0

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.)

stoic-santiago
  • 3,822
  • 2
  • 29
  • 68

1 Answers1

2

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.

AChem
  • 40,037
  • 2
  • 63
  • 126
  • 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. – Poutnik Apr 27 '19 at 04:53
  • "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. – AChem Apr 27 '19 at 14:36
  • I have thought the needed voltage is bigger, due hydrogen overvoltage. It may also depend on electrolyte. – Poutnik Apr 27 '19 at 14:50
  • 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. – AChem Apr 27 '19 at 14:53
  • @Poutnik, your electrochemistry is very good. You must have had excellent teachers! – AChem Apr 27 '19 at 14:55
  • Thanks. It is from 80s, so it gets little rusty. I know the theoretical potential difference for water electrolysis is 1.23V, as as far as I know, the Daniel cell has EMF 1.1V. – Poutnik Apr 27 '19 at 15:10
  • according to https://en.m.wikipedia.org/wiki/Overpotential on the shiny platinum, hydrogen has overpotential -0.16V oxygen +0.95V. For platinezed(black) Pt, it is -0.07V and +0.77V. – Poutnik Apr 27 '19 at 16:03