-1
$\begingroup$

According to my book (and various other sources), the salt bridge in a galvanic cell serves to keep the solution electrically neutral. Yet I was not able to find any information on why the solution needs to be electrically neutral. As long as the oxidation reaction takes place in the anode (thereby dumping electrons and rendering the anode negatively charged) and the reduction reaction continues in the cathode, the charge disparity should create a potential difference right?

What happens when the solution isn't electrically neutral that prevents the oxidation and reduction reactions from taking place, because if they were taking place the electrodes should have a non-zero potential difference right?

So my question basically is, why does the electrolytic solution need to be electrically neutral in order for a potential difference to develop between the electrodes of a galvanic cell?

$\endgroup$
  • $\begingroup$ Then your book does a poor job at explaining this phenomenon. Remove the salt bridge, and the potential difference will develop all right. The problem is, there will be no current. $\endgroup$ – Ivan Neretin Apr 13 '17 at 6:33
  • $\begingroup$ Then if you have a solution of soluble copper salt with solid copper rod dipped into it and solution of a soluble silver salt with solid silver rod dipped into it, and you try to measure the potential difference between the two metal cathodes with a volt meter, you would read a non-zero potential difference? There are no salt bridges, only cathode and anode $\endgroup$ – E7_82_8E Apr 13 '17 at 7:12
1
$\begingroup$

Take a cell, for example, Zn/ZnSO4 & Cu/CuSO4

Initially, Zn is at more negative potential (high electron density) than at Cu. So electrons move through external circuit (the conducting wire, if the 2 electrodes are connected) from Zn to Cu. At the same time Zn2+ ions pass into solution, making it successively more +ve; while at Cu electrode - Cu2+ ions from solution combine with these electrons (came from Zn) leaving behind excess SO4 2- ions in the solution which will make the originally neutral CUSO4 successively -ve

Zn losses electron = anode (oxidation) Cu2+ gains electron = cathode (reduction)

There will be a situation (if no other provisions like salt bridge are made) that +vely charged ZnSO4 solution will not be ready to accept further Zn2+ ion from Zn rod & at the same time -vely charged CuSO4 solution to loose Cu2+ ions

This is when we say, the potential difference will soon be zero, as the electron density at the two electrodes will be zero.

$\endgroup$
0
$\begingroup$

The oxidation and reduction will take place. But remember we want the electrons to pass through the wire only, not through solution. The salt bridge won't let the electrons to reach the other side by going through the solution itself. Thus, we will have electric current through the wire and can see the potential difference.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.