nitroboy
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H2SO4 as an oxidizer
in the reaction:
H2SO4 + KI => ??? i cant seem to work out the products. i know from the smell it was H2S, and the fact it turned brown that it was also I.
H2SO4 + KI => H2S + I2 + ? can anybody help me with the remaining products and maybe the reaction mechanics please?
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BromicAcid
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Normally when H2SO4 acts as an oxidizing agent, your reaction being a very typical reaction, the sulfate ion is reduced to sulfur dioxide not H2S. I
would be very suprised if you produced H2S with this reaction, it normally proceeds as shown below.
2H2SO4 + 2KI ---> I2 + SO2 + 2H2O + K2SO4
But as you know H2SO4 is not normally known as an oxidizing agent, is usually only oxidizes when highly concentrated and at high temperature such as
dissolving Cu in hot conc H2SO4. However in this case with the iodide anion being a decent reducing agent to begin with the reaction proceeds at room
temp. Moving one up the periodic table to bromine, and the anologous case with potassium bromide only a comparatively small precentate of the bromine
is oxidized in my experience and if done under aqueous conditions almost none is oxidized to free bromine.
[Edited on 5/23/2004 by BromicAcid]
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hodges
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I'm not sure of the equations involved, but I have personally witnessed that adding concentrated H2SO4 to KI solution produces not only H2S but
even some free S. In fact (not sure where) I've seen this listed as a quantitative test to distinguish between iodide and other halides.
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BromicAcid
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Well this could be one of those all too common situations where the books don't sync up with reality. I really was under the impression that the
reaction with KI would result in H2SO4 being reduced and that H2SO4 normally gets reduced as shown above. However since you mentioned that even
sulfur may be a product there are probably a number of steps and products possible (Also the sulfate could be mildly reduced to the SO3 2- anion).
2H2SO4 + 2e- ----> SO2 + 2H2O + SO4 2-
4H2SO4 + 3e- ----> S + 4H2O + 3SO4 2-
5H2SO4 + 4e- ----> H2S + 4H2O + 4SO4 2-
Depending on the reaction conditions, however like I've already expressed I believe the first reaction is the most common however with the
reducing properties of I- in the mix I guess from what your saying the other two reactions are prevelent as well.
Also nitroboy, what were your reaction conditions? Hot concentrated H2SO4 is going to react in a completely different manner then cold dillute H2SO4
although I'm assuming a comprimise between those extremes being room temperature concentrated H2SO4, and also, was your potassium iodide a solid
or was it dissolved in water? Also was your solution efficently stirred, I believe iodine will react with H2S like the other halogens although I
could be wrong:
H2S + I2 ----> 2HI + S
However the reaction:
I2 + SO2 + 2H2O --X--> 2HI + H2SO4
Does not proceed for iodine so if the first reaction involving H2S reacting with iodine does occur and your reaction solution is efficently stirred
then your final solution will have a mix of HI and sulfur particulate and some SO2 may form but H2S will be minimal as it would be consumed.
[Edited on 5/24/2004 by BromicAcid]
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S.C. Wack
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It can happen. Excess I and hot H2SO4: 8KI + 9H2SO4 = 4I2 + H2S + 8KHSO4 + 4H2O.
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nitroboy
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we were at room temperature, using cold, concentrated H2SO4.
i know the smell was H2S, but i am going to confirm with my teacher the exact conditions and what SHOULD have happened.
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