Sciencemadness Discussion Board

H2O2 + I2 + Na2CO3

ktcd - 30-1-2013 at 20:14

iodine dissolved past solubility in 77%triethanolamine 23%isopropanol then dissolved in large excess of water makes a slightly yellow/brown solution with precipitated iodine. Adding sodium carbonate, weight several times of iodine does not change the color. Adding some 3%hydrogen peroxide makes color change quickly to colorless with bubbles of course. Was wondering why.

ktcd - 30-1-2013 at 22:06

This time no alcohol...

Iodine dropped into water, lots anhydrous Na2CO3 used... Na2CO3 dissolves, that's it
Iodine dropped into H2O2, lots anhydrous Na2CO3 used... bubbling, iodine completely antiprecipitates. bubbling for long time afterward.

We all know NaI catalyzes decomposition of H2O2.

Perhaps Na2CO3 + I2 -> 2NaI + CO2 + 0.5O2 requires heat
NaI + H2O2 -> NaI + H2O + 0.5O2 is exothermic

So with H2O2 the heat provided helps heat the first equation.

Confirmed, upon heating a lack of H2O2 also works. Iodine antiprecipitates here also.

You can delete this if you want.

[Edited on 31-1-2013 by ktcd]

[Edited on 31-1-2013 by ktcd]

[Edited on 31-1-2013 by ktcd]

[Edited on 31-1-2013 by ktcd]

[Edited on 31-1-2013 by ktcd]

Boffis - 2-2-2013 at 06:31

H2O2 can act both as an oxidizing and reducing agent. In the latter case oxygen is usually liberated, hence the bubbles.

H2O2 will reduce an iodine solution to iodide hence the colour change.

Another example is the dissolution of manganese dioxide in say HCl; you can either boil conc HCl and Mn oxide to get manganous chloride and clouds of chlorine gas or you can use dilute HCl and H2O2 in which case solution takes place rapidly, no heat required and no chlorine just a little oxygen!

haloform

quantumcorespacealchemyst - 16-12-2014 at 07:23

i don't know about the TEA, that is interesting. with isopropanol dissolving I2 and mixing with water. the I2 can interact (i dont know quite how yet) with water it seems more than normal, or can dump in I2 as it is used up to the water faster than from a solid form in water.
[i haven't tested out iodine powder, and crystal in water with sodium carbonate to see reaction rates.]

so if I2 + H2O <=> 2H+ +OI- + I-
and Na2CO3 + H2O <=> 2Na+ + CO3-2 + H+ +OH-
I2 + H2O + Na2CO3 <=> 2Na+ + CO3-2 + 2H+ +OI- + I-
and using the product side [Na+ + OI-] + [Na+ + I-] + [2H+ + CO3-2]
as i stated, i haven't done this alone,
from experience, adding isopropanol to the mix seems to allow the haloform reaction

i do not remember the details but that isopropanol in the haloform reaction takes an extra step in regards to using acetone, the acetone i believe turns to acetaldehyde and undergoes haloformation, while i believe isopropanol turns to acetone first. i think it uses a mol more of a reactant, i forgot if it is the hypo[ halogen ]ite or if it is a catalyst type reaction, moving on though.

shortened from acetone and not writing the aldehyde part,
CH3COCH3 + 3NaOI => CHI3 [Iodoform] + 3NaOH + NaOAc [NaCH3COO, sodium acetate]

but what puzzles me is if the reaction needs a Hydroxide somewhere formed as a catalyst, or so i thought, or a high pH, i believe if left overnight, it Iodoforms. perhaps the sodium does other reactions to a lesser extant and more slowly, making hydrides or oxides which form hydroxides in water and/or just straight hydroxides. i don't know., no that wouldn't make sense as they couldn't react and be regenerated aqua regia style, as they would need to build up to cause a high pH.

if it only needs a Hydroxl ion it may work slowly that way or even with water's OH- group.

i am intrigued by the mechanics, it does form Iodoform though when left overnight and if not that a bit longer, without adding a hydroxide. this from experience and a recollection of that [it was overnight or i forgot it for a couple days/a week, i believe it was overnight though:D].
the precipitate was yellow and smelt sweet, volatile, evaporates slowly in air over days

perhaps the there was an exess of TEA or isopropanol and the Iodoform was in solution causeing the yellow/brown color? i don't know being there being precipitated iodine. puzzled i am.

i think what boffis said about the H2O2 being both an oxidizing and reducing agent having something to do with it has something to do with it too, weird being the haloform seems to be happening or have happened. perhaps equilibriums? the haloform doesn't seem to reverse.

[Edited on 16-12-2014 by quantumcorespacealchemyst]

woelen - 16-12-2014 at 23:29

I think that in the alkaline environment the H2O2 acts as a reductor which reduces hypoiodite ion to iodide, itself being oxidized to water and oxygen. Hydrogen peroxide reacts similarly with hypochlorite (try adding some H2O2 to bleach, you get a very similar reaction and in the end you are left with chloride only).

Hypoiodite is formed from iodine and base: I2 + 2OH(-) --> I(-) + IO(-) + H2O.