What would happen, if in a galvanic cell, the salt bridge is replaced by an inert platinum sheet?

Would charges accumulate on the sheet? Would the cell still work, and produce less emf than earlier, or stop working at all?

P.S. I know why a salt bridge is necessary for the functioning of a galvanic cell - my question is rather specific to what would happen if I were to use a platinum sheet instead, and why we can't use it. (Hence, not a duplicate.)

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    $\begingroup$ Possible duplicate of Why is it important to use a salt bridge in a voltaic cell? Can a wire be used? $\endgroup$ – andselisk Apr 26 at 2:51
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    $\begingroup$ You would have 2 cells in a serie, instead of one cell, each having Pt as one of electrodes. $\endgroup$ – Poutnik Apr 26 at 3:14
  • $\begingroup$ So, according to you, the cell would operate? $\endgroup$ – arya_stark Apr 26 at 3:36
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    $\begingroup$ You will not have that cell any more. You would have 2 cells different from the original one, with behaviour implicitly undefined. $\endgroup$ – Poutnik Apr 26 at 3:45

In continuation of the discussion above, I have another opinion:

What would happen, if in a galvanic cell, the salt bridge is replaced by an inert platinum sheet?

The cell would stop working in most cases. Opposite charges would accumulate in the two > half cells, opposing the reduction and oxidation half reaction.

I am afraid the statement is not correct. The set up in which Cu and Zn are partitioned by a platinum sheet, is equivalent to two galvanic cells connected in series.

The two cells (a) Cu-Pt cell (b) Pt-Zn cell

which are connected via conductor. In this case, the conductor is also platinum.

I just was reading LeBlanc's historical textbook on electrochemistry (lovely book from German translation) for another purpose. He shows a nice picture of two cells connected in series.

enter image description here

Consider a similar situation in case of Pt. However we may not be able to calculate the potentials because Pt is not dipping in Pt ions in either beaker. Instead, one end is in copper ions and the other end in the zinc ions. It still may show a positive potential (Ecell) just like how a lemon battery does (=Zn and Copper rods immersed in a citric acid solution).

This has been discussed before in physics. https://www.physicsforums.com/threads/copper-wire-as-salt-bridge.146868/ The cell still works! Charges can never accumulate like static electricity in a galvanic cell in the presence of a conductor.

  • $\begingroup$ But the both Compartments would have just single solution, I=CuSO4, II=ZnSO4, and there would be no separation. There would be, from left to right: Cu/CuSO4, Pt/CuSO4, Pt/ZnSO4, Zn/ZnSO4. The Pt in CuSO4 would have to oxidize water to oxygen, right to reduce either Zn2+ either H+, depending on their potentials. But Cu/Zn Cell IMHO does not have enough EMF to electrolyze water. $\endgroup$ – Poutnik Apr 27 at 4:53
  • $\begingroup$ "But Cu/Zn Cell IMHO does not have enough EMF to electrolyze water", yes a single cell. LeBlanc gives an example that the setup shown above can electrolyze water. This book is more than 100 years old and still a classic. Excellent German work. $\endgroup$ – M. Farooq Apr 27 at 14:36
  • $\begingroup$ I have thought the needed voltage is bigger, due hydrogen overvoltage. It may also depend on electrolyte. $\endgroup$ – Poutnik Apr 27 at 14:50
  • $\begingroup$ This Daniel cell was a standard workhorse of electrochemical studies in the 19th and 20th century. Hydrogen overpotential is not that great otherwise Faraday would not have been able to electrolyze water with voltaic cells. $\endgroup$ – M. Farooq Apr 27 at 14:53
  • $\begingroup$ @Poutnik, your electrochemistry is very good. You must have had excellent teachers! $\endgroup$ – M. Farooq Apr 27 at 14:55

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