## 14.23

Moderators: Chem_Mod, Chem_Admin

Katelyn Li 2J
Posts: 27
Joined: Wed Sep 21, 2016 3:00 pm

### 14.23

The questions asks: For each reaction that is spontaneous under standard conditions, write a cell diagram, determine the standard cell potential, and calculate G for the reaction.

I'm confused about part c, which is the reaction (c) Cr2O72 (aq) + 14 H (aq) + 6 Pu3 (aq) ---> 6 Pu4 (aq) + 2 Cr3 (aq) + 7 H2O(l). I understand which is the anode and cathode, but I don't know why the platinum is written on both sides of the cell diagram, especially in contrast to part (a), where it is only written on the side of the cathode. Could someone please explain?

Chem_Mod
Posts: 17949
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 406 times

### Re: 14.23

Platinum is an inert conductor that must be in the cell with the half reaction that has no conducting solids. Platinum does not participate in the reaction. A good trick if you are unsure is to check the states of the participants in the half reactions. If there are no solids and everything is (g), (l), or (aq), you definitely do not have a metal conductor and need to add Pt (s). You don't need Pt (s) if the electrode is a conducting, solid metal such as Zn (s) and Cu (s).

In problem 14.23 (c), neither the anode nor the cathode have conducting solids. Thus, you require a Pt (s) electrode on both sides of the cell diagram.

Cam Bear 2F
Posts: 60
Joined: Thu Jul 27, 2017 3:01 am

### Re: 14.23

Why doesn't part a) have platinum then on the oxidation side?
The answer in the solutions manual is $Hg(l)|Hg_{2}^{2+} (aq)||NO_{3}^{-}(aq), H^{+}(aq)|NO(g)|Pt(s)$

Jasmin Tran 1J
Posts: 54
Joined: Thu Jul 27, 2017 3:00 am

### Re: 14.23

Part a doesn't have platinum on the oxidation side because liquid mercury (Hg) can also be used as an electrode.

Return to “Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams”

### Who is online

Users browsing this forum: No registered users and 1 guest