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

Posted: Mon Feb 13, 2017 1:49 am
by Michael_Johanis_2K
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)?

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

Posted: Mon Feb 13, 2017 7:54 am
by Maggie Bui 1H
To be an electrode, it's not enough that something is solid; it also needs to conduct electricity, as electrons are transferred via electrode.

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

Posted: Mon Feb 13, 2017 11:19 am
by Amy_Bugwadia_3I
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.