# Effect of salt bridge on Nernst potential

I learned 2 things:

1. A salt bridge is necessary in a galvanic cell so charge does not build up in the halfcells countering the potential difference created by the separate redox pairs. Indeed, this would stop the flow of electrons eventually.

2. The potential of the galvanic cell depends on the concentrations of ions of both redox pairs through the Nernst equation.

Now looking up the salt bridge on Wikipedia, I get 'It maintains electrical neutrality within the internal circuit, preventing the cell from rapidly running its reaction to equilibrium.'

So if the salt bridge influences the equilibrium, it must affect the Nernst equation, I was wondering how you could work this in the equation? Is the original Nernst valid when the salt bridge is there? And how do you write it in absence of the salt bridge?

You read that wrong. The salt bridge doesn't influence the equilibrium (provided you're using normal salts and not something insane). If you start transferring electrons from anode to the cathode, you're going to quickly build up excess negative charge that opposes the potential of the cell. Having a salt bridge simply closes the circuit so that charges can continue to move.

You have the wrong concept.

(1) below is an image of a Cu/Zn galvanic cell (a battery...). The reaction is:

$$\ce{Zn + Cu^{2+} -> Zn^{2+} + Cu}$$

The problem is that if the copper ions, $\ce{Cu^{2+}}$, get into the Zn half-cell then the copper would plate directly on the zinc electrode. So the salt bridge is necessary to keep the the copper ions away from the zinc electrode. That way the electrons flow through the wire.

(2) In measuring the voltage with the Nernst equation it is dependent on the fact that "no" current is flowing. Of course a very very tiny amount must flow, say something on the order of microamps or smaller.

When there is significant current flow then all sorts of nasty things start happening inside the galvanic cell and the voltage drops. So the Nernst equation by itself isn't sufficient to calculate the cell voltage of a galvanic cell when it has a significant current flow.

• Thanks for that! Although I see your point that the Nernst equation fails when significant current flows from a practical standpoint, it does predict what will happen when you connect the 2 half cells from your example-> DeltaE=1.10-(0,059/2)*Log[(Zn2+)/Cu2+)]. This means that if we start with 2 solutions of 1M the initial voltage will be 1,10V. Once the reaction starts the concentration of Zn2+ will increase and that of Cu2+ will decrease so the voltage will drop until equilibrium is reached en DeltaE=0. Now what would happen then in absence of a salt bridge? How far could the reaction go? – Stikke Jul 23 '18 at 20:22
• @Stikke - If you took away the salt bridge on the cell shown the electrons will not flow through the wire. Electricity is about charge flow. Electrons flow through the wire, but ions flow through the salt bridge. You also have to realize that you can't create charged solutions. So $\ce{Cu^{2+}}$ ions lost have to be balanced by losing some anion. Likewise "excess" $\ce{Zn^{2+}}$ ions have to go somewhere. – MaxW Jul 24 '18 at 13:45
• My idea was that charges would build up creating a field exactly counteracting the potential created by the Redox couple, so a small flow of electrons would be allowed for a short while, after which the reaction would stop. – Stikke Jul 25 '18 at 7:50
• @Stikke - Actually you're right. An itty bitty teeny weeny amount of current would flow for an even ittier bittier teenier weenier amount of time polarizing each electrode. I don't know if anyone has ever been able to measure that infinitesimal current pulse. You have to realize that solutions don't want to have a net charge. The key point however is that current flow would not form a galvanic cell. In a true galvanic cell a very very small amount of current could flow at a constant rate and at a constant voltage for a long time. – MaxW Jul 25 '18 at 14:05