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Katelyn Li 2J
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Postby Katelyn Li 2J » Thu Feb 09, 2017 1:06 am

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?

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Re: 14.23

Postby Chem_Mod » Thu Feb 09, 2017 11:11 am

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
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Re: 14.23

Postby Cam Bear 2F » Wed Mar 14, 2018 10:34 pm

Why doesn't part a) have platinum then on the oxidation side?
The answer in the solutions manual is

Jasmin Tran 1J
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Re: 14.23

Postby Jasmin Tran 1J » Fri Mar 16, 2018 4:15 pm

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

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