Page 1 of 1

Platinum in cell diagrams

Posted: Mon Feb 09, 2015 5:05 pm
by Casey Collet 1I
I keep getting confused on the conditions in which you use Platinum as an inert conductor in cell diagrams. If a metal is present, but is liquid instead of solid, can that metal still act as the electrode? Is it only with gases and nonmetals that platinum must be present?

Re: Platinum in cell diagrams

Posted: Mon Feb 09, 2015 6:40 pm
by Regina Chi 2K
You must have a solid present. If both species are either gases, liquids, or aqueous solutions, you must add an electrode.

Re: Platinum in cell diagrams

Posted: Mon Feb 09, 2015 6:45 pm
by Casey Collet 1I
Okay that's what I thought, but then why on problem 13.21 part a does the cell diagram for the anode look like Hg(l) l Hg2^(2+) (aq), where there isn't a solid present, but it indicates that mercury is acting as the electrode?

Re: Platinum in cell diagrams

Posted: Tue Feb 10, 2015 12:23 am
by Justin Le 2I
I think Hg (l) is an exception and can be an electrode because I see it being used as one in 19c and 43a.

Re: Platinum in cell diagrams

Posted: Tue Feb 26, 2019 4:37 pm
by Eva Guillory 2E
This makes sense, but I'm confused, because for the answer of 6L.5 part b in the 7th edition (which i listed below), despite Iodine being a solid, platinum is still included? In the solution manual, it does say that "An inert electrode such as Pt is necessary when both oxidized and reduced species are in the same solution," but how would we know that they are in the same solution?

Pt(s)|I-(aq)|I2 (s) || Ce4+ (aq), Ce3- (aq) | Pt(s)