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.

  • $\begingroup$ 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. $\endgroup$ – Poutnik Jun 10 at 9:08

If there is no bridge in a voltaic cell, this will be no cell any more. The arrangement will just be a metallic connection between two different containers. It will not deliver any continuous current. And replacing the salt bridge by a wire will simply add a second metallic connection between the solutions. There will be two parallel but different metallic connections between the two solutions. It is not a cell.

To get a cell, a liquid connection must exist between the two solutions through which ions can move.

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

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