrons could happen continuously without a salt bridge:

• The zinc half cell would become positively charged, since there would be more $$\ce{Zn^2+}$$ ions than $$\ce{SO4^2-}$$ ions.
• The the copper half cell would become negatively charged since there would be less $$\ce{Cu^2+}$$ ions than $$\ce{SO4^2-}$$ ions.

You can't generate a continuous current like this since it would create "charged solutions", so sulfate ions must flow, from the copper half cell to the zinc half cell, through the salt bridge to balance the charges and create neutral solutions.

Now what does the phrase "Open Circuit voltage 1.1018 V" mean?

You can think of a battery being like this:

So the battery has an internal resistance $$R_s$$. Now we add an external load (be it voltmeter or resistor...) on the battery so that current flows.

Now the current, $$i$$ is flowing through both the internal resistance of the battery, $$R_s$$ as well as the external load, $$R_l$$. So the Voltage measured across the load of the Daniel Cell is $$V_{out} = iR_l$$, but for the whole cell:

$$V_s = 1.1018 V = iR_s + iR_l$$

Rearranging a bit we get $$V_{out}$$ as related to the "open circuit" voltage of 1.1018 V.

$$V_{out} = 1.1018 V - iR_s$$

So if any current flows in the cell then the load voltage, $$V_{out}$$ will be less than 1.1018 V. However if the current is small so that $$iR_s < 0.0001 V$$ , then for all practical purposes we can ignore the internal voltage drop of the Daniel cell and claim that the load voltage measured is the "open circuit voltage."

• Thanks for the detailed answer, but I'm afraid the question remains, if the current is small enough that the reaction could be practically considered to not be happening at all, then no half cell will become positively or negatively charged and the salt bridge will no longer have a purpose, right? – G. Ayala Sep 22 '18 at 20:04
• "You can't generate a continuous current like this since it would create 'charged solutions', so sulfate ions must flow from the copper half cell to the zinc half cell to balance the charges and create neutral solutions." // I quit. I don't know how to explain this any better. – MaxW Sep 22 '18 at 20:09
• But how would the solutions get "charged" if there is practically no current going on? I'm not trying to generate a continuous current, Im just meassuring the potential of a cell to create one, without it practially happening. I'm sorry for being so dense, but I can't wrap my head around it, you tried, man, thanks a lot. – G. Ayala Sep 22 '18 at 20:29
• I'll try again... (1) With no salt bridge there is no continuous current flow so the solutions don't get "charged." The point that I was trying to make is that if a continuous current did in fact flow then the zinc solution would become more and more positively charged and the copper solution would become more and more negatively charged. (2) You got to think about the numbers involved here. 1.00 microamp of current for 1 second is $1\times10^{-6}$ coulombs. A coulomb is $6.24\times10^{18}$ electrons, so that would be $6.24\times10^{12}$ electrons. That is a lot of electrons! – MaxW Sep 23 '18 at 3:00
• (3) A real voltmeter can't measure a "theoretical voltage." It works by measuring a small current flow. The overall notion is that the internal resistance of the voltmeter is much much greater than the resistance of the load. So practically you can ignore the current flowing though the voltmeter. That isn't to say that there is 0.00... amps flowing through the voltmeter. – MaxW Sep 23 '18 at 3:09