Why Pt(s) when we already have I2 (s)? HW 14.13

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Michael_Johanis_2K
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Why Pt(s) when we already have I2 (s)? HW 14.13

Postby Michael_Johanis_2K » Mon Feb 13, 2017 1:49 am

2I- (aq) + 2Ce4+ (aq) ----> I2 (s) + 2Ce3+ (aq)

I thought the cell diagram should be:
I2(s)| I- (aq) || Ce4+ (aq), Ce3+ (aq) | Pt(s)

However, it is:
Pt(s)| I- (aq) | I2 (s) || Ce4+ (aq), Ce3+ (aq) | Pt(s).

Why is Pt the solid on the left side even though we already have I2 (s)?

Maggie Bui 1H
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Re: Why Pt(s) when we already have I2 (s)? HW 14.13

Postby Maggie Bui 1H » Mon Feb 13, 2017 7:54 am

To be an electrode, it's not enough that something is solid; it also needs to conduct electricity, as electrons are transferred via electrode.

Amy_Bugwadia_3I
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Re: Why Pt(s) when we already have I2 (s)? HW 14.13

Postby Amy_Bugwadia_3I » Mon Feb 13, 2017 11:19 am

Generally speaking, we need to add Pt(s) (in order to conduct electricity) for reactions with only aqueous or gases on one side of the equation (you would add Pt to that side). However, there are two exceptions to this rule: if you have solid Iodine, you still need to add in Pt (because Iodine is not a good conductor). Also, if you have mercury, you do not need to add in Pt.


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