Homework 14.13b

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Karen Ung 2H
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Joined: Fri Sep 29, 2017 7:04 am
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Homework 14.13b

Postby Karen Ung 2H » Sat Mar 10, 2018 11:22 pm

Reaction: Ce4+(aq) + I-(aq) -> I2(s) + Ce3+(aq)
I separated into half reactions, identified the reduction and oxidation reaction, and write the cell diagram to be:
I2(s) | I-(aq) || Ce3+(aq), Ce4+(aq) | Pt(s)
But, the correct cell diagram is
Pt(s) | I2(s) | I-(aq) || Ce3+(aq), Ce4+(aq) | Pt(s)

Why is there a Pt on the left if there is already a solid element (I2) that can be the metal anode?

Nisarg Shah 1C
Posts: 54
Joined: Sat Jul 22, 2017 3:00 am

Re: Homework 14.13b

Postby Nisarg Shah 1C » Sat Mar 10, 2018 11:24 pm

I think it is because I2 is not a good conductor, ie. because it is a nonmetal it would not make sense to use it in an electrochemical cell. Instead, platinum is a better element that would facilitate the reaction.

Nisarg Shah 1C
Posts: 54
Joined: Sat Jul 22, 2017 3:00 am

Re: Homework 14.13b

Postby Nisarg Shah 1C » Sat Mar 10, 2018 11:25 pm

You could also refer to this earlier post: viewtopic.php?f=140&t=19452.


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