I will try to rephrase my confusion:

I am not sure if the redox reactions occur in Galvanic cell because of the wire connecting the 2 metals or because of the solution causing electrons to leave the zinc atoms

I have 2 questions:

  1. What is the purpose of the solution in a Galvanic cell? Why in both halves of the cell, the solution contains positively charged ions?

  2. Do electrons leave a Zinc atom and a positively Zinc ion is added to the solution even without connecting both halves of the cell with a wire, or this process (the oxidation) occurs only when both electrodes are being connected with a wire


  • $\begingroup$ This is a homework question. We have a policy which states that ‎you should show your thoughts and/or efforts into solving the problem. It'll make us certain that ‎we aren't doing your homework for you. Otherwise, this question may get closed.‎ Please edit in your full reasoning or thoughts on this. $\endgroup$ Mar 26, 2017 at 15:14
  • $\begingroup$ I recently started to research this subject. About the first question I have no thoughts as I have no idea what is the purpose of it. About the second question, I understand that electrons leave the zinc and want to 'join' the copper, because zinc is more active metal than the copper but I am missing the understanding if the electrons leave zinc atom and a positive zinc ion is added to the solution because of the solution or because of the two metals being connected with a wire $\endgroup$
    – yaniv
    Mar 26, 2017 at 15:32

1 Answer 1


Lets start with the answer to the first question:-

(1)The Purpose of the solution in the galvanic cell is to facilitate the conversion of $M^{2+}$/$M$ or $M$/$M^{2+}$ as the case may be. Basically the solution contains Metal Ions which when required may convert to the Neutral Metal Atom. Similarly the electrode contains Neutral Metal Atoms which when required may convert into ions.

For Example, in the $Cu$ Cathode:- $$\ce{Cu_{(aq)}^{2+} + 2e ->Cu_{(s)} }$$

Over here an atom from the aqueous phase loses two electrons and moves onto the cathode. BUT THIS WOULD NOT HAVE BEEN POSSIBLE HAD THERE NOT BEEN A SOLUTION AROUND THE CATHODE CONTAINING $Cu^{2+}$ IONS AROUND IT. Thus we need a soultion around the electrodes.

Moreover the solution in both the halves contains both positive cations(like $Zn^{2+}$ and $Cu^{2+}$) as well as negative anions like $Cl^-$ or $SO_4^{2-}$ if the solutions are made up of $\ce{ZnSO4}$ and $\ce{CuCl2}$ .

(2) Electrons do not leave the $Zn$ atom unless and until the two electrodes are connected by a wire, since when electrons leave the metal atom they need to get used by another metal ion, which is what happens in the $Cu$ cathode $$\ce{Cu_{(aq)}^{2+} + 2e ->Cu_{(s)} }$$.

Moreover current(electrons) does not flow in a broken circuit thus the two electrodes need to be connected by a wire, and the two solutions need to be in contact too, possibly through a salt bridge or a permeable membrane.

  • $\begingroup$ What a detailed answer. Did the solution contained Metal Ions before being in contact with the cathode (the metal) or when they first mixed and touched each other then the solution started to "absorb" Metal Ions from the cathode? $\endgroup$
    – yaniv
    Mar 28, 2017 at 12:16
  • $\begingroup$ A solution of an ionic metal always exists in the form of ions, so yeah the solution contained ions even before it touched the electrodes. Any solution of an ionic metal like Zn Cu Na or Li exists in the form of ions. However, once it comes in contact with the electrode the conversion of ions->metal atoms or metal atoms->ions depends upon the nature of the metal and the nature of the metal of the other electrode. $\endgroup$
    – asds_asds
    Mar 29, 2017 at 21:38
  • $\begingroup$ I get it. So if I understand well, without connecting the wire between the two halves of the cell no reaction will happen between the solution and the metal of the electrode right? No oxidation nor reduction will happen without connecting the wire, is that right? $\endgroup$
    – yaniv
    Mar 30, 2017 at 14:15
  • $\begingroup$ Yes that's correct. The wire is the reason the electrons flow and facilitate the reactions. However, there's a catch, I don't know whether you need to know this right now or not but when you dip an electrode into a solution some reduction/oxidation takes place which leads to the formation of a potential on the electrode which stops further oxidation/reduction from happening. And this happens without the connecting wire. However, DO NOT research about this if you don't need to know this RIGHT NOW. Learn only what you have to right now, Don't get confused. $\endgroup$
    – asds_asds
    Mar 30, 2017 at 17:22

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