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Topic: Electrochem with elemental iodine and iodate  (Read 1096 times)

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Offline Zalzul

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Electrochem with elemental iodine and iodate
« on: December 30, 2018, 04:04:33 PM »
[IO3]-(aq) + 6H+ + 6e- ::equil:: I-(aq) + 3H2O(l) with
I2(aq) + 2e- ::equil:: 2I-(aq)

The question asks me to write a balanced equation for the spontaneous reaction that occurs when this pair of half-cells is combined.  From a table of E° (298 K) values, +1.09 V is reported for the top reaction and +0.54 V for the one below.  This tells me that, following the counterclockwise rule, the spontaneous reaction would progress from H+ (an oxidized species) to I- (a reduced species) but iodide's presence in both half-cells confuses me.  Do I need to combine its two appearances and end up with 3I- in the final balanced equation?  I know cell is independent of the amount of material present, but I see stoichiometry in half-cells all the time.

I determined oxidation numbers for each: in [IO3]-, +5 for I and -2 for each O.  +1 for each H+.  -1 for each I-.  +1 for each hydrogen and -2 for oxygen in H2O(l).  (I used colors to segment each species, for ease in vision.)  To my mind, however, this doesn't help me.

In short, I'm lost.  Can anyone offer any clarity on steps to take next or steps for me to re-examine if I was wrong in any of my assumptions?

Offline AWK

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Re: Electrochem with elemental iodine and iodate
« Reply #1 on: December 30, 2018, 04:53:43 PM »
Note, there are two reduction reaction for iodate(V)
2IO3- + 12H+ + 10e- → I2 + 6H2O  with Eo 1.195 V
and
IO3-  + 6H+ + 6e- → I- + 3H2O with Eo 1.085 V

Which one fits better to the second reaction?
AWK

Offline Zalzul

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Re: Electrochem with elemental iodine and iodate
« Reply #2 on: December 30, 2018, 05:00:23 PM »
I would say the 1.195 V reaction, from looking at the second reaction

Offline Zalzul

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Re: Electrochem with elemental iodine and iodate
« Reply #3 on: December 30, 2018, 09:45:16 PM »
Okay, in light of that, I thought I should flip the second half-cell to get I2 on the right.  Overall,

[IO3]-(aq) + 6H+(aq) + 5I-(aq) :rarrow: 3H2O(l) + 3I2(aq)

Anyone else agree with this?

Offline mjc123

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Re: Electrochem with elemental iodine and iodate
« Reply #4 on: January 03, 2019, 05:01:41 AM »
Yes.

The key thing you need to do is to eliminate electrons from the overall equation. So if your first half-equation has +6e- on the left, the second must have +6e- on the right. So you flip your second equation and multiply by 3.

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