# How does the charge imbalance affect the cell reaction when there is no salt bridge in a galvanic cell?

As mentioned in the answers to this question Why is it important to use a salt bridge in a voltaic cell? Can a wire be used? a salt bridge is used to keep the two half cells neutral. i understand how the imbalance occurs without a salt bridge, but i don't understand the way this imbalance affects the cell reaction. Can someone please explain? Much thanks.

• The charge disbalance caused by the current 1 A per 1 min is big enough to create 10 km long GC+ lightning with potential diffference up to 1 gigavolt. I guess it is strong enough force for a cell going against it. Potential difference would stop cell net reactions very soon, with external cell voltage quickly converging to zero. Jun 10, 2020 at 9:08

Let's consider the famous Daniell cell, without this bridge. In the anode solution, a piece of zinc metal is dipping in water or in a zinc sulfate solution. Zinc has a tendency to loose electrons and produce ions $$\ce{Zn^{2+}}$$ which pass in water. The electrons are following a long wire connected to the zinc plate up to the cathode (Cu in $$\ce{CuSO_4}$$ solution). But this process will be stopped after some microseconds, because the first $$\ce{Zn^{2+}}$$ ions appearing in the anodic solution are not counter balanced by any negative ions. The solution becomes positively charged and repels any other $$\ce{Zn^{2+}}$$ ions form being produced. $$\ce{Zn -> Zn^{2+} + 2 e^-}$$
The same thing happens in the cathodic solution ($$\ce{CuSO_4 solution}$$). The electrons produced at the zinc anode discharge $$\ce{Cu^{2+}}$$ ions : a deposit of metallic copper appears on the cathode. But this may last a couple of microseconds, and not more, because the solution around the copper electrode becomes negatively charged due to the remaining sulfate ions. These negative ions are not counter-balanced by positive ions : they will preventl new electrons from coming to the copper electrode. $$\ce{Cu^{2+} + 2 e^- -> Cu}$$
So the chemical reactions will be stopped in both electrodes after a couple of microseconds. To transform this arrangement in a cell, the remaining sulfate ions $$\ce{SO_4^{2-}}$$ must be allowed to move to the anodic cell where there is an excess of positive charges (the $$\ce{Zn^{2+}}$$ ions). The only way of allowing this migration is to build a wet bridge between the two solutions. This bridge should allow too numerous negative ions from the cathode to migrate to the anodic solution where $$\ce{Zn^{2+}}$$ are attracting them.
But the solutions should not be mixed. If mixed, the anodic and cathodic reactions occurs in the solution on the zinc plate : no electrons will be emitted in the outer circuit. $$\ce{Zn + Cu^{2+} -> Zn^{2+} + Cu}$$