J.R. S. answered • 05/07/24
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Ag in AgCl(s) is being reduced and ending up as Ag(s) .. cathode
Pb(s) is being oxidized and ending up as PbF2(s) .. anode
From these observations, it seems that the line diagram might be the following, but hopefully another reader will weigh in, as I'm not positive of this answer. Just want to get the discussion started.
Pb(s) | PbF2(s) | F-(aq) || Cl-(aq) | AgCl(s) | Ag(s)
J.R. S.
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Thank you A.T. Good to know, and much appreciated.
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05/08/24
Anthony T.
I looked up a Wikepedia article on the AgCl reference electrode, and it gives the Ag/AgCl half cell as you have written it. It makes sense that the Pb/PbF2 half-cell is also as you have it written.05/07/24