14.13 b

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sofiakavanaugh
Posts: 58
Joined: Thu Jul 13, 2017 3:00 am
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14.13 b

Postby sofiakavanaugh » Thu Mar 15, 2018 12:11 pm

Hi,

In this problem you are suppose to find the 1/2 rxns, balance, write the full chemical equation, calculate the Eo of the cell and write the cell diagram. For the reaction Ce4+(aq) + I-(aq)--> I2(s) +Ce3+(aq), I got the correct cell diagram 2I-(aq)|I2(s)||Ce4+, Ce3+(aq)|Pt(s) except I didn't have platinum on the left like it does on the answers. I thought that you only had to write platinum id there was no solid on a side of the cell diagram? So why wouldn't you write it just on the right?

Thanks!

Vincent Kim 2I
Posts: 50
Joined: Fri Sep 29, 2017 7:04 am
Been upvoted: 2 times

Re: 14.13 b

Postby Vincent Kim 2I » Thu Mar 15, 2018 2:24 pm

I'm pretty sure you're correct, it shouldn't need to be written on the left side

Ishan Saha 1L
Posts: 60
Joined: Fri Sep 29, 2017 7:03 am

Re: 14.13 b

Postby Ishan Saha 1L » Thu Mar 15, 2018 2:28 pm

an inert metal conductor like Pt(s) needs to be added to the left side because I2(s) is a nonconductive nonmetal solid.

Adrian Lim 1G
Posts: 88
Joined: Fri Sep 29, 2017 7:03 am

Re: 14.13 b

Postby Adrian Lim 1G » Thu Mar 15, 2018 4:08 pm

How would we know that I2 is not conductive?

andrewr2H
Posts: 29
Joined: Fri Sep 29, 2017 7:05 am

Re: 14.13 b

Postby andrewr2H » Thu Mar 15, 2018 5:17 pm

Going back to the characteristics of each group on the periodic table we know that metals tend to be conductive whereas non metals such as the halogens are non conductive.


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