Elemental iodine from Potassium Iodide - Reaction Equations

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The discussion centers on the chemical reaction to produce elemental iodine (I2) from potassium iodide (KI) using concentrated hydrochloric acid and hydrogen peroxide (H2O2). The participants clarify the reaction equations, emphasizing the oxidation of iodide ions (I-) to iodine and the reduction of hydrogen peroxide. The net equation is confirmed as correct, with the presence of hydrogen ions from hydrochloric acid being crucial for the reaction. There is uncertainty about whether H2O2 would react with KI in the absence of hydrogen ions, but it is noted that water's autoprotolysis ensures some hydrogen ions are always present. The conversation also touches on the behavior of H2O2 in alkaline conditions, suggesting it can still oxidize iodide due to its higher redox potential compared to iodine.
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In the reaction to get I2 from an aqueous solution of KI.

The process. Take the aqeuous KI, and pour concentrated hydrochloric acid, followed by hydrogen peroxide.

K++I-+H++Cl-+H2O\rightarrow KCl + HI + H2O

The reaction with H2O2 is perplexing.

I'm a physics student trying to do chemistry, I'm sure I made some mistake in writing out that chemical equation (I forgot if ions have to be separated in the equation, when in a solute).

What are the specifics of this chain of reactions? The elemental iodine will precipitate out of solution. The H2O2 must be oxidizing something in the K+ + Cl- + H+ + I- + H2O
 
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The Iodine gets oxidized:
\rm 2 I^-\rightarrow I_2+2e^-
Hydrogen peroxide gets reduced:
\rm H_2O_2 +2e^-+2H^+\rightarrow 2 H_2O
The protons on the LHS stem from the hydrochloric acid, that's why you add acid.
 
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I see, hydrogen peroxide is a oxidizing agent, I seemed to have missed that.

Would H_{2}O_{2} react with KI in the absence of any hydrogen ions?

The net equation would be 2I^{-}+2K^{+}+2H^{+}+2Cl^{-}+H_{2}O_{2}\rightarrow I_{2} + 2KCl + 2H_{2}O ?
 
Astrum said:
Would H_{2}O_{2} react with KI in the absence of any hydrogen ions?

The net equation is correct. As soon as you use water as a solvent there will always be hydrogen ions around due to the autoprotolysis of water. I am not sure whether the reaction would also take place in an alkaline medium.
 
DrDu said:
I am not sure whether the reaction would also take place in an alkaline medium.

In alkaline solution of ##H_{2}O_{2}## the actual oxidant is the perhydroxyl ion ##HO^{-}_{2}##, for which we have the redox half-reaction

##HO^{-}_{2}+H_{2}O+2e^{-} \rightarrow 3OH^{-}## ##E^{0}## = +0.87 V

The redox potential is higher than that of iodine (+0.59 V), so the reaction should also happen in alkaline medium.
 
I'm wondering why the H_2O_2 doesn't oxidize the chlorine anion as an equal amount to the oxidation of iodide.

This process probably produces some amount of Cl_2.
 
Chloride is significantly harder to oxidize than iodide. Look up the redox potentials.
 
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